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Tamarin: Principles of
Genetics, Seventh Edition

Front Matter

Preface

©TheMcGraw-Hil
Companies, 2001

PREFACE

The twentieth century began with the redis-
covery of Mendel's rules of inheritance and
ended with the complete sequence of the hu-
man genome, one of the most monumental
scientific accomplishments of all time. What
lies in the future? What will the twenty-first century, the
century of genomics, bring? Will geneticists a hundred
years from now speak of a complete cure for cancer,
heart disease, and mental illness? Will we have a cure for
autoimmune diseases such as diabetes and arthritis? Will
aging be slowed or even prevented? Will we have a com-
plete understanding of the process of development and a
concurrent elimination of birth defects and developmen-
tal problems? Will genetics put an end to world hunger?
How will we live, and what will be the quality of our
lives? The students who now are taking genetics will
learn the answers to these questions as time progresses.
Some students will contribute to the answers.

The science of genetics includes the rules of inheri-
tance in cells, individuals, and populations and the mo-
lecular mechanisms by which genes control the growth,
development, and appearance of an organism. No area of
biology can truly be appreciated or understood without
an understanding of genetics because genes not only
control cellular processes, they also determine the
course of evolution. Genetic concepts provide the frame-
work for the study of modern biology.

This text provides a balanced treatment of the ma-
jor areas of genetics in order to prepare the student for
upper-level courses and to help share in the excitement
of research. Most readers of this text will have taken a
general biology course and will have had some back-
ground in cell biology and organic chemistry. For an un-
derstanding of the concepts in this text, however, the
motivated student will need to have completed only an
introductory biology course and have had some chem-
istry and algebra in high school.

Genetics is commonly divided into three areas: classi-
cal, molecular, and population, although molecular ad-
vancements have blurred these distinctions. Many genetics
teachers feel that a historical approach provides a sound
introduction to the field and that a thorough grounding
in Mendelian genetics is necessary for an understanding
of molecular and population genetics — an approach this
text follows. Other teachers, however, may prefer to be-
gin with molecular genetics. For this reason, the chapters
have been grouped as units that allow for flexibility

in their use. A comprehensive glossary and index will
help maintain continuity if the instructor chooses to
change the order of the chapters from the original.

An understanding of genetics is crucial to advance-
ments in medicine, agriculture, and many industries. Ge-
netic controversies — such as the pros and cons of the
Human Genome Project, the potential ethical and med-
ical risks of recombinant DNA and cloning of mammals,
and human behavioral genetic issues such as the degree
of inheritance of homosexuality, alcoholism, and intelli-
gence — have captured the interest of the general public.
Throughout this text, we examine the implications for
human health and welfare of the research conducted
in universities and research laboratories around the
world; boxed material in the text gives insight into ge-
netic techniques, controversies, and breakthroughs.

Because genetics is the first analytical biology course
for many students, some may have difficulty with its
quantitative aspects. There is no substitute for work with
pad and pencil. This text provides a larger number of
problems to help the student learn and retain the mate-
rial. All problems within the body of the text and a selec-
tion at the end of the chapters should be worked through
as they are encountered. After the student has worked
out the problems, he or she can refer to the answer sec-
tion in Appendix A. We provide solved problems at the
end of each chapter to help.

In this text, we stress critical thinking, an approach
that emphasizes understanding over memorization, ex-
perimental proof over the pronouncements of authori-
ties, problem solving over passive reading, and active
participation in lectures. The latter is best accomplished
if the student reads the appropriate text chapter before
coming to lecture rather than after. That way the student
can use the lecture to gain insight into difficult material
rather than spending the lecture hectically transcribing
the lecturer's comments onto the notebook page.

For those students who wish to pursue particular
topics, a reference section in the back of the text pro-
vides chapter-by-chapter listings of review articles and ar-
ticles in the original literature. Although some of these
articles might be difficult for the beginner to follow, each
is a landmark paper, a comprehensive summary, or a pa-
per with some valuable aspect. Some papers may contain
an insightful photograph or diagram. Some magazines
and journals are especially recommended for the student
to look at periodically, including Scientific American,

Xlll

Tamarin: Principles of
Genetics, Seventh Edition

Front Matter

Preface

©TheMcGraw-Hil
Companies, 2001

XIV

Preface

Science, and Nature, because they contain nontechnical
summaries as well as material at the cutting edge of ge-
netics. Some articles are included to help the instructor
find supplementary materials related to the concepts in
this book. Photographs of selected geneticists also are in-
cluded. Perhaps the glimpse of a face from time to time
will help add a human touch to this science.

The World Wide Web also can provide a valuable re-
source. The textbook has its own website: www.
mhhe.com/tamarin7. In addition, the student can find
much material of a supplemental nature by "surfing" the
web. Begin with a search engine such as: www.
yahoo.com, or www.google.com and type in a key word.
Follow the links from there. Remember that the material
on the web is "as is"; it includes a lot of misinformation.
Usually, content from academic, industrial, and organiza-
tional sources is relatively reliable; however, caveat emp-
tor — buyer beware. Often in surfing for scientific key
words, the student will end up at a scientific journal or
book that does not have free access. Check with the uni-
versity librarian to see if access might be offered to that
journal or book. The amount of information that is accu-
rate and free is enormous. Be sure to budget the amount
of time spent on the Internet.

NEW TO THIS EDITION

Since the last edition of this text, many exciting discover-
ies have been made in genetics. All chapters have been
updated to reflect those discoveries. In particular:

• The chapter on Recombinant DNA Technology has
been revised to be a chapter on Genomics, Biotech-
nology, and Recombinant DNA (sixth edition chapter
12 has become chapter 13 in this edition). The chap-
ter includes new material on the completion of the
Human Genome Project, bioinformatics, proteomics,
and the latest techniques in creating cDNA and
knockout mice.

• The chapter on Control of Transcription in Eukary-
otes (sixth edition chapter 15 has become chapter
16 in this edition) has been completely reorganized
and rewritten to emphasize signal transduction, spe-
cific transcription factors, methylation, and chro-
matin remodeling in control of gene expression; as in
the last edition, there are specific sections on
Drosophila and plant development, cancer, and im-
munogenetics.

• For better continuity, the chapter on Mutation, Re-
combination, and DNA Repair has been moved to fol-
low the chapters on Transcription and Translation
(sixth edition chapter 16 has become chapter 12 in
this edition).

The material in chapter 3 on Genetic Control of the
Cell Cycle has been upgraded to a chapter section on
the Cell Cycle.

Molecular material throughout the book has been
completely updated to include such subjects as nu-
merous DNA repair polymerases and their function-
ing; base-flipping; TRAP control of attenuation; and
chromatosomes .

LEARNING AIDS FOR
THE STUDENT

To help the student learn genetics, as well as enjoy the
material, we have made every effort to provide pedagog-
ical aids. These aids are designed to help organize the ma-
terial and make it understandable to students.

• Study Objectives Each chapter begins with a set of
clearly defined, page-referenced objectives. These ob-
jectives preview the chapter and highlight the most
important concepts.

• Study Outline The chapter topics are provided in
an outline list. These headings consist of words or
phrases that clearly define what the various sections
of the chapter contain.

• Boldface Terms Throughout the chapter, all new
terms are presented in boldface, indicating that each
is defined in the glossary at the end of the book.

• Boxed Material In most chapters, short topics
have been set aside in boxed readings, outside the
main body of the chapter. These boxes fall into four
categories: Historical Perspectives, Experimental
Methods, Biomedical Applications, and Ethics
and Genetics. The boxed material is designed to
supplement each chapter with entertaining, interest-
ing, and relevant topics.

• Full Color Art and Graphics Many genetic con-
cepts are made much clearer with full-color illustra-
tions and the latest in molecular computer models to
help the student visualize and interpret difficult
concepts. We've added thirty new photographs and
over a hundred new and modified line drawings to
this edition.

• Summary Each chapter summary recaps the study
objectives at the beginning of the chapter. Thus, the
student can determine if he or she has gained an un-
derstanding of the material presented in the study ob-
jectives and reinforce them with the summary.

• Solved Problems From two to four problems are
worked out at the end of each chapter to give the stu-
dent practice in solving and understanding basic
problems related to the material.

• Exercises and Problems At the end of the chap-
ter are numerous problems to test the student's

Tamarin: Principles of
Genetics, Seventh Edition

Front Matter

Preface

©TheMcGraw-Hil
Companies, 2001

Preface

understanding of the material. These problems are
grouped according to the sections of the chapter. An-
swers to the odd-numbered problems are presented
in Appendix A, with the even-numbered problems an-
swered only in the Student Study Guide so that the
student and instructor can be certain that the student
is gaining an understanding of the material.
Critical Thinking Questions Two critical think-
ing questions at the end of each chapter are designed
to help the student develop an ability to evaluate and
solve problems. The answer to the first critical think-
ing question can be found in Appendix A, and the an-
swer to the second question is in the Student Study
Guide.

ANCILLARY MATERIALS

For the Instructor

• Website. Visit us at www.mhhe.com/tamarin7.
Here instructors will find jpeg files of the line draw-
ings and tables suitable for downloading into Power-
Point, quizzes for study support, and links to genetic
sites. In addition, instructors will also find a link to
our hugely successful PageOut: The Course Web-
site Development Center, where instructors can
create a professional-looking, customized course
website. It's incredibly easy to use, and you need not
know html coding.

• Visual Resource Library (VRL). This Windows- and
Macintosh-compatible CD-ROM has all the line draw-
ings and tables from the text suitable for PowerPoint
presentations. (ISBN 0072334266)

• Instructor's Manual with Test Item File. Available on
the website, the Instructor's Manual contains out-
lines, key words, summaries, instructional hints, and
supplemental aids. The Test Item File contains 35 to
50 objective questions with answers for each chap-
ter. (ISBN 0072334215)

• Test Item File on MicroTest III Classroom Testing
Software is an easy-to-use CD-ROM test generator also
offered free upon request to adopters of this text. The
software requires no programming experience and is
compatible with Windows or Macintosh systems.
(ISBN 0072334231).

For the Student

• Website. Visit us at www.mhhe.com/tamarin7.
Here the student will find quizzes for study support,
web exercises and resources, and links to genetic sites.

• Genetics: From Genes to Genomes CD-ROM, by Ann
E. Reynolds, University of Washington. Packaged free
with every text, this CD-ROM covers the most chal-

lenging concepts in the course and makes them more
understandable through the presentation of full-
color, narrated animations and interactive exercises.
The text indicates related topics on the CD with the
following icon: ^

Student Study Guide. This study guide features key
concepts, problem-solving hints, practice problems,
terms, study questions, and answers to even-numbered
questions in the text. (ISBN 0072334207)
Laboratory Manual of Genetics 4/e, by A. M. Win-
chester and P. J. Wejksnora, University of Wisconsin-
Milwaukee. This manual for the genetics laboratory
features classical and molecular biology exercises
that give students the opportunity to apply the scien-
tific method to "real" — not simulated — lab investiga-
tions. (ISBN 0697122875)

Case Workbook in Human Genetics, 2/e, by Ricki
Lewis, SUNY- Albany. The Workbook includes
thought-provoking case studies in human genetics,
with many examples gleaned from the author's expe-
riences as a practicing genetic counselor. (ISBN
0072325305) Also included is the Answer Key. (ISBN
0072439009)

ACKNOWLEDGMENTS

I would like to thank many people for their encourage-
ment and assistance in the production of this Seventh
Edition. I especially thank Brian Loehr, my Developmen-
tal Editor, for continuous support, enthusiasm, and help
in improving the usability of the text. It was also a plea-
sure to work with many other dedicated and creative
people at McGraw-Hill during the production of this
book, especially James M. Smith, Thomas Timp, Gloria
Schiesl, David Hash, Sandy Ludovissy, Carrie Burger, and
Jodi Banowetz. I wish to thank Dr. Michael Gaines of the
University of Miami for many comments that helped me
improve the textbook and Marion Muskiewicz, Refer-
ence Librarian at the University of Massachusetts Lowell,
who was an enormous help in my efforts to use the uni-
versity's electronic library. Many reviewers greatly
helped improve the quality of this edition. I specifically
wish to thank the following:

Reviewers of the Seventh Edition

John Belote

Syracuse University

Douglas Coulter

Saint Louis University

Tamarin: Principles of
Genetics, Seventh Edition

Front Matter

Preface

©TheMcGraw-Hil
Companies, 2001

XVI

Preface

James M. Freed

Ohio Wesleyan University

Elliott S. Goldstein

Arizona State University

Keith Hartberg

Baylor University

Vincent Henrich

University of North Carolina at Greensboro

Mitrick A. Johns

Northern Illinois University

Philip Mathis

Middle Tennessee State University

Bruce McKee

University of Tennessee

Elbert Myles

Tennessee State University

John Osterman

University of Nebraska-Lincoln

Uwe Pott

University of Wisconsin -Green Bay

Ken Spitze

University of Miami

Randall G. Terry

University of Montana

Michael Wooten

Auburn University

Reviewers of the Sixth Edition

Edward Berger

Dartmouth

Deborah C. Clark

Middle Tennessee State University

John R. Ellison

Texas A&M University

Elliott S. Goldstein

Arizona State University

Keith Hartberg

Baylor University

David R. Hyde

University of Notre Dame

Pauline A. Lizotte

Northwest Missouri State University

James J. McGivern

Gannon University

Gregory J. Phillips

Iowa State University

Mark Sanders

University of California-Davis

Ken Spitze

University of Miami

Joan M. Stoler

Massachusetts General Hospital, Harvard Medical
School

Robert J. Wiggers

Stephen F. Austin State University

Ronald B. Young

University of Alabama

Lastly thanks are due to the many students, particu-
larly those in my Introductory Genetics, Population Biol-
ogy, Evolutionary Biology, and Graduate Seminar courses,
who have helped clarify points, find errors, and discover
new and interesting ways of looking at the many topics
collectively called genetics.

ROBERT H. TAMARIN
Lowell, Massachusetts

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

INTRODUCTION

STUDY OBJECTIVES

1. To examine a brief overview of the modern history
of genetics 3

2. To gain an overview of the topics included in this book — the
syllabus of genetics 4

3. To analyze the scientific method 5

4. To look at why certain organisms and techniques have been
used preferentially in genetics research 7

STUDY OUTLINE

A Brief Overview of the Modern History of Genetics

Before I860 3

1860-1900 3

1900-1944 3

1944-Present 4
The Three General Areas of Genetics 4
How Do We Know? 5
Why Fruit Flies and Colon Bacteria? 7
Techniques of Study 8
Classical, Molecular, and Evolutionary Genetics 9

Classical Genetics 9

Molecular Genetics 10

Evolutionary Genetics 13
Summary 14
Box 1.1 The Lysenko Affair 6

Chameleon, Cameleo pardalis.

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

A Brief Overview of the Modern History of Genetics

Genetics is the study of inheritance in all of its
manifestations, from the distribution of hu-
man traits in a family pedigree to the bio-
chemistry of the genetic material in our
chromosomes — deoxyribonucleic acid, or
DNA. It is our purpose in this book to introduce and de-
scribe the processes and patterns of inheritance. In this
chapter, we present a broad outline of the topics to be
covered as well as a summary of some of the more im-
portant historical advancements leading to our current
understanding of genetics.

A BRIEF OVERVIEW OF
THE MODERN HISTORY
OF GENETICS

For a generation of students born at a time when incred-
ible technological advances are commonplace, it is valu-
able to see how far we have come in understanding the
mechanisms of genetic processes by taking a very brief,
encapsulated look at the modern history of genetics. Al-
though we could discuss prehistoric concepts of animal
and plant breeding and ideas going back to the ancient
Greeks, we will restrict our brief look to events begin-
ning with the discovery of cells and microscopes. For our
purposes, we divide this recent history into four periods:
before I860, 1860-1900, 1900-1944, and 1944 to the
present.

Before 1860

Before I860, the most notable discoveries paving the
way for our current understanding of genetics were
the development of light microscopy, the elucidation of
the cell theory, and the publication in 1859 of Charles
Darwin's The Origin of Species. In 1665, Robert Hooke
coined the term cell in his studies of cork. Hooke saw, in
fact, empty cells observed at a magnification of about
thirty power. Between 1674 and 1683, Anton van
Leeuwenhoek discovered living organisms (protozoa and
bacteria) in rainwater. Leeuwenhoek was a master lens
maker and produced magnifications of several hundred
power from single lenses (fig. 1.1). More than a hundred
years passed before compound microscopes could equal
Leeuwenhoek's magnifications. In 1833, Robert Brown
(the discoverer of Brownian motion) discovered the nu-
clei of cells, and between 1835 and 1839, Hugo von Mohl
described mitosis in nuclei. This era ended in 1858, when
Rudolf Virchow summed up the concept of the cell the-
ory with his Latin aphorism omnis cellula e cellula: all
cells come from preexisting cells. Thus, by 1858, biolo-
gists had an understanding of the continuity of cells and
knew of the cell's nucleus.

1860-1900

The period from I860 to 1900 encompasses the publica-
tion of Gregor Mendel's work with pea plants in 1866 to
the rediscovery of his work in 1900. It includes the dis-
coveries of chromosomes and their behavior — insights
that shed new light on Mendel's research.

From 1879 to 1885, with the aid of new staining tech-
niques, W. Flemming described the chromosomes — first
noticed by C. von Nageli in 1842 — including the way they
split during division, and the separation of sister chromatids
and their movement to opposite poles of the dividing cell
during mitosis. In 1888, W. Waldeyer first used the term
chromosome. In 1875, 0. Hertwig described the fusion of
sperm and egg to form the zygote. In the 1880s, Theodor
Boveri, as well as K. Rabl and E. van Breden, hypothesized
that chromosomes are individual structures with continuity
from one generation to the next despite their "disappear-
ance" between cell divisions. In 1885, August Weismann
stated that inheritance is based exclusively in the nucleus.
In 1887, he predicted the occurrence of a reductional di-
vision, which we now call meiosis. By 1890, 0. Hertwig and
T Boveri had described the process of meiosis in detail.

1900-1944

From 1900 to 1944, modern genetics flourished with the
development of the chromosomal theory, which showed

Lens
Specimen holder

Focus screw

Handle

Figure 1.1 One of Anton van Leeuwenhoek's microscopes,
ca. 1680. This single-lensed microscope magnifies up to 200x.
(© Kathy Talaro/Visuals Unlimited, Inc.)

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

Chapter One Introduction

that chromosomes are linear arrays of genes. In addition,
the foundations of modern evolutionary and molecular
genetics were derived.

In 1900, three biologists working independently —
Hugo de Vries, Carl Correns, and Erich von Tschermak —
rediscovered Mendel's landmark work on the rules of in-
heritance, published in 1866, thus beginning our era of
modern genetics. In 1903, Walter Sutton hypothesized
that the behavior of chromosomes during meiosis ex-
plained Mendel's rules of inheritance, thus leading to the
discovery that genes are located on chromosomes. In
1913, Alfred Sturtevant created the first genetic map, us-
ing the fruit fly. He showed that genes existed in a lin-
ear order on chromosomes. In 1927, L. Stadler and
H.J. Muller showed that genes can be mutated artificially
by X rays.

Between 1930 and 1932, R. A. Fisher, S. Wright, and
J. B. S. Haldane developed the algebraic foundations for
our understanding of the process of evolution. In 1943,
S. Luria and M. Delbriick demonstrated that bacteria have
normal genetic systems and thus could serve as models
for studying genetic processes.

1944-Present

The period from 1944 to the present is the era of molec-
ular genetics, beginning with the demonstration that
DNA is the genetic material and culminating with our
current explosion of knowledge due to recombinant
DNA technology.

In 1944, O. Avery and colleagues showed conclu-
sively that deoxyribonucleic acid — DNA — was the ge-
netic material. James Watson and Francis Crick worked
out the structure of DNA in 1953. Between 1968 and
1973, W Arber, H. Smith, and D. Nathans, along with their
colleagues, discovered and described restriction endonu-

cleases, the enzymes that opened up our ability to ma-
nipulate DNA through recombinant DNA technology. In
1972, Paul Berg was the first to create a recombinant
DNA molecule.

Since 1972, geneticists have cloned numerous genes.
Scientists now have the capability to create transgenic
organisms, organisms with functioning foreign genes. For
example, we now have farm animals that produce phar-
maceuticals in their milk that are harvested easily and in-
expensively for human use. In 1997, the first mammal
was cloned, a sheep named Dolly. The sequence of the
entire human genome was determined in 2000; we will
spend the next century mining its information in the
newly created field of genomics, the study of the com-
plete genetic complement of an organism. Although no
inherited disease has yet been cured by genetic interven-
tion, we are on the verge of success in numerous dis-
eases, including cancer.

The material here is much too brief to convey any of
the detail or excitement surrounding the discoveries of
modern genetics. Throughout this book, we will expand
on the discoveries made since Darwin first published his
book on evolutionary theory in 1859 and since Mendel
was rediscovered in 1900.

THE THREE GENERAL AREAS
OF GENETICS

Historically, geneticists have worked in three different ar-
eas, each with its own particular problems, terminology,
tools, and organisms. These areas are classical genetics,
molecular genetics, and evolutionary genetics. In classi-
cal genetics, we are concerned with the chromosomal
theory of inheritance; that is, the concept that genes are

TablG 1.1 The Three Major Areas of Genetics — Classical, Molecular, and Evolutionary —
and the Topics They Cover

Classical Genetics

Molecular Genetics

Evolutionary Genetics

Mendel's principles

Structure of DNA

Quantitative genetics

Meiosis and mitosis

Chemistry of DNA

Hardy-Weinberg equilibrium

Sex determination

Transcription

Assumptions of equilibrium

Sex linkage

Translation

Evolution

Chromosomal mapping

DNA cloning and genomics

Speciation

Cytogenetics (chromosomal changes)

Control of gene expression
DNA mutation and repair
Extrachromosomal inheritance

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

How Do We Know?

located in a linear fashion on chromosomes and that the
relative positions of genes can be determined by their
frequency in offspring. Molecular genetics is the study of
the genetic material: its structure, replication, and ex-
pression, as well as the information revolution emanating
from the discoveries of recombinant DNA techniques
(genetic engineering, including the Human Genome Proj-
ect). Evolutionary genetics is the study of the mecha-
nisms of evolutionary change, or changes in gene fre-
quencies in populations. Darwin's concept of evolution
by natural selection finds a firm genetic footing in this
area of the study of inheritance (table 1.1).

Today these areas are less clearly defined because of
advances made in molecular genetics. Information com-
ing from the study of molecular genetics allows us to un-
derstand better the structure and functioning of chromo-
somes on the one hand and the mechanism of natural
selection on the other. In this book we hope to bring to-
gether this information from a historical perspective.
From Mendel's work in discovering the rules of inheri-
tance (chapter 2) to genetic engineering (chapter 13) to
molecular evolution (chapter 21), we hope to present a
balanced view of the various topics that make up
genetics.

HOW DO WE KNOW?

Observation

Hypothesis

Prediction

Refute

Support

Experiment

New hypothesis

Figure 1.2 A schematic of the scientific method. An
observation leads the researcher to propose a hypothesis, and
then to make predictions from the hypothesis and to test these
predictions by experiment. The results of the experiment either
support or refute the hypothesis. If the experiment refutes the
hypothesis, a new hypothesis must be developed. If the
experiment supports the hypothesis, the researcher or others
design further experiments to try to disprove it.

Genetics is an empirical science, which means that our
information comes from observations of the natural
world. The scientific method is a tool for understanding
these observations (fig. 1.2). At its heart is the experi-
ment, which tests a guess, called a hypothesis, about how
something works. In a good experiment, only two types
of outcomes are possible: outcomes that support the hy-
pothesis and outcomes that refute it. Scientists say these
outcomes provide strong inference.

For example, you might have the idea that organisms
can inherit acquired characteristics, an idea put forth by
Jean-Baptiste Lamarck (1744-1829), a French biologist.
Lamarck used the example of short-necked giraffes evolv-
ing into the long-necked giraffes we know of today. He
suggested that giraffes that reached higher into trees to
get at edible leaves developed longer necks. They passed
on these longer necks to their offspring (in small incre-
ments in each generation), leading to today's long-necked
giraffes. An alternative view, evolution by natural selec-
tion, was put forward in 1859 by Charles Darwin. Ac-
cording to the Darwinian view, giraffes normally varied
in neck length, and these variations were inherited.
Giraffes with slightly longer necks would be at an advan-
tage in reaching edible leaves in trees. Therefore, over

time, the longer-necked giraffes would survive and
reproduce better than the shorter-necked ones. Thus,
longer necks would come to predominate. Any genetic
mutations (changes) that introduced greater neck length
would be favored.

To test Lamarck's hypothesis, you might begin by de-
signing an experiment. You could do the experiment on
giraffes to test Lamarck's hypothesis directly; however, gi-
raffes are difficult to acquire, maintain, and breed. Re-
member, though, that you are testing a general hypothe-
sis about the inheritance of acquired characteristics
rather than a specific hypothesis about giraffes. Thus, if
you are clever enough, you can test the hypothesis with
almost any organism. You would certainly choose one
that is easy to maintain and manipulate experimentally.
Later, you can verify the generality of any particular con-
clusions with tests on other organisms.

You might decide to use lab mice, which are relatively
inexpensive to obtain and keep and have a relatively
short generation time of about six weeks, compared with
the giraffe's gestation period of over a year. Instead of
looking at neck length, you might simply cut off the tip of
the tail of each mouse (in a painless manner), using short-
ened tails as the acquired characteristic. You could then

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

Chapter One Introduction

BOX 1 . 1

As the pictures of geneticists
throughout this book indi-
cate, science is a very human
activity; people living within soci-
eties explore scientific ideas and
combine their knowledge. The soci-
ety in which a scientist lives can
affect not only how that scientist
perceives the world, but also what
that scientist can do in his or her
scholarly activities. For example, the
United States and other countries
decided that mapping the entire hu-
man genome would be valuable (see
chapter 13). Thus, granting agencies
have directed money in this direc-
tion. Since much of scientific re-
search is expensive, scientists often
can only study areas for which fund-
ing is available. Thus, many scientists
are working on the Human Genome
Project. That is a positive example of
society directing research. Examples
also exist in which a societal decision
has had negative consequences for
both the scientific establishment
and the society itself. An example is

Ethics and Genetics

The Lysenko Affair

the Lysenko affair in the former
Soviet Union during Stalin's and
Krushchev's reigns.

Trofim Denisovich Lysenko was a
biologist in the former Soviet Union
researching the effects of temperature
on plant development. At the same
time, the preeminent Soviet geneticist
was Nikolai Vavilov Vavilov was inter-
ested in improving Soviet crop yields
by growing and mating many vari-
eties and selecting the best to be the
breeding stock of the next generation.
This is the standard way of improving
a plant crop or livestock breed (see
chapter 18, "Quantitative Inheri-
tance"). The method conforms to ge-
netic principles and therefore is suc-
cessful. However, it is a slow process
that only gradually improves yields.

Lysenko suggested that crop
yields could be improved quickly by
the inheritance of acquired charac-
teristics (see chapter 21, "Evolution
and Speciation"). Although doomed
to fail because they denied the true
and correct mechanisms of inheri-
tance, Lysenko's ideas were greeted
with much enthusiasm by the politi-
cal elite. The enthusiasm was due not
only to the fact that Lysenko prom-
ised immediate improvements in
crop yields, but also to the fact that
Lysenkoism was politically favored.
That is, Lysenkoism fit in very well
with communism; it promised that
nature could be manipulated easily
and immediately. If people could ma-
nipulate nature so easily, then com-
munism could easily convert people
to its doctrines.

Not only did Stalin favor Lysenko-
ism, but Lysenko himself was favored
politically over Vavilov because Ly-
senko came from peasant stock,
whereas Vavilov was from a wealthy
family. (Remember that communism

mate these short-tailed mice to see if their offspring have
shorter tails. If they do not, you could conclude that a
shortened tail, an acquired characteristic, is not inher-
ited. If, however, the next generation of mice have tails
shorter than those of their parents, you could conclude
that acquired characteristics can be inherited.

One point to note is that every good experiment has
a control, a part of the experiment that ensures that
some unknown variable, often specific to a particular
time and place, is not causing the observed changes. For
example, in your experiment, the particular food the
mice ate may have had an effect on their growth, result-
ing in offspring with shorter tails. To control for this, you
could handle a second group of mice in the exact same
way that the experimental mice are handled, except you
would not cut off their tails. Any reduction in the lengths
of the tails of the offspring of the control mice would in-
dicate an artifact of the experiment rather than the in-
heritance of acquired characteristics.

The point of doing this experiment (with the control
group), as trivial as it might seem, is to determine the an-

swer to a question using data based on what happens in
nature. If you design your experiment correctly and
carry it out without error, you can be confident about
your results. If your results are negative, as ours would be
here, then you would reject your hypothesis. Testing hy-
potheses and rejecting those that are refuted is the
essence of the scientific method.

In fact, most of us live our lives according to the sci-
entific method without really thinking about it. For ex-
ample, we know better than to step out into traffic with-
out looking because we are aware, from experience
(observation, experimentation), of the validity of the
laws of physics. Although from time to time anti-
intellectual movements spread through society, few peo-
ple actually give up relying on their empirical knowledge
of the world to survive (box 1.1).

Nothing in this book is inconsistent with the scien-
tific method. Every fact has been gained by experiment
or observation in the real world. If you do not accept
something said herein, you can go back to the original
literature, the published descriptions of original experi-

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
Companies, 2001

Why Fruit Flies and Colon Bacteria?

was a revolution of the working class
over the wealthy aristocracy.) Sup-
ported by Stalin, and then Krushchev,
Lysenko gained inordinate power in
his country. All visible genetic re-
search in the former Soviet Union
was forced to conform to Lysenko 's
Lamarckian views. People who dis-
agreed with him were forced out of
power; Vavilov was arrested in 1940
and died in prison in 1943. It was not
until Nikita Krushchev lost power
in 1964 that Lysenkoism fell out of
favor. Within months, Lysenko 's
failed pseudoscience was repudiated
and Soviet genetics got back on track.
For thirty years, Soviet geneticists
were forced into fruitless endeavors,
forced out of genetics altogether, or
punished for their heterodox views.
Superb scientists died in prison while
crop improvement programs failed,
all because the Soviet dictators fa-
vored Lysenkoism. The message of
this affair is clear: Politicians can sup-
port research that agrees with their
political agenda and punish scientists

Trofim Denisovich Lysenko (1 898-1 976) shows branched wheat to collective
farmers in the former Soviet Union. (© SOVFOTO.)

doing research that disagrees with
this agenda, but politicians cannot
change the truth of the laws of na-
ture. Science, to be effective, must be

done in a climate of open inquiry and
free expression of ideas. The scien-
tific method cannot be subverted by
political bullies.

merits in scientific journals (as cited at the end of the
book) and read about the work yourself. If you still don't
believe a conclusion, you can repeat the work in ques-
tion either to verify or challenge it. This is in keeping
with the nature of the scientific method.

As mentioned, the results of experimental studies are
usually published in scientific journals. Examples of jour-
nals that many geneticists read include Genetics, Pro-
ceedings of the National Academy of Sciences, Science,
Nature, Evolution, Cell, American Journal of Human
Genetics, Journal of Molecular Biology, and hundreds
more. The reported research usually undergoes a process
called peer review in which other scientists review an ar-
ticle before it is published to ensure its accuracy and its
relevance. Scientific articles usually include a detailed jus-
tification for the work, an outline of the methods that al-
lows other scientists to repeat the work, the results, a dis-
cussion of the significance of the results, and citations of
prior work relevant to the present study.

At the end of this book, we cite journal articles de-
scribing research that has contributed to each chapter.

(In chapter 2, we reprint part of Gregor Mendel's
work, and in chapter 9, we reprint a research article by
J. Watson and F. Crick in its entirety.) We also cite sec-
ondary sources, that is, journals and books that publish
syntheses of the literature rather than original contribu-
tions. These include Scientific American, Annual Re-
view of Biochemistry, Annual Review of Genetics,
American Scientist, and others. You are encouraged to
look at all of these sources in your efforts both to im-
prove your grasp of genetics and to understand how sci-
ence progresses.

WHY FRUIT FLIES AND
COLON BACTERIA?

As you read this book, you will see that certain organisms
are used repeatedly in genetic experiments. If the goal of
science is to uncover generalities about the living world,
why do geneticists persist in using the same few organisms

Tamarin: Principles of
Genetics, Seventh Edition

I. Genetics and the
Scientific Method

1. Introduction

©TheMcGraw-Hil
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Chapter One Introduction

.^-

Figure 1.3 Adult female fruit fly, Drosophila melanogaster.
Mutations of eye color, bristle type and number, and wing
characteristics are easily visible when they occur.

in their work? The answer is probably obvious: the or-
ganisms used for any particular type of study have certain
attributes that make them desirable model organisms for
that research.

In the early stages of genetic research, at the turn of
the century no one had yet developed techniques to
do genetic work with microorganisms or mammalian
cells. At that time, the organism of preference was the
fruit fly, Drosophila melanogaster, which developmen-
tal biologists had used (fig. 1.3). It has a relatively short
generation time of about two weeks, survives and
breeds well in the lab, has very large chromosomes in
some of its cells, and has many aspects of its pheno type
(appearance) genetically controlled. For example, it is
easy to see the external results of mutations of genes
that control eye color, bristle number and type, and
wing characteristics such as shape or vein pattern in
the fruit fly.

At the middle of this century, when geneticists devel-
oped techniques for genetic work on bacteria, the com-
mon colon bacterium, Escherichia colt, became a fa-
vorite organism of genetic researchers (fig. 1.4). Because
it had a generation time of only twenty minutes and only
a small amount of genetic material, many research groups
used it in their experiments. Still later, bacterial viruses,
called bacteriophages, became very popular in genetics
labs. The viruses are constructed of only a few types of
protein molecules and a very small amount of genetic
material. Some can replicate a hundredfold in an hour.
Our point is not to list the major organisms geneticists
use, but to suggest why they use some so commonly.

Figure 1.4 Scanning electron micrograph of Escherichia coli
bacteria. These rod-shaped bacilli are magnified 18,000x.
(© K. G. Murti/Visuals Unlimited, Inc.)

Comparative studies are usually done to determine
which generalities discovered in the elite genetic organ-
isms are really scientifically universal.

TECHNIQUES OF STUDY

Each area of genetics has its own particular techniques of
study. Often the development of a new technique, or an
improvement in a technique, has opened up major new
avenues of research. As our technology has improved
over the years, geneticists and other scientists have been
able to explore at lower and lower levels of biological or-
ganization. Gregor Mendel, the father of genetics, did
simple breeding studies of plants in a garden at his
monastery in Austria in the middle of the nineteenth cen-
tury. Today, with modern biochemical and biophysical
techniques, it has become routine to determine the se-
quence of nucleotides (molecular subunits of DNA and
RNA) that make up any particular gene. In fact, one of the
most ambitious projects ever carried out in genetics is the
mapping of the human genome, all 33 billion nucleotides
that make up our genes. Only recently was the technol-
ogy available to complete a project of this magnitude.

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Scientific Method

1. Introduction

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Classical, Molecular, and Evolutionary Genetics

CLASSICAL, MOLECULAR,
AND EVOLUTIONARY
GENETICS

In the next three sections, we briefly outline the general
subject areas covered in the book: classical, molecular,
and evolutionary genetics.

Classical Genetics

Gregor Mendel discovered the basic rules of transmis-
sion genetics in 1866 by doing carefully controlled
breeding experiments with the garden pea plant, Pisum

Alternative forms

Seeds (1) Round

Wrinkled

Pods (2) Full

Constricted

(3) Yellow

Green

Figure 1.5 Mendel worked with garden pea plants. He
observed seven traits of the plant — each with two discrete
forms — that affected attributes of the seed, the pod, and the
stem. For example, all plants had either round or wrinkled
seeds, full or constricted pods, or yellow or green pods.

Diploid parents

Haploid
gametes

TT
Tall

tt
Dwarf

Diploid offspring

Tt
Tall

Figure 1.6 Mendel crossed tall and dwarf pea plants,
demonstrating the rule of segregation. A diploid individual with
two copies of the gene for tallness (7~) per cell forms gametes
that all have the T allele. Similarly, an individual that has two
copies of the gene for shortness (f) forms gametes that all
have the t allele. Cross-fertilization produces zygotes that have
both the T and t alleles. When both forms are present {Tt), the
plant is tall, indicating that the T allele is dominant to the
recessive t allele.

sativum. He found that traits, such as pod color, were
controlled by genetic elements that we now call genes
(fig. 1.5). Alternative forms of a gene are called alleles.
Mendel also discovered that adult organisms have two
copies of each gene (diploid state); gametes receive just
one of these copies (haploid state). In other words, one
of the two parental copies segregates into any given ga-
mete. Upon fertilization, the zygote gets one copy from
each gamete, reconstituting the diploid number (fig.
1.6). When Mendel looked at the inheritance of several

13.0

44.0

48.5

N_x

dumpy wings

ancon wings

black body

72.0
75.5

91.5

104.5
107.0

Tuft bristles
/y spiny legs
purple eyes
apterous (wingless)
tufted head
cinnabar eyes
arctus oculus eyes

Lobe eyes
curved wings

smooth abdomen

brown eyes
orange eyes

Figure 1.7 Genes are located in linear order on chromosomes,
as seen in this diagram of chromosome 2 of Drosophila
melanogaster, the common fruit fly. The centromere is a
constriction in the chromosome. The numbers are map units.

Tamarin: Principles of
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Scientific Method

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Chapter One Introduction

ATP

ADP

Hexokinase

Glucose-6-phosphate

Phosphoglucose
isomerase

Fructose-6-phosphate

ATP

ADP

Phosphofructo-kinase

Fructose-1 ,6-bisphosphate

Figure 1.8 Biochemical pathways are the sequential changes
that occur in compounds as cellular reactions modify them. In
this case, we show the first few steps in the glycolytic pathway
that converts glucose to energy. The pathway begins when
glucose + ATP is converted to glucose-6-phosphate + ADP
with the aid of the enzyme hexokinase. The enzymes are the
products of genes.

traits at the same time, he found that they were inherited
independently of each other. His work has been distilled
into two rules, referred to as segregation and indepen-
dent assortment. Scientists did not accept Mendel's
work until they developed an understanding of the seg-
regation of chromosomes during the latter half of the
nineteenth century At that time, in the year 1900, the
science of genetics was born.

During much of the early part of this century, geneti-
cists discovered many genes by looking for changed or-
ganisms, called mutants. Crosses were made to deter-
mine the genetic control of mutant traits. From this
research evolved chromosomal mapping, the ability to
locate the relative positions of genes on chromosomes
by crossing certain organisms. The proportion of recom-
binant offspring, those with new combinations of
parental alleles, gives a measure of the physical separa-
tion between genes on the same chromosomes in dis-
tances called map units. From this work arose the chro-
mosomal theory of inheritance: Genes are located at
fixed positions on chromosomes in a linear order (fig.
1.7, p. 9). This "beads on a string" model of gene

arrangement was not modified to any great extent until
the middle of this century, after Watson and Crick
worked out the structure of DNA.

In general, genes function by controlling the synthe-
sis of proteins called enzymes that act as biological cata-
lysts in biochemical pathways (fig. 1.8). G. Beadle and
E. Tatum suggested that one gene controls the formation
of one enzyme. Although we now know that many pro-
teins are made up of subunits — the products of several
genes — and that some genes code for proteins that are
not enzymes and other genes do not code for proteins,
the one-gene-one-enzyme rule of thumb serves as a gen-
eral guideline to gene action.

Molecular Genetics

With the exception of some viruses, the genetic material
of all cellular organisms is double-stranded DNA, a dou-
ble helical molecule shaped like a twisted ladder. The
backbones of the helices are repeating units of sugars
(deoxyribose) and phosphate groups. The rungs of the

G---C

Figure 1.9 A look at a DNA double helix, showing the sugar-
phosphate units that form the molecule's "backbone" and the
base pairs that make up the "rungs." We abbreviate a
phosphate group as a "P" within a circle; the pentagonal ring
containing an oxygen atom is the sugar deoxyribose. Bases are
either adenine, thymine, cytosine, or guanine (A, T, C, G).

Tamarin: Principles of
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Classical, Molecular, and Evolutionary Genetics

DNA

ITTTTTTTm

:> Adenine

> Thymine

> Guanine

Cytosine

Figure 1.10 The DNA double helix unwinds during replication,
and each half then acts as a template for a new double helix.
Because of the rules of complementarity, each new double
helix is identical to the original, and the two new double helices
are identical to each other. Thus, an AT base pair in the original
DNA double helix replicates into two AT base pairs, one in
each of the daughter double helices.

ladder are base pairs, with one base extending from
each backbone (fig. 1.9). Only four bases normally occur
in DNA: adenine, thymine, guanine, and cytosine, abbre-
viated A, T, G, and C, respectively There is no restriction
on the order of bases on one strand. However, a rela-
tionship called complementarity exists between bases
forming a rung. If one base of the pair is adenine, the
other must be thymine; if one base is guanine, the other

RNA

DNA

1 1 1 1 1 1 1 1 1 1 1
A A T C C G C C T A T,

TTAGGCGGATA

RNA
transcript

UUAGGCGGAUA

Transcribed
from

Figure 1.11 Transcription is the process that synthesizes RNA
from a DNA template. Synthesis proceeds with the aid of the
enzyme RNA polymerase. The DNA double helix partially
unwinds during this process, allowing the base sequence of
one strand to serve as a template for RNA synthesis. Synthesis
follows the rules of DNA-RNA complementarity: A, T, G, and C
of DNA pair with U, A, C, and G, respectively, in RNA. The
resulting RNA base sequence is identical to the sequence that
would form if the DNA were replicating instead, with the
exception that RNA replaces thymine (T) with uracil (U).

must be cytosine. James Watson and Francis Crick de-
duced this structure in 1953, ushering in the era of mo-
lecular genetics.

The complementary nature of the base pairs of DNA
made the mode of replication obvious to Watson and
Crick: The double helix would "unzip," and each strand
would act as a template for a new strand, resulting in two
double helices exactly like the first (fig. 1.10). Mutation, a
change in one of the bases, could result from either an
error in base pairing during replication or some damage
to the DNA that was not repaired by the time of the next
replication cycle.

Information is encoded in DNA in the sequence of
bases on one strand of the double helix. During gene ex-
pression, that information is transcribed into RNA, the
other form of nucleic acid, which actually takes part in
protein synthesis. RNA differs from DNA in several re-
spects: it has the sugar ribose in place of deoxyribose; it
has the base uracil (U) in place of thymine (T); and it usu-
ally occurs in a single-stranded form. RNA is transcribed
from DNA by the enzyme RNA polymerase, using DNA-
RNA rules of complementarity: A, T, G, and C in DNA pair
with U, A, C, and G, respectively, in RNA (fig. 1.11). The
DNA information that is transcribed into RNA codes for
the amino acid sequence of proteins. Three nucleotide
bases form a codon that specifies one of the twenty

Tamarin: Principles of
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Chapter One Introduction

Table 1 .2

The Genetic Code Dictionary of RNA

Codon

Amino Acid

Codon

Amino Acid

Codon

Amino Acid

Codon

Amino Acid

uuu

Phe

ucu

Ser

UAU

Tyr

UGU

Cys

uuc

Phe

ucc

Ser

UAC

Tyr

UGC

Cys

UUA

Leu

UCA

Ser

UAA

STOP

UGA

STOP

UUG

Leu

UCG

Ser

UAG

STOP

UGG

Trp

CUU

Leu

ecu

Pro

CAU

His

CGU

Arg

cue

Leu

CCC

Pro

CAC

His

CGC

Arg

CUA

Leu

CCA

Pro

CAA

Gin

CGA

Arg

CUG

Leu

CCG

Pro

CAG

Gin

CGG

Arg

AUU

ACU

Thr

AAU

Asn

AGU

Ser

AUC

ACC

Thr

AAC

Asn

AGC

Ser

AUA

ACA

Thr

AAA

Lys

AGA

Arg

AUG

Met (START)

ACG

Thr

AAG

Lys

AGG

Arg

GUU

Val

GCU

Ala

GAU

Asp

GGU

Gly

GUC

Val

GCC

Ala

GAC

Asp

GGC

Gly

GUA

Val

GCA

Ala

GAA

Glu

GGA

Gly

GUG

Val

GCG

Ala

GAG

Glu

GGG

Gly

Note: A codon, specifying one amino acid, is three bases long (read in RNA bases in which U replaced the T of DNA). There are sixty-four different codons, speci-
fying twenty naturally occurring amino acids (abbreviated by three letters: e.g., Phe is phenylalanine — see fig. 11.1 for the names and structures of the amino acids).
Also present is stop (UAA, UAG, UGA) and start (AUG) information.

Ribosomes

RNA

Nascent protein

Figure 1.12 In prokaryotes, RNA translation begins shortly
after RNA synthesis. A ribosome attaches to the RNA and
begins reading the RNA codons. As the ribosome moves along
the RNA, amino acids attach to the growing protein. When the
process is finished, the completed protein is released from the
ribosome, and the ribosome detaches from the RNA. As the
first ribosome moves along, a second ribosome can attach at
the beginning of the RNA, and so on, so that an RNA strand
may have many ribosomes attached at one time.

naturally occurring amino acids used in protein synthe-
sis. The sequence of bases making up the codons are re-
ferred to as the genetic code (table 1.2).

The process of translation, the decoding of nu-
cleotide sequences into amino acid sequences, takes
place at the ribosome, a structure found in all cells that is
made up of RNA and proteins (fig. 1.12). As the RNA
moves along the ribosome one codon at a time, one
amino acid attaches to the growing protein for each
codon.

The major control mechanisms of gene expression
usually act at the transcriptional level. For transcription
to take place, the RNA polymerase enzyme must be able
to pass along the DNA; if this movement is prevented,
transcription stops. Various proteins can bind to the
DNA, thus preventing the RNA polymerase from continu-
ing, providing a mechanism to control transcription. One
particular mechanism, known as the operon model, pro-
vides the basis for a wide range of control mechanisms in
prokaryotes and viruses. Eukaryotes generally contain no
operons; although we know quite a bit about some con-
trol systems for eukaryotic gene expression, the general
rules are not as simple.

In recent years, there has been an explosion of infor-
mation resulting from recombinant DNA techniques.
This revolution began with the discovery of restriction
endonucleases, enzymes that cut DNA at specific se-

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Classical, Molecular, and Evolutionary Genetics

quences. Many of these enzymes leave single-stranded
ends on the cut DNA. If a restriction enzyme acts on both
aplasmid, a small, circular extrachromosomal unit found
in some bacteria, and another piece of DNA (called for-
eign DNA), the two will be left with identical single-
stranded free ends. If the cut plasmid and cut foreign
DNA are mixed together, the free ends can re-form dou-
ble helices, and the plasmid can take in a single piece of
foreign DNA (fig. 1.13). Final repair processes create a
completely closed circle of DNA. The hybrid plasmid is
then reinserted into the bacterium. When the bacterium
grows, it replicates the plasmid DNA, producing many
copies of the foreign DNA. From that point, the foreign
DNA can be isolated and sequenced, allowing re-
searchers to determine the exact order of bases making
up the foreign DNA. (In 2000, scientists announced the
complete sequencing of the human genome.) That se-
quence can tell us much about how a gene works. In ad-
dition, the foreign genes can function within the bac-
terium, resulting in bacteria expressing the foreign genes
and producing their protein products. Thus we have, for
example, E. colt bacteria that produce human growth
hormone.

This technology has tremendous implications in med-
icine, agriculture, and industry. It has provided the oppor-
tunity to locate and study disease-causing genes, such as
the genes for cystic fibrosis and muscular dystrophy, as
well as suggesting potential treatments. Crop plants and
farm animals are being modified for better productivity by
improving growth and disease resistance. Industries that
apply the concepts of genetic engineering are flourishing.

One area of great interest to geneticists is cancer re-
search. We have discovered that a single gene that has
lost its normal control mechanisms (an oncogene) can
cause changes that lead to cancer. These oncogenes exist
normally in noncancerous cells, where they are called
proto-oncogenes, and are also carried by viruses, where
they are called viral oncogenes. Cancer-causing viruses
are especially interesting because most of them are of the
RNA type. AIDS is caused by one of these RNA viruses,
which attacks one of the cells in the immune system.
Cancer can also occur when genes that normally prevent
cancer, genes called anti-oncogenes, lose function. Dis-
covering the mechanism by which our immune system
can produce millions of different protective proteins
{antibodies) has been another success of modern mo-
lecular genetics.

Evolutionary Genetics

From a genetic standpoint, evolution is the change in
allelic frequencies in a population over time. Charles
Darwin described evolution as the result of natural selec-
tion. In the 1920s and 1930s, geneticists, primarily Sewall

Plasmid

Foreign DNA

Treat with a

restriction

endonuclease

ITTTT

I Circle opens

End pieces lost

Hybrid
plasmid

Figure 1.13 Hybrid DNA molecules can be constructed from
a plasmid and a piece of foreign DNA. The ends are made
compatible by cutting both DNAs with the same restriction
endonuclease, leaving complementary ends. These ends will
re-form double helices to form intact hybrid plasmids when the
two types of DNA mix. A repair enzyme, DNA ligase, finishes
patching the hybrid DNA within the plasmid. The hybrid
plasmid is then reinjected into a bacterium, to be grown into
billions of copies that will later be available for isolation and
sequencing, or the hybrid plasmid can express the foreign DNA
from within the host bacterium.

Wright, R. A. Fisher, and J. B. S. Haldane, provided alge-
braic models to describe evolutionary processes. The
marriage of Darwinian theory and population genetics
has been termed neo-Darwinism.

In 1908, G. H. Hardy and W. Weinberg discovered that a
simple genetic equilibrium occurs in a population if the
population is large, has random mating, and has negligible
effects of mutation, migration, and natural selection. This
equilibrium gives population geneticists a baseline for
comparing populations to see if any evolutionary

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Chapter One Introduction

processes are occurring. We can formulate a statement to
describe the equilibrium condition: If the assumptions are
met, the population will not experience changes in allelic
frequencies, and these allelic frequencies will accurately
predict the frequencies of genotypes (allelic combinations
in individuals, e.g., AA, Aa, or ad) in the population.

Recently, several areas of evolutionary genetics have
become controversial. Electrophoresis (a method for sep-
arating proteins and other molecules) and subsequent
DNA sequencing have revealed that much more poly-
morphism (variation) exists within natural populations
than older mathematical models could account for. One
of the more interesting explanations for this variability is
that it is neutral. That is, natural selection, the guiding
force of evolution, does not act differentially on many, if
not most, of the genetic differences found so commonly
in nature. At first, this theory was quite controversial, at-
tracting few followers. Now it seems to be the view the

majority accept to explain the abundance of molecular
variation found in natural populations.

Another controversial theory concerns the rate of
evolutionary change. It is suggested that most evolution-
ary change is not gradual, as the fossil record seems to in-
dicate, but occurs in short, rapid bursts, followed by long
periods of very little change. This theory is called punc-
tuated equilibrium.

A final area of evolutionary biology that has generated
much controversy is the theory of sociobiology Sociobi-
ologists suggest that social behavior is under genetic
control and is acted upon by natural selection, as is any
morphological or physiological trait. This idea is contro-
versial mainly as it applies to human beings; it calls altru-
ism into question and suggests that to some extent we
are genetically programmed to act in certain ways. Peo-
ple have criticized the theory because they feel it justifies
racism and sexism.

SUMMARY

The purpose of this chapter has been to provide a brief
history of genetics and a brief overview of the following
twenty chapters. We hope it serves to introduce the ma-
terial and to provide a basis for early synthesis of some of
the material that, of necessity, is presented in the discrete
units called chapters. This chapter also differs from all
the others because it lacks some of the end materials that

should be of value to you as you proceed: solved prob-
lems, and exercises and problems. These features are pre-
sented chapter by chapter throughout the remainder of
the book. At the end of the book, we provide answers to
exercises and problems and a glossary of all boldface
words throughout the book.

Suggested Readings for chapter 1 are on page B-l.

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MENDEL'S
PRINCIPLES

STUDY OBJECTIVES

1. To understand that genes are discrete units that control the
appearance of an organism 17

2. To understand Mendel's rules of inheritance: segregation and
independent assortment 18

3. To understand that dominance is a function of the interaction
of alleles; similarly, epistasis is a function of the interaction of
nonallelic genes 22

4. To define how genes generally control the production of
enzymes and thus the fate of biochemical pathways 37

The garden pea plant, Pisum sativum.

(©Adam Hart-Davis/SPL/Photo Researchers, Inc.)

STUDY OUTLINE

Mendel's Experiments 17
Segregation 18

Rule of Segregation 18

Testing the Rule of Segregation 21
Dominance Is Not Universal 22
Nomenclature 23
Multiple Alleles 25
Independent Assortment 26

Rule of Independent Assortment 27

Testcrossing Multihybrids 30
Genotypic Interactions 30

Epistasis 32

Mechanism of Epistasis 34
Biochemical Genetics 37

Inborn Errors of Metabolism 37

One-Gene-One-Enzyme Hypothesis 38
Summary 40
Solved Problems 40
Exercises and Problems 41
Critical Thinking Questions 45
Box 2.1 Excerpts from Mendel's Original Paper
Box 2.2 Did Mendel Cheat? 30

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Mendel's Experiments

Genetics is concerned with the transmission,
expression, and evolution of genes, the mol-
ecules that control the function, develop-
ment, and ultimate appearance of individu-
als. In this section of the book, we will look
at the rules of transmission that govern genes and affect
their passage from one generation to the next. Gregor
Johann Mendel discovered these rules of inheritance; we
derive and expand upon his rules in this chapter (fig. 2.1).
In 1900, three botanists, Carl Correns of Germany,
Erich von Tschermak of Austria, and Hugo de Vries of
Holland, defined the rules governing the transmission of
traits from parent to offspring. Some historical contro-
versy exists as to whether these botanists actually redis-
covered Mendel's rules by their own research or whether
their research led them to Mendel's original paper. In any
case, all three made important contributions to the early
stages of genetics. The rules had been published previ-
ously, in 1866, by an obscure Austrian monk, Gregor Jo-
hann Mendel. Although his work was widely available af-
ter 1866, the scientific community was not ready to
appreciate Mendel's great contribution until the turn of
the century. There are at least four reasons for this lapse
of thirty-four years.

First, before Mendel's experiments, biologists were
primarily concerned with explaining the transmission of
characteristics that could be measured on a continuous
scale, such as height, cranium size, and longevity. They
were looking for rules of inheritance that would explain
such continuous variations, especially after Darwin
put forth his theory of evolution in 1859 (see chap-
ter 21). Mendel, however, suggested that inherited char-
acteristics were discrete and constant (discontinuous):
peas, for example, were either yellow or green. Thus, evo-
lutionists were looking for small changes in traits with
continuous variation, whereas Mendel presented them
with rules for discontinuous variation. His principles did
not seem to apply to the type of variation that biologists
thought prevailed. Second, there was no physical ele-
ment identified with Mendel's inherited entities. One
could not say, upon reading Mendel's work, that a certain
subunit of the cell followed Mendel's rules. Third, Mendel
worked with large numbers of offspring and converted
these numbers to ratios. Biologists, practitioners of a very
descriptive science at the time, were not well trained in
mathematical tools. And last, Mendel was not well known
and did not persevere in his attempts to convince the ac-
ademic community that his findings were important.

Between 1866 and 1900, two major changes took
place in biological science. First, by the turn of the cen-
tury, not only had scientists discovered chromosomes,
but they also had learned to understand chromosomal
movement during cell division. Second, biologists were
better prepared to handle mathematics by the turn of the
century than they were during Mendel's time.

Figure 2.1 Gregor Johann Mendel (1822-84). (Reproduced by
permission of the Moravski Museum, Mendelianum.)

MENDEL'S EXPERIMENTS

Gregor Mendel was an Austrian monk (of Briinn, Austria,
which is now Brno, Czech Republic). In his experiments,
he tried to crossbreed plants that had discrete, nonover-
lapping characteristics and then to observe the distribu-
tion of these characteristics over the next several genera-
tions. Mendel worked with the common garden pea
plant, Pisum sativum. He chose the pea plant for at least
three reasons: (1) The garden pea was easy to cultivate
and had a relatively short life cycle. (2) The plant had dis-
continuous characteristics such as flower color and pea
texture. (3) In part because of its anatomy, pollination of
the plant was easy to control. Foreign pollen could be
kept out, and cross-fertilization could be accom-
plished artificially.

Figure 2.2 shows a cross section of the pea flower
that indicates the keel, in which the male and female
parts develop. Normally, self-fertilization occurs when
pollen falls onto the stigma before the bud opens.
Mendel cross-fertilized the plants by opening the keel of

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Chapter Two Mendel's Principles

Filament

Stigma

(half cut
away)

Figure 2.2 Anatomy of the garden pea plant flower. The female
part, the pistil, is composed of the stigma, its supporting style,
and the ovary. The male part, the stamen, is composed of the
pollen-producing anther and its supporting filament.

a flower before the anthers matured and placing pollen
from another plant on the stigma. In the more than ten
thousand plants Mendel examined, only a few were fer-
tilized other than the way he had intended (either self- or
cross-pollinated).

Mendel used plants obtained from suppliers and
grew them for two years to ascertain that they were ho-
mogeneous, or true-breeding, for the particular charac-
teristic under study He chose for study the seven charac-
teristics shown in figure 2.3. Take as an example the
characteristic of plant height. Although height is often
continuously distributed, Mendel used plants that dis-
played only two alternatives: tall or dwarf. He made the
crosses shown in figure 2.4. In the parental, or P l5 gener-
ation, dwarf plants pollinated tall plants, and, in a recip-
rocal cross, tall plants pollinated dwarf plants, to deter-
mine whether the results were independent of the
parents' sex. As we will see later on, some traits follow in-
heritance patterns related to the sex of the parent carry-
ing the traits. In those cases, reciprocal crosses give dif-
ferent results; with Mendel's tall and dwarf pea plants,
the results were the same.

Offspring of the cross of P 1 individuals are referred to
as the first filial generation, or F x . Mendel also referred
to them as hybrids because they were the offspring of
unlike parents (tall and dwarf). We will specifically refer
to the offspring of tall and dwarf peas as monohybrids
because they are hybrid for only one characteristic
(height). Since all the ¥ 1 offspring plants were tall,
Mendel referred to tallness as the dominant trait. The al-
ternative, dwarfness, he referred to as recessive. Differ-

ent forms of a gene that exist within a population are
termed alleles. The terms dominant and recessive are
used to describe both the relationship between the al-
leles and the traits they control. Thus, we say that both
the allele for tallness and the trait, tall, are dominant.
Dominance applies to the appearance of the trait when
both a dominant and a recessive allele are present. It
does not imply that the dominant trait is better, is more
abundant, or will increase over time in a population.

When the F : offspring of figure 2.4 were self-
fertilized to produce the F 2 generation, both tall and
dwarf offspring occurred; the dwarf characteristic reap-
peared. Among the F 2 offspring, Mendel observed 787
tall and 277 dwarf plants for a ratio of 2.84:1. It is an in-
dication of Mendel's insight that he recognized in these
numbers an approximation to a 3:1 ratio, a ratio that sug-
gested to him the mechanism of inheritance at work in
pea plant height.

SEGREGATION

Rule of Segregation

Mendel assumed that each plant contained two determi-
nants (which we now call genes) for the characteristic
of height. For example, a hybrid F : pea plant possesses
the dominant allele for tallness and the recessive allele
for dwarfness for the gene that determines plant height.
A pair of alleles for dwarfness is required to develop the
recessive phenotype. Only one of these alleles is passed
into a single gamete, and the union of two gametes to
form a zygote restores the double complement of alleles.
The fact that the recessive trait reappears in the F 2 gen-
eration shows that the allele controlling it was hidden in
the V 1 individual and passed on unaffected. This explana-
tion of the passage of discrete trait determinants, or
genes, comprises Mendel's first principle, the rule of
segregation. The rule of segregation can be summarized
as follows: A gamete receives only one allele from the
pair of alleles an organism possesses; fertilization (the
union of two gametes) reestablishes the double number.
We can visualize this process by redrawing figure 2.4 us-
ing letters to denote the alleles. Mendel used capital let-
ters to denote alleles that control dominant traits and
lowercase letters for alleles that control recessive traits.
Following this notation, T refers to the allele controlling
tallness and t refers to the allele controlling shortness
(dwarf stature). From figure 2.5, we can see that Mendel's
rule of segregation explains the homogeneity of the V 1
generation (all tall) and the 3:1 ratio of tall-to-dwarf off-
spring in the F 2 generation.

Let us define some terms. The genotype of an organ-
ism is the gene combination it possesses. In figure 2.5,

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Segregation

Alternative forms

Seeds

(1) Round

Wrinkled

(2) Yellow

cotyledons

Green
cotyledons

(3) Gray coat
(violet flowers)

White coat
(white flowers)

Pods

(4) Full

Constricted

(5) Green

Yellow

Stem

(6) Axial pods
and flowers
along stem

Terminal pods
and flowers on
top of stem

(7)

Tall
(6-7 ft)

Dwarf
(3/4-1 ft)

Figure 2.3 Seven characteristics that Mendel observed in peas. Traits in the left column
are dominant.

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Tall

Dwarf

X Self

Tall

Dwarf

3 : 1

Figure 2.4 First two offspring generations from the cross of tall plants with dwarf plants.

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Segregation

Gametes

X Self

Gametes

T or t

i 77

i
3/4

1/4

ill

arf

3:1

Figure 2.5 Assigning genotypes to the cross in figure 2.4.

the genotype of the parental tall plant is TT; that of the ¥ 1
tall plant is Tt. Phenotype refers to the observable at-
tributes of an organism. Plants with either of the two
genotypes TT or Tt are phenotypically tall. Genotypes
come in two general classes: homozygotes, in which
both alleles are the same, as in TT or tt, and heterozy-
gotes, in which the two alleles are different, as in Tt.
William Bateson coined these last two terms in 1901.
Danish botanist Wilhelm Johannsen first used the word
gene in 1909.

If we look at figure 2.5, we can see that the TT
homozygote can produce only one type of gamete, the
T-bearing kind, and the tt homozygote can similarly pro-
duce only ^-bearing gametes. Thus, the F : individuals are
uniformly heterozygous Tt, and each F : individual can
produce two kinds of gametes in equal frequencies, T- or
^-bearing. In the F 2 generation, these two types of ga-
metes randomly pair during fertilization. Figure 2.6
shows three ways of picturing this process.

Testing the Rule of Segregation

We can see from figure 2.6 that the F 2 generation has a
phenotypic ratio of 3:1, the classic Mendelian ratio.
However, we also see a genotypic ratio of 1:2:1 for domi-
nant homozygote :heterozygote: recessive homozygote.
Demonstrating this genotypic ratio provides a good test
of Mendel's rule of segregation.

The simplest way to test the hypothesis is by prog-
eny testing, that is, by self-fertilizing F 2 individuals to

Schematic

Tt X Tt

(as in fig. 2.5)

Pollen
Tt

T T t

Ovule
Tt

f + +

T t T

T T
TT Tt

1:2:

Diagrammatic

(Punnett square)

Pollen

T t

+
t

tt
1

TT
1

W T

Tt tt

o t

: 2 : 1

Probabilistic

(Multiply; see rule 2, chapter 4.)

Pollen

Ovules

1/2 T

1/2
1/2

1/4 77
1/4 Tt

1/2 t

1/2
1/2

7 =

1/4 7*
1/4 tt

Figure 2.6 Methods of determining F 2 genotypic combinations
in a self -fertilized monohybrid. The Punnett square diagram is
named after the geneticist Reginald C. Punnett.

produce an F 3 generation, which Mendel did (fig. 2.7).
Treating the rule of segregation as a hypothesis, it is pos-
sible to predict the frequencies of the phenotypic classes
that would result. The dwarf F 2 plants should be reces-
sive homozygotes, and so, when selfed (self-fertilized),
they should produce only ^-bearing gametes and only
dwarf offspring in the F 3 generation. The tall F 2 plants,
however, should be a heterogeneous group, one-third of
which should be homozygous TT and two-thirds het-
erozygous Tt. The tall homozygotes, when selfed, should
produce only tall F 3 offspring (geno typically TT). How-
ever, the F 2 heterozygotes, when selfed, should produce

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Tall

TT X Self

Tall
100%

Tt X Self

Tall

Dwarf
1

Dwarf

tt X Self

t
Dwarf
100%

Figure 2.7 Mendel self-fertilized F 2 tall and dwarf plants. He found that
all the dwarf plants produced only dwarf progeny. Among the tall plants,
72% produced both tall and dwarf progeny in a 3:1 ratio.

Genotype to be tested

Gamete of aa

Offspring

„ . Gamete

AA >- ( a ) X (a

(dominant phenotype)

Gamete

X a

(dominant phenotype)

(recessive phenotype)

Figure 2.8 Testcross. In a testcross, the phenotype of an offspring is
determined by the allele the offspring inherits from the parent with the
genotype being tested.

tall and dwarf offspring in a ratio identical to that the
selfed Fj plants produced: three tall to one dwarf off-
spring. Mendel found that all the dwarf (homozygous) F 2
plants bred true as predicted. Among the tall, 28%
(28/100) bred true (produced only tall offspring) and
72% (72/100) produced both tall and dwarf offspring.
Since the prediction was one-third (33. 3%) and two-
thirds (66.7%), respectively, Mendel's observed values
were very close to those predicted. We thus conclude
that Mendel's progeny-testing experiment confirmed his
hypothesis of segregation. In fact, a statistical test —
developed in chapter 4 — would also the support this
conclusion.

Another way to test the segregation rule is to use the
extremely useful method of the testcross, that is, a cross
of any organism with a recessive homozygote. (Another
type of cross, a backcross, is the cross of a progeny with

Tall (two classes)

TT X tt = all Tt
Tt X tt =Tt : tt
1 :1

Figure 2.9 Testcrossing the dominant phenotype of the F 2
generation from figure 2.5.

an individual that has a parental genotype. Hence, a test-
cross can often be a backcross.) Since the gametes of the
recessive homozygote contain only recessive alleles, the
alleles that the gametes of the other parent carry will de-
termine the phenotypes of the offspring. If a gamete
from the organism being tested contains a recessive al-
lele, the resulting F : organism will have a recessive phe-
notype; if it contains a dominant allele, the ¥ 1 organism
will have a dominant phenotype. Thus, in a testcross, the
genotypes of the gametes from the organism being
tested determine the phenotypes of the offspring
(fig. 2.8). A testcross of the tall F 2 plants in figure 2.5
would produce the results shown in figure 2.9. These re-
sults further confirm Mendel's rule of segregation.

DOMINANCE IS NOT
UNIVERSAL

If dominance were universal, the heterozygote would al-
ways have the same phenotype as the dominant ho-
mozygote, and we would always see the 3:1 ratio when
heterozygotes are crossed. If, however, the heterozygote
were distinctly different from both homozygotes, we

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Nomenclature

Red

White
R 2 R 2

Pink x Self
R^R 2

Red

Pink
R^R 2

1:2:1

White

RpRn

Figure 2.10 Flower color inheritance in the four-o'clock plant:
an example of partial, or incomplete, dominance.

would see a 1:2:1 ratio of phenotypes when heterozy-
gotes are crossed. In partial dominance (or incom-
plete dominance), the phenotype of the heterozygote
falls between those of the two homozygotes. An example
occurs in flower petal color in some plants.

Using four-o'clock plants (Mirabilis jalapa), we can
cross a plant that has red flower petals with another that
has white flower petals; the offspring will have pink
flower petals. If these pink-flowered F : plants are
crossed, the F 2 plants appear in a ratio of 1:2:1, having
red, pink, or white flower petals, respectively (fig. 2.10).
The pink-flowered plants are heterozygotes that have a
petal color intermediate between the red and white col-
ors of the homozygotes. In this case, one allele (R^ spec-
ifies red pigment color, and another allele specifies no
color (i? 2 ; the flower petals have a white background

color). Flowers in heterozygotes (i?ii? 2 ) have about half
the red pigment of the flowers in red homozygotes
(RiR^ because the heterozygotes have only one copy of
the allele that produces color, whereas the homozygotes
have two copies.

As technology has improved, we have found more
and more cases in which we can differentiate the het-
erozygote. It is now clear that dominance and recessive-
ness are phenomena dependent on which alleles are in-
teracting and on what phenotypic level we are studying.
For example, in Tay-Sachs disease, homozygous recessive
children usually die before the age of three after suffering
severe nervous system degeneration; heterozygotes seem
to be normal. As biologists have discovered how the dis-
ease works, they have made the detection of the het-
erozygotes possible.

As with many genetic diseases, the culprit is a defec-
tive enzyme (protein catalyst). Afflicted homozygotes
have no enzyme activity, heterozygotes have about half
the normal level, and, of course, homozygous normal in-
dividuals have the full level. In the case of Tay-Sachs dis-
ease, the defective enzyme is hexoseaminidase-A, needed
for proper lipid metabolism. Modern techniques allow
technicians to assay the blood for this enzyme and to
identify heterozygotes by their intermediate level of en-
zyme activity. Two heterozygotes can now know that
there is a 25% chance that any child they bear will have
the disease. They can make an educated decision as to
whether or not to have children.

The other category in which the heterozygote is dis-
cernible occurs when the heterozygous phenotype is
not on a scale somewhere between the two homozy-
gotes, but actually expresses both phenotypes simulta-
neously. We refer to this situation as codominance. For
example, people with blood type AB are heterozygotes
who express both the A and B alleles for blood type (see
the section entitled "Multiple Alleles" for more informa-
tion about blood types). Electrophoresis (a technique de-
scribed in chapter 5) lets us see proteins directly and also
gives us many examples of codominance when we can
see the protein products of both alleles.

NOMENCLATURE

Throughout the last century, botanists, zoologists, and
microbiologists have adopted different methods for nam-
ing alleles. Botanists and mammalian geneticists tend to
prefer the capital-lowercase scheme. Drosophila geneti-
cists and microbiologists have adopted schemes that re-
late to the wild-type. The wild-type is the phenotype of
the organism commonly found in nature. Though other
naturally occurring phenotypes of the same species may
also be present, there is usually an agreed-upon common

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Table 2.1 Some Mutants of Drosophila

Adult male

Adult female

Figure 2.11 Wild-type fruit fly, Drosophila melanogaster.

phenotype that is referred to as the wild-type. For fruit
flies (Drosophila), organisms commonly used in genetic
studies, the wild-type has red eyes and round wings
(fig. 2.11). Alternatives to the wild-type are referred to as
mutants (fig. 2.12). Thus, red eyes are wild-type, and
white eyes are mutant. Fruit fly genes are named after the
mutant, beginning with a capital letter if the mutation is
dominant and a lowercase letter if it is recessive.
Table 2.1 gives some examples. The wild-type allele often
carries the symbol of the mutant with a + added as a

Dominance

Mutant

Relationship

Designation

Description

to Wild-Type

abrupt (ab)

Shortened, longitudinal,
median wing vein

Recessive

amber iamb)

Pale yellow body

Recessive

black (b)

Black body

Recessive

Bar (E)

Narrow, vertical eye

Dominant

dumpy (dp)

Reduced wings

Recessive

Hairless (H)

Various bristles absent

Dominant

white (w)

White eye

Recessive

white-apricot

Apricot-colored eye

Recessive

(w a )

(allele of white eye)

superscript; by definition, every mutant has a wild-type
allele as an alternative. For example, w stands for the
white-eye allele, a recessive mutation. The wild-type (red
eyes) is thus assigned the symbol w + . Hairless is a domi-
nant allele with the symbol H. Its wild-type allele is de-
noted as H + . Sometimes geneticists use the + symbol
alone for the wild-type, but only when there will be no
confusion about its use. If we are discussing eye color
only, then + is clearly the same as w + : both mean red
eyes. However, if we are discussing both eye color and
bristle morphology, the + alone could refer to either of
the two aspects of the phenotype and should be avoided.

Figure 2.12 Wing mutants of Drosophila melanogaster and their allelic designations: Cy, curly; sd, scalloped; ap, apterous; vg,
vestigial; dp, dumpy; D, Dichaete; c, curved.

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Multiple Alleles

Table 2.2 ABO Blood Types with Immunity Reactions

Blood Type Corresponding

Reaction of Red

to Antigens on Red

Cells to Anti-A

Cells to Anti-B

Blood Cells

Antibodies in Serum

Genotype

Antibodies

Anti-A and anti-B

—

Anti-B

I A I A or I A i

—

Anti-A

ff or A

—

None

I A f

MULTIPLE ALLELES

A given gene can have more than two alleles. Although
any particular individual can have only two, many alleles
of a given gene may exist in a population. The classic ex-
ample of multiple human alleles is in the ABO blood
group, which Karl Landsteiner discovered in 1900. This is
the best known of all the red-cell antigen systems pri-
marily because of its importance in blood transfusions.
There are four blood-type phenotypes produced by three
alleles (table 2.2). The I A and I B alleles are responsible for
the production of the A and B antigens found on the sur-
face of the erythrocytes (red blood cells). Antigens are
substances, normally foreign to the body, that induce the
immune system to produce antibodies (proteins that
bind to the antigens). The ABO system is unusual because
antibodies can be present (e.g., anti-B antibodies can ex-
ist in a type A person) without prior exposure to the anti-
gen. Thus, people with a particular ABO antigen on their
red cells will have in their serum the antibody against the

other antigen: type A persons have A antigen on their red
cells and anti-B antibody in their serum; type B persons
have B antigen on their red cells and anti-A antibody in
their serum; type O persons do not have either antigen
but have both antibodies in their serum; and type AB
persons have both A and B antigens and form neither
anti-A or anti-B antibodies in their serum.

The I A and I B alleles, coding for glycosyl transferase
enzymes, each cause a different modification to the ter-
minal sugars of a mucopolysaccharide (H structure)
found on the surface of red blood cells (fig. 2.13). They
are codominant because both modifications (antigens)
are present in a heterozygote. In fact, whichever enzyme
(product of the I A or I B allele) reaches the H structure
first will modify it. Once modified, the H structure will
not respond to the other enzyme. Therefore, both A and
B antigens will be produced in the heterozygote in
roughly equal proportions. The i allele causes no change
to the H structure: because of a mutation it produces a
nonfunctioning enzyme. The / allele and its phenotype
are recessive; the presence of the I A or I B allele, or both,

H structure

Fucose

/allele
(no change in
H structure)

Gal

Glunac

/ A allele
(Galnac added
to H structure)

/ B allele
(Gal added to
H structure)

Fucose

Gal

Glunac

Fucose

Gal

Galnac

Glunac

Fucose

Gal

Glunac

Gal

Gal = Galactose

Galnac = N-Acetylgalactosamine

Glunac = N-Acetylglucosamine

Figure 2.13 Function of the l A , l B , and / alleles of the ABO gene. The gene products of the / A and / B alleles of the ABO gene affect
the terminal sugars of a mucopolysaccharide (H structure) found on red blood cells. The gene products of the / A and / B alleles are
the enzymes alpha-3-N-acetyl-D-galactosaminyltransferase and alpha-3-D-galactosyltransferase, respectively.

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will modify the H product, thus masking the fact that
the i allele was ever there.

Adverse reactions to blood transfusions primarily occur
because the antibodies in the recipient's serum react with
the antigens on the donor's red blood cells. Thus, type A
persons cannot donate blood to type B persons. Type B
persons have anti-A antibody, which reacts with the A anti-
gen on the donor red cells and causes the cells to clump.

Since both I A and I B are dominant to the i allele, this
system not only shows multiple allelism, it also demon-
strates both codominance and simple dominance. (As
with virtually any system, intense study yields more in-
formation, and subgroups of type A are known. We will
not, however, deal with that complexity here.) According
to the American Red Cross, 46% of blood donors in the
United States are type O, 40% are type A, 10% are type B,
and 4% are type AB.

Many other genes also have multiple alleles. In some
plants, such as red clover, there is a gene, the S gene, with
several hundred alleles that prevent self-fertilization. This
means that a pollen grain is not capable of forming a suc-
cessful pollen tube in the style if the pollen grain or its
parent plant has a self-incompatibility allele that is also
present in the plant to be fertilized. Thus, pollen grains
from a flower falling on its own stigma are rejected. Only
a pollen grain with either a different self-incompatibility
allele or from a parent plant with different self-
incompatibility alleles is capable of fertilization; this
avoids inbreeding. Thus, over evolutionary time, there
has been selection for many alleles of this gene. Presum-
ably, a foreign plant would not want to be mistaken for
the same plant, providing the selective pressure for many
alleles to survive in a population. Recent research has in-
dicated that the products of the S alleles are ribonuclease
enzymes, enzymes that destroy RNA. Researchers are in-
terested in discovering the molecular mechanisms for
this pollen rejection.

In Drosophila, numerous alleles of the white-eye gene
exist, and people have numerous hemoglobin alleles. In
fact, multiple alleles are the rule rather than the exception.

INDEPENDENT ASSORTMENT

Mendel also analyzed the inheritance pattern of traits ob-
served two at a time. He looked, for instance, at plants
that differed in the form and color of their peas: he
crossed true-breeding (homozygous) plants that had
seeds that were round and yellow with plants that pro-
duced seeds that were wrinkled and green. Mendel's re-
sults appear in figure 2.14. The F : plants all had round,
yellow seeds, which demonstrated that round was domi-
nant to wrinkled and yellow was dominant to green.
When these F 1 plants were self-fertilized, they produced

an F 2 generation that had all four possible combinations
of the two seed characteristics: round, yellow seeds;
round, green seeds; wrinkled, yellow seeds; and wrin-
kled, green seeds. The numbers Mendel reported in these
categories were 315, 108, 101, and 32, respectively. Di-
viding each number by 32 gives a 9.84 to 3.38 to 3.16 to
1.00 ratio, which is very close to a 9:3:3:1 ratio. As you
will see, this is the ratio we would expect if the genes
governing these two traits behaved independently of
each other.

In figure 2.14, the letter R is assigned to the dominant
allele, round, and r to the recessive allele, wrinkled; Fand
y are used for yellow and green color, respectively. In fig-
ure 2.15, we have rediagrammed the cross in figure 2.14.
The P : plants in this cross produce only one type of ga-
mete each, RY for the parent with the dominant traits
and ry for the parent with the recessive traits. The result-
ing F : plants are heterozygous for both genes (dihy-
brid). Self-fertilizing the dihybrid (RrYy) produces the
F 2 generation.

In constructing the Punnett square in figure 2.15 to
diagram the F 2 generation, we make a critical assump-
tion: The four types of gametes from each parent will be
produced in equal numbers, and hence every offspring
category, or "box," in the square is equally likely. Thus, be-
cause sixteen boxes make up the Punnett square (named
after its inventor, Reginald C. Punnett), the ratio of F 2 off-
spring should be in sixteenths. Grouping the F 2 offspring
by phenotype, we find there are 9/16 that have round,
yellow seeds; 3/16 that have round, green seeds; 3/16
that have wrinkled, yellow seeds; and 1/16 that have
wrinkled, green seeds. This is the origin of the expected
9:3:3:1 F 2 ratio.

Reginald C. Punnett (1875-1967).
From Genetics, 58 (1968): frontispiece.
Courtesy of the Genetics Society of
America.

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Independent Assortment

Round, yellow
{RRYY)

Wrinkled, green
(rryy)

XSelf

Round, yellow
{RrYy)

Round, yellow
(315)

(RRYY; RRYy;
RrYY;RrYy)

Round, green
(108)
(RRyy; Rryy)

Round, yellow
RRYY

Wrinkled, green
rryy

Gametes

RrYy

Gametes

Qty) (V)

1:1:1:1

Pollen

F 2 RY

CO
_CD

RRYY

RRYy

RrYY

RrYy

RRYy

RRyy

RrYy

Rryy

RrYY

RrYy

rrYY

wfi/

rrYy

RrYy

Rryy

rrYy

rryy

Wrinkled, yellow

(101)

(rrYY; rrYy)

Figure 2.15 Assigning genotypes to the cross in figure 2.14.

Wrinkled, green

(32)

(rryy)

Figure 2.14 Independent assortment in garden peas.

Rule of Independent Assortment

This ratio comes about because the two characteristics
behave independently. The F : plants produce four types
of gametes (check fig. 2.15): RY, Ry, rY, and ry These ga-
metes occur in equal frequencies. Regardless of which
seed shape allele a gamete ends up with, it has a 50:50

chance of getting either of the alleles for color — the two
genes are segregating, or assorting, independently This is
the essence of Mendel's second rule, the rule of inde-
pendent assortment, which states that alleles for one
gene can segregate independently of alleles for other
genes. Are the alleles for the two characteristics of color
and form segregating properly according to Mendel's
first principle?

If we look only at seed shape (see fig. 2.14), we find
that a homozygote with round seeds was crossed with a
homozygote with wrinkled seeds in the V x generation
(RR X rr). This cross yields only heterozygous plants
with round seeds (Rr) in the ¥ 1 generation. When these

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Chapter Two Mendel's Principles

BOX 2 . 1

In February and March of 1865,
Mendel delivered two lectures to
the Natural History Society of
Briinn. These were published as a
single forty-eight-page article hand-
written in German. The article ap-
peared in the 1865 Proceedings of
the Society, which came out in 1866.
It was entitled "Versuche iiber
Pflanzen-Hybriden," which means
"Experiments in Plant Hybridization."
Following are some paragraphs from
the English translation to give us
some sense of the original.

In his introductory remarks,
Mendel writes:

That, so far, no generally applicable
law governing the formation and
development of hybrids has been
successfully formulated can hardly
be wondered at by anyone who is
acquainted with the extent of the
task, and can appreciate the difficul-
ties with which experiments of this
class have to contend. A final deci-
sion can only be arrived at when
we shall have before us the results
of detailed experiments made on
plants belonging to the most diverse
orders.

Those who survey the work
done in this department will arrive
at the conviction that among all the
numerous experiments made, not
one has been carried out to such an
extent and in such a way as to make
it possible to determine the number
of different forms under which the
offspring of hybrids appear, or to
arrange these forms with certainty
according to their separate genera-

Historical
Perspectives

Excerpts from Mendel's
Original Paper

tions, or definitely to ascertain their
statistical relations. . . .

The paper now presented
records the results of such a detailed
experiment. This experiment was
practically confined to a small plant
group, and is now, after eight years'
pursuit, concluded in all essentials.
Whether the plan upon which the
separate experiments were con-
ducted and carried out was the best
suited to attain the desired end is
left to the friendly decision of the
reader.

After discussing the origin of his
seeds and the nature of the experi-
ments, Mendel discusses the F 1} or hy-
brid, generation:

This is precisely the case with the
Pea hybrids. In the case of each
of the seven crosses the hybrid-
character resembles that of one
of the parental forms so closely that
the other either escapes observa-
tion completely or cannot be
detected with certainty. This circum-
stance is of great importance in the
determination and classification of
the forms under which the offspring
of the hybrids appear. Henceforth in
this paper those characters which

are transmitted entire, or almost un-
changed in the hybridization, and
therefore in themselves constitute
the characters of the hybrid, are
termed the dominant, and those
which become latent in the process,
recessive. The expression "reces-
sive" has been chosen because the
characters thereby designated with-
draw or entirely disappear in the hy-
brids, but nevertheless reappear un-
changed in their progeny, as will be
demonstrated later on.

He then writes about the F 2 genera-
tion:

In this generation there reappear, to-
gether with the dominant charac-
ters, also the recessive ones with
their peculiarities fully developed,
and this occurs in the definitely ex-
pressed average proportion of three
to one, so that among each four
plants of this generation three dis-
play the dominant character and
one the recessive. This relates with-
out exception to all the characters
which were investigated in the ex-
periments. The angular wrinkled
form of the seed, the green colour of
the albumen, the white colour of
the seed-coats and the flowers, the
constrictions of the pods, the yel-
low colour of the unripe pod, of
the stalk, of the calyx, and of the
leaf venation, the umbel-like form
of the inflorescence, and the
dwarfed stem, all reappear in the nu-
merical proportion given, without
any essential alteration. Transi-
tional forms were not observed in
any experiment. . . .

¥ 1 plants are self-fertilized, the result is 315 + 108 round
seeds (RR or Rf) and 101 + 32 wrinkled seeds (rr) in the
F 2 generation. This is a 423:133 or a 3.18:1.00 pheno-
typic ratio — very close to the expected 3:1 ratio. So the
gene for seed shape is segregating normally. In a similar
manner, if we look only at the gene for color, we see that
the F 2 ratio of yellow to green seeds is 4 16: 140, or

2.97:1.00 — again, very close to a 3:1 ratio. Thus, when
two genes are segregating normally according to the rule
of segregation, their independent behavior demonstrates
the rule of independent assortment (box 2.1).

From the Punnett square in figure 2.15, you can see
that because of dominance, all phenotypic classes ex-
cept the homozygous recessive one — wrinkled, green

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Independent Assortment

Expt. 1. Form of seed. — From
253 hybrids 7,324 seeds were ob-
tained in the second trial year.
Among them were 5,474 round or
roundish ones and 1,850 angular
wrinkled ones. Therefrom the ratio
2.96 to 1 is deduced.

If A be taken as denoting one of
the two constant characters, for in-
stance the dominant, a the reces-
sive, and Aa the hybrid form in
which both are conjoined, the ex-
pression

A + 2Aa + a

shows the terms in the series for the
progeny of the hybrids of two differ-
entiating characters.

Mendel used a notation system
different from ours. He designated
heterozygotes with both alleles
(e.g., Ad) but homozygotes with only
one allele or the other (e.g., A for our
AA). Thus, whereas he recorded A +
2Aa + a, we would record AA +
2Aa + aa. Mendel then went on to
discuss the dihybrids. He mentions
the genotypic ratio of 1:2:1:2:4:
2:1:2:1 and the principle of inde-
pendent assortment:

The fertilized seeds appeared round
and yellow like those of the seed
parents. The plants raised therefrom
yielded seeds of four sorts, which
frequently presented themselves in
one pod. In all, 556 seeds were
yielded by 15 plants, and of those
there were:

315 round and yellow,
101 wrinkled and yellow,

108 round and green,
32 wrinkled and green.

Consequently the offspring of
the hybrids, if two kinds of differen-
tiating characters are combined
therein, are represented by the ex-
pression

AB + Ab + aB + ab + 2ABb +
2aBb + 2AaB + 2Aab + 4AaBb.

(In today's notation, we would write:
AABB + AAbb + aaBB + aabb +
2AABb + 2aaBb + 2AaBB + 2Aabb
+ 4AaBb.)

This expression is indisputably a
combination series in which the
two expressions for the characters^
and a, B and b are combined. We ar-
rive at the full number of the classes
of the series by the combination of
the expressions

A + 2Aa + a

B + 2Bb + b

Table 1 Mendel's Data

(In today's notation we would write

AA + 2Aa + aa

BB + 2Bb + bb.)

There is therefore no doubt that for
the whole of the characters in-
volved in the experiments the prin-
ciple applies that the offspring of
the hybrids in which several essen-
tially different characters are com-
bined exhibit the terms of a series
of combinations, in which the de-
velopmental series for each pair of
differentiating characters are
united. It is demonstrated at the
same time that the relation of each
pair of different characters in hy-
brid union is independent of the
other differences in the two origi-
nal parental stocks.

Table 1 is a summary of all the data
Mendel presented on monohybrids
(the data from only one dihybrid and
one trihybrid cross were presented):

Dominant Phenotype Recessive Phenotype Ratio

Seed form

5,474

Cotyledon color

6,022

Seed coat color

705

Pod form

882

Pod color

428

Flower position

651

Stem length

787

Total

14,949

1,850

2.96

2,001

3.01

224

3.15

299

2.95

152

2.82

207

3.14

277

2.84

5,010

2.98

Source: Copyright The Royal Horticultural Society. Taken from the Journal of the Royal Horti-
cultural Society, vol. 26. Pg. 1-32. 1901.

seeds — are actually genetically heterogeneous, with phe-
notypes made up of several genotypes. For example, the
dominant phenotypic class, with round, yellow seeds,
represents four genotypes: RRYY, RRYy, RrYY, and RrYy.
When we group all the genotypes by phenotype, we ob-
tain the ratio shown in figure 2.16. Thus, with complete
dominance, a self-fertilized dihybrid gives a 9:3:3:1 phe-

notypic ratio in its offspring (F 2 ). A 1:2:1:2:4:2:1:2:1
genotypic ratio also occurs in the F 2 generation. If the
two genes exhibited incomplete dominance or codomi-
nance, the latter would also be the phenotypic ratio.
What ratio would be obtained if one gene exhibited dom-
inance and the other did not? An example of this case ap-
pears in figure 2.17.

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Chapter Two Mendel's Principles

BOX 2.2

Overwhelming evidence gath-
ered during this century has
proven the correctness of
Mendel's conclusions. However, close
scrutiny of Mendel's paper has led
some to suggest that (1) Mendel
failed to report the inheritance of
traits that did not show independent
assortment and (2) Mendel fabricated
numbers. Both these claims are, on
the surface, difficult to ignore; both
have been countered effectively.

The first claim — that Mendel
failed to report crosses involving
traits that did not show independent
assortment — arises from the observa-
tion that all seven traits that Mendel
studied do show independent assort-
ment and that the pea plant has pre-
cisely seven pairs of chromosomes.
For Mendel to have chosen seven
genes, one located on each of the
seven chromosomes, by chance
alone seems extremely unlikely. In
fact, the probability would be

7/7 X 6/7 X 5/7 X 4/7 X 3/7
X 2/7 X 1/7 = 0.006

Historical
Perspectives

Did Mendel Cheat?

That is, Mendel had less than one
chance in one hundred of randomly
picking seven traits on the seven dif-
ferent chromosomes. However, L.
Douglas and E. Novitski in 1977 ana-
lyzed Mendel's data in a different
way. To understand their analysis, you
have to know that two genes suffi-
ciently far apart on the same chromo-
some will appear to assort indepen-
dently (to be discussed in chapter 6).
Thus, Mendel's choice of characters
showing independent assortment has
to be viewed in light of the lengths of
the chromosomes. That is, Mendel
could have chosen two genes on the
same chromosome that would still
show independent assortment. In
fact, he did. For example, stem length
and pod texture (wrinkled or

smooth) are on the fourth chromo-
some pair in peas. In their analysis,
Douglas and Novitski report that the
probability of randomly choosing
seven characteristics that appear to
assort independently is actually be-
tween one in four and one in three.
So it seems that Mendel did not have
to manipulate his choice of charac-
ters in order to hide the failure of in-
dependent assortment. He had a one
in three chance of naively choosing
the seven characters that he did,
thereby uncovering no deviation
from independent assortment.

The second claim — that Mendel
fabricated data — comes from a care-
ful analysis of Mendel's paper by R. A.
Fisher, a brilliant English statistician
and population geneticist. In a paper
in 1936, Fisher pointed out two prob-
lems in Mendel's work. First, all of
Mendel's published data taken to-
gether fit their expected ratios better
than chance alone would predict.
Second, some of Mendel's data fit in-
correct expected ratios. This second
"error" on Mendel's part came about
as follows.

Testcrossing Multihybrids

A simple test of Mendel's rule of independent assortment
is the testcrossing of the dihybrid plant. We would pre-
dict, for example, that if we crossed an RrYy F x individual
with an rryy individual, the results would include four
pheno types in a 1:1:1:1 ratio, as shown in figure 2.18.
Mendel's data verified this prediction (box 2.2). We will
proceed to look at a trihybrid cross in order to develop
general rules for multihybrids.

A trihybrid Punnett square appears in figure 2.19.
From this we can see that when a homozygous dominant
and a homozygous recessive individual are crossed in the
P : generation, plants in the F : generation are capable of
producing eight gamete types. When these Fj individuals
are selfed, they in turn produce F 2 offspring of twenty-
seven different genotypes in a ratio of sixty-fourths. By
extrapolating from the monohybrid through the trihy-
brid, or simply by the rules of probability, we can con-
struct table 2.3, which contains the rules for V 1 gamete

production and F 2 zygote formation in a multihybrid
cross. For example, from this table we can figure out the
F 2 offspring when a dodecahybrid (twelve segregating
genes: AA BB CC . . .LL X aabb cc . . . //) is selfed. The ¥ 1
organisms in that cross will produce gametes with 2 12 , or
4,096, different genotypes. The proportion of homozy-
gous recessive offspring in the F 2 generation is l/(2^) 2
where n = 12, or 1 in 16,777,216. With complete domi-
nance, there will be 4,096 different phenotypes in the F 2
generation. If dominance is incomplete, there can be 3 12 ,
or 531,441, different phenotypes in the F 2 generation.

GENOTYPIC INTERACTIONS

Often, several genes contribute to the same phenotype.
An example occurs in the combs of fowl (fig. 2.20). If we
cross a rose-combed hen with a pea-combed rooster (or
vice versa), all the V 1 offspring are walnut-combed. If we

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Genotypic Interactions

Mendel determined whether a
dominant phenotype in the F 2 gener-
ation was a homozygote or a het-
erozygote by self-fertilizing it and
examining ten offspring. In an F 2 gen-
eration composed of \AA:2Aa\\aa,
he expected a 2:1 ratio of heterozy-
gotes to homozygotes within the
dominant phenotypic class. In fact,
this ratio is not precisely correct be-
cause of the problem of misclassifica-
tion of heterozygotes. It is probable
that some heterozygotes will be clas-
sified as homozygotes because all
their offspring will be of the domi-
nant phenotype. The probability that
one offspring from a selfed Aa indi-
vidual has the dominant phenotype is
3/4, or 0.75: the probability that ten
offspring will be of the dominant
phenotype is (0.75) 10 or 0.056. Thus,
Mendel misclassified heterozygotes
as dominant homozygotes 5.6% of
the time. He should have expected a
1.89:111 ratio instead of a 2:1 ratio
to demonstrate segregation. Mendel
classified 600 plants this way in one
cross and got a ratio of 201 homozy-
gous to 399 heterozygous offspring.

This is an almost perfect fit to the pre-
sumed 2:1 ratio and thus a poorer fit
to the real 1.89: 111 ratio. This bias is
consistent and repeated in Mendel's
trihybrid analysis.

Fisher, believing in Mendel's basic
honesty, suggested that Mendel's data
do not represent an experiment but
more of a hypothetical demonstra-
tion. In 1971, F. Weiling published a
more convincing case in Mendel's de-
fense. Pointing out that the data of
Mendel's rediscoverers are also sus-
pect for the same reason, he sug-
gested that the problem lies with the
process of pollen formation in plants,
not with the experimenters. In znAa
heterozygote, two A and two a cells
develop from a pollen mother cell.
These cells tend to stay together on
the anther. Thus, pollen cells do not
fertilize in a strictly random fashion.
A bee is more likely to take equal
numbers of A and a pollen than
chance alone would predict. The re-
sult is that the statistics Fisher used
are not applicable. By using a differ-
ent statistic, Weiling showed that, in
fact, Mendel need not have manipu-

lated any numbers (nor would have
his rediscoverers) in order to get data
that fit the expected ratios well. By
the same reasoning, very little mis-
classification of heterozygotes would
have occurred.

More recently, Weiling and others
have made several additional points.
First, for Mendel to be sure of ten off-
spring, he probably examined more
than ten, and thus he probably kept
his misclassification rate lower than
5.6%. Second, despite Fisher's bril-
liance as a statistician, several have
made compelling arguments that
Fisher's statistical analyses were in-
correct. In other words, for subtle sta-
tistical reasons, many of his analyses
involved methods and conclusions
that were in error.

We conclude that there is no com-
pelling evidence to suggest that
Mendel in any way manipulated his
data to demonstrate his rules. In fact,
taking into account what is known
about him personally, it is much more
logical to believe that he did not
"cheat."

cross the hens and roosters of this heterozygous F : group,
we will get, in the F 2 generation, walnut-, rose-, pea-, and
single-combed fowl in a ratio of 9:3:3:1. Can you figure
out the genotypes of this F 2 population before reading
further? An immediate indication that two allelic pairs are
involved is the fact that the 9:3:3:1 ratio appeared in the
F 2 generation. As we have seen, this ratio comes about

when we cross dihybrids in which both genes have alleles
that control traits with complete dominance.

Figure 2.21 shows the analysis of this cross. When
dominant alleles of both genes are present in an individ-
ual (R- P-), the walnut comb appears. (The dash indicates
any second allele; thus, R- P- could be RRPP, RrPP, RRPp,
or RrPp.) A dominant allele of the rose gene (R-) with

Table 2.3 Multihybrid Self-Fertilization, Where n Equals Number of Genes Segregating Two Alleles Each

Monohybrid

Dihybrid

Trihybrid

n = 1

n = 2

n = 3

General Rule

Number of F x gametic genotypes

2 n

Proportion of recessive homozygotes among the F 2 individuals

1/4

1/16

1/64

l/(2 n f

Number of different F 2 phenotypes, given complete dominance

2 n

Number of different genotypes (or phenotypes, if no

dominance exists)

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Chapter Two Mendel's Principles

Genotype

RRYY

RRYy

RRyy

RrYY

RrYy

Rryy

rrYY

rrYy

rryy

Ratio of
genotype
in F„

Ratio of
phenotype
in F n

Figure 2.16 The phenotypic and genotypic ratios of the
offspring of dihybrid peas.

recessive alleles of the pea gene (pp) gives a rose comb.
A dominant allele of the pea gene (P-) with recessive al-
leles of the rose gene (rr) gives pea-combed fowl. When
both genes are homozygous for the recessive alleles, the
fowl are single-combed. Thus, a 9:3:3:1 F2 ratio arises
from crossing dihybrid individuals even though different
expressions of the same phenotypic characteristic, the
comb, are involved. In our previous 9:3:3:1 example (see
fig. 2.15), we dealt with two separate characteristics:
shape and color of peas.

In corn (or maize, Zea mays), several different field
varieties produce white kernels on the ears. In certain
crosses, two white varieties will result in an ¥ 1 genera-
tion with all purple kernels. If plants grown from these
purple kernels are selfed, the F 2 individuals have both
purple and white kernels in a ratio of 9:7. How can we
explain this? We must be dealing with the offspring of di-
hybrids with each gene segregating two alleles, because
the ratio is in sixteenths. Furthermore, we can see that
the F 2 9:7 ratio is a variation of the 9:3:3:1 ratio. The 3, 3,
and 1 categories here are producing the same phenotype
and thus make up 7/16 of the F 2 offspring. Figure 2.22
outlines the cross. We can see from this figure that the
purple color appears only when dominant alleles of both
genes are present. When one or both genes have only re-
cessive alleles, the kernels will be white.

Epistasis

The color of corn kernels illustrates the concept of epis-
tasis, the interaction of nonallelic genes in the formation

Figure 2.17 Independent
assortment of two blood
systems in human beings. In the
ABO system, the / A and / B
alleles are codominant. In a
simplified view of the Rhesus
system, the Rh + phenotype (D
allele) is dominant to the Rh~
phenotype (d allele).

I A D

l A d

l a D

l a d

Phenotype

l A D

l A l A DD X l B l B dd

l A l B Dd
( F 1 x F 1)

Male

l A d l B D

l a d

l A l A DD

l A l A Dd

l A l B DD

l A l B Dd

l A l A Dd

l A l A dd

l A l B Dd

l A l B dd

l A l B DD

l A l B Dd

l B l B DD

l B l B Dd

l A l B Dd

l A l B dd

l B l B Dd

l B l B dd

Summary Frequency

A Rh + A Rh" B Rh + B Rrr AB Rh + AB Rh

3 13 16 2

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Genotypic Interactions

Gametes

RrYy

rryy

RY Ry

rY ry

RrYy

Rryy

rrYy

rryy

(55)

(51)

(49)

(52)

1:1:1:1

Figure 2.18 Testcross of a dihybrid. A 1:1:1:1 ratio is
expected in the offspring.

of the phenotype. This is a process analogous to domi-
nance among alleles of one gene. For example, the reces-
sive apterous (wingless) gene in fruit flies is epistatic to
any gene that controls wing characteristics; hairy wing is
hypostatic to apterous (that is, the recessive apterous
gene, when homozygous, masks the presence of the
hairy wing gene, because, obviously, without wings, no

wing characteristics can be expressed). Note that the ge-
netic control of comb type in fowl does not involve epis-
tasis. There are no allelic combinations at one locus that
mask genotypes at another locus: the 9:3:3:1 ratio is not
an indication of epistasis. To illustrate further the princi-
ple of epistasis, we can look at the control of coat color
in mice.

In one particular example, a pure-breeding black
mouse is crossed with a pure-breeding albino mouse
(pure white because all pigment is lacking); all of the off-
spring are agouti (the typical brownish-gray mouse
color). When the F : agouti mice are crossed with each
other, agouti, black, and albino offspring appear in the F 2
generation in a ratio of 9:3:4. What are the genotypes in
this cross? The answer appears in figure 2.23. By now it
should be apparent that the F 2 ratio of 9:3:4 is also a vari-
ant of the 9:3:3:1 ratio; it indicates epistasis in a dihybrid
cross. What is the mechanism producing this 9:3:4 ratio?
Of a potential 9:3:3:1 ratio, one of the 3/16 classes and
the 1/16 class are combined to create a 4/16 class. Any
genotype that includes c a c a will be albino, masking the
A gene, but as long as at least one dominant C allele is
present, the A gene can express itself. Mice with domi-
nant alleles of both genes (A-C-) will have the agouti
color, whereas mice that are homozygous recessive at the
A gene (aaC-) will be black. So, at the A gene, A for agouti

AA BB CC X aa bb cc
Aa Bb Cc X Self

ABC

ABc

AbC

Abe

aBC

aBc

abC

a be

ABC

ABc

AbC

Abe

aBC

aBc

abC

a b c

AA BB CC

AA BB Cc

AA Bb CC

AA Bb Cc

Aa BB CC

Aa BB Cc

Aa Bb CC

Aa Bb Cc

AA BB Cc

AA BB cc

AA Bb Cc

AA Bb cc

Aa BB Cc

Aa BB cc

Aa Bb Cc

Aa Bb cc

AA Bb CC

AA Bb Cc

AA bb CC

AA bb Cc

Aa Bb CC

Aa Bb Cc

Aa bb CC

Aa bb Cc

AA Bb Cc

AA Bb cc

AA bb Cc

AA bb cc

Aa Bb Cc

Aa Bb cc

Aa bb Cc

Aa bb cc

Aa BB CC

Aa BB Cc

Aa Bb CC

Aa Bb Cc

aa BB CC

aa BB Cc

aa Bb CC

aa Bb Cc

Aa BB Cc

Aa BB cc

Aa Bb Cc

Aa Bb cc

aa BB Cc

aa BB cc

aa Bb Cc

aa Bb cc

Aa Bb CC

Aa Bb Cc

Aa bb CC

Aa bb Cc

aa Bb CC

aa Bb Cc

aa bb CC

aa bb Cc

Aa Bb Cc

Aa Bb cc

Aa bb Cc

Aa bb cc

aa Bb Cc

aa Bb cc

aa bb Cc

aa bb cc

Figure 2.19 Trihybrid cross.

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Rose

Pea

Walnut Single

Figure 2.20 Four types of combs in fowl.

First white variety X Second white variety
AAbb aaBB

Purple
AaBb X Self

AABB

AABb

AaBB

AaBb

AABb

AAbb

AaBb

Aabb

AaBB

AaBb

aaBB

aaBb

AaBb 1

Aabb

aaBb

aabb

Purple : White
9 : 7

Summary

Figure 2.22 Color production in corn

Rose comb Pea comb

RRpp X rrPP

is dominant to a for black. The albino gene (c ), when
homozygous, is epistatic to the A gene; the A gene is hy-
postatic to the gene for albinism.

Ftp

Summary

Walnut comb
RrPp

F 1 X F 1

RRPP
Walnut

RRPp
Walnut

RrPP
Walnut

RrPp
Walnut

RRPp
Walnut

RRpp
Rose

RrPp
Walnut

Rrpp
Rose

RrPP
Walnut

RrPp
Walnut

rrPP
Pea

rrPp
Pea

RrPp
Walnut

Rrpp
Rose

rrPp
Pea

rrpp
Single

Walnut : Rose : Pea : Single
9:3:3:1

Figure 2.21 Independent assortment in the determination of
comb type in fowl.

Mechanism ofEpistasis

In this case, the physiological mechanism of epistasis is
known. The pigment melanin is present in both the black
and agouti pheno types. The agouti is a modified black
hair in which yellow stripes (the pigment phaeomelanin)
have been added. Thus, with melanin present, agouti is
dominant. Without melanin, we get an albino regardless
of the genotype of the agouti gene because both agouti
and black depend on melanin. Albinism is the result of
one of several defects in the enzymatic pathway for the
synthesis of melanin (fig. 2.24).

Knowing that epistatic modifications of the 9:3:3:1
ratio come about through gene interactions at the bio-
chemical level, we can look for a biochemical explana-
tion for the 9:7 ratio in corn kernel color (fig. 2.22). Two
possible mechanisms for a 9:7 ratio are shown in figure
2.25. Either a two-step process takes a precursor mole-
cule and turns it into purple pigment, or two precursors
that must be converted to final products then combine to
produce purple pigment. The dominant alleles from the
two genes control the two steps in the process. Reces-
sive alleles are ineffective. Thus, dominants are necessary
for both steps to complete the pathways for a purple pig-

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Genotypic Interactions

Albino
AAc a c a

Black hair

White hair

Agout
AaCc a

Black hair with
yellow stripes

AACC

Agouti

AACcf

Agouti

AaCC

Agouti

AaCcf

Agouti

ac c

AACcf

Agouti

AAcfcf

Albino

AaCcf

Agouti

Aacfcf

Albino

AaCC

Agouti

AaCcf

Agouti

aaCC

Black

aaCcf

AaCcf

Agouti

Aacfcf

Albino

aaCcf

Black

aacfcf

Albino

Agouti : Black : Albino
9:3:4

Figure 2.23 Epistasis in the coat color of mice.

ment. Stopping the process at any point prevents the
production of purple color.

Another example of epistasis occurs in the snap-
dragon (Antirrhinum majus^.There, a gene called nivea
has alleles that determine whether any pigment is pro-
duced; the nn genotype prevents pigment production,
whereas the NN or Nn genotypes permit pigment color
genes to express themselves. The eosinea gene controls
the production of a red anthocyanin pigment. In the

presence of the N allele of the nivea gene, the genotypes
EE or Ee of the eosinea gene produce red flowers; the ee
genotype produces pink flowers. When dihybrids are
self-fertilized, red-, pink-, and white-flowered plants are
produced in a ratio of 9:3:4 (fig. 2.26). The epistatic inter-
action is the nn genotype masking the expression of
alleles at the eosinea gene. In other words, regardless
of the genotypes of the eosinea gene (EE, Ee, or ee),
the flowers will be white if the nivea gene has the nn

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Chapter Two Mendel's Principles

Phenylalanine

Tyrosine

3,4-Dihydroxy-

phenylalanine

(DOPA)

ChL— CH — COO"

<t>.

<t>

Thyroxine

P-Hydroxyphenyl
pyruvate

Homogentisic
acid

CH 2 — CH — COCT

NH.

C0 2 + H 2

■> Melanins

CH 2 — CH — COO"
NH+

Enzyme defect conditions
(?) Phenylketonuria (PKU)
(2) Genetic goitrous cretinism
(S) Tyrosinosis
(?) Alkaptonuria
(5) Albinism

Figure 2.24 In humans, errors in melanin synthesis produce different physical conditions
and diseases, depending on which part of the tyrosine (an amino acid) metabolic pathway
is disrupted. The broken arrows indicate that there is more than one step in the pathways;
the conditions listed occur only in homozygous recessives.

Pathway 1

Control by
gene A

Colorless intermediate

Control by

gene B

>►

Purple pigment

Pathway 2

Control by
gene A

Purple pigment

Control by
gene B

Figure 2.25 Possible metabolic pathways of color production that would yield 9:7
ratios in the F 2 generation of a self -fertilized dihybrid.

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Biochemical Genetics

combination of alleles. Thus, nivea is epistatic to
eosinea, and eosinea is hypostatic to nivea. (We should
add that at least seven major colors occur in snap-
dragons, along with subtle shade differences, all genetically
controlled by the interactions of at least seven genes.)

Other types of epistatic interactions occur in other
organisms. Table 2.4 lists several. We do not know the ex-
act physiological mechanisms in many cases, especially
when developmental processes are involved (e.g., size
and shape). However, from an analysis of crosses, we can
know the number of genes involved and the general na-
ture of their interactions.

BIOCHEMICAL GENETICS

Inborn Errors of Metabolism

The examples of mouse coat color, corn kernel color, and
snapdragon flower petal color demonstrate that genes con-
trol the formation of enzymes, proteins that control the
steps in biochemical pathways. For the most part, domi-
nant alleles control functioning enzymes that catalyze bio-
chemical steps. Recessive alleles often produce nonfunc-
tioning enzymes that cannot catalyze specific steps. Often
a heterozygote is normal because one allele produces a
functional enzyme; usually only half the enzyme quantity of
the dominant homozygote is enough. The study of the rela-
tionship between genes and enzymes is generally called
biochemical genetics because it involves the genetic
control of biochemical pathways. A. E. Garrod, a British
physician, pointed out this general concept of human gene
action in Inborn Errors of Metabolism, published in 1909.
Only nine years after Mendel was rediscovered, Garrod de-
scribed several human conditions, such as albinism and
alkaptonuria, that occur in individuals who are homozy-
gous for recessive alleles (see fig. 2.24).

Red
NNEE

White
nnee

Red X
NnEe

Self

Red

Pink

White

NNEE

NNee

nnEE

NNEe

nnEe

NnEE

Nnee

nnee

NnEe

9:3:4

Figure 2.26 Flower color inheritance in snapdragons. This is
an example of epistasis: an nn genotype masks the expression
of alleles {EE, Ee, or ee) at the eosinea gene.

Table 2.4 Some Examples of Epistatic Interactions Among Alleles of Two Genes

Characteristic

Phenotype

of F a

Dihybrid (AaBb^

Phenotypic F 2 Ratio

Corn and sweet pea color

Purple

Purple: white
9:7

Mouse coat color

Agouti

Agouti:black:albino
9:3:4

Shepherd's purse seed capsule shape

Triangular

Triangular: oval
15:1

Summer squash shape

Disk

Disk : sphere : elongate
9:6:1

Fowl color

White

White: colored
13:3

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For example, people normally degrade homogentisic
acid (alkapton) into maleylacetoacetic acid. Persons with
the disease alkaptonuria are homozygous for a nonfunc-
tional form of the enzyme essential to the process: ho-
mogentisic acid oxidase, found in the liver. Absence of this
enzyme blocks the degradation reaction so that homogen-
tisic acid builds up. This acid darkens upon oxidation.
Thus, affected persons can be identified by the black color
of their urine after its exposure to air. Eventually, alkap-
tonuria causes problems in the joints and a darkening of
cartilage that is visible in the ears and the eye sclera.

One-Gene-One-Enzyme Hypothesis

Pioneering work in the concept that genes control the
production of enzymes, which in turn control the steps in
biochemical pathways, was done by George Beadle and
Edward Tatum, who eventually shared the Nobel Prize for
their work. They not only put forth the one-gene-one-
enzyme hypothesis, but also used mutants to work out
the details of biochemical pathways. In 1941, Beadle and
Tatum were the first scientists to isolate mutants with nu-
tritional requirements that defined steps in a biochemical
pathway. In the early 1940s, they united the fields of bio-
chemistry and genetics by using strains of a bread mold
with specific nutritional requirements to discover the
steps in biochemical pathways in that organism.

Through this century, the study of mutations has
been the driving force in genetics. The process of muta-
tion produces alleles that differ from the wild-type and
shows us that a particular aspect of the phenotype is un-
der genetic control. Beadle and Tatum used mutants to
work out the steps in the biosynthesis of niacin (vitamin
B 3 ) in pink bread mold, Neurospora crassa.

Normally, Neurospora synthesizes niacin via the path-
way shown in figure 2.27. Beadle and Tatum isolated mu-
tants that could not grow unless niacin was provided in

George W. Beadle

(1 903-89). Courtesy of the

Archives, California Institute of

Technology.

Edward L. Tatum
(1 909-75). Courtesy of the
Proceedings for the National
Academy of Sciences.

the culture medium; these mutants had enzyme deficien-
cies in the synthesizing pathway that ends with niacin.
Thus, although wild-type Neurospora could grow on a
medium without additives, the mutants could not. Beadle
and Tatum had a general idea, based on the structure of
niacin, as to what substances would be in the niacin
biosynthesis pathway. They could thus make educated
guesses as to what substances they might add to the cul-
ture medium to enable the mutants to grow. Mutant B
(table 2.5), for example, could grow if given niacin or, al-
ternatively, 3-hydroxyanthranilic acid. It could not grow
if given only kynurenine. Thus, Beadle and Tatum knew
that the B mutation affected the pathway between
kynurenine and 3-hydroxyanthranilic acid. Similarly, mu-
tant A could grow if given 3-hydroxyanthranilic acid or
kynurenine instead of niacin. Therefore, these two prod-

■w

Anthranilic

acid

Kvni irpnin^

I *w

Mutant B

Tryptophan

Mutant A

3-Hydroxy-

anthranilic

acid

Figure 2.27 Pathway of niacin synthesis in Neurospora. Each arrow represents an
enzyme-mediated step. Each question mark represents a presumed but (at the time
Beadle and Tatum were working) unknown compound.

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Table 2.5 Growth Performance of Neurospora Mutants (plus sign indicates growth;
minus sign indicates no growth)

Additive

Tryptophan

Kynurenine

3-Hydroxyanthranilic Acid

Niacin

Wild-type
Mutant A
Mutant B

+
+

+
+
+

ucts must be in the pathway after the step interrupted in
mutant A. Conversely, since neither of these mutant or-
ganisms could grow when given only tryptophan, Beadle
and Tatum knew that tryptophan occurred in the path-
way before the steps with the deficient enzymes. By this
type of analysis, they discovered the steps in several bio-
chemical pathways of Neurospora. Many biochemical
pathways are similar in a huge range of organisms, and
thus Beadle and Tatum 's work was of general impor-
tance. (We will spend more time studying Neurospora in
chapter 6.)

Beadle and Tatum could further verify their work by
observing which substances accumulated in the mutant
organisms. If a biochemical pathway is blocked at a cer-
tain point, then the substrate at that point cannot convert
into the next product, and it builds up in the cell. For ex-
ample, in the niacin pathway (fig. 2.27), if a block occurs
just after 3-hydroxyanthranilic acid, that substance will
build up in the cell because it cannot convert into the
next substance on the way to niacin.

This analysis could be misleading, however, if the
built-up substance is being "siphoned off" into other bio-
chemical pathways in the cell. Also, the cell might at-
tempt to break down or sequester toxic substances. This
would mean there might not be an obvious buildup of
the substance just before the blocked step.

Beadle and Tatum concluded from their studies that
one gene controls the production of one enzyme. The
one-gene-one-enzyme hypothesis is an oversimplification
that we will clarify later in the book. As a rule of thumb,
however, the hypothesis is valid, and it has served to di-
rect attention to the functional relationship between
genes and enzymes in biochemical pathways.

Although a change in a single enzyme usually disrupts
a single biochemical pathway, it frequently has more than
one effect on phenotype. Multiple effects are referred to
as pleiotropy. A well-known example is sickle-cell ane-
mia, caused by a mutation in the gene for the p chain of
the hemoglobin molecule. In a homo zygote, this muta-
tion causes a sickling of red blood cells (fig. 2.28). The
sickling of these cells has two major ramifications.

First, the liver destroys the sickled cells, causing ane-
mia. The phenotypic effects of this anemia include phys-
ical weakness, slow development, and hypertrophy of
the bone marrow, resulting in the "tower skull" seen in
some of those afflicted with the disease. The second ma-
jor effect of sickle-cell anemia is that the sickled cells in-
terfere with capillary blood flow, clumping together and
resulting in damage to every major organ. The individual
can suffer pain, heart failure, rheumatism, and other ill ef-
fects. Hence, a single mutation shows itself in many as-
pects of the phenotype.

Figure 2.28 Sickle-shaped red blood cells from a person with
sickle-cell anemia. Red blood cells are about 7 to 8 |xm in

diameter. (Courtesy of Dr. Patricia N. Farnsworth.)

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Chapter Two Mendel's Principles

SUMMARY

STUDY OBJECTIVE 1: To understand that genes are dis-
crete units that control the appearance of an organism
17-18

Genes control phenotypic traits such as size and color. They
are inherited as discrete units.

STUDY OBJECTIVE 2: To understand Mendel's rules of
inheritance: segregation and independent assortment
18-22

Higher organisms contain two alleles of each gene, but only
one allele enters each gamete. Zygote formation restores
the double number of alleles in the cell. This is Mendel's
rule of segregation.

Alleles of different genes segregate independently of
each other. Mendel was the first to recognize the 3:1 phe-
notypic ratio as a pattern of inheritance; the 9:3:3:1 ratio
demonstrates independent assortment in hybrids. Mendel
was successful in his endeavor because he performed care-
ful experiments using discrete characteristics, large num-
bers of offspring, and an organism (the pea plant) amenable
to controlled fertilizations.

STUDY OBJECTIVE 3: To understand that dominance is a
function of the interaction of alleles; similarly, epistasis is
a function of the interaction of nonallelic genes 22-37

There can be many alleles for one gene, although each indi-
vidual organism has only two alleles for each gene. A phe-
notype is dominant if it is expressed when one or two
copies of its allele are present (heterozygote or homozy-
gote). Dominance depends, however, on the level of the
phenotype one looks at.

Genes usually control the production of enzymes,
which control steps in metabolic pathways. Many human
metabolic diseases are due to homozygosity of an allele that
produces a nonfunctioning enzyme.

Nonallelic genes can interact in producing a phenotype
so that alleles of one gene mask the expression of alleles of
another gene. This process, termed epistasis, alters the ex-
pected phenotypic ratios.

STUDY OBJECTIVE 4: To define how genes generally con-
trol the production of enzymes and thus the fate of bio-
chemical pathways 37-39

Beadle and Tatum used mutants with mutations in the
niacin biosynthesis pathway to work out the steps in the
pathway. A single mutation can have many phenotypic ef-
fects (pleiotropy).

SOLVED PROBLEMS

PROBLEM 1: In corn, rough sheath (rs) is recessive
to smooth sheath (Rs), midrib absent (mrl) is recessive
to midrib present (Mrl), and crinkled leaf (cr) is reces-
sive to smooth leaf (Cf). (Alleles are named for the mu-
tants, which are all recessive.) What are the results of
testcrossing a trihybrid?

Answer: The trihybrid has the genotype Rsrs Mrlmrl
Crcr. This parent is capable of producing eight different
gamete types in equal frequencies, all combinations of
one allele from each gene (Rs Mrl Cr, Rs Mrl cr, Rs mrl Cr,
Rs mrl cr, rs Mrl Cr, rs Mrl cr, rs mrl Cr, and rs mrl cr). In
a testcross, the other parent is a recessive homozygote
with the genotype rsrs mrlmrl crcr, capable of producing
only one type of gamete, with the alleles rs mrl cr Thus,
this cross can produce zygotes of eight different geno-
types (and phenotypes), one for each of the gamete types
of the trihybrid parent: Rsrs Mrlmrl Crcr (smooth sheath,
midrib present, smooth leaf); Rsrs Mrlmrl crcr (smooth
sheath, midrib present, crinkled leaf); Rsrs mrlmrl Crcr
(smooth sheath, midrib absent, smooth leaf); Rsrs mrlmrl

crcr (smooth sheath, midrib absent, crinkled leaf); rsrs
Mrlmrl Crcr (rough sheath, midrib present, smooth leaf);
rsrs Mrlmrl crcr (rough sheath, midrib present, crinkled
leaf); rsrs mrlmrl Crcr (rough sheath, midrib absent,
smooth leaf); and rsrs mrlmrl crcr (rough sheath, midrib
absent, crinkled leaf). Each should make up one-eighth of
the total number of offspring.

PROBLEM 2: Summer squash come in three shapes: disk,
spherical, and elongate. In one experiment, researchers
crossed two squash plants with disk-shaped fruits. The
first 160 seeds planted from this cross produced plants
with fruit shapes as follows: 89 disk, 61 sphere, and 10
elongate. What is the mode of inheritance of fruit shape
in summer squash?

Answer: The numbers are very close to a ratio of
90:60:10, or 9:6:1, an epistatic variant of the 9:3:3:1, with
the two 3/l6ths categories combined. If this is the case,
then the parent plants with disk-shaped fruits were dihy-
brids (AaBb). Among the offspring, 9/l6ths had disk-

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Exercises and Problems

shaped fruit, indicating that it takes at least one dominant
allele of each gene to produce disk-shaped fruits (A-B-:
AABB, AaBB, AABb, or AaBb). The l/l6th category of
plants with elongate fruits indicates that this fruit shape
occurs in homozygous recessive plants (aabb). The
plants with spherical fruit are thus plants with a domi-
nant allele of one gene but a homozygous recessive com-
bination at the other gene (AAbb, Aabb, aaBB, or aaBb).
In summary, then, two genes combine to control fruit
shape in summer squash. The epistatic interactions be-
tween the two genes produce a 9:6:1 ratio of offspring
pheno types when dihybrids are crossed.

PROBLEM 3: A geneticist studying the pathway of synthesis
of phenylalanine in Neurospora isolated several mutants
that require phenylalanine to grow. She tested whether

Additive

,e*

sfy Jr S?

Wild-type
Mutant 1
Mutant 2
Mutant 3

+
+

+
+
+
+

each mutant would grow when provided additives that she
believed were in the pathway of phenylalanine synthesis
(see table); a plus indicates growth and minus indicates the
lack of growth in the three mutants tested.

Where in the pathway to phenylalanine synthesis
does each of the additives belong, if at all?

Answer: The wild-type grows in the presence of all addi-
tives. This is not surprising since the wild-type can grow,
by definition, in the absence of all the additives because
it can synthesize phenylalanine de novo. Mutant 1 cannot
grow in the presence of any additive except phenylala-
nine, indicating that its mutation affects the step just be-
fore the end of the pathway at phenylalanine. In other
words, each of the other additives occurs in the phenyl-
alanine pathway before the point of the mutation in mu-
tant 1 . Mutant 2 can grow if given any additive but cho-
rismate, indicating that chorismate is at the beginning of
the pathway, and the mutation affects the pathway just
after that step. Finally, mutant 3 can grow if given
phenylpyruvate or phenylalanine, indicating that its mu-
tation affects the step before phenylpyruvate and phenyl-
alanine, but after the earlier part of the pathway. Putting
all of this information together indicates that the path-
way to phenylalanine, with mutants indicated, is:

2 3 1

chorismate — > prephenate — ► phenylpyruvate — ►

phenylalanine

EXERCISES AND PROBLEMS

SEGREGATION

1. Mendel crossed tall pea plants with dwarf ones. The
F x plants were all tall. When these F : plants were
selfed to produce the F 2 generation, he got a 3:1 tall-
to-dwarf ratio in the offspring. Predict the genotypes
and phenotypes and relative proportions of the F 3
generation produced when the F 2 generation was
selfed.

2. What properties of fruit flies and corn made them
the organisms of choice for geneticists during most
of the first half of the twentieth century? (Molecular
geneticists have made great strides working with
bacteria and viruses. You could begin thinking at this
point about the properties that have made these or-
ganisms so valuable to geneticists.)

3. State precisely the rules of segregation and inde-
pendent assortment. (See also the Exercises and
Problems section on Independent Assortment.)

4. In Drosophila, a cross between a dark-bodied fly and
a tan-bodied fly yields seventy-six tan and eighty
dark flies. Diagram the cross.

5. If two black mice are crossed, ten black and three
white mice result.

a. Which allele is dominant?

b. Which allele is recessive?

c. What are the genotypes of the parents?

6. In Drosophila, two red-eyed flies mate and yield 110
red-eyed and 35 brown-eyed offspring. Diagram the
cross and determine which allele is dominant.

DOMINANCE IS NOT UNIVERSAL

7. Explain how Tay-Sachs disease can be both a reces-
sive and an incomplete dominant trait. What are the
differences between incomplete dominance and
codominance?

8. How does the biochemical pathway in figure 2.13
explain how alleles I A and I B are codominant, yet
both dominant to allele i?

* Answers to selected exercises and problems are on page A-l.

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Chapter Two Mendel's Principles

9. Two short-eared pigs are mated. In the progeny,
three have no ears, seven have short ears, and four
have long ears. Explain these results by diagramming
the cross.

10. A plant with red flowers is crossed with a plant with
white flowers. All the progeny are pink. When the
plants with pink flowers are crossed, the progeny
are eleven red, twenty-three pink, and twelve white.
What is the mode of inheritance of flower color?

NOMENCLATURE

11. In fruit flies, a new dominant trait, washed eye, was
discovered. Describe different ways of naming the
alleles of the washed-eye gene.

12. The following is a list of ten genes in fruit flies, each
with one of its alleles given. Are the alleles shown
dominant or recessive? Are they mutant or wild-
type? What is the alternative allele for each? Is the al-
ternative allele dominant or recessive in each case?

Name of Gene

Allele

yellow

Hairy wing

Abruptex

Confluens

raven

downy

Minute(2)e

Jammed

tufted

burgundy

Ax +

Co
rv

dow
M(2)e

J
tuf +

bur

MULTIPLE ALLELES

13. In the ABO blood system in human beings, alleles I A
and I B are codominant, and both are dominant to the
i allele. In a paternity dispute, a type AB woman
claimed that one of four men, each with different
blood types, was the father of her type A child.
Which of the following could be the blood type of
the father of the child on the basis of the evidence
given?

a. Type A

b. Type B

c. Type O

d. Type AB

14. Under what circumstances can the pheno types of
the ABO system be used to refute paternity?

15. In blood transfusions, one blood type is called the "uni-
versal donor" and one the "universal recipient" be-
cause of their ABO compatibilities. Which is which?

16. Among the genes having the greatest number of al-
leles are those involved in self-incompatibility in
plants. In some cases, hundreds of alleles exist for a

single gene. What types of constraints might exist
to set a limit on the number of alleles a gene can have?

17. In the human ABO blood system, the alleles I A and I B
are dominant to /. What possible phenotypic ratios
do you expect from a mating between a type A indi-
vidual and a type B individual?

18. In screech owls, crosses between red and silver indi-
viduals sometimes yield all red; sometimes 1/2
red: 1/2 silver; and sometimes 1/2 red: 1/4 white: 1/4
silver offspring. Crosses between two red owls yield
either all red, 3/4 red: 1/4 silver, or 3/4 red: 1/4 white
offspring. What is the mode of inheritance?

19. A premed student, Steve, plans to marry the daughter
of the dean of nursing. The dean's husband was ster-
ile, and the daughter was conceived by artificial in-
semination. Steve's father puts pressure on Steve to
marry someone else. Having served as an anonymous
sperm donor, he is concerned that Steve and his fi-
ance may be half brother and sister. Given the fol-
lowing information, deduce whether Steve and his fi-
ance could be related. (The MN and Ss systems are
two independent, codominant blood-type systems.)

Blood Type

Dean

Her daughter

Steve's father

Steve

Steve's mother

A, MN, Ss
0,M, S

A, MN, Ss
0,N, s
B,N,s

INDEPENDENT ASSORTMENT

20. Mendel self-fertilized dihybrid plants (RrYy) with
round and yellow seeds and got a 9:3:3:1 ratio in the
F 2 generation. As a test of Mendel's hypothesis of in-
dependent assortment, predict the kinds and num-
bers of progeny produced in testcrosses of these F 2
offspring.

21. Four o'clock plants have a gene for color and a gene
for height with the following pheno types:

RR: red flower TT: tall plant

Rr: pink flower Tt: medium height plant

rr: white flower tt: dwarf plant

Give the proportions of genotypes and phenotypes
produced if a dihybrid plant is self-fertilized.

22. A particular variety of corn has a gene for kernel
color and a gene for height with the following phe-
notypes:

CC, Cc: purple kernels TT: tall stem

cc: white kernels Tt: medium height stem

tt: dwarf stem

Give the proportions of genotypes and phenotypes
produced if a dihybrid plant is selfed.

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Exercises and Problems

23. To determine the genotypes of the offspring of a
cross in which a corn trihybrid (Aa Bb Cc) was
selfed, a geneticist has three choices. He or she can
take a sample of the progeny and (a) self-fertilize the
individual plants, (b) testcross the plants, or (c) cross
the individuals with a trihybrid (backcross). Which
method is preferable?

24. In figure 2.17, the F 2 phenotypic ratio is 3:1:3:1:6:2.
What are the phenotypic segregation ratios for each
blood system (AB, Rh) separately? Are they segregat-
ing properly? What phenotypic ratio in the F 2 gener-
ation would indicate interference with independent
assortment?

25. Assume that Mendel looked simultaneously at four
traits of his pea plants (and each trait exhibited dom-
inance). If he crossed a homozygous dominant plant
with a homozygous recessive plant, all the ¥ 1 off-
spring would be of the dominant phenotype. If he
then selfed the ¥ 1 plants, how many different types
of gametes would these F : plants produce? How
many different phenotypes would appear in the F 2
generation? How many different genotypes would
appear? What proportion of the F 2 offspring would
be of the fourfold recessive phenotype?

26. A geneticist crossed two corn plants, creating an F :
decahybrid (ten segregating loci). He then self-
fertilized this decahybrid. How many different kinds
of gametes did the F : plant produce? What propor-
tion of the F 2 offspring were recessive homozy-
gotes? How many different kinds of genotypes and
phenotypes were generated in the F 2 offspring?
What would your answer be if the geneticist test-
crossed the decahybrid instead?

27. Consider the following crosses in pea plants and de-
termine the genotypes of the parents in each cross.
Yellow and green refer to seed color; tall and short
refer to plant height.

Progeny

Cross

Yellow, Yellow, Green, Green,
Tall Short Tall Short

Yellow, tall X
yellow, tall

Yellow, short X
yellow, short

Green, tall X
yellow, short

28. A brown-eyed, long-winged fly is mated with a red-
eyed, long-winged fly. The progeny are

5 1 long, red
53 long, brown

What are the genotypes of the parents?

29. True-breeding flies with long wings and dark
bodies are mated with true-breeding flies with short
wings and tan bodies. All the V 1 progeny have long
wings and tan bodies. The F : progeny are allowed to
mate and produce:

44 tan, long 14 tan, short

16 dark, long 6 dark, short

What is the mode of inheritance?

30. In peas, tall (7 1 ) is dominant to short (f), yellow (F)
is dominant to green ( y), and round (R) is dominant
to wrinkled (r). From a cross of two triple heterozy-
gotes, what is the chance of getting a plant that is

a. tall, yellow, round?

b. short, green, wrinkled?

c. short, green, round?

31. In corn, the genotype A- C- R- is colored. Individuals
homozygous for at least one recessive allele are col-
orless. Consider the following crosses involving col-
ored plants, all with the same genotype. Based on the
results, deduce the genotypes of the colored plants.

colored X aa cc RR — ► 1/2 colored; 1/2 colorless
colored X aa CC rr -> 1/4 colored; 3/4 colorless
colored X AA cc rr — » 1/2 colored; 1/2 colorless

32. Consider the following crosses in Drosophila. Based
on the results, deduce which alleles are dominant
and the genotypes of the parents. Orange and red are
eye colors; crossveins occur on the wings.

Progeny

Parents

O ,o

a. Orange,

crossveins

X orange,

crossveins

b. Red,

crossveins

X red,

crossveinless

c. Red,

crossveinless

X red,

crossveins

d. Red,

crossveins

X red,

crossveins

18 short, red
16 short, brown

33. In Drosophila melanogaster, a recessive autosomal
gene, ebony, produces a dark body color when ho-
mozygous, and an independently assorting autosomal

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Chapter Two Mendel's Principles

gene, black, has a similar effect. If homozygous ebony
flies are crossed with homozygous black flies,

a. what will be the phenotype of the ¥ 1 flies?

b. what phenotypes and what proportions would
occur in the F 2 generation?

c. what phenotypic ratios would you expect to find
in the progeny of the backcrosses of F : X ebony?
F : X black?

34. A, B, and C are independently assorting Mendelian
factors (genes) controlling the production of black
pigment in a rodent species. Alleles of these genes
are indicated as a, b, and c, respectively. Assume that
A, B, and C act in this pathway:

control by A

colorless

control by B

colorless

red
I control by C
+ > black

t
red

A black AA BB CC individual is crossed with a color-
less aa bb cc to give black ¥ 1 individuals. The F : in-
dividuals are selfed to give F 2 progeny.

a. What proportion of the F 2 generation is color-
less?

b. What proportion of the F 2 generation is red?

35. In a particular Drosophila species, there are four
strains differing in eye color: wild-type, orange- 1,
orange-2, and pink. The following matings of true-
breeding individuals were performed.

Cross

wild-type X orange- 1
wild-type X orange-2
orange- 1 X orange-2
orange-2 X pink
F x (orange- 1 X
orange-2) X pink

all wild-type

all orange-2

1/4 orange-2:

1/4 pink: 1/4 orange-1

1/4 wild-type

What F 2 ratio would you expect if the ¥ 1 progeny
from orange-1 X orange-2 were selfed?

GENOTYPIC INTERACTIONS

36. In a variety of onions, three bulb colors segregate:
red, yellow, and white. A plant with a red bulb is
crossed to a plant with a white bulb, and all the off-
spring have red bulbs. When these are selfed, the fol-
lowing plants are obtained:

Red-bulbed

119

Yellow-bulbed

White-bulbed

What is the mode of inheritance of bulb color, and
how do you account for the ratio?

37. When studying an inherited phenomenon, a geneti-
cist discovers a phenotypic ratio of 9:6:1 among off-
spring of a given mating. Give a simple, genetic ex-
planation for this result. How would you test this
hypothesis?

38. You notice a rooster with a pea comb and a hen with
a rose comb in your chicken coop. Outline how you
would determine the nature of the genetic control
of comb type. How would you proceed if both your
rooster and hen had rose combs?

39. Suggest possible mechanisms for the epistatic ratios
given in table 2.4. Can you add any further ratios?

40. What are the differences among dominance, epista-
sis, and pleiotropy? How can you determine that
pleiotropic effects, such as those seen in sickle-cell
anemia, are not due to different genes?

41. You are working with the exotic organism Phobia
laboris and are interested in obtaining mutants that
work hard. Normal phobes are lazy. Perseverance fi-
nally pays off, and you successfully isolate a true-
breeding line of hard workers. You begin a detailed
genetic analysis of this trait. To date you have ob-
tained the following results:

hard worker

nonworker

Fi:

all nonworkers of both sexes
Fj female X worker male

3/4 hard workers: 1/4 nonworkers of both sexes

From these results, predict the expected phenotypic
ratio from crossing two F 1 nonworkers.

BIOCHEMICAL GENETICS

42. The following is a pathway from substance Q to sub-
stance U, with each step numbered:

12 3 4

Q^R^S^T^U

Which product should build up in the cell and
which products should never appear if the pathway
is blocked at point 1? At 2? At 3? At 4?

43. The following chart shows the growth (+) or lack of
growth (— ) of four mutant strains of Neurospora
with various additives. The additives are in the path-
way of niacin biosynthesis. Diagram the pathway
and show which steps the various mutants block.
Which compound would each mutant accumulate?
When you complete this problem, compare your re-
sults with figure 2.27. What effect on growth would
you observe following a mutation in the pathway of
serine biosynthesis?

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Critical Thinking Questions

Mutants

Additives

Nothing

—

Niacin

Tryptophan

Kynurenine

3-Hydroxyanthranilic

acid

Indole

—

44. The following shows the growth (+) or lack of
growth (— ) of various mutants in another biosyn-
thesis pathway. Determine this pathway, the point of
blockage for each mutant, and the substrate each
mutant accumulates.

Mutants

Additives

Nothing

+
+
+

—

45. Maple sugar urine disease is a rare inborn error of
human metabolism in which the urine of affected in-
dividuals smells like maple sugar.

a. If two unaffected individuals have an affected
child, what is the probable mode of inheritance
of the disease?

b. What is the chance that the second child will be
unaffected?

CRITICAL THINKING QUESTIONS

1. In the shepherd's purse plant, the seed capsule comes
in two forms, triangular and rounded. If two dihybrids
are crossed, the resulting ratio of capsules is 15:1 in fa-
vor of triangular seed capsules. What type of biochemi-
cal pathway might generate that ratio?

2. Assume Mendel made the cross of two true-breeding
plants that differed in all seven traits under study, one
with all dominant traits, the other with all recessive
traits. What would the ratio of phenotypes be in the F 2
generation?

Suggested Readings for chapter 2 are on page B-l.

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Chromosomal Theory

3. Mitosis and Meiosis

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MITOSIS AND

MEIOSIS

STUDY OBJECTIVES

1. To observe the morphology of chromosomes 48

2. To understand the processes of mitosis and meiosis 50

3. To analyze the relationships between meiosis
and Mendel's rules 61

STUDY OUTLINE

Chromosomes 48
The Cell Cycle 50
Mitosis 52

The Mitotic Spindle 52

Prophase 53

Metaphase 54

Anaphase 54

Telophase 54

The Significance of Mitosis 55
Meiosis 55

Prophase I 56

Metaphase I and Anaphase I 59

Telophase I and Prophase II 59

Meiosis II 60

The Significance of Meiosis 61
Meiosis in Animals 63
Life Cycles 64

Chromosomal Theory of Heredity 66
Summary 66
Solved Problems 61
Exercises and Problems 61
Critical Thinking Questions 69

Onion (Allium cepa) cells in various stages of mitosis.

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Chromosomes

The zygote, or fertilized egg of higher organisms,
is the starting point of most life cycles. This
zygote then divides many times to produce an
adult organism. In animals, the adults then
produce gametes that combine to start the cy-
cle again. In higher plants, the adult is a sporophyte that
produces spores by genetic reduction. These spores de-
velop into gametophytes, which may or may not be inde-
pendent, and gametophytes produce gametes that fuse
to form the zygote (fig. 3.1). (Numerous variations on
these themes exist, some of which we will discuss later
in this chapter or others.) The process of cell division in-
cludes a nuclear and a cytoplasmic component. Nuclear
division (karyokinesis) has two forms, a nonreduc-
tional mitosis in which the mother and daughter cells
have exactly the same genetic complement, and a reduc-
tional meiosis in which the products, gametes in ani-
mals and spores in higher plants, have approximately half

Fertilization

Growth

(a)

Genetic
reduction

Fertilization

Gametes

Growth

Gametophyte

(b)

Growth

Genetic reduction

Figure 3.1 Generalized life cycle of (a) animals and (b) plants.

the genetic material as the parent cell. Halving the
amount ensures that, when the gametes recombine, the
amount of genetic material in a zygote is the same from
generation to generation. The division of the cytoplasm,
resulting in two cells from one original cell, is termed
cytokinesis. In this chapter, we examine the processes
of mitosis and meiosis, which allow chromosomes, the
gene vehicles, to properly apportion among daughter
cells. We will discuss the engineering difficulties these
processes pose and the relationship of meiosis to
Mendel's rules.

Mendel's work was rediscovered at the turn of the
century after being ignored for thirty-four years. One of
the major reasons scientists could appreciate it in 1900
was that many of the processes that chromosomes under-
go had been described. With those discoveries, a physical
basis for genes had been found. That is, chromosomal be-
havior during gamete formation precisely fits Mendel's
predictions for gene behavior during gamete formation.
In this chapter, we look at the morphology of chromo-
somes and their behavior during somatic-cell division
and gamete and spore formation.

Modern biologists classify organisms into two major
categories: eukaryotes, organisms that have true,
membrane-bound nuclei, and prokaryotes, organisms
that lack true nuclei (table 3.1). Bacteria and blue-green
algae are prokaryotes. All other organisms are eukary-
otes. In most prokaryotes, the genetic material is a circle
of double-stranded DNA (deoxyribonucleic acid) with
some associated proteins; ancillary circles of double-
stranded DNA called plasmids are also found frequently
(see chapters 13 and 17). In eukaryotes, the genetic ma-
terial, located in the nucleus (fig. 3.2), is linear, double-
stranded DNA highly complexed with protein (nucleo-
protein). In this chapter, we concentrate on the nuclear
division processes of eukaryotes.

TablG 3.1 Differences Between Prokaryotic and Eukaryotic

Cells

Prokaryotic Cells

Eukaryotic Cells

Taxonomic groups Bacteria

All plants, fungi, animals, protists

Size* Usually less than 5 |Jim in greatest dimension

Usually greater than 5 |xm in smallest dimension

Nucleus No true nucleus, no nuclear membrane

Nuclear membrane

Genetic material One circular molecule of DNA, little protein

Linear DNA molecules complexed with histones

Mitosis and meiosis Absent

Present

See table 3.2 on page 48.

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Chapter Three Mitosis and Meiosis

Table 3.2

Metric Units of Linear Measurement

Unit

Abbreviation

Size

meter

39.37 U.S. inches

centimeter

10~ 2 meter

millimeter

10~ 3 meter

micrometer

|jim

10~ 6 meter

nanometer

10~ 9 meter

Angstrom

10" 10 meter

Ribosomes on
endoplasmic
reticulum .

B Lamellar body

Mitochondrion

Cell membrane Jil?

*T*-~-

Figure 3.2 Mouse lung cell magnified 4,270x. (Courtesy of
Wayne Rosenkrans.)

CHROMOSOMES

Chromosomes were discovered by C. von Nageli in 1842.
The term chromosome, which W. Waldeyer coined in
1888, means "colored body." Von Nageli discovered chro-
mosomes after staining techniques were developed that
made them visible. The nucleoprotein material of the
chromosomes is referred to as chromatin. When dif-
fuse, chromatin is referred to as euchromatin; when
condensed and readily visible, as heterochromatin.

Although all eukaryotes have chromosomes, in the
interphase between divisions, they are spread out or
diffused throughout the nucleus and are usually not iden-
tifiable. Each chromosome, with very few exceptions, has
a distinct attachment point for fibers (microtubules)
that make up the mitotic and meiotic spindle appara-
tuses. The attachment point occurs at a constriction in

the chromosome termed the centromere, which is
composed of several specific DNA sequences (see
chapter 15). The kinetochore is the proteinaceous
structure on the surface of the centromere to which mi-
crotubules of the spindle attach. Chromosomes can be
classified according to whether the centromere is in the
middle of the chromosome (metacentric), at the end of
the chromosome (telocentric), very near the end of the
chromosome (acrocentric), or somewhere in between
(subtelocentric or submetacentric; figs. 33 and 3.4).
For any particular chromosome, the position of the cen-
tromere is fixed. In various types of preparations, dark
bands (chromomeres) are visible (see chapter 15).

Most higher eukaryotic cells are diploid; that is, all
their chromosomes occur in pairs. One member of each
pair came from each parent. Haploid cells, which in-
clude the reproductive cells (gametes), have only one
copy of each chromosome. In the diploid state, members
of the same chromosome pair are referred to as homol-
ogous chromosomes (homologues); the two make up
a homologous pair.

The total chromosomal complement of a cell, the
karyotype, can be photographed during mitosis and re-
arranged in pairs to make a picture called a karyotype or
idiogram (fig. 3.5). From the idiogram it is possible to
see whether the chromosomes have any abnormalities
and to identify the sex of the organism. As you can see
from figure 3.5, all of the homologous pairs are made up
of identical partners, and are thus referred to as homo-
morphic chromosome pairs. A potential exception is
the sex chromosomes, which in some species are of un-

Short arm

Centromere

Long arm

(a)
Chromosome

(b)

Sister chromatids

Figure 3.3 (a) Submetacentric chromosome and
{b) submetacentric chromosome in mitosis. The chromosome is
best seen after it has duplicated but before the identical halves
(sister chromatids) separate.

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Chromosomes

♦

Figure 3.4 (a) Metacentric, (b) submetacentric, and
(c) acrocentric chromosomes in human beings. Except in
telocentric chromosomes, the centromere divides the
chromosome into two arms. (Reproduced courtesy of Dr. Thomas G.
Brewster, Foundation for Blood Research, Scarborough, Maine.)

equal size and are therefore called a heteromorphic
chromosome pair.

The number of chromosomes individuals of a partic-
ular species possess is constant. Some species exist
mostly in the haploid state or have long haploid intervals
in their life cycle. For example, pink bread mold,
Neurospora crassa, a fungus, has a chromosome number
of seven (n = 7) in the haploid state. Its diploid number
is, of course, fourteen (2n = 14). The diploid chromo-
some numbers of several species appear in table 33.

In eukaryotes, two processes partition the genetic
material into offspring, or daughter, cells. One is the sim-
ple division of one cell into two. In this process, the two
daughter cells must each receive an exact copy of the ge-
netic material in the parent cell. The cellular process is
simple cell division, and the nuclear process accompany-
ing it is mitosis. In the other partitioning process, the ge-
netic material must precisely halve so that fertilization
will restore the diploid complement. The cellular process
is gamete formation in animals and spore formation in
higher plants, and the nuclear process is meiosis. The
term mitosis comes from the Greek word for "thread," re-
ferring to a chromosome. The term meiosis comes from
the Greek meaning "to lessen."

Chromosomes separate in both processes of nuclear di-
vision. The division of the cytoplasm of the cell, cytokinesis,

> ft

-?,

U 11 It M U

U il if

»*

Ifi

> •

' t

20
F

« »

Figure 3.5 Idiogram or karyotype of a human female (two X chromosomes, no Y
chromosome). A male would have one X and one Y chromosome. The chromosomes are
grouped into categories (A-G, X, Y) by length and centromere position. Similar
chromosomes are often distinguished by their chromomeres. (Reproduced courtesy of Dr.
Thomas G. Brewster, Foundation for Blood Research, Scarborough, Maine.)

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Chromosomal Theory

3. Mitosis and Meiosis

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Chapter Three Mitosis and Meiosis

Table 3.3 Chromosome Number for Selected
Species

Species

Human being (Homo sapiens)

Garden pea (Pisum sativum)

Fruit fly (Drosophila melanogaster)

House mouse (Mus musculus)

Roundworm (Ascaris sp.)

Pigeon (Columba livia)

Boa constrictor (Constrictor constrictor)

Cricket (Gryllus domesticus)

Lily (Lilium longiflorum)

Indian fern (Ophioglossum reticulatum)

1,260

Note: In is the diploid complement. The fern has the highest known diploid
chromosome number.

is less organized. In animals, a constriction of the cell
membrane distributes the cytoplasm. In plants, the
growth of a cell plate accomplishes the same purpose.

THE CELL CYCLE

The continuity of life depends on cells growing, replicat-
ing their genetic material, and then dividing, a process
called the cell cycle (fig. 3.6). Although cells usually di-
vide when they have doubled in volume, the control of
this process is very complex and precise. Not only do all
the steps have to occur in sequence, but the cell must
also "know" when to proceed and when to wait. Contin-
uing at inappropriate moments — for example, before the
DNA has replicated or when the chromosomes or spin-
dle are damaged — could have catastrophic conse-
quences to a cell or a whole organism. Numerous stops
occur during the cycle to assess whether the next step
should proceed.

Early research into the cell cycle involved fusing cells
in different stages of the cycle (such as the G 1} S, and G 2
phases; see fig. 3.6) to determine whether the cytoplas-
mic components of one cell would affect the behavior of
the other. Results of these experiments led to the discov-
ery of a protein complex called the maturation-
promoting factor (MPF) because of its role in causing
oocytes to mature. It is now also referred to as the
mitosis-promoting factor since it initiates the mitosis
phase of the cell cycle. Further research has shown that
MPF is made of two proteins, one that oscillates in quan-
tity during the cell cycle and one whose quantity is con-

stant. The oscillating component is referred to as cyclin;
the constant gene product is an enzyme controlled by
the cdc2 gene (cdc stands for cell division cycle) called
Cdc2p. Cdc2p is a kinase, an enzyme that phosphorylates
other proteins, transferring a phosphate group from ATP
to an amino acid of the protein it is acting on. (Phosphor-
ylation controls many of the processes in mitosis and in
metabolism in general; for example, the nuclear mem-
brane begins to break down when its subunits are phos-
phorylated.) Because the Cdc2p kinase works when
combined with cyclin, it is referred to as a cyclin-
dependent kinase (CDK). Several of these kinase-
cyclin combinations control stages of the cell cycle; the
cyclin of the mitosis-promoting factor is called cyclin B.
In general, cylin-dependent kinases are regulated by
phosphorylation and dephosphorylation, cyclin levels,
and activation or deactivation of inhibitors.

Normally, Cdc2p remains at high levels in the cell but
does not initiate mitosis for two reasons. First, phosphate
groups block its active site, the place on the enzyme that
actually does the phosphorylating. Second, the enzyme
can only function when it combines with a molecule of
cyclin B, the protein that oscillates during the cell cycle.
Cyclin B is at very low levels when mitosis ends. During
ensuing cell growth, numbers of cyclin B molecules
increase, combining with Cdc2p proteins until a critical
quantity is reached. However, Cdc2p-cyclin B complexes
are still not active. That requires the product of another
gene to dephosphorylate the Cdc2p-cyclin B complex. At
that point, the Cdc2p-cyclin B complex goes into action,
initiating the changes that begin mitosis (fig. 3.7). Pre-
sumably the cell is ready for mitosis at that point, having

Figure 3.6 Cell cycle in the broad bean, Vicia faba. Total time
in the cycle is under twenty hours. The DNA content of the cell
doubles during the S phase and is then reduced back to its
original value by mitosis.

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The Cell Cycle

gone through Gl, S, and G2 phases (which we will dis-
cuss in detail later in the chapter).

Once mitosis has been initiated, cyclin B, along with
other proteins that have served their purpose by this
point in the cell cycle, breaks down with the help of a
protein complex called the anaphase-promoting com-
plex (APC), also called the cyclosome. The cyclosome
works by attaching a ubiquitin molecule to the proteins
that are to be broken down. (Ubiquitin is a polypeptide of
76 amino acids; it directs the attached protein into a
breakdown pathway discussed in chapter 16.) Cdc2p is
then phosphorylated to block its active site. The cell now
completes mitosis and enters G x ; quantities of cyclin B
are very low, and virtually no functioning Cdc2p-cyclin B
remains (fig. 3.7). Thus, active Cdc2p is the kinase that
controls the initiation of mitosis.

Some points in the cell cycle, such as the initiation of
mitosis, can be delayed until all necessary conditions are

in place. These checkpoints allow the cell to make sure
that various events have been "checked off" as com-
pleted before the next phase begins. Surveillance
mechanisms that involve dozens of proteins, many just
discovered, oversee these checkpoints. In the cell cycle,
three checkpoints involve cyclin-dependent kinases;
each has its own specific cyclin that initiates either the
G 1; S, or mitosis phase. In addition, other checkpoints
that don't involve cyclin-dependent kinases occur at
other transition phases in the cell cycle.

Cell cycle control is of particular interest because the
cell cycle routinely halts if there is genetic damage, giv-
ing the cell a chance to repair the damage before com-
mitting to cell division. If the damage is too extreme, the
cell can enter a programmed cell death sequence, dis-
cussed in chapter 16. If these mechanisms fail, cancer
may result. The genetic control of the cell cycle is one of
the most active areas of current research.

Inactive
complex

Phosphates block
active site

Dephosphorylation
P

Phosphorylation

Cdc2p

Active
complex

Active
site

Breakdown

cyclin B

Inactive
Cdc2p

Buildup

of
cyclin B

Figure 3.7 The proteins Cdc2p (CDK1) and cyclin B combine to form the maturation-promoting (or
mitosis-promoting) factor. During mitosis, cyclin B is broken down. During G 1 and S phases, cyclin B
builds up and combines with Cdc2P, which is then phosphorylated at the active site to render it
inactive. Dephosphorylation, a process that begins to take place only after DNA replication is finished,
produces an active maturation-promoting factor.

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Chapter Three Mitosis and Meiosis

MITOSIS

Consider the engineering problem that mitosis must solve.
Identical chromatids, called sister chromatids, the re-
sult of chromosomal replication, must separate so that
each goes into a different daughter cell (see fig. 3.3). These
chromatids are the visible manifestation of the chromo-
somal replication that has taken place in the S phase of the
cell cycle. The chromatids are initially held together; each
will be called a chromosome when it separates and be-
comes independent. Each of the two daughter cells then
ends up with a chromosome complement identical to that
of the parent cell. Mitosis is nature's elegant process to
achieve that end — surely an engineering marvel.

Mitosis is a continuous process. However, for descrip-
tive purposes, we can break it into four stages: prophase,
metaphase, anaphase, and telophase (Greek: pro-, be-
fore; meta-, mid; ana-, back; telo-, end). Replication (du-
plication) of the genetic material occurs during the S
phase of the cell cycle (see fig. 3.6). The timing of the
four stages varies from species to species, from organ to
organ within a species, and even from cell to cell within
a given cell type.

Tubulin

Minus
end

■cxd

.cf=>

Plus
end

Figure 3.8 Microtubules are hollow tubes made of a and (3
tubulin subunits that are constantly being added or removed.

Microtubules are formed from active centers called
microtubule organizing centers. Centrioles, com-
posed of two cylinders — themselves composed of
microtubules — are microtubule organizing centers for
cilia and flagella. Under those circumstances, the centri-
oles are referred to as basal bodies. The centrioles were
also originally believed to organize spindles. However,
for most organisms, the microtubule organizing center is
called the centrosome. In some organisms, such as
fungi, a different cell organelle, the spindle pole body,
serves this function. In most animals, the centrosome
contains a centriole (fig. 39). However, the centriole is
absent in most higher plants. Moreover, innovative ex-

The Mitotic Spindle ^C*

The process of mitosis involves an apparatus called the
spindle. This structure is composed of microtubules, hol-
low cylinders made of protein subunits; each subunit is
composed of one molecule of a tubulin and one of p tubu-
lin; and each tubulin is the product of a different gene.
(The spindle is named for the rounded rods, tapered at
each end, once commonly used to hold yarn or thread.)
Microtubules provide shape and structure to a eukaryotic
cell as well as allow the cell to move its internal compo-
nents and to move the cell itself with cilia and flagella. Mo-
tion occurs as the microtubules slide past each other, a
vesicle of some kind slides along the microtubules, and the
microtubules shorten. Two proteins make up the micro-
tubule motors that allow motion: kinesin and dynein.
Scientists have studied microtubules through protein
chemistry, through mutant organisms, and through inno-
vative methods such as by coupling tubulin subunits with
fluorescing dyes to observe the microtubules in action.

Microtubules are in a dynamic equilibrium, with
subunits constantly being added or removed at both ends.
On any microtubule, more activity occurs at one end than
the other. The more active end of the tubule is called the
plus end, the less active end the minus end (fig. 38). Both
ends may be adding or removing subunits, or the plus end
may be adding while the minus end is removing subunits.
Generally, dynein causes movement toward the minus
end, whereas kinesin causes movement toward the plus
end of a microtubule, although exceptions exist.

Figure 3.9 A centriole is composed of two barrels at right
angles to each other. Each barrel is composed of nine tripartite
units and a central cartwheel. Each of the three parts of a
tripartite unit is a microtubule. Magnification 111,800x.
(Reproduced from The Journal of Cell Biology, 1968, Vol. 37, p. 381 , by
copyright permission of The Rockefeller University Press.)

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Mitosis

Centrosome

\\ i//

Aster

Nucleus

Centriole

Centrosome

divides

Interpolar microtubules

Nuclear
membrane,

breaks
down

Figure 3.10 Early in mitosis, the centrosome divides, and the separating halves move to opposite poles of the cell. This creates a
spindle in the middle of the cell after the nuclear membrane breaks down.

periments that removed the centrioles from cells that
normally had them demonstrated that the centriole is not
necessary for spindle formation. So, although we used to
believe that the centriole formed the spindle in many or-
ganisms, we now know that the spindle is usually orga-
nized around the centrosome, which can function in this
capacity without a centriole.

The centriole, when present, replicates during the
S and G 2 phases. When mitosis begins, the centrosome
divides and moves to opposite poles of the cell, around
the nucleus (fig. 3.10). The centrosomes trail micro-
tubules, forming the spindle, that at this point begin at
each centrosome and overlap in the middle of the cell.
These are called interpolar microtubules. Micro-
tubules also spread out from the centrosome in the op-
posite direction from the spindle itself, forming an aster
(see fig. 3.10). The minus ends of microtubules emanate
from the centrosome and the plus ends overlap in the
middle of the cell. A third form of tubulin, 7 tubulin, is
needed to begin the formation of a microtubule.

Prophase

This stage of mitosis is characterized by the formation
of the spindle and a shortening and thickening of the
chromosomes so that individual chromosomes become
visible. (We will discuss details of the molecular structure
of the eukaryotic chromosome and the processes of coil-
ing and shortening in chapter 15.) At this time also, the
nuclear envelope (membrane) disintegrates and the nu-
cleolus disappears (fig. 3.11). The nucleolus is a darkly
stained body in the nucleus that is involved in ribosome
construction and that forms around a nucleolar orga-
nizer locus on one of the chromosome pairs. The num-
ber of nucleoli varies in different species, but in the sim-
plest case there are two nucleolar organizers per
nucleus, one each on the two members of a homologous
pair of chromosomes. Nucleoli re-form after mitosis.

As prophase progresses, each chromosome is com-
posed of two identical (sister) chromatids (see fig. 33); the

chromosomes continue to shorten and thicken. The cen-
tromeres have already divided, and no new DNA synthesis
is needed for the process to be completed. At this point,
the sister chromatids are kept together by a complex,
called cohesin, made up of at least four different protiens.
Spindle fibers are initially nucleated at the centrosome
and grow outward into the cytoplasm (fig. 3.12). Some of
these fibers "capture" a kinetochore, the proteinaceous
complex at the centromere of each sister chromatid;
these fibers are called kinetochore microtubules. At
first, one kinetochore or the other randomly attaches to a
spindle fiber. As the microtubules further move the chro-
mosomes and as new microtubules attach and old micro-
tubules break, each sister kinetochore eventually attaches
to microtubules emanating from different poles. This en-
sures that sister chromatids move to opposite poles dur-
ing anaphase. The number of microtubules that attach to
each kinetochore differs in different species. It seems that
1 attaches to each kinetochore in yeast, 4 to 7 attach to
each kinetochore in the cells of a rat fetus, and 70 to 150
attach in the plant Haemanthus (see fig. 3.12).

Interphase

Early
prophase

Late
prophase

Nucleolus

Nuclear
membrane

Figure 3.11 Nuclear events during interphase and prophase of
mitosis. In this cell, 2n = 4, consisting of one pair of long and
one pair of short metacentric chromosomes. Maternal
chromosomes are red; paternal chromosomes are blue. Note
that each chromosome consists of two chromatids when the
cell enters mitosis.

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Chapter Three Mitosis and Meiosis

Figure 3.12 Scanning electron micrograph of the centromeric
region of a metaphase chromosome from the plant Haemanthus
katherinae. Spindle fiber bundles on either side of the
centromere extend in opposite directions. A fiber not
connected to the kinetochore is visible lying over the
centromere. These fibers are 60 to 70 nm in diameter. (Waheeb
K. Heneen, "The centromeric region in the scanning electron microscope,"
Hereditas, 97 (1982): 311-14. Reproduced by permission.)

Metaphase

During metaphase, the chromosomes move to the equa-
tor of the cell. With the attachment of the spindle fibers
and the completion of the spindle itself, the chromo-
somes jockey into position in the equatorial plane of the
spindle, called the metaphase plate. This happens as
kinetochore microtubules exert opposing tension on the
two sister kinetochores. Alignment of the chromosomes
on this plate marks the end of metaphase (fig. 3.13).

Anaphase ^l*

During anaphase, the sister chromatids separate and
move toward opposite poles on the spindle. The physical
separation of the sister chromatids and their movement
to opposite poles are two separate activities. Chromatid
separation represents a checkpoint in the process of mi-
tosis; a surveillance mechanism will not allow the
process to continue until all chromosomes are lined up
on the metaphase plate with their sister kinetochores
held by microtubules from opposite poles. The surveil-
lance mechanism somehow checks the physical tension
the spindle fibers exert on a pair of sister chromatids; an
unpaired chromatid can delay or stop the process. Ini-
tially, an inhibitory protein called securin binds an en-
zyme called separin that can break down cohesin, the
complex holding the chromatids together. At the correct

moment, the cyclosome ubiquitinates the inhibitor, caus-
ing it to break down and freeing the separin to break
down cohesin. This liberates the sister chromatids from
each other (and is the instant when chromatids become
chromosomes) .

The spindle then separates the sister chromatids in
two stages, called anaphase A and anaphase B. In
anaphase A, the chromosomes move toward the poles
(fig. 3.14). During this process, the kinetochore itself acts
as a microtubule motor, disassembling microtubules as it
moves down them, pulling the chromosomes along
(fig. 3.15). Thus, metacentric chromosomes appear
V-shaped (as in fig. 3.15), subtelocentrics appear
J-shaped, and telocentrics appear rod-shaped. In
anaphase B, the spindle itself elongates as overlapping in-
terpolar microtubules slide apart. The general elongation
of the spindle pulls the chromosomes apart.

Telophase

At the end of anaphase (fig. 3.16), the separated sister
chromatids (now full-fledged chromosomes) have been
pulled to opposite poles of the cell. The cell now re-
verses the steps of prophase to return to the interphase
state (fig. 3. 17). The chromosomes uncoil and begin to di-
rect protein synthesis. A nuclear envelope re-forms
around each set of chromosomes, nucleoli re-form, and
cytokinesis takes place. The spindle breaks down into
tubulin subunits; a residual of microtubules remains at

Metaphase plate

Figure 3.13 Metaphase of mitosis. In this cell, 2n = 4. Maternal
chromosomes are red; paternal chromosomes are blue.

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Meiosis

the center of the cell and seems to be involved in the for-
mation of a constricting ring in animal cells or in
the growth of a cell plate in plant cells. The cell has now
entered the G 1 phase of the cell cycle (see fig. 3.6).
Figure 3.18 summarizes mitosis.

life functions; with mitosis, they will produce offspring
cells with these same capabilities. With this stability as-
sured, single-celled organisms could thrive and multicel-
lular organisms could evolve.

The Significance of Mitosis

Cytokinesis and mitosis result in two daughter cells, each
with genetic material identical to that of the parent cell.
This exact distribution of the genetic material, in the
form of chromosomes, to the daughter cells, ensures the
stability of cells and the inheritance of traits from one
cell generation to the next. Cells have evolved complex

(a)

(b)

Figure 3.14 (a) The mitotic spindle during anaphase. In this
cell, 2/i = 4. Maternal chromosomes are red; paternal
chromosomes are blue, (b) Fluorescent microscope image of a
cultured cell in anaphase. Microtubles are red; chromosomes
(DNA) are Stained yellow, {[b] John M. Murray, Department of
Anatomy, University of Pennsylvania. Cover of BioTechniques, volume 7,
number 3, March 1989. Reproduced with permission.)

MEIOSIS Q

Gamete formation presents an entirely new engineering
problem to be solved. To form gametes in animals (and,
for the most part, to form spores in plants), a diploid or-
ganism with two copies of each chromosome must form
daughter cells that have only one copy of each chromo-
some. In other words, the genetic material must be re-
duced by half so that when gametes recombine to form
zygotes, the original number of chromosomes is re-
stored, not doubled.

If we were to try to engineer this task, we would first
need to be able to recognize homologous chromosomes.
We could then push one member of each pair into one
daughter cell and the other into the other daughter cell. If
we were unable to recognize homologues, we would not
be able to ensure that each daughter cell received one and
only one member of each pair. The cell solves this problem
by pairing up homologous chromosomes during an ex-
tended prophase. The spindle apparatus then separates
members of the homologous chromosome pairs, just as it
separates sister chromatids during mitosis. But there is one
complication. As in mitosis, cells entering meiosis have al-
ready replicated their chromosomes. Therefore, two nu-
clear divisions without an intervening chromosome repli-
cation are necessary to produce haploid gametes or

Kinetochore

Motor fibers

Kinetochore

microtubule

Chromosome

Disassembled f ^
tubulin *

Direction of chromosomal movement

Figure 3.15 The kinetochore acts as a microtubule motor,
pulling the chromosome along the kinetochore microtubules
toward the pole. One microtubule is shown, although many
may be present.

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Chapter Three Mitosis and Meiosis

Constriction

Late anaphase

Figure 3.16 Late anaphase of mitosis; 2n = 4. A constriction begins to form
in the middle of the cell (in animals). Maternal chromosomes are red; paternal
chromosomes are blue.

spores. Meiosis is, then, a two-division process that pro-
duces four cells from each original parent cell. The two di-
visions are known as meiosis I and meiosis II.

Unlike mitosis, meiosis occurs only in certain kinds of
cells. In animals, meiosis begins in the primary gameto-
cytes; in higher plants, the process takes place only in the
spore-mother cells of the sporophyte generation (see

Constriction

Telophase

Nucleolus

Interphase

Figure 3.17 Telophase and interphase of mitosis; 2n = 4.
Maternal chromosomes are red; paternal chromosomes are blue.

fig. 31). At the end of this chapter, we review the
processes of gamete and spore formation in animals and
plants, respectively.

Prophase I

Cytogeneticists have divided the prophase of meiosis I
into five stages: leptonema, zygonema, pachynema,
diplonema, and diakinesis (Greek: lepto-, thin; zygo-,
yoke-shaped \pachy-, thick; diplo-, double; dia-, across). A
cell entering prophase I (leptotene stage) behaves simi-
larly to one entering prophase of mitosis, with the cen-
trosome duplicated and the spindle forming around the
intact nucleus. (Note the adjectival forms — leptotene —
versus the noun forms — leptonema — of the stage
names.) As the chromosomes coil down in size during
leptonema, they are visible as individual threads: sister
chromatids are in such close apposition that they are not
distinct. The chromosomes are more spread out than
they are in mitosis, with dark spheres or bands called
chromomeres interspersed.

The tips of the chromosomes are attached to the nu-
clear membrane in the leptotene stage (fig. 3.19). In the
leptotene to zygotene transition, the tips of the chromo-
somes move until most end up in a limited region near
each other. This forms an arrangement called a bouquet
stage. Presumably, this arrangement helps homologous
chromosomes find each other and begin the pairing
process without becoming entangled.

The pairing of homologous chromosomes marks the
zygotene stage. Initial contact between identical regions
of homologous chromosomes leads to a point-for-point
pairing along their lengths. This process is referred to as
synapsis. A proteinaceous complex, referred to as a

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^^^p."™

(a) Interphase

(b) Early prophase

(d) Metaphase

(e) Anaphase

(f) Telophase

Figure 3.18 Cells in interphase and in various
stages of mitosis in the onion root tip. The
average cell is about 50 |im long. (© The
McGraw-Hill Companies, Inc./Kingsley Stern, photographer.)

(g) Daughter cells

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Chapter Three Mitosis and Meiosis

Nuclear
membrane

Nucleolus

Leptonema

Synaptonemal
complex

Bouquet

Early zygonema

Diplonema

Diakinesis

Figure 3.19 Prophase I of meiosis; 2n = 4 (nuclei
shown). Maternal chromosomes are red; paternal
chromosomes are blue. Note that crossing over is
evident at diplonema.

(a)

Tripartite

synaptonemal

complex

Central element
Lateral element

Laterally displaced DNA

0-2|im

(b)

Attachment plate formed by
swollen end of lateral element

Cytoplasm
Nuclear membrane

Figure 3.20 The synaptonemal complex, (a) In the electron micrograph,
M is the central element, La are lateral elements, and F are
chromosome fibers. Magnification 400,000x. (b) Diagram of the
Structure, ([a] R. Wettstein and J. R. Sotelo, "The molecular architecture of synap-
tonemal complexes," in E. J. DuPraw, ed., Advances in Cell and Molecular Biology, vol.
1 (New York: Academic Press, 1971), p. 118. Reproduced by permission, [b] From B.
John and K. R. Lewis, Chromosome Hierarchy. Copyright © 1 975 Oxford University
Press, London, England. Reprinted by permission of the Oxford University Press.)

synaptonemal complex (fig. 3.20), appears between
the homologous chromosomes and mediates synapsis in
an unknown way. At this point, the chromosome figures
are referred to as bivalents, one bivalent per homolo-
gous pair. The synapsis of all chromosomes marks the
end of zygonema.

The chromosomes now continue to shorten and
thicken, giving pachynema its name. During the entire
prophase, crossing over takes place. When two chro-
matids come to lie in close proximity, enzymes can break
both chromatid strands and reattach them differently
(fig. 321). Thus, although genes have a fixed position on
a chromosome, alleles that started out attached to a pa-
ternal centromere can end up attached to a maternal cen-

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Meiosis

_s\_

Figure 3.21 Crossing over in a tetrad during prophase of
meiosis I. Maternal chromosomes are red; paternal
chromosomes are blue. Note the exchange of chromosome
pieces after the process is completed.

tromere. Crossing over can greatly increase the genetic
variability in gametes by associating alleles that were not
previously joined. (We examine the molecular mecha-
nism of this process in chapter 12.) Before crossing over
takes place, densely staining nodules are visible, first in
zygonema and lasting through pachynema. These are
called recombination nodules (fig. 3. 22a); they are
correlated with crossing over and presumably represent
the enzymatic machinery present on the chromosomes.

As the chromosomes shorten and thicken further in
diplonema, each chromosome can be seen to be made of
two sister chromatids. Now the chromosome figures are
referred to as tetrads because each is made up of four
chromatids (see fig. 3.19). At about this time, the synap-
tonemal complex disintegrates in all but the areas of the
chiasmata (singular: chiasma), the X-shaped configura-
tions marking the places of crossing over (fig. 3. 22£>). Vir-
tually all tetrads exhibit chiasmata; in cases in which no
crossing over occurs, the tetrads tend to fall apart and
segregate randomly. Thus, crossing over not only in-
creases genetic diversity but also ensures the proper sep-
aration of homologous chromosomes. A meiosis-specific
form of cohesin keeps sister chromatids together.

During the diplotene stage, chromosomes can again
uncondense and become active. This is especially obvi-
ous in amphibians and birds, which produce a great
amount of cytoplasmic nutrient for the future zygote. Re-
condensation of the chromosomes takes place at the end
of diplonema. This stage can be very long; in human fe-
males, it begins in the fetus and does not complete until
the egg is shed during ovulation, sometimes more than
fifty years later. As prophase I moves into diakinesis, the
chromosomes become very condensed (see fig. 3.19).

Metaphase I and Anaphase I

Metaphase I is marked by the breakdown of the nuclear
membrane and the attachment of kinetochore micro-
tubules to the tetrads. Unlike in mitosis, in which sister
chromatids are pulled apart because each sister kineto-
chore is attached to a different pole, both sister kineto-
chores become attached to spindle microtubules coming
from the same pole in metaphase I (fig. 3.23). During
anaphase I, cohesin breaks down every place but at the
centromeres, allowing sister chromatids to be pulled to

(a)

(b)

Figure 3.22 (a) Recombination nodules {arrowhead) in
spermatocytes of the pigeon, Columba livia. (Bar = 1 |xm.)
{b) A tetrad from the grasshopper, Chorthippus parallelus, at
diplonema with five chiasmata. ([a] From M. I. Pigozzi and

A. J. Solari, "Recombination Nodule Mapping and Chiasma Distribution in
Spermatocytes of the Pigeon, Columba livia," in Genome, 42: 308-314,
1999. Reprinted by permission, [b] Courtesy of Bernard John.)

the same pole: homologous chromosomes are separated
(fig. 324). This meiotic division is therefore called a re-
ductional division because it reduces the number of
chromosomes to half the diploid number in each daugh-
ter cell. For every tetrad there is now one chromosome
in the form of a chromatid pair, known as a dyad or
monovalent, at each pole of the cell. The initial objec-
tive of meiosis, separating homologues into different
daughter cells, is accomplished. However, since each
dyad consists of two sister chromatids, a second, mitosis-
like division is required to reduce each chromosome to a
single chromatid.

Telophase I and Prophase II

Depending on the organism, telophase I may or may not
be greatly shortened in time. In some organisms, all the
expected stages take place; chromosomes enter an
interphase configuration as cytokinesis takes place. How-
ever, no chromosome duplication (DNA replication) oc-
curs during this abbreviated interphase, termed interki-
nesis. Next, in these organisms, prophase II begins and

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Chapter Three Mitosis and Meiosis

Aster

Kinetochore

Kinetochore
microtubule

Figure 3.23 Metaphase of meiosis I; 2n = 4. Maternal chromosomes
are red; paternal chromosomes are blue. Sister kinetochores (effectively
single, merged kinetochores) are attached to microtubules from the
same pole.

Anaphase I

Figure 3.24 Anaphase of meiosis I; 2n = 4. Maternal chromosomes are red;
paternal chromosomes are blue. Homologous chromosomes separate and
move to opposite poles.

meiosis II proceeds. In still other organisms, the late
anaphase I chromosomes go almost directly into
metaphase II, virtually skipping telophase I, interphase,
and prophase II.

Meiosis II

Meiosis II is basically a mitotic division in which the
chromatids of each chromosome are pulled to opposite
poles. For each original cell entering meiosis I, four cells
emerge at telophase II. Meiosis II is an equational divi-

sion; although it reduces the amount of genetic material
per cell by half, it does not further reduce the chromo-
some number per cell (fig. 3.25). (Sometimes it is simpler
to concentrate on the behavior of centromeres during
meiosis than on the chromosomes and chromatids. Meio-
sis I separates maternal from paternal centromeres, and
meiosis II separates sister centromeres.) Figure 3.26 sum-
marizes meiosis in corn (Zea mays).

In terms of chromosomes, meiosis begins with a
diploid cell and produces four haploid cells. In terms of
DNA, the process is a bit more complex but has the same

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Meiosis

Metaphase II

1 >

Anaphase II

Telophase II

#•♦♦♦♦♦♦ .•'••^ &i

Interphase
Figure 3.25 Meiosis II; 2/7 = 4. Maternal chromosomes are red; paternal chromosomes are blue.

result. Let us call the quantity of DNA in a gamete "C."
A diploid cell before S phase has 2C DNA, and the same
cell after S phase, but before mitosis, has 4C DNA. Mitosis
reduces the quantity of DNA to 2C. A cell entering meio-
sis also has 4C DNA. After the first meiotic division, each
daughter cell has 2C DNA, and after the second meiotic
division, each daughter cell has C DNA, the quantity
appropriate for a gamete.

The Significance of Meiosis ^CT

Meiosis is significant for several reasons. First, it reduces
the diploid number of chromosomes so that each of

four daughter cells has one complete haploid chromo-
some set. Second, because of the randomness of the
process of chromosomal separation, a very large num-
ber of different chromosomal combinations can form in
the gametes. For example, in human beings, if each ga-
mete could get either the maternal or paternal chromo-
some, and we have twenty-three chromosomal pairs,
2 23 or 8,388,608 different combinations can occur. Third,
because of crossing over, even more allelic combinations
are possible. The process of creating new arrangements,
either by crossing over or by independent segregation of
homologous pairs of chromosomes, is called recombina-
tion. Assuming one hundred thousand genes in a human

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Chromosomal Theory

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Chapter Three Mitosis and Meiosis

\ v**»

Leptotene

,- f

* if

Diplotene

Pachytene

Diakinesis

Metaphase I

•./*

i»v

Telophase I

Anaphase I

Interphase

Prophase II
(early)

■

Prophase II
(late)

•
***

Metaphase II

Anaphase II ,

Figure 3.26 Meiosis in corn (Zea mays). (Courtesy of Dr. M. M. Rhoades. "Meiosis in maize," Journal of Heredity, 41 : 59-67, 1950. Reproduced by
permission.)

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Meiosis in Animals

being with two alleles each, 2 100000 different gametes
could potentially arise by meiosis.

The behavior of any tetrad follows the pattern of
Mendel's rule of segregation. At spore or gamete forma-
tion (meiosis), the diploid number of chromosomes is
halved; each gamete receives only one chromosome
from a homologous pair. This process, of course, explains
Mendel's rule of segregation. Chromosomal behavior at
meiosis also explains independent assortment (fig. 3. 27).
In anaphase I, the direction of separation is independent
in different tetrads. Whereas one pole may get the mater-
nal centromere from chromosomal pair number 1, it
could get either the maternal or the paternal centromere
from chromosomal pair number 2, and so on (see fig.

3.27). Alleles of one gene segregate independently of al-
leles of other genes. Very shortly after the rediscovery of
Mendel's principles in 1900, geneticists were quick to
realize this.

MEIOSIS IN ANIMALS

In male animals, each meiosis produces four equal-sized
sperm cells in a process called spermatogenesis (fig.
3.28). In vertebrates, a cell type in the testes known as
a spermatogonium produces primary spermato-
cytes, as well as additional spermatogonia, by mitosis.

Meiosis
II

I«
I'

Figure 3.27 Relationship of meiosis to the rule of independent assortment. Maternal {reef)
and paternal (blue) chromosomes separate independently in different tetrads.

First

meiotic

division

Second
meiotic
division

Differentiation

Spermatogonium

Primary
spermatocyte

Secondary
spermatocytes

Spermatids

Sperm
cells

Figure 3.28 Spermatogenesis; 2n = 4. Maternal chromosomes are red; paternal chromosomes are blue.

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Chromosomal Theory

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Chapter Three Mitosis and Meiosis

The primary spermatocytes then undergo meiosis. After
the first meiotic division, these cells are known as sec-
ondary spermatocytes; after the second meiotic divi-
sion, they are known as spermatids. The spermatids
mature into spermatozoa by a process called spermio-
genesis — with four sperm cells resulting from each pri-
mary spermatocyte. In human beings and other verte-
brates without a specific mating season, the process of
spermatogenesis is continuous throughout adult life. A
normal human male may produce several hundred mil-
lion sperm cells per day.

During embryonic development in human females,
cells in the ovary, known as oogonia, proliferate by nu-
merous mitotic divisions to form primary oocytes.
About one million form per ovary. These begin the first
meiotic division and then stop before the birth of the fe-
male in a prolonged diplonema, called the dictyotene
stage. A primary oocyte does not resume meiosis until
the female is past puberty, when, under hormonal con-
trol, ovulation takes place. This process usually occurs for
only one oocyte per month during the female's repro-
ductive life span (from about twelve to fifty years of age).
Meiosis only then proceeds in the ovulated oocyte. In the
female, the two cells formed by meiosis I are of unequal
size. One, termed the secondary oocyte, contains al-
most all the nutrient-rich cytoplasm; the other, a polar
body, receives very little cytoplasm. The second meiotic
division in the larger cell yields another polar body and
an ovum. The first polar body may or may not divide to
form two other polar bodies, which ultimately disinte-
grate. Thus, oogenesis produces cells of unequal size —
an ovum and two or three polar bodies (fig. 3.29). Cells of
unequal size are produced because the oocyte nucleus
and meiotic spindle reside very close to the surface of
this large cell.

LIFE CYCLES

For eukaryotes, the basic pattern of the life cycle alter-
nates between a diploid and a haploid state (see fig. 3.1).
With the exception of the life cycles of bacteria and
viruses, all life cycles are modifications of this general
pattern. Bacteria, including blue-green algae, have a sin-
gle circular chromosome; with exceptions described
later, they are always in the haploid state. They divide by
replicating their DNA and having the two copies separate
into two daughter cells by simple cell division (see chap-
ter 7). Viruses, on the border of being called alive, insert
their genetic material into the cells of other organisms,
and then manufacture new copies of themselves (see
chapter 7).

Most animals are diploids that form gametes by meio-
sis, then restore the diploid number by fertilization. Ex-
ceptions, however, are numerous. For example, in the
bees, wasps, and ants (hymenoptera), males are haploid
and produce gametes by mitosis; females are diploid.
Some fishes exist by parthenogenesis, in which the
offspring come from unfertilized eggs that do not under-
go meiosis. And, in some copepods, the sexual and
parthenogenetic stages of their life cycles alternate.

The general pattern of the life cycle of plants alternates
between two distinct generations, each of which, depend-
ing on the species, may exist independently. In lower
plants, the haploid generation predominates, whereas in
higher plants, the diploid generation is dominant. In flow-
ering plants (angiosperms), the plant you see is the
diploid sporophyte (see fig. 3.1). It is referred to as a
sporophyte because, through meiosis, it will give rise to
spores. The spores germinate into the alternate genera-
tion, the haploid gametophyte, which produces gametes
by mitosis. Fertilization then produces the next generation

First meiotic
division

Second meiotic
division

S
^

Polar,
body

•

Oogonium

Primary oocyte

Secondary oocyte

Ovum and polar bodies

Figure 3.29 Oogenesis; 2n = 4. Maternal chromosomes are red; paternal chromosomes are blue.

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Life Cycles

of diploid sporophytes. In lower plants, the gametophyte
has an independent existence; in angiosperms, this gener-
ation is radically reduced. For example, in corn (fig. 3. 30),
an angiosperm, the mature corn plant is the sporophyte.
The male flowers produce microspores by meiosis. After
mitosis, three cells exist in each spore, a structure that we
call a pollen grain, the male gametophyte. In female
flowers, meiosis produces megaspores. Mitosis within a
megaspore produces an embryo sac of seven cells with
eight nuclei. This is the female gametophyte. A sperm cell
fertilizes the egg cell. The two polar nuclei of the embryo
sac are fertilized by a second sperm cell, producing
triploid On) nutritive endosperm tissue. The sporophyte
grows from the diploid fertilized egg.

Many fungi and protista are haploid. Fertilization pro-
duces a diploid stage, which almost immediately under-
goes meiosis to form haploid cells. These cells, in turn, in-
crease in number by mitosis. We will analyze organisms
such as Neurospora, the pink bread mold, in more detail
later (see chapter 6).

Much of our knowledge of genetics derives from the
study of specific organisms with unique properties.
Mendel found pea plants useful because he could control
matings carefully, their generation time was only a year,
he could easily grow them in his garden, and they had
the discrete traits that he was seeking. Our interest in hu-
man beings is obvious. However, we are members of a
very difficult species to study experimentally. We have a
long generation time and a small number of offspring
from matings that we cannot tailor for research pur-
poses. The fruit fly, Drosophila melanogaster, is one of
the organisms geneticists have studied most extensively.
Fruit flies have a short generation time (twelve to four-
teen days), which means that many matings can be car-
ried out in a reasonable amount of time. In addition, they
do exceptionally well in the laboratory, they have many
easily observable mutants, and in several organs they
have giant banded chromosomes of great interest to cy-
togeneticists.

Tube nucleus
Sperm cells

Endosperm (3n): 2 polar + 1 sperm nuclei
Embryo (2n)\ egg + 1 sperm nuclei

Mature

microgametophyte
(three cells)

Mature

megagametophyte
(eight nuclei)

6
Flower

Sporophyte

Meiosis

9
Flower

Microspores Megaspores

Pollen ( O

(four per meiosis)

Three degenerate
cells

One functional
cell

Figure 3.30 Life cycle of the corn plant.

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Chromosomal Theory

3. Mitosis and Meiosis

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Chapter Three Mitosis and Meiosis

Note that species used in food production tend to be
intermediate in their life cycles. That is, many crop
plants, such as peas and corn, have only one generation
interval per year under normal circumstances. (We use
the term generation interval here in the broadest sense,
as the time it takes to complete an entire life cycle; see
also chapter 19) Crop plants are easier to work with
from a genetic standpoint than people, but much more
difficult than, say, Drosophila or bacteria (table 3.4). Be-
cause of their relatively long generation interval, crop
plants are limited in their utility for studying basic ge-
netic concepts or applying genetic technology to agri-
culture.

As you make your way through this book and through
other readings on genetics, and as you come across stud-
ies involving new organisms, ask yourself the question,
What are the properties of this organism that make it
ideal for this type of research?

CHROMOSOMAL THEORY
OF HEREDITY

In a paper in 1903, cytologist Walter Sutton firmly stated
the concepts we have developed here: The behavior of
chromosomes during meiosis explains Mendel's princi-
ples. Genes, then, must be located on chromosomes. This
idea, which several other biologists were also developing
at the time, was immediately accepted, ushering in the
era of the chromosomal theory of inheritance. Dur-

ing this era, intensive effort was devoted to studying the
relationships between genes and chromosomes. The ma-
jor portion of the first section of this book is devoted to
classical studies of linkage and mapping. Linkage deals
with the association of genes to each other and to spe-
cific chromosomes. Mapping deals with the sequence of
genes on a chromosome and the distances between
genes on the same chromosome. This is basic informa-
tion for a study of the structure and function of genes.
Here we introduce a new term for the gene. The term lo-
cus (plural: loci), meaning "place" in Latin, refers to the
location of a gene on the chromosome.

Table 3.4 Approximate Generation Intervals of
Some Organisms of Genetic Interest

Approximate

Organism

Generation Interval

Intestinal bacterium {Escherichia colt)

20 minutes

Bacterial virus {lambda)

1 hour

Pink bread mold (Neurospora crassa)

2 weeks

Fruit fly (Drosophila melanogaster)

2 weeks

House mouse (Mus musculus)

2 months

Corn (Zea mays)

6 months

Sheep (Ovus aries)

1 year

Cattle (Bos taurus)

2 years

Human being (Homo sapiens)

14 years

SUMMARY

STUDY OBJECTIVE 1: To observe the morphology of
chromosomes 48-50

Chromosomes are made of chromatin and divided by cen-
tromeres. Within centromeres are kinetochores, attachment
points for spindle fibers. Structure within the chromosomes
is evident from bands on chromosomes called chromomeres.

STUDY OBJECTIVE 2: To understand the processes of mi-
tosis and meiosis 50-61

During eukaryotic cell division, the processes of mitosis
and meiosis apportion the chromosomes to daughter cells.
Both processes are preceded by chromosome replication
during the S phase of the cell cycle, which is under genetic
control. In mitosis, the two sister chromatids making up
each replicated chromosome separate into two daughter
cells. Sex cells — gametes in animals and spores in plants —
are produced by the two-stage process of meiosis. In meio-

sis, homologous chromosomes are first separated into two
daughter cells, and then the sister chromatids making up
each chromosome are distributed to two new daughter
cells. We end up with four cells, each with the haploid chro-
mosomal complement. The spindle is the apparatus that
separates chromosomes in both mitosis and meiosis.

STUDY OBJECTIVE 3: To analyze the relationships
between meiosis and Mendel's rules 61-65

The behavior of chromosomes during meiosis explains
Mendel's two principles, segregation and independent as-
sortment.

At the end of this chapter, we define the chromosomal
theory of inheritance, the concept that shapes the first sec-
tion of this book. This theory states that genes are located
on chromosomes; their positions and order on the chromo-
somes can be discovered by mapping techniques described
in later chapters.

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Exercises and Problems

SOLVED PROBLEMS

PROBLEM 1: What are the differences between chromo-
somes and chromatids?

Answer: In higher organisms, a chromosome is a linear
DNA molecule complexed with protein and, generally,
with a centromere somewhere along its length. During the
cell cycle, in the S phase, the DNA replicates and each chro-
mosome is duplicated. The duplication is visible in the early
stages of mitosis and meiosis when chromosomes shorten.
At this point, each duplicated chromosome is made up of
two chromatids. The chromatids are called chromosomes
when their centromeres are pulled to opposite poles of the
spindle and each chromatid becomes independent.

PROBLEM 2: What are the relationships between mitosis
and meiosis and Mendel's rules of segregation and inde-
pendent assortment?

Answer: The process of mitosis does not relate directly
to Mendel's rules. The behavior of chromosomes during
meiosis, however, explains both segregation and inde-
pendent assortment. Segregation is explained by the fact

that only one chromosome from each homologous pair
goes into a gamete; this is also true for the maternal and
paternal alleles of a given gene. Independent assortment
is explained by the independent behavior of each tetrad
at meiosis. That is, the separation of maternal and pater-
nal alleles in one tetrad is independent of the separation
of maternal and paternal alleles in any other tetrad.

PROBLEM 3: A hypothetical organism has six chromo-
somes (2n = 6). How many different combinations of
maternal and paternal chromosomes can appear in the
gametes?

Answer: You could figure this empirically by listing all
combinations. For example, let A, B, and C = maternal
chromosomes and A', B', and C = paternal chromo-
somes. Two combinations in the gametes could be A B C
and A' B' C; obviously, several other combinations exist.
It is easier to recall that 2 n = number of combinations,
where n = the number of chromosome pairs. In this
case, n = 3, so we expect 2 3 = 8 different combinations.

EXERCISES AND PROBLEMS

CHROMOSOMES

1. What are the major differences between prokaryotes
and eukaryotes?

2. What is the difference between a centromere and a
kinetochore?

3. What is the difference between sister and nonsister
chromatids? Between homologous and nonhomolo-
gous chromosomes?

4. In human beings, 2n = 46. How many chromosomes
would you find in a

a. brain cell? d. sperm cell?

b. red blood cell? e. secondary oocyte?

c. polar body?

(See also MEIOSIS IN ANIMALS)

MITOSIS

5. You are working with a species with 2n = 6, in
which one pair of chromosomes is telocentric, one
pair sub telocentric, and one pair metacentric. The A,
B, and C loci, each segregating a dominant and re-

cessive allele (A and a, B and b, C and c), are each lo-
cated on different chromosome pairs. Draw the
stages of mitosis.

6. Identify stages a- fin the nuclear division shown in
figure 1 (on the next page). Include the process,
stage, and diploid number (e.g., meiosis I, prophase,
2n = 10). Keep in mind that one picture could rep-
resent more than one process and stage. Chromo-
somes are drawn as threads, with circles represent-
ing kinetochores. (See also MEIOSIS)

7. When during the cell cycle does chromosome repli-
cation take place?

8. A mature human sperm cell has c amount of DNA.
How much DNA (c, 2c, Ac, etc.) will a somatic cell
have if it is in

a. G : ?

b. G 2 ?

How much DNA will be in a cell at the end of
meiosis I?

* Answers to selected exercises and problems are on page A-3.

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Chapter Three Mitosis and Meiosis

MEIOSIS

9. Given the same information as in problem 5, diagram
one of the possible meioses. How many different ga-
metes can arise, absent crossing over? What varia-
tion in gamete genotype is introduced by a
crossover between the A locus and its centromere?

10. How many bivalents, tetrads, and dyads would you
find during meiosis in human beings? in fruit flies? in
the other species of table 33?

11. Can you devise a method of chromosome partitioning
during gamete formation that would not involve
synapsis — that is, can you reengineer meiosis without
passing through a synapsis stage?

12. What are the differences between a reductional and
an equational division? What do these terms refer to?

13. How does the process of meiosis explain Mendel's
two rules of inheritance?

14. Drosophila has four pairs of chromosomes. Let chro-
mosomes from the male parent be A, B, C, and D, and

those from the female parent be A', B', C, and D'.
What fraction of the gametes from an AA' BB' CC
DD' individual will be

a. all of paternal origin?

b. all of maternal origin?

c. half of maternal origin and half of paternal origin?

15. Wheat has 2n = 42 and rye has 2n = 14 chromo-
somes. Explain why a wheat-rye hybrid is usually
sterile.

16. The arctic fox has fifty small chromosomes, and the
red fox has thirty-eight larger chromosomes. Hybrids
of these two species are sterile, but cytological stud-
ies during meiosis in these hybrids reveal both
paired and unpaired chromosomes.

a. Account for the sterility of the hybrids.

b. How can you explain the paired chromosomes?

17. An organism has six pairs of chromosomes. In the
absence of crossing over, how many different chro-
mosomal combinations are possible in the gametes?

(a)

irr.

•%h 1 1

(b)

(c)

s I/,

i \

(f)

\ i

i \

Figure 1 Stages in nuclear division.

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Chromosomal Theory

3. Mitosis and Meiosis

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Critical Thinking Questions

MEIOSIS IN ANIMALS

18. How many sperm come from ten primary spermato-
cytes? How many ova from ten primary oocytes?

19. How do the quantity of genetic material and the
ploidy change from stage to stage of spermatogene-
sis and oogenesis (see figs. 3.28 and 3.29)? (Consider
the spermatogonium and the oogonium to be
diploid, with the chromosome number arbitrarily set
at two.)

20. How many sperm cells will form from

a. fifty primary spermatocytes?

b. fifty secondary spermatocytes?

c. fifty spermatids?

21. In human beings, how many eggs will form from

a. fifty primary oocytes?

b. fifty secondary oocytes?

LIFE CYCLES

22. In corn (see fig. 3.30), the diploid number is twenty.
How many chromosomes would you find in a(n)

a. sporophyte leaf cell? d. pollen grain?

b. embryo cell? e. polar nucleus?

c. endosperm cell?

23. If a dihybrid corn plant is self-fertilized, what geno-
types of the triploid endosperm can result? If you
know the endosperm genotype, can you determine
the genotype of the embryo?

24. Change the generalized life cycles of figure 3.1 so
they describe the life cycles of human beings, peas,
and Neurospora.

25. If the cytoplasm rather than nuclear genes con-
trolled inheritance, what might be the relationship

in phenotype and genotype between an organism
and its parents in

a. Drosophila?

b. corn?

c. Neurospora?

26. A drone (male) honeybee is haploid (arising from
unfertilized eggs), and a queen (female) is diploid.
Draw a testcross between a dihybrid queen and a
drone. How many different kinds of sons and daugh-
ters might result from this cross?

27. The plant Arabidopsis thaliana has five pairs
of chromosomes: AA, BB, CC, DD, and EE. If this
plant is self-fertilized, what chromosome comple-
ment would be found in a root cell of the offspring?

a. ABCDE

b. AA BB CC DD EE

c. AAA BBB CCC DDD EEE

d. AAAA BBBB CCCC DDDD EEEE

28. In wheat, the haploid number is twenty-one. How
many chromosomes would you expect to find in

a. the tube nucleus?

b. a leaf cell?

c. the endosperm?

CHROMOSOMAL THEORY OF HEREDITY

29. A hypothetical organism has two distinct chromo-
somes (2n = 4) and fifty known genes, each with
two alleles. If an individual is heterozygous at all
known loci, how many gametes can be produced if

a. all genes behave independently?

b. all genes are completely linked?

CRITICAL THINKING QUESTIONS

1. Can meiosis occur in a haploid cell? can mitosis?

2. What is the minimum number of chromosomes that an
organism can have? the maximum number?

Suggested Readings for chapter 3 are on page B-l.

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Chromosomal Theory

4. Probability and Statistics

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PROBABILITY
AND STATISTICS

STUDY OBJECTIVES

1. To understand the rules of probability and how they apply
to genetics 71

2. To understand the use of the chi-square statistical test
in genetics 74

An agricultural worker studies variability in plants in
a greenhouse. Probability influences the differences
among organisms. (© David Joel/Tony Stone Images.)

STUDY OUTLINE

Probability 71

Types of Probabilities 71

Combining Probabilities 71
Use of Rules 72
Statistics 74

Hypothesis Testing 74

Chi-Square 76

Failing to Reject Hypotheses
Summary 78
Solved Problems 78
Exercises and Problems 79
Critical Thinking Questions 81

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Chromosomal Theory

4. Probability and Statistics

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Probability

In an experimental science, such as genetics, scien-
tists make decisions about hypotheses on the basis
of data gathered during experiments. Geneticists
must therefore have an understanding of probabil-
ity theory and statistical tests of hypotheses. Proba-
bility theory allows geneticists to construct accurate pre-
dictions of what to expect from an experiment. Statistical
testing of hypotheses, particularly with the chi-square
test, allows geneticists to have confidence in their inter-
pretations of experimental data.

PROBABILITY

Part of Gregor Mendel's success was due to his ability to
work with simple mathematics. He was capable of turn-
ing numbers into ratios and deducing the mechanisms of
inheritance from them. Taking numbers that did not ex-
actly fit a ratio and rounding them off to fit lay at the
heart of Mendel's deductive powers. The underlying
rules that make the act of "rounding to a ratio" reason-
able are the rules of probability.

In the scientific method, scientists make predic-
tions, perform experiments, and gather data that they
then compare with their original predictions (see chap-
ter 1). However, even if the bases for the predictions are
correct, the data almost never exactly fit the predicted
outcome. The problem is that we live in a world perme-
ated by random, or stochastic, events. A bright new
penny when flipped in the air twice in a row will not al-
ways give one head and one tail. In fact, that penny, if
flipped one hundred times, could conceivably give one
hundred heads. In a stochastic world, we can guess how
often a coin should land heads up, but we cannot know
for certain what the next toss will bring. We can guess
how often a pea should be yellow from a given cross,
but we cannot know with certainty what the next pod
will contain. Thus, we need probability theory to tell
us what to expect from data. This chapter closes with
some thoughts on statistics, a branch of mathematics
that helps us with criteria for supporting or rejecting
our hypotheses.

Types of Probabilities

The probability (P) that an event will occur is the num-
ber of favorable cases (a) divided by the total number of
possible cases (n)\

P = a/n

The probability can be determined either by observation
(empirical) or by the nature of the event (theoretical).
For example, we observe that about one child in ten
thousand is born with phenylketonuria. Therefore, the

probability that the next child born will have phenylke-
tonuria is 1/10,000. The odds based on the geometry of
an event are, for example, like the familiar toss of dice. A
die (singular of dice) has six faces. When that die is
tossed, there is no reason one face should land up more
often than any other. Thus, the probability of any one of
the faces being up (e.g., a four) is one-sixth:

P = a/n= 1/6

Similarly, the probability of drawing the seven of clubs
from a deck of cards is

P = 1/52

The probability of drawing a spade from a deck of
cards is

P = 13/52 = 1/4

The probability (assuming a 1 : 1 sex ratio, though the ac-
tual ratio is about 1.06 males per female born in the
United States) of having a daughter in any given preg-
nancy is

P= 1/2

And the probability that an offspring from a self-fertilized
dihybrid will show the dominant phenotype is

P = 9/16

From the probability formula, we can say that an event
with certainty has a probability of one, and an event that
is an impossibility has a probability of zero. If an event has
the probability of P, all the other alternatives combined
will have a probability of Q = 1 — P; thus P + Q = 1 .
That is, the probability of the completely dominant phe-
notype in the F 2 of a selfed dihybrid is 9/16. The proba-
bility of any other phenotype is 7/16, and when the two
are added together, they equal 16/16, or 1.

Combining Probabilities

The basic principle of probability can be stated as fol-
lows: If one event has c possible outcomes and a second
event has d possible outcomes, then there are cd possible
outcomes of the two events. From this principle, we ob-
tain three rules that concern us as geneticists.

To understand these rules of probability requires a
few definitions. Mutually exclusive events are events in
which the occurrence of one possibility excludes the
occurrence of the other possibilities. In the throwing of
a die, for example, only one face can land up. Thus, if it
comes up a four, it precludes the possibility of any of the
other faces. Similarly, a blue-eyed daughter is mutually
exclusive of a brown-eyed son or any other combination
of gender and eye color. Independent events, however,
are events whose outcomes do not influence one an-
other. For example, if two dice are thrown, the face

Tamarin: Principles of
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Chapter Four Probability and Statistics

value of one die is not able to affect the face value of the
other; they are thus independent of each other. Similarly,
the gender of one child in a family is generally inde-
pendent of the gender of the children who have come
before or might come after. Finally unordered events are
events whose probability of outcome does not depend
on the order in which the events occur; the probabilities
combine both mutual exclusivity and independence. For
example, when two dice (one red, one green) are
thrown at the same time, we generally do not specify
which die has which value; a seven can occur whether
the green die is the four or the red die is the four. Simi-
larly, the probability that a family of several children will
have two boys and one girl is the same irrespective of
their birth order. If the family has two boys and one girl,
it does not matter whether the daughter is born first,
second, or third. In general, probabilities differ depend-
ing on whether order is specified. With these definitions
in mind, let us look at three rules of probability that af-
fect genetics.

1. Sum Rule

When events are mutually exclusive, the sum rule is
used: The probability that one of several mutually exclu-
sive events will occur is the sum of the probabilities of
the individual events. This is also known as the either-or
rule. For example, what is the probability, when we
throw a die, of its showing either a four or a six? Accord-
ing to the sum rule,

P = 1/6 + 1/6 = 2/6 = 1/3

2. Product Rule

When the occurrence of one event is independent of the
occurrence of other events, the product rule is used:
The probability that two independent events will both
occur is the product of their separate probabilities. This
is known as the and rule. For example, the probability of
throwing a die two times and getting a four and then a
six, in that order, is

P = 1/6 X 1/6 = 1/36

3. Binomial Theorem

The binomial theorem is used for unordered events:
The probability that some arrangement will occur in
which the final order is not specified is defined by the bi-
nomial theorem. For example, what is the probability
when tossing two pennies simultaneously of getting a
head and a tail? We will look more closely at how to use
the rules of probability to answer this question.

USE OF RULES

There are several ways to calculate the probability just
asked for. To put the problem in the form for rule 3 is
the quickest method, but this problem can also be
solved by using a combination of rules 1 and 2. For
each penny, the probability of getting a head (H) or a
tail (T) is

for H: P = 1/2
for T: Q = 1/2

Tossing the pennies one at a time, it is possible to get a
head and a tail in two ways:

first head, then tail (HT)

or
first tail, then head (TH)

Within a sequence (HT or TH), the probabilities apply to
independent events. Thus, the probability for any one of
the two sequences involves the product rule (rule 2):

1/2 X 1/2 = 1/4 for HT or TH

The two sequences (HT or TH) are mutually exclusive.
Thus, the probability of getting either of the two se-
quences involves the sum rule (rule 1):

1/4 + 1/4 = 1/2

Thus, for unordered events, we can obtain the probabil-
ity by combining rules 1 and 2. The binomial theorem
(rule 3) provides the shorthand method.

To use rule 3, we must state the theorem as follows: If
the probability of an event (X) is^? and an alternative (F)
is q, then the probability in n trials that event X will oc-
cur 5 times and F will occur t times is

n\
s\t\

P s q*

In this equation, s + t = n, andp + q = l.The symbol !,
as in n\, is called factorial, as in "n factorial," and is the
product of all integers from n down to one. For example,
7! = 7X6X5X4X3X2X1. Zero factorial equals
one, as does anything to the power of zero (0! = n° = 1).
Now, what is the probability of tossing two pennies
at once and getting one head and one tail? In this case,
n = 2,s and t = 1, and^? and q = 1/2. Thus,

1!1!

(l/2)\l/2y = 2(l/2) 2 = 1/2

This is, of course, our original answer. Now on to a few
more genetically relevant problems. What is the proba-
bility that a family with six children will have five girls
and one boy? (We assume that the probability of either a

Tamarin: Principles of
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II. Mendelism and the
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4. Probability and Statistics

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Use of Rules

son or a daughter equals 1/2.) Since the order is not spec-
ified, we use rule 3:

5!1!

(l/2) 5 a/2y = 6(l/2) 6 = 6/64 = 3/32

What would happen if we asked for a specific family or-
der, in which four girls were born, then one boy, and then
one girl? This would entail rule 2; for a sequence of six in-
dependent events:

P = 1/2 X 1/2 X 1/2 X 1/2 X 1/2 X 1/2 = 1/64

When no order is specified, the probability is six times
larger than when the order is specified; the reason is that
there are six ways of getting five girls and one boy, and
the sequence 4-1-1 is only one of them. Rule 3 tells us
that there are six ways. These are (letting B stand for boy
and G for girl) as follows:

Birth Order

Let us look at yet another problem. If two persons,
heterozygous for albinism (a recessive condition), have
four children, what is the probability that all four chil-
dren will be normal? The answer is simply (3/4) 4 by rule
2. What is the probability that three will be normal and
one albino? If we specify which of the four children will
be albino (e.g., the fourth), then the probability is
(3/4) 3 (l/4) 1 = 27/256. If, however, we do not specify
order,

P =

(3/4) 3 (l/4) 1 = 4(3/4) 3 (l/4) 1

3!1!
= 4(27/256) = 108/256

This is precisely four times the ordered probability be-
cause the albino child could have been born first, sec-
ond, third, or last.

The formula for rule 3 is the formula for the terms of
the binomial expansion. That is, if (p + q) n is ex-
panded (multiplied out), the formula (n\/s\t\)p s q l gives
the probability for any one of these terms, given that
p + q = 1 and that s + t = n. Since there are (n + 1)
terms in the binomial, the formula gives the probability
for the term numbered (t + 1). Two bits of useful infor-
mation come from recalling that rule 3 is in reality the

binomial expansion formula. First, if you have difficulty
calculating the term, you can use Pascal's triangle to
get the coefficients:

1 1

1 2 1

13 3 1
14 6 4 1

1 5 10 10 5 1

Pascal's triangle is a triangular array made up of coeffi-
cients in the binomial expansion. It is calculated by start-
ing any row with a 1, proceeding by adding two adjacent
terms from the row above, and then ending with a 1 . For
example, the next row would be

1, (1 + 5), (5 + 10), (10 + 10), (10 + 5), (5 + 1), 1

or 1, 6, 15, 20, 15, 6, 1

These numbers give us the combinations for any p s q t
term. That is, in our previous example, n = 4; so we use
the (n + 1), or fifth, row of Pascal's triangle. (The sec-
ond number in any row of the triangle gives the power
of the expansion, or n. Here, 4 is the second number in
the row.) We were interested in the case of one albino
child in a family of four children, ovp^q 1 , where ^? is the
probability of the normal child (3/4) and q is the prob-
ability of an albino child (1/4). Hence, we are inter-
ested in the (t + 1) — that is, the (1 + 1) — or the sec-
ond term of the fifth row of Pascal's triangle, which
will tell us the number of ways of getting a four-child
family with one albino child. That number is 4. Thus, us-
ing Pascal's triangle, we see that the solution to the
problem is

4(3/4) 3 (l/4) 1 = 108/256

This is the same as the answer we obtained the conven-
tional way.

The second advantage from knowing that rule 3 is
the binomial expansion formula is that we can now gen-
eralize to more than two outcomes. The general form for
the multinomial expansion is (p + q + r + ...) n and
the general formula for the probability is

P =

s\t\u\

p qr .

where s + t + u + ... = n and p + q + r + ... = 1 For
example, our albino-carrying heterozygous parents may
want an answer to the following question: If we have five
children, what is the probability that we will have two
normal sons, two normal daughters, and one albino son?
(This family will have no albino daughters.) By rule 2, the
probability of

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4. Probability and Statistics

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Chapter Four Probability and Statistics

a normal son = (3/4)(l/2) = 3/8
a normal daughter = (3/4)(l/2) = 3/8

an albino son = (l/4)(l/2) = 1/8
an albino daughter = (l/4)(l/2) = 1/8

Thus:

P =

2!2!1!0!

(3/8) 2 (3/8) 2 (l/8) 1 (l/8)°

= 30(3/8) 4 (l/8y = 30(3)7(8) 5 = 2,430/32,768
= 0.074

STATISTICS

In one of Mendel's experiments, F : heterozygous pea
plants, all tall, were self-fertilized. In the next generation
(F 2 ), he recorded 787 tall offspring and 277 dwarf off-
spring for a ratio of 2.84: 1 . Mendel saw this as a 3: 1 ratio,
which supported his proposed rule of inheritance. In
fact, is 787:277 "roundable" to a 3:1 ratio? From a brief
discussion of probability, we expect some deviation from
an exact 3:1 ratio (798:266), but how much of a deviation
is acceptable? Would 786:278 still support Mendel's rule?
Would 785:279 support it? Would 709:355 (a 2:1 ratio) or
532:532 (a 1:1 ratio)? Where do we draw the line? It is at
this point that the discipline of statistics provides help.

We can never speak with certainty about stochastic
events. For example, take Mendel's cross. Although a 3:1
ratio is expected on the basis of Mendel's hypothesis,
chance could mean that the data yield a 1:1 ratio
(532:532), yet the mechanism could be the one that
Mendel suggested. In other words, we could flip an hon-
est coin and get ten heads in a row. Conversely, Mendel
could have gotten exactly a 3:1 ratio (798:266) in his F 2
generation, yet his hypothesis of segregation could have
been wrong. The point is that any time we deal with
probabilistic events there is some chance that the data
will lead us to support a bad hypothesis or reject a good
one. Statistics quantifies these chances. We cannot say
with certainty that a 2.84:1 ratio represents a 3:1 ratio;
we can say, however, that we have a certain degree of
confidence in the ratio. Statistics helps us ascertain these
confidence limits.

Statistics is a branch of probability theory that helps
the experimental geneticist in three ways. First, part of
statistics deals with experimental design. A bit of
thought applied before an experiment may help the in-
vestigator design the experiment in the most efficient
way. Although he did not know statistics, Mendel's ex-
perimental design was very good. The second way in
which statistics is helpful is in summarizing data. Familiar
terms such as mean and standard deviation are part of
the body of descriptive statistics that takes large masses

of data and reduces them to one or two meaningful
values. We examine further some of these terms and
concepts in the chapter on quantitative inheritance
(chapter 18).

Hypothesis Testing

The third way that statistics is valuable to geneticists is in
the testing of hypotheses: determining whether to
support or reject a hypothesis by comparing the data to
the predictions of the hypothesis. This area is the most
germane to our current discussion. For example, was the
ratio of 787:277 really indicative of a 3:1 ratio? Since we
know now that we cannot answer with an absolute yes,
how can we decide to what level the data support the
predicted 3:1 ratio?

Statisticians would have us proceed as follows. To be-
gin, we need to establish how much variation to expect.
We can determine this by calculating a sampling distri-
bution: the frequencies with which various possible
events could occur in a particular experiment. For exam-
ple, if we self-fertilized a heterozygous tall plant, we
would expect a 3:1 ratio of tall to dwarf plants among the
progeny. (The 3:1 ratio is our hypothesis based on the as-
sumption that height is genetically controlled by one lo-
cus with two alleles.) If we looked at the first four off-
spring, what is the probability we would see three tall
and one dwarf plant? We can calculate the answer using
the formula for the terms of the binomial expansion:

P =

3!1!

(3/4) 3 (l/4) 1 = 108/256 = 0.42

Similarly, we can calculate the probability of getting all
tall (81/256 = 0.32), two tall and two dwarf (54/256 =
0.21), one tall and three dwarf (12/256 = 0.05), and all
dwarf (1/256 = 0.004) in this first set of four. Table 4.1
shows this distribution, as well as the distributions for
samples of eight and forty progeny. Figure 4.1 shows
these distributions in graph form.

As sample sizes increase (from four to eight to forty
in fig. 4.1), the sampling distribution takes on the shape
of a smooth curve with a peak at the true ratio of 3:1
(75% tall progeny) — that is, there is a high probability of
getting very close to the true ratio. However, there is
some chance the ratio will be fairly far off, and a very
small part of the time our ratio will be very far off. It is
important to see that any ratio could arise in a given ex-
periment even though the true ratio is 3:1. At what
point do we decide that an experimental result is not in-
dicative of a 3:1 ratio?

Statisticians have agreed on a convention. When all
the frequencies are plotted, as in figure 4.1, we can treat
the area under the curve as one unit, and we can draw
lines to mark 95% of this area (fig. 4.2). Any ratios in-

Tamarin: Principles of
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Chromosomal Theory

4. Probability and Statistics

©TheMcGraw-Hil
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Statistics

Table 4.1 Sampling Distribution for Sample Sizes of Four, Eight, and Forty, Given a 3:1
Ratio of Tall and Dwarf Plants

= 4

w = 8

= 40

No. Tall Plants

Probability*

No.

Tall Plants

Probability*

No.

Tall Plants

Probability*

256

- 0.32

0.10

0.00001

108
256

= 0.42

7
6

0.27
0.31

39
38

0.0001
0.0009

54
256

= 0.21

0.21
0.09

0.14

256

= 0.05

2
1

0.02

0.004

0.0004

2
1

0.59 X 10" 20
0.10 X 10" 21

256

= 0.004

0.00002

0.83 X 10" 24

* Probabilities are calculated from the binomial theorem.

probability = (n\/s\t\)p s q t

where n — number of progeny observed

5 = number of progeny that are tall
t — number of progeny that are dwarf
p — probability of a progeny plant being tall (3/4)
q — probability of a progeny plant being dwarf (1/4)

1.-

0.9 -
S 0.8 -

n = 4

-r

\ i i r

25 50 75 100
Percent tall

1r-

0.9 -
S 0.8 -

0.7 -

2 0.6 -

0.5 -
0.4 -

o- 0.3 h

£ 0.2 h

A7 = 8

i i i r

25 50 75 100
Percent tall

0.9

& 0.8

1 0.7

A7 = 40

2 0.6

- 0.5

S 0- 4

o- 0.3

lL 0.2

0.1

I I I I I

25 50 75 100

Percent tall

Figure 4.1 Sampling distributions from an experiment with an expected ratio of three tall to one dwarf plant. As
the sample size, n, gets larger, the distribution curve becomes smoother. These distributions are plotted terms of
the binomial expansion (see table 4.1). Note also that as n gets larger, the peak of the curve gets lower because as
more points (possible ratios) are squeezed in along the x-axis, the probability of producing any one ratio decreases.

eluded within the 95% limits are considered supportive
of (failing to reject) the hypothesis of a 3:1 ratio. Any ra-
tio in the remaining 5% area is considered unacceptable.
(Other conventions also exist, such as rejection within
the outer 10% or 1% limits; we consider these at the end

of the chapter.) Thus, it is possible to see whether the
experimental data support our hypothesis (in this case,
the hypothesis of 3:1). One in twenty times (5%) we will
make a type I error: We will reject a true hypothesis. (A
type II error is failing to reject a false hypothesis.)

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Chapter Four Probability and Statistics

To determine whether to reject a hypothesis, we
must derive a frequency distribution for each type of ex-
periment. Mendel could have used the distribution
shown in figure 4.1 for seed coat or seed color, as long as
he was expecting a 3:1 ratio and had a similar sample
size. What about independent assortment, which pre-
dicts a 9:3:3:1 ratio? A geneticist would have to calculate
a new sampling distribution based on a 9:3:3:1 ratio and
a particular sample size. Statisticians have devised short-
cut methods by using standardized probability distribu-
tions. Many are in use, such as the ^-distribution, binomial
distribution, and chi-square distribution. Each is useful
for particular kinds of data; geneticists usually use the
chi-square distribution to test hypotheses regarding
breeding data.

Chi-Square

When sample subjects are distributed among discrete
categories such as tall and dwarf plants, geneticists fre-
quently use the chi-square distribution to evaluate
data. The formula for converting categorical experimen-
tal data to a chi-square value is

(Q - By

where x is the Greek letter chi, O is the observed number
for a category, E is the expected number for that cate-
gory, and 2 means to sum the calculations for all cate-
gories.

A chi-square (x ) value of 0.60 is calculated in table
4.2 for Mendel's data on the basis of a 3:1 ratio. If Mendel
had originally expected a 1:1 ratio, he would have calcu-
lated a chi-square of 244.45 (table 4.3). However, these x
values have little meaning in themselves: they are not
probabilities. We can convert them to probabilities by de-
termining where the chi-square value falls in relation to
the area under the chi-square distribution curve. We usu-
ally use a chi-square table that contains probabilities that
have already been calculated (table 4.4). Before we can
use this table, however, we must define the concept of
degrees of freedom.

Reexamination of the chi-square formula and tables
4.2 and 4.3 reveals that each category of data contributes
to the total chi-square value, because chi-square is a
summed value. We therefore expect the chi-square value
to increase as the total number of categories increases.
That is, the more categories involved, the larger the chi-
square value, even if the sample fits relatively well against
the hypothesized ratio. Hence, we need some way of
keeping track of categories. We can do this with degrees
of freedom, which is basically a count of independent cat-
egories. With Mendel's data, the total number of offspring
is 1,064, of which 787 had tall stems. Therefore, the

short-stem group had to consist of 277 plants (1,064 —
787) and isn't an independent category. For our purposes
here, degrees of freedom equal the number of categories
minus one. Thus, with two phenotypic categories, there
is only one degree of freedom.

Table 4.4, the table of chi-square probabilities, is read
as follows. Degrees of freedom appear in the left column.
We are interested in the first row, where there is one de-
gree of freedom. The numbers across the top of the table
are the probabilities. We are interested in the next-to-the-
last column, headed by the 0.05. We thus gain the fol-
lowing information from the table: The probability is
0.05 of getting a chi-square value of 3-841 or larger by
chance alone, given that the hypothesis is correct. This
statement formalizes the information in our discussion of
frequency distributions. Hence, we are interested in how
large a chi-square value will be found in the 5% unac-
ceptable area of the curve. For Mendel's plant experi-
ment, the critical chi-square (at^? = 0.05, one degree
of freedom) is 3. 841. This is the value to which we com-
pare the calculated x 2 values (0.60 and 244.45). Since the
chi-square value for the 3:1 ratio is 0.60 (table 4.2), which
is less than the critical value of 3. 841, we do not reject
the hypothesis of a 3: 1 ratio. But since x 2 for the 1 : 1 ratio
(table 4.3) is 244.45, which is greater than the critical
value, we reject the hypothesis of a 1:1 ratio. Notice that
once we did the chi-square test for the 3:1 ratio and failed
to reject the hypothesis, no other statistical tests were
needed: Mendel's data are consistent with a 3:1 ratio.

A word of warning when using the chi-square: If the
expected number in any category is less than five, the
conclusions are not reliable. In that case, you can repeat
the experiment to obtain a larger sample size, or you can

Figure 4.2 Sampling distribution of figure 4.1, n = 40. By
convention, 5% of the area is marked off (2.5% at each end).

Tamarin: Principles of
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Chromosomal Theory

4. Probability and Statistics

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Statistics

combine categories. Note also that chi-square tests are al-
ways done on whole numbers, not on ratios or percent-
ages.

Vailing to Reject Hypotheses

Hypothesis testing, in general, involves testing the as-
sumption that there is no difference between the ob-
served and the expected samples. Therefore, the hypoth-
esis against which the data are tested is referred to as the
null hypothesis. If the null hypothesis is not rejected,
then we say that the data are consistent with it, not that
the hypothesis has been proved. (As previously dis-
cussed, it is always possible we are not rejecting a false
hypothesis or are rejecting the true one.) If, however, the

hypothesis is rejected, as we rejected a 1:1 ratio for
Mendel's data, we fail to reject the alternative hypothesis:
that there is a difference between the observed and the
expected values. We may then retest the data against
some other hypothesis. (We don't say "accept the hy-
pothesis" but rather "fail to reject the hypothesis," be-
cause supportive numbers could arise for many reasons.
Our failure to reject is tentative acceptance of a hypothe-
sis. However, we are on stronger ground when we reject
a hypothesis.)

The use of the 0.05 probability level as a cutoff for re-
jecting a hypothesis is a convention called the level of
significance. When a hypothesis is rejected at that level,
statisticians say that the data depart significantly from
the expected ratio. Other levels of significance are also

Table 4.2 Chi-Square Analysis of One of Mendel's
Experiments, Assuming a 3:1 Ratio

Tall

Dwarf

Plants

Total

Observed numbers (O)

787

277

1,064

Expected ratio

3/4

1/4

Expected numbers (£)

798

266

1,064

O- E

-11

(O - Ef

121

(O - Ef/E

0.15

0.45

0.60 = x 2

Table 4.3 Chi-Square Analysis of One of Mendel's
Experiments, Assuming a 1:1 Ratio

Tall

Dwarf

Plants

Total

Observed numbers (O)

787

277

1,064

Expected ratio

1/2

Expected numbers (i?)

532

1,064

O-E

255

-255

(O - Ef

65,025

(O - Ef/E

122.23

244.45 = x 2

Table 4.4

Chi-Square Values

Probabilities

Degrees of

Freedom

0.99

0.95

0.80

0.50

0.20

0.05

0.01

0.000

0.004

0.064

0.455

1.642

3.841

6.635

0.020

0.103

0.446

1.386

3.219

5.991

9.210

0.115

0.352

1.005

2.366

4.642

7.815

11.345

0.297

0.711

1.649

3.357

5.989

9.488

13-277

0.554

1.145

2.343

4.351

7.289

11.070

15.086

0.872

1.635

3.070

5.348

8.558

12.592

16.812

1.239

2.167

3.822

6.346

9.803

14.067

18.475

1.646

2.733

4.594

7.344

11.030

15.507

20.090

2.088

3.325

5.380

8.343

12.242

16.919

21.666

2.558

3.940

6.179

9.342

13-442

18.307

23.209

5.229

7.261

10.307

14.339

19.311

24.996

30.578

8.260

10.851

14.578

19.337

25.038

31.410

37.566

11.524

14.611

18.940

24.337

30.675

37.652

44.314

14.953

18.493

23.364

29.336

36.250

43.773

50.892

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Chapter Four Probability and Statistics

used, such as O.Ol. If a calculated chi-square is greater
than the critical value in the table at the 0.01 level, we say
that the data depart in a highly significant manner from
the null hypothesis. Since the chi-square value at the 0.01
level is larger than the value at the 0.05 level, it is more
difficult to reject a hypothesis at this level and hence
more convincing when it is rejected. Other levels of
rejection are also set. In clinical trials of medication, for
example, experimenters attempt to make it very easy to
reject the null hypothesis: a level of significance of 0.10
or higher is set. The rationale is that it is not desirable to

discard a drug or treatment that may well be beneficial.
Since the null hypothesis states that the drug has no
effect — that is, the control and drug groups show the
same response — clinicians would rather be overly con-
servative. Not rejecting the hypothesis means concluding
that the drug has no effect. Rejecting the hypothesis
means that the drug has some effect and should be tested
further. It is much better to have to retest some drugs
that are actually worthless than to discard drugs that
have potential value.

SUMMARY

STUDY OBJECTIVE 1: To understand the rules of proba-
bility and how they apply to genetics 71-74

We have examined the rules of probability theory rele-
vant to genetic experiments. Probability theory allows us
to predict the outcomes of experiments. The probability
(P) of independent events occurring is calculated by
multiplying their separate probabilities. The probability
of mutually exclusive events occurring is calculated by
adding their individual probabilities. And the probability
of unordered events is defined by the polynomial expan-
sion (p + q + r + ...y

„«.

P = p s q f r u .

To assess whether data gathered during an experiment ac-
tually support a particular hypothesis, it is necessary to de-
termine what the probability is of getting a particular data
set when the null hypothesis is correct. One way to do this
is through the chi-square test:

= 2

(Q - Ey

This test gives us a method of quantifying the confidence
we can place in the results obtained from typical genetic
experiments. The rules of probability and statistics allow us
to devise hypotheses about inheritance and to test these
hypotheses with experimental data.

STUDY OBJECTIVE 2: To understand the use of the chi-
square statistical test in genetics 74-78

SOLVED PROBLEMS

PROBLEM 1: Mendel self-fertilized a dihybrid plant that
had round, yellow peas. In the offspring generation:
What is the probability that a pea picked at random will
be round and yellow? What is the probability that five
peas picked at random will be round and yellow? What is
the probability that of five peas picked at random, four
will be round and yellow, and one will be wrinkled and
green?

Answer: The offspring peas will be round and yellow,
round and green, wrinkled and yellow, and wrinkled and
green in a ratio of 9:3:3:1. Thus, the probabilities that a
pea picked at random will be one of these four categories
are 9/16, 3/16, 3/16, and 1/16, respectively. Thus, the
probability that a pea picked at random will be round

and yellow is 9/16, or 0.563. The probability of picking
five of these peas in a row is (9/16) 5 or 0.056. The proba-
bility that of five peas picked at random, four will be
round and yellow, and one will be wrinkled and green,
is (substituting into the binomial equation): (5!/4!l!)
(9/16) 4 (1/16) 1 = 5(9 4 )/(l6 5 ) = 5(0.006) = 0.031.

PROBLEM 2: On a chicken farm, walnut-combed fowl
that were crossed with each other produced the follow-
ing offspring: walnut-combed, 87; rose-combed, 31; pea-
combed, 30; and single-combed, 12. What hypothesis
might you have (based on chapter 2) about the control
of comb shape in fowl? Do the data support that
hypothesis?

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Exercises and Problems

Answer: The numbers 87, 31, 30, and 12 are very similar
to 90, 30, 30, and 10, which would be a perfect fit to a
9:3:3:1 ratio. We might expect that ratio, having previ-
ously learned something about how comb type is inher-
ited in fowl. Thus, we hypothesize that inheritance of
comb type is by two loci, and that dominant alleles at
both result in walnut combs, a dominant allele at one
locus and recessives at the other result in rose or pea
combs, and the recessive alleles at both loci result in a
single comb. The results of the cross of dihybrids should
produce fowl with the four comb types in a 9:3:3:1 ratio
of walnut-, rose-, pea-, and single-combed fowl, respec-
tively. Therefore, our observed numbers are 87, 31, 30,
and 10 (sum = 160). Our expected ratio is 9:3:3:1, or 90,
30, 30, and 10 fowl, which are 9/16, 3/16, 3/16, and 1/16,
respectively, of the sum of 160. We therefore set up the
following chi-square table:

Comb

Type

Walnut

Rose

Pea

Single

Total

Observed

160

Numbers (O)

Expected Ratio

9/16

3/16

1/16

Expected

160

Numbers (E)

O- E

-3

(O - Ef

(O - Ef/E

0.1

0.033

0.4

0.533 = x 2

There are three degrees of freedom since there are
four categories of combs. (4—1 = 3). The critical chi-
square value with three degrees of freedom and proba-
bility of 0.05 = 7.815 (table 4.4). Since our calculated
chi-square value (0.533) is less than this critical value, we
cannot reject our hypothesis. In other words, our data
are consistent with the hypothesis of a 9:3:3:1 pheno-
typic ratio, indicative of a two-locus genetic model with
dominance at each locus.

EXERCISES AND PROBLEMS

PROBABILITY

1. Assuming a 1 : 1 sex ratio, what is the probability that
five children produced by the same parents will
consist of

a. three daughters and two sons?

b. alternating sexes, starting with a son?

c. alternating sexes?

d. all daughters?

e. all the same sex?

f. at least four daughters?

g. a daughter as the eldest child and a son as the
youngest?

(See also USE OF RULES)

2. Phenylthiocarbamide (PTC) tasting is dominant (T)
to nontasting (f). If a taster woman with a nontaster
father produces children with a taster man, and the
man previously had a nontaster daughter, what
would be the probability that

a. their first child would be a nontaster?

b. their first child would be a nontaster girl?

c. if they had six children, they would have two
nontaster sons, two nontaster daughters, and two
taster sons?

d. their fourth child would be a taster daughter?

(See also USE OF RULES)

3. Albinism is recessive; assume for this problem that
blue eyes are also recessive (albinos have blue eyes).
What is the probability that two brown-eyed per-
sons, heterozygous for both traits, would produce
(remembering epistasis)

a. five albino children?

b. five albino sons?

c. four blue-eyed daughters and a brown-eyed son?

d. two sons genotypically like their father and two
daughters genotypically like their mother?

(See also USE OF RULES)

4. On the average, about one child in every ten thou-
sand live births in the United States has phenylke-
tonuria (PKU). What is the probability that

a. the next child born in a Boston hospital will have
PKU?

b. after that child with PKU is born, the next child
born will have PKU?

c. two children born in a row will have PKU?

5. In fruit flies, the diploid chromosome number is
eight.

a. What is the probability that a male gamete will
contain only paternal centromeres or only mater-
nal centromeres?

* Answers to selected exercises and problems are on page A-4.

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Chapter Four Probability and Statistics

b. What is the probability that a zygote will contain
only centromeres from male grandparents? (Dis-
regard the problems that the sex chromosomes
may introduce.)

6. How many seeds should Mendel have tested to de-
termine with complete certainty that a plant with a
dominant phenotype was heterozygous? With 99%
certainty? With 95% certainty? With "pretty reliable"
certainty?

7. What chance do a man and a woman have of pro-
ducing one son and one daughter?

8. PKU and albinism are two autosomal recessive dis-
orders, unlinked in human beings. If two people,
each heterozygous for both traits, produce a child,
what is the chance of their having a child with

a. PKU?

b. either PKU or albinism?

c. both traits?

9. In human beings, the absence of molars is inherited
as a dominant trait. If two heterozygotes have four
children, what is the probability that

a. all will have no molars?

b. three will have no molars and one will have
molars?

c. the first two will have molars and the second two
will have no molars?

(See also USE OF RULES)

10. Galactosemia is inherited as a recessive trait. If two
normal heterozygotes produce children, what is the
chance that

a. one of four children will be affected?

b. three children will be born in this order: normal
boy, affected girl, affected boy?

(See also USE OF RULES)

11. A normal man (A) whose grandfather had galac-
tosemia and a normal woman (B) whose mother was
galactosemic want to produce a child. What is the
probability that their first child will be galactosemic?

12. A city had nine hundred deaths during the year, and
of these, three hundred were from cancer and two
hundred from heart disease. What is the probability
that the next death will be from

a. cancer?

b. either cancer or heart disease?

13. A plant that has the genotype AA bb cc DD EE is
mated with one that is aa BB CC dd ee. ¥ 1 individu-
als are selfed. What is the chance of getting an F 2
plant whose genotype exactly matches the genotype
of one of the parents?

USE OF RULES

14. The ability to taste phenylthiocarbamide is domi-
nant in human beings. If a heterozygous taster mates
with a nontaster, what is the probability that of their
five children, only one will be a taster?

15. In mice, coat color is determined by two indepen-
dent genes, A and C, as indicated here: A-C-, agouti;
aaC-, black; A-c a c a and aac a c a , albino. If the follow-
ing two mice are crossed AaCc a X Aac a c a , what is
the probability that among the first six offspring,
two will be agouti, two will be black, and two will be
albino?

STATISTICS

16. The following data are from Mendel's original exper-
iments. Suggest a hypothesis for each set and test
this hypothesis with the chi-square test. Do you
reach different conclusions with different levels of
significance?

a. Self-fertilization of round-seeded hybrids pro-
duced 5,474 round seeds and 1,850 wrinkled
ones.

b. One particular plant from a yielded 45 round
seeds and 12 wrinkled ones.

c. Of the 565 plants raised from F 2 round-seeded
plants, 372, when self-fertilized, gave both round
and wrinkled seeds in a 3:1 proportion, whereas
193 yielded only round seeds.

d. A violet-flowered, long-stemmed plant was
crossed with a white-flowered, short-stemmed
plant, producing the following offspring:

47 violet, long-stemmed plants

40 white, long-stemmed plants
38 violet, short-stemmed plants

41 white, short-stemmed plants

17. Mendel self-fertilized pea plants with round and yel-
low peas. In the next generation he recovered the
following numbers of peas:

315 round and yellow peas

108 round and green peas

101 wrinkled and yellow peas

32 wrinkled and green peas

What is your hypothesis about the genetic control of
the phenotype? Do the data support this hypothesis?

18. Two agouti mice are crossed, and over a period of a
year they produce 48 offspring with the following
pheno types:

28 agouti mice

7 black mice

13 albino mice

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Critical Thinking Questions

What is your hypothesis about the genetic control of
coat color in these mice? Do the data support that
hypothesis?

19. Two curly-winged flies, when mated, produce sixty-
one curly and thirty-five straight-winged progeny
Use a chi-square test to determine whether these
numbers fit a 3:1 ratio.

20. A short-winged, dark-bodied fly is crossed with a
long-winged, tan-bodied fly All the F : progeny are

long- winged and tan-bodied. ¥ 1 flies are crossed
among themselves to yield 84 long-winged, tan-
bodied flies; 27 long-winged dark-bodied flies; 35
short-winged, tan-bodied flies; and 14 short-winged,
dark-bodied flies.

a. What ratio do you expect in the progeny?

b. Use the chi-square test to evaluate your hypothe-
sis. Is the observed ratio within the expected
range?

CRITICAL THINKING QUESTIONS

1. A friend shows you three closed boxes, one of which
contains a prize, and asks you to choose one. Your
friend then opens one of the two remaining boxes, a
box she knows is empty. At that point, she gives you the
opportunity to change your choice to the last remain-
ing box. Should you?

2. If all couples wanted at least one child of each sex, ap-
proximately what would the average family size be?

Suggested Readings for chapter 4 are on page B-2.

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SEX

)ETERMINATION,
SEX LINKAGE,
AND PEDIGREE

STUDY OBJECTIVES

1. To analyze the causes of sex determination in various
organisms 83

2. To understand methods of dosage compensation 90

3. To analyze the inheritance patterns of traits that loci
on the sex chromosomes control 95

4. To use pedigrees to infer inheritance patterns 97

ANALYSIS STUDY OUTLINE

Sex Determination 83

Patterns 83

Sex Chromosomes 83

Sex Determination in Flowering Plants
Dosage Compensation 90

Proof of the Lyon Hypothesis 90

Dosage Compensation for Drosophila 94
Sex Linkage 95

X Linkage in Drosophila 95

Nonreciprocity 96

Sex-Limited and Sex-Influenced Traits 96
Pedigree Analysis 97

Penetrance and Expressivity

Family Tree 98

Dominant Inheritance

Recessive Inheritance

Sex-Linked Inheritance
Summary 103
Solved Problems 103
Exercises and Problems
Critical Thinking Questions 108
Box 5.1 Why Sex and Why Y? 88
Box 5.2 Electrophoresis 92

99
99
100

104

Three generations of a family.

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Sex Determination

We ended chapter 3 with a discussion of
the chromosomal theory of heredity,
stated lucidly in 1903 by Walter Sutton,
that genes are located on chromosomes.
In 1910, T. H. Morgan, a 1933 Nobel lau-
reate, published a paper on the inheritance of white eyes
in fruit flies. The mode of inheritance for this trait, dis-
cussed later in this chapter, led inevitably to the conclu-
sion that the locus for this gene is on a chromosome that
determines the sex of the flies: when a white-eyed male
was mated with a red-eyed female, half of the F 2 sons were
white-eyed and half were red-eyed; all F 2 daughters were
red-eyed. Not only was this the first evidence that localized
a particular gene to a particular chromosome, but this
study also laid the foundation for our understanding of the
genetic control of sex determination.

SEX DETERMINATION

Patterns

At the outset, we should note that the sex of an organism
usually depends on a very complicated series of develop-
mental changes under genetic and hormonal control.
However, often one or a few genes can determine which
pathway of development an organism takes. Those
switch genes are located on the sex chromosomes, a
heteromorphic pair of chromosomes, when those chro-
mosomes exist.

However, sex chromosomes are not the only determi-
nants of an organism's sex. The ploidy of an individual, as
in many hymenoptera (bees, ants, wasps), can determine
sex; males are haploid and females are diploid. Allelic
mechanisms may determine sex by a single allele or mul-
tiple alleles not associated with heteromorphic chromo-
somes; even environmental factors may control sex. For
example, temperature determines the sex of some
geckos, and the sex of some marine worms and gas-
tropods depends on the substrate on which they land. In
this chapter, however, we concentrate on chromosomal
sex-determining mechanisms.

Sex Chromosomes

Basically, four types of chromosomal sex-determining
mechanisms exist: the XY, ZW, XO, and compound chro-
mosomal mechanisms. In the XY case, as in human beings
or fruit flies, the females have a homomorphic pair of chro-
mosomes (XX) and males are heteromorphic (XY). In the
ZW case, males are homomorphic (ZZ), and females are
heteromorphic (ZW). (XY and ZW are chromosome nota-
tions and imply nothing about the sizes or shapes of these
chromosomes.) In the XO case, the organism has only one

sex chromosome, as in some grasshoppers and beetles; fe-
males are usually XX and males XO. And in the compound
chromosome case, several X and Y chromosomes combine
to determine sex, as in bedbugs and some beetles. We
need to emphasize that the chromosomes themselves do
not determine sex, but the genes they carry do. In general,
the genotype determines the type of gonad, which then
determines the phenotype of the organism through male
or female hormonal production.

The XY System

The XY situation occurs in human beings, in which fe-
males have forty-six chromosomes arranged in twenty-
three homologous, homomorphic pairs. Males, with the
same number of chromosomes, have twenty-two homo-
morphic pairs and one heteromorphic pair, the XY pair
(fig. 5.1). During meiosis, females produce gametes that
contain only the X chromosome, whereas males produce
two kinds of gametes, X- and Ybearing (fig. 5.2). For this
reason, females are referred to as homogametic and
males as heterogametic. As you can see from figure 5.2,
in people, fertilization has an equal chance of producing
either male or female offspring. In Drosophila, the sys-
tem is the same, but the Y chromosome is almost 20%
larger than the X chromosome (fig. 5.3).

Since both human and Drosophila females normally
have two X chromosomes, and males have an X and a Y
chromosome, it seems impossible to know whether male-
ness is determined by the presence of a Y chromosome or
the absence of a second X chromosome. One way to re-
solve this problem would be to isolate individuals with
odd numbers of chromosomes. In chapter 8, we examine
the causes and outcomes of anomalous chromosome
numbers. Here, we consider two facts from that chapter.
First, in rare instances, individuals form, although they are
not necessarily viable, with extra sets of chromosomes.
These individuals are referred to as polyploids (triploids
with $n, tetraploids with 4n, etc.). Second, also infre-
quently, individuals form that have more or fewer than the
normal number of any one chromosome. These aneu-
ploids usually come about when a pair of chromosomes
fails to separate properly during meiosis, an occurrence
called nondisjunction. The existence of polyploid and
aneuploid individuals makes it possible to test whether
the Y chromosome is male determining. For example, a
person or a fruit fly that has all the proper nonsex chro-
mosomes, or autosomes (forty-four in human beings, six
in Drosophila), but only a single X without a Y would an-
swer our question. If the Y were absolutely male deter-
mining, then this XO individual should be female. How-
ever, if the sex-determining mechanism is a result of the
number of X chromosomes, this individual should be a
male. As it turns out, an XO individual is a Drosophila
male and a human female.

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

U {( U \\

II II It

\t.

1 9 if

» 4 IB

X t

! 1

* «

t«

Figure 5.1 Human male karyotype. Note the X and Y chromosomes. A female would have a second X chromosome in place of the Y.
(Reproduced courtesy of Dr. Thomas G. Brewster, Foundation for Blood Research, Scarborough, Maine.)

Genie Balance in Drosophila

When geneticist Calvin Bridges, working with Dro-
sophila, crossed a triploid (3w) female with a normal
male, he observed many combinations of autosomes and
sex chromosomes in the offspring. From his results,
Bridges suggested in 1921 that sex in Drosophila is deter-
mined by the balance between (ratio of) autosomal al-
leles that favor maleness and alleles on the X chromosomes
that favor femaleness. He calculated a ratio of X chromo-
somes to autosomal sets to see if this ratio would predict
the sex of a fly. An autosomal set (A) in Drosophila con-
sists of one chromosome from each autosomal pair, or
three chromosomes. (An autosomal set in human beings
consists of twenty-two chromosomes.) Table 5.1, which
presents his results, shows that Bridges's genie balance

Sperm

One autosomal
set plus

Ovum x Y

One autosomal
set plus

Two autosomal

sets plus

Daughter

Son

Figure 5.2 Segregation of human sex chromosomes during
meiosis, with subsequent zygote formation.

Calvin B. Bridges (1889-1938).
(From Genetics 25 (1940): frontis-
piece. Courtesy of the Genetics
Society of America.)

Figure 5.3 Chromosomes of Drosophila melanogaster.

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Sex Determination

Table 5.1

Data

Supporting Bridges's

Theory

of Sex Determination

Genie Balance in

Drosophila

Number of

X Chromosomes

Number of
Autosomal Sets (A)

Total Number
of Chromosomes

— Ratio
A

Sex

1.50

Metafemale

1.33

Female

1.00

Female

1.00

Female

1.00

Female

1.00

Female

0.67

Intersex

0.50

Male

0.33

Metamale

theory of sex determination was essentially correct.
When the X:A ratio is 1.00, as in a normal female, or
greater than 1.00, the organism is a female. When this ra-
tio is 0.50, as in a normal male, or less than 0.50, the or-
ganism is a male. At 0.67, the organism is an intersex.
Metamales (X/A = 0.33) and metafemales (X/A = 1.50)
are usually very weak and sterile. The metafemales usually
do not even emerge from their pupal cases.

A sex-switch gene has been discovered that directs fe-
male development. This gene, Sex-lethal (SxO, is located
on the X chromosome. (It was originally called female-
lethal because mutations of this gene killed female em-
bryos.) Apparently, Sxl has two states of activity. When it is
"on," it directs female development; when it is "off," male-
ness ensues. Other genes located on the X chromosome
and the autosomes regulate this sex-switch gene. Genes on
the X chromosome that act to regulate Sxl into the on state
(female development) are called numerator elements
because they act on the numerator of the X/A genie
balance equation. Genes on the autosomes that act to
regulate Sxl into the off state (male development) are
called denominator elements. Geneticists have discov-
ered four numerator elements — genes named sisterless-a,
sisterless-b, sisterless-c, and runt. Sxl "counts" the number
of X chromosomes; it turns on when two are present. It
counts by measuring the level of the numerator genes' pro-
tein product. If the level is high, Sxl turns on, and the or-
ganism develops as a female. If the level is relatively low,
Sxl does not turn on, and development proceeds as a male.

Sex Determination in Human Beings

Since the X0 genotype in human beings is a female
(having Turner syndrome), it seems reasonable to
conclude that the Y chromosome is male determining in
human beings. The fact that persons with Klinefelter syn-
drome (XXY, XXXY, XXXXY) are all male, and XXX,

XXXX, and other multiple-X karyotypes are all female,
verifies this idea. (More details on these anomalies are
presented in chapter 8.) For a long time, researchers have
sought a single gene, a testis-determining factor
(TDF), located on the Y chromosome that acts as a sex
switch to initiate male development. Human embryolo-
gists had discovered that during the first month of em-
bryonic development, the gonads that develop are nei-
ther testes nor ovaries, but instead are indeterminate. At
about six or seven weeks of development, the indetermi-
nate gonads become either ovaries or testes.

In the 1950s, Ernst Eichwald found that males had a
protein on their cell surfaces not found in females; he dis-
covered that female mice rejected skin grafts from geneti-
cally identical brothers, whereas the brothers accepted
grafts from sisters. This implies that an antigen exists on the
surface of male cells that is not found on female cells. This
protein was called the histocompatibility Y antigen (H-Y
antigen). The gene for this protein was found on the Y
chromosome, near the centromere. At first, scientists be-
lieved it to be the sex switch: if the gene were present, the
gonads would begin development as testes. Further male
development, as in male secondary sexual characteristics,
came about through the testosterone the functional testes
produced. If the gene were absent, the gonads would de-
velop into ovaries. Recently, however, by studying "sex-
reversed" individuals, biologists refuted this theory.

Sex-reversed individuals are XX males or XY females.
David Page, at the Whitehead Institute for Biomedical
Research, found twenty XX males who had a small piece
of the short arm of the Y chromosome attached to one of
their X chromosomes. He found six XY females in whom
the Y chromosome was missing the same small piece
at the end of its short arm. This region, which did not
contain the HYA gene, must carry the testis-determining
factor. The first candidate gene from this region believed
to code for the testis-determining factor was named the

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

David Page (1956- ).

(Courtesy of Dr. David Page.)

ZFY gene, for zinc finger on the Y chromosome. Zinc fin-
gers are protein configurations known to interact with
DNA (discussed in detail in chapter 16). Thus, re-
searchers believed that the ZFY gene, coding for the
testis-determining factor, worked by directly interacting
with DNA. (Later in the book we look at the way regula-
tory genes, whose proteins interact with DNA, work.)
However, men who lack the ZFY gene have been found,
suggesting that the testis-determining factor is very close
to, but not, the ZFFgene. From work in mice, it has been
suggested that the ZFY gene controls the initiation of
sperm cell development, but not maleness.

In 1991, Robin Lo veil-Badge and Peter Goodfellow
and their colleagues in England isolated a gene called
Sex-determining region Y (SRY) — Sry in mice — adja-
cent to the ZFY gene. Sry has been positively identified
as the testis-determining factor because, when injected
into normal (XX) female mice, it caused them to develop
as males (fig. 5.4). Although these XX males are sterile,
they appear as normal males in every other way. (We dis-
cuss in chapter 13 how scientists introduce new genes
into an organism.) Note also that the mouse and human
systems are very similar genetically, and the homologous
genes have been isolated from both. However, at present,

Figure 5.4 Normal male mouse (left) and female littermate
given the Sry gene [right). Both mice are indistinguishably male.
(Courtesy of Robin Lovell-Badge.)

Robin Lovell-Badge (1953- ).
(Courtesy of Robin Lovell-Badge.)

Peter Goodfellow (1951- ).
(Courtesy of Peter Goodfellow.)

the human SRY gene does not convert XX female mice
into males. Like the ZFY gene product, Sry protein (the
protein the 5i?Fgene produces) also binds to DNA.

The Sry protein appears to bind to at least two genes.
One, the p450 aromatase gene, has a protein product that
converts the male hormone testosterone to the female
hormone estradiol; the Sry protein inhibits production of
p450 aromatase. The second gene the Sry protein affects
is the gene for the Mullerian-inhibiting substance, which
induces testicular development and the digression of fe-
male reproductive ducts; the Sry protein enhances this
gene's activity. Thus, the Sry protein points an indifferent
embryo toward maleness and the maintenance of testos-
terone production. The sex switch initiates a develop-
mental sequence involving numerous genes. Eva Eicher
and Linda Washburn have developed a model in which
two pathways of coordinated gene action help determine
sex, one pathway for each sex. The first gene in the ovary-
determining pathway is termed ovary determining (Od).
The first gene in the testis-determining pathway must
function before the Od gene begins, in order to allow XY
individuals to develop as males. Once the steps of a path-
way are initiated, the other pathway is inhibited (fig. 5.5).

Other Chromosomal Systems

The XO system, sometimes referred to as an XO-XX system,
occurs in many species of insects. It functions just as the
XY chromosomal mechanism does, except that instead of
a Y chromosome, the heterogametic sex (male) has only
one X chromosome. Males produce gametes that contain
either an X chromosome or no sex chromosome, whereas
all the gametes from a female contain the X chromosome.
The result of this arrangement is that females have an even
number of chromosomes (all in homomorphic pairs) and
males have an odd number of chromosomes.

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Sex Determination

Time

TDF gene functions,
if present

Inhibition

Gonad becomes testis
Male

■>■ Od gene functions

Gonad becomes ovary
Female

Figure 5.5 A model for the initiation of gonad determination
in mammals.

The ZW system is identical to the XY system except
that males are homogametic and females are heteroga-
metic. This situation occurs in birds, some fishes, and
moths.

Compound chromosomal systems tend to be com-
plex. For example, Ascaris incurva, a nematode, has
eight X chromosomes and one Y. The species has twenty-
six autosomes. Males have thirty-five chromosomes
(26A + 8X + Y), and females have forty-two chromo-
somes (26A + 16X). During meiosis, the X chromosomes
unite end to end and so behave as one unit.

The Y Chromosome

In both human beings and fruit flies, the Y chromosome
has very few functioning genes. In human beings, two
homologous regions exist, one at either end of the X and
Y chromosomes, allowing the chromosomes to pair dur-
ing meiosis. These regions are termed pseudoautoso-
mal. Mapping the Y chromosome (see chapters 6 and
13) has shown us the existence of about thirty-five genes
(fig. 5.6). Other, nonfunctioning genes are present, too,
remnants of a time in the evolutionary past when those
genes were probably active (box 5.1). The Drosophila Y
chromosome is known to carry genes for at least six fer-
tility factors, two on the short arm (ks-1 and ks-2) and
four on the long arm (kl-1, kl-2, kl-3, and kl-5). The Y
chromosome carries two other known genes: bobbed,
which is a locus of ribosomal RNA genes (the nucleolar
organizer), and Suppressor of Stellate or Su(Ste), a gene
required for RNA splicing (see chapter 10). The fertility
factors code for proteins needed during spermatogene-
sis. For example, kl-5 codes for part of the dynein motor
needed for sperm flagellar movement.

Sex Determination in Flowering Plants

Flowering plant species (angiosperms) generally have
three kinds of flowers: hermaphroditic, male, and female.

Hermaphroditic flowers have both male and female
parts. The male parts are the anthers and filaments, mak-
ing up the stamen, and the female parts are the stigma,
style, and ovary, making up the pistil (see fig. 2.2). Ninety
percent of angiosperms have hermaphroditic flowers. Of
the 10% of the species that have unisexual flowers, some
are monoecious (Greek, one house), bearing both male
and female flowers on the same plant (e.g., walnut); and
some are dioecious (Greek, two houses), having plants
with just male or just female flowers (e.g., date palm).

Within the group of plant species with unisexual
flowers, sex-determining mechanisms vary. Some species
have a single locus determining sex, some have two or
more loci involved in sex determination, and some have
X and Y chromosomes. In most of the species with X and
Y chromosomes, the sex chromosomes are indistinguish-
able. Among these species, most have heterogametic
males, although in some species, such as the strawberry,
females are heterogametic. In the very few species that
have distinguishable X and Y chromosomes — only thir-
teen are known — two sex-determination mechanisms
are found. One is similar to the system in mammals, in
which the Y chromosome has a gene or genes present

Pseudoautosomal
region

Centromere

MIC2Y
IL3RAY

SRY

RPS4

ZFY

-AMELY

Condensed region

Pseudoautosomal
region

-HYA

AZF1

RBM1

RBM2

Figure 5.6 The human Y chromosome. In addition to the genes
shown, the Y chromosome carries other genes, homologous to
X chromosome genes, that do not function because of accu-
mulated mutations. Some of these are in multiple copies. Note
the two pseudoautosomal regions that allow synapsis between
the Y and X chromosomes. The gene symbols shown include
MIC2Y, T cell adhesion antigen; IL3RAY, interleukin-3 receptor;
RPS4, a ribosomal protein; AMELY, amelogenin; HYA, histo-
compatibility Y antigen; AZF1, azoospermia factor 1 (mutants
result in tailless sperm); and RBM1, RBM2, RNA binding pro-
teins 1 and 2. (Adapted from Online Mendelian Inheritance in Man
website, http://www3.ncbi.nlm.nih.gov/omim/. Reprinted with permission.)

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

BOX 5. 1

Evolutionary biologists have
asked, Why does sex exist? A
haploid, asexual way of life
seems like a very efficient form of ex-
istence. Haploid fungi can produce
thousands of haploid spores, each of
which can grow into a new colony.
What evolutionary benefit do organ-
isms gain by developing diploidy and
sexual processes? Although this may
not seem like a serious question, evo-
lutionary biologists look for com-
pelling answers.

In chapter 21, we discuss evolu-
tionary thinking in some detail. For
the moment, accept that evolution-
ary biologists look for an adaptive ad-
vantage in most evolutionary out-
comes. Thus they ask, What is better
about the combining of gametes to
produce a new generation of off-
spring? Why would a diploid organ-
ism take a random sample of its
genome and combine it with a ran-
dom sample of someone else's
genome to produce offspring? Why
not simply produce offspring by mi-
tosis? If offspring are produced by
mitosis, all of an individual's genes
pass into the next generation with
every offspring. Not only does just
half the genome of an individual pass
into the next generation with every
offspring produced sexually, but that
half is a random jumble of what
might be a very highly adapted
genome. In addition, males are dou-
bly expensive to produce because
males themselves do not produce off-
spring: males fertilize females who
produce offspring. Thus, on the sur-
face, evolutionary biologists need to
find very strong reasons for an organ-
ism to turn to sexual reproduction
when an individual might be at an ad-
vantage evolutionarily to reproduce
asexually

There have been numerous sug-
gestions as to the advantage of sex,
nicely summarized in a 1994 article
by James Crow, of the University of

Experimental
Methods

Why Sex and Why Y?

Wisconsin, in Developmental Genet-
ics, and more recently in a special
section of the 25 September 1998
issue of Science magazine. We aren't
really sure what the true evolutionary
reasons for sex are, but at least three
explanations seem reasonable to evo-
lutionary biologists:

• Adjusting to a changing envi-
ronment. Sexual reproduction al-
lows for much more variation in
organisms. A haploid, asexual or-
ganism collects variation over
time by mutation. A sexual organ-
ism, on the other hand, can
achieve a tremendous amount of
variation by recombination and
fertilization. Remember that a hu-
man being can produce poten-
tially 2 100000 different gametes.
In a changing environment, a sex-
ually reproduced organism is
much more likely than an asexual
organism to produce offspring
that will be adapted to the
changes.

• Combining beneficial muta-
tions. As mentioned, a haploid,
asexual organism accrues muta-
tions as they happen over time in
a given individual. A sexual or-
ganism can combine beneficial
mutations each generation by re-
combination and fertilization.
Thus, sexually reproducing or-
ganisms can adapt at a much
more rapid rate than asexual or-
ganisms.

• Removing deleterious muta-
tions. Mutation is more likely to
produce deleterious changes

than beneficial ones. An asexual
organism gathers more and more
deleterious mutations as time
goes by (a process referred to
as Mutter's ratchet, in honor of
Nobel Prize-winning geneticist
H. J. Muller and referring to a
ratchet wheel that can only go
forward). Sexually reproducing
organisms can eliminate deleteri-
ous mutations each generation
by forming recombined off-
spring that are relatively free of
mutation.

Hence, this list provides three of
the generally assumed advantages of
sexual reproduction that offset its dis-
advantages.

Another subtle question about
sexual reproduction that evolution-
ary biologists ask is, Why is there a Y
chromosome? In other words, why
do we have, in some species (e.g.,
people), a heteromorphic pair of
chromosomes involved in sex deter-
mination, with one of the chromo-
somes having the gene for that sex
and very few other loci? In people,
the Y chromosome is basically a de-
generate chromosome with very few
loci. This morphological difference
between the members of the sex
chromosome pair is puzzling. After
all, chromosome pairs that do not
carry sex-determining loci do not
tend to be morphologically heteroge-
neous. Consider the following possi-
ble scenario that Virginia Morell pre-
sented in the 14 January 1994 issue
of Science.

In a particular species in the
past — evolutionarily speaking — a
sex-determining gene arises on a par-
ticular chromosome. One allele at
this locus confers maleness on its
bearer. The absence of this allele
causes the carrier to be female. At
this point, millions of years ago, the
sex chromosomes are not morpho-
logically heterogeneous: the X and Y
chromosomes are identical. In time,

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Sex Determination

however, the Y chromosome comes
to carry a gene that is beneficial to
the male but not the female. For ex-
ample, there might be a gene with an
allele for a colorful marking; this al-
lele confers a reproductive advantage
for the male but also confers a preda-
tory risk on the bearer, whether male
or female. Males have a reproductive
advantage to outweigh the predation
risk, whereas females have none.
Thus, the allele is favored in males
and selected against in females.

An evolutionary solution to this
situation is to isolate the gene for this
marking on the Y chromosome and
keep it off the X chromosome so that
males have it but females do not. This
can take place if the two chromo-
somes do not recombine over most
of their lengths. Assume then, that
some mechanism evolves to prevent
recombination of the X and Y chro-
mosomes. Thereafter, the Y chromo-
some degenerates, losing most of its
genes but retaining the sex-determin-
ing locus and the loci conferring an
advantage on males but a disadvan-
tage on females.

What evidence do we have that
any of these links in this complex line
of logic are true? To begin with, when
we look at evolutionary lineages, we
usually see a spectrum of species

with sex chromosomes in all stages
of differentiation. Evolutionary biolo-
gists generally accept the notion that
the similar sex chromosomes are the
original condition and the morpho-
logically heterogeneous sex chromo-
somes are the more evolved condi-
tion. In addition, as reported in the
same issue of Science, William Rice
of the University of California at
Santa Cruz has shown experimen-
tally with fruit flies that if recombina-
tion is prevented between sex chro-
mosomes, the Y chromosome
degenerates; it loses the function of
many loci that are also found on the
X chromosome. Rice showed this
with an ingenious set of experiments
that successfully prevented a nascent
Y chromosome from recombining
with the X. The results confirmed the
prediction that the Y chromosome
degenerates (fig. 1).

More recently, in an October 1999
article in Science, Bruce Lahn and
David Page, at the Massachusetts In-
stitute of Technology, reported re-
search findings indicating that degen-
eration of the human Y chromosome
has taken place in four stages, start-
ing as long as 320 million years ago in
our mammalian ancestors. Using
DNA sequence data and methods dis-
cussed in chapter 21, they showed

that the 19 genes known from both
the X and Y chromosomes are
arranged as if the Y chromosome has
undergone four rearrangements, each
preventing further recombination of
the X and Y According to their calcu-
lations, this process began shortly af-
ter the mammals split from the birds,
which themselves went on to evolve
a ZW pair of sex chromosomes.

Clearly, much more work is
needed to validate all the steps in this
logical, evolutionary argument. How-
ever, at this point, enough empirical
support exists to make the idea at-
tractive to evolutionary biologists.

Although we have gotten a bit
ahead of ourselves by talking about
subtle evolutionary arguments before
reaching that material in the book, it
is a good idea to keep an evolutionary
perspective on processes and struc-
tures. Presumably, evolution has
shaped us and the biological world in
which we live. If that is so, we should
be able to figure out how evolution
was working. That thinking should
hold from the level of the molecule
(e.g., enzymes and DNA) to that of
the whole organism. Behind every
process and structure should be a
hint of the evolutionary pressures
that caused that structure or process
to evolve.

Evolution of
male-
determining

gene

Evolution of
homology and
crossover

>■

limitation

Homomorphic

chromosome

pair

X Y

(nascent)

>< Degeneration
of the Y

chromosome

Figure 1 Evolution of a hypothetical Y chromosome. Red represents homologous
regions, blue shows the male-determining gene, and white marks evolved areas of the
Y chromosome that no longer recombine with the X chromosome.

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

that actively determine male-flowering plants. The other
system is similar to that found in fruit flies, in which the
X:A ratio determines sex.

In the mammalian-type system, the Y chromosome
carries genes needed for the development of male flower
parts while suppressing the development of female parts.
An example of this is in the white campion (Silene latifo-
lia). In the Drosopbila-type system, found in the sorrel
(Rumex acetosa), the ratios determine sex exactly as in
the flies. That is, an X:A ratio of 0.5 or lower results in a
male; a ratio of 1 .0 or higher results in a female; and an in-
termediate ratio results in a plant with hermaphroditic
flowers. It seems that all flowers have the potential to be
hermaphroditic. That is, flower primordia for hermaphro-
ditic, male, and female flowers look identical during early
development. The simplest mechanism of sex determina-
tion would involve repressing the development of the fe-
male flower parts in male flowers and repressing the male
flower parts in female flowers. Current research indicates
that this repression of one component or another is prob-
ably involved in most flower sex determination and is un-
der genetic and hormonal control. (We discuss further
the genetic control of flower development in chapter 16.)

DOSAGE COMPENSATION

In the XY chromosomal system of sex determination,
males have only one X chromosome, whereas females
have two. Thus, disregarding pseudoautosomal regions,
males have half the number of X-linked alleles as females
for genes that are not primarily related to gender. A ques-
tion arises: How does the organism compensate for this
dosage difference between the sexes, given the potential
for serious abnormality? In general, an incorrect number
of autosomes is usually highly deleterious to an organism
(see chapter 8). In human beings and other mammals,
the necessary dosage compensation is accomplished
by the inactivation of one of the X chromosomes in fe-
males so that both males and females have only one func-
tional X chromosome per cell.

In 1949, M. Barr and E. Bertram first observed a con-
densed body in the nucleus that was not the nucleolus.
Noting that normal female cats show a single condensed
body, while males show none, these researchers referred
to the body as sex chromatin, since known as a Barr
body (fig 5.7). Mary Lyon then suggested that this Barr
body represented an inactive X chromosome, which in
females becomes tightly coiled into heterochromatin, a
condensed, and therefore visible, form of chromatin.

Various lines of evidence support the Lyon hypoth-
esis that only one X chromosome is active in any cell.
First, XXY males have a Barr body, whereas XO females
have none. Second, persons with abnormal numbers of X

Mary F. Lyon (1925- ).
(Courtesy of Dr. Mary F. Lyon.)

chromosomes have one fewer Barr body than they have
X chromosomes per cell: XXX females have two Barr
bodies and XXXX females have three.

Proof of the Lyon Hypothesis

Direct proof of the Lyon hypothesis came when cytolo-
gists identified the Barr body in normal females as an X
chromosome. Genetic evidence also supports the Lyon
hypothesis: Females heterozygous for a locus on the X
chromosome show a unique pattern of phenotypic ex-
pression. We now know that in human females, an X
chromosome is inactivated in each cell on about the
twelfth day of embryonic life; we also know that the in-
activated X is randomly determined in a given cell. From
that point on, the same X remains a Barr body for future
cell generations. Thus, heterozygous females show mo-
saicism at the cellular level for X-linked traits. Instead of
being typically heterozygous, they express only one or
the other of the X chromosomal alleles in each cell.

Glucose-6-phosphate dehydrogenase (G-6-PD) is an
enzyme that a locus on the X chromosome controls. The

Figure 5.7 Barr body {arrow) in the nucleus of a cheek
mucosal cell of a normal woman. This visible mass of hetero-
chromatin is an inactivated X chromosome. (Thomas G. Brewster
and Park S. Gerald, "Chromosome disorders associated with mental retarda-
tion," Pediatric Annals, 7, no. 2, 1978. Reproduced courtesy of Dr. Thomas G.
Brewster, Foundation for Blood Research, Scarborough, Maine.)

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Dosage Compensation

enzyme occurs in several different allelic forms that differ
by single amino acids. Thus, both forms (A and B) will de-
hydrogenate glucose-6-phosphate — both are fully func-
tional enzymes — but because they differ by an amino
acid, they can be distinguished by their rate of migration
in an electrical field (one form moves faster than an-
other). This electrical separation, termed electrophore-
sis, is carried out by placing samples of the enzymes in a
supporting gel, usually starch, polyacrylamide, agarose, or
cellulose acetate (fig. 5.8 and box 5.2). After the electric
current is applied for several hours, the enzymes move in
the gel as bands, revealing the distance each enzyme trav-
eled. Since blood serum is a conglomerate of proteins
from many cells, the serum of a female heterozygote (fig.
5.8, lane 3) has both A and B forms (bands), whereas any
single cell (lanes 4-10) has only one or the other. Since
the gene for glucose-6-phosphate dehydrogenase is car-
ried on the X chromosome, this electrophoretic display
indicates that only one X is active in any particular cell.

Another aspect of the glucose-6-phosphate dehydrog-
enase system provides further proof of the Lyon hypoth-
esis. If a cell has both alleles functioning, both A and B
proteins should be present. Since the functioning
glucose-6-phosphate dehydrogenase enzyme is a dimer
(made up of two protein subunits), 50% of the enzymes
should be heterodimers (AB). These would form a third,
intermediate band between the A form (AA dimer) and
the B form (BB dimer; fig. 5.9). The lack of heterodimers
in the blood of heterozygotes is further proof that both
G-6-PD alleles are not active within the same cells. That
is, in any one cell, only AA or BB dimers can form, be-
cause no single cell has both the A and B forms.

The Lyon hypothesis has been demonstrated with
many X-linked loci, but the most striking examples are
those for color pheno types in some mammals. For exam-
ple, the tortoiseshell pattern of cats is due to the inacti-
vation of X chromosomes (fig. 5.10). Tortoiseshell cats
are normally females heterozygous for the yellow and
black alleles of the X-linked color locus. They exhibit
patches of these two colors, indicating that at a certain
stage in development, one or the other of the X chromo-
somes was inactivated and all of the ensuing daughter
cells in that line kept the same X chromosome inactive.
The result is patches of coat color.

The X chromosome is inactivated starting at a point
called the X inactivation center (XIC). That region
contains a gene called XIST (for X mactive-specific ton-
scripts, referring to the transcriptional activity of this
gene in the inactivated X chromosome). The XIST gene
has been putatively identified as the gene that initiates
the inactivation of the X chromosome. This gene is
known to be active only in the inactive X chromosome in
a normal XX female. Another aspect of "Lyonization" is
that several other loci are known to be active on the in-
activated X chromosome; they are active in both X chro-
mosomes, even though one is heterochromatic (inacti-
vated). Although several of these loci are in the
pseudoautosomal region of the short arm of the X chro-
mosome, several other of the thirty or more genes
known to be active are on other places on the mam-
malian X chromosome. Active genes on the inactive X in-
clude the gene for the enzyme steroid sulphatase; the
red-cell antigen Xg a ; MIC2; a ZFF-like gene termed ZFX;
the gene for Kallmann syndrome; and several others.

8 9 10

Sample
inserts

A form

B form

Figure 5.8 Electrophoretic gel stained for glucose-6-phosphate
dehydrogenase. Lanes 1-3 contain blood from an AA homozy-
gote, a BB homozygote, and an AB heterozygote, respectively.
Lanes 4-70 contain homogenates of individual cells of an AB
heterozygote.

Sample
inserts

AA homodimer

AB heterodimer

BB homodimer

Figure 5.9 Electrophoretic gel stained for glucose-6-phosphate
dehydrogenase. Lanes 1 and 2 contain blood serum from AA
and BB women, respectively, and lane 3 contains serum from
an AB heterozygote. Lane 4 shows the pattern expected if
both the A and B alleles were active within the same cell.

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

BOX 5.2

Electrophoresis, a technique for
separating relatively similar
types of molecules (for exam-
ple, proteins and nucleic acids), has
opened up new and exciting areas of
research in population, biochemical,
and molecular genetics. It has al-
lowed us to see variations in large
numbers of loci, previously difficult
or impossible to sample. In biochem-
ical genetics, electrophoretic tech-
niques can be used to study enzyme
pathways. In molecular genetics,

Experimental
Methods

Electrophoresis

electrophoresis is used to sequence
nucleotides (see chapter 13) and to
assign various loci to particular chro-

mosomes. In population genetics
(see chapter 21), electrophoresis has
made it possible to estimate the
amount of variability that occurs in
natural populations.

Here we discuss protein elec-
trophoresis, a process that entails
placing a sample — often blood
serum or a cell homogenate — at the
top of a gel prepared from a suitable
substrate (e.g., hydrolyzed starch,
polyacrylamide, or cellulose acetate)
and a buffer. An electrical current is

Figure 1 Vertical starch gel apparatus. Current
flows from the upper buffer chamber to the lower
one by way of the paper wicks and the starch
gel. Cooling water flows around the system.
(R. P. Canham, "Serum protein variations and selection in
fluctuating populations of cricetid rodents," Ph.D. thesis, Uni-
versity of Alberta, 1969. Reproduced by permission.)

2
M

3
M

4
J

5
J

6
H

7
H

Figure 2 Ten samples of deer mouse (Peromyscus man-
iculatus) blood studied for general protein. Al is albumin
and 77 is transferrin, the two most abundant proteins in
mammalian blood. The six 77 allozymes are labeled G, H,
J, L, M, and Q. (R. P. Canham, "Serum protein variations and
selection in fluctuating populations of cricetid rodents," Ph.D. thesis,
University of Alberta, 1969. Reproduced by permission.)

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Dosage Compensation

passed through the gel to cause
charged molecules to move (fig. 1),
and the gel is then treated with a dye
that stains the protein. In the sim-
plest case, if a protein is homoge-
neous (usually the product of a ho-
mozygote), it forms a single band on
the gel. If it is heterogeneous (usually
the product of a hetero zygote), it
forms two bands. This is because the
two allelic protein products differ by
an amino acid; they have different
electrical charges and therefore
travel through the gel at different
rates (see fig. 5.8). The term
allozyme refers to different elec-
trophoretic forms of an enzyme con-
trolled by alleles at the same locus.

Figure 2 shows samples of mouse
blood serum that have been stained
for protein. Most of the staining re-
veals albumins and |3-globulins
(transferrin). Because they are pres-
ent in very small concentrations,
many enzymes present in the serum
are not visible, but a stain that is spe-
cific for a particular enzyme can
make that enzyme visible on the gel.

For example, lactate dehydrogenase
(LDH) can be located because it cat-
alyzes this reaction:

LDH

lactic acid + NAD + <-» pyruvic acid + NADH

Thus, we can stain specifically for the
lactate dehydrogenase enzyme by
adding the substrates of the enzyme
(lactic acid and nicotinamide adenine
dinucleotide, NAD + ) and a suitable
stain specific for a product of the en-
zyme reaction (pyruvic acid or nico-
tinamide adenine dinucleotide, re-
duced form, NADH). That is, if lactic
acid and NAD + are poured on the
gel, only lactate dehydrogenase con-
verts them to pyruvic acid and
NADH. We can then test for the pres-
ence of NADH by having it reduce
the dye, nitro blue tetrazolium, to the
blue precipitate, formazan, an elec-
tron carrier. We then add all the pre-
ceding reagents and look for blue
bands on the gel (fig. 3).

In addition to its uses in popula-
tion genetics and chromosome map-
ping, electrophoresis has been ex-

tremely useful in determining the
structure of many proteins and for
studying developmental pathways. As
we can see from the lactate dehy-
drogenase gel in figure 3, five bands
can occur. In some tissues of a ho-
mozygote, these bands occur roughly
in a ratio of 1:4:6:4:1. This pattern
can come about if the enzyme is a
tetramer whose four subunits are ran-
dom mixtures of two gene products
(from the A and B loci). Thus we
would get

AAAA (1/16)
AAAB (4/16)
AABB (6/16)
ABBB (4/16)
BBBB (1/16)

(Note that the ratio 1:4:6:4:1 is the
expansion of [A + B] 4 , and the rela-
tive "intensity" of each band — the
number of protein doses — is calcu-
lated from the rule of unordered
events described in chapter 4.)

continued

Breast muscle

Heart

Thigh muscle

Liver

(-)

Origin

III

(+)

Figure 3 Lactate dehydrogenase isozyme patterns in pigeons. Note the five bands for some individual samples. Lanes /, //,
and /// under each tissue type indicate the range Of individual variation. (W. H. Zinkham, et al., "A Variant of Lactate Dehydrogenase in
Somatic Tissues of Pigeons" in Journal of Experimental Zoology 162, no. 1 (June):45-46, 1966. Reproduced by permission of the Wistar Institute.)

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

Protein chemists have verified this
tetramer model. In this way, elec-
trophoresis has helped us determine
the structure of several enzymes.
(The term isozymes refers to multi-
ple electrophoretic forms of an en-
zyme due to subunit interaction
rather than allelic differences.)

BOX 5.2 (CONTINUED)

Chemists have also discovered that
the five forms differ in concentration
in different tissues of the body (fig.
4). This has led to various hypotheses
as to how the production of enzymes
is controlled developmentally

Electrophoresis is also valuable in
clinical diagnosis. In various diseases,

cell destruction causes the release of
proteins into the bloodstream. Thus,
the lactate dehydrogenase pattern is
found in the blood in certain disease
states (fig. 5). This is why examination
of the blood LDH is often a diagnostic
test used to pick up early signs of heart
and liver disease (among others).

Normal
serum

Heart
muscle

Liver

Skeletal
muscle

LDH.

LDH,

Figure 4 LDH patterns found in different tissues in
human beings.

Normal Myocardial Infectious Acute
serum infarction hepatitis leukemia

LDH.

LDH,

Figure 5 LDH patterns from normal human serum and
from serum affected by various disease states.

The gene product of XIST is an RNA that does not
seem to be translated into a protein. Rather, using local-
ization techniques, geneticists have found this RNA is as-
sociated with Barr bodies, coating the inactive chromo-
some. Current research is aimed at determining the
details of this interaction.

Dosage Compensation for Drosophila

Dosage compensation also occurs in fruit flies, and it ap-
pears that the gene activity of X chromosome loci is also
about equal in males and females. The mechanism is dif-
ferent from that in mammals since no Barr bodies are
found in fruit flies. Instead, the male's single X chromo-
some is hyperactive, approaching the level of transcrip-
tional activity of both of the female's X chromosomes com-
bined. Researchers have discovered a multisubunit protein
complex called MSL (for male-specific lethal) that binds to

hundreds of sites on the single X chromosome in males.
Presumably, the binding mediates the hyperactivity of the
genes on the X chromosome. (We discuss control of tran-
scription later in the book.) At least five genes contribute
products to this protein complex: msll, msl2, msl3, mle,
and mof (Mle comes from male/ess, and mo/ comes from
males absent on theyirst.) Along with this protein com-
plex are RNAs that also bind to the male X chromosome.
These RNAs, also implicated in dosage compensation, are
the products of the roxl and rox2 genes (for i?NA on the
X). Together, the MSL protein complex and the RNAs com-
prise a compensasome.

Mutant alleles of the male-specific lethal (ms/) genes
disrupt dosage compensation in males and are, as their
names imply, lethal. However, they appear to have no ef-
fect in females. Expression of at least one of these genes,
msl2, is repressed by the protein product of the Sxl gene.
Thus, sex determination and dosage compensation are

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Sex Linkage

Figure 5.10 Tortoiseshell cat. A female heterozygous for the
X-linked yellow and black alleles. (Courtesy of Donna Bass.)

ultimately under the control of the same master switch
gene, Sxl This should not be surprising since the ability
of Sxl to count the number of X chromosomes in a cell
makes it the most efficient initiator of both sexual devel-
opment and dosage compensation.

SEX LINKAGE

In an XY chromosomal system of sex determination, the
pattern of inheritance for loci on the heteromorphic sex
chromosomes differs from the pattern for loci on the ho-
momorphic autosomal chromosomes because alleles of
the sex chromosome are inherited in association with the
sex of the offspring. Alleles on a male's X chromosome go
to his daughters but not to his sons, because the presence
of his X chromosome normally determines that his off-
spring is a daughter For example, the inheritance pattern
of hemophilia (failure of blood to clot), the common form
of which is caused by an allele located on the X chromo-
some, has been known since the end of the eighteenth cen-
tury It was known that mostly men had the disease,
whereas women could pass on the disease without actually
having it. (In fact, the general nature of the inheritance of
this trait was known in biblical times. The Talmud — the

Jewish book of laws and traditions — specified exemptions
to circumcision on the basis of hemophilia among relatives
consistent with an understanding of who was at risk.)

Before we continue, we need to make a small distinc-
tion. Since both X and Y are sex chromosomes, three dif-
ferent patterns of inheritance are possible, all sex linked
(for loci found only on the X chromosome, only on the Y
chromosome, or on both). However, the term sex-
linked usually refers to loci found only on the X chro-
mosome; the term Y-linked is used to refer to loci found
only on the Y chromosome, which control holandric
traits (traits found only in males). Loci found on both the
X and Y chromosomes are called pseudoautosomal. In
human beings, at least four hundred loci are known to be
on the X chromosome; only a few are known to be on the

Y chromosome.

X Linkage in Drosophila

T. H. Morgan demonstrated the X-linked pattern of inheri-
tance in Drosophila in 1910, when a white-eyed male ap-
peared in a culture of wild-type (red-eyed) flies (fig. 5.11).
This male was crossed with a wild-type female. All of the
offspring were wild-type. However, when these F : individ-
uals were crossed with each other, their offspring fell into
two categories (fig. 5.12). All the females and half the males
were wild-type, whereas the remaining half of the males
were white-eyed. Ultimately, Morgan and others inter-
preted this to mean that the white-eye locus was on the X
chromosome. We can redraw figure 5.12 to include the sex
chromosomes of Morgan's flies (fig. 5.13). We denote the X
chromosome with the white-eye allele as X w . Similarly X +
is the X chromosome with the wild-type allele, and Y is the

Y chromosome, which does not have this locus.

Another property of sex linkage appears in figure 5.13.
Since females have two X chromosomes, they can have
normal homozygous and heterozygous allelic combina-
tions. But males, with only one copy of the X chromo-
some, can be neither homozygous nor heterozygous.

Thomas Hunt Morgan
(1866-1945). (From
Genetics 32 (1947): frontis-
piece. Courtesy of the
Genetics Society of America.)

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

Figure 5.11 (a) Wild-
type (red -eyed) and
(b) white-eyed fruit flies.

(Carolina Biological Supply
Company.)

(a)

(b)

Instead, the term hemizygous describes the presence of
X-linked genes (and other genes present in only one copy)
in males. Since only one allele is present, a single copy of a
recessive allele determines the phenotype in a phenome-
non called pseudodominance. Thus, a male with one w
allele is white-eyed, the allele acting in a dominant fashion.
This is the same way one copy of a dominant autosomal al-
lele would determine the phenotype of a normal diploid
organism. Hence the term pseudodominance.

Nonreciprocity

The X-linked pattern has long been known as the
crisscross pattern of inheritance because the father
passes a trait to his daughters, who pass it to their sons.
Figure 5.14 shows why this analysis is correct and the in-
heritance pattern is not reciprocal through a cross be-
tween a white-eyed female and a wild-type male. Here the
F : males are white-eyed, the ¥ 1 females are wild-type, and
50% of each sex in the F 2 generation are white-eyed. Such
nonreciprocity and different ratios in the two sexes sug-
gest sex linkage, which the crisscross pattern confirms.

Figure 5.15 shows the inheritance pattern of a sex-
linked trait in chickens, in which the male is the homoga-
metic sex (22). The gene for barred plumage is 2 linked,
and barred plumage is dominant to nonbarred plumage. If
we substitute white-eyed for nonbarred and male for fe-
male, we get the same pattern as in fruit flies (fig. 5.13) —
in which, of course, females are homogametic.

The Y chromosome in fruit flies carries the pseudoau-
tosomal bobbed locus (bb), the nucleolar organizer. In
the homozygous recessive state, it causes bristles to
shorten. Figures 5.16 and 5.17 show the results of recip-
rocal crosses involving bobbed. In both cases, one quar-
ter of the F 2 individuals are bobbed. In one cross it is
males, and in the other it is females.

9
Wild-type

6
White eye

6* and 9
Wild-type

Wild-type

1/2 Wild-type
1/2 White eye

Figure 5.12 Pattern of inheritance of the white-eye trait in
Drosophila.

Sex-Limited and Sex-Influenced Traits

Aside from X-linked, holandric, and pseudoautosomal in-
heritance, two inheritance patterns show nonreciproc-
ity without necessarily being under the control of loci
on the sex chromosomes. Sex-limited traits are traits
expressed in only one sex, although the genes are pres-
ent in both. In women, breast and ovary formation are
sex-limited traits, as are facial hair distribution and
sperm production in men. Nonhuman examples are
plumage patterns in birds — in many species, the male is
brightly colored — and horns found only in males of cer-
tain sheep species. Milk yield in mammals is expressed
phenotypically only in females. Sex- influenced, or
sex-conditioned, traits appear in both sexes but occur
in one sex more than the other. Pattern, or premature, bald-
ness in human beings is an example of a sex-influenced
trait. In women, it is usually expressed as a thinning of
hair rather than as balding. Apparently testosterone, the
male hormone, is required for the full expression of the
allele.

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Pedigree Analysis

9 6

Wild-type X White eye

x+x+ x wy

9 6

White eye X Wild-type
X w x w X+Y

x+x w

Wild-type 9

X + Y

Wild-type S

F, »

x +

x+x w

Wild-type 9

X^Y

White eye 6*

X +

F 2 9

9 6

Wild-type X Wild-type

x + x +

Wild-type 9

x+x w

Wild-type 9

X + Y

Wild-type 6

White eye 6

X +

F 2 9

9 6

Wild-type X White eye

x+x w

Wild-type 9

White eye 9

X + Y

Wild-type 6

X wy

White eyed

Figure 5.13 Crosses of figure 5.12 redrawn to include the sex
chromosomes.

Figure 5.14 Reciprocal cross to that in figure 5.13.

PEDIGREE ANALYSIS

Inheritance patterns in many organisms are relatively easy
to determine, because crucial crosses can test hypotheses
about the genetic control of a particular trait. Many of these
same organisms produce an abundance of offspring so that
investigators can gather numbers large enough to compute
ratios. Recall Mendel's work with garden peas; his 3:1 ratio
in the F 2 generation led him to suggest the rule of segrega-
tion. If Mendel's sample sizes had been smaller, he might
not have seen the ratio. Think of the difficulties Mendel
would have faced had he decided to work with human be-
ings instead of pea plants. Human geneticists face the same
problems today. The occurrence of a trait in one of four
children does not necessarily indicate a true 3:1 ratio.

To determine the inheritance pattern of many human
traits, human geneticists often have little more to go on
than a pedigree that many times does not include criti-
cal mating combinations. Frequently uncertainties and
ambiguities plague pedigree analysis, a procedure
whereby conclusions are often a product of the process
of elimination. Other difficulties human geneticists en-
counter are the lack of penetrance and different de-
grees of expressivity in many traits. Both are aspects of
the expression of a phenotype.

Penetrance and Expressivity

Penetrance refers to the appearance in the phenotype of
genotypically determined traits. Unfortunately for geneti-
cists, not all genotypes "penetrate" the phenotype. For ex-
ample, a person could have the genotype that specifies
vitamin-D-resistant rickets and yet not have rickets (a bone
disease). This disease is caused by a sex-linked dominant al-
lele and is distinguished from normal vitamin D deficiency
by its failure to respond to low levels of vitamin D. It does,
however, respond to very high levels of vitamin D and is
thus treatable. In any case, in some family trees, affected
children are born to unaffected parents. This would violate
the rules of dominant inheritance because one of the par-
ents must have had the allele yet did not express it. The fact
that the parent actually had the allele is demonstrated by
the occurrence of low levels of phosphorus in the blood, a
pleiotropic effect of the same allele. The low-phosphorus
aspect of the phenotype is always fully penetrant.

Thus, certain genotypes, often those for developmen-
tal traits, are not always fully penetrant. Most genotypes,
however, are fully penetrant. For example, no known
cases exist of individuals homozygous for albinism who
do not actually lack pigment. Vitamin-D-resistant rickets
illustrates another case in which a phenotype that is not
genetically determined mimics a phenotype that is. This

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

Barred
z b z b

Nonbarred
Z b W

Wild-type

X + X +

6
Bobbed

vbbybb

X H

X bb

ybb

X+X bb

Wild-type 9

X+Y bb

Wild-type 6

x +

<bb

Z B z b

Barred $

Z B W
Barred 9

6
Barred

Z B d

Barred

> Z b

Z B Z B

Barred $

Z B z b

Barred S

Z B W
Barred 9

Z b W
Nonbarred 9

Figure 5.15 Inheritance pattern of barred plumage in chickens
in which males are homogametic (ZZ) and females are het-
erogametic (ZW).

phenocopy is the result of dietary deficiency or envi-
ronmental trauma. A dietary deficiency of vitamin D, for
example, produces rickets that is virtually indistinguish-
able from genetically caused rickets.

Many developmental traits not only sometimes fail to
penetrate, but also show a variable pattern of expression,
from very mild to very extreme, when they do. For exam-
ple, cleft palate is a trait that shows both variable pene-
trance and variable expressivity. Once the genotype pene-
trates, the severity of the impairment varies considerably,
from a very mild external cleft to a very severe clefting of
the hard and soft palates. Failure to penetrate and variable
expressivity are not unique to human traits but are charac-
teristic of developmental traits in many organisms.

Family Tree

One way to examine a pattern of inheritance is to draw a
family tree. Figure 5.18 defines the symbols used in con-

X H

x + x +

Wild-type 9

X+Y bb

Wild-type 6*

X + X^

Wild-type 9

•ybb ybb

Bobbed <S

Figure 5.16 Inheritance pattern of the bobbed locus in
Drosophila.

structing a family tree, or pedigree. The circles represent
females, and the squares represent males. Symbols that
are filled in represent individuals who have the trait under
study; these individuals are said to be affected. The open
symbols represent those who do not have the trait. Direct
horizontal lines between two individuals (one male, one
female) are called marriage lines. Children are attached to
a marriage line by a vertical line. All the brothers and sis-
ters (siblings or sibs) from the same parents are con-
nected by a horizontal line above their symbols. Siblings
are numbered below their symbols according to birth or-
der (fig. 5.19), and generations are numbered on the right
in Roman numerals. When the sex of a child is unknown,
the symbol is diamond-shaped (e.g., the children of III-l
and III-2 in fig. 5.19). A number within a symbol repre-
sents the number of siblings not separately listed. Individ-
uals IV-7 and IV-8 in figure 5.19 are fraternal (dizygotic or
nonidentical) twins: they originate from the same point.
Individuals III-3 and III-4 are identical (monozygotic)
twins: they originate from the same short vertical line.

When other symbols occur in a pedigree, they are usu-
ally defined in the legend. Individual V-5 in figure 5.19 is
called a proband or propositus (female, proposita).
The arrow pointing to individual V-5 indicates that the
pedigree was ascertained through this individual, usually
by a physician or clinical investigator.

On the basis of the information in a pedigree, ge-
neticists attempt to determine the mode of inheritance

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Pedigree Analysis

Bobbed

ybbybb

Wild-type
X + Y +

X bb x+

Wild-type 9

x bb Y+

Wild-type 6

Y +

X bb x+

Wild-type 9

X + Y +

Wild-type 6

•ybb x bb

Bobbed 9

x bb Y+

Wild-type 6

Figure 5.17 Reciprocal cross to that in figure 5.16.

of a trait. There are two types of questions the pedigree
might be used to answer. First, are there patterns
within the pedigree that are consistent with a particu-
lar mode of inheritance? Second, are there patterns
within the pedigree that are inconsistent with a partic-
ular mode of inheritance? Often, it is not possible to
determine the mode of inheritance of a particular trait
with certainty. McKusick has reported that, as of 2001,
the mode of inheritance of over nine thousand loci in
human beings was known with some confidence, in-
cluding autosomal dominant, autosomal recessive, and
sex-linked genes.

Dominant Inheritance

If we look again at the pedigree in figure 5.19, several
points emerge. First, Polydactyly (fig. 5.20) occurs in
every generation. Every affected child has an affected
parent — no generations are skipped. This suggests
dominant inheritance. Second, the trait occurs about
equally in both sexes; there are seven affected males
and six affected females in the pedigree. This indicates
autosomal rather than sex-linked inheritance. Thus, so
far, we would categorize Polydactyly as an autosomal
dominant trait. Note also that individual IV-11, a male,
passed on the trait to two of his three sons. This would
rule out sex linkage. (Remember that a male gives his X
chromosome to all of his daughters but none of his

OK)

Male
Female
Affected male

Affected female

Mating (marriage
line)

Identical twins

Parents

Siblings

Fraternal twins

Sex unknown

Four sisters

Marriage
among relatives

Figure 5.18 Symbols used in a pedigree.

sons. His sons receive his Y chromosome.) Consistency
in many such pedigrees, has confirmed that an autoso-
mal dominant gene causes Polydactyly

Polydactyly shows variable penetrance and expressiv-
ity. The most extreme manifestation of the trait is an extra
digit on each hand (fig. 5.20) and one or two extra toes
on each foot. However, some individuals have only extra
toes, some have extra fingers, and some have an asym-
metrical distribution of digits such as six toes on one foot
and seven on the other.

Recessive Inheritance

Figure 5.21 is a pedigree with a different pattern of in-
heritance. Here affected individuals are not found in each
generation. The affected daughters, identical triplets,
come from unaffected parents. They represent, in fact,
the first appearance of the trait in the pedigree. A telling
point is that the parents of the triplets are first cousins; a
mating between relatives is referred to as consan-
guineous. If the degree of relatedness is closer than law
permits, the union is called incestuous. In all states,
brother-sister and mother-son marriages are forbidden;
and in all states except Georgia, father-daughter mar-
riages are forbidden. Georgia did not intend to permit
father-daughter marriages. However, the law was drafted
using biblical terminology that inadvertently did not pro-
hibit a man from marrying his daughter or his grand-
mother. Thirty states prohibit the marriage of first
cousins.

Consanguineous matings often produce offspring that
have rare recessive, and often deleterious, traits. The rea-
son is that through common ancestry (e.g., when first
cousins have a pair of grandparents in common), a rare al-
lele can be passed on both sides of the pedigree and be-
come homozygous in a child. The occurrence of a trait in
a pedigree with common ancestry is often good evidence

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

Ofl

I
II

ChH

2 3 4

3tO

III

» o

8 9

C12

1 2-3 4

Figure 5.19 Part of a pedigree for Polydactyly.

for an autosomal recessive mode of inheritance. Con-
sanguinity by itself does not guarantee that a trait has an
autosomal recessive mode of inheritance; all modes of
inheritance appear in consanguineous pedigrees. Con-
versely, recessive inheritance is not confined to consan-
guineous pedigrees. Hundreds of recessive traits are
known from pedigrees lacking consanguinity.

Sex-Linked Inheritance

Figure 5.22 is the pedigree of Queen Victoria of England.
Through her children, hemophilia was passed on to
many of the royal houses of Europe. Several interesting
aspects of this pedigree help to confirm the method of
inheritance. First, generations are skipped. Although
Alexis (1904-18) was a hemophiliac, neither his parents
nor his grandparents were. This pattern occurs in several
other places in the pedigree and indicates a recessive
mode of inheritance. From other pedigrees and from the
biochemical nature of the defect, scientists have deter-
mined that hemophilia is a recessive trait.

Further inspection of the pedigree in figure 5.22 re-
veals that all the affected individuals are sons, strongly
suggesting sex linkage. Since males are hemizygous for
the X chromosome, more males than females should
have the phenotype of a sex-linked recessive trait be-
cause males do not have a second X chromosome that
might carry the normal allele. If this is correct, we can
make several predictions. First, since all males get their X
chromosomes from their mothers, affected males should
be the offspring of carrier (heterozygous) females. A fe-
male is automatically a carrier if her father had the dis-
ease. She has a 50% chance of being a carrier if her
brother, but not her father, has the disease. In that case,
her mother was a carrier. The pedigree in figure 5.22 is
consistent with these predictions.

Figure 5.20 Hands of a person with Polydactyly. Manifesta-
tions range in severity from one extra finger or toe to one or
more extra digits on each hand and foot. (© L.v. Bergman/

The Bergman Collection.)

f°

JOO

)^]( 1 )6 J

III

1 23456789 10

Figure 5.21 Part of a pedigree of hypotrichosis (early hair loss).

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Pedigree Analysis

101

Leopold Duke
of Albany
(1853-84)

GhO

6 6

Edward Duke
of Kent
(1767-1820)

Victoria

Princess of Saxe-Coburg

(1786-1861)

U-)

Queen Victoria
of England
(1819-1901)

Emperor Frederick III
of Germany (1831-88)

Alice

(1843-78)

6 6

King Edward VII of England
(1841-1910)+

Victoria
(1840-1901)

©■

O O
I

Alix (Alexandra)
(1872-1918)

6
t

Olga '

(1895-1918)

6 (0iH

Tsar Nicholas II of Russia
(1868-1918)

o o o

/ 1 \

Marie
(1899-1918)

Alexis
(1904-18)

Tatiana Anastasia

(1897-1918) (1901-18)

Beatrice
(1857-1944)

Irene
(1866-1953)

Victoria
(1887-1969)

King Alfonso
XIII of Spain
(1886-1941)

6 6

(^J Normal female

©Normal, but known
carrier (heterozygous) female

Normal male

Affected male

+ Descendants include present
British royal family

Figure 5.22 Hemophilia in the pedigree of Queen Victoria of England. In the photograph of the Queen and
some of her descendants, three carriers — Queen Victoria {center), Princess Irene of Prussia (right), and Princess
Alix (Alexandra) Of Hesse (left) — are indicated. (Photo © Mary Evans Picture Library/Photo Researchers, Inc.)

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

In no place in the pedigree is the trait passed from fa-
ther to son. This would defy the route of an affected X
chromosome. We can conclude from the pedigree that he-
mophilia is a sex-linked recessive trait. (Several different
inherited forms of hemophilia are known, each deficient
in one of the steps in the pathway that forms fibrinogen,
the blood clot protein. Two of these forms, "classic" he-
mophilia A and hemophilia B, also called Christmas dis-
ease, are sex linked. Other hemophilias are autosomal.)

One other interesting point about this pedigree is
that there is no evidence of the disease in Queen Victo-
ria's ancestors, yet she was obviously a heterozygote, hav-
ing one affected son and two daughters who were
known carriers. Thus, though she was born to what ap-
pears to be a homozygous normal mother and a hemizy-
gous normal father, one of Queen Victoria's X chromo-
somes had the hemophilia allele. This could have
happened if a change (mutation) had occurred in one of
the gametes that formed Queen Victoria. (We explore
the mechanisms of mutation in chapter 12.)

Figure 5.23 is another pedigree that points to domi-
nant inheritance because the trait skips no generations.
The pedigree shows the distribution of low blood-
phosphorus levels, the fully penetrant aspect of vitamin-
D-resistant rickets, among the sexes. Affected males pass
on the trait to their daughters but not their sons. This pat-
tern follows that of the X chromosome: a male passes it
on to all of his daughters but to none of his sons. Although
this pedigree accords with a sex-linked dominant mode of
inheritance, it does not rule out autosomal inheritance.
The pedigree shown is a small part of one involving hun-
dreds of people, all with phenotypes consistent with the
hypothesis of sex-linked dominant inheritance.

In figure 5.23, there is the slight possibility that the
trait is recessive. This could be true if the male in genera-
tion I and the mates of II-5 and II-7 were all heterozy-
gotes. Since this is a rare trait, the possibility that all these
conditions occurred is small. For example, if one person
in fifty (0.02) is a heterozygote, then the probability of
three heterozygotes mating within the same pedigree is
(0.02) 3 , or eight in one million. The rareness of this event
further supports the hypothesis of dominant inheritance.
The expected patterns for the various types of inheri-
tance in pedigrees can be summarized in the following
four categories:

Autosomal Recessive Inheritance

1. Trait often skips generations.

2. An almost equal number of affected males and females
occur.

3. Traits are often found in pedigrees with consan-
guineous matings.

4. If both parents are affected, all children should be af-
fected.

5. In most cases when unaffected people mate with af-
fected individuals, all children are unaffected. When at
least one child is affected (indicating that the unaf-
fected parent is heterozygous), approximately half the
children should be affected.

6. Most affected individuals have unaffected parents.

Autosomal Dominant Inheritance

1. Trait should not skip generations (unless trait lacks
full penetrance).

2. When an affected person mates with an unaffected
person, approximately 50% of their offspring should
be affected (indicating also that the affected individ-
ual is heterozygous).

3. The trait should appear in almost equal numbers
among the sexes.

Sex-Linked Recessive Inheritance

1. Most affected individuals are male.

2. Affected males result from mothers who are affected or
who are known to be carriers (heterozygotes) because
they have affected brothers, fathers, or maternal uncles.

3. Affected females come from affected fathers and af-
fected or carrier mothers.

4. The sons of affected females should be affected.

5. Approximately half the sons of carrier females should
be affected.

Sex-Linked Dominant Inheritance

1 . The trait does not skip generations.

2. Affected males must come from affected mothers.

3. Approximately half the children of an affected het-
erozygous female are affected.

4. Affected females come from affected mothers or fa-
thers.

5. All the daughters, but none of the sons, of an affected
father are affected.

Ch^

2 3

OrO 6

5 6

III

Figure 5.23 Part of a pedigree of vitamin-D-resistant rickets.
Affected individuals have low blood-phosphorus levels.
Although the sample is too small for certainty, dominance is
indicated because every generation was affected, and sex
linkage is suggested by the distribution of affected individuals.

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Chromosomal Theory

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Linkage, and Pedigree
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Solved Problems

103

SUMMARY

This chapter begins a four-chapter sequence that ana-
lyzes the relationship of genes to chromosomes. We be-
gin with the study of sex determination.

STUDY OBJECTIVE 1: To analyze the causes of sex deter-
mination in various organisms 83-90

Sex determination in animals is often based on chromosomal
differences. In human beings and fruit flies, females are ho-
mogametic (XX) and males are heterogametic (XY). In hu-
man beings, a locus on the Y chromosome, SRY, determines
maleness; in Drosophila, sex is determined by the balance
between genes on the X chromosome and genes on the au-
tosomes that regulate the state of the sex-switch gene, Sxl.

STUDY OBJECTIVE 2: To understand methods of dosage
compensation 90-95

Different organisms have different ways of solving problems
of dosage compensation for loci on the X chromosome. In hu-
man beings, one of the X chromosomes in cells in a woman is
Lyonized, or inactivated. Lyonization in women leads to cellu-

lar mosaicism for most loci on the X chromosome. In
Drosophila, the X chromosome in males is hyperactive.

STUDY OBJECTIVE 3: To analyze the inheritance patterns
of traits that loci on the sex chromosomes control
95-97

Since different chromosomes are normally associated with
each sex, inheritance of loci located on these chromosomes
shows specific, nonreciprocal patterns. The white-eye lo-
cus in Drosophila was the first case when a locus was as-
signed to the X chromosome. Over four hundred sex-linked
loci are now known in human beings.

STUDY OBJECTIVE 4: To use pedigrees to infer inheri-
tance patterns 97-102

Human genetic studies use pedigree analysis to determine in-
heritance patterns because it is impossible to carry out large-
scale, controlled human crosses. However, not all traits deter-
mined by genotype are apparent in the phenotype, and this
lack of penetrance can pose problems in genetic analysis.

SOLVED PROBLEMS

PROBLEM 1: A Female fruit fly with a yellow body is dis-
covered in a wild-type culture. The female is crossed with
a wild-type male. In the F : generation, the males are

fellow

Wild-type

X*X*

X + Y

xyx +

Wild-type 9

X^Y

Yellowed

xyx +

Wild-type 9

Yellow 9

X + Y

Wild-type S

xyy

Yellowed

Figure 1 Cross between yellow-bodied and wild-type fruit flies.

yellow-bodied and the females are wild-type. When these
flies are crossed among themselves, the F 2 produced are
both yellow-bodied and wild-type, equally split among
males and females (see fig. 1). Explain the genetic control
of this trait.

Answer: Since the results in the F : generation differ be-
tween the two sexes, we suspect that a sex-linked locus is
responsible for the control of body color. If we assume
that it is a recessive trait, then the female parent must have
been a recessive homozygote, and the male must have
been a wild-type hemizygote. If we assign the wild-type al-
lele as X + , the yellow-body allele as X^, and the Y chromo-
some as Y, then figure 1, showing the crosses into the F 2
generation, is consistent with the data. Thus, a recessive X-
linked gene controls yellow body color in fruit flies.

PROBLEM 2: The affected individuals in the pedigree in
figure 2 are chronic alcoholics (data from the National In-
stitute of Alcohol Abuse and Alcoholism). What can you
say about the inheritance of this trait?

Answer: We begin by assuming 100% penetrance. If that
is the case, then we can rule out either sex-linked or au-
tosomal recessive inheritance because both parents had
the trait, yet they produced some unaffected children.

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Chromosomal Theory

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

o o o o o

12 3 4 5

Figure 2 A pedigree for alcoholism.

Nor can the mode of inheritance be by a sex-linked dom-
inant gene because an affected male would have only af-
fected daughters, since his daughters get copies of his
single X chromosome. We are thus left with autosomal
dominance as the mode of inheritance. If that is the case,
then both parents must be heterozygotes; otherwise, all
the children would be affected. If both parents are het-
erozygotes, we expect a 3:1 ratio of affected to unaf-
fected offspring (a cross of Aa X Aa produces offspring
of A-\ aa in a 3:1 ratio); here, the ratio is 6:4. If we did a
chi-square test, the expected numbers would be 7.5:2.5
(3/4 and 1/4, respectively, of 10). Although the expected
value of 2.5 makes it inappropriate to do a chi-square test
(the expected value is too small), we can see that the ob-
served and expected numbers are very close. Thus, from
the pedigree we would conclude that an autosomal dom-
inant allele controls chronic alcoholism. (Although the
analysis is consistent, we actually cannot draw that con-
clusion about alcoholism because other pedigrees are
not consistent with 100% penetrance, a one-gene model,
or the lack of environmental influences. In fact, scientists
are currently debating whether alcoholism is inherited at
all. These types of problems related to complex human
traits are discussed in chapter 18.)

PROBLEM 3: A female fly with orange eyes is crossed
with a male fly with short wings. The ¥ 1 females have
wild-type (red) eyes and long wings; the V 1 males have or-
ange eyes and long wings. The ¥ 1 flies are crossed to yield

47 long wings, red eyes

45 long wings, orange eyes

17 short wings, red eyes

14 short wings, orange eyes

with no differences between the sexes. What is the ge-
netic basis of each trait?

Answer: In the F : flies, we see a difference in eye color
between the sexes, indicating some type of sex linkage.
Since the females are wild-type, wild-type is probably
dominant to orange. We can thus diagram the cross for
eye color as

(female) X°X°

r Y (male)

orange

red

X + X°

X°Y

red

orange

x + x° x°x°

X + Y

X°Y

red orange

red

orange

We would thus expect to see equal numbers of red-
eyed and orange-eyed males and females, which is what
we observe. Now look at long versus short wings. If we
disregard eye color, wing length seems to be under auto-
somal control with short wings being recessive. Thus, the
parents are homozygotes (ss and s + s + ), the F : offspring
are heterozygotes (s + s), and the F 2 progeny have a phe-
notypic ratio of 3:1, wild-type (long) to short wings.

EXERCISES AND PROBLEMS

SEX DETERMINATION

1. What is the difference between an X and a 2 chro-
mosome?

2. Transformer (tra) is an autosomal recessive gene
that converts chromosomal females into sterile
males. A female Drosophila heterozygous for the
transformer allele (tra) is mated with a normal male
homozygous for transformer. What is the sex ratio of
their offspring? What is the sex ratio of their off-
spring's offspring?

3. The autosomal recessive doublesex (dsx) gene con-
verts males and females into developmental inter-
sexes. Two fruit flies, both heterozygous for the
doublesex (dsx) allele, are mated. What are the sexes
of their offspring?

4. What is a sex switch? What genes serve as sex
switches in human beings and Drosophila?

* Answers to selected exercises and problems are on page A-4.

Tamarin: Principles of
Genetics, Seventh Edition

II. Mendelism and the
Chromosomal Theory

5. Sex Determination, Sex
Linkage, and Pedigree
Analysis

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Exercises and Problems

105

DOSAGE COMPENSATION

5. The diagram of an electrophoretic gel in figure 3
shows activity for a particular enzyme. Lane 1 is a
sample from a "fast" homozygote. Lane 2 is a sample
from a "slow" homozygote. In lane 3, the blood from
the first two was mixed. Lane 4 comes from one of
the children of the two homozygotes.

Figure 3 The activity of a
particular enzyme as revealed
on an electrophoretic gel.

Can you guess the structure of the enzyme? If this
were an X-linked trait, what pattern would you ex-
pect from a heterozygous female's

a. whole blood?

b. individual cells?

6. How many different zones of activity (bands) would
you see on a gel stained for lactate dehydrogenase
(LDH) activity from a person homozygous for the A
protein gene but heterozygous for the B protein
gene? Are the bands due to the activity of allozymes
or isozymes?

7. How many Barr bodies would you see in the nuclei
of persons with the following sex chromosomes?

a. XO e. XXX

b. XX f. XXXXX

c. XY g. XX/XY mosaic

d. XXY

What would the sex of each of these persons be? If
these were the sex chromosomes of individual
Drosophila that were diploid for all other chromo-
somes, what would their sexes be?

8. Calico cats have large patches of colored fur. What
does this indicate about the age of onset of Lyoniza-
tion (is it early or late)? Tortoiseshell cats have very
small color patches. Explain the difference between
the two pheno types.

SEX LINKAGE

9. In Drosophila, the lozenge phenotype, caused by a
sex-linked recessive allele Qz), is narrow eyes. Dia-
gram to the F 2 generation a cross of a lozenge male
and a homozygous normal female. Diagram the re-
ciprocal cross.

10. Sex linkage was originally detected in 1906 in moths
with a ZW sex-determining mechanism. In the cur-
rant moth, a pale color (p) is recessive to the wild-
type and located on the 2 chromosome. Diagram re-
ciprocal crosses to the F 2 generation in these moths.

11. What family history of hemophilia would indicate to
you that a newborn male baby should be exempted
from circumcision?

12. What is the difference between pseudodominance
and phenocopy?

13. In Drosophila, cut wings are controlled by a reces-
sive sex-linked allele (cf), and fuzzy body is con-
trolled by a recessive autosomal allele (fy). When a
fuzzy female is mated with a cut male, all the mem-
bers of the V 1 generation are wild-type. What are the
proportions of F 2 phenotypes, by sex?

14. Consider the following crosses in canaries:

Parents

a. pink-eyed female
X pink-eyed male

b. pink-eyed female
X black-eyed male

c. black-eyed female
X pink-eyed male

Progeny

all pink-eyed

all black-eyed

all females pink-eyed,
all males black-eyed

Explain these results by determining which allele is
dominant and how eye color is inherited.

15. Consider the following crosses involving yellow and
gray true-breeding Drosophila:

Cross

gray female X
yellow male

yellow female
X gray male

all males gray,
all females gray

all females gray,
all males yellow

97 gray females,
42 yellow males,
48 gray males

a. Is color controlled by an autosomal or an X-linked
gene?

b. Which allele, gray or yellow, is dominant?

c. Assume 100 F 2 offspring are produced in the
second cross. What kinds and what numbers of
progeny do you expect? List males and females
separately.

16. A man with brown teeth mates with a woman with
normal white teeth. They have four daughters,
all with brown teeth, and three sons, all with white
teeth. The sons all mate with women with white
teeth, and all their children have white teeth.
One of the daughters (A) mates with a man with
white teeth (B), and they have two brown-toothed
daughters, one white-toothed daughter, one brown-
toothed son, and one white-toothed son.

a. Explain these observations.

b. Based on your answer to a, what is the chance
that the next child of the A-B couple will have
brown teeth?

17. In human beings, red-green color blindness is inher-
ited as an X-linked recessive trait. A woman with nor-
mal vision whose father was color-blind marries a
man with normal vision whose father was also color-

Tamarin: Principles of
Genetics, Seventh Edition

II. Mendelism and the
Chromosomal Theory

5. Sex Determination, Sex
Linkage, and Pedigree
Analysis

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

blind. This couple has a color-blind daughter with a
normal complement of chromosomes. Is infidelity
suspected? Explain.

18. A white-eyed male fly is mated with a pink-eyed fe-
male. All the Fj offspring have wild-type red eyes. F x
individuals are mated among themselves to yield:

Females

Males

red-eyed

450

red-eyed

231

pink-eyed

155

white-eyed

301

pink-eyed

Provide a genetic explanation for the results.

19. In Drosophila, white eye is an X-linked recessive
trait, and ebony body is an autosomal recessive trait.
A homozygous white-eyed female is crossed with a
homozygous ebony male.

a. What phenotypic ratio do you expect in the ¥ 1
generation?

b. What phenotypic ratio do you expect in the F 2
generation?

c. Suppose the initial cross was reversed: ebony fe-
male X white-eyed male. What phenotypic ratio
would you expect in the F 2 generation?

20. In Drosophila, abnormal eyes can result from muta-
tions in many different genes. A true-breeding wild-
type male is mated with three different females, each
with abnormal eyes. The results of these crosses are
as follows:

Females

Males

male X

-> all normal

all normal

abnormal- 1

male X

-> 1/2 normal,

1/2 normal,

abnormal-2

1/2 abnormal

male X

-> all abnormal

all abnormal

abnormal-3

Explain the results by determining the mode of in-
heritance for each abnormal trait.

21. A black and orange female cat is crossed with a black
male, and the progeny are as follows:

females: two black, three orange and black

males: two black, two orange

Explain the results.

22. Based on the following Drosophila crosses, explain
the genetic basis for each trait and determine the
genotypes of all individuals:

white-eyed, dark-bodied female X red-eyed, tan-
bodied male

F x : females are all red-eyed, tan-bodied; males are all
white-eyed, tan-bodied

F 2 : 27 red-eyed, tan-bodied

24 white-eyed, tan-bodied

9 red-eyed, dark-bodied

7 white-eyed, dark-bodied

(No differences between males and females in the F 2
generation.)

PEDIGREE ANALYSIS

23. What is the difference between penetrance and ex-
pressivity?

24. What are the possible modes of inheritance in pedi-
grees a-c in figure 4? What modes of inheritance are
not possible for a given pedigree?

ChO

(a)

ChU

O O DK30

O D O

(b)

ChO

6 ChD

ChO

(c)

6 6

Figure 4 Three pedigrees showing different modes of inheritance.

Tamarin: Principles of
Genetics, Seventh Edition

II. Mendelism and the
Chromosomal Theory

5. Sex Determination, Sex
Linkage, and Pedigree
Analysis

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Exercises and Problems

107

25. In pedigrees a-d in figure 5, which show the inheri-
tance of rare human traits, including twin produc-
tion, determine which modes of inheritance are
most probable, possible, or impossible.

26. Hairy ears, a human trait expressed as excessive hair
on the rims of ears in men, shows reduced pene-
trance (less than 100% penetrant). Mechanisms pro-
posed include Y linkage, autosomal dominance, and

autosomal recessiveness. Construct a pedigree con-
sistent with each of these mechanisms.

27. Construct pedigrees for traits that could not be

a. autosomal recessive.

b. autosomal dominant.

c. sex-linked recessive.

d. sex-linked dominant.

(a)

JX^ J\^

71 / A

1 2

7 8

3 4 5

3 4

(b)

aK3

-o

1 2

■o

6ft

(c)

hOODW

12 3 4

J O

8 9 10 11

ChU

IK) O O

12 3 4 5 6

D O

CM3

o □ o

8 9 10 11 12

ChU

a o o

OODODOOODO

(d) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

a o o o o o

Figure 5 Pedigrees of rare human traits, including twin production (a).

Tamarin: Principles of
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Chromosomal Theory

5. Sex Determination, Sex
Linkage, and Pedigree
Analysis

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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis

28. Determine the possible modes of inheritance for
each trait in pedigrees a-c in figure 6.

cna

6 6

(a)

Dkj> 6 D

(b)

O-o (y-B

6 6 6

6 6 6 6

\-o o

Figure 6 Varying modes of inheritance.

(c)

CRITICAL THINKING QUESTIONS

1. What effects do null alleles, alleles that produce no pro-
tein product, have in electrophoretic systems? How
could you tell if a null allele were present?

2. In 1918, the Bolsheviks killed Tsar Nicholas II of Russia
and his family (fig. 5.22). However, the remains of one

daughter, Princess Anastasia, were never recovered.
At one point, a woman appeared who claimed to be
Anastasia. How could you validate her claim geneti-
cally?

Suggested Readings for chapter 5 are on page B-2.

Tamarin: Principles of
Genetics, Seventh Edition

II. Mendelism and the
Chromosomal Theory

6. Linkage and Mapping in
Eukaryotes

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LINKAGE AND

MAPPING IN

EUKARYOTES

STUDY OBJECTIVES

1. To learn about analytical techniques for locating the relative
positions of genes on chromosomes in diploid eukaryotic
organisms 110

2. To learn about analytical techniques for locating the relative
positions of genes on chromosomes in ascomycete fungi 122

3. To learn about analytical techniques for locating the relative
positions of genes on human chromosomes 132

STUDY OUTLINE

Diploid Mapping 110

Two-Point Cross 110

Three-Point Cross 114

Cytological Demonstration of Crossing Over 120
Haploid Mapping (Tetrad Analysis) 122

Phenotypes of Fungi 124

Unordered Spores (Yeast) 124

Ordered Spores (Neurospord) 125
Somatic (Mitotic) Crossing Over 132
Human Chromosomal Maps 132

X Linkage 132

Autosomal Linkage 1 34
Summary 140
Solved Problems 140
Exercises and Problems 142
Critical Thinking Questions 147
Box 6. 1 The Nobel Prize 112
Box 6.2 The First Chromosomal Map 121
Box 6.3 Lod Scores 135

Scanning electron micrograph (false color) of a fruit
fly, Drosophila melanogaster.

109

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II. Mendelism and the
Chromosomal Theory

6. Linkage and Mapping in
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110

Chapter Six Linkage and Mapping in Eukaryotes

After Sutton suggested the chromosomal the-
ory of inheritance in 1903, evidence accu-
mulated that genes were located on chromo-
somes. For example, Morgan showed by an
analysis of inheritance patterns that the
white-eye locus in Drosophila is located on the X chro-
mosome. Given that any organism has many more genes
than chromosomes, it follows that each chromosome has
many loci. Since chromosomes in eukaryotes are linear, it
also follows that genes are arranged in a linear fashion on
chromosomes, like beads on a string. Sturtevant first
demonstrated this in 1913. In this chapter, we look at an-
alytical techniques for mapping chromosomes — tech-
niques for determining the relationship between differ-
ent genes on the same chromosome. These techniques
are powerful tools that allow us to find out about the
physical relationships of genes on chromosomes without
ever having to see a gene or a chromosome. We deter-
mine that genes are on the same chromosome when the
genes fail to undergo independent assortment, and then
we use recombination frequencies to determine the dis-
tance between genes.

If loci were locked together permanently on a chro-
mosome, allelic combinations would always be the same.
However, at meiosis, crossing over allows the alleles of
associated loci to show some measure of independence.
A geneticist can use crossing over between loci to deter-
mine how close one locus actually is to another on a
chromosome and thus to map an entire chromosome and
eventually the entire genome (genetic complement) of
an organism.

Loci carried on the same chromosome are said to be
linked to each other. There are as many linkage groups
(/) as there are autosomes in the haploid set plus sex
chromosomes. Drosophila has five linkage groups (2n =
8; / = 3 autosomes + X + Y), whereas human beings
have twenty-four linkage groups (2n = 46; / = 22 auto-
somes + X + Y). Prokaryotes and viruses, which usually
have a single chromosome, are discussed in chapter 7.

Historically, classical mapping techniques, as de-
scribed in this chapter and the next, gave researchers
their only tools to determine the relationships of particu-
lar genes and their chromosomes. When geneticists
know the locations of specific genes, they can study
them in relation to each other and begin to develop a
comprehensive catalogue of the genome of an organism.
Knowing the location of a gene also helps in isolating the
gene and studying its function and structure. And map-
ping the genes of different types of organisms (diploid,
haploid, eukaryotic, prokaryotic) gives geneticists insight
into genetic processes. More recently, recombinant DNA
technology has allowed researchers to sequence whole
genomes, including the human and fruit fly genomes; this
means they now know the exact locations of all the
genes on all the chromosomes of these organisms (see

chapter 13). Geneticists are now creating massive data-
bases containing this information, much of it available for
free or by subscription on the World Wide Web. Until in-
vestigators mine all this information for all organisms of
interest, they will still use analytical techniques in the
laboratory and field to locate genes on chromosomes.

DIPLOID MAPPING

Two-Point Cross

In Drosophila, the recessive band gene (bn) causes a
dark transverse band on the thorax, and the detached
gene {def) causes the crossveins of the wings to be either
detached or absent (fig. 6.1). A banded fly was crossed
with a detached fly to produce wild-type, dihybrid off-
spring in the F : generation. F : females were then test-
crossed to banded, detached males (fig. 6.2). (There is no
crossing over in male fruit flies; in experiments designed
to detect linkage, heterozygous females — in which cross-
ing over occurs — are usually crossed with homozygous
recessive males.) If the loci were assorting indepen-
dently, we would expect a 1:1:1:1 ratio of the four possi-
ble pheno types. However, of the first one thousand off-
spring examined, experimenters recorded a ratio of

2:483:512:3.

Several points emerge from the data in figure 6.2.
First, no simple ratio is apparent. If we divide by two,
we get a ratio of 1:241:256:1.5. Although the first and
last categories seem about equal, as do the middle two,
no simple numerical relation seems to exist between
the middle and end categories. Second, the two cate-

Wild-type

Detached

Figure 6.1 Wild-type {det + ) and detached (def) crossveins in
Drosophila.

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Chromosomal Theory

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Diploid Mapping

111

Banded

bn bn det + det +

(homozygous for the bn
allele and the det + allele)

Detached

bn + bn + det det
(homozygous for the bn +
allele and the det allele)

Teste ross

Wild-type

Banded, detached

bn + bn det + det

bn bn det det

bn det

bn + det

+w^* +

bn + det

bn bn det det

bn bn det + det

bn + bn det det

bn + bn det + det

S bn det

Phenotype Banded,

detached

Number 2

Banded

483

Figure 6.2 Testcrossing a dihybrid Drosophila.

Detached

512

Wild-type

gories in very high frequency have the same pheno-
types as the original parents in the cross C 1 P 1 of fig. 6.2).
That is, banded flies and detached flies were the original
parents as well as the great majority of the testcross off-
spring. We call these phenotypic categories parentals,
or nonrecombinants. On the other hand, the testcross
offspring in low frequency combine the phenotypes of
the two original parents (P : ). These two categories are re-
ferred to as nonparentals, or recombinants. The sim-
plest explanation for these results is that the banded and
detached loci are located near each other on the same
chromosome (they are a linkage group), and therefore
they move together as associated alleles during meiosis.

We can analyze the original cross by drawing the loci
as points on a chromosome (fig. 6.3). This shows that
99.5% of the testcross offspring (the nonrecombinants)
come about through the simple linkage of the two loci.
The remaining 0.5% (the recombinants) must have arisen
through a crossover of homologues, from a chiasma at
meiosis, between the two loci (fig. 6.4). Note that since it
is not possible to tell from these crosses which chromo-
some the loci are actually on or where the centromere is
in relation to the loci, the centromeres are not included
in the figures. The crossover event is viewed as a break-
age and reunion of two chromatids lying adjacent to each
other during prophase I of meiosis. Later in this chapter,
we find cytological proof for this; in chapter 12, we ex-
plore the molecular mechanisms of this breakage and re-
union process.

From the testcross in figure 6.3, we see that 99. 5% of
the gametes produced by the dihybrid are nonrecombi-
nant, whereas only 0.5% are recombinant. This very
small frequency of recombinant offspring indicates that

the two loci lie very close to each other on their partic-
ular chromosome. In fact, we can use the recombina-
tion percentages of gametes, and therefore of testcross
offspring, as estimates of distance between loci on a
chromosome: 1% recombinant offspring is referred to
as one map unit (or one centimorgan, in honor of ge-
neticist T H. Morgan, the first geneticist to win the No-
bel Prize; box 6.1). Although a map unit is not a physi-
cal distance, it is a relative distance that makes it
possible to know the order of and relative separation
between loci on a chromosome. In this case, the two
loci are 0.5 map units apart. (From sequencing various
chromosomal segments — see chapter 13 — we have
learned that the relationship between centimorgans
and DNA base pairs is highly variable, depending on
species, sex, and region of the chromosome. For exam-
ple, in human beings, 1 centimorgan can vary between
100,000 and 10,000,000 base pairs. In the fission yeast,
Schizosaccharomyces pombe, 1 centimorgan is only
about 6,000 base pairs.)

The arrangement of the bn and det alleles in the di-
hybrid of figure 6.3 is termed the trans configuration,
meaning "across," because the two mutants are across
from each other, as are the two wild-type alleles. The al-
ternative arrangement, in which one chromosome car-
ries both mutants and the other chromosome carries
both wild-type alleles (fig. 6.5), is referred to as the cis
configuration. (Two other terms, repulsion and cou-
pling, have the same meanings as trans and cis, respec-
tively.)

A cross involving two loci is usually referred to as a
two-point cross; it gives us a powerful tool for dissect-
ing the makeup of a chromosome. The next step in our

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Chapter Six Linkage and Mapping in Eukaryotes

BOX 6.1

On 10 December each year,
the king of Sweden awards
the Nobel Prizes at the
Stockholm Concert Hall. The date is
the anniversary of Alfred Nobel's
death. Awards are given annually in
physics, chemistry, medicine and
physiology, literature, economics, and
peace. In 2000, each award was
worth $900,000, although an award
sometimes goes to two or three re-
cipients. The prestige is priceless.

Winners of the Nobel Prize are
chosen according to the will of Alfred
Nobel, a wealthy Swedish inventor
and industrialist, who held over three
hundred patents when he died in
1896 at the age of sixty-three. Nobel
developed a detonator and processes
for detonation of nitroglycerine, a
substance invented by Italian chemist
Ascanio Sobrero in 1847. In the form
Nobel developed, the explosive was
patented as dynamite. Nobel also in-
vented several other forms of explo-
sives. He was a benefactor of Sobrero,
hiring him as a consultant and paying
his wife a pension after Sobrero died.
Nobel believed that dynamite
would be so destructive that it would
serve as a deterrent to war. Later, real-
izing that this would not come to
pass, he instructed that his fortune be
invested and the interest used to fund
the awards. The first prizes were
awarded in 1901 . Each award consists
of a diploma, medal, and check.

Historical
Perspectives

The Nobel Prize

American, British, German,
French, and Swedish citizens have
earned the most prizes (table 1).
Table 2 features some highlights of
Nobel laureate achievements in ge-
netics.

The Nobel medal. The medal is half a pound of 23-karat
gold, measures about 2 1/2 inches across, and has Nobel's
face and the dates of his birth and death on the front. The
diplomas that accompany the awards are individually designed.

(Reproduced by permission of the Nobel Foundation.)

Table 1 Distribution of Nobel Awards to the Top Five Recipient Nations (Including 2000 Winners)

Medicine and

Physics

Chemistry

Physiology

Peace

Literature

Economics

Total

United States

270

Britain

Germany

France

Sweden

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Table 2 Some Nobel Laureates in

Genetics (Medicine and Physiology; Chemistry)

Name

Year

Nationality

Cited for

Thomas Hunt Morgan

1933

USA

Discovery of how chromosomes govern heredity

Hermann J. Muller

1946

USA

X-ray inducement of mutations

George W. Beadle

1958

USA

Genetic regulation of biosynthetic pathways

Edward L. Tatum

1958

USA

Joshua Lederberg

1958

USA

Bacterial genetics

Severo Ochoa

1959

USA

Discovery of enzymes that synthesize nucleic acids

Arthur Kornberg

1959

USA

Francis H. C. Crick

1962

British

Discovery of the structure of DNA

James D. Watson

1962

USA

Maurice Wilkins

1962

British

Francois Jacob

1965

French

Regulation of enzyme biosynthesis

Andre Lwoff

1965

French

Jacques Monod

1965

French

Peyton Rous

1966

USA

Tumor viruses

Robert WHolley

1968

USA

Unraveling of the genetic code

H. Gobind Khorana

1968

USA

Marshall W. Nirenberg

1968

USA

Max Delbriick

1969

USA

Viral genetics

Alfred Hershey

1969

USA

Salvador Luria

1969

USA

Renato Dulbecco

1975

USA

Tumor viruses

Howard Temin

1975

USA

Discovery of reverse transcriptase

David Baltimore

1975

USA

Werner Arber

1978

Swiss

Discovery and use of restriction endonucleases

Hamilton Smith

1978

USA

Daniel Nathans

1978

USA

Walter Gilbert

1980

USA

Techniques of sequencing DNA

Frederick Sanger

1980

British

Paul Berg

1980

USA

Pioneer work in recombinant DNA

Baruj Benacerraf

1980

USA

Genetics of immune reactions

Jean Dausset

1980

French

George Snell

1980

USA

Aaron Klug

1982

British

Crystallographic work on protein-nucleic acid
complexes

Barbara McClintock

1983

USA

Transposable genetic elements

Cesar Milstein

1984

British/Argentine

Immunogenetic s

Georges Koehler

1984

German

Niels K. Jerne

1984

British/Danish

Susumu Tonegawa

1987

Japanese

Antibody diversity

J. Michael Bishop

1989

USA

Proto-oncogenes

Harold E.Varmus

1989

USA

Thomas R. Cech

1989

USA

Enzymatic properties of RNA

Sidney Altman

1989

Canada

Kary Mullis

1993

USA

Polymerase chain reaction

Michael Smith

1993

Canada

Site-directed mutagenesis

Richard Roberts

1993

British

Discovery of intervening sequences in RNA

Phillip Sharp

1993

USA

E. B. Lewis

1995

USA

Genes control development

continued

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BOX 6. 1

CONTINUED

Table 2 continued

Name

Year

Nationality

Cited for

Christiane Niisslein-Volhard
Eric Wieschaus
Stanley B. Prusiner
Gunter Blobel

1995
1995
1997
1999

German
USA
USA
German

Discovery of prions

Signal recognition during protein synthesis

analysis is to look at three loci simultaneously so that we
can determine their relative order on the chromosome.
More important, we can also analyze the effects of multi-
ple crossovers, which cannot be detected in a two-point
cross, on map distances. Two crossovers between two
loci can cause the chromosome to look as if no
crossovers took place, causing us to underestimate map
distances. Thus we need a third locus, between the first
two, to detect multiple crossover events.

Three-Point Cross

Analysis of three loci, each segregating two alleles, is re-
ferred to as a three-point cross. We will examine wing
morphology, body color, and eye color in Drosophila.
Black body (£>), purple eyes (pr), and curved wings (c) are
all recessive genes. Since the most efficient way to study
linkage is through the testcross of a multihybrid, we will
study these three loci by means of the crosses shown in

Testcross

8
bn det

Banded

Detached

bn det +

bn + det

bn det +

— i 1 —

bn + det

— i 1 —

Wild

-type

det*

bn +

— i —

det

— i —

-I 1-

bn det +

— i 1 —

bn det

Banded, detached

bn det

bn det* bn + det

bn det

bn + det

-i 1-

i + de* +

-i 1-

bn + det

— i 1 —

bn det

— i 1 —

bn det

bn + det

i + de*-

-i 1-

bn det

— i 1 —

Phenotype

Banded Detached Banded, Wild-type

detached

Number

483

512

99.5%

0.5%

Figure 6.3 Chromosomal arrangement of the two loci in the crosses of
figure 6.2. A line arbitrarily represents the chromosomes on which these
loci are actually situated.

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bn
bn
bn +

bn-

det +
det +
det

det

det +
det +
det

det

det +
det +
det

det

Gametes

det +

det

bn +

det +

bn +

det

Nonrecombinant

- Recombinant

Nonrecombinant

Figure 6.4 Crossover of homologues during meiosis between
the bn and det loci in the tetrad of the dihybrid female.

figure 6.6. One point in this figure should be clarified.
Since the organisms are diploid, they have two alleles at
each locus. Geneticists use various means to present this
situation. For example, the recessive homozygote can be
pictured as

1. bb prpr cc

2. b/b pr/pr c/c

3. b pr c/b pr c

b pr c

trans
(repulsion)

det'

det

bn +

CIS

(coupling)

det

det +

Figure 6.5 Trans (repulsion) and cis (coupling) arrangements of
dihybrid chromosomes.

A slash (also called a rule line) is used to separate alleles
on homologous chromosomes. Thus (2) is used tenta-
tively, when we do not know the linkage arrangement of
the loci, (2) is used to indicate that the three loci are on
different chromosomes, and (3) indicates that all three
loci are on the same chromosome.

In figure 6.6, the trihybrid organism is testcrossed. If
independent assortment is at work, the eight types of re-
sulting gametes should appear with equal frequencies,
and thus the eight phenotypic classes would each make
up one-eighth of the offspring. However, if there were
complete linkage, so that the loci are so close together
on the same chromosome that virtually no crossing over
takes place, we would expect the trihybrid to produce
only two gamete types in equal frequency and to yield
two phenotypic classes identical to the original parents.
This would occur because, under complete linkage, the
trihybrid would produce only two chromosomal types in
gametes: the b pr c type from one parent and the b + pr +
c + type from the other. Crossing over between linked
loci would produce eight phenotypic classes in various
proportions depending on the distances between loci.
The actual data appear in table 6.1.

The data in the table are arranged in reciprocal classes.
Two classes are reciprocal if between them they contain
each mutant phenotype just once. Wild-type and black,
purple, curved classes are thus reciprocal, as are the pur-
ple, curved and the black classes. Reciprocal classes occur
in roughly equal numbers: 5,701 and 5,617; 388 and 367;
1,412 and 1,383; and 60 and 72. As we shall see, a single
meiotic recombinational event produces reciprocal
classes. Wild-type and black, purple, curved are the two
nonrecombinant classes. The purple, curved class of 388 is
grouped with the black class of 367. These two would be
the products of a crossover between the b and the^?r loci
if we assume that the three loci are linked and that the
gene order is b pre (fig. 6.7). The next two classes, of 1,412
and 1,383 flies, would result from a crossover between pr
and c, and the last set, 60 and 72, would result from two
crossovers, one between b and pr and the other between

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Black, purple, curved Wild-type

b b prpr cc

Testcross the trihybrid

b + b + pr + pr + c + c +

Wild-type (trihybrid) X Black, purple, curved
b + b pr + pr c + c bb prpr cc

If unlinked If completely linked

1 18 b/b prlpr clc M2bprclb pre

1 18 b/b prlpr c + /c 1/2 b + pr + c + /b pr c

1/8 M? prVprdc
1/8 M? pr + lprc + lc
M8b + lb prlpr clc
M8b + lb prlpr c + lc
M8b + /b prVprdc
M8b + lb prVpr c + lc

Figure 6.6 Possible results in the testcross progeny of the b pr c trihybrid.

pr and c (fig. 6.8). Groupings according to these recombi-
nant events are shown at the right in table 6.1.

In the final column of table 6.1, recombination be-
tween b and c is scored. Only those recombinant classes
that have a new arrangement of b and c alleles, as com-
pared with the parentals, are counted. This last column
shows us what a b-c, two-point cross would have revealed
had we been unaware of the^?r locus in the middle.

Map Distances

The percent row in table 6.1 reveals that 5.9%
(887/15,000) of the offspring in the Drosophila trihybrid

testcross resulted from recombination between b and pr,
195% between pr and c, and 23.7% between b and c.
These numbers allow us to form a tentative map of the
loci (fig. 6.9). There is, however, a discrepancy. The dis-
tance between b and c can be calculated in two ways.
By adding the two distances, b-pr and pr-c, we get
5.9 + 19.5 = 25.4 map units; yet by directly counting the
recombinants (the last column of table 6.1), we get a dis-
tance of only 23.7 map units. What causes this discrep-
ancy of 1.7 map units?

Returning to the last column of table 6.1, we observe
that the double crossovers (60 and 72) are not counted,
yet each actually represents two crossovers in this re-

Table 6.1 Results of Testcrossing Female Drosophila Heterozygous for Black Body Color,
Purple Eye Color, and Curved Wings (b + b pr + pr c + c X bb prpr cc)

Number Recombinant

Alleles from

Between

Phenotype

Genotype

Number

Trihybrid Female

b and pr

pr and c

b and c

Wild-type

b + b pr + pr c + c

5,701

j + ^ + +
b pr c

Black, purple, curved

bb prpr cc

5,617

b pr c

Purple, curved

b + b prpr cc

388

b + pr c

388

Black

bb pr + pr c + c

367

b pr + c +

367

Curved

b + b pr + pr cc

1,412

b + pr + c

1,412

Black, purple

bb prpr c + c

1,383

b pr c +

1,383

Purple

b + b prpr c + c

b pr c

Black, curved

bb pr + pr cc

b pr + c

Total

15,000

887

2,927

3,550

Percent

5.9

19.5

23.7

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Meiotic tetrad
pr

Gametes
pr

Non recombinant

Gametes

pr +

c +

u +

pr +

c +

"fc L +

- Recombinant

> b

b +

pr +

c +

b +

pr +

c +

Nonrecombinant

Figure 6.7 Results of a crossover between the black and purple loci in Drosophila.

Gametes

b +

c
c

c +
c +

Nonrecombinant

Double
recombinant

Nonrecombinant

Figure 6.8 Results of a double crossover in the b pr c region of the Drosophila chromosome.

gion.The reason they are not counted is simply that if we
observed only the end loci of this chromosomal segment,
we would not detect the double crossovers; the first one
of the two crossovers causes a recombination between
the two end loci, whereas the second one returns these
outer loci to their original configuration (see fig. 6.8). If
we took the 3,550 recombinants between b and c and
added in twice the total of the double recombinants, 264,
we would get a total of 3,814. This is 25.4 map units,
which is the more precise figure we calculated before.
The farther two loci are apart on a chromosome, the
more double crossovers occur between them. Double
crossovers tend to mask recombinants, as in our exam-
ple, so that distantly linked loci usually appear closer
than they really are. Thus, the most accurate map dis-

tances are those established on very closely linked loci.
In other words, summed short distances are more accu-
rate than directly measured larger distances.

The results of the previous experiment show that we
can obtain at least two map distances between any two
loci: measured and actual. Measured map distance be-
tween two loci is the value obtained from a two-point
cross. Actual map distance is an idealized, more accurate
value obtained from summing short distances between
many intervening loci. We obtain the short distances
from crosses involving other loci between the original
two. When we plot measured map distance against actual
map distance, we obtain the curve in figure 6.10. This
curve is called a mapping function. This graph is of
both practical and theoretical value. Pragmatically, it allows

25.4 (sum of two shorter distances; best
estimate of true map distance)

5.9

-\—

19.5

-h

23.7 (measured map distance in a two-point cross; double
recombinants between b and c are masked)

Figure 6.9 Tentative map of the black, purple, and curved chro-
mosome in Drosophila. Numbers are map units (centimorgans).

Figure 6.10 Mapping function.

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us to convert a measured map distance into a more accu-
rate one. Theoretically, it shows that measured map dis-
tance never exceeds 50 map units in any one cross. Mul-
tiple crossovers reduce the apparent distance between
two loci to a maximum of 50 map units, the value that in-
dependent assortment produces (50% parentals, 50% re-
combinants).

Gene Order

Although we performed the previous analysis merely as-
suming that^?r was in the middle, the data in table 6.1
confirm our original assumption that the gene order is
b pr c. Of the four pairs of reciprocal phenotypic classes
in table 6.1, one pair has the highest frequency (5,701
and 5,617) and one pair has the lowest (60 and 72). The
pair with highest frequency is the nonrecombinant
group. The one with the lowest frequency is the double
recombinant group, the one in which only the middle lo-
cus has been changed from the parental arrangement. A
comparison of either of the double recombinant classes
with either of the nonrecombinant classes shows the
gene that is in the middle and, therefore, the gene order.
In other words, the data allow us to determine gene or-
der. Since b + pr + c + was one of the nonrecombinant
gametes, and b + pr c + was one of the double recombi-
nant gametes, pr stands out as the changed locus, or the
one in the middle. In a similar manner, comparing bpr +
c with bprc would also point to pr as the inside locus (or
inside marker). So would comparing b + pr c + with b
pr c or bpr + c with b + pr c + . In each case, the middle lo-
cus, pr, displays the different pattern, whereas the allelic
arrangements of the outside markers, b and c, behave in
concert.

If this seems confusing, simply compare the double
crossovers and nonrecombinants to find one of each in
which two alleles are identical. For example, the double
recombinant b + pr c + and the nonrecombinant b + pr +
c + share the b + and c + alleles. The pr locus is mutant in
one case and wild-type in the other. Hence, pr is the lo-
cus in the middle.

From the data in table 6.1, we can confirm the associ-
ation of alleles in the trihybrid parent. That is, since the
data came from testcrossing a trihybrid, the allelic con-
figuration in that trihybrid is reflected in the nonrecom-
binant classes of offspring. In this case, one is the result
of a b + pr + c + gamete, the other, of a bpr c gamete. Thus,
the trihybrid had the genotype b pr c/b + pr + c + : all al-
leles were in the cis configuration.

Coefficient of Coincidence

The next question in our analysis of this three-point
cross is, are crossovers occurring independently of
each other? That is, does the observed number of dou-

ble recombinants equal the expected number? In the
example, there were 132/15,000 double crossovers, or
0.88%. The expected number is based on the inde-
pendent occurrence of crossing over in the two re-
gions measured. That is, 5.9% of the time there is a
crossover in the b-pr region, which we can express as
a probability of occurrence of 0.059. Similarly, 19.5% of
the time there is a crossover in the pr-c region, or a
probability of occurrence of 0.195. A double crossover
should occur as a product of the two probabilities:
0.059 X 0.195 = 0.0115. This means that 1.15% of the
gametes (1.15% of 15,000 = 172.5) should be double
recombinants. In our example, the observed number of
double recombinant offspring is lower than expected
(132 observed, 172.5 expected). This implies a posi-
tive interference, in which the occurrence of the first
crossover reduces the chance of the second. We can ex-
press this as a coefficient of coincidence, defined as

observed number of double recombinants
expected number of double recombinants

In the example, the coefficient of coincidence is
132/172.5 = 0.77. In other words, only 77% of the ex-
pected double crossovers occurred. Sometimes we ex-
press this reduced quantity of double crossovers as the
degree of interference, defined as

interference = 1 — coefficient of coincidence

In our example, the interference is 23%.

It is also possible to have negative interference, in
which we observe more double recombinants than ex-
pected. In this situation, the occurrence of one crossover
seems to enhance the probability that crossovers will oc-
cur in adjacent regions.

Another Example

Let us work out one more three-point cross, in which
neither the middle gene nor the cis-trans relationship
of the alleles in the trihybrid F : parent is given. On the
third chromosome of Drosophila, hairy (h) causes extra
bristles on the body, thread (th) causes a thread-shaped
arista (antenna tip), and rosy (ry) causes the eyes to be
reddish brown. All three traits are recessive. Trihybrid
females were testcrossed; the phenotypes from one
thousand offspring are listed in table 6.2. At this point,
it is possible to use the data to determine the parental
genotypes (the P : generation, assuming that they were
homozygotes), the gene order, the map distances, and
the coefficient of coincidence. The table presents the
data in no particular order, as an experimenter might
have recorded them. Phenotypes are tabulated and,
from these, the genotypes can be reconstructed. Notice
that the data can be put into the form found in table 6.1;

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Table 6.2 Offspring from a Trihybrid (h h ry ry
th + th) Testcross (x hh ryry thth) in
Drosophila

Table 6.3 Data from Table 6.2 Arranged to Show
Recombinant Regions

Genotype

Phenotype

(order unknown)

Number

Thread

h ry th/h ry th

359

Rosy, thread

h + ry th/h ry th

Hairy, rosy, thread

h ry th/h ry th

Hairy, thread

h ry + th/h ry th

Rosy

h + ry th + /h ry th

Hairy, rosy

h ry th /h ry th

351

Wild-type

h ry th /h ry th

Hairy

h ry + th + /h ry th

Trihybrids

Gamete

Number

h-th

th—ry

h—ry

h th ry

359

h th ry

351

h th ry

h + th + ry

h + th ry

h th + ry +

h th ry

Total

1,000

200

100

280

we see a large reciprocal set (359 and 351), a small re-
ciprocal set (4 and 6), and large and small intermediate
sets (98 and 92, 47 and 43).

From the data presented, is it obvious that the three
loci are linked? The pattern, as just mentioned, is identi-
cal to that of the previous example, in which the three
loci were linked. (What pattern would appear if two of
the loci were linked and one assorted independently?
See problem 6 at the end of the chapter.) Next, what is
the allelic arrangement in the trihybrid parent? The off-
spring with the parental, or nonrecombinant, arrange-
ments are the reciprocal pair in highest frequency. Table
6.2 shows that thread and hairy rosy offspring are the
nonrecombinant s. Thus, the nonrecombinant gametes of
the trihybrid Y x parent were h ry th + and h + ry + th,
which is the allelic arrangement of the trihybrid with
the actual order still unknown — h ry th + /h + ry + th.
(What were the genotypes of the parents of this trihy-
brid, assuming they were homozygotes?) Continuing,
which gene is in the middle? From table 6.2, we know
that h ry th and h + ry + th + are the double recombinant
gametes of the trihybrid parent because they occur in
such low numbers. Comparison of these chromosomes
with either of the nonrecombinant chromosomes (h
ry + th or h ry th + ) shows that the thread (th) locus is in
the middle. We now know that the original trihybrid had
the following chromosomal composition: h th ry/h th
ry + . The h and ry alleles are in the cis configuration,
with th in the trans configuration.

We can now compare the chromosome from the tri-
hybrid in each of the eight offspring categories with the
parental arrangement and determine the regions that had
crossovers. Table 6.3 does this. We can see that the h-th
distance is 20 map units, the th-ry distance is 10 map
units, and the apparent h—ry distance is 28 map units

(fig. 6.11). As in the earlier example, the h—ry discrep-
ancy is from not counting the double crossovers twice
each: 280 + 2(10) = 300, which is 30 map units and the
more accurate figure. Last, we wish to know what the co-
efficient of coincidence is. The expected occurrence of
double recombinants is 0.200 X 0.100 = 0.020, or 2%.
Two percent of 1,000 = 20. Thus

coefficient of coincidence =
observed number of double recombinants
expected number of double recombinants
= 10/20 = 0.50

Only 50% of the expected double crossovers occurred.

Geneticists have mapped the chromosomes of many
eukaryotic organisms from three-point crosses of this
type — those of Drosophila are probably the most ex-
tensively studied. Drosophila and other species of flies
have giant polytene salivary gland chromosomes,
which arise as a result of endomitosis. In this process,

30 (all recombinants; best
estimate of true map distance)

20.0

10.0

— i—

28.0 (measured map distance)

Figure 6.11 Map of the h th ry region of the Drosophila
chromosome, with numerical discrepancy in distances. Num-
bers are map units (centimorgans).

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Figure 6.12 Giant salivary gland chromosomes of Drosophila.
X, 2, 3, and 4 are the four nonhomologous chromosomes. L
and R indicate the left and right arms (in relation to the cen-
tromere). The dark bands are chromomeres. (B. P. Kaufman,
"Induced Chromosome Rearrangements in Drosophila melanogaster," Journal
of Heredity, 30:178-90, 1939. Reproduced by permission of Oxford Univer-
sity Press.)

the chromosomes replicate, but the cell does not divide.
In the salivary gland of the fruit fly, homologous chromo-
somes synapse and then replicate to make about one
thousand copies, forming very thick structures with a
distinctive pattern of bands called chromomeres (fig.
6.12). Using methods chapter 8 will discuss, scientists
have mapped many loci to particular bands. Part of the
Drosophila chromosomal map is presented in figure 6.13
(see also box 6.2). Locate the loci we have mapped so far
to verify the map distances.

In summary, we know that two or more loci are
linked if offspring do not fall into simple Mendelian ra-
tios. Map distances are the percentage of recombinant
offspring in a testcross. With three loci, determine the
parental (nonrecombinant) and double recombinant
groups first. Then establish the locus in the middle, and
recast the data in the correct gene order. The most accu-
rate map distances are those obtained by summing
shorter distances. Determine a coefficient of coinci-
dence by comparing observed number of double recom-
binants to expected number.

Cytological Demonstration of Crossing Over

If we are correct that a chiasma during meiosis is the visi-
ble result of a physical crossover, then we should be able
to demonstrate that genetic crossing over is accompanied
by cytological crossing over. That is, the recombination

event should entail the exchange of physical parts of ho-
mologous chromosomes. This can be demonstrated if we
can distinguish between two homologous chromosomes,
a technique Creighton and McClintock first used in maize
(corn) and Stern first applied to Drosophila, both in 1931.
We will look at Creighton and McClintock's experiment.

Harriet Creighton and Barbara McClintock worked
with chromosome 9 in maize (n = 10). In one strain,
they found a chromosome with abnormal ends. One end
had a knob, and the other had an added piece of chro-
matin from another chromosome (fig. 6.14). This
knobbed chromosome was thus clearly different from its
normal homologue. It also carried the dominant colored
(6) allele and the recessive waxy texture (wx) allele. Af-
ter mapping studies showed that C was very close to the
knob and wx was close to the added piece of chromatin,
Creighton and McClintock made the cross shown in fig-
ure 6.14. The dihybrid plant with heteromorphic chro-
mosomes was crossed with the normal homomorphic
plant (only normal chromosomes) that had the genotype
of c Wx/c wx (colorless and nonwaxy phenotype). If a
crossover occurred during meiosis in the dihybrid in the
region between C and wx, a physical crossover, visible cy-
tologically (under the microscope), should also occur,
causing the knob to become associated with an other-
wise normal chromosome and the extra piece of chromo-
some 9 to be associated with a knobless chromosome.
Four types of gametes would result (fig. 6.14).

Barbara McClintock
(1902-1992). (Courtesy of
Cold Spring Harbor Research
Library Archives. Photographer,
David Miklos.)

Harriet B. Creighton
(1909- ). (Courtesy of
Harriet B. Creighton.)

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Diploid Mapping

121

The first chromosomal map
ever published included just
five loci on the X chromo-
some of Drosophila melanogaster
(fig. 1). It was published in 1913 by
Alfred H. Sturtevant, who began
working in Thomas Hunt Morgan's
"fly lab" while an undergraduate stu-
dent at Columbia University. The fly
lab included H.J. Muller, later to win a
Nobel Prize, and Calvin B. Bridges,
whose work on sex determination in
Drosophila we discussed in the last
chapter.

Sturtevant worked with six mu-
tants: yellow body (j); white (w),
eosin (^ e ), and vermilion eyes QiS);
and miniature Qri) and rudimentary
wings (r). (White and eosin are actu-
ally allelic; Sturtevant found no cross-
ing over between the two "loci.") Us-
ing crosses similar to the ones we
outline in this chapter, he con-
structed the map shown in figure 1.
The map distances we accept today
are very similar to the ones he ob-
tained.

Sturtevant's work was especially
important at this point because his
data supported several basic con-
cepts, including the linear arrange-
ment of genes, which argued for the

y w

0.0 1.0

— i 1 —

BOX 6.2

Historical
Perspectives

The First Chromosomal Map

placement of genes on chromosomes
as the only linear structures in the nu-
cleus. Sturtevant also pointed out
crossover interference. His summary
is clear and succinct:

It has been found possible to
arrange six sex-linked factors in
Drosophila in a linear series, using
the number of crossovers per one
hundred cases as an index of the dis-
tance between any two factors. This
scheme gives consistent results, in
the main.

A source of error in predicting
the strength of association between
untried factors is found in double
crossing over. The occurrence of
this phenomenon is demonstrated,
and it is shown not to occur as often
as would be expected from a purely
mathematical point of view, but the
conditions governing its frequency
are as yet not worked out.

30.7

m
33.7

These results . . . form a new ar-
gument in favor of the chromosome
view of inheritance, since they
strongly indicate that the factors in-
vestigated are arranged in a linear
series, at least mathematically.

Alfred H. Sturtevant (1891-1970).
(Courtesy of the Archives, California Institute
of Technology.)

57.6

(0.0 1.5)

(33.0)

(36.1)

(54.5)

Figure 1 The first chromosomal linkage map. Five loci in Drosophila melanogaster are mapped to the X chro-
mosome. The numbers in parentheses are the more accurately mapped distances recognized today. We also show
today's allelic designations rather than Sturtevant's original nomenclature. (Data from sturtevant. "The linear arrangement
of six sex-linked factors in Drosophila, as shown by their mode of association," Journal of Experimental Zoology, 14:43-59, 1913.)

Of twenty-eight offspring examined, all were consis-
tent with the predictions of the Punnett square in figure
6.14. Those of class 8 (lower right box) with the col-
ored, waxy phenotype all had a knobbed interchange
chromosome as well as a normal homologue. Those
with the colorless, waxy phenotype (class 4) had a knob-
less interchange chromosome. All of the colored, non-

waxy phenotypes (classes 5, 6, and 7) had a knobbed,
normal chromosome, which indicated that only classes
5 and 6 were in the sample. Of the two that were tested,
both were WxWx, indicating that they were of class 5.
The remaining classes (1,2, and 3) were of the color-
less, nonwaxy phenotype. All were knobless. Of those
that contained only normal chromosomes, some were

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Chapter Six Linkage and Mapping in Eukaryotes

1(X)

0.0 ^ _ ^ yellow body

abnormal eyes
white eyes

echinus eyes
ruby eyes

5.5
7.5

13.7

18.9
20.1
21.0

/~\

27.7

32.0
33.0

36.1

44.4

55.0
56.7
57.0
59.4
59.5
62.5
64.4
66.0

crossveinless wings

carmine eyes
bistre eyes
singed bristles

lozenge eyes

notchy wings
vermilion eyes

miniature wings

garnet eyes

inflated wings
y forked bristles
Bar eyes
Beadex wings
fused wing veins
carnation eyes
warty eyes
bobbed bristles

3
A

13.0 — dumpy wings

26.0
26.5

44.0 _ ancon wings

black body

7tyft bristles
/ y spiny legs
y purple eyes

apterous (wingless)

ftyffec/ head

cinnabar eyes
v arcfrvs oculus eyes

sep/a eyes
/?a/Ay body

59.5

— Lobe eyes

— curved wings

91.5 — smooth abdomen

104.5
107.0

brown eyes
orange eyes

90.0
91.1

100.7

Hairless body
'-ebony body

banc/ thorax
^detached veins

thread aristae
scarlet eyes
maroon eyes
.dwarf body
curled wings
rosy eyes

//tytec/ wings

1.4
2.0

groucho bristles
roug/? eyes

c/aref eyes

oenf wings
eyeless

Figure 6.13 Partial map of the chromosomes of Drosophila melanogaster. The centromere is marked by an open circle.
(From C. Bridges, "Salavary Chromosome Maps," Journal of Heredity, 26:60-64, 1935. Reprinted with permission of Oxford University Press.)

WxWx (class 1) and some were heterozygotes (Wxwx,
class 2). Of those containing interchange chromosomes,
two were heterozygous, representing class 3 Two were
homozygous, WxWx, yet interchange-normal hetero-
morphs. These represent a crossover in the region be-
tween the waxy locus and the extra piece of chromatin,
producing a knobless-c-Wx-extra-piece chromosome.
When combined with a c-Wx-normal chromosome,
these would give these anomalous genotypes. The sam-
ple size was not large enough to pick up the reciprocal
event. Creighton and McClintock concluded: "Pairing
chromosomes, heteromorphic in two regions, have

been shown to exchange parts at the same time they ex-
change genes assigned to these regions."

HAPLOID MAPPING
(TETRAD ANALYSIS)

For Drosophila and other diploid eukaryotes, the genetic
analysis considered earlier in this chapter is referred to as
random strand analysis. Sperm cells, each of which
carry only one chromatid of a meiotic tetrad, unite with

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c
C

Knob

Meiosis
Wx

wx
wx

Normal chromosome 9

Knobbed interchange chromosome 9

Extra piece

( Wx

^^ ^^H

Gametes

Nonrecombinant

Recombinant

Nonrecombinant

Colorless

Nonwaxy

Colorless

Nonwaxy

Colorless

Nonwaxy

Colorless

Waxy

C^^^^^^^^H

<~~^^^^^^^m

Colored

Nonwaxy

Colored

Nonwaxy

Colored

Nonwaxy

Colored

Waxy

Figure 6.14 Creighton and McClintock's experiment in maize demonstrated that genetic crossover corre-
lates with cytological crossing over.

eggs, which also carry only one chromatid from a tetrad.
Thus, zygotes are the result of the random uniting of
chromatids.

Fungi of the class Ascomycetes retain the four haploid
products of meiosis in a sac called an ascus. These or-
ganisms provide a unique opportunity to look at the total
products of meiosis in a tetrad. Having the four products

of meiosis allowed geneticists to determine such basics as
the reciprocity of crossing over and the fact that DNA
replication occurs before crossing over. Different tech-
niques are used for these analyses. We will look at two
fungi, the common baker's yeast, Saccharomyces cere-
visiae, and pink bread mold, Neurospora crassa, both of
which retain the products of meiosis as ascospores.

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Chapter Six Linkage and Mapping in Eukaryotes

Phenotypes of Fungi

At this point, you might wonder what phenotypes
fungi such as yeast and Neurospora express. In general,
microorganisms have phenotypes that fall into three
broad categories: colony morphology, drug resistance,
and nutritional requirements. Many microorganisms
can be cultured in petri plates or test tubes that con-
tain a supporting medium such as agar, to which vari-
ous substances can be added (fig. 6.15). Wild-type Neu-
rospora, the familiar pink bread mold, generally grows
in a filamentous form, whereas yeast tends to form
colonies. Various mutations exist that change colony
morphology. In yeast, the ade gene causes the colonies
to be red. In Neurospora, fluffy (/7), tuft (tu), dirty
(dir), and colonial (col4) are all mutants of the basic
growth form. In addition, wild-type Neurospora is sen-
sitive to the sulfa drug sulfonamide, whereas one of its
mutants (Sfo) actually requires sulfonamide in order to
survive and grow. Yeast shows similar sensitivities to
antifungal agents.

Nutritional-requirement phenotypes provide great in-
sight not only into genetic analysis but also into the bio-
chemical pathways of metabolism, as mentioned in
chapter 2. Wild-type Neurospora can grow on a medium
containing only sugar, a nitrogen source, some organic
acids and salts, and the vitamin bio tin. This is referred to
as minimal medium. However, several different mutant
types, or strains, of Neurospora cannot grow on this
minimal medium until some essential nutrient is added.
For example, one mutant strain will not grow on minimal
medium, but will grow if one of the amino acids, argi-
nine, is added (fig. 6.16). From this we can infer that the
wild-type has a normal, functional enzyme in the syn-
thetic pathway of arginine. The arginine-requiring mu-
tant has an allele that specifies an enzyme that is inca-
pable of converting one of the intermediates in the
pathway directly into arginine or into one of the precur-
sors to arginine. We can see that if the synthetic pathway
is long, many different loci may have alleles that cause
the strain to require arginine (fig. 6. 17). This, in fact, hap-
pens, and the different loci are usually named arg 1 , arg 2 ,
and so on. There are numerous biosynthetic pathways in
yeast and Neurospora, and mutants exhibit many differ-
ent nutritional requirements. Mutants can be induced ex-
perimentally by radiation or by chemicals and other
treatments. These, then, are the tools we use to analyze
and map the chromosomes of microorganisms, including
yeast and Neurospora. These techniques are expanded
on in the next chapter.

Unordered Spores (Yeast)

Baker's, or budding, yeast, Saccharomyces cerevisiae, ex-
ists in both a haploid and diploid form (fig. 6.18). The

Glass needle

Agar

Petri plate

Individual spores are pressed
out of the ascus. The spores
are lined up and the agar is
sliced into sections.

Knife

Squares are lifted out and
placed in individual tubes
to grow.

Spore
Agar square

Figure 6.15 Spore isolation technique in Neurospora.

haploid form usually forms under nutritional stress (star-
vation). When better conditions return, haploid cells of
the two sexes, called a and a mating types, fuse to form
the diploid. (Mating types are generally the result of a
one-locus, two-allele genetic system that determines that
only opposite mating types can fuse. We discuss this sys-
tem in more detail in chapter 16.) The haploid is again es-
tablished by meiosis under starvation conditions. In
yeast, all the products of meiosis are contained in the as-
cus. Let us look at a mapping problem, using the a and b
loci for convenience.

When an ab spore (or gamete) fuses with an a + b +
spore (or gamete), and the diploid then undergoes meio-
sis, the spores can be isolated and grown as haploid
colonies, which are then observed for the phenotypes the
two loci control. Only three patterns can occur (table 6.4).

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Spore isolation

Complete medium

Minimal medium

Selective medium

With vitamins

\\\\\\\\

With amino acids
(nutrient required
by strain)

With supplements other
than amino acids

Figure 6.16 Isolation of nutritional-requirement mutants in Neurospora.

Class 1 has two types of spores, which are identical to the
parental haploid spores. This ascus type is, therefore, re-
ferred to as a parental ditype (PD). The second class
also has only two spore types, but they are recombinants.
This ascus type is referred to as a nonparental ditype
(NPD). The third class has all four possible spore types
and is referred to as a tetratype (TT).

All three ascus types can be generated whether or not
the two loci are linked. As figure 6.19 shows, if the loci
are linked, parental ditypes come from the lack of a
crossover, whereas nonparental ditypes come about
from four-strand double crossovers (double crossovers
involving all four chromatids). We should thus expect
parental ditypes to be more numerous than nonparental
ditypes for linked loci. However, if the loci are not linked,
both parental and nonparental ditypes come about
through independent assortment — they should occur in
equal frequencies. We can therefore determine whether
the loci are linked by comparing parental ditypes and
nonparental ditypes. In table 6 A, the parental ditypes
greatly outnumber the nonparental ditypes; the two loci

are, therefore, linked. What is the map distance between
the loci?

A return to figure 6.19 shows that in a nonparental di-
type, all four chromatids are recombinant, whereas in a
tetratype, only half the chromatids are recombinant. Re-
membering that 1% recombinant offspring equals 1 map
unit, we can use the following formula:

map units =
(1/2) the number of TT asci +the number of NPD asci

total number of asci

Thus, for the data of table 6.4,

X 100

map = d/2)2Q + 5
units 100

10 + 5

X 100 = X 100 = 15

100

Ordered Spores (Neurospora,)

Unlike yeast, Neurospora has ordered spores; Neu-
rospora's life cycle is shown in figure 6.20. Fertilization
takes place within an immature fruiting body after a

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Chapter Six Linkage and Mapping in Eukaryotes

Stepl

Gene 1

Precursor

■>► Enzyme 1

H H

t H
H NH

:— C— C— C

HN— C— C-

H H H H
Ornithine

Step 2

Gene 2

■^ Enzyme 2

H H

H I I

HN HN— C— C

\ / II

C H H

Y H
H NH

— C— C— C

H H

Citrulline

Step 3

Gene 3

■>- Enzyme 3

H H

V H
H NH

HN HN— C— C

\ / II

C H H

— C— C— C

H H

Arginine

Figure 6.17 Arginine biosynthetic pathway of Neurospora.

Table 6.4 The Three Ascus Types in Yeast

Resulting from Meiosis in a Dihybrid,

aa + bb +

(PD)

(NPD)

(TT)

ab +

a b

a + b

a + b +

a + b

a + b +

Spores or
gametes

Fertilization a / a (2n)

"Bud"

Vegetative

reproduction

Ascus (n)

(n) colony

Figure 6.18 Life cycle of yeast. Mature cells are mating types
a or a; n is the haploid stage; 2n is diploid.

spore or filament of one mating type contacts a special
filament extending from the fruiting body of the oppo-
site mating type (mating types are referred to as A and a).
The zygote's nucleus undergoes meiosis without any in-
tervening mitosis. Unlike yeast, Neurospora does not
have a diploid phase in its life cycle. Rather, it undergoes
meiosis immediately after the diploid nuclei form.

Since the Neurospora ascus is narrow, the meiotic
spindle is forced to lie along the cell's long axis. The two
nuclei then undergo the second meiotic division, which is
also oriented along the long axis of the ascus. The result is
that the spores are ordered according to their cen-
tromeres (fig. 6.21). That is, if we label one centromere^
and the other a, for the two mating types, a tetrad at meio-
sis I will consist of one A and one a centromere. At the
end of meiosis in Neurospora, the four ascospores are in
the order A A aaoraaAAin regard to centromeres. (We
talk more simply of centromeres rather than chromo-
somes or chromatids because of the complications that

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If linked

a +

b +

a +

b +

If not linked

-,+

o
o

-,+

o
o

ab
ab

a + b +
a + b +

Nonrecombinant

Independent assortment

NPD

ab +
ab +

a + b
a + b

O
O

a
a

b +

-i+

Independent assortment

ab +

a + b

a + b +

a
a

a +

b
b

a
a

b
b

-i+

a +

6 +

a
a

5
6

a
a

b +

Figure 6.19 Formation of parental ditype (PD), nonparental ditype (NPD), and tetratype (TT) asci in a dihybrid yeast by
linkage or independent assortment at meiosis. Open circles are centromeres.

crossing over adds. A type A centromere is always a type
A centromere, whereas, due to crossing over, a chromo-
some attached to that centromere may be partly from the
type A parent and partly from the type a parent.)

Before the ascospores mature in Neurospora, a mito-
sis takes place in each nucleus so that four pairs rather

than just four spores are formed. In the absence of phe-
nomena such as mutation or gene conversion, to be dis-
cussed later in the book, pairs are always identical (fig.
6.21). As we will see in a moment, because of the or-
dered spores, we can map loci in Neurospora in relation
to their centromeres.

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Chapter Six Linkage and Mapping in Eukaryotes

Fertilization

(2n) Aa

Vegetative

hyphae

(n)

Asexual spore a

Vegetative \|j
hyphae ^

Figure 6.20 Life cycle of Neurospora. A and a are mating types; n is a haploid stage; 2n
is diploid.

First and Second Division Segregation

Recall that there is a 4:4 segregation of the centromeres
in the ascus of Neurospora. Two kinds of patterns appear
among the loci on these chromosomes. These patterns
depend on whether there was a crossover between the
locus and its centromere (fig. 6.22). If there was no
crossover between the locus and its centromere, the al-
lelic pattern is the same as the centromeric pattern,
which is referred to as first-division segregation
(FDS), because the alleles separate from each other at
meiosis I. If, however, a crossover has occurred between
the locus and its centromere, patterns of a different type
emerge (2:4:2 or 2:2:2:2), each of which is referred to as
second-division segregation (SDS). Because the
spores are ordered, the centromeres always follow a first-
division segregation pattern. Hence, we should be able to
map the distance of a locus to its centromere. Under the
simplest circumstances (fig. 6.22), every second-division
segregation configuration has four recombinant and four
nonrecombinant chromatids (spores). Thus, half of the
chromatids (spores) in a second-division segregation as-

cus are recombinant. Therefore, since 1% recombinant
chromatids equal 1 map unit,

(1/2) the number of SDS asci

map distance = X 100

total number of asci

An example using this calculation appears in table 6.5.

Three-point crosses in Neurospora can also be exam-
ined. Let us map two loci and their centromere. For sim-
plicity, we will use the a and b loci. Dihybrids are formed
from fused mycelia (ab X a + b + ), which then undergo
meiosis. One thousand asci are analyzed, keeping the
spore order intact. Before presenting the data, we should
consider how to group them. Since each locus can show
six different patterns (fig. 6.22), two loci scored together
should give thirty-six possible spore arrangements (6 X
6). Some thought, however, tells us that many of these
patterns are really random variants of each other. The
tetrad in meiosis is a three-dimensional entity rather than
a flat, four-rod object, as it is usually drawn. At the first
meiotic division, either centromere can go to the left or
the right, and when centromeres split at the second mei-

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Meiosis I

Meiosis II

Mitosis

Eight spores formed

Figure 6.21 Meiosis in Neurospora. Although Neurospora has seven pairs of chromosomes at meiosis, only one
pair is shown. A and a, the two mating types, represent the two centromeres of the tetrad.

otic division, movement within the future half-ascus (the
four spores to the left or the four spores to the right) is
also random. Thus, one genetic event can produce up to
eight "different" patterns. For example, consider the
arrangements figure 6.23 shows, in which a crossover oc-
curs between the a and b loci. All eight arrangements,
producing the ascus patterns of table 6.6, are equally
likely. The thirty-six possible patterns then reduce to only
the seven unique patterns shown in table 6.7. Note also
that these asci can be grouped into the three types of asci
found in yeast with unordered spores: parental ditypes,

nonparental ditypes, and tetratypes. Had we not had the
order of the spores from the asci, that would, in fact, be
the only way we could score the asci (see the bottom of
table 6.7).

Gene Order

We can now determine the distance from each locus to
its centromere and the linkage arrangement of the loci if
they are both linked to the same centromere. We can es-
tablish by inspection that the two loci are linked to each

Table 6.5

Genetic Patterns

Following

Meiosis in

an a + a

Heterozygous

Neurospora (Ten

Asci Examined)

Ascus Number

Spore

Number

+
a

a +

+
a

a +

+
a

a +

+
a

a +

+
a

a +

+
a

a +

+
a

a +

+
a

FDS

SDS

FDS

SDS

FDS

Note: Map distance (a locus to centromere) = (1/2)% SDS

= (1/2) 40%
= 20 map units

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a +

O O

a + a

First-division segregation with
no crossover between a locus
and centromere

/ \

6 6 6 6

a +

a + a

a^+Ya+Ya+YaXaYaYal 4:4

alalalala + la + la + la + )) 4:4

a + \a + a\ a

/ \

Second-division segregation with
crossover between a locus
and centromere

i i <!> i

a +

a a

ia + Xa + X a la Ia + la + \ a\a

aiaia + Ya + YaYaYaira*

ana^ia la la laia^a

2:2:2:2

2:4:2

aYaYaTa + Ya + Ya + YaYa^ 2:4:2

Figure 6.22 The six possible Neurospora ascospore patterns
in respect to one locus.

other — and therefore to the same centromere — by ex-
amining classes 1 (parental ditype) and 2 (nonparental di-
type) in table 6.7. If the two loci are unlinked, these two
categories would represent two equally likely alternative
events when no crossover takes place. Since category 1
represents almost 75% of all the asci, we can be sure the
two loci are linked.

To determine the distance of each locus to the
centromere, we calculate one-half the percentage of
second-division segregation patterns for each locus.
For the a locus, classes 4, 5, 6, and 7 are second-division
segregation patterns. For the b locus, classes 3,5,6, and
7 are second-division segregation patterns. Therefore,

Table 6.6

Eight of the Thirty-Six Possible Spore
Patterns in Neurospora Scored for Two
Loci, a and b (All Random Variants of
the Same Genetic Event)

Ascus Number

Spore
Number

ab +

a b

a + b

ab +

a b

a + b

ab +

a + b

a + b +

ab +

a + b

a + b +

a + b

a + b +

ab +

a + b

a + b +

ab +

a b

a + b

ab +

a b

a + b

ab +

Table 6.7 The Seven Unique Classes of Asci

Resulting from Meiosis in a Dihybrid
Neurospora, ab/a + b +

Ascus Number

Spore

Number

ab +

a + b

a b

a + b

a b

ab +

a + b

a b

a + b

a b

a + b +

a + b

a + b +

a + b

a + b +

a + b

a + b +

a + b

a + b +

a + b

ab +

a + b +

a + b

ab +

729

101

150

SDS for a locus

150

SDS for b locus

101

150

Unordered:

NPD

the distances to the centromere, in map units, for each
locus are

9+150+1 + 8
for locus a: (1/2) X 100

1,000

8.4 centimorgans

101 + 150+1+8
for locus b: (1/2) X 100

1,000

= 130 centimorgans

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a a +

(1)

b + b

b +

(2)

a a +
b b

■s+

(3)

a a +

b + b +

C3 C3

(4)

a +

6 tf

a'\aay a

6 6

(5)

6 +

a + a

6 b +

a +

a a

a + a

(6)

(7)

a + a

b + b +

C3 C3

(8)

a +

a + a

to + to

to +

Figure 6.23 The eight random arrangements possible when a
single crossover occurs between the a and b loci in Neurospora
(see table 6.6). Circular arrows represent the rotation of a
centromere from its position in the original configuration.

It should now be possible to describe exactly what type
of crossover event produced each of the seven classes in
table 6.7.

Unfortunately these two distances do not provide a
unique determination of gene order. In figure 6.24, we
see that two alternatives are possible: one has a map dis-
tance between the loci of 21.4 map units; the other has
4.6 map units between loci. How do we determine
which of these is correct? The simplest way is to calcu-
late the a-b distance using the unordered spore infor-
mation. That is, the map distance is

map units =
(1/2) the number of TT asci +the number of NPD asci

total number of asci

X 100

(1/2)118 + 3
1,000

X 100 = 6.2

Since 6.2 map units is much closer to the a-b distance
expected if both loci are on the same side of the cen-
tromere, we accept alternative 2 in figure 6.24.

A second way to choose between the alternatives in
figure 6.24 is to find out what happens to the b locus
when a crossover occurs between the a locus and its
centromere. If the order in alternative 1 is correct,
crossovers between the a locus and its centromere
should have no effect on the b locus; if 2 is correct, most
of the crossovers that move the a locus in relation to its
centromere should also move the b locus.

Asci classes 4, 5, 6, and 7 include all the SDS patterns
for the a locus. Of 168 asci, 150 (class 5) have similar SDS
patterns for the b locus. Thus, 89% of the time, a
crossover between the a locus and its centromere is also
a crossover between the b locus and its centromere —
compelling evidence in favor of alternative 2. (What form
would the data take if alternative 1 were correct?)

In summary, mapping by tetrad analysis proceeds as
follows. For both ordered and unordered spores, linkage
is indicated by an excess of parental ditypes over non-
parental ditypes. For unordered spores (yeast), the dis-
tance between two loci is one-half the number of

-i-

Centromere
I

O —

(1)

8.4

r -

21.4

Centromere
I

13.0

-h

8.4

4.6

(2)

13.0

-h

-i-

Figure 6.24 Two possible arrangements of the a and b loci and their centromere.
Distances are in map units.

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Chapter Six Linkage and Mapping in Eukaryotes

tetratypes plus the number of nonparental ditypes, all di-
vided by the total number of asci, expressed as a percent-
age. For ordered spores (Neurospora), the distance from a
locus to its centromere is one-half the percentage of
second-division segregants. Mapping the distance between
two loci is similar to the process in unordered spores.

// ////// !\\\ WW \\

y. ////// > i \ \ \ \\\\ v

SOMATIC (MITOTIC)
CROSSING OVER

Crossing over is known to occur in somatic cells as well
as during meiosis. It apparently occurs when two homol-
ogous chromatids come to lie next to each other and
breakage and reunion follow, most likely as a conse-
quence of DNA repair (see chapter 12). Unlike in meio-
sis, no synaptonemal complex forms. The occurrence of
mitotic crossing over is relatively rare. In the fungus As-
pergillus nidulans, mitotic crossing over occurs about
once in every one hundred cell divisions.

Mitotic recombination was discovered in 1936 by Curt
Stern, who noticed the occurrence of twin spots in fruit
flies that were dihybrid for the yellow allele for body color
(y) and the singed allele (sn) for bristle morphology (fig.
6.25). A twin spot could be explained by mitotic crossing
over between the sn locus and its centromere (fig. 6.26). A
crossover in the sn-y region would produce only a yellow
spot, whereas a double crossover, one between y and sn
and the other between sn and the centromere, would pro-
duce only a singed spot. (Verify this for yourself.) These
three phenotypes were found in the relative frequencies
expected. That is, given that the gene locations are drawn
to scale in figure 6.26, we would expect double spots to be
most common, followed by yellow spots, with singed
spots rarest of all because they require a double crossover.
This in fact occurred, and no other obvious explanation
was consistent with these facts. Mitotic crossing over has
been used in fungal genetics as a supplemental, or even a
primary, method for determining linkage relations. Al-
though gene orders are consistent between mitotic and
meiotic mapping, relative distances are usually not, which
is not totally unexpected. We know that neither meiotic
nor mitotic crossing over is uniform along a chromosome.
Apparently, the factors that cause deviation from unifor-
mity differ in the two processes.

'i 'l K

7, l.

f I
t f

i I I

1 <*
\ \
\

Yellow spot
Singed spot

Figure 6.25 Yellow and singed twin spots on the thorax of a
female Drosophila.

Curt Stern (1902-1981)
(Courtesy of the Science Council
of Japan.)

cific crosses coupled with the relatively small number of
offspring) make these techniques of human chromo-
some mapping very difficult. However, some progress
has been made based on pedigrees, especially in assign-
ing genes to the X chromosome. As the pedigree analysis
in the previous chapter has shown, X chromosomal traits
have unique patterns of inheritance, and loci on the X
chromosome are easy to identify. Currently over four
hundred loci are known to be on the X chromosome. It
has been estimated, by several different methods, that be-
tween fifty and one hundred thousand loci exist on hu-
man chromosomes. In later chapters, we will discuss
several additional methods of human chromosomal map-
ping that use molecular genetic techniques.

HUMAN CHROMOSOMAL
MAPS

In theory, we can map human chromosomes as we
would those of any other organism. Realistically, the
problems mentioned earlier (the inability to make spe-

X Linkage

After determining that a human gene is X linked, the next
problem is to determine the position of the locus on the
X chromosome and the map units between loci. Some-
times we can do this with the proper pedigrees, if cross-
ing over can be ascertained. An example of this "grand-

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133

O
O

sn+
sn+

y
y

Dihybrid at prophase
(mitosis)

Yellow-spot
cell

Singed-spot
cell

Crossover

sn+
sn+
sn

▼ sn+
sn

sn +
sn

▼ sn+

y
y
y+

y
y+

y
y±_

y
y

Figure 6.26 Formation of twin spots by somatic crossing over.

No crossover

o
o

sn+
sn+
sn

t
sn+

sn+
sn

y
y
y+

Wild-
-type
cells

father method" appears in figure 6.27. In this example, a
grandfather has one of the traits in question (here, color
blindness). We then find that he has a grandson who is
glucose-6-phosphate dehydrogenase (G-6-PD) deficient.
From this we can infer that the mother (of the grandson)
was dihybrid for the two alleles in the trans configura-
tion. That is, she received her color-blindness allele on
one of her X chromosomes from her father, and she must
have received the G-6-PD-deficiency allele on the other X
chromosome from her mother (why?). Thus, the two
sons on the left in figure 6.27 are nonrecombinant, and
the two on the right are recombinant. Theoretically, we
can determine map distance by simply totaling the re-
combinant grandsons and dividing by the total number
of grandsons. Of course, the methodology would be the
same if the grandfather were both color-blind and G-6-PD
deficient. The mother would then be dihybrid in the cis
configuration, and the sons would be tabulated in the re-
verse manner. The point is that the grandfather's pheno-

type gives us information that allows us to infer that the
mother was dihybrid, as well as telling us the cis-trans
arrangement of her alleles. We can then score her sons as
either recombinant or nonrecombinant.

Grandfather

Mother

Grandsons

Color-blind

G-6-PD deficient

Recombinant

Figure 6.27 "Grandfather method" of determining crossing
over between loci on the human X chromosome. G-6-PD is
glucose-6-phosphate dehydrogenase.

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Chapter Six Linkage and Mapping in Eukaryotes

Autosomal Linkage

From this we can see that it is relatively easy to map the
X chromosome. The autosomes are another story. Since
there are twenty-two autosomal linkage groups (twenty-
two pairs of nonsex chromosomes), it is virtually impos-
sible to determine from simple pedigrees which chromo-
some two loci are located on. Pedigrees can tell us if two
loci are linked to each other, but not on which chromo-
some. In figure 6.28, the nail-patella syndrome includes,
among other things, abnormal nail growth coupled with
the absence or underdevelopment of kneecaps. It is a
dominant trait. The male in generation II is dihybrid, with
the A allele of the ABO blood type system associated
with the nail-patella allele (NPS1) and the B allele with
the normal nail-patella allele {npsT). Thus only one child
in eight (III-5) is recombinant. Actually, the map distance
is about 10%. In general, map distances appear greater in
females than in males because more crossing over occurs
in females (box 6.3).

We now turn our attention to the localization of loci
to particular human chromosomes. The first locus that
was definitely established to be on a particular autosome
was the Duffy blood group on chromosome 1 . This was
ascertained in 1968 from a family that had a morphologi-
cally odd, or "uncoiled," chromosome 1. Inheritance in
the Duffy blood group system followed the pattern of in-
heritance of the "uncoiled" chromosome. Real strides
have been made since then. Two techniques, chromo-
somal banding and somatic-cell hybridization, have been
crucial to autosomal mapping.

Chromosomal Banding

Techniques were developed around 1970 that make use
of certain histochemical stains that produce repeatable
banding patterns on the chromosomes. For example,
Giemsa staining is one such technique; the resulting
bands are called G-bands. More detail on these tech-
niques is presented in chapter 15. Before these tech-

III

3 4

B \k

O O

7 8

Nail-patella syndrome

A, AB, B, O Blood types

Recombinant

Figure 6.28 Linkage of the nail-patella syndrome and ABO loci.

niques, human and other mammalian chromosomes
were grouped into general size categories because of the
difficulty of differentiating many of them. With banding
techniques came the ability to identify each human chro-
mosome in a karyotype (see fig. 5.1).

Somatic-Cell Hybridization

The ability to distinguish each human chromosome is re-
quired to perform somatic-cell hybridization, in which hu-
man and mouse (or hamster) cells are fused in culture to
form a hybrid. The fusion is usually mediated chemically
with polyethylene glycol, which affects cell membranes;
or with an inactivated virus, for example the Sendai virus,
that is able to fuse to more than one cell at the same time.
(The virus is able to do this because it has a lipid mem-
brane derived from its host cells that easily fuses with
new host cells. Because of this properly, the virus can fuse
to two cells close together, forming a cytoplasmic bridge
between them that facilitates their fusion.) When two
cells fuse, their nuclei are at first separate, forming a het-
erokaryon, a cell with nuclei from different sources.
When the nuclei fuse, a hybrid cell is formed, and this hy-
brid tends to lose human chromosomes preferentially
through succeeding generations. Upon stabilization, the
result is a cell with one or more human chromosomes in
addition to the original mouse or hamster chromosomal
complement. Banding techniques allow the observer to
recognize the human chromosomes. A geneticist looks
for specific human phenotypes, such as enzyme products,
and can then assign the phenotype to one of the human
chromosomes in the cell line.

When cells are mixed together for hybridization,
some cells do not hybridize. It is thus necessary to be able
to select for study just those cells that are hybrids. One
technique, originally devised by J. W Littlefield in 1964,
makes use of genetic differences in the way the cell lines
synthesize DNA. Normally, in mammalian cells,
aminopterin acts as an inhibitor of enzymes involved in
DNA metabolism. Two enzymes, hypoxanthine phospho-
ribosyl transferase (HPRT) and thymidine kinase (TK), can
bypass aminopterin inhibition by making use of second-
ary, or salvage, pathways in the cell. If hypoxanthine is
provided, HPRT converts it to a purine, and if thymidine is
provided, TK converts it to the nucleotide thymidylate.
(Purines are converted to nucleotides and nucleotides are
the subunits of DNA — see chapter 9) Thus, normal cells
in the absence of aminopterin synthesize DNA even if
they lack HPRT activity (HPRT) or TK activity (TK). In
the presence of aminopterin, HPRT" or TK~ cells die.
However, in the presence of aminopterin, HPRT + TK +
cells can synthesize DNA and survive. Using this informa-
tion, the following selection system was developed.

Mouse cells that have the phenotype of HPRT + TK~
are mixed with human cells that have the phenotype of

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Human Chromosomal Maps

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BOX 6.3

Human population geneticists
can increase the accuracy of
their linkage analysis by us-
ing a probability technique, devel-
oped by Newton Morton, called the
lod score method (log Odds). The
geneticist asks what the probability is
of getting a particular pedigree as-
suming a particular recombination
frequency (0), as compared with get-
ting the same pedigree assuming in-
dependent assortment (0 = 0.50). In
other words, he or she calculates the
ratio of the probability of genotypes
in a family given a certain crossover
frequency compared with the proba-
bility of those genotypes if the loci
are unlinked. Logarithms are used for
ease of calculation, and the parame-
ter is called z, the lod score. Using
this method, a researcher can try dif-
ferent crossover frequencies until the
one giving the highest lod score is
found.

For example, take the pedigree in
figure 6.28. The father in generation

Newton E. Morton (1929- ).

(Courtesy of Dr. Newton E. Morton.)

Experimental
Methods

Lod Scores

II can have one of two allelic arrange-
ments: A NPS1/B npsl or A/B
NPSl/npsl. The former assumes link-
age, whereas the latter does not. Our
initial estimate of recombination, as-
suming linkage, was (1/8) X 100, or
12.5 map units. We now need to cal-
culate the ratio of two probabilities:

probability of birth sequence

assuming 12.5 map units

z = log

probability of birth sequence

assuming independent

assortment

Assuming 12.5 map units (or a
probability of 0.125 of a crossover;
= 0.125), the probability of child
III-l is 0.4375. This child would be a
nonrecombinant, so his probability of
having the nail-patella syndrome and
type A blood is half the probability of
no crossover during meiosis, or (1 —
0.125)/2. We divide by two because
there are two nonrecombinant types.
This is the same probability for all
children except III-5, whose probabil-
ity of occurrence is 0.125/2 =
0.0625, since he is a recombinant.
Thus, the numerator of the previous
equation is (0.4375) 7 (0.0625).

If the two loci are not linked, then
any genotype has a probability of 1/4,
or 0.25. Thus, the sequence of the
eight children has the probability of

(0.25)°. This is the denominator of
the equation. Thus,

z = log

(0.4375) 7 (0.0625)
(0.25) 8

z = log [12.566] = 1.099

Any lod score greater than zero fa-
vors linkage. A lod score less than
zero suggests that has been under-
estimated. A lod of 30 or greater (10 3
or one thousand times more likely
than independent assortment) is con-
sidered a strong likelihood of linkage.
Thus, in our example, we have an in-
dication of linkage with a recombina-
tion frequency of 0. 125. Now we can
calculate lod scores assuming other
values of recombination, as table 1
does. You can see that the recombi-
nation frequency as calculated, 0.125
(12.5 map units), gives the highest
lod score.

Table 1 Lod Scores for the
Cross in Figure 6.28

Recombination

Frequency (0)

Lod Score

0.05

0.951

0.10

1.088

0.125

1.099

0.15

1.090

0.20

1.031

0.25

0.932

0.30

0.801

0.35

0.643

0.40

0.457

0.45

0.244

0.50

0.000

HPRT TK T in the presence of Sendai virus or polyethyl-
ene glycol. Fusion takes place in some of the cells, and
the mixture is grown in a medium containing hypoxan-
thine, aminopterin, and thymidine (called HAT
medium). In the presence of aminopterin, unfused
mouse cells (TK~) and unfused human cells (HPRT)
die. Hybrid cells, however, survive because they are

HPRT + TK + . Eventually, the hybrid cells end up with ran-
dom numbers of human chromosomes. There is one re-
striction: All cell lines selected are TK + .This HAT method
(using the HAT medium) not only selects for hybrid
clones, but also localizes the TK gene to human chromo-
some 17, the one human chromosome found in every
successful cell line.

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Chapter Six Linkage and Mapping in Eukaryotes

TablG 6.8 Assignment of the Gene for Blood Coagulating Factor III to Human Chromosome 1
Using Human-Mouse Hybrid Cell Lines

Hybrid

Tissue/

Human Chromosome Present

Cell Line

Factor

Designation

Score

win

—

WIL6

—

WIL7

—

WIL14

—

SIR3

—

SIR8

—

SIR 11

—

REW7

—

REW15

—

DUA1A

DUAlCsAzE

DUAlCsAzH

TSI1

—

TSL2

—

TSL2CsBF

XTR1

—

XTR2

—

XTR3BsAgE

—

XTR22

—

XER9

—

XER11

—

REX12

—

JSR29

JVR22

—

JWR22H

—

AIR2

—

ICL15

—

ICL15CsBF

—

MH21

% Discord

Source: Reprinted with permission from S.D. Carson, et al., "Tissue Factor Gene Localized to Human Chromosome 1 (after lp21)," Science, 229:229-291.

* A translocation in which only part of the chromosome is present.

+ Discord refers to cases in which the tissue factor score is plus, and the human chromosome is absent, or in which the score is minus and the chromosome is present.

After successful cell hybrids are formed, two particu-
lar tests are used to map human genes. A synteny test
(same linkage group) determines whether two loci are in
the same linkage group if the phenotypes of the two loci
are either always together or always absent in various hy-
brid cell lines. An assignment test determines which
chromosome a particular locus is on by the concordant

appearance of the phenotype whenever that particular
chromosome is in a cell line, or by the lack of the particu-
lar phenotype when a particular chromosome is absent
from a cell line. The first autosomal synteny test, per-
formed in 1970, demonstrated that the B locus of lactate
dehydrogenase (LDHb) was linked to the B locus of pep-
tidase (PEPb). (Both enzymes are formed from subunits

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Human Chromosomal Maps

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controlled by two loci each. In addition to the B locus,
each protein has subunits controlled by an A locus.) Later,
these loci were shown to reside on chromosome 12.

In another example, a blood-coagulating glycoprotein
(a protein-polysaccharide complex) called tissue factor
III was localized by assignment tests to chromosome 1 .
Table 6.8 shows twenty-nine human-mouse hybrid cell
lines, or clones, the human chromosomes they contain,
and their tissue factor score, the results of an assay for the
presence of the coagulating factor. (Clones are cells aris-
ing from a single ancestor.) It is obvious from table 6.8
that the gene for tissue factor III is on human chromo-
some 1 . Every time human chromosome 1 is present in a
cell line, so is tissue factor III. Every time human chro-
mosome 1 is absent, so is the tissue factor (zero discor-
dance or 100% concordance). No other chromosome
showed that pattern.

The human map as we know it now (compiled by Vic-
tor McKusick at Johns Hopkins University), containing
over six thousand assigned loci of over twelve thousand
known to exist, is shown in table 6.9 and figure 6.29. At

Victor A. McKusick
(1921- ). (Courtesy
of Victor A. McKusick.)

Table 6.9 Definition of Selected Loci of the Human Chromosome Map (figure 6.29)

Locus

Protein Product Chromosome

ABO

ABO blood group

Alpha globin gene family

ALB

Albumin

AMY1

Amylase, salivary

AMY2

Amylase, pancreatic

BCS

Breast cancer susceptibility

Complement component-2

CAT

Catalase

CBD

Color blindness, deutan

CBP

Color blindness, protan

CML

Chronic myeloid leukemia

DMD

Duchenne muscular dystrophy

FES

Feline sarcoma virus oncogene

Duffy blood group

GLB1

Beta-galactosidase- 1

Histone-1

HBB

Hemoglobin beta chain

HEMA

Classic hemophilia

HEXA

Hexosaminidase A

HLA

Human leukocyte antigens

Haptoglobin

HYA

Y histocompatibility antigen, locus A

IDDM

Insulin-dependent diabetes mellitus

IFF

Interferon, fibroblast

Locus Protein Product

Chromosome

IGH

Immunoglobulin heavy-chain gene
family

IGK

Immunoglobulin kappa-chain gene
family

INS

Insulin

LDHA

Lactate dehydrogenase A

MDI

Manic depressive illness

MHC

Major histocompatibility complex

MN blood group

MYB

Avian myeloblastosis virus oncogene

NHCP1

Nonhistone chromosomal protein- 1

NPS1

Nail-patella syndrome

PEPA

Peptidase A

PVS

Polio virus sensitivity

Rhesus blood group

RN5S

5S RNA gene(s)

RNTMI

Initiator methionine tRNA

RWS

Ragweed sensitivity

Surface antigen 1

SIS

Simian sarcoma virus oncogene

STA

Stature

Transferrin

Xg blood group

XRS

X-ray sensitivity

Note: A more complete list appears in V. A. McKusick, Mendelian Inheritance in Man: A Catalog of Human Genes (Baltimore: Johns Hopkins University Press,
1994).

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AMY1 AMY2

^NGFB TSHB

CTSE
LAMC1

_ RCAC

REN

C4BPA

C4BPB

CR1

CR2

DAF

HF1

ACP1
POMC

APOB MYCN
MDH1

TGFA

IGK-

CD8A
CD8B

ILRA
PROC

— LCT

V . (many)
J I (>4)

— SCN1A SCN2A SCN3A

COL3A1 COL5A2
-NEB

CRYG

HGCG
^ALPI ALPP

RAF1
THRB

ACY1
PTPRG

RHO

FIM3
SST

pter

25 |

2 22

1 12

IHF13A1 ME2

MHC
GSTA2

ME1 PGM3

COL9A1

CNR

— MYB

D-ARG1
ESR

rSOD2

^PLG
^VIP

HLA-A

HLA-B

HLA-C

C4A

C4B

CYP21

HLA-DR

HLA-DX/DQ

HLA-DO

HLA-DP/DZ

centromere

PDGFA
IL6

TCRG HOXA
EGFR

COL1A2 GUSB PGY1
EPO

— PLANH1

CFTR

MET
CPA1 TRY1
KEL
TCRB
NOS

23
2 22

1 11

>^DEF1
1 — CTSB LPL
rGSR

1 12

2 22

— CRH

CYP11B1 CYP11B2

CA1
-CACHCA2

CA3

MYC

>TG

^ATP1AL1
-ESD

q 2

31
3 32

^, /

COL4A1
-ICOL4A2
LAMP1

TCRA

MYH6
MYH7

centromere
3' a CA2

CG4

CG2

CA1

CEP1

CG1

CG3

J H (<4)

D (many)
5' t V H (-250)
qter

1 -1

21
22

r GABRA5 MANA1
-EPB42
-FBN1

LI PC

^PKM2
^B2M
-MPI
-IDH2
-FES
ANPEP
IGF1R

P 1

q 1 13

EF2

FUT1 FUT3

MANB

LU FUT2

APOE
APOLP2 ^APOC1
FTL APOC2

CGB

LHB

PRNP
CHGB

CST3

1 12 =
11 "

21
O

_ r

"22

SOD1

BCEI

COL6A1

COL6A2

CRYA1

ETS2

Figure 6.29 Human G-banded chromosomes with their accompanying assigned loci. The p and q refer to the short
and long arms of the chromosomes, respectively. A key to the loci is given in McKusick (1994). (From Victor A. McKusick,
Mendelian inheritance in man, 11th edition, 1994. Reprinted by permission of Johns Hopkins University Press, Baltimore, MD.)

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Human Chromosomal Maps

139

THE HUMAN GENE MAP
(selected 'anchor' loci)

A confirmed assignment EN01

A provisional assignment DHPR

Gene cluster. MHC

100

150

SCALE
(in megabases)

— ADD1 PDEB

— QDPR

~\- GABRA2 GABRB1

PEPS

-AFP ALB r-

ADH1

-ADHC1-^3
^EGF ADH4

-IL2
FGC

FGA
FGB
FGG
GYPA GYPB

GRIA2

P 1

1 13

3 33~

DAT1 NHE3
SRD5A1

HMGCS1
\—PRLR
-TARS

-LARS
-ARSB DHFR

HEXB
HMGCR

CAMK4

DTS

-CD14 CSF2 IL3 IL4 IL5 IL9
PDGFRB
CSF2 EGR1
CSF1R
DRD1

ANT1 F11 KLK3

EAAC1 OVC
IFNBI
IFN1
AC01

]—ALDH1

ALDOB
TMOD

AK1 ABL1 C5
ABO DBH GSN
ORM1 ORM2

-IL2RA
VIM

-ADK

THK1
^ r GLUD1
L DNTT

-D

^PLAU GOT1
PAX2 VAT2

ACP2

GSTP1

PGA-

PGR

APOLP1

ETS1

PGA2
PGA4
PGA5

APOA1
APOC3
APOA4

i — CD4
GAPD
PRB

J^^LDHB
L-KRAS2

n WNT1

>{HOXC
SHMT2

^_y

■PEPB

-IFNG
-NOS1 TCF1

1 12

1 11

2 231

24 1

■

PGP

~PRKCB1
^PN
CETP
MT1
MT2
-DIA4 HP
— CTRB

MYH1
TP53

MYH1
MYH2
MYH3
MYH4
MYH8

HOX2 -

NGFR

TK1

CACNLG SSTR2

P4HB

1 11

11 P

12 |

2 22 |

-YES1

\—JK

-GRP
KMBP
-PEPA

/ \

centromere
PVALB 5' a IGLV (many)

^vY IGLC-' -9 J " C du P lexes

>PDGFB
— MB

qter

P 1 11 nT M|C2Y

n TDF

-XG

h— POLA

— ARAF1
- — MAOA MAOB

Kar

-PGK1

^PLP
PRPS1

HPRT

FRAXA

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Chromosomal Theory

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140

Chapter Six Linkage and Mapping in Eukaryotes

present, geneticists studying human chromosomes are
hampered not by a lack of techniques but by a lack of
marker loci. When a new locus is discovered, it is now
relatively easy to assign it to its proper chromosome.

The problem still exists of determining exactly where
a particular locus belongs on a chromosome. This is facil-
itated by developing particular cell lines with broken
chromosomes, so that parts are either missing or have

moved to other chromosomes. These processes reveal
new linkage arrangements and make it possible to deter-
mine the region in which a locus is situated on a particu-
lar chromosome. In chapter 13, we describe additional
techniques used to locate genes on human chromosomes,
including a description of the Human Genome Project,
the program that sequenced the entire human genome as
well as the genomes of other model organisms.

SUMMARY

STUDY OBJECTIVE 1: To learn about analytical tech-
niques for locating the relative positions of genes on
chromosomes in diploid eukaryotic organisms
110-122

The principle of independent assortment is violated when
loci lie near each other on the same chromosome. Recom-
bination between these loci results from the crossing over
of chromosomes during meiosis. The amount of recombi-
nation provides a measure of the distance between these
loci. One map unit (centimorgan) equals 1% recombinant
gametes. Map units can be determined by testcrossing a di-
hybrid and recording the percentage of recombinant
offspring. If three loci are used (a three-point cross), double
crossovers will be revealed. A coefficient of coincidence,
the ratio of observed to expected double crossovers, can be
calculated to determine if one crossover changes the prob-
ability that a second one will occur nearby.

A chiasma seen during prophase I of meiosis represents
both a physical and a genetic crossing over. This can be
demonstrated by using homologous chromosomes with
morphological distinctions.

Because of multiple crossovers, the measured percent-
age recombination underestimates the true map distance,
especially for loci relatively far apart; the best map esti-
mates come from summing the distances between closely
linked loci. A mapping function can be used to translate ob-
served map distances into more accurate ones.

STUDY OBJECTIVE 2: To learn about analytical tech-
niques for locating the relative positions of genes on
chromosomes in ascomycete fungi 122-132

Organisms that retain all the products of meiosis lend them-
selves to chromosome mapping by haploid mapping
(tetrad analysis). With unordered spores, such as in yeast,
we use

map units =
(1/2) the number of TT asci + the number of NPD asci
total number or asci

X 100

Map units between a locus and its centromere in organ-
isms with ordered spores, such as Neurospora, can be cal-
culated as

(1/2) the number of SDS asci

map units = X 100

total number of asci

Crossing over also occurs during mitosis, but at a much re-
duced rate. Somatic (mitotic) crossing over can be used to
map loci.

STUDY OBJECTIVE 3: To learn about analytical tech-
niques for locating the relative positions of genes on hu-
man chromosomes 132-140

Human chromosomes can be mapped. Recombination dis-
tances can be established by pedigrees, and loci can be at-
tributed to specific chromosomes by synteny and assign-
ment tests in hybrid cell lines.

SOLVED PROBLEMS

PROBLEM 1: A homozygous claret (ca, claret eye color),
curled (cu, upcurved wings), fluted {fl, creased wings)
fruit fly is crossed with a pure-breeding wild-type fly The
F x females are testcrossed with the following results:

fluted 4

claret 173

curled

fluted, claret

fluted, curled

167

claret, curled

fluted, claret, curled

298

wild-type

302

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Chromosomal Theory

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Solved Problems

141

a. Are the loci linked?

b. If so, give the gene order, map distances, and co-
efficient of coincidence.

Answer: The pattern of numbers among the eight
offspring classes is the pattern we are used to seeing for
linkage of three loci. We can tell from the two groups in
largest numbers (the nonrecombinants — fluted, claret,
curled and wild-type) that the alleles are in the coupling
(cis) arrangement. If we compare either of the
nonrecombinant classes with either of the double
crossover classes (fluted and claret, curled), we see that
the fluted locus is in the center. For example, compare
fluted, a double crossover offspring, with the wild-type, a
nonrecombinant; clearly, fluted has the odd pattern. Thus
the trihybrid female parent had the following arrange-
ment of alleles:

cafl cu

ca + fl + cu +

A crossover in the ca-fl region produces claret and
fluted, curled offspring, and a crossover in the fl-cu re-
gion produces fluted, claret and curled offspring. Count-
ing the crossovers in each region, including the double
crossovers in each, and then converting to percentages,
yields a claret-to-fluted distance of 35.0 map units (173 +
167 + 6 + 4) and a fluted-to-curled distance of 6.0 map
units (26 + 24 + 6 + 4). We expect 0.35 X 0.06 X 1,000
= 21 double crossovers, but we observed only 6 + 4 =
10. Thus, the coefficient of coincidence is 10/21 = 0.48.

PROBLEM 2: The ad5 locus in Neurospora is a gene for
an enzyme in the synthesis pathway for the DNA base
adenine. A wild-type strain (ad5 + ) is crossed with an
adenine-requiring strain, ad5 . The diploid undergoes
meiosis, and one hundred asci are scored for their segre-
gation patterns with the following results:

ad5 + ad5 + ad5 + ad5 + ad5 ad5 ad5 ad5

ad^ ad^ ad^ ad5 ad5 + ad^ ad5 + ad5

ad^ ad5 + ad^ ad5 ad5 ad^ ad5 + ad5

ad5 ad5 ad5 + ad5 + ad5 + ad5 + ad*) ad5

ad5 ad5 ad5 + ad5 + ad5 ad*) ad5 + ad5

ad5 ad5 ad5 ad5 ad5 ad5 ad5 ad5

40
46

5
3

4
2

What can you say about the linkage arrangements at this
locus?

Answer: You can see that 14 (5 + 3 + 4 + 2) asci are of
the second-division segregation type (SDS) and 86 (40 +
46) are of the first-division segregation type (FDS). To
map the distance of the locus to its centromere, we di-
vide the percentage of SDS types by 2: 14/100 = 14%; di-
vided by 2 is 7%. Thus, the ad5 locus is 7 map units from
its centromere.

PROBLEM 3: In yeast, the his 5 locus is a gene for an en-
zyme in the synthesis pathway for the amino acid histi-
dine, and the lysll locus is a gene for an enzyme in the
synthesis pathway for the amino acid lysine. A haploid
wild-type strain (his5 + lysll+) is crossed with the dou-
ble mutant (his5~ lysll). The diploid is allowed to un-
dergo meiosis, and 100 asci are scored with the following
results:

his5
his5
his5
his5

lysll
lysll +
lysll~
lysir

his5
his5
his5
his5

lysll
lysll
lysll
lysll

his5
his5
his5
his5 lysll
8

lysll~
lysll'
lysll +

What is the linkage arrangement of these loci?

Answer: Of the 100 asci analyzed, 62 were parental
ditypes (PD), 30 were tetratypes (TT), and 8 were
nonparental ditypes (NPD). To map the distance
between the two loci, we take the percentage of NPD
(8%) plus half the percentage of TT (1/2 of 30 = 15%) =
23% or 23 centimorgans between loci.

PROBLEM 4: A particular human enzyme is present only
in clone B.The human chromosomes present in clones A,
B, and C appear as pluses in the following table. Deter-
mine the probable chromosomal location of the gene for
the enzyme.

Human Chromosome

Clone

—

Answer: If a gene is located on a chromosome, the gene
must be present in the clones with the chromosome (+).
Chromosomes 1, 2, 5, 6 are present in B. If the gene in
question were located on chromosome 1, the enzyme
should have been present in all three clones. A similar ar-
gument holds for chromosome 2, in which the enzyme
should have been present in clones A and B, and so on for
the rest of the chromosomes. The only chromosome that
is unique to clone B is 6. Therefore, the gene is located on
chromosome 6.

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142

Chapter Six Linkage and Mapping in Eukaryotes

EXERCISES AND PROBLEMS

DIPLOID MAPPING

1. A homozygous groucho fly (gro, bristles clumped
above the eyes) is crossed with a homozygous rough
fly (ro, eye abnormality). The ¥ 1 females are
testcrossed, producing these offspring:

groucho

518

rough

471

groucho, rough

wild-type

1,000

a. What is the linkage arrangement of these loci?

b. What offspring would result if the V 1 dihybrids
were crossed among themselves instead of being
testcrossed?

2. A female fruit fly with abnormal eyes (abe) of a
brown color (bis, bistre) is crossed with a wild-type
male. Her sons have abnormal, brown eyes; her
daughters are of the wild-type. When these F : flies
are crossed among themselves, the following off-
spring are produced:

Sons

Daughters

abnormal, brown

219

197

abnormal

brown

wild-type

201

223

What is the linkage arrangement of these loci?

3. In Drosophila, the loci inflated (if, small, inflated
wings) and warty (wa, abnormal eyes) are about 10
map units apart on the X chromosome. Construct a
data set that would allow you to determine this link-
age arrangement. What differences would be in-
volved if the loci were located on an autosome?

4. A geneticist crossed female fruit flies that were het-
erozygous at three electrophoretic loci, each with
fast and slow alleles, with males homozygous for the
slow alleles. The three loci were gotl (glutamate ox-
aloacetate transaminase- 1), amy (alpha-amylase),
and sdh (succinate dehydrogenase). The first 1,000
offspring isolated had the following genotypes:

Class 1
Class 2
Class 3
Class 4
Class 5
Class 6
Class 7
Class 8

got s got s amy s amy s sdh s sdh
got { got s amy f amy s sdh f sdh s
got { got s amy s amy s sdh s sdh f
got s got s amy f amy s sdh f sdh 5
got { got s amy f amy s sdh s sdh s
got s got s amy s amy s sdh f sdh f
got { got s amy s amy s sdh f sdh 5
got s got s amy f amy s sdh s sdh f

441

421

58
53

1
1

What are the linkage arrangements of these three
loci, including map units? If the three loci are
linked, what is the coefficient of coincidence?

5. The following three recessive markers are known in
lab mice: h, hotfoot; o, obese; and wa, waved. A tri-
hybrid of unknown origin is testcrossed, producing
the following offspring:

hotfoot, obese, waved

357

hotfoot, obese

waved

obese

wild-type

343

hotfoot, waved

obese, waved

hotfoot

1,000

a. If the genes are linked, determine the relative or-
der and the map distances between them.

b. What was the cis-trans allele arrangement in the
trihybrid parent?

c. Is there any crossover interference? If yes, how
much?

6. The following three recessive genes are found in
corn: btl, brittle endosperm; gll 7, glossy leaf; rgdl,
ragged seedling. A trihybrid of unknown origin is
testcrossed, producing the following offspring:

brittle, glossy, ragged 236

brittle, glossy 241

ragged 219

glossy 23

wild-type 224

brittle, ragged 17

glossy, ragged 2 1

brittle 19

1,000

a. If the genes are linked, determine the relative or-
der and map distances.

b. Reconstruct the chromosomes of the trihybrid.

c. Is there any crossover interference? If yes, how
much?

7. In Drosophila, ancon (an, legs and wings short),
spiny legs (sple, irregular leg hairs), and arctus ocu-
lus (at, small narrow eyes) have the following link-
age arrangement on chromosome 3:

an sple at

10.0

6.1

* Answers to selected exercises and problems are on page A-6.

Tamarin: Principles of
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Chromosomal Theory

6. Linkage and Mapping in
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Exercises and Problems

143

a. Devise a data set with no crossover interference
that would yield these map units.

b. What data would yield the same map units but
with a coefficient of coincidence of 0.60?

8. Ancon (an) and spiny legs (sple), from problem 7,
are 10 map units apart on chromosome 3. Notchy
(ny, wing tips nicked) is on the X chromosome
(chromosome 1). Create a data set that would result
if you were making crosses to determine the linkage
arrangement of these three loci. How would you
know that the notchy locus is on the X chromo-
some?

9. In the house mouse, the autosomal alleles Trembling
and Rex (short hair) are dominant to not trembling
(normal) and long hair, respectively. Heterozygous
Trembling, Rex females were crossed with normal,
long-haired males and yielded the following off-
spring:

Trembling, Rex 42

Trembling , long-haired 105

normal, Rex 109

normal, long-haired 44

a. Are the two genes linked? How do you know?

b. In the heterozygous females, were Trembling and
Rex in cis or trans position? Explain.

c. Calculate the percent recombination between
the two genes.

10. In corn, a trihybrid Tunicate (T), Glossy (G), Liguled
(Z) plant was crossed with a nontunicate, nonglossy,
liguleless plant, producing the following offspring:

Tunicate, liguleless, Glossy 58

Tunicate, liguleless, nonglossy 15

Tunicate, Liguled, Glossy 55

Tunicate, Liguled, nonglossy 13

nontunicate, Liguled, Glossy 16

nontunicate, Liguled, nonglossy 53

nontunicate, liguleless, Glossy 14

nontunicate, liguleless, nonglossy 59

a. Determine which genes are linked.

b. Determine the genotype of the heterozygote; be
sure to indicate which alleles are on which chro-
mosome.

c. Calculate the map distances between the linked
genes.

11. In Drosophila, kidney-shaped eye (k), cardinal eye
(cd), and ebony body (e) are three recessive genes. If
homozygous kidney, cardinal females are crossed
with homozygous ebony males, the ¥ 1 offspring are
all wild-type. If heterozygous F : females are mated
with kidney, cardinal, ebony males, the following
2,000 progeny appear:

880 kidney, cardinal

887 ebony

64 kidney, ebony

67 cardinal

49 kidney

46 ebony, cardinal

3 kidney, ebony, cardinal

4 wild-type

a. Determine the chromosomal composition of the
V 1 females.

b. Derive a map of the three genes.

12. Following is a partial map of the third chromosome
in Drosophila.

19.2 javelin bristles (jv)
43.2 thread arista (th)
66.2 Delta veins (D/)
70.7 ebony body (e)

a. If flies heterozygous in cis position for javelin
and ebony are mated among themselves, what
phenotypic ratio do you expect in the progeny?

b. A true-breeding thread, ebony fly is crossed with
a true-breeding Delta fly. An ¥ 1 female is test-
crossed to a thread, ebony male. Predict the ex-
pected progeny and their frequencies for this
cross. Assume no interference.

c. Repeat b, but assume a coefficient of coinci-
dence of 0.4.

13. Suppose that you have determined the order of
three genes to be a, c, b, and that by doing two-point
crosses you have determined map distances as a-c
= 10 and c-b = 5. If interference is -1.5, and the
three-point cross is

ACB acb

acb acb

what frequency of double crossovers do you expect?

HAPLOID MAPPING (TETRAD ANALYSIS)

14. Given the following cross in Neurospora: ab X
a + b + , construct results showing that crossing over
occurs in two of the four chromatids of a tetrad at
meiosis. What would the results be if crossing over
occurred during interphase before each chromo-
some became two chromatids? if each crossover
event involved three or four chromatids?

15. A strain of yeast requiring both tyrosine (tyr~) and
arginine (arg~) is crossed to the wild-type. After
meiosis, the following ten asci are dissected. Classify
each ascus as to segregational type (PD, NPD, TT).
What is the linkage relationship between these two
loci?

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Chapter Six Linkage and Mapping in Eukaryotes

2
3

arg~ tyr~
arg + tyr +
arg~ tyr +

arg + tyr +
arg + tyr +
arg~ tyr +

arg + tyr +
arg~ tyr~
arg + tyr~

arg" tyr
arg~ tyr
arg + tyr

arg~ tyr~

arg + tyr +

arg + tyr

5
6

7
8

arg~ tyr~
arg + tyr +
arg~ tyr~
arg + tyr +

arg~ tyr +
arg + tyr +
arg + tyr +
arg + tyr +

arg + tyr~
arg~ tyr~
arg~ tyr +
arg~ tyr~

arg + tyr
arg~ tyr
arg + tyr
arg" tyr

9
10

arg + tyr +
arg~ tyr~

arg~ tyr~
arg + tyr +

arg + tyr
arg~ tyr

16. A certain haploid strain of yeast was deficient for the
synthesis of the amino acids tryptophan (try - ) and
methionine (met~). It was crossed to the wild-type,
and meiosis occurred. One dozen asci were analyzed
for their tryptophan and methionine requirements.
The following results, with the inevitable lost
spores, were obtained:

try

met

■?

try met

try~

met~

try +

met +

try + met

try"

met +

try~

met~

try +

met~

try + met

try"

met~

try +

met +

■?

try + met

try~

met +

try + met

try

met +

try

met +

try~

met~

try~ met

try +

met +

try +

met~

■?

try~ met

try

met +

try~

met~

■?

try met

try~

met +

try +

met~

try"

met +

try + met

try~

met~

try +

met +

try~

met~

try + met

try

met +

try

met +

■?

try +

met~

try~ met

a. Classify each ascus as to segregational type (note
that some asci may not be classifiable).

b. Are the genes linked?

c. If so, how far apart are they?

17. In Neurospora, a haploid strain requiring arginine
(arg~) is crossed with the wild-type (arg + ). Meiosis
occurs, and ten asci are dissected with the following
results. Map the arg locus.

arg
arg
arg
arg

arg
arg
arg

5
6
7
8
9
10

arg
arg
arg
arg
arg
arg

arg
arg
arg
arg
arg
arg
arg

arg
arg
arg
arg

arg
arg
arg
arg
arg
arg

18. A haploid strain of Neurospora with fuzzy colony
morphology (/) was crossed with the wild-type
(/ + ). Twelve asci were scored. The following results,
with the inevitable lost spores were obtained:

/ +

■?

/ +

a. Classify each ascus as to segregational type, and
note which asci cannot be classified.

b. Map the chromosome containing the / locus
with all the relevant measurements.

19. Draw ten of the remaining twenty-eight ascus pat-
terns not included in table 6.6. To which of the seven
major categories of table 6.7 does each belong?

20. In yeast, the a and b loci are 12 map units apart. Con-
struct a data set to demonstrate this.

21. In Neurospora, the a locus is 12 map units from its
centromere. Construct a data set to show this.

22. An ab Neurospora was crossed with ana + & + form.
Meiosis occurred, and 1,000 asci were dissected. Us-
ing the classes of table 6.7, the following data re-
sulted:

Class 1

700

Class 5

Class 2

Class 6

Class 3

190

Class 7

Class 4

What is the linkage arrangement of these loci?

23. Given the following linkage arrangement in Neu-
rospora, construct a data set similar to that in table
6.7 that is consistent with it (cm is centromere).

-e-

-h

15 15

24. Determine crossover events that led to each of the
seven classes in table 6.7.

25. In Neurospora, a cross is made between ab + and
a + b individuals. The following one hundred ordered
tetrads are obtained:

Spores I

III

VII VIII

1,2

a + b

a + b +

a + b

ab +

3,4

a + b

a b

a + b

ab +

a + b

5,6

ab +

a + b

ab +

a + b

7,8

ab +

a + b

ab +

a. Are genes a and b linked? How do you know?

b. Calculate the gene-to-centromere distances for a
and b.

Tamarin: Principles of
Genetics, Seventh Edition

II. Mendelism and the
Chromosomal Theory

6. Linkage and Mapping in
Eukaryotes

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Exercises and Problems

145

26. Neurospora has four genes — a, b, c, and d — that
control four different phenotypes. Your job is to map
these genes by performing pairwise crosses. You ob-
tain the following ordered tetrads:

ab + Xa + b

bc + Xb + c

Spores

III

Spores

III

1,2

ab +

1,2

bc +

b + c +

b + c

3,4

ab +

a b

3,4

bc +

b c

5,6

a + b

a + b +

a + b

5,6

b + c

7,8

a + b

a b

7,8

b + c

bc +

+ Xc + d

Spores

III

VII

1,2

cd +

3,4

cd +

c + d

c + d +

c + d

5,6

c + d

c + d + c + d +

c + d +

c + d

c + d +

7,8

c + d

c + d + c + d

cd +

a. Calculate the gene-to-centromere distances.

b. Which genes are linked? Explain.

c. Derive a complete map for all four genes.

27. You have isolated a new fungus and have obtained a
strain that requires both arginine (arg~} and adenine
(ad~). You cross these two strains and collect four
hundred random spores that you plate on minimal
medium. If twenty-five spores grow, what is the dis-
tance between these two genes?

28. Three distinct genes,pab, pk, and ad, were scored in a
cross of Neurospora. From the cross pabpk + ad + X
pab + pk ad, the following ordered tetrads were re-
covered:

Spores

III

VII

VIII

1,2

pab pk + ad +

pab pk + ad

3,4

pab pk + ad +

pab + pk ad

pab pk ad

pab pk + ad

pab + pk ad

pab + pk ad +

5,6

pab + pk ad

pab pk + ad +

pab + pk + ad +

pab + pk ad +

pab pk ad

pab pk + ad

pab + pk ad +

pab pk + ad +

7,8

pab + pk ad

pab+pk ad

pab + pk ad

pab + pk + ad +

pab + pk ad +

pabpk + ad +

pab+pk ad

Based on the data, construct a map of the three
genes. Be sure to indicate centromeres.

HUMAN CHROMOSOMAL MAPS

29. The Duffy blood group with alleles FY* and FY h was
localized to chromosome 1 in human beings when
an "uncoiled" chromosome was associated with it.
Construct a pedigree that would verify this.

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Chromosomal Theory

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146

Chapter Six Linkage and Mapping in Eukaryotes

30. What pattern of scores would you expect to get, us-
ing the hybrid clones in table 6.8, for a locus on hu-
man chromosome 6? 14? X?

31. A man with X-linked color blindness and X-linked
Fabry disease (alpha-galactosidase-A deficiency)
mates with a normal woman and has a normal daugh-
ter. This daughter then mates with a normal man and
produces ten sons (as well as eight normal daugh-
ters). Of the sons, five are normal, three are like their
grandfather, one is only color-blind, and one has
Fabry disease. From these data, what can you say
about the relationship of these two X-linked loci?

32. In people, the ABO system (I A , I 3 , i alleles) is linked
to the aldolase-B locus (ALDOE), a gene that func-
tions in the liver. Deficiency, which is recessive, re-
sults in fructose intolerance. A man with blood type
AB has a fructose-intolerant, type B father and a nor-
mal, type AB mother. He and a woman with blood
type O and fructose intolerance have ten children.
Five are type A and normal, three are fructose intol-
erant and type B, and two are type A and intolerant
to fructose. Draw a pedigree of this family and deter-
mine the map distances involved. (Calculate a lod
score to determine the most likely recombination
frequency between the loci.)

33. Hemophilia and color-blindness are X-linked reces-
sive traits. A normal woman whose mother was
color-blind and whose father was a hemophiliac
mates with a normal man whose father was color-
blind. They have the following children:

4 normal daughters

1 normal son

2 color-blind sons

2 hemophiliac sons

1 color-blind, hemophiliac son

Estimate the distance between the two genes.

34. The results of an analysis of five human-mouse hy-
brids for five enzymes are given in table along with
the human chromosomal content of each clone
(+ = enzyme or chromosome present; — = absent).
Deduce which chromosome carries which gene.

Clone

Human Enzyme

glutathione reductase

—

malate dehydrogenase

—

adenosine deaminase

—

galactokinase

—

hexosaminidase

—

Human Chromosome

6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

Clone A

—

Clone B

—

Clone C

—

Clone D

—

Clone E

—

+ + +

—

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Critical Thinking Questions

147

35. You have selected three mouse-human hybrid clones
and analyzed them for the presence of human chro-
mosomes. You then analyze each clone for the pres-
ence or absence of particular human enzymes (+ =
presence of human chromosome or enzyme activ-
ity). Based on the following results indicate the prob-
able chromosomal location for each enzyme.

Human Chromosomes

Clone

—

Enzyme

Clone

—

36. Three mouse-human cell lines were scored for the
presence (+) or absence (— ) of human chromo-
somes, with the results as follows:

Human Chromosomes

Clone

—

+
+

If a particular gene is located on chromosome 3,
which clones should be positive for the enzyme
from that gene?

CRITICAL THINKING QUESTIONS

Do three-point crosses in fruit flies capture all the mul-
tiple crossovers in a region?

If 4% of all tetrads have a single crossover between
two loci: (a) What is the map distance between these

loci if these are fruit flies? (b) What is the proportion
of second-division segregants if these are Neurospora?
(c) What is the proportion of nonparental ditypes if
these are yeast?

Suggested Readings for chapter 6 are on page B-3.

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LINKAGE AND

MAPPING IN

PROKARYOTES

AND BACTERIAL

STUDY OBJECTIVES

1. To define bacteria and bacterial viruses and learn about
methods of studying them 149

2. To study life cycles and sexual processes in bacteria and
bacteriophages 154, 163

3. To make use of the sexual processes of bacteria and their
viruses to map their chromosomes 155, 166

VIRIISFS STUDY OUTLINE

Scanning electron micrograph (color enhanced) of an

Escherichia co// bacterium with adsorbed T-family

bacteriophages (36,000x). (© Oliver Meckes/MPI-

Tubingen/Photo Researchers.)

Bacteria and Bacterial Viruses in Genetic Research 149
Techniques of Cultivation 1 50
Bacterial Phenotypes 151

Colony Morphology 151

Nutritional Requirements 151

Resistance and Sensitivity 153
Viral Phenotypes 154
Sexual Processes in Bacteria and Bacteriophages 1 54

Transformation 154

Conjugation 157
Life Cycles of Bacteriophages 163

Recombination 163

Lysogeny 165
Transduction 165

Specialized Transduction 165

Generalized Transduction 166

Mapping with Transduction 1 66
Summary 171
Solved Problems 172
Exercises and Problems 172
Critical Thinking Questions 176

148

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Bacteria and Bacterial Viruses in Genetic Research

149

All organisms and viruses have genes located
sequentially in their genetic material; and al-
most all can undergo recombination be-
tween homologous (equivalent) pieces of ge-
netic material. Because recombination can
occur, it is possible to map, by analytical methods, the lo-
cations and sequence of genes along the chromosomes
of all organisms and almost all viruses. In this chapter, the
viruses we look at are those that attack bacteria. Through
work with bacteria and viruses, we have entered the
modern era of molecular genetics, the subject of the next
section of this book.

Bacteria (including the cyanobacteria, the blue-green
algae) are prokaryotes. The prokaryotes also include the
archaea, or archaebacteria, a kingdom recognized in
1980. These highly specialized organisms (previously
classified as bacteria), along with the bacteria and eu-
karyotes, make up the three domains of life on Earth.

The true bacteria can be classified according to shape:
a spherical bacterium is called a coccus; a rod-shaped
bacterium is called a bacillus; and a spiral bacterium is
called a spirillum. Prokaryotes do not undergo mitosis
or meiosis but simply divide in two after their chromo-
some (usually only one), most often a circle of DNA, has
replicated (see chapter 9). Bacterial viruses do not even
divide; they are mass-produced within a host cell.

BACTERIA AND BACTERIAL
VIRUSES IN GENETIC
RESEARCH

Several properties of bacteria and viruses make them es-
pecially suitable for genetic research. First, bacteria and
their viruses generally have a short generation time.
Some viruses increase three-hundredfold in about a half
hour; an Escherichia coli cell divides every twenty min-
utes. In contrast, generation time is fourteen days in fruit
flies, a year in corn, and twenty years or so in human be-
ings. (E. coli, the common intestinal bacterium, was dis-
covered byTheodor Escherich in 1885.)

Another reason bacteria and bacterial viruses are so
well-suited for genetic research is because they have
much less genetic material than eukaryotes do, and the or-
ganization of this material is much simpler. The term
prokaryote arises from the lack of a true nucleus (pro
means before and karyon means kernel or nucleus); they
have no nuclear membranes (see fig. 3.2) and usually have
only a single, relatively "naked" chromosome, so they are
haploid. Bacteria may, however, contain small, auxiliary
circles of DNA, called plasmids. Bacterial viruses are
even simpler. Although animal and plant viruses, dis-
cussed in more detail later in the book (chapters 13 and
16), can be more complicated, the viruses we are inter-

ested in studying in this chapter — the bacterial viruses,
bacteriophages, or just phages (Greek: eating) — are ex-
clusively genetic material surrounded by a protein coat.

Bacteriophages are usually classified first by the type
of genetic material (nucleic acid) they have (DNA or
RNA, single- or double-stranded), then by structural fea-
tures of their protein surfaces (capsids) such as type or
symmetry and number of discrete protein subunits (cap-
someres) in the capsid, and general size. Most bacterio-
phages are complex, like T2 (fig. 7.1), or made up of a
headlike capsule like T2 without the tail appendages, or
filamentous. Most contain double-stranded DNA. Bacte-
riophages are obligate parasites; outside of a host, they
are inert molecules. Once their genetic material pene-
trates a host cell, they can take over the metabolism of
that cell and construct multiple copies of themselves.
We will discuss details of this and alternative infection

(a)

Head (contains DNA)

Collar
Tail core

Tail sheath
Base plate

Tail fiber -

(b)

Figure 7.1 Phage T2 and its chromosome, (a) The
chromosome, which is about 50 iaiti long, has burst from the
head, (b) The intact phage. The phage attaches to a bacterium
using its tail fibers and base plate and then injects its genetic

material into the host cell, ([a] A. K. Kleinschmidt, et al., "Darstellung
und Langen messungen des gesamten Deoxyribose-nucleinsaure Inhaltes
von T2-Bacteriophagen" Biochemica et Biophysica Acta, 61:857-64, 1962.
Reproduced by permission of Elsevier Science Publishers.)

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

pathways later in the chapter. The smallest bacterio-
phages (e.g., R17) have RNA as their genetic material and
contain just three genes, one each for a coat protein, an
attachment protein, and an enzyme to replicate their
RNA. The larger bacteriophages (T2, T4) have DNA as
their genetic material and contain up to 130 genes.

A third reason for the use of bacteria and viruses in
genetic study is their ease of handling. A researcher can
handle millions of bacteria in a single culture with a min-
imal amount of work compared with the effort required
to grow the same number of eukaryotic organisms such
as fruit flies or corn. (Some eukaryotes, such as yeast or
Neurospora, can, of course, be handled using prokary-
otic techniques, as we saw in chapter 6.) Let us look at an
expansion of the techniques, introduced in chapter 6,
that geneticists use in bacterial and viral studies.

TECHNIQUES

OF CULTIVATION

All organisms need an energy source, a carbon source,
nitrogen, sulfur, phosphorus, several metallic ions, and
water. Those that require an organic form of carbon are
termed heterotrophs. Those that can utilize carbon as
carbon dioxide are termed autotrophs. All bacteria ob-
tain their energy either by photosynthesis or chemical
oxidation. Bacteria are usually grown in or on a chemi-
cally defined synthetic medium, either in liquid in
flasks or test tubes, or on petri plates using an agar base
to supply rigidity. When one cell is placed on the medium
in the plate, it will begin to divide. After incubation, often
overnight, a colony, or clone, will exist where previously
was only one cell. Overlapping colonies form a confluent
growth (fig. 7.2). A culture medium that has only the
minimal necessities required by the bacterial species is
referred to as minimal medium (table 7.1).

Alternatively, bacteria can grow on a medium that sup-
plies, in addition to their minimal requirements, the more

Table 7.1 Minimal Synthetic Medium for Growing
E. colt, a Heterotroph

Component

Quantity

NH 4 H 2 P0 4

Glucose

NaCl

MgS0 4 • 7H 2

0.2 g

K 2 HP0 4

1 g

H 2

1,000 ml

complex substances that the bacteria normally synthesize,
including amino acids, vitamins, and so on. A medium of
this kind allows the growth of strains of bacteria, called
auxotrophs, that have particular nutritional require-
ments. (The parent, or wild-type, strain is referred to as a
prototroph.) For example, a strain that has an enzyme de-
fect in the pathway that produces the amino acid histidine
will not grow on a minimal medium because it has no
way of obtaining histidine; it is a histidine-requiring auxo-
troph. If, however, histidine were provided in the medium,
the organisms could grow. This type of mutant is called a
conditional-lethal mutant. The organism would nor-
mally die, but under appropriate conditions, such as the
addition of histidine, the organism can survive.

This histidine-requiring auxotrophic mutant can grow
only on an enriched or complete medium, whereas
the parent prototroph could grow on a minimal medium.
Media are often enriched by adding complex mixtures of
organic substances such as blood, beef extract, yeast ex-
tract, or peptone, a digestion product. Many media, how-
ever, are made up of a minimal medium with the addition
of only one other substance, such as an amino acid or a vi-
tamin. These are called selective media; we will discuss
their uses later in the chapter. In addition to minimal,
complete, and selective media, other media exist for spe-
cific purposes such as aiding in counting colonies, help-
ing maintain cells in a nongrowth phase, and so on.

Source: Data from M. Rogosa, et al., Journal of Bacteriology, 54:13, 1947.

Figure 7.2 Confluent growth of bacterial colonies on a petri
plate. Bacteria were streaked on the petri plate with an
inoculation loop — a metal wire with a looped end — covered with
bacteria. Streaks began at the upper right and continued around
clockwise. With a heavy inoculation on the loop, bacterial growth
is confluent. Eventually, only a few bacteria are left; they form
Single colonies at the upper left. (Photo by Robert Tamarin.)

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Bacterial Phenotypes

151

Figure 7.3 Viral plaques (phage T1) on a bacterial lawn of
E. COli. (© Bruce Iverson, BSc.)

The experimental cultivation of viruses is somewhat
different. Since viruses are obligate parasites, they can
grow only in living cells. Thus, for the cultivation of
phages, petri plates of appropriate media are inoculated
with enough bacteria to form a continuous cover, or bac-
terial lawn. This bacterial culture serves as a medium for
the growth of viruses added to the plate. Since the virus
attack usually results in rupture, or lysis, of the bacterial
cell, addition of the virus usually produces clear spots,
known as plaques, on the petri plates (fig. 7.3). Large
quantities of viruses can be grown in flasks of bacteria.

BACTERIAL PHENOTYPES

Bacterial phenotypes fall into three general classes:
colony morphology, nutritional requirements, and drug
or infection resistance.

Colony Morphology

The first of these classes, colony morphology, relates sim-
ply to the form, color, and size of the colony that grows
from a single cell. A bacterial cell growing on a petri plate
in an incubator at 37° C divides as frequently as once
every twenty minutes. Each cell gives rise to a colony, or
clone, at its original position. In a relatively short amount
of time (e.g., overnight), the colonies will consist of
enough cells to be seen with the unaided eye. The differ-

ent morphologies observed among the colonies are usu-
ally under genetic control (fig. 7.4).

Nutritional Requirements

The second basis for classifying bacteria — by their nutri-
tional requirements — reflects the failure of one or more
enzymes in the bacteria's biosynthetic pathways. If an
auxotroph has a requirement for the amino acid cysteine
that the parent strain (prototroph) does not have, then
that auxotroph most likely has a nonfunctional enzyme
in the pathway for the synthesis of cysteine. Figure 7.5
shows five steps in cysteine synthesis; a different enzyme
controls each step. All enzymes are proteins, and the in-
formation in one or more genes determines the se-
quences in the strings of amino acids that make up
those proteins (chapter 1 1). A normal or wild-type allele

(a)

(b)

(c)

(d)

Figure 7.4 Various bacterial colony forms on agar petri plates.

(a) Red and white colonies of Serratia marcescens. (b) Irregular

raised folds of Streptomyces griseus. (c) Round colonies with

concentrated centers and diffuse edges of Mycoplasma.

(d) Irregularly folded raised colonies of Streptomyces antibioticus.

{[a] © Dr. E. Buttone/Peter Arnold, Inc., [b] © C. Case/Visuals Unlimited,

[c] © Michael G. Gabridge/Visuals Unlimited, [d] © Cabisco/Visuals Unlimited.)

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Adenine

H
O
O
C-

Methionine
adenosyltransferase

CH CrL CrL S CH,

NH,

ATP + H 2

H
O
O
■>- C— CH — CHU-

P.+ PP.

NH,

CH2 S CHo

Methionine

S-Adenosylmethionine

H
Methyltransferase Q

rv — * c -

Methyl-group Methylated
acceptor acceptor

Adenine

CH,

CH — CH 2 — CH 2 — S

NH,

S-Adenosylhomocysteine

Adenosylhomocysteinase O

-► C— CH — CH 2 — CH 2 — SH

H 2 Adenosine NH

Homocysteine

OH
OH

H
Cystathionine p-synthase ^

m — * °

NH,

Serine

H 2

CH — CH 2 — CH 2 — S — CH 2 — C-- COOH

NH,

L-Cystathionine

Cystathionine y-Lyase

H 2 NH 3 oc-Ketobutyrate

NH,

SH — CH 2 — C — COOH
H

Cysteine

Figure 7.5 Five-step conversion of methionine to cysteine. Each step is controlled by a different enzyme {red).

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Joshua Lederberg (1925- ).
(Courtesy of Dr. Joshua Lederberg.)

produces a normal, functional enzyme. The alternative al-
lele may produce a nonfunctional enzyme. Recall the
one-gene-one-enzyme hypothesis from chapter 2.

A technique known as replica-plating, devised by
Joshua Lederberg, is a rapid screening technique that
makes it possible to determine quickly whether a given
strain of bacteria is auxotrophic for a particular metabolite.
In this technique, a petri plate of complete medium is in-
oculated with bacteria. The resulting growth will have a
certain configuration of colonies. This plate of colonies is
pressed onto a piece of sterilized velvet. Then any number
of petri plates, each containing a medium that lacks some
specific metabolite, can be pressed onto this velvet to pick
up inocula in the same pattern as the growth on the origi-
nal plate (fig. 7.6). If a colony grows on the complete
medium but does not grow on a plate with a medium miss-
ing a metabolite, the inference is that the colony is made of
auxotrophic cells that require the absent metabolite. Sam-
ples of this bacterial strain can be obtained from the colony
growing on complete medium for further study. The nutri-
tional requirement of this strain is its phenotype. The
methionine-requiring auxotroph of figure 7.6 would be
designated as Met" (methionine-minus or Met-minus).

In terms of energy sources, the plus or minus notation
has a different meaning. For example, a strain of bacteria
that can utilize the sugar galactose as an energy source

Sterile
velvet

Complete
medium

No colony

Incubate

Medium

without

methionine

(c)

Figure 7.6 Replica-plating technique, (a) A pattern of colonies
from a plate of complete medium is transferred (b) to a second
plate of medium that lacks methionine, (c) In the locations
where colonies fail to grow on the second plate, we can infer
that the original colony was a methionine-requiring auxotroph.

would be Gal . If it could not utilize galactose, it would
be called Gal. The latter strain will not grow if galactose
is its sole carbon source. It will grow if a sugar other than
galactose is present. Note that a Met" strain needs me-
thionine to grow, whereas a Gal~ strain needs a carbon
source other than galactose; it cannot use galactose.

Resistance and Sensitivity

The third common classification of phenotypes in bacte-
ria involves resistance and sensitivity to drugs, phages,
and other environmental insults. For example, penicillin,
an antibiotic that prevents the final stage of cell-wall con-
struction in bacteria, will kill growing bacterial cells.
Nevertheless, we frequently find a number of cells that
do grow in the presence of penicillin. These colonies are
resistant to the drug, and this resistance is under simple
genetic control. The phenotype is penicillin resistant
(Pen r ) as compared with penicillin sensitive (Pen s ), the
normal condition, or wild-type. Numerous antibiotics are
used in bacterial studies (table 7.2).

Table 7.2 Some Antibiotics and Their Antibacterial Mechanisms

Antibiotic

Microbial Origin

Mode of Action

Penicillin G

Penicillium chrysogenum

Blocks cell-wall synthesis

Tetracycline

Streptomyces aureofaciens

Blocks protein synthesis

Streptomycin

Streptomyces griseus

Interferes with protein synthesis

Terramycin

Streptomyces rimosus

Blocks protein synthesis

Erythromycin

Streptomyces erythraeus

Blocks protein synthesis

Bacitracin

Bacillus subtilis

Blocks cell-wall synthesis

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Drug sensitivity provides another screening tech-
nique for isolating nutritional mutations. For example, if
we were looking for mutants that lacked the ability to
synthesize a particular amino acid (e.g., methionine), we
could grow large quantities of bacteria (prototrophs) and
then place them on a medium that lacked methionine
but contained penicillin. Here, any growing cells would
be killed. But methionine auxotrophs would not grow,
and, therefore, they would not be killed. The penicillin
could then be washed out and the cells reinoculated
onto a complete medium. The only colonies that form
should be composed of methionine auxotrophs (Met - ).

Screening for resistance to phages is similar to screen-
ing for drug resistance. When bacteria are placed in a
medium containing phages, only those bacteria that are
resistant to the phages will grow and produce colonies.
They can thus be easily isolated and studied.

VIRAL PHENOTYPES

Bacteriophage phenotypes fall generally into two cate-
gories: plaque morphology and growth characteristics on
different bacterial strains. For example, T2, an E. colt
phage (see fig. 7.1), produces small plaques with fuzzy
edges (genotype r + ). Rapid-lysis mutants (genotype r)
produce large, smooth-edged plaques (fig. 7.7). Similarly,

T4, another E. colt phage, has rapid-lysis mutants that pro-
duce large, smooth-edged plaques on E. colt B but will not
grow at all on E. colt K12, a different strain. Here, rapid-lysis
mutants illustrate both the colony morphology pheno-
types and the growth-restriction phenotypes of phages.

Figure 7.7 Normal (r + ) and rapid-lysis (r) mutants of phage T2.
Mottled plaques occur when r and r + phages grow together.
(From Molecular Biology of Bacterial Viruses by Gunther S. Stent. © 1 963,
1978 by W. H. Freeman and Company. Used with permission.)

SEXUAL PROCESSES _
IN BACTERIA AND **
BACTERIOPHAGES

Although bacteria and viruses are ideal subjects for bio-
chemical analysis, they would not be useful for genetic
study if they did not have sexual processes. If we define
a sexual process as combining genetic material from two
individuals, then the life cycles of bacteria and viruses in-
clude sexual processes. Although they do not undergo
sexual reproduction by the fusion of haploid gametes,
bacteria and viruses do undergo processes that incorpo-
rate genetic material from one cell or virus into another
cell or virus, forming recombinants. Actually, bacteria
have three different methods to gain access to foreign
genetic material: transformation, conjugation, and
transduction (fig. 7.8).

Phages can exchange genetic material when a bac-
terium is infected by more than one virus particle
(virion). During the process of viral infection, the ge-
netic material of different phages can exchange parts (or
recombine; see fig. 7.8). We will examine the exchange
processes in bacteria and then in bacteriophages, and
then proceed to the use of these methods for mapping
bacterial and viral chromosomes. {Chromosome refers to
the structural entity in the cell or virus made up of the ge-
netic material. In eukaryotes, it is double-stranded DNA
complexed with proteins [chapter 15]. Staining of this
eukaryotic organelle led to the term chromosome, which
means "colored body." In prokaryotes, the chromosome is
a circle [usually] of double-stranded DNA. In viruses, it is
virtually any combination of linear or circular, single- or
double-stranded RNA or DNA. Sometimes the term
genophore is used for the prokaryotic and viral genetic
material, limiting the word chromosome to the eukary-
otic version. We will use the term chromosome for the in-
tact genetic material of any organism or virus.)

Transformation

Transformation was first observed in 1928 by F. Griffith
and was examined at the molecular level in 1944 by
O. Avery and his colleagues, who used the process to
demonstrate that DNA was the genetic material of bacte-
ria. Chapter 9 presents the details of these experiments.
In transformation, a cell takes up extraneous DNA found
in the environment and incorporates it into its genome

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Bacterial sexual processes

Exogenous DNA

E. coli DNA Transformation

Conjugation

E. coli
DNA
from
phage

Transduction

Viral sexual process

Viral recombination

Figure 7.8 Summary of bacterial and viral sexual processes.

(genetic material) through recombination. Not all bacte-
ria are competent to be transformed, and not all extracel-
lular DNA is competent to transform. To be competent to
transform, the extracellular DNA must be double-
stranded. To be competent to be transformed, a cell must
have the surface protein, competence factor, which
binds to the extracellular DNA in an energy-requiring re-
action. However, bacteria that are not naturally compe-
tent can be treated to make them competent, usually by
treatment with calcium chloride, which makes them
more permeable.

Mechanisms of Transformation

Under natural conditions, only one of the strands of ex-
tracellular DNA is brought into the cell. The single strand
brought into the cell can then be incorporated into the
host genome by two crossovers (fig. 7.9). (The molecular
mechanisms of crossing over are presented in chapter

12.) Note that unlike eukaryotic crossing over, this is not
a reciprocal process. The bacterial chromosome incorpo-
rates part of the foreign DNA. The remaining single-
stranded DNA, originally part of the bacterial chromo-
some, is degraded by host enzymes called exonucleases;
linear DNA is degraded rapidly in prokaryotes.

Transformation is a very efficient method of mapping
in some bacteria, especially those that are inefficient in
other mechanisms of DNA intake (such as transduction,
discussed later in this chapter). For example, a good deal
of the mapping of the soil bacterium, Bacillus subtilis,
has been done through the process of transformation;
E. coli, however, is inefficient in transformation, so other
methods are used to map its chromosome.

Transformation Mapping

The general idea of transformation mapping is to add
DNA from a bacterial strain with known genotype to an-
other strain, also with known genotype, but with differ-
ent alleles at two or more loci. We then look for incorpo-
ration of the donor alleles into the recipient strain of
bacteria. The more often alleles from two loci are incor-
porated together into the host, the closer together these
loci must be to each other. Thus, we can use an index of
co-occurrence that is in inverse relationship to map dis-
tance: the larger the co-occurrence of alleles of two loci,
the closer together the loci must be. This is another way
of looking at the mapping concepts we discussed in

Transforming
DNA

Degraded

Bacterial
chromosome

Cell
division

Transformed
cell

Figure 7.9 A single strand of transforming DNA (blue with a +
allele) enters a bacterial cell {red chromosome with a allele). Two
crossovers bring the foreign DNA into the bacterial chromosome.
After DNA replication and cell division, one cell has the a allele
and the other the a + allele. The chromosome is drawn as a
double circle, symbolizing the double-stranded structure of DNA.

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

chapter 6, where we discovered that the closer two loci
are, the fewer the recombinations between them and
thus the higher the co-occurrence.

Now, we also must look at another concept, that is,
selecting for recombinant cells. In fruit flies, every off-
spring of a mated pair represents a sampling of the mei-
otic tetrad, and thus a part of the total, whether or not re-
combination took place. Here, however, many cells are
present that do not take part in the transformation
process. In a bacterial culture, for example, only one cell
in a thousand might be transformed. We must thus always
be sure when working with bacterial gene transfer that
we count only those cells that have taken part in the
process. Let us look at an example.

A recipient strain of B. subtilis is auxotrophic for the
amino acids tyrosine (tyrA~} and cysteine (cysC~}. We
are interested in how close these loci are on the bacterial
chromosome. We thus isolate DNA from a prototrophic
strain of bacteria (tyrA + cysC + ). We add this donor DNA
to the auxotrophic strain and allow time for transforma-
tion to take place (fig. 7.10). If the experiment is suc-
cessful, and the loci are close enough together, then
some of the recipient bacteria may incorporate donor
DNA that has either both donor alleles or one or the

other donor allele. Thus, some of the recipient cells will
now have the tyrA + and cysC + alleles, some will have
just the donor tyrA + allele, some will have just the donor
cysC + allele, and the overwhelming majority will be of
the untransformed auxotrophic genotype, tyrA~ cysC~ '.
We thus need to count the transformed cells.

We do this by removing any extraneous transforming
DNA and then pouring the cells out onto a complete
medium so that all cells can grow. These cells are then
replica-plated onto three plates — a minimal medium
plate, a minimal medium plus tyrosine plate, and a minimal
medium plus cysteine plate — and allowed to grow
overnight in an incubator at 37° C. We then count colonies
(fig. 7. 1 1). Those growing on minimal medium are of geno-
type tyrA + cysC + ; those growing on minimal medium
with tyrosine but not growing on minimal medium are
tyrA~ cysC + ; and those growing on minimal medium with
cysteine but not growing on minimal medium are tyrA +
cysC~ .The overwhelming majority will grow on complete
medium, but not on minimal medium or minimal media
with just tyrosine or cysteine added. This majority is made
up of the nontransformants, that is, auxotrophs that were
not involved in a transformation event — they took up no
foreign DNA.

Isolate

DNA

Transform

>■

tyr/£__2££* ^\

Growth

tyrA + cysC*

Complete medium

Minimal medium

Minimal + tyrosine

Minimal + cysteine

tyrA + cysC

tyrA cysC*

Figure 7.10 Transformation experiment with B. subtilis. A tyrA~ cysC~ strain is transformed with DNA from a tyrA + cysC + strain.
Nontransformants as well as three types of transformants (two single and one double) result. Genotypes are determined by growth
characteristics on four different types of petri plates (see fig. 7.11).

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Complete medium

tyrA + cysC

tyrA cysC +

tyrA + cysC

tyrA cysC

Replica plating

Minimal
medium

+
tyrosine

Minimal
medium

+
cysteine

Figure 7.11 Four patterns of growth on different media reveal
the genotypes of transformed and untransformed cells. Only
four colonies are shown, and a grid is added for ease of
identification. After transformation (see fig. 7.10), cells are
plated on complete medium and then replica-plated onto
minimal medium with either tyrosine or cysteine added.

As a control against reversion, the normal mutation of
tyrA~ to tyrA + or cysC~ to cysC + , we grow several plates
of auxotrophs in minimal medium and minimal medium
with tyrosine or cysteine added. These are auxotrophs
that were not exposed to prototrophic donor DNA. We
then count the number of natural revertants and correct
our experimental numbers by the natural reversion rate.
Thus, we are sure that what we measure is the actual
transformation rate rather than just a mutation rate that
we mistake for transformation. This control should
always be carried out.

From the experiment (see figs. 7.10 and 7.11), we
count twelve double transformants (tyrA + cysC + ), thirty-
one tyrA + cysC~ , and twenty-seven tyrAT cysC + . From
these data, we calculate the co-occurrence, or cotransfer
index, (r) as

r =

number of double transformants

number of double transformants
+ number of single transformants

From our data

r = 12/(12 + 31 + 27) = 0.17.

This is a relative number indicating the co-occurrence of
the two loci and thus their relative distance apart on the

bacterial chromosome. Remember that as this number in-
creases for different pairs of loci, the loci are closer and
closer together.

By systematically examining many loci, we can estab-
lish their relative order. For example, if locus A is closely
linked to locus B and B to C, we can establish the order
A B C. It is not possible by this method to determine ex-
act order for very closely linked genes. For this informa-
tion we need to rely upon transduction, which we will
consider shortly. However, transformation has allowed us
to determine that the map of B. subtilis is circular, a phe-
nomenon found in all prokaryotes and many phages.
(The E. colt map is shown later.)

Conjugation

In 1946, Joshua Lederberg and Edward L.Tatum (later to
be Nobel laureates) discovered that E. colt cells can ex-
change genetic material through the process of conju-
gation. They mixed two auxotrophic strains of E. coll
One strain required methionine and biotin (Met - Bio),
and the other required threonine and leucine (Thr~
Leu). This cross is shown in figure 7.12. Remember
that if a strain is Met" Bio~, it is, without saying, wild-
type for all other loci. Thus, a cell with the Met" Bio~
phenotype actually has the genotype of met~ bio~ tbr +
leu + . Similarly, the Thr~ Leu~ strain is actually met +
bio + thr~ leu~ . (Note that symbols such as "Thr~" rep-
resent phenotypes; symbols such as "thr~" represent
genotypes.)

Lederberg and Tatum used multiple auxotrophs in or-
der to rule out spontaneous reversion (mutation). About
one in 10 Met" cells will spontaneously become pro-
totrophic (Met + ) every generation. However, with multi-
ple auxotrophs, the probability that several loci will si-
multaneously and spontaneously revert (e.g., met~ — >
met + ) becomes vanishingly small. (In fact, the control
plates in the experiment, illustrated in fig. 7.12, showed
no growth for parental double mutants.) After mixing the
strains, Lederberg and Tatum found that about one cell in
10 7 was prototrophic (met + bio + tbr + leu + ).

To rule out transformation, one strain was put in
each arm of a U-tube with a sintered glass filter at the
bottom, (fig. 7.13). The liquid and large molecules, in-
cluding DNA, were mixed by alternate application of
pressure and suction to one arm of the tube; whole cells
did not pass through the filter. The result was that the
fluids surrounding the cells, as well as any large mole-
cules (e.g., DNA), could be freely mixed while the cells
were kept separate. After cell growth stopped in the
two arms (in complete medium), the contents were
plated out on minimal medium. There were no pro-
totrophs in either arm. Therefore, cell-to-cell contact
was required for the genetic material of the two cells to
recombine.

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

<^>

Strain A

met~ bio~ thr + leu +

Complete medium

^=^>

^^ Centrifuge and wash cells

Strain B
met + bio + thr'

leu'

2X 10 8 cells

1 X 10 8 cells

Minimal medium

2X 10 8 cells

No growth

met + bio + thr + leu +

No growth

Figure 7.12 Lederberg and Tatum's cross showing that E. coli undergoes genetic
recombination.

Strain A

< >■

Strain B

Filter

Figure 7.13 The U-tube experiment. Alternating suction and
pressure force liquid and macromolecules back and forth
across the filter.

F Factor

In bacteria, conjugation is a one-way transfer, with one
strain acting as donor and the other as recipient. Some-
times donor cells, if stored for a long time, lose the ability
to be donors, but they can regain the ability if they are
mated with other donor strains. This discovery led to the
hypothesis that a fertility factor, F, made any strain that
carried it a male (donor) strain, termed F + .The strain that
did not have the F factor, referred to as a female or F"
strain, served as a recipient for genetic material during
conjugation. Research supports this hypothesis.

The F factor is aplasmid, a term originally coined by
Lederberg to refer to independent, self-replicating genetic
particles. Plasmids are usually circles of double-stranded
DNA. (Plasmids are at the heart of recombinant DNA
technology, which is discussed in detail in chapter 13.)
They are auxiliary circles of DNA that many bacteria
carry. They are usually much smaller than the bacterial
chromosome.

Researchers found that the transfer of the F factor oc-
curred far more frequently than the transfer of other
genes from the donor. That is, during conjugation, about
one recombinant occurred in 10 7 cells, whereas transfer
of the F factor occurred at a rate of about one conversion
of F" to F + in every five conjugations. An E. coli strain
was then discovered that transferred its genetic material
at a rate about one thousand times that of the normal F +
strain. This strain was called Hfr, for Mgh frequency of
recombination. Several other phenomena occurred with
this high rate of transfer. First, the ability to transfer the F
factor itself dropped to almost zero in this strain. Second,

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not all loci were transferred at the same rate. Some loci
were transferred much more frequently than others.

Escherichia coli cells are normally coated with hair-
like pili (fimbriae). F + and Hfr cells have one to three
additional pili (singular: pilus) called F-pili, or sex pili.
During conjugation, these sex pili form a connecting
bridge between the F + (or Hfr) and F" cells (fig. 7.14).
Once a connection is made, the sex pilus then contracts
to bring the two cells into contact. DNA transfer takes
place through a nick in either the plasmid (in F + cells) or
the bacterial chromosome (in Hfr cells). A single strand
of the DNA double-stranded donor DNA then passes
from the F + or Hfr cell to the F~ cell across the cell mem-
branes. DNA replication in both the donor and recipient
cells reestablishes double-stranded DNA in both. The
F factor itself has the genes for sex-pilus formation and
DNA transfer to a conjugating F~ cell. At least twenty-two
genes are involved in the transfer process, including
genes for the pilus protein, nicking the DNA, and regula-
tion of the process.

In the transfer process of conjugation, the donor cell
does not lose its F factor or its chromosome because only

a single strand of the DNA double helix is transferred; the
remaining single strand is quickly replicated. (The
process of DNA replication is described in chapter 9)
For a short while, the F~ cell that has conjugated with an
Hfr cell has two copies of whatever chromosomal loci
were transferred: one copy of its own and one trans-
ferred in. With these two copies, the cell is a partial
diploid, or a merozygote. The new foreign DNA
(exogenote) can be incorporated into the host chromo-
some (endogenote) by an even number of breakages
and reunions between the two, just as in transformation.
The unincorporated linear DNA is soon degraded by
enzymes. The conjugation process is diagrammed in
figure 7.15.

Interrupted Mating

To demonstrate that the transfer of genetic material
from the donor to the recipient cell during conjugation
is a linear event, F. Jacob and E. Wollman devised the
technique of interrupted mating. In this technique, F"
and Hfr strains were mixed together in a food blender.

Figure 7.14 Electron micrograph of conjugation between an
F + {upper right) and an F~ {lower left) cell with the F-pilus
between them. Magnification 3,700x. (Courtesy of Wayne
Rosenkrans and Dr. Sonia Guterman.)

Interruption

Figure 7.15 Bacterial conjugation, (a) The F-pilus draws an Hfr
and an F~ cell close together, {b) The Hfr chromosome then
begins to pass into the F~ cell, beginning at the F region of
the Hfr chromosome but in the direction away from the
F factor. Only a single strand passes into the F~ cell; this
strand and the single strand remaining in the Hfr cell are
replicated. After the process is interrupted (c), two crossovers
bring the a + allele into the F~ a~ chromosome (c/).

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Elie Wollman (1917- ).
(Courtesy of Dr. Elie Wollman
and the Pasteur Institute.)

After waiting a specific amount of time, Jacob and Woll-
man turned the blender on. The spinning motion sepa-
rated conjugating cells and thereby interrupted their
mating. Then the researchers tested the F~ cells for vari-
ous alleles originally in the Hfr cell. In an experiment
like this, the Hfr strain is usually sensitive to an antibiotic
such as streptomycin. After conjugation is interrupted,
the cells are plated onto a medium containing the an-
tibiotic, which kills all the Hfr cells. Then the genotypes
of the F~ cells can be determined by replica-plating
without fear of contamination by Hfr cells.

The mating outlined in table 7.3 was carried out. In
the food blender, an Hfr strain sensitive to streptomycin
(s£r s ) but resistant to azide (azi r ), resistant to phage Tl
(tonA r ), and prototrophic for the amino acid leucine
Qeu + ) and the sugars galactose (galB + ) and lactose
(/ac + ) was added to an F~ strain that was resistant to
streptomycin (str r ), sensitive to azide (azf^), sensitive to
Tl (tonA s }, and auxotrophic for leucine, galactose, and
lactose (leu~ , galB~, and lac~). After a specific number
of minutes (ranging from zero to sixty), the food blender

Table 7.3 Genotypes of Hfr and F Cells Used
in an Interrupted Mating Experiment

Hfr

str s

str r

azf

tonA v

tonA s

leu +

leu~

galB +

galB~

lac +

lac"

was turned on. To kill all the Hfr cells, the cell suspension
was plated on a medium containing streptomycin. The
remaining cells were then plated on medium without
leucine. The only colonies that resulted were F~ recom-
binants. They must have received the leu + allele from the
Hfr in order to grow on a medium lacking leucine.
Hence, all colonies had been selected to be F~ recombi-
nants. By replica-plating onto specific media, investiga-
tors were able to determine the azi, tonA, lac, and galB
alleles and the percentage of recombinant colonies that
had the original Hfr allele (/ez/ + ). (Note that by trial and
error, it was determined that leucine should be the locus
to use to select for recombinants. As we will see, the
leucine locus entered first.)

Figure 7.16 shows that as time of mating increases,
two things happen. First, new alleles enter the F~ cells
from the Hfr cells. The tonA r allele first appears among
recombinants after about ten minutes of mating, whereas
galB + first enters the F~ cells after about twenty-five
minutes. This suggests a sequential entry of loci into the
F" cells from the Hfr (fig. 7.17). Second, as time pro-
ceeds, the percentage of recombinants with a given allele
from the Hfr increases. At ten minutes, tonA r is first
found among recombinants. After fifteen minutes, about
40% of recombinants have the tonA r allele from the Hfr;
and after about twenty-five minutes, about 80% of the re-
combinants have the tonA r allele. This limiting percent-

CD
O

CT
CD

100 i-

80 -

60 -

40 -

20 -

► .

J I I I lb I I L

galB

J I L

20 30 40

Time (minutes)

'Limiting percentage for az, tonA x , lac, and galB loci.

Figure 7.16 Frequency of Hfr genetic characters among
recombinants after interrupted mating. As time proceeds,
new alleles appear and then increase in frequency. Interruption
of the mating limits the frequency of successful passage.

(From F. Jacobs and E. L. Wollman, Sexuality and the Genetics of Bacte-
ria, Academic Press, 1961.)

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Hfr
str s

10 minutes

str r

15 minutes

20 minutes

30 minutes

Figure 7.17 Conjugation in E. coll. Hfr chromosome is blue;
F~ chromosome is red; and new DNA replication is black.
As time proceeds, alleles from the Hfr enter the F~ cell in an
orderly, sequential fashion. After the cells separate, two
crossovers can bring Hfr alleles into the F~ chromosome.
The F factor {orange) is the last part of the Hfr chromosome
to enter the F" cell.

age does not increase with additional time. The limiting
percentage is lower for loci that enter later, a fact ex-
plained by the assumption that even without the food
blender, mating is usually interrupted before completion
by normal agitation alone.

Mapping and Conjugation

Jacob and Wollman, working with several different Hfr
strains, collected data that indicated that the bacterial chro-
mosome was circular. The strains were of independent
origin, and the results were quite striking (table 7.4).

If we ponder this table for a short while, one fact be-
comes obvious: The relative order of the loci is always
the same. What differs is the point of origin and the di-
rection of the transfer. Jacob and Wollman proposed that
normally the F factor is an independent circular DNA en-
tity in the F + cell, and that during conjugation only the F
factor is passed to the F~ cell. Since it is a small fragment
of DNA, it can be passed entirely in a high proportion of
conjugations before the cells separate. Every once in a
while, however, the F factor becomes integrated into the
chromosome of the host, which then becomes an Hfr
cell. The point of integration can be different in different
strains. However, once the F factor is integrated, it deter-
mines the initiation point of transfer for the E. colt chro-
mosome, as well as the direction of transfer.

The F factor is the last part of the E. colt chromosome
to be passed from the Hfr cell. This explains why an Hfr,
in contrast to an F + , rarely passes the F factor itself. In the
original work of Lederberg and Tatum, the one recombi-
nant in 10 7 cells most likely came from a conjugation
between an F" cell and an Hfr that had formed sponta-
neously from an F + cell. Integration of the F factor is dia-
grammed in figure 7. 18. The F factor can also reverse this
process and loop out of the E. colt chromosome. (Some-
times the F factor loops out incorrectly, as in figure 7.19,
forming an F' [F-prime] factor. The passage of this F' fac-
tor to an F~ cell is called F-duction or sexduction. Not
really useful in mapping, the process has proved excep-
tionally useful in studies of gene expression because of
the formation of stable merozygotes, which we will ex-
amine in chapter 14.)

We could now diagram the E. colt chromosome and
show the map location of all known loci. The map units
would be in minutes, obtained by interrupted mating.
However, at this point, the map would not be complete.
Interrupted mating is most accurate in giving the relative
position of loci that are not very close to each other. With
this method alone, a great deal of ambiguity would arise
as to the specific order of very close genes on the chro-
mosome. The remaining sexual process in bacteria, trans-
duction, provides the details that interrupted mating or
transformation don't explain.

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Table 7.4

Gene Order of Various Hfr Strains Determined

by Means of Interrupted Mating

Types of Hfr Order of Transfer of Genetic Characters*

HfrH

Pro

Lac

Gal

Try

S-G

Mai

Xyl

Mtl

Isol

B x

Isol

Mtl

Xyl

Mai

S-G

Try

Gal

Lac

Pro

T x

Isol

Mtl

Xyl

Mai

S-G

Try

Gal

Lac

Pro

Isol

Mtl

Xyl

Mai

S-G

Try

Gal

Isol

Mtl

Xyl

Mai

S-G

Try

Gal

Lac

Pro

Lac

Gal

Try

S-G

Mai

Xyl

Mtl

Isol

Pro

Lac

Gal

Try

S-G

Mai

Xyl

Mtl

B x

Isol

Mtl

Xyl

Mai

S-G

Try

Gal

Lac

Pro

AB311

Try

Gal

Lac

Pro

T x

Isol

Mtl

Xyl

Mai

S-G

AB312

Mai

Xyl

Mtl

Isol

T x

Pro

Lac

Gal

Try

S-G

AB313

Mtl

Xyl

Mai

S-G

Try

Gal

Lac

Pro

Isol

Source: From F. Jacobs and E. L. Wollman, Sexuality and the Genetics of Bacteria, Academic Press, 1961.
* The refers to the origin of transfer.

E. co//' chromosome

Hfr chromosome

Figure 7.18 Integration of the F factor by a single crossover.
After a simultaneous breakage in both the F factor and the
E. coli chromosome, the two broken circles reunite to make
one large circle, the Hfr chromosome. In this case, the
integration occurs between the ton A and lac loci.

Figure 7.19 Occasionally, the F factor loops out imprecisely,
taking part of the cell's genome in the loop. The circular F
factor is freed by a single recombination (crossover) at the
loop point.

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Life Cycles of Bacteriophages

163

LIFE CYCLES

OF BACTERIOPHAGES

Phages are obligate intracellular parasites. Phage genetic
material enters the bacterial cell after the phage has ad-
sorbed to the cell surface. Once inside, the viral genetic
material takes over the metabolism of the host cell. Dur-
ing the infection process, the cell's genetic material is
destroyed, while the viral genetic material is replicated
many times. The viral genetic material then controls the
mass production of various protein components of the
virus. New virus particles are assembled within the host
cell, which bursts open (is lysed), releasing a lysate of
hundreds of viral particles to infect other bacteria. This
life cycle appears in figure 7.20.

Recombination

Much genetic work on phages has been done with a
group of seven E. colt phages called the T series (T-odd:
Tl, T3, T5, T7; T-even: T2, T4, and T6) and several others,
including phage X (lambda; fig. 7.21). Figure 7.1 dia-
grammed the complex structure of T2. Phages can
undergo recombination processes when a cell is infected

Adsorbed phage

.Cell's DNA

Bacterium

Injection of phage
genetic material

with two genetically distinct virions. Hence, the phage
genome can be mapped by recombination. As an exam-
ple, consider the host-range and rapid-lysis loci. Rapid-
lysis mutants (r) of the T-even phages produce large,
sharp-edged plaques. The wild-type produces a smaller,
more fuzzy-edged plaque (see fig. 7.7).

Alternative alleles are known also for host-range loci,
phage loci that determine the strains of bacteria the
phage can infect. For example, T2 can infect E. colt cells.
These phages can be designated as T2h + for the normal
host range. The E. colt is then called Tto s , referring to
their sensitivity to the T2 phage. In the course of evolu-
tion, an E. colt mutant arose that is resistant to the normal
phage. This mutant strain is named Tto r forT2 resistance.
In the further course of evolution, the phages have pro-
duced mutant forms that can grow on the Tto r strain of
E. colt These phage mutants are designated as T2h for
host-range mutant. Remember, host-range signifies a mu-
tation in the phage genome, whereas phage resistance
indicates a mutation in the bacterial genome.

In 1945, Max Delbriick (a 1969 Nobel laureate) de-
veloped mixed indicators, which can be used to demon-
strate four phage phenotypes on the same petri plate
(fig. 7.22). A bacterial lawn of mixed Tto r and Tto s is
grown. On this lawn, the rapid-lysis phage mutants (r)
produce large plaques, whereas the wild-type (r + ) pro-
duce smaller plaques. Phages with host-range mutation
(h) lyse both Tto r and Tto s bacteria. They produce the
plaques that are clear (but appear dark) in figure 7.22.
Since phages with the wild-type host-range allele (Z? + )
can only infect the Tto s bacteria, they produce turbid
plaques. The Tto r bacteria growing within these plaques
(which appear light-colored in fig. 7.22) produce the
turbidity.

Replication of phage
genetic material and
breakdown of host's
genetic material

Assembly of
new phage

Manufacture of
phage proteins

Lysis
of cell

Figure 7.20 The viral life cycle, using T4 infection of E. coli as
an example.

Figure 7.21 Phage X. Magnification 167,300x. Note that
phage X lacks the tail fibers and base plate of phage T2
(see fig. 7.1). (Courtesy of Robley Williams.)

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

.+„+

Iff

Figure 7.22 Four types of plaques produced by mixed phage
T2 on a mixed lawn of E. COW. (From Molecular Biology of Bacterial
Viruses by Gunther S. Stent, © 1963, 1978 by W. H. Freeman & Company.
Used with permission.)

From the wild stock of phages, we can isolate host-
range mutants by looking for plaques on a Tto r bacterial
lawn. Only h mutants will grow. These phages can then
be tested for the r phenotype and the double mutants
isolated. Once the two strains (double mutant and wild-
type) are available, they can be added in large numbers to
sensitive bacteria (fig. 7.23). Large numbers of phages are
used to ensure that each bacterium is infected by at least
one of each phage type, creating the possibility of re-
combination within the host bacterium. After a round of
phage multiplication, the phages are isolated and plated
out on Delbriick's mixed-indicator stock. From this
growth, the phenotype (and, hence, genotype) of each
phage can be recorded. The percentage of recombinants
can be read directly from the plate. For example, on a
given petri plate (e.g., fig. 7.22) there might be

hr
h + r

h + r +

52
26

+ „+

The first two, hr and h r , are the original, or parental,
phage genotypes. The second two categories result from
recombination between the h and r loci on the phage
chromosome. A single crossover in this region produces
the recombinants. Note that with phage recombination,
parental phages are counted, since every opportunity
was provided for recombination within each bacterium.
Thus, every progeny phage arises from a situation in

E. coli
Tto s

hr +

h + r

h + r +

E. coli

mixed-indicator
lawn (Tto r + Tto s )

Figure 7.23 Crossing hr and h + r + phage. Enough of both
types are added to sensitive bacterial cells (Tto s ) to ensure
multiple infections. The lysate, consisting of four genotypes,
is grown on a mixed-indicator bacterial lawn (Tto s and Tto^.
Plaques of four types appear (see fig. 7.22), indicating the
genotypes of the parental and recombinant phages.

which recombination could have taken place. The pro-
portion of recombinants is

(34 + 26)/(46 + 52 + 34 + 26)
= 60/158 = 0.38 or 38% or 38 map units

This percentage recombination is the map distance,
which (as in eukaryotes) is a relative index of distance
between loci: The greater the physical distance, the
greater the amount of recombination, and thus the larger
the map distance. One map unit (1 centimorgan) is equal
to 1% recombinant offspring.

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Transduction

165

Curing (loss of prophage)

Vegetative growth

Lysogenic growth

Figure 7.24 Alternative life-cycle stages of a temperate phage (lysogenic and vegetative growth).

Lysogeny

Certain phages are capable of two different life-cycle
stages. Some of the time, they replicate in the host cyto-
plasm and destroy the host cell. At other times, these
phages are capable of surviving in the host cell. The host
is then referred to as lysogenic and the phage as tem-
perate. (The term lysogeny means "giving birth to lysis."
A lysogenic bacterium can be induced to initiate the vir-
ulent phase of the phage life cycle.)

The majority of research on lysogeny has been done
on phage X (see fig. 7.21). The X prophage integrates into
the host chromosome; other prophages, like PI, exist as
independent plasmids. Phage X, unlike the F factor, at-
taches at a specific point, termed attX. This locus can be
mapped on the E. coli chromosome; it lies between the
galactose {gal} and biotin (bio) loci. When the phage is
integrated, it protects the host from further infection (su-
perinfection) by other X phages. The integrated phage is
now termed a prophage. Presumably it becomes inte-
grated by a single crossover between itself and the host
after apposition at the attX site. (This process resembles
the F-factor integration shown in fig. 7.18.)

A prophage can enter the lytic cycle of growth by a
process of induction, which involves the excision of the
prophage followed by the virulent or lytic stage of the vi-
ral life cycle. We consider the interesting and complex
control mechanisms of life cycle in detail in chapter 14.
Induction can take place through a variety of mecha-
nisms, including UV irradiation and passage of the inte-
grated prophage during conjugation (zygotic induc-

tion). The complete life cycle of a temperate phage is
shown in figure 7.24.

TRANSDUCTION

Before lysis, when phage DNA is being packaged into
phage heads, an occasional error occurs that causes bac-
terial DNA to be incorporated into the phage head in-
stead. When this happens, bacterial genes can be trans-
ferred to another bacterium via the phage coat. This
process, called transduction, has been of great use in
mapping the bacterial chromosome. Transduction occurs
in two patterns: specialized and generalized.

Specialized Transduction

The process of specialized or restricted transduction

was first discovered in phage X by Lederberg and his stu-
dents. Specialized transduction is analogous to sexduc-
tion — it depends upon a mistake made during a looping-
out process. In sexduction, the error is in the F factor. In
specialized transduction, the error is in the X prophage.
Figure 7.25 shows the X prophage looping out incor-
rectly to create a defective phage carrying the adjacent
gal locus. Since only loci adjacent to the phage attach-
ment site can be transduced in this process, specialized
transduction has not proven very useful for mapping the
host chromosome.

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166

Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Bacterial chromosome

X prophage

Exogenote

Endogenote

'/o

Defective X phage

gal

Figure 7.25 Imprecise excision, or looping out, of the X
prophage, resulting in a defective phage carrying the gal locus.

Generalized Transduction

Generalized transduction, which Zinder and Leder-
berg discovered, was the first mode of transduction dis-
covered. The bacterium was Salmonella typhimurium
and the phage was P2 2. Virtually any locus can be trans-
duced by generalized transduction. The mechanism,
therefore, does not depend on a faulty excision, but
rather on the random inclusion of a piece of the host
chromosome within the phage protein coat. A defective
phage, one that carries bacterial DNA rather than phage
DNA, is called a transducing particle. Transduction is
complete when the genetic material from the transduc-
ing particle is injected into a new host and enters the
new host's chromosome by recombination.

For P22, the rate of transduction is about once for
every 10 5 infecting phages. Since a transducing phage
can carry only 2 to 2.5% of the host chromosome, only
genes very close to each other can be transduced to-
gether (cotransduced). Cotransduction can thus help
to fill in the details of gene order over short distances af-
ter interrupted mating or transformation is used to ascer-
tain the general pattern. Transduction is similar to trans-
formation in that cotransduction, like co-occurrence in
transformation, is a relative indicator of map distance.

Mapping with Transduction

Transduction can be used to establish gene order and
map distance. Gene order can be established by two-
factor transduction. For example, if gene A is cotrans-
duced with gene B and B with gene C, but A is never co-
transduced with C, we have established the order ABC

B H

C + C

Two crossovers

Four crossovers

A +

Figure 7.26 The rarest transductant requires four crossovers.

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Transduction

167

Table 7.5 Gene Order Established by Two-Factor
Cotransduction*

Transductants

Number

A + B +

A + C +

B + C +

A + B + C +

* An A + B + C + strain of bacteria is infected with phage. The lysate is used to
infect an^4~ B~ C~ strain. The transductants are scored for the wild-type alleles
they contain. These data include only those bacteria transduced for two or more
of the loci. Since AB cotransductants and BC cotransductants occur, but no AC
types, we can infer the ABC order.

(table 7.5). This would also apply to quantitative differ-
ences in cotransduction. For example, if E is often co-
transduced with F and F often with G, but E is very rarely
cotransduced with G, then we have established the order
FFG.

However, even more valuable is three-factor trans-
duction, in which we can simultaneously establish
gene order and relative distance. Three-factor transduc-
tion is especially valuable when the three loci are so
close that it is very difficult to make ordering decisions
on the basis of two-factor transduction or interrupted
mating. For example, if genes A, B, and C are usually co-
transduced, we can find the order and relative dis-
tances by taking advantage of the rarity of multiple
crossovers. Let us use the prototroph (A + B + C + ) to
make transducing phages that then infect the A~ B~
C~ strain of bacteria.

To detect cells that have been transduced for one,
two, or all three of the loci, we need to eliminate the

nontransduced cells. In other words, after transduction,
there will be A~ B~ C~ cells in which no transduction
event has taken place. There will also be seven classes of
bacteria that have been transduced for one, two, or all
three loci (A + B + C + ,A + B + C~,A + B~ C + ,A~ B + C + ,

A + B

B + C , and A B C + ). The simplest way

C ,A

to select for transduced bacteria is to select bacteria in
which the wild-type has replaced at least one of the loci.
For example, if, after transduction, we grow the bacteria
in minimal media with the requirements of B~ and C~
added, all the bacteria that are^4 + will grow. (Without the
requirement of A~ bacteria, no A~ bacteria will grow.)
Hence, although we lose the A~ B + C + ,A~ B + C~, and
A~ B~ C + categories, we also lose the A~ B~ C~, un-
transduced bacteria. In this example, the A locus is the
selected locus; we must keep in mind that we have an in-
complete, although informative, data set. Replica-plating
allows us to determine genotypes at the B and C loci for
the^4 + bacteria.

In this example, colonies that grow on complete
medium without the requirement of the A mutant are
replica-plated onto complete medium without the re-
quirement of the B mutant and then onto complete
medium without the requirement of the C mutant. In this
way, each transductant can be scored for the other two
loci (table 7.6). Now let us take all these selected trans-
ductants in which the A + allele was incorporated. These
can be of four categories: A + B + C + ,A + B + C~,A + B~
C + , and A + B~ C~ . We can now compare the relative
numbers of each of these four categories. The rarest cat-
egory will be caused by the event that brings in the outer
two markers, but not the center one, because this event
requires four crossovers (fig. 7.26). Thus, by looking at
the number of transductants in the various categories,
we can determine that the gene order isAB C (table 7.7),
since the^4 + B~ C + category is the rarest.

Table 7.6 Method of Scoring Three-Factor Transductants

Minimal Medium

Genotype

Colony Number

Without ,4

Requirement

Without B Requirement

Without C Requirement

—

A + B + C~

—

A + B~ C~

—

A + B~ C~

A + B + C +

A + B~ C +

•

Note: The plus indicates growth, the minus lack of growth. An ABC strain was transduced by phage from ani B C strain.

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168

Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Table 7.7 Numbers of Transductants and Relative
Cotransduction Frequencies in the
Experiment Used to Determine the
ABC Gene Order (Table 7.6)

Class

Number

A + B + C +

A + B + C~

A + B~ C +

A + B~ C~

300
426

Relative Cotransductance

A-B: (50 +

75)/426

= 0.29

A-C: (50 +

l)/426 =

0.12

Table 7.7 also includes calculations of the relative co-
transduction frequencies. Remember that in all organ-
isms and viruses, the higher the frequency of co-
occurrence between the alleles of two loci, the closer
those loci are on the chromosome. We usually measure
the separation of loci by crossing over between them;
the closer together, the lower those crossing-over values
are and, hence, the smaller the measure of map units

apart. Here, as with transformation, we are measuring the
co-occurrence directly; therefore, the measure — cotrans-
ductance — is the inverse of map distance. In other
words, the greater the cotransduction rate, the closer the
two loci are; the more frequently two loci are transduced
together, the closer they are and the higher the cotrans-
duction value will be.

The data in table 7.7 should not be used to calculate
the B-C cotransduction rate because the data are selected
values, all of which are A + ; they do not encompass the to-
tal data. Missing is the A~ B + C + group that would con-
tribute to the B-C cotransductance rate. The A~ B + C~
and A~ B~ C + groups, also missing, would contribute
only to the totals in the denominator, not the numerator,
of the cotransductance index.

From these sorts of transduction experiments, it is
possible to round out the details of map relations in
E. colt after obtaining the overall picture by interrupted
mating. The partial map of E. colt appears in figure 7.27.
Definitions of loci can be found in table 7.8. Unlike the
measurements in eukaryotic mapping, prokaryotic map
distances are not generally thought of in map units (cen-
timorgans). Rather, the general distance between loci is
determined in minutes with cotransduction values used
for loci that are very close to each other. (In chapter 13,
we discuss mapping methods that rely on directly se-
quencing the DNA.)

ma/7"

Figure 7.27 Selected loci on a circular
map of E. coli. Definitions of loci not
found in the text can be found in table
7.8. Units on the map are in minutes.
Arrows within the circle refer to Hfr-strain
transfer starting points, with directions
indicated. The two thin regions on the
outer circle are the only areas not covered
by P1 transducing phages. (From B. J.
Bachmann et al., "Recalibrated linkage map of
Escherichia coli K-12," Bacteriological Reviews,
40:116-17. Copyright © 1976 American Society for
Microbiology, Washington, D.C. Reprinted by
permission.)

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Transduction

169

Table 7.8

Symbols Used in the Gene Map of the E. coli Chromosome

Genetic Symbols Mutant Character

Enzyme or Reaction Affected

araD Cannot use the sugar arabinose as a carbon source

L-Ribulose-5-phosphate-4-epimerase

araA

L-Arabinose isomerase

araB

L-Ribulokinase

araC

argB

Requires the amino acid arginine for growth

N-Acetylglutamate synthetase

argC

N-Acetyl-7-glutamokinase

argH

N-Acetylglutamic-7-semialdehyde

dehydrogenase

argG

Acetylornithine-^-transaminase

argA

Acetylornithinase

argD

Ornithine transcarbamylase

argE

Argininosuccinic acid synthetase

argF

Argininosuccinase

argR Arginine operon regulator

aroA, B, C

Requires several aromatic amino acids

Shikimic acid to

> and vitamins for growth

3-Enolpyruvyl-shikimate-5-phosphate

aroD

Biosynthesis of shikimic acid

azi Resistant to sodium azide

bio Requires the vitamin biotin for growth

carA Requires uracil and arginine

Carbamate kinase

carB

chlA-E Cannot reduce chlorate

Nitrate-chlorate reductase and hydrogen

lysase

cysA

Requires the amino acid cysteine for growth

3-Phosphoadenosine-5-phosphosulfate

to sulfide

cysB

Sulfate to sulfide; four known enzymes

cysC >

dapA 1 Requires the cell-wall component diaminopimelic acid

Dihydrodipicolinic acid synthetase

dapB

N-Succinyl-diaminopimelic acid deacylase

dap + horn Requires the amino acid precursor homoserine and the

Aspartic semialdehyde dehydrogenase

cell-wall component diaminopimelic acid for growth

dnaA-Z Mutation, DNA replication

DNA biosynthesis

Dsd Cannot use the amino acid D-serine as a nitrogen

D-Serine deaminase

source

fla Flagella are absent

galA

Cannot use the sugar galactose as a carbon source

Galactokinase

galB j

Galactose- 1 -phosphate uridyl transferase

galD J

Uridine-diphosphogalactose-4-epimerase

glyA Requires glycine

Serine hydroxymethyl transferase

gua Requires the purine guanine for growth

H The H antigen is present

his Requires the amino acid histidine for growth

Ten known enzymes*

hsdR Host restriction

Endonuclease R

He Requires the amino acid isoleucine for growth

Threonine deaminase

ilvA >

Requires the amino acids isoleucine and valine

a-Hydroxy-p-keto acid rectoisomerase

<
ilvB

for growth

a, (3-Dihydroxyisovaleric dehydrase*

ilvC >

Transaminase B

ind (indole) Cannot grow on tryptophan as a carbon source

Tryptophanase

A (attX) Chromosomal location where prophage X is

normally inserted

lad Lac operon regulator

lacY Unable to concentrate (3-galactosides

Galactoside permease

lacZ Cannot use the sugar lactose as a carbon source

(3-Galactosidase

continued

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

Table 7.8 continued

Genetic Symbols Mutant Character

Enzyme or Reaction Affected

lacO Constitutive synthesis of lactose operon proteins

Defective operator

leu Requires the amino acid leucine for growth

Three known enzymes*

lip Requires lipoate

Ion (long form) Filament formation and radiation sensitivity are

affected

lys Requires the amino acid lysine for growth

Diaminopimelic acid decarboxylase

lys + met Requires the amino acids lysine and methionine

for growth

Xrec, malT Resistant to phage X and cannot use the sugar maltose

Regulator for two operons

malK Cannot use the sugar maltose as a carbon source

Maltose permease

man Cannot use mannose sugar

Phosphomannose isomerase

melA Cannot use melibiose sugar

Alpha-galactosidase

met A-M Requires the amino acid methionine for growth

Ten or more genes

mtl Cannot use the sugar mannitol as a carbon source

Two enzymes

muc Forms mucoid colonies

Regulation of capsular polysaccharide

synthesis

nalA Resistant to nalidixic acid

O The O antigen is present

pan Requires the vitamin pantothenic acid for growth

pabB Requires />-aminobenzoate

phe A, B Requires the amino acid phenylalanine for growth

pho Cannot use phosphate esters

Alkaline phosphatase

pil Has filaments (pili) attached to the cell wall

plsB Deficient phospholipid synthesis

Glycerol 3-phosphate acyltransferase

polA Repairs deficiencies

DNA polymerase I

proA

Requires the amino acid proline for growth

proB

proC

ptsl Defective phosphotransferase system

Pts-system enzyme I

purA

Requires certain purines for growth

Adenylosuccinate synthetase

purB

Adenylosuccinase

purC, E

5-Aminoimidazole ribotide (AIR) to
5-aminoimidazole-4-(N-succino carboximide)
ribotide

purD

Biosynthesis of AIR

pyrB

Requires the pyrimidine uracil for growth

Aspartate transcarbamylase

pyrC

Dihydroorotase

pyrD

Dihydroorotic acid dehydrogenase

pyrE

Orotidylic acid pyrophosphorylase

pyrF ,

Orotidylic acid decarboxylase

R gal Constitutive production of galactose

Repressor for enzymes involved in

galactose production

R 1 pho, R2 pho Constitutive synthesis of phosphatase

Alkaline phosphatase repressor

R try Constitutive synthesis of tryptophan

Repressor for enzymes involved in

tryptophan synthesis

RC (RNA control) Uncontrolled synthesis of RNA

recA Cannot repair DNA radiation damage or recombine

rhaA-D Cannot use the sugar rhamnose as a carbon source

Isomerase, kinase, aldolase, and regulator

rpoA-D Problems of transcription

Subunits of RNA polymerase

serA Requires the amino acid serine for growth

3-Phosphoglycerate dehydrogenase

serB J

Phosphoserine phosphatase

str Resistant to or dependent on streptomycin

sue Requires succinic acid

continued

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7. Linkage and Mapping in
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Summary

111

Table 7.8 continued

Genetic Symbols Mutant Character

Eniyme or Reaction Affected

supB

Suppresses ochre mutations

t-RNA

tonA

Resistant to phages Tl and T5 (mutants called B/l, 5)

Tl, T5 receptor sites absent

tonB

Resistant to phage Tl (mutants called B/l)

Tl receptor site absent

T6, colK rec Resistant to phage T6 and colicine K

T6 and colicine receptor sites absent

T4 rec

Resistant to phage T4 (mutants called B/4)

T4 receptor site absent

tsx

T6 resistance

thi

Requires the vitamin thiamine for growth

tolC

Tolerance to colicine El

thr

Requires the amino acid threonine for growth

thy

Requires the pyrimidine thymine for growth

Thymidylate synthetase

trpA

Requires the amino acid tryptophan for growth

Tryptophan synthetase, A protein

trpB

Tryptophan synthetase, B protein

trpC

Indole-3-glycerolphosphate synthetase

trpD

Phosphoribosyl anthranilate transferase

trpE >

Anthranilate synthetase

tyrA

Requires the amino acid tyrosine for growth

Chorismate mutase T-prephenate

dehydrogenase

tyrR

Regulates three genes

uvrA-l

J Resistant to ultraviolet radiation

Ultraviolet-induced lesions in DNA are
reactivated

valS

Cannot charge Valyl-tRNA

Valyl-tRNA synthetase

xyl

Cannot use the sugar xylose as a carbon source

Source: B. J. Bachmann and K. B. Low, "Linkage map of Escherichia coli K-12," Microbiological Reviews, 44:1-56. Copyright © 1990 American Society for Microbi-
ology, Washington, D.C. Reprinted by permission.

* Denotes enzymes controlled by the homologous gene loci of Salmonella typhimurium.

SUMMARY

STUDY OBJECTIVE 1: To define bacteria and bacterial
viruses and learn about methods of studying them
149-154

Prokaryotes (bacteria) usually have a single circular chro-
mosome of double-stranded DNA. A bacteriophage consists
of a chromosome wrapped in a protein coat. Its chromo-
some can be DNA or RNA. Phenotypes of bacteria include
colony morphology, nutritional requirements, and drug re-
sistance. Phage phenotypes include plaque morphology
and host range. Replica-plating is a rapid screening tech-
nique for assessing the phenotype of a bacterial clone.

STUDY OBJECTIVE 2: To study life cycles and sexual
processes in bacteria and bacteriophages 154-166

In transformation, a competent bacterium can take up rela-
tively large pieces of DNA from the medium. This DNA can
be incorporated into the bacterial chromosome.

During the process of conjugation, the fertility factor, F,
is passed from an F + to an F~ cell. If the F factor integrates

into the host chromosome, an Hfr cell results that can pass
its entire chromosome into an F~ cell. The F factor is the
last region to cross into the F~ cell.

In transduction, a phage protein coat containing some
of the host chromosome passes to a new host bacterium.
Again, recombination with this new chromosomal segment
can take place.

STUDY OBJECTIVE 3: To make use of the sexual
processes of bacteria and their viruses to map their chro-
mosomes 155-171

We can map the phage chromosome by measuring recom-
bination after a bacterium has been simultaneously infected
by two strains of the virus carrying different alleles. In
E. coli, mapping is most efficiently accomplished via inter-
rupted mating and transduction. The former provides infor-
mation on general gene arrangement and the latter pro-
vides finer details.

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

SOLVED PROBLEMS

PROBLEM 1: A wild-type strain of B. subtilis is trans-
formed by DNA from a strain that cannot grow on galac-
tose (gal~) and also needs biotin for growth (bio ~).
Transformants are isolated by exposing the transformed
cells to minimal medium with penicillin, killing the wild-
type cells. After the penicillin is removed, replica-plating
is used to establish the genotypes of 30 transformants:

Class 1 gal~ bio~ 17

Class 2 gal ~ bio + 4

Class 3 gal + bio ~ 9

What is the relative co-occurrence of these two loci?

Answer: The three classes of colonies represent the
three possible transformant groups. Classes 2 and 3 are
single transformants and class 1 is the double transfor-
mant. We are interested in the relative co-occurrence of
the two loci. Therefore we divide the number of double
transformants by the total: r = 17/(17 + 4 + 9) = 0.57.
This is a relative value inverse to a map distance; the
larger it is, the closer the loci are to each other.

PROBLEM 2: A gal~ bio~ att\~ strain of E. colt is trans-
duced by P22 phages from a wild-type strain. Transduc-
tants are selected for by growing the cells with galactose
as the sole energy source. Replica-plating and testing for
lysogenic ability gives the genotypes of 106 transformants:

Class 1 gal bio~ att\~ 71

Class 2 gal + bio + atik~

Class 3 gal + bio ~ atfk + 9

Class 4 gal + bio + att\ + 26

What is the gene order, and what are the relative cotrans-
duction frequencies?

Answer: We have selected all transductants that are
gal + . Class 2 is in the lowest frequency (0) and therefore
represents the quadruple crossover between the trans-
ducing DNA and the host chromosome. From this, we see
that att\ must be in the middle because this low-
probability event is the one that would have switched
only the middle locus. In other words, the two end loci
would be recombinant, and the middle locus would have
the host allele. We can only calculate two cotransduction
frequencies because these are selected data. Note that in
class 1 , there is no cotransduction between gal and ei-
ther of the other two loci; class 2 would show the co-
transduction of gal and bio; class 3 represents the co-
transduction of gal and att\; and class 4 represents the
cotransduction of gal and both other loci. Therefore,
cotransduction values are

gal-attX = (9 + 26)/106 = 35/106 = 0.33
gal-bio = (0 + 26)/106 = 26/106 = 0.25.

EXERCISES AND PROBLEMS

BACTERIA AND BACTERIAL VIRUSES
IN GENETIC RESEARCH

1. What is the nature and substance of prokaryotic chro-
mosomes and viral chromosomes? Are viruses alive?

TECHNIQUES OF CULTIVATION

2. What are the differences between a heterotroph
and an auxotroph? a minimal and a complete
medium? an enriched and a selective medium?

3. What are the differences between a plaque and a
colony?

BACTERIAL PHENOTYPES

4. What genotypic notation indicates alleles that make
a bacterium

a. resistant to penicillin?

b. sensitive to azide?

c. require histidine for growth?

d. unable to grow on galactose?

e. able to grow on glucose?

f. susceptible to phage Tl infection?

5. An E. colt cell is placed on a petri plate containing X
phages. It produces a colony overnight. By what
mechanisms might it have survived?

6. An E. colt lawn is formed on a petri plate containing
complete medium. Replica-plating is used to trans-
fer material to plates containing minimal medium
and combinations of the amino acids arginine and
histidine (see the figure). Give the genotype of the
original strain as well as the genotypes of the odd
colonies found growing on the plates.

*Answers to selected exercises and problems are on page A-7.

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Exercises and Problems

173

Complete medium Minimal and arginine
\ if and histidine

Minimal and histidine Minimal and arginine

7. Prototrophic Hfr E. colt strain Gil, sensitive to
streptomycin and malT^ (can use maltose) is used
in a conjugation experiment. The str locus is one of
the last to be transferred, whereas the malT locus is
one of the first. This strain is mated to an F~ strain
resistant to streptomycin, malT (cannot utilize
maltose), and requiring five amino acids (histidine,
arginine, leucine, lysine, and methionine). Recombi-
nants are selected for by plating on a medium with
streptomycin, with maltose as the sole carbon
source, and all five amino acids present. Thus, all re-
combinant F~ cells will grow irrespective of their
amino acid requirements. Five colonies are grown
on the original plate with streptomycin, maltose,
and all five amino acids in question (see the figure).
These colonies are replica-plated onto minimal
medium containing various amino acids. What are
the genotypes of each of the five colonies?

Medium with amino acids, streptomycin, and maltose

Minimal medium +

Histidine and arginine Leucine and lysine

8. A petri plate with complete medium has six
colonies growing on it after one of the conjugation
experiments described earlier. The colonies are
numbered, and the plate is used as a master to repli-
cate onto plates of glucose-containing selective
(minimal) medium with various combinations of ad-
ditives. From the following data, which show the
presence (+) or absence (— ) of growth, give your
best assessment of the genotypes of the six
colonies.

Colony

On Minimal Medium +

Nothing

Xylose + arginine
Xylose + histidine
Arginine + histidine
Galactose + histidine
Threonine + isoleucine

+ valine
Threonine + valine

+ lactose

—

+
+
+

—

- - +

VIRAL PHENOTYPES

9. Give possible genotypes of an E. co/i-phage Tl sys-
tem in which the phage cannot grow on the bac-
terium. Give genotypes for aTl phage that can grow
on the bacterium.

SEXUAL PROCESSES IN BACTERIA
AND BACTERIOPHAGES

10. What is a plasmid? How does one integrate into a
host's chromosome? How does it leave?

11. In conjugation experiments, one Hfr strain should
carry a gene for some sort of sensitivity (e.g., azf or
str s ) so that the Hfr donors can be eliminated on se-
lective media after conjugation has taken place.
Should this locus be near to or far from the origin of
transfer point of the Hfr chromosome? What are the
consequences of either alternative?

12. How does a geneticist doing interrupted mating
experiments know that the locus for the drug-
sensitivity allele, used to eliminate the Hfr bacteria
after conjugation, has crossed into the F~ strain?

13. Diagram the step-by-step events required to integrate
foreign DNA into a bacterial chromosome in each of
the three processes outlined in the chapter (transfor-
mation, conjugation, transduction). Do the same for
viral recombination. (See also TRANSDUCTION)

Histidine and methionine Arginine and leucine Arginine and lysine

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

14. The DNA from a prototrophic strain of E. coli is iso-
lated and used to transform an auxotrophic strain
deficient in the synthesis of purines (purB~), pyrim-
idines (pyrC~), and the amino acid tryptophan
(trp~). Tryptophan was used as the marker to deter-
mine whether transformation had occurred (the se-
lected marker). What are the gene order and the rel-
ative co-occurrence frequencies between loci, given
these data:

trp + pyrC + purB + 86

trp + pyrC + purB~ 4

trp + pyrC~ purB + 61

trp + pyrC~ purB~ 14

15. Using the data in figure 7.16, draw a tentative map
of the E. coli chromosome.

16. Three Hfr strains ofE. coli (P4X, KL98, and Ra-2) are
mated individually with an auxotrophic F~ strain us-
ing interrupted mating techniques. Using the fol-
lowing data, construct a map of the E. coli chromo-
some, including distances in minutes.

Approximate Time of Entry

Donor Loci

Hfr P4X

Hfr KL98

gat

thr +

xyl +

lac +

his +

ilv +

argG +

Hfr Ra-2

70
87

8
79

How many different petri plates and selective media
are needed?

17. Design an experiment using interrupted mating and
create a resulting possible data set that would cor-
rectly map five of the loci on the E. coli chromo-
some (fig. 7.27).

18. Lederberg and his colleagues (Nester, Schafer, and
Lederberg, 1963, Genetics 48:529) determined gene
order and relative distance between genes using
three markers in the bacterium Bacillus subtilis.
DNA from a prototrophic strain (trp^ his + tyr + )
was used to transform the auxotroph. The seven
classes of transformants, with their numbers, are
tabulated as follows:

trp
his~
tyr~
2,600

trp~
his +
tyr~
418

trp~
his~
tyr
685

trp
his +
tyr~
1,180

trp
his~
tyr
107

trp

his

trp
his +
tyr' tyr

3,660 11,940

Outline the techniques used to obtain these data. Tak-
ing the loci in pairs, calculate co-occurrences. Con-
struct the most consistent linkage map of these loci.

19. In a transformation experiment, ana + b + c + strain
is used as the donor and ana~ b~ c~ strain as the re-
cipient. One hundred a + transformants are selected
and then replica-plated to determine whether b +
and c + are present. What can you conclude about
the relative positions of the genes, based on the
listed genotypes?

a b c

a + b~ c +

a + b + c~

a b c

+ 1 +

20. In a transformation experiment, an a b c strain

is used as donor and ana b c strain as recipient

If you select for a transformants, the least frequent

+ T + +

class is a b c . What is the order of the genes?

21. A mating between his + , leu + , tbr + ,pro + , str s cells
(Hfr) and his~ , leu~ ', thr~ ,pro~ , str r cells (F~) is al-
lowed to continue for twenty-five minutes. The mat-
ing is stopped, and the genotypes of the recombi-
nants are determined. What is the first gene to enter,
and what is the probable gene order, based on the
following data?

Genotype

Number of Colonies

his
leu A
thr^
pro

27
6

22. a. In a transformation experiment, the donor is

trp + leu^ arg + , and the recipient is trp~ leu~
arg~. The selection process is for trp + transfor-
mants, which are then further tested. Forty per-
cent are trp + arg + ; 5% are trp + leu + . In what two
possible orders could the genes be arranged?
b. You can do only one more transformation to de-
termine gene order. You must use the same
donor and recipient, but you can change the se-
lection procedure for the initial transformants.
What should you do, and what results should you
expect for each order you proposed in a?.

23. DNA from a bacterial strain that is a + b + c + is used
to transform a strain that is a~ b~ c.The numbers
of each transformed genotype appear. What can we
say about the relative position of the genes?

Genotype

Number

b +

+
a

b +

a +

b +

a +

b +

214

231

206

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Exercises and Problems

175

.+ ».+ _+ j+

24. An Hfr strain that is a b^ c^ d^ e^ is mated with an
F" strain that is a~ b~ c~ d~ e~ . The mating is in-
terrupted every five minutes, and the genotypes of
the F~ recombinants are determined. The results ap-
pear following. (A plus indicates appearance; a
minus the lack of the locus.) Draw a map of the
chromosome and indicate the position of the F
factor, the direction of transfer, and the minutes be-
tween genes.

Time

—

+
+
+
+
+
+
+
+

+
+
+

+
+
+
+
+
+
+
+
+
+

25. A bacterial strain that is lys + bis + val + is used as a
donor, and lys~ his~ val~ as the recipient. Initial
transformants are isolated on minimal medium +
histidine + valine.

a. What genotypes will grow on this medium?

b. These colonies are replicated to minimal medium
+ histidine, and 75% of the original colonies
grow. What genotypes will grow on this medium?

c. The original colonies are also replicated to mini-
mal medium + valine, and 6% of the colonies
grow. What genotypes will grow on this medium?

d. Finally, the original colonies are replicated to
minimal medium. No colonies grow. From this
information, what genotypes will grow on mini-
mal medium + histidine and on minimal
medium + valine?

e. Based on this information, which gene is closer
to lys?

f. The original transformation is repeated, but the
original plating is on minimal medium + lysine
+ histidine. Fifty colonies appear. These colonies
are replicated to determine their genotypes, with
these results:

val his lys
val + his~ lys

37
3

val his lys

Based on all the results, what is the most likely gene
order?

LIFE CYCLES OF BACTERIOPHAGES

26. Define prophage, lysate, lysogeny and temperate
phage.

27. Outline an experiment to demonstrate that two
phages do not undergo recombination until a bac-
terium is infected simultaneously with both.

Doermann (1953, Cold Spr. Harb. Symp. Quant.
Biol. 18:3) mapped three loci of phage T4: minute,
rapid lysis, and turbid. He infected E. coli cells with
both the triple mutant (m r tu) and the wild-type
(m + r + tu + ) and obtained the following data:

28.

+
r

tu +

5,467

474

162

853

965

172

520

3,729

What is the linkage relationship among these loci?
In your answer include gene order, relative distance,
and coefficient of coincidence.

29. Wild-type phage T4 (r + ) produce small, turbid
plaques, whereas rll mutants produce large, clear
plaques. Four rll mutants (a—d) are crossed. (As-
sume, for the purposes of this problem, that a-d are
four closely linked loci. The actual structure of the
rll region is presented in chapter 12. Here, assume

that a X b means a b + c + d + X a* b

c + d + .)

These percentages of wild-type plaques are ob-
tained in crosses:

aX b

0.3

a X c

1.0

a X d

0.4

b X c

0.7

b X d

0.1

cX d

0.6

Deduce a genetic map of these four mutants.

30. A phage cross is performed between a + b + c + and
a b c phage. Based on these results, derive a com-
plete map:

a + b + c
a + b + c
a be
a + b c
a b c
a b + c
a b c +
a b c

1,801
954
371
160
178
309
879

1,850

6,502

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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses

31.

The rll mutants of T4 phage will grow and produce
large plaques on strain B; rll mutants will not grow
on strain K12. Certain crosses are performed in
strain B. (As with question 29, assume that the three
mutants are of three separate loci in the rll region.)
By diluting and plating on strain B, it is determined
that each experiment generates about 250 X 10 7
phage. By dilution, approximately 1/10,000 of the
progeny are plated on K12 to generate these wild-
type recombinants (plaques on K12):

1 X 2

l x 3
2X3

50
25

Draw a map of these three mutants (1,2, and 3) and
indicate the distances between them.

TRANSDUCTION

32. Define and illustrate specialized and generalized
transduction.

33. In E. coli, the three loci ara, leu, and ilvH are within
1/2-minute map distance apart. To determine the ex-
act order and relative distance, the prototroph (ara +
leu + ilvH + ) was infected with transducing phage
PI. The lysate was used to infect the auxotroph
(ara~ leu~ ilvH}. The ara^ classes of transductants
were selected to produce the following data:

ara

leu~

leu +

leu~

leu +

ilvH~

ilvH +

ilvH

340

Outline the specific techniques used to isolate the
various transduced classes. What is the gene order
and what are the relative cotransduction frequen-
cies between genes? Why do some classes occur so
infrequently?

34. Consider this portion of an E. coli chromosome:

thr ara leu

Three ara loci, ara-\,ara-2, and ara-3, are located in
the ara region. A mutant of each locus (ara-\ ~, ara-
2~, and ara-3 ) was isolated, and their order with
respect to thr and leu was analyzed by transduction.
The donor was always thr + leu + and the recipient
was always thr~ leu~ ' . Each ara mutant was used as
a donor in one cross and as a recipient in another;
ara + transductants were selected in each case. The
ara^ transductants were then scored for leu^ and
thr + . Based on the following results, determine the
order of the ara~ mutants with respect to thr and
leu.

Cross

Recipient

Donor

Ratio:

thr~

' ara + leu +

thr + ara + leu~

ara-\~

ara-2~

48.5

ara-2~

ara-\~

2.4

ara-\~

ara-3

4.0

ara-^~

ara-\~

19.1

ara-2~

ara-^~

1.5

ara-3

ara-2~

25.5

35.

An E. coli strain that is leu + tbr + azf is used as a
donor in a transduction of a strain that is leu~ thr~

azi . Either leu or thr transductants are selected
and then scored for unselected markers. The results
are obtained:

Selected Marker

Unselected Markers

leu
leu
thr
thr

48% azf
2% thr +
3% leu +
0% azf

What is the order of the three loci?

CRITICAL THINKING QUESTIONS

1. Consider the data from table 7.4. Is there another way 2. Why might transformation have evolved, given that
to interpret the data other than coming from a circular the bacterium is importing DNA from a dead organ-

bacterial chromosome? ism?

Suggested Readings for chapter 7 are on page B-4.

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CYTOGENETICS

STUDY OBJECTIVES

1. To observe the nature and consequences of chromosomal
breakage and reunion 178

2. To observe the nature and consequences of variation in
chromosome numbers in human and nonhuman
organisms 190

STUDY OUTLINE

Variation in Chromosomal Structure 178

Single Breaks 178

Two Breaks in the Same Chromosome 179

Two Breaks in Nonhomologous Chromosomes 182

Centromeric Breaks 185

Duplications 185

Chromosomal Rearrangements in Human Beings 186
Variation in Chromosome Number 190

Aneuploidy 190

Mosaicism 190

Aneuploidy in Human Beings 192

Euploidy 197
Summary 199
Solved Problems 200
Exercises and Problems 200
Critical Thinking Questions 202
Box 8.1 A Case History of the Use of Inversions to
Determine Evolutionary Sequence 182

Chromosomes of an individual with trisomy 21 ,
Down syndrome. (© Dr. Ram Verma /Phototake, NYC.)

177

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Chapter Eight Cytogenetics

Our understanding of the chromosomal the-
ory of genetics grew primarily through
mapping loci, using techniques that require
alternative allelic forms, or mutations, of
these loci. Changes in the genetic material
also occur at a much coarser level — the level of cytoge-
netics, which is a level visible under the light microscope.
The word cytogenetics combines the words cytology and
genetics; cytology is the study of cells. Cytogenetics is
thus defined as the study of cells from the perspective of
genetics. In practice, it is the study of changes in the
gross structure and number of chromosomes in cells. In
this chapter, we investigate how these alterations happen
and what their consequences are to the organism.

VARIATION IN

CHROMOSOMAL

STRUCTURE

In general, chromosomes can break due to ionizing radi-
ation, physical stress, or chemical compounds. When a
break occurs in the chromosome before DNA replica-
tion, during the S phase of the cell cycle (see fig. 3.6), the
break itself is replicated. After the S phase, any breaks
that occur affect single chromatids.

Every break in a chromatid produces two ends. These
ends have been described as "sticky," meaning simply that
enzymatic processes of the cell tend to reunite them.
Broken ends do not attach to the undamaged terminal
ends of other chromosomes. (Normal chromosomal ends
are capped with structures called telomeres — see chap-
ter 15.) If broken ends are not brought together, they can
remain broken. But, if broken chromatid ends are
brought into apposition, they may rejoin in any of several
ways. First, the two broken ends of a single chromatid
can reunite. Second, the broken end of one chromatid
can fuse with the broken end of another chromatid, re-
sulting in an exchange of chromosomal material and a
new combination of alleles. Multiple breaks can lead to a
variety of alternative recombinations. These chromo-
somal aberrations have major genetic, evolutionary, and
medical consequences. The types of breaks and reunions
discussed in this chapter can be summarized as follows:

I. Noncentromeric breaks

A. Single breaks

1. Restitution

2. Deletion

3. Dicentric bridge

B. Two breaks (same chromosome)

1. Deletion

2. Inversion

C. Two breaks (nonhomologous chromosomes)

II. Centromeric breaks

A. Fission

B. Fusion

Single Breaks

If a chromosome breaks, the broken ends may rejoin.
When the broken ends of a single chromatid rejoin (in a
process called restitution), there is no consequence to
the break. If they do not rejoin, the result is an acentric
fragment, without a centromere, and a centric frag-
ment, with a centromere. The centric fragment migrates
normally during the division process because it has a
centromere. The acentric fragment, however, is soon lost.
It is subsequently excluded from the nuclei formed and
eventually degrades. In other words, the viable, centric
part of the chromosome has suffered a deletion. After mi-
tosis, the daughter cell that receives the deletion chro-
mosome may show several effects.

Pseudodominance is one possible effect. (This term
was used in chapter 5 when we described alleles located
on the X chromosome. With only one copy of the locus
present, a recessive allele in males shows itself in the
phenotype as if it were dominant — hence the term pseu-
dodominance.} The normal chromosome homologous
to the deletion chromosome has loci in the region, and
recessive alleles show pseudodominance. A second pos-
sible effect is that, depending on the length of the
deleted segment and the specific loci lost, the imbalance
the deletion chromosome creates in the daughter cell
may be lethal. If the deletion occurs before or during
meiosis, it may be observed under the microscope. We
discuss this event later in the chapter.

A single break can have yet another effect. Occasion-
ally, the two centric fragments of a single chromosome may
join, forming a two-centromere, or dicentric, chromo-
some and leaving the two acentric fragments to join or, al-
ternatively, remain as two fragments (fig. 8.1). The acentric
fragments are lost, as mentioned before. Because the cen-
tromeres are on sister chromatids, the dicentric fragment is
pulled to opposite ends of a mitotic cell forming a bridge
there; or, if meiosis is occurring, the dicentric fragment is
pulled apart during the second meiotic division. The ulti-
mate fate of this bridge is breakage as the spindle fibers pull
the centromeres to opposite poles (or possibly exclusion
from a new nucleus if the bridge is not broken).

The dicentric chromosome does not necessarily break
in the middle, and subsequent processes exacerbate the im-
balance created by an off-center break: duplications occur on
one strand, whereas more deletions occur on the other (fig.
8.2). In addition, the "sticky" ends produced on both frag-
ments increase the likelihood of repeating this breakage-
fusion-bridge cycle in each generation. The great imbal-
ances resulting from the duplications and deletions usually
cause the cell line to die within several generations.

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Variation in Chromosomal Structure

179

a b c

d e f g h

i J

Centric
fragments

a b c d e f g h

i J

Acentric

fragments

' J

a b c d e f g h i j

Dicentric O Acentric

chromosome O fragment

a b c d e f g h i j

Figure 8.1 Chromosomal break with subsequent reunion to
form a dicentric chromosome and an acentric fragment.

Two Breaks in the Same Chromosome
Deletion

Figure 8.3 shows two of the possible results when two
breaks occur in the same chromosome. One alternative is
a reunion that omits an acentric fragment, which is then
lost. The centric piece, missing the acentric fragment
(e-f-g in fig. 8.3), is a deletion chromosome. An organism

gh duplication

f e d

gh deletion

Figure 8.2 Breakage of a dicentric bridge causes duplications
and additional deficiencies.

having this chromosome and a normal homologue will
have, during meiosis, a bulge in the tetrad if the deleted
section is large enough (fig. 8.4). The bulge also appears
in the paired, polytene giant salivary gland chromosomes
of Drosophila. (Note that when a bulge like that illus-
trated in figure 8.4 is seen in paired chromosomes, it in-
dicates that one chromosome has a piece that is missing
in the other. In our example, the bulge resulted from a
deletion in one chromosome; it could also result from an
insertion of a piece in the other chromosome.)

Inversion

Two breaks in the same chromosome can also lead to in-
version, in which the middle section is reattached but in
the inverted configuration (see fig. 8.3). An inversion has
several interesting properties. To begin with, fruit flies
homozygous for an inversion show new linkage relations
when their chromosomes are mapped. One outcome of
this new linkage arrangement is the possibility of a posi-
tion effect, a change in the expression of a gene due to
a changed linkage arrangement. Position effects are ei-
ther stable, as in Bar eye of Drosophila (to be discussed),
or variegated, as with Drosophila eye color. A normal fe-
male fly that is heterozygous (X^X + ) has red eyes. If,
however, the white locus is moved through an inversion
so that it comes to lie next to heterochromatin (fig. 8.5),
the fly shows a variegation — patches of the eye are
white. This is presumably caused by a spread of the tight
coiling of the heterochromatin, "turning off" the expres-
sion of the locus. In a heterozygote, if the turned-off al-
lele is the wild-type, the cell will express the normally re-
cessive white-eye allele. Depending on what happens in
each cell, patches of red and white eye color result.

When synapsis occurs in an inversion heterozygote,
either at meiosis or in the Drosophila salivary gland dur-
ing endomitosis, a loop often forms to accommodate the
point-for-point pairing process (figs. 8.6 and 8.7). An out-
come of this looping tendency is crossover suppres-
sion. That is, an inversion heterozygote shows very little
recombination of alleles within the inverted region.
The reason is usually not that crossing over is actually

/ J

Inversion
chromosome

Deletion
chromosome

Acentric
fragment

Figure 8.3 Two possible consequences of a double break {top arrows) in the same chromosome.

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Chapter Eight Cytogenetics

c d

O
O

Normal chromatids

Deletion chromatids

Figure 8.4 A bulge can occur in a meiotic tetrad if a large
deletion has occurred.

Heterochromatin

/ — o

suppressed, but rather that the products of recombina-
tion within a loop are usually lost. (Suppression can also
occur in small inversions where loops don't form.) Fig-
ure 8.8 shows a crossover within a loop. The two nonsis-
ter chromatids not involved in a crossover in the loop
will end up in normal gametes (carrying either the nor-
mal chromosome or the intact inverted chromosome).
The products of the crossover, rather than being a simple
recombination of alleles, are a dicentric and an acentric
chromatid. The acentric chromatid is not incorporated
into a gamete nucleus, whereas the dicentric chromatid
begins a breakage-fusion-bridge cycle that creates a ge-
netic imbalance in the gametes. The gametes thus carry
chromosomes with duplications and deficiencies.

The inversion pictured in figure 8.8 is a paracentric
inversion, one in which the centromere is outside the
inversion loop. A pericentric inversion is one in which
the inverted section contains the centromere. It, too,
suppresses crossovers, but for slightly different reasons

\ o

Heterochromatin-

Figure 8.5 An inversion in the X chromosome of Drosophila
produces a variegation in eye color in a female if her other
chromosome is normal and carries the white-eye allele (K w ).

(fig. 8.9). All four chromatid products of a single crossover
within the loop have centromeres and are thus incorpo-
rated into the nuclei of gametes. However, the two re-
combinant chromatids are unbalanced — they both have
duplications and deficiencies. One has a duplication for

a b c

a b c d a

Synapsis
occurs

a b c

a be

o
o

h i j

Figure 8.6 Tetrad at meiosis showing the loop characteristic of an inversion heterozygote.

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Variation in Chromosomal Structure

181

Figure 8.7 A Drosophila heterozygous for an inversion will
show a loop in the salivary gland chromosomes. (Compare with
figure 8.6.)

a-b-c-d and is deficient for h-i-j, whereas the other is the
reciprocal — deficient for a-b-c-d and duplicated for h-i-j
(in fig. 8.9). These duplication-deletion gametes tend to
form inviable zygotes. The result, as with the paracentric
inversion, is the apparent suppression of crossing over.

Results of Inversion

Crossing over within inversion loops results in semi-
sterility. Almost all gametes that contain dicentric or im-
balanced chromosomes form inviable zygotes. Thus, a
certain proportion of the progeny of inversion heterozy-
gotes are not viable.

Inversions have several evolutionary ramifications.
Those alleles originally together in the noninversion
chromosome and those found together within the inver-

sion loop tend to stay together because of the low rate of
successful recombination within the inverted region. If
several loci affect the same trait, the alleles are referred to
as a supergene. Until careful genetic analysis is done,
the loci in a supergene could be mistaken for a single lo-
cus; they affect the same trait and are inherited appar-
ently as a single unit. Examples include shell color and
pattern in land snails and mimicry in butterflies (see
chapter 21). Supergenes can be beneficial when they in-
volve favorable gene combinations. However, at the same
time, their inversion structure prevents the formation of
new complexes. Supergenes, therefore, have evolution-
ary advantages and disadvantages. Chapter 21 discusses
these evolutionary topics in more detail.

Sometimes the inversion process produces a record of
the evolutionary history of a group of species. As species
evolve, inversions can occur on preexisting inversions.
This leads to very complex arrangements of loci. We can
readily study these patterns in Diptera by noting the
changed patterns of bands in salivary gland chromo-
somes. Since certain arrangements can only come about
by a specific sequence of inversions, it is possible to know
which species evolved from which. The same series of
events can occur within the same species (box 8.1).

In summary then, inversions result in suppressed
crossing over, semisterility, variegation position effects,
and new linkage arrangements. All of these events have
evolutionary consequences.

a b c

o
o

h i i

a b c d e

a b c d g

g d c b a

i J

Nonrecombinant chromosome

Dicentric chromosome

Nonrecombinant inversion
chromosome

Acentric chromosome

Figure 8.8 Consequences of a crossover in the loop region of a paracentric inversion heterozygote.

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Chapter Eight Cytogenetics

BOX 8. 1

In 1966, David Futch published a
study of the chromosomes of a
fruit fly, Drosophila ananassae,
an organism widely distributed
throughout the tropical Pacific. The
study was designed to determine
something about the species status of
various melanic forms of the fly. In
the course of his work, Futch looked
at the salivary gland chromosomes of
flies from twelve different localities.
He discovered twelve paracentric in-
versions, three pericentric inver-
sions, and one translocation. Because
of the precise banding patterns of
these chromosomes, it was possible

Experimental
Methods

A Case History of the Use of

Inversions to Determine

Evolutionary Sequence

to determine the breakage points for
each inversion.

Observation of several popula-
tions that have had sequential
changes in their chromosomes makes

it possible to determine the sequence
of successive changes. Once one
knows the sequence of changes in
different populations of Drosophila
ananassae, along with the geo-
graphic locations of the populations,
it is possible to determine the history
of the way the flies colonized these
tropical islands. D. ananassae is par-
ticularly suited to this type of work
because it is believed to be a recent
invader to most of the Pacific Islands
that it occupies. It is of interest to
know about the spread of this
species as an adjunct to studies of hu-
man migration in the Pacific Islands

Figure 1 Photomicrographs of the left arm of chromosome
2 (2L) from larval Drosophila ananassae heterozygous for vari-
ous complex gene arrangements, (a) Pairing when heterozy-
gous for standard gene sequence and overlapping inversions
(2LC; 2LD) and inversion 2LB (Standard x Tutuila light).
(b) Pairing when heterozygous for standard gene sequence
and single inversion 2LC and overlapping inversions (2LE;
2LB: Standard x New Guinea), (c) Pairing when heterozygous
for overlapping inversions (2LD; 2LE; 2LF: Tutuila light x New
Guinea). (From David G. Futch, "A study of speciation in South Pacific
populations of Drosophila ananassae," in Marshall R. Wheeler, ed., Studies
in Genetics, no. 6615 [Austin: University of Texas Press, 1966].
Reproduced by permission.)

(b)

i#V*/i

ft*flfitf

(a)

(c)

Two Breaks in Nonhomologous Chromosomes

Breaks can occur simultaneously in two nonhomologous
chromosomes. Reunion can then take place in various
ways. The most interesting case occurs when the ends of
two nonhomologous chromosomes are translocated to
each other in a reciprocal translocation (fig. 8. 10). The
organism in which this has happened, a reciprocal translo-

cation heterozygote, has all the genetic material of the nor-
mal homozygote.Two outcomes of a reciprocal transloca-
tion, like those of an inversion, are new linkage arrange-
ments in a homozygote — an organism with translocated
chromosomes only — and variegation position effects.

During synapsis, either at meiosis or endomitosis, a
point-for-point pairing in the translocation heterozygote

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Variation in Chromosomal Structure

183

(a)

— 38

^ 36

_^ 35

34
33

32
31
30

= 29

28
27

g 25

=■ 24

^ 23

= 22

~ 21

— 19

— 18

= 16

™ 15

— 13
^ 12

- 10

_9_
8_

_6_
_5_

_3_

_2_

— — -

39
38
37

— -

35
34

CO
I

==si

I -

28
27

11 —

I- 12
(b)

g 25

=■ 24

^ 23

= 22

~ 21

— 19

— 18

—

16
15
14

sSJSi

= 2

— 38

30 =

33 =

15 __

16
10

11 =

(c)

™ 35

= 29

28
27

H 25
24

7
6
5
4

■*

13 ■

14 -

12 = 2

—

23 °

^^^

20
19
18

— 9

— 8

39
38

.r**~~

^—

35
34

31 :::::

15 —

18 —

19 =

23
24

25 =

26
27

28
29

= 33

= 14

= 13

3 —

5
6

11 =

(d)

Figure 2 Chromosomal maps of 2L. (a) Standard gene sequence.
(b) Ponape: breakpoints of 2LC and 2LB are indicated and the segments
are shown inverted, (c) Tutuila light: breakpoints of 2LD are indicated. 2LC
and 2LB are inverted. 2LD, which overlaps 2LC, is also shown inverted.
(d) New Guinea: breakpoints of 2LE and 2LG are indicated. 2LC, 2LB,
and 2LE are shown inverted. Note: only the breakpoints of 2LF and 2LG
are shown; neither of these is inverted in the map. (From David G. Futch, "A
study of speciation in South Pacific populations of Drosophila ananassae," in Marshall R.
Wheeler, ed., Studies in Genetics, no. 6615 [Austin: University of Texas Press, 1966].
Reproduced by permission.)

(a)

l^ i | M ^ | pM^ l <^

! *

(b)

•%^^P

(c)

Figure 3 Photomicrographs of the right arm of
chromosome 2 (2R) from larvae heterozygous for
various complex gene arrangements, (a) Pairing
when heterozygous for standard gene sequence
and overlapping inversions (2RA; 2RB: Standard
x Tutuila light), (b) Pairing when heterozygous
for standard gene sequence and overlapping
inversions (2RA; 2RC) and inversion 2RD (Stan-
dard x New Guinea), (c) Pairing when heterozy-
gous for overlapping inversions 2RB, 2RC, and
2RD. Inversion 2RA is homozygous (Tutuila light
X New Guinea). (From David G. Futch, "A study of
speciation in South Pacific populations of Drosophila
ananassae," in Marshall R. Wheeler, ed., Studies in Genet-
ics, no. 6615 [Austin: University of Texas Press, 1966].

Reproduced by permission.)

continued

can be accomplished by the formation of a cross-shaped
figure (fig. 8.10). Such a figure is diagnostic of a recipro-
cal translocation. A single crossover in a reciprocal
translocation heterozygote will not produce chromatids
that are further imbalanced, as it does in an inversion
heterozygote. However, reciprocal translocation het-
erozygotes do produce nonviable progeny Problems

can arise when centromeres separate at the first mei-
otic division.

Segregation After Translocation

Since two homologous pairs of chromosomes are
involved, we have to keep track of the independent

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Chapter Eight Cytogenetics

because D. ananassae is commensal
with people.

Some of Futch's results are shown
in figures 1-4, which diagram the left
and right arms of the fly's second
chromosome, as well as the synaptic
patterns. We can see vividly the se-
quence of change in which one in-
version occurs after a previous inver-

= 31
gi 30
= 29

m 27

— 26

= 24

— 21

— 20

— 16

= 15

_ ^

~ 13
= 12

= 10

— 9

= 6
= 5

(a)

= 3

--» ■•

%l 2

BOX 8.1 CONTINUED

sion has already taken place. In fig-
ures 2 and 4, the standard (a) gave
rise to (b), which then gave rise inde-
pendently to (c) and (d). The stan-
dard is from Majuro in the Marshall
Islands and is believed to be in the an-
cestral group of the species. Ponape
is the home of (b), (c) is from Tutuila
(eastern Samoa), and (d) is from New

DC
CO

(b)

^rrrrrrr.

^ 30

= 29

=: 28

= 27

= 26
= 25

= 24

m. 23

= 21

; 18

_ 1?

= 15
14

3 —

5 =

6 =

7 =

8 =

9 —
10 =

| I III:
CO CO

2
■ 1

-_^-

^ 30

= 29

:=: 28

= 27
—.. 26

= 25

= 24

= 23

" 22

::::::::: 8

= 7

— 5

CQ
DC
CO

15 =

17 —

18 —

19 _
20

21 =

9 =

10 =

11 _

12 —

13 - 3
'l£ 2

(c)

—

■ 1

Guinea. Thus, the sequence is Majuro
to Ponape, and from there the same
stock was transferred to Tutuila and
New Guinea. This type of analysis has
been useful in the Drosophila group
throughout its range but especially in
the Pacific Island populations and in
the southwestern United States.

= 31

= 30

= 29

[NO

3 27

= 26

= ?5

20 —

21 =

Q
DC

22^

23 ^

24 ^

= 19

;;= 1 7

— 16

= 15

= 14

3 ■=

4 =l
h — :
6 -

lull 1
!|i>i|

_ 13

— 12

= 10

= 8

= 7

(d)

Figure 4 Chromosomal maps of 2R. (a) Standard gene sequence, (b) Ponape: break-
points of 2RA are indicated and the segment is shown inverted, (c) Tutuila light: break-
points of 2RB are indicated. 2RA is inverted and 2RB, which overlaps it, is also shown
inverted, (d) New Guinea: breakpoints of 2RC and 2RD are indicated. 2RA is inverted;
2RC, which overlaps 2RA, and 2RD are shown inverted. (From David G. Futch, "A study of
speciation in South Pacific populations of Drosophila ananassae," in Marshall R. Wheeler, ed., Studies in
Genetics, no. 6615 [Austin: University of Texas Press, 1966]. Reproduced by permission.)

segregation of the centromeres of the two tetrads. There
are two common possibilities and one that occurs less of-
ten (fig. 8.11). The first, called alternate segregation,
occurs when the first centromere assorts with the fourth
centromere, leaving the second and third centromeres to
go to the opposite pole. The result will be balanced

gametes, one with normal chromosomes and the other
with a reciprocal translocation. Also likely is the
adjacent-1 type of segregation, in which the first and
third centromeres segregate together in the opposite di-
rection from the second and fourth centromeres. Here,
both types of gametes are unbalanced, carrying duplica-

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Variation in Chromosomal Structure

185

c d

abed

a b c
j i h

e f g h i i

f 9

b a

Nonrecombinant chromosome
Imbalanced chromosome

Nonrecombinant inversion
chromosome

Imbalanced chromosome

Figure 8.9 Consequences of a crossover in the loop region of a pericentric inversion
heterozygote.

tions and deficiencies that are usually lethal. Since
adjacent- 1 segregation occurs at a relatively high fre-
quency a significant amount of sterility results from the
translocation (as much as 50%).

An adjacent-2 type of segregation (fig. 8. 1 1), in which
homologous centromeres go to the same pole (first with
second, third with fourth), is a third possibility. This can re-
sult when the cross-shaped double tetrad opens into a cir-
cle in late prophase I. In the German cockroach, adjacent-
2 patterns have been observed in 10 to 25% of meioses,
depending upon which chromosomes are involved.

In summary, then, reciprocal translocations result in
new linkage arrangements, variegated position effects, a
cross-shaped figure during synapsis, and semisterility.

Centromeric Breaks

Another interesting variant of the simple reciprocal
translocation occurs when two acrocentric chromo-
somes join at or very near their centromeres. The
process, called a Robertsonian fusion after cytologist
W. Robertson, produces a decrease in the number of
chromosomes, although virtually the same amount of ge-
netic material is maintained. Often, closely related
species undergo Robertsonian fusions and end up with

markedly different chromosome numbers without any
significant difference in the quantity of their genetic ma-
terial. Therefore, cytologists frequently count the number
of chromosomal arms rather than the number of chro-
mosomes to get a more accurate picture of species affini-
ties. The number of arms is referred to as the
fundamental number, or NF (French: nombre fonda-
mentale). In a similar fashion, centromeric fission in-
creases the chromosome number without changing the
fundamental number.

Duplications

Duplications of chromosomal segments can occur, as we
have just seen, by the breakage-fusion-bridge cycle or by
crossovers within the loop of an inversion. There is an-
other way that duplications arise in small adjacent regions
of a chromosome. We illustrate this with a particularly in-
teresting example, the Bar eye phenotype in Drosophila
(fig. 8.12). The wild-type fruit fly has about 800 facets in
each eye. The Bar (B) homozygote has about 70 (a range
of 20-120 facets). Another allele, Doublebar (BB: some-
times referred to as Ultrabar, B u ), brings the facet num-
ber of the eye down to about 45 when heterozygous and
to about 25 when homozygous. Around 1920, researchers

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Chapter Eight Cytogenetics

c d

c° d

c d

c° d

1 2 3 4 5 6

h Q f 4 5 6

1 2 3 4 5 6

Synapses occur

a b c d e

O
O

o
o

a b c d e

~N\

Figure 8.10 A reciprocal translocation heterozygote forms after breaks occur in
nonhomologous chromosomes. Synapsis at meiosis forms a cross-shaped figure.

showed that about one progeny in 1,600 from homozy-
gous Bar females is Doublebar. This is much more fre-
quent than we expect from mutation.

Alfred Sturtevant found that in every Doublebar fly, a
crossover had occurred between loci on either side of
the Bar locus. He suggested that the change to Double-
bar was due to unequal crossing over rather than to a
simple mutation of one allele to another (fig. 8.13). If the
homologous chromosomes do not line up exactly during
synapsis, a crossover produces an unequal distribution of
chromosomal material. Later, an analysis of the banding
pattern of the salivary glands confirmed Sturtevant's hy-
pothesis. It was found that Bar is a duplication of several
bands in the 16A region of the X chromosome (fig. 8.14).
Doublebar is a triplication of the segment.

A position effect also occurs in the Bar system. A Bar
homozygote (B/E) and a Doublebar/wild-type heterozy-
gote (BB/B + ) both have four copies of the 16A region. It
would therefore be reasonable to expect that both geno-
types would produce the same phenotype. However, the
Bar homozygote has about seventy facets in each eye,
whereas the heterozygote only has about forty-five. Thus,
not only the amount of genetic material, but also its con-

figuration, determines the extent of the phenotype. Bar
eye was the first position effect discovered.

Chromosomal Rearrangements
in Human Beings

Several human syndromes and abnormalities are the re-
sult of chromosomal rearrangements, including deletions
and translocations. The most common are described
here. Keep three points in mind as you read. First, all of
these disorders are rare. Second, the deletion syndromes
are often caused by a balanced translocation in one of
the parents. And third, about one in five hundred live
births contains a balanced rearrangement of some kind,
either a reciprocal translocation or inversion.

Fragile-X Syndrome

The most common cause of inherited mental retardation
is the fragile-X syndrome. It occurs in about one in
every 1,250 males and about one in every 2,000 females.
Symptoms include mental retardation, altered speech
patterns, and other physical attributes. The condition is

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Variation in Chromosomal Structure

187

First
Second

O
O

o
o

Alternate segregation

Adjacent -1 segregation

Third

X Fourth

Adjacent -2 segregation

First with
fourth

a
b

d 6
e 15

9 |3

h 12

Normal

Second with
third

e
3
2

6
5

Reciprocal
translocation

First with
third

o o

6
5
4
f

Duplication
deficiency

Second with
fourth

o o

d
e
3
2

6
5
4
3

J L

Duplication
deficiency

First with
second

a
b

c c

o o

d d

f 3

9 2

Duplication
deficiency

Third with
fourth

o o

6
5
4
f

6
5
4
3

J L

Duplication
deficiency

Figure 8.11 Three possible results of chromatid separation during meiosis in a reciprocal
translocation heterozygote.

called the fragile-X syndrome because it is related to a re-
gion at the X chromosome tip that breaks more fre-
quently than other chromosomal regions. However, the
break is not required for the syndrome to occur, and the
fragile-X chromosome is usually identified by the lack of
chromatin condensation at the site; in fact, under the mi-
croscope, it appears that the tip of the chromosome is
being held in place by a thread (fig. 8.15). The gene re-
sponsible for the syndrome is called FMR-1, for fragile-X
mental retardation- 1.

Fragile-X syndrome has a highly unusual pattern of
inheritance: the chance of inheriting the disease in-
creases through generations. This is so unusual a pat-
tern that it was termed the Sherman Paradox. Approxi-
mately 20% of males with the fragile-X chromosome do
not have symptoms but have grandchildren who do
have the symptoms. The daughters of the symptomatic
males also don't have symptoms, but obviously, they
have another X chromosome to mask the symptoms. As
generations proceed, the percentage of affected sons of

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Chapter Eight Cytogenetics

Wild-type

Heterozygous Bar Homozygous Bar Heterozygous

Doublebar

B^/E?

B/B*

B/B

BB/B*

800 facets

350 facets

Figure 8.12 Bar eye in Drosophila females.

70 facets

45 facets

0.0

Bar

Doublebar

56.7
f

—\—

: +

57.0
B B

59.5
fu

— I—

66.0

Crossover point

B B

Mismatch

B B B

fu +

-r-

Wild-type

f +

Figure 8.13 Unequal crossing over in a female Bar-eyed Drosophila homozygote as a result
of improper pairing. A Doublebar chromosome (and concomitant wild-type chromosome) is
produced by a crossover between forked (f) and fused (fu), two flanking loci.

carrier mothers increases. Molecular techniques, dis-
cussed in chapter 13, revealed the odd nature of this
syndrome.

Basically, the FMR-1 gene normally has between 6 and
50 copies of a three-nucleotide repeat, CCG. Chromo-
somes that have the fragile-site appearance have be-
tween 230 and 2,000 copies of the repeat. The number of
repeats is very unstable; when carrier women transmit
the chromosome, the number of repeats usually goes up.
Repeat numbers above 230 inactivate the gene and thus
cause the syndrome in men, who have only one copy of
the X chromosome. The function of the gene is not cur-
rently known. This unusual form of inheritance, with un-

stable repeats in a gene, seems to be the mechanism in
several other diseases as well, including muscular dystro-
phy and Huntington disease. We will discuss other un-
usual modes of inheritance in chapter 17.

Cri du Chat Syndrome, 46,XX orXY,5p-

The syndrome known as cri du chat (French: cry of the
cat) is so called because of the catlike cry that about half
the affected infants make. Microcephaly (an abnormally
small head), congenital heart disease, and severe mental
retardation are also common symptoms. This disorder
arises from a deletion in chromosome 5 (fig. 8.16); most

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Variation in Chromosomal Structure

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Wild-type

Bar

Doublebar

Figure 8.14 Bar region of the X chromosome of Drosophila.

Figure 8.15 Human metaphase chromosomes with the
fragile-X site indicated by an arrow. (From lan Craig, "Methyiation
and the Fragile X," Nature [1991] 349:742. Copyright © 1991 Macmillan
Magazines, Ltd.)

•*\

* »«

ift U

II IS II

* 1

1 1

14 »*

* *

* &

• *

I I

♦ *

k ft

• #

Figure 8.16 Karyotype of individual with cri du chat syndrome, due to a partial deletion of the short arm of chromosome 5 (Sp-
arrow). (Reproduced courtesy of Dr. Thomas G. Brewster, Foundation for Blood Research, Scarborough, Maine.)

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Chapter Eight Cytogenetics

other deletions studied (4p-, 13q-, 18p-, 18q-) also re-
sult in microcephaly and severe mental retardation. The
rarity of viable deletion heterozygotes is consistent with
the fact that viable monosomies (having a single chro-
mosome of a pair) are rare. An individual heterozygous
for a deletion is, in effect, monosomic for the deleted re-
gion of the chromosome. Evidently, monosomy or het-
erozygosity for larger deleted regions of a chromosome is
generally lethal in human beings.

VARIATION IN
CHROMOSOME NUMBER

Anomalies of chromosome number occur as either eu-
ploidy or aneuploidy. Euploidy involves changes in
whole sets of chromosomes; aneuploidy involves
changes in chromosome number by additions or dele-
tions of less than a whole set.

Aneuploidy

An explanation for the terminology of aneuploid change
appears in table 8.1. A diploid cell missing a single chro-
mosome is monosomic. A cell missing both copies of
that chromosome is nullisomic. A cell missing two non-
homologous chromosomes is a double monosomic. A
similar terminology exists for extra chromosomes. For
example, a diploid cell with an extra chromosome is tri-
somic. Aneuploidy results from nondisjunction in meio-
sis or by chromosomal lagging whereby one chromo-
some moves more slowly than the others during
anaphase, is excluded from the telophase nucleus, and is
thus lost. Here, nondisjunction is illustrated using the sex
chromosomes in XY organisms such as human beings or
fruit flies. Four examples are shown (fig. 8.17): nondis-
junction in either the male or female at either the first or
second meiotic divisions. Figure 8.18 shows the types of

zygotes that can result when these nondisjunctional ga-
metes fuse with normal gametes. All of the offspring pro-
duced are chromosomally abnormal. The names and
kinds of these imbalances in human beings are detailed
later in this chapter.

Bridges first showed the occurrence of nondisjunc-
tion in Drosophila in 1916 with crosses involving the
white-eye locus. When a white-eyed female was crossed
with a wild-type male, typically the daughters were wild-
type and the sons were white-eyed. However, occasion-
ally (one or two per thousand), a white-eyed daughter or
a wild-type son appeared. This could be explained most
easily by a nondisjunctional event in the white-eyed fe-
males, where X W X W and eggs (without sex chromo-
somes) were formed. Under this hypothesis, if a Y- bearing
sperm fertilized an X W X W egg, the offspring would be an
X W X™Y white-eyed daughter. If a normal X + -bearing
sperm fertilized the egg without sex chromosomes, the
result would be an X + wild-type son. Subsequently,
these exceptional individuals were found by cytological
examination to have precisely the predicted chromo-
somes (XXY daughters and XO sons). The other types
produced by this nondisjunctional event are the XX egg
fertilized by an X-bearing sperm and the egg fertilized
by the Ybearing sperm. The XXX zygotes are genotypi-
cally X W X W X + , or wild-type daughters (which usually
die), and YO flies (which always die).

Mosaicism

Rarely, an individual is made up of several cell lines, each
with different chromosome numbers. These individuals
are referred to as mosaics or chimeras, depending on
the sources of the cell lines. Such conditions can be the
result of nondisjunction or chromosomal lagging during
mitosis in the zygote or in nuclei in the early embryo
(mosaic). This is demonstrated, again for sex chromo-
somes, in figure 8. 19. A lagging chromosome is shown in
figure 8.20; in the figure, the X chromosome is lost in one

Table 8.1 Partial List of Terms to Describe Aneuploidy, Using Drosophila as an Example
(Eight Chromosomes: X, X, 2, 2, 3, 3, 4, 4)

Type

Formula

Number of Chromosomes

Example

Normal

X, X, 2, 2, 3, 3, 4, 4

Monosomic

2n - 1

X, X, 2, 2, 3, 4, 4

Nullisomic

2n — 2

X, X, 2, 2, 4, 4

Double monosomic

2n - 1 - 1

X, X, 2, 3, 4, 4

Trisomic

2n + 1

X, X, 2, 2, 3, 3, 4, 4, 4

Tetrasomic

2n + 2

X, X, 2, 2, 3, 3, 3, 3, 4, 4

Double trisomic

2n + 1 + 1

X, X, 2, 2, 2, 3, 3, 3, 4, 4

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Variation in Chromosome Number

191

At meiosis I

At meiosis II

At meiosis I

XX YY

XX XX

At meiosis II

)) xx

I YY

)) xx

Figure 8.17 Nondisjunction of the sex chromosomes in Droso-
phila or human beings. "0" refers to the lack of sex chromosomes.

of the dividing somatic cells, resulting in an XX cell line
and an XO cell line. In Drosophila, if this chromosomal
lagging occurs early in development, an organism that is
part male (XO) and part female (XX) develops.
Figure 8.21 shows a fruit fly in which chromosomal lag-
ging has occurred at the one-cell stage, causing the fly to
be half male and half female. A mosaic of this type, involv-
ing male and female phenotypes, has a special name —
gynandromorph. (A hermaphrodite is an individual,
not necessarily mosaic, with both male and female repro-
ductive organs.) Many sex-chromosomal mosaics are
known in humans, including XX/X, XY/X, XX/XY, and
XXX/X. At least one case is known of a human XX/XY
chimera that resulted from the fusion of two zygotes, one

Nondisjunction

S 9

XXY

XXX

XYY

£ Sx

XXX

0*Y

XXY

Figure 8.18 Results of fusion of a nondisjunction gamete (top)
with a normal gamete {side).

XYY

x ; x

x r
\

XXX

Figure 8.19 Mitotic nondisjunction of the sex chromosomes.

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Metaphase

Anaphase

Telophase

Figure 8.20 Chromosomal lagging at mitosis in the X chromosomes of a female Drosophila.

Sex comb.

Figure 8.21 Drosophila gynandromorph. The left side is wild-
type XX female; the right side is XO male, hemizygous for white
eye and miniature wing.

formed by a sperm fertilizing an ovum and the other formed
by a second sperm fertilizing a polar body of that ovum.

Aneuploidy in Human Beings

Approximately 50% of spontaneous abortions (miscar-
riages) among women in the United States involve fe-
tuses with some chromosomal abnormality; about half of
these are autosomal trisomies. About one in 160 live hu-
man births has some sort of chromosomal anomaly; most
are balanced translocations, autosomal trisomies, or sex-
chromosomal aneuploids.

In the standard system of nomenclature, a normal hu-
man chromosome complement is 46, XX for a female and
46,XY for a male. The total chromosome number appears
first, then the description of the sex chromosomes, and,
finally, a description of autosomes if some autosomal
anomaly is evident. For example, a male with an extra X
chromosome would be 47,XXY. A female with a single X
chromosome would be 45, X. Since all the autosomes are
numbered, we describe their changes by referring to
their addition (+) or deletion (— ). For example, a female
with trisomy 2 1 would be 47,XX, + 21. The short arm of a
chromosome is designated p, the longer arm, q. When a
change in part of the chromosome occurs, a + after the
arm indicates an increase in the length of that arm,
whereas a minus sign (— ) indicates a decrease in its
length. For example, a translocation (t) that transfers part
of the short arm of chromosome 9 to the short arm of
chromosome 18 would be 46, XX, t(9p-;18p+). The
semicolon indicates that both chromosomes kept their
centromeres.

Following are descriptions of viable human aneu-
ploids who survive long enough after birth to have a
named syndrome.

Trisomy 21 (Down Syndrome), 47,XX or XY,+21

Down syndrome (figs. 8.22 and 8.23) affects about one in
seven hundred live births. Most affected individuals are
mildly to moderately mentally retarded and have congen-
ital heart defects and a very high (1/100) risk of acute
leukemia. They are usually short and have a broad, short
skull; hyperflexibility of joints; and excess skin on the
back of the neck. The physician John Langdon Down first
described this syndrome in 1866. (Modern convention is
to avoid the possessive form of a name in referring to a
syndrome.) Down syndrome was the first human syn-
drome attributed to a chromosomal disorder; Jerome
Lejeune, a physician in Paris, published this finding in

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ft ft It IS II \* II

ft ft* II

Figure 8.22 Karyotype of an individual with trisomy 21 , Down
syndrome. (Reproduced courtesy of Dr. Thomas G. Brewster, Foundation
for Blood Research, Scarborough, Maine.)

* f » t «ft

• •

Jerome Lejeune (1926-1994).
(Courtesy of Dr. Jerome Lejeune, Institut de
Progenese, Paris.)

1959. An interesting aspect of this syndrome is its in-
creased incidence among children of older mothers
(fig. 8.24), a fact known more than twenty-five years be-
fore the discovery of the cause of the syndrome. Since
the future ova are in prophase I of meiosis (dictyotene)
since before the mother's birth, all ova are the same age
as the female. Presumably, older ova are more susceptible
to nondisjunction of chromosome 2 1 .

Ik «

•

Recently, techniques of molecular genetics (chapter
13) have been used to identify the origins of the three
copies of chromosome 2 1 in a large sample of individuals
with Down syndrome. As expected, the overwhelming
majority of the extra copies of chromosome 21 (95%)
were of maternal origin. About 5% of the cases of Down
syndrome were of mitotic origin, occurring either in the
gonad of one of the parents (evenly split between moth-
ers and fathers) or possibly postzygotically in the fetus.

Familial Down Syndrome

Down syndrome (trisomy 21), as described, is usually the
result of either a nondisjunctional event during gameto-
genesis or, rarely, a mitotic event. It is a function of ma-
ternal age and is not inherited. (Although about half the
children of a person with trisomy 2 1 will have trisomy 2 1
because of aneuploid gamete production, the possibility
that an unaffected relative of the person will have abnor-
mal children is no greater than for a person of the same
age chosen at random from the general population.)
However, about 4% of those with Down syndrome have
been found to have a translocation of chromosome 2 1 ,

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Chapter Eight Cytogenetics

Figure 8.23 Individual with trisomy 21
Young/SPL/ Photo Researchers.)

16-24 25-29 30-34 35-39 40-44

Mother's age

45 +

16-24
25-29
30-34
35-39
40-44
45 +

Number per thousand

1/1700 0.58

1/1100 0.91

1/770 1.30

1/250 4

1/80 12.5

1/25 40

Figure 8.24 Increased risk of trisomy 21 attributed to the age
Of the mother. (From E. Hook, "Estimates of Maternal Age-Specific Risks
of a Down-Syndrome Birth in Women age 34-41 ," Lancet, 2:33-34,
Copyright © 1976 by The Lancet Ltd.)

usually associated with chromosome 14, 15, or 22. The
translocational and nontranslocational types of Down
syndrome have identical symptoms; however, a balanced
translocation can be passed on to offspring (see fig.
8.11). Alternate segregation of centromeres in the
translocation heterozygote produces either a normal
gamete or one carrying the balanced translocation. Adja-
cent segregation causes partial trisomy for certain chromo-
somal parts. When this occurs for most of chromosome
21, Down syndrome results.

It is worth mentioning that aside from trisomy and
translocation, Down syndrome can come about through
mosaicism, as mentioned earlier, or a centromeric event.
About 2% of individuals with Down syndrome are mosaic
for cells with both two and three copies of chromosome
21. Some evidence suggests that the original zygotes
were trisomic, but then a daughter cell lost one of the
copies of chromosome 2 1 . The severity of the symptoms
in these individuals relates to the percentage of trisomic
cells they possess. Mosaicism increases with maternal
age, just as trisomy in general does. In extremely rare
cases, Down syndrome is caused by an abnormal chro-
mosome 21 that has, rather than a short and long arm,
two identical long arms attached to the centromere. This

type of chromosome, called an isochromosome, pre-
sumably occurs by an odd centromeric fission (fig. 8.25).
Hence, a person with a normal chromosome 21 and an
isochromosome 2 1 has three copies of the long arm of
the chromosome and has Down syndrome.

Trisomy 18 (Edward Syndrome), 47, XX or XY,+ 18

Edward syndrome affects one in ten thousand live births
(fig. 8.26). Most affected individuals are female, with 80
to 90% mortality by two years of age. The infant usually
has an elfin appearance with small nose and mouth, a re-
ceding lower jaw, abnormal ears, and a lack of distal flex-
ion creases on the fingers. The distal joints have limited
motion, and the fingers display a characteristic posturing
in which the little and index fingers overlap the middle
two. The syndrome is usually accompanied by severe
mental retardation.

Trisomy 13 (Patau Syndrome), 47, XX or XY, + 13,
and Other Trisomic Disorders

Patau syndrome affects one in twenty thousand live
births. Diagnostic features are cleft palate, cleft lip, con-

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Variation in Chromosome Number

195

Isochromosome

Split

Fragments

Isochromosome

Figure 8.26 Child with trisomy 18, Edward syndrome.
(Reproduced courtesy of Dr. Jerome Lejeune, Institut de Progenese, Paris.)

Figure 8.25 If the centromere of chromosome 21 breaks
perpendicular to the normal division axis, it can form an
isochromosome of the long arms and either an isochromosome
of the short arms or two separate fragments. This can happen
during anaphase of mitosis or meiosis II.

genital heart defects, Polydactyly, and severe mental re-
tardation. Mortality is very high in the first year of life.

Other autosomal trisomies are known but are ex-
tremely rare. These include trisomy 8 (47, XX or XY, + 8)
and cat's eye syndrome, a trisomy of an unidentified,
small acrocentric chromosome (47,XX or XY,[ + acrocen-
tric]). Several aneuploids involving sex chromosomes are
also known.

Turner Syndrome, 45,X

About one in ten thousand live female births is of an infant
with Turner syndrome. This and 4 5, XX or XY, — 21 and

45,XX or XY, — 22 are the only nonmosaic, viable
monosomies recorded in human beings (fig. 8.27), indicat-
ing the severe consequences monosomy has on all but the
two smallest autosomes and a sex chromosome. In-
dividuals with Turner syndrome usually have normal intel-
ligence but underdeveloped ovaries, abnormal jaws,
webbed necks, and shieldlike chests.

The symptoms of Turner syndrome have been logi-
cally deduced to be caused by a single dosage of genes
that are normally present and active in two dosages.
Thus, these genes would be located on both the X and Y
chromosomes (pseudoautosomal) to provide two dosages
in normal XY males and also be active in both X chromo-
somes in normal XX females. Therefore, they should be
located on regions of the X chromosome that escape in-
activation (see chapter 5). Studies of persons with small
X-chromosomal deletions and molecular analyses of the
X and Y chromosomes (outlined in chapter 13) have
caused two genes to emerge as candidates: ZFY (on the Y
chromosome, termed ZFX on the X chromosome) and
RPS4Y (on the Y chromosome, termed RPS4X on the X
chromosome). ZFY (zinc finger on the Y chromosome)
was once believed to be the male-determining gene in
mammals. RPS4Y encodes a ribosomal protein, one of
the many proteins making up the ribosome.

It is interesting to note a dosage-compensation
difference in people and mice, which have analogous
genes termed Zfx and Rps4x. In mice, unlike in people,
these genes are inactivated in the "Lyonized" X chromo-
some in females and have restricted activity in the Y

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11 *) II u w

M 4| II

It * * %l

t I

* ft 4 *

21 22

Figure 8.27 Karyotype of a person with Turner syndrome (X0). (Reproduced courtesy of Dr. Thomas G. Brewster, Foundation for Blood Research,
Scarborough, Maine.)

chromosome. Hence, mouse cells seem normally to have
only one copy of these genes functioning in normal XY
males and XX females. Therefore, we would predict that
the XO genotype in mice would produce few if any nega-
tive effects as compared with a human XO genotype, since
mouse cells of both sexes normally only have one func-
tional copy of each gene. In fact, human Turner syndrome
fetuses have a 99% prenatal mortality rate, but virtually no
prenatal mortality affects mouse fetuses with the XO geno-
type (born of XX mothers). This confirms our predictions
and points to differences between people and mice in
dosage-compensation mechanisms for specific genes.

XYY Karyotype, 47,XYY

About one in one thousand live male births is of an indi-
vidual with an XYY karyotype. (We avoid the term
syndrome here because XYY men have no clearly de-
fined series of attributes, other than often being taller

than normal.) Some controversy has surrounded this
karyotype because it was once reported that it occurred
in abundance in a group of mentally subnormal males in
a prison hospital. Seven XYY males were found among
197 inmates, whereas only one in about two thousand
control men were XYY. This study has subsequently been
expanded and corroborated. Although it is now fairly
well established that the incidence of XYY males in
prison is about twentyfold higher than in society at large,
the statistic is somewhat misleading: the overwhelming
number of XYY men seem to lead normal lives. At most,
about 4% of XYY men end up in penal or mental institu-
tions, where they make up about 2% of the population.

There is some indication that the XYY men in prison
had lower intelligence test levels. Thus, criminal tendency
may be attributed to lower intelligence rather than a pre-
disposition toward criminality caused by an extra Y chro-
mosome. For the most part, expanded studies have indi-
cated that XYY criminals do not commit violent crimes.

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Variation in Chromosome Number

197

Research on this karyotype has produced its own
problems. A research project at Harvard University on
XYY males came under intense public pressure and was
eventually terminated. The project, under the direction
of Stanley Walzer (a psychiatrist) and Park Gerald (a ge-
neticist), involved screening all newborn boys at the
Boston Hospital for Women and following the develop-
ment of those with chromosomal anomalies. The criti-
cism of this work centered mainly on the necessity of in-
forming parents that their sons had an XYY karyotype
that might be associated with behavioral problems. Op-
ponents of this work claimed that telling the parents
could trigger a self-fulfilling prophecy; that is, parents
who heard that their children were not normal and
might cause trouble might then behave toward their chil-
dren in a manner that would increase the probability that
their children would cause trouble. The opponents
claimed that the risks of this research outweighed the
benefits. The project was terminated in 1975 primarily
because of the harassment Walzer faced.

Klinefelter Syndrome, 47,XXY

The incidence of Klinefelter syndrome is about one in
one thousand live births. Tall stature and infertility are
common symptoms. Diagnosis is usually by buccal
(cheek tissue) smear to ascertain the presence of a Barr
body in a male, indicating an XXY karyotype. Some prob-
lems with behavior and speech development are associ-
ated with this syndrome.

Triple-X Female, 47,XXX, and Other Aneuploid
Disorders of Sex Chromosomes

A triple-X female appears in about one in one thousand
female live births. Fertility can be normal, but these indi-
viduals are usually mildly mentally retarded. Delayed
growth, as well as congenital malformations, are also
sometimes present. Other sex-chromosomal aneuploids,
including XXXX, XXXXX, and XXXXY, are extremely rare.
All seem to be characterized by mental retardation and
growth deficiencies.

organ systems. Second, if there is a chromosomal sex-
determining mechanism, it may be disrupted by poly-
ploidy. And third, meiosis produces unbalanced gametes
in many polyploids.

If the polyploid has an odd number of sets of chro-
mosomes, such as triploid 0ri), two of the three homo-
logues will tend to pair at prophase I of meiosis, produc-
ing a bivalent and a univalent. The bivalent separates
normally, but the third chromosome goes independently
to one of the poles. This separation results in a 50%
chance of aneuploidy in each of the ^-different chromo-
somes, rapidly decreasing the probability of a balanced
gamete as n increases. Therefore, as n increases, so does
the likelihood of sterility. An alternative to the bivalent-
univalent type of synapsis is the formation of trivalents,
which have similar problems (fig. 8.28). Even-numbered
polyploids, such as tetraploids (4n), can do better during
meiosis. If the centromeres segregate two by two in each
of the n meiotic figures, balanced gametes can result. Of-
ten, however, the multiple copies of the chromosomes
form complex figures during synapsis, including mono-
valents, bivalents, trivalents, and quadrivalent s, tending
to result in aneuploid gametes and sterility.

Meiosis I and II

Euploidy

Euploid organisms have varying numbers of complete
haploid chromosomal sets. We are already familiar with
haploids (n) and diploids (2n). Organisms with higher
numbers of sets, such as triploids On) and tetraploids
(4n), are called polyploids. Three kinds of problems
plague polyploids. First, the potential exists for a general
imbalance in the organism due to the extra genetic mate-
rial in each cell. For example, a triploid human fetus has
about a one in a million chance to survive to birth, at
which time death usually occurs due to problems in all

Disomies

Double disomies

Figure 8.28 Meiosis in a triploid (3n = 9) and one possible
resulting arrangement of gametes. The probability of a "normal"
gamete is (1/2) n where n equals the haploid chromosome
number. Here, n = 3 and (1/2) 3 = 1/8.

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Chapter Eight Cytogenetics

Some groups of organisms, primarily plants, have
many polyploid members. An estimated 30 to 80% of all
flowering plant species (angiosperms) are polyploids, as
are 95% of ferns. (Polyploidy is apparently rare in gym-
nosperms and fungi.) For example, the genus of wheat,
Triticum, has members with fourteen, twenty-eight, and
forty-two chromosomes. Because the basic Triticum
chromosome number is n = 7, these forms are 2n, 4n,
and 6n species, respectively. Chrysanthemums have
species of eighteen, thirty-six, fifty-four, seventy-two, and
ninety chromosomes. With a basic number of n = 9,
these species represent a 2n, 4n, 6n, 8n, and lOn series.
In both these examples, the even-numbered polyploids
are viable and fertile, but the odd-numbered polyploids
are not.

Autopolyploidy

Polyploidy can come about in two different ways. In
autopolyploidy, all of the chromosomes come from
within the same species. In allopolyploidy, the chro-
mosomes come from the hybridization of two different
species (fig. 8.29). Autopolyploidy occurs in several dif-
ferent ways. The fusion of nonreduced gametes creates
polyploidy. For example, if a diploid gamete fertilizes a
normal haploid gamete, the result is a triploid. Simi-
larly, if a diploid gamete fertilizes another diploid ga-
mete, the result is a tetraploid. The equivalent of a
nonreduced gamete comes about in meiosis if the par-
ent cell is polyploid to begin with. For example, if one
branch of a diploid plant is tetraploid, its flowers pro-
duce diploid gametes. These gametes are not the result
of a failure to reduce chromosome numbers meioti-
cally, but rather the result of successful meiotic reduc-
tion in a polyploid flower. The tetraploid tissue of the

Autopolyploidy

V
AAAA

V
AABB

Allopolyploidy

Figure 8.29 Autopolyploidy and allopolyploidy. If A and B are
the haploid genomes of species A and B, respectively, then
autopolyploidy produces a species with an AAAA karyotype,
and allopolyploidy (with chromosome doubling) produces a
species with AABB karyotype. If A represents seven
chromosomes, then an AA diploid has fourteen chromosomes
and an AAAA tetraploid has twenty-eight chromosomes. If B
represents five chromosomes, then a BB diploid has ten
chromosomes and an AABB allotetraploid has twenty-four
chromosomes.

plant in this example can originate by the somatic
doubling of diploid tissues.

Somatic doubling can come about spontaneously or
be caused by anything that disrupts the normal se-
quence of a nuclear division. For example, colchicine
induces somatic doubling by inhibiting microtubule for-
mation. This prevents the formation of a spindle and
thus prevents the chromosomes from moving apart dur-
ing either mitosis or meiosis. The result is a cell with
double the chromosome number. Other chemicals, tem-
perature shock, and physical shock can produce the
same effect.

Allopolyploidy

Allopolyploidy comes about by cross-fertilization be-
tween two species. The resulting offspring have the sum
of the reduced chromosome number of each parent
species. If each chromosome set is distinctly different,
the new organisms have difficulty in meiosis because no
two chromosomes are sufficiently homologous to pair.
Then every chromosome forms a univalent (unpaired)
figure, and they separate independently during meiosis,
producing aneuploid gametes. However, if an organism
can survive by vegetative growth until somatic doubling
takes place in gamete precursor cells (2n — ► 4ri), or al-
ternatively, if the zygote was formed by two unreduced
gametes (2n + 2n), the resulting offspring will be fully
fertile because each chromosome has a pairing partner at
meiosis. We can draw an example from the work of Rus-
sian geneticist G. D. Karpechenko.

In 1928, Karpechenko worked with the radish
(Raphanus sativus, 2n = 18, n = 9) and cabbage (Bras-
sica oleracea, 2n = 18, n = 9). When these two plants
are crossed, an F : results with n + n = 18 (9 + 9). This
plant, which is an allodiploid, has characteristics inter-
mediate between the two parental species (fig. 8.30). If
somatic doubling takes place, the chromosome number
is doubled to thirty-six, and the plant becomes an al-
lopolyploid (an allotetraploid of 4ri). Since each chro-
mosome has a homologue, this allotetraploid is also re-
ferred to as an amphidiploid. If we did not know its
past history, this plant would simply be classified as a
diploid with 2n = 36. In this case, the new amphidiploid
cannot successfully breed with either parent because
the offspring are sterile triploids. It is, therefore, a new
species and has been named Raphanobrassica. As an
agricultural experiment, however, it was not a success
because it did not combine the best features of the cab-
bage and radish.

Polyploidy in Plants and Animals

Although polyploids in the animal kingdom are known
(in some species of lizards, fish, invertebrates, and a

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Summary

199

Radish
{Raphanus,
2n=18)

Gametes
(n = 9)

Cabbage
X {Brass ica,

2n=18)

Gametes

(/7 = 9)

F 1 hybrid (sterile,
2n= 18)

Chromosome
doubling

Raphanobrassica

(fertile,

4/7 = 36)

tetraploid mammal, the red viscacha rat), polyploidy as
a successful evolutionary strategy is primarily a plant
phenomenon. There are several reasons for this. To be-
gin with, many more animals than plants have chromo-
somal sex-determining mechanisms. Polyploidy severely
disrupts these mechanisms. For example, Bridges dis-
covered a tetraploid female fruit fly, but it has not been
possible to produce a tetraploid male. The tetraploid fe-
male's progeny were triploids and intersexes. A second
reason why polyploidy is more common in plants is be-
cause plants can generally avoid the meiotic problems
of polyploidy longer than most animals. Some plants
can exist vegetatively, allowing more time for the rare
somatic doubling event to occur that will produce an
amphidiploid; animal life spans are more precisely de-
fined, allowing less time for a somatic doubling. And
third, many plants depend on the wind or insect pollina-
tors to fertilize them and thus have more of an opportu-
nity for hybridization. Many animals have relatively elab-
orate courting rituals that tend to restrict hybridization.
Polyploidy has been used in agriculture to produce
"seedless" as well as "jumbo" varieties of crops. Seedless
watermelon, for example, is a triploid. Its seeds are
mostly sterile and do not develop. It is produced by
growing seeds from the cross between a tetraploid vari-
ety and a diploid variety. Jumbo Macintosh apples are
tetraploid.

Figure 8.30 Hybridization of cabbage and radish, showing the
resulting hybrid fruiting structures.

SUMMARY

STUDY OBJECTIVE 1: To observe the nature and conse-
quences of chromosomal breakage and reunion
178-190

Variation can occur in the structure and number of chro-
mosomes in the cells of an organism. When chromosomes
break, the ends become "sticky"; they tend to reunite with
other broken ends. A single break can lead to deletions or
the formation of acentric or dicentric chromosomes. Di-
centrics tend to go through breakage-fusion-bridge cycles,
which result in duplications and deficiencies.

Two breaks in the same chromosome can yield dele-
tions and inversions. Variegation position effects, as well as
new linkage arrangements, can result. Inversion heterozy-
gotes produce loop figures during synapsis, which can form
either at meiosis or in polytene chromosomes. Heterozy-
gosity for an inversion suppresses crossovers; organisms
that are heterozygotes are semisterile.

Reciprocal translocations can result from single breaks
in nonhomologous chromosomes. These produce cross-
shaped figures at synapsis and result in semisterility. The
Bar eye phenotype of Drosophila is an example of a dupli-
cation that causes a position effect.

STUDY OBJECTIVE 2: To observe the nature and conse-
quences of variation in chromosome numbers in human
and nonhuman organisms 190-199

Changes in chromosome number can involve whole sets
(euploidy) or partial sets (aneuploidy) of chromosomes.
Aneuploidy usually results from nondisjunction or chromo-
somal lagging. Several medical syndromes, such as Down,
Turner, and Klinefelter syndromes, and the XYY karyotype
are caused by aneuploidy.

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Chapter Eight Cytogenetics

Polyploidy leads to difficulties in chromosomal sex-
determining mechanisms, general chromosomal imbal-
ance, and problems during meiotic segregation. It has been
more successful in plants than in animals because plants
generally lack chromosomal sex-determining mechanisms.
Plants can also avoid meiotic problems by propagating veg-
etatively In both animals and plants, even-numbered poly-

ploids do better than odd-numbered polyploids because
they have a better chance of producing balanced gametes
during meiosis. Somatic doubling provides each chromo-
some in a hybrid organism with a homologue, and thus
makes possible tetrad formation at meiosis. New species
have arisen by polyploidy.

SOLVED PROBLEMS

PROBLEM 1: What are the consequences of an inversion?

Answer: In an inversion homo zygote, the consequences
are change in linkage arrangements, including new or-
ders and map distances, and the possibility of position ef-
fects if a locus is placed into or near heterochromatin. In
an inversion heterozygote, crossover suppression causes
semisterility because zygotes that carry genie imbalances
are lost. Inversion heterozygotes can be seen as meiotic
loop structures or loops formed in endomitotic chromo-
somes such as those found in the salivary glands of fruit
flies. In an evolutionary sense, inversions result in super-
genes, locking together allelic combinations.

PROBLEM 2: What are the consequences of a monosomic
chromosome in human beings?

Answer: In human beings, monosomy is rare, meaning
that, with few exceptions, it is lethal. In fact, monosomies
are also rare in spontaneous abortions, indicating that
most monosomic fetuses are lost before the woman is
aware of the pregnancy. The only monosomies known to

be viable in human beings are Turner syndrome (45,X)
and monosomies of chromosomes 21 and 22, the two
smallest autosomal chromosomes.

PROBLEM 3: Ebony body (e) in flies is an autosomal re-
cessive trait. A true-breeding ebony female (ee) is mated
with a true-breeding wild-type male that has been irradi-
ated. Among the wild-type progeny is a single ebony
male. Explain this observation.

Answer: The cross is ee X e + e + , and all F : s should be
e + e (wild-type). The use of irradiation alerts us to the
possibility of chromosomal breaks, as well as simple mu-
tations. What type of chromosomal aberration would al-
low a recessive trait to appear unexpectedly? A deletion,
which creates pseudodominance when there is no sec-
ond allele, is a good possibility. The male in question
could have gotten the ebony allele from its mother and
no homologous allele from its father. Alternatively, the
wild-type allele from the father could have mutated to an
ebony allele.

EXERCISES AND PROBLEMS

VARIATION IN CHROMOSOMAL STRUCTURE

1. What kind of figure is observed in meiosis of a re-
ciprocal translocation homozygote?

2. Can a deletion result in the formation of a variega-
tion position effect? If so, how?

3. Does crossover suppression occur in an inversion
homozygote? Explain.

4. Which rearrangements of chromosomal structure
cause semisterility?

5. What are the consequences of single crossovers dur-
ing tetrad formation in a reciprocal translocation
heterozygote?

Answers to selected exercises and problems are on page A-9.

6. Give the gametic complement, in terms of acentrics,
dicentrics, duplications, and deficiencies, when a
three-strand double crossover occurs within a para-
centric inversion loop.

7. In studying a new sample of fruit flies, a geneticist
noted phenotypic variegation, semisterility, and the
nonlinkage of previously linked genes. What proba-
bly caused this, and what cytological evidence would
strengthen your hypothesis?

8. In a second sample of flies, the geneticist found a po-
sition effect and semisterility. The linkage groups
were correct, but the order was changed and cross-
ing over was suppressed. What probably caused this,
and what cytological evidence would strengthen
your hypothesis?

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8. Cytogenetics

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Exercises and Problems

201

9. Diagram the results of alternate segregation for a
three-strand double crossover between a cen-
tromere and the cross center in a reciprocal translo-
cation heterozygote.

10. A heterozygous plant A B C D E/a b c d e is
testcrossed with an a b c d e/a b c d e plant. Only
the following progeny appear.

AB CDE/abcde
abed e/a b c d e
Abed e/a b c d e
a B C D E/a b c d e
AB CD e/a b c d e
abed E/a b c d e

What is unusual about the results? How can you ex-
plain them?

11. White eye color in Drosophila is an X-linked reces-
sive trait. A wild-type male is irradiated and mated
with a white-eyed female. Among the progeny is a
white-eyed female.

a. Why is this result unexpected, and how could
you explain it?

b. What type of progeny would you expect if this
white-eyed female is crossed with a normal, non-
irradiated male?

12. You are trying to locate an enzyme-producing gene
in Drosophila, which you know is located on the
third chromosome. You have five strains with dele-
tions for different regions of the third chromosome

(a slash — /

Normal
Strain A
Strain B
Strain C
Strain D
Strain E

//////-

indicates a deleted region):

20 30 40 50 60 map units

_//////////////////.

.////////////.

.////////////

You cross each strain with wild-type flies and mea-
sure the amount of enzyme in the ¥ 1 progeny. The re-
sults appear as follows. In what region is the gene lo-
cated?

Percentage of Wild-Type

Enzyme ]

Produced in F a

Strain Crossed

Progeny

100

101

13. Consider the following table, which shows the num-
ber of viable progeny produced by a plant under
standard conditions. Provide an explanation for the
results.

Pi:

Strain A X
Strain A

Strain B X
Strain B

Strain A X
Strain B

Fi:
F 2 :

765
712

750
783

775
416

14. The map position for three X-linked recessive genes
in Drosophila (v, vermilion eyes; m, miniature
wings; and s, sable body) is:

33.0

36.1

43.0

A wild-type male is X-rayed and mated to a vermil-
ion, miniature, sable female. Among the progeny is a
single vermilion-eyed, long-winged, tan-bodied fe-
male. The following shows the progeny when this
female is mated with avms hemizygous male.

Females

Males

87 vermilion,
miniature, sable
93 vermilion

89 vermilion, miniature,

sable

1 vermilion

15.

Explain these results by drawing a genetic map.

In Drosophila, recessive genes clot (ct) and black
body (b) are located at 16.5 and 48.5 map units, re-
spectively, on the second chromosome. In one
cross, wild-type females that are ct + b + /ct b are
mated with ct b/ct b males. They produce these
progeny:

wild-type

1,250

clot, black

1,200

black

clot

What is unusual about the results? How can you ex-
plain them?

16. You have four strains of Drosophila (1 -4) that were
isolated from different geographic regions. You com-
pare the banding patterns of the second chromo-
some and obtain these results (each letter corre-
sponds to a band):

(l)mnrqpostuv
(2)mnopqrstuv
(3)mnrqtsupov
(4)mnrqtsopuv

If (3) is presumed to be the ancestral strain, in what
order did the other strains arise?

17. In Drosophila, the recessive gene for white eyes is
located near the tip of the X chromosome. A wild-
type male is irradiated and mated with a white-eyed
female. Among the progeny is one red-eyed male.
How can you explain the red-eyed male, and how
could you test your hypothesis?

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202

Chapter Eight Cytogenetics

VARIATION IN CHROMOSOME NUMBER

18. Is a tetraploid more likely to show irregularities in
meiosis or mitosis? Explain. What about these
processes in a triploid?

19. How many chromosomes would a human tetraploid
have? How many chromosomes would a human
monosomic have?

20. Do autopolyploids or allopolyploids experience
more difficulties during meiosis? Do amphidiploids
have more or less trouble than auto- or allopoly-
ploids?

21. If a diploid species of 2n = 16 hybridizes with one
of 2n = 12, and the resulting hybrid doubles its
chromosome number to produce an allotetraploid
(amphidiploid), how many chromosomes will it
have? How many chromosomes will an allotetra-
ploid have if both parent species had 2n = 20?

22. If nondisjunction of the sex chromosomes occurs in
a female at the second meiotic division, what type of
eggs will arise?

23. How might an XO/XYY human mosaic arise? An
XX/XXY mosaic? How might a trisomy 2 1 individual
arise?

24. Plant species P has 2n = 18, and species U has 2n =
14. A fertile hybrid is found. How many chromo-
somes does it have?

25. A woman with normal vision whose father was
color-blind mates with a man with normal vision.
They have a color-blind daughter with Turner syn-
drome. In which parent did nondisjunction occur?

26. A color-blind man mates with a woman with normal
vision whose father was color-blind. They have a
color-blind son with Klinefelter syndrome. In which
parent did nondisjunction occur?

27. Describe a genetic event that can produce an XYY
man.

28. Chromosomal analysis of a spontaneously aborted
fetus revealed that the fetus was 92,XXYY. Propose
an explanation to account for this unusual kary-
otype.

CRITICAL THINKING QUESTIONS

1. Various species in the grass genus Bromus have chro-
mosome numbers of 14, 28, 42, 56, 70, 84, 98, and 112.
What can you tell about the genetic relationships
among these species and how they might have arisen?

2. There was a humorous television commercial in which
someone accidentally discovered the desirability of com-
bining chocolate and peanut butter. Could this combina-
tion be achieved by crossing peanut and cocoa plants?

Suggested Readings for chapter 8 are on page B-4.

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9. Chemistry of the Genel

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CHEMISTRY
OF THE GENE

A computer-generated image of deoxyribonucleic
acid, DNA. (© Professor K. Seddon & Dr. T. Evans/
Queen's University Belfast/SPL/Photo Researchers.)

STUDY OBJECTIVES

1. To understand the properties that a genetic material
must have 205

2. To examine the structure of DNA, the genetic material 211

3. To investigate the way in which DNA replicates 220

STUDY OUTLINE

In Search of the Genetic Material 205

Required Properties of a Genetic Material 205

Evidence for DNA as the Genetic Material 209
Chemistry of Nucleic Acids 211

Biologically Active Structure 214

Requirements of Genetic Material 218

Alternative Forms of DNA 219
DNA Replication— The Process 220

The Meselson and Stahl Experiment 220

Autoradiographic Demonstration of DNA Replication 222
DNA Replication — The Enzymology 225

Continuous and Discontinuous DNA Replication 225

The Origin of DNA Replication 229

Events at the Y- Junction 232

Supercoiling 234

Termination of Replication 236

DNA Partitioning in E. coli 238
Replication Structures 238

Rolling-Circle Model 238

D-Loop Model 238
Eukaryotic DNA Replication 238
Summary 240
Solved Problems 240
Exercises and Problems 241
Critical Thinking Questions 242
Box 9.1 Molecular Structure of Nucleic Acids:

A Structure for Deoxyribose Nucleic Acid 206
Box 9.2 Prions: The Biological Equivalent of

Ice-Nine 213
Box 9.3 Multiple-Stranded DNA 221

204

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In Search of the Genetic Material

205

In 1953, James Watson and Francis Crick pub-
lished a two-page paper in the journal Nature en-
titled "Molecular Structure of Nucleic Acids: A
Structure for Deoxyribose Nucleic Acid." It began
as follows: "We wish to suggest a structure for the
salt of deoxyribose nucleic acid (D.N.A.).This structure
has novel features which are of considerable biological
interest." This paper, which first put forth the correct
model of DNA structure, is a milestone in the modern
era of molecular genetics, compared by some to the
work of Mendel and Darwin (box 9.1). (Watson, Crick,
and X-ray crystallographer Maurice Wilkins won Nobel
Prizes for this work; Rosalind Franklin, also an X-ray
crystallographer, was acknowledged, posthumously, to
have played a major role in the discovery of the struc-
ture of DNA.) Once the structure of the genetic mate-
rial had been determined, an understanding of its
method of replication and its functioning quickly fol-
lowed.

James D. Watson Francis Crick

(1928- ). (Cold Spring (1916- ). (Reproduced
Harbor Laboratory Research by permission of Herb
Library Archives. Margot Weitman, Washington

Bennet, photographer.) University, St. Louis,

Missouri.)

Maurice H. F. Wilkins
(1916- ). (Courtesy of
Dr. Maurice H. F. Wilkins
and Biophysics Department,
King's College, London.)

IN SEARCH OF THE GENETIC
MATERIAL

This chapter begins a sequence of nine chapters on the
molecular structure of the genetic material, its replica-
tion, its expression, and the control of its expression. In
this chapter, we look at the evidence that DNA is the ge-
netic material, the chemistry of DNA, and the way in
which DNA replicates, including the general enzymatic
processes. We look first at prokaryotic, then at eukary-
otic, DNA replication. Note that we concentrate on the
molecular structure of DNA because, generally, structure
reveals function: molecules have shapes that define how
they work.

Required Properties of a Genetic Material

We begin with a look at the properties that a genetic ma-
terial must have and review the evidence that nucleic
acids make up the genetic material. To comprise the
genes, DNA must carry the information to control the
synthesis of the enzymes and proteins within a cell or or-
ganism; self-replicate with high fidelity, yet show a low
level of mutation; and be located in the chromosomes.

Control of the Proteins

The growth, development, and functioning of a cell are
controlled by the proteins within it, primarily its en-
zymes. Thus, the nature of a cell's phenotype is con-
trolled by the protein synthesis within that cell. The ge-
netic material must therefore determine the need for and
effective amounts of the enzymes in a cell. For example,
given inorganic salts and glucose, an E. colt cell can syn-
thesize, through its enzyme-controlled biochemical path-
ways, all of the compounds it needs for growth, survival,
and reproduction. In contrast, a mammalian red blood
cell primarily produces hemoglobin.

At this point we need to review some basic informa-
tion regarding enzymes. An enzyme is a protein that acts
as a catalyst for a specific metabolic process without it-
self being markedly altered by the reaction. Most reac-
tions that enzymes catalyze could occur anyway, but only
under conditions too extreme to take place within living
systems. For example, many oxidations occur naturally at
high temperatures. Enzymes allow these reactions to oc-
cur within the cell by lowering the free energy of acti-
vation (AG*) of a particular reaction. In other words, an
enzyme allows a reaction to take place without needing
the boost in energy that heat usually supplies (fig. 9.1).

CD
>

_CD

CD
LU

No enzyme

Enzyme

Substrate

Final product

Time

Figure 9.1 An enzyme lowers the free energy of activation
(AG*) for a particular reaction.

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206

Chapter Nine Chemistry of the Gene

BOX 9.1

We wish to suggest a struc-
ture for the salt of de-
oxyribose nucleic acid
(D.N.A.). This structure has novel fea-
tures which are of considerable bio-
logical interest.

A structure for nucleic acid has al-
ready been proposed by Pauling and
Corey. l They kindly made their man-
uscript available to us in advance of
publication. Their model consists of
three intertwined chains, with the
phosphates near the fibre axis, and
the bases on the outside. In our opin-
ion, this structure is unsatisfactory
for two reasons: (1) We believe that
the material which gives the X-ray di-
agrams is the salt, not the free acid.
Without the acidic hydrogen atoms it
is not clear what forces would hold
the structure together, especially as
the negatively charged phosphates
near the axis will repel each other.
(2) Some of the van der Waals dis-
tances appear to be too small.

Another three-chain structure has
also been suggested by Fraser (in the
press). In his model the phosphates
are on the outside and the bases on
the inside, linked together by hydro-
gen bonds. This structure as described
is rather ill-defined, and for this reason
we shall not comment on it.

We wish to put forward a radically
different structure for the salt of de-
oxyribose nucleic acid. This structure
has two helical chains each coiled
round the same axis (see diagram

Historical
Perspectives

Molecular Structure

of Nucleic Acids:

A Structure for Deoxyribose

Nucleic Acid

[fig. 1]). We have made the usual
chemical assumptions, namely, that
each chain consists of phosphate di-
ester groups joining (3-D-deoxyribofu-
ranose residues with 3', 5' linkages.
The two chains (but not their bases)
are related by a dyad perpendicular
to the fibre axis. Both chains follow
right-handed helices, but owing to
the dyad the sequences of the atoms
in the two chains run in opposite di-
rections. Each chain loosely resem-
bles Furberg's 2 model No. 1; that is,
the bases are on the inside of the he-
lix and the phosphates on the out-
side. The configuration of the sugar
and the atoms near it is close to
Furberg's 'standard configuration,' the
sugar being roughly perpendicular to
the attached base. There is a residue
on each chain every 3.4 A in the
z-direction. We have assumed an an-
gle of 36° between adjacent residues
in the same chain, so that the struc-
ture repeats after 10 residues on each
chain, that is, after 34 A. The distance

Figure 1 This figure is purely
diagrammatic. The two ribbons
symbolize the two phosphate-sugar
chains, and the horizontal rods
represent the pairs of bases holding
the chains together. The vertical line
marks the fiber axis. (Reprinted with
permission from Nature, Vol. 171, No. 4356.
Watson and Crick, "Molecular Structure of
Nucleic Acids," pp. 737-738. Copyright ©
1953 Macmillan Magazines Limited.)

of a phosphorus atom from the fibre
axis is 10 A. As the phosphates are on
the outside, cations have easy access
to them.

Most metabolic processes, such as the biosynthesis or
degradation of molecules, occur in pathways, with en-
zyme facilitating each step in the pathway (see chapter 2).
The metabolic pathway for the conversion of threonine
into isoleucine (two amino acids) appears in figure 9.2.
Each reaction product in the pathway is altered by an en-
zyme that converts it to the next product. The enzyme
threonine dehydratase, for example, converts threonine
into a-ketobutyric acid. Enzymes are composed of folded
polymers of amino acids. The average protein is three
hundred to five hundred amino acids long; only twenty
naturally occurring amino acids are used in constructing

these proteins. The sequence of amino acids determines
the final structure of an enzyme. (We discuss the struc-
ture of proteins in more detail in chapter 11.) The genetic
material determines the sequence of the amino acids.

The three-dimensional structure of enzymes permits
them to perform their function. An enzyme combines
with its substrate or substrates (the molecules it works
on) at a part of the enzyme called the active site (fig. 9.3).
The substrates "fit" into the active site, which has a shape
that allows only the specific substrates to enter. This view
of the way an enzyme interacts with its substrates is
called the lock-and-key model of enzyme functioning.

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In Search of the Genetic Material

207

The structure is an open one, and
its water content is rather high. At
lower water contents we would ex-
pect the bases to tilt so that the struc-
ture could become more compact.

The novel feature of the structure is
the manner in which the two chains
are held together by the purine and
pyrimidine bases. The planes of the
bases are perpendicular to the fibre
axis. They are joined together in pairs,
a single base from one chain being
hydrogen-bonded to a single base from
the other chain, so that the two lie side
by side with identical z-co-ordinates.
One of the pair must be a purine and
the other a pyrimidine for bonding to
occur. The hydrogen bonds are made
as follows: purine position 1 to pyrimi-
dine position 1; purine position 6 to
pyrimidine position 6.

If it is assumed that the bases only
occur in the structure in the most
plausible tautomeric forms (that is,
with the keto rather than the enol
configurations), it is found that only
specific pairs of bases can bond to-
gether. These pairs are: adenine
(purine) with thymine (pyrimidine),
and guanine (purine) with cytosine
(pyrimidine).

In other words, if an adenine
forms one member of a pair, on either
chain, then on these assumptions the
other member must be thymine; sim-
ilarly for guanine and cytosine. The
sequence of bases on a single chain
does not appear to be restricted in

any way. However, if only specific
pairs of bases can be formed, it fol-
lows that if the sequence of bases on
one chain is given, then the sequence
on the other chain is automatically
determined.

It has been found experimen-

2 A

tally ' that the ratio of the amounts
of adenine to thymine, and the ratio
of guanine to cytosine, are always
very close to unity for deoxyribose
nucleic acid. It is probably impossible
to build this structure with a ribose
sugar in place of the deoxyribose, as
the extra oxygen atom would make
too close a van der Waals contact.

The previously published X-ray
data 5 ' 6 on deoxyribose nucleic acid
are insufficient for a rigorous test of
our structure. So far as we can tell, it is
roughly compatible with the experi-
mental data, but it must be regarded
as unproved until it has been checked
against more exact results. Some of
these are given in the following com-
munications. We were not aware of
the details of the results presented
there when we devised our structure,
which rests mainly though not en-
tirely on published experimental data
and stereo-chemical arguments.

It has not escaped our notice that
the specific pairing we have postu-
lated immediately suggests a possible
copying mechanism for the genetic
material.

Full details of the structure, in-
cluding the conditions assumed in

building it, together with a set of co-
ordinates for the atoms, will be pub-
lished elsewhere.

We are much indebted to Dr.
Jerry Donohue for constant advice
and criticism, especially on inter-
atomic distances. We have also been
stimulated by a knowledge of the
general nature of the unpublished
experimental results and ideas of Dr.
M. H. F. Wilkins, Dr. R. E. Franklin and
their coworkers at King's College,
London. One of us (J. D. W) has
been aided by a fellowship from the
National Foundation for Infantile
Paralysis.

J. D. Watson
EH. C Crick

Medical Research Council Unit for
the Study of the Molecular Structure
of Biological Systems, Cavendish
Laboratory, Cambridge. April 2.

Reprinted by permission from Nature,
171:737-38. Copyright © 1953 Macmillan Mag-
azines Ltd.

1. Pauling, L., and Corey, R. B., Nature, 111, 346
(1953); Proc. U.S. Nat. Acad. Sci., 39, 84 (1953).

2. Furberg, S., Acta Chem. Scand., 6, 634 (1952).

3. Chargaff, E., for references see Zamenhof, S.,
Brawerman, G., and Chargaff, E., Biochim. et
Biophys. Acta, 9, 402 (1952).

4. Wyatt, G. R.J. Gen. Physiol, 36, 201 (1952).

5. Astbury, W. T., Symp. Soc. Exp. Biol. 1,
Nucleic Acid 66 (Camb. Univ. Press, 1947).

6. Wilkins, M. H. F., and Randall, J. T., Biochim.
et Biophys. Acta, 10, 192 (1953).

When the substrates are in their proper position in the ac-
tive site of the enzyme, the particular reaction that the en-
zyme catalyzes takes place. The reaction products then
separate from the enzyme and leave it free to repeat the
process. Enzymes can work at phenomenal speeds. Some
can catalyze as many as a million reactions per minute.

Not all of the cell's proteins function as catalysts.
Some are structural proteins, such as keratin, the main
component of hair. Other proteins are regulatory — they
control the rate at which other enzymes work. Still oth-
ers are involved in different functions; albumins, for ex-
ample, help regulate the osmotic pressure of blood.

Replication

The genetic material must be capable of precisely direct-
ing its own replication so that every daughter cell re-
ceives an exact copy. Some mutability, or the ability to
change, is also required, because we know that the ge-
netic material has changed, or evolved, over the history
of life on earth. In their 1953 paper, Watson and Crick
had already worked out the replication process based on
the structure of DNA. The fidelity of the replication
process is so great that the error rate is only about one in
a billion.

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Chapter Nine Chemistry of the Gene

Structure Name of

of compound enzyme

CH 3 — CH -CHCOOH

OH NH,

Threonine dehydratase

CH 3 — CH 2 — C- -COOH

Pyruvate ->v

CO,

Acetolactate synthase

C 2 H 5

CH 3 — C- -C- -COOH

O OH

Acetolactate mutase

NAD(P)H-^
NAD(P) + <-^

Reductase

CH 3

CH 3 CH 2 — C- -C- -COOH

OH OH

H o 0<^

Name of
compound

Threonine

a-ketobutyric acid

a-aceto-a-hydroxy butyric
acid

oc,p-dihydroxy-
p-methylvaleric acid

Dihydroxyacid dehydratase

CH,

CH 3 CH 2 — C- -C- -COOH

a-keto-p-methylvaleric
acid

H O
Glutamate -\

a-ketoglutarate ^

Valine transaminase

CH,

CH 3 — CH 2 — CH — C- - COOH

NH,

Isoleucine

ATP

Enzyme

Enzyme-substrate
complex

V Enzyme

Glucose-6-phosphate

Figure 9.3 The active site of an enzyme
recognizes a specific substance. In this
case, ATP plus glucose is converted into
ADP and glucose-6-phosphate by the
enzyme hexokinase. The active site is
diagrammed in red. The terminal
phosphate group of ATP is tan.

Figure 9.2 Metabolic pathway of conversion of the amino acid threonine into isoleucine.

Location

It has been known since the turn of the century that
genes, the discrete functional units of genetic material,
are located in chromosomes within the nuclei of eukary-
otic cells: the way chromosomes behave during the cel-

lular division stages of mitosis and meiosis mimics the be-
havior of genes. Thus, the genetic material in eukaryotes
must be a part of the chromosomes.

For a long time, proteins were considered the most
probable genetic material because they have the neces-

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In Search of the Genetic Material

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Oswald T. Avery (1877-1955).
(Courtesy of the National Academy of
Sciences.)

sary molecular complexity. The twenty naturally occur-
ring amino acids can be combined in an almost unlimited
variety, creating thousands and thousands of different
proteins. The first proof that the genetic material is de-
oxyribonucleic acid (DNA) came in 1944 from Oswald
Avery and his colleagues. The Watson and Crick model in
1953 ended a period when many thought DNA was the
genetic material, but its structure was unknown.

Figure 9.4 Petri plate with smooth and rough colonies of
Streptococcus pneumoniae. R (rough) strain colonies appear
on the left and S (smooth) colonies on the right on the same
agar. Magnification 3.5 x. (0. T. Avery, C. M. Macleod, and
M. McCarty, "Studies on the chemical nature of the substance inducing
transformation of pneumococcal types." Reproduced from the Journal of
Experimental Medicine 79 (1944):1 37-58, fig. 1 by copyright permission of
the Rockefeller University Press. Reproduced by permission. Photograph
made by Mr. Joseph B. Haulenbeek.)

Evidence for DNA as the Genetic Material

Transformation

In 1928, F. Griffith reported that heat-killed bacteria of
one type could "transform" living bacteria of a different
type. Griffith demonstrated this transformation using two
strains of the bacterium Streptococcus pneumoniae. One
strain (S) produced smooth colonies on media in a petri
plate because the cells had polysaccharide capsules. It
caused a fatal bacteremia (bacterial infection) in mice.
Another strain (R), which lacked polysaccharide capsules,
produced rough colonies on petri plates (fig. 9.4); it did
not have a pathological effect on mice. Bacteria of the
rough strain are engulfed by the mice's white blood cells;
bacteria of the virulent smooth strain survive because
their polysaccharide coating protects them.

Griffith found that neither heat-killed S-type nor live
R-type cells, by themselves, caused bacteremia in mice.
However, if he injected a mixture of live R-type and heat-
killed S-type cells into mice, the mice developed a bac-
teremia identical to that caused by living S-type cells
(fig. 95). Thus, something in the heat-killed S cells
transformed the R-type bacteria into S-type cells.

In 1944, Oswald Avery and two of his associates, C.
MacLeod and M. McCarty, reported the nature of the
transforming substance. Avery and his colleagues did
their work in vitro (literally, in glass), using colony mor-
phology on culture media rather than bacteremia in mice
as evidence of transformation. They ruled out proteins,
carbohydrates, and lipids by their extraction procedure,
by the chemical analysis of the transforming material,
and by demonstrating that the only enzymes that de-
stroyed the transforming ability were enzymes that de-

stroyed DNA. This study provided the first experimental
evidence that DNA was the genetic material: DNA trans-
formed R-type bacteria into S-type bacteria.

Phage Labeling

Valuable information about the nature of the genetic ma-
terial has also come from viruses. Of particular value are
studies of bacterial viruses — the bacteriophages, or
phages. Since phages consist only of nucleic acid sur-
rounded by protein, they lend themselves nicely to the
determination of whether the protein or the nucleic acid
is the genetic material.

A. D. Hershey and M. Chase published, in 1952, the
results of research that supported the notion that DNA
is the genetic material and, in the process, helped to

A. D. Hershey
(1908-1997). (Courtesy of
Dr. A. D. Hershey.)

Martha Chase. (Courtesy
of Cold Springs Harbor
Laboratory Archives.)

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Chapter Nine Chemistry of the Gene

O
R-type cells

(Dead)

(No change)

Heat-killed
S-type cells

OoV

R-type plus
heat-killed S-type cells

(Dead)

Figure 9.5 Griffith's experiment with Streptococcus. S-type cells will kill mice; so will heat-killed S-type
cells injected with live R-type cells. S-type cells are recovered from dead mice in both cases.

explain the nature of the viral infection process. Since all
nucleic acids contain phosphorus, whereas proteins do
not, and since most proteins contain sulfur (in the amino
acids cysteine and methionine), whereas nucleic acids do
not, Hershey and Chase designed an experiment using ra-
dioactive isotopes of sulfur and phosphorus to keep sep-
arate track of the viral proteins and nucleic acids during
the infection process. They used the T2 bacteriophage
and the bacterium Escherichia colt The phages were la-
beled by having them infect bacteria growing in culture

35,

32,

medium containing the radioactive isotopes S or P.
Hershey and Chase then proceeded to identify the mate-
rial injected into the cell by phages attached to the bac-
terial wall.

When 32 P-labeled phages were mixed with unlabeled
E. coli cells, Hershey and Chase found that the 32 P label
entered the bacterial cells and that the next generation of
phages that burst from the infected cells carried a signif-
icant amount of the 32 P label. When 35 S-labeled phages

were mixed with unlabeled E. coli, the researchers found
that the 35 S label stayed outside the bacteria for the most
part. Hershey and Chase thus demonstrated that the
outer protein coat of a phage does not enter the bac-
terium it infects, whereas the phage's inner material, con-
sisting of DNA, does enter the bacterial cell (fig. 9.6).
Since the DNA is responsible for the production of the
new phages during the infection process, the DNA, not
the protein, must be the genetic material.

RNA as Genetic Material

In some viruses, RNA (ribonucleic acid) is the genetic
material. The tobacco mosaic virus that infects tobacco
plants consists only of RNA and protein. The single, long
RNA molecule is packaged within a rodlike structure
formed by over two thousand copies of a single protein.
No DNA is present in tobacco mosaic virus particles
(fig. 9.1a). In 1955, H. Fraenkel-Conrat and R. Williams

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Bacterium in 35 S medium

Bacterium in 32 P medium

Labeled protein
stays outside,
unlabeled DNA
enters cell

Phage (unlabeled)

Progeny phage with f
35 S- labeled protein
and unlabeled DNA

Labeled phage
progeny released

Infects unlabeled
bacteria

Progeny phage
virtually unlabeled
(less than 1%of
original 35 S
recovered)

Progeny phage
strongly labeled
(more than 30%
of original 32 P
recovered)

Progeny phage
with unlabeled
protein and
32 P- labeled DNA

Unlabeled
protein stays
outside, labeled
DNA enters cell

Figure 9.6 The Hershey and Chase experiments using 35 S-labeled and 32 P-labeled T2 bacteriophages.
The nucleic acid label ( 32 P) enters the E. coli bacteria during infection; the protein label ( 35 S) does not.

showed that a virus can be separated, in vitro, into its
component parts and reconstituted as a viable virus.
This finding led Fraenkel-Conrat and B. Singer to recon-
stitute tobacco mosaic virus with parts from different
strains (fig. 9.1b). For example, they combined the RNA
from the common tobacco mosaic virus with the pro-
tein from the masked (M) strain of tobacco mosaic
virus. They then made the reciprocal combination of
common-type protein and M-type RNA. In both cases,
the tobacco mosaic virus produced during the process
of infection was the type associated with the RNA, not
with the protein. Thus, it was the nucleic acid (RNA in
this case) that was the genetic material. Subsequently,
scientists rubbed pure tobacco mosaic virus RNA into
plant leaves. Normal infection and a new generation of
typical, protein-coated tobacco mosaic virus resulted,
confirming RNA as the genetic material for this virus.

We thus conclude that DNA is the genetic material. In
the few viruses that do not have DNA, RNA serves as the

genetic material. The only exception to these statements
is one type of disease that is transmitted by a protein
without accompanying DNA or RNA (box 9.2).

CHEMISTRY OF
NUCLEIC ACIDS

Having identified the genetic material as the nucleic acid
DNA (or RNA), we proceed to examine the chemical
structure of these molecules. Their structure will tell us a
good deal about how they function.

Nucleic acids are made by joining nucleotides in a
repetitive way into long, chainlike polymers. Nucleotides
are made of three components: phosphate, sugar, and a
nitrogenous base (table 9.1 and fig. 9.8). When incorpo-
rated into a nucleic acid, a nucleotide contains one of
each of the three components. But, when free in the cell

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Chapter Nine Chemistry of the Gene

(a)

Common type

Type M

Degradation

o Q /o I

Protein

RNA

Reconstituted mixed virus

Common-type RNA

Type M protein

Infection of tobacco leaf

Common-type offspring

(b)

Figure 9.7 (a) Electron micrograph of tobacco mosaic virus.
Magnification 37,428x. (b) Reconstitution experiment of
Fraenkel-Conrat and Singer. The nucleic acid (RNA), not the
protein component of the virus, controls inheritance.
([a] © Biology Media/Photo Researchers, Inc.)

Table 9.1 Components of Nucleic Acids

Phosphate

Sugar

Base

Purines

Pyrimidines

DNA
RNA

Present
Present

Deoxyribose
Ribose

Guanine
Adenine

Cytosine
Thymine

Cytosine
Uracil

pool, nucleotides usually occur as triphosphates. The en-
ergy held in the extra phosphates is used, among other
purposes, to synthesize the polymer. A nucleoside is a
sugar-base compound. Nucleotides are therefore nucleo-
side phosphates (fig. 9.9). (Note that ATP, adenosine
triphosphate, the energy currency of the cell, is a nucle-
oside triphosphate.)

Phosphate

O ...also

represented ^^

Sugar

CD
CO
O
-Q

5'
HOCH 2 OH

4'l/°<l ,

C C ■■■ a ' so

iViWtyi represented

OH OH

5'
HOCK

cd 4' y \i

£ ? X H H/V
§ H c — C H

Q I I

OH H

Base

NH,

^6\ (

O
,C.

CO
CD

_c
Q_

5
4

HINT

-N

8CH

Adenine

9/ c^

H H 2 N

:ch

NT
Guanine

N
H

CD Ng

NH,
5

O
,C S

CH HN'

Cytosine

HN'

^N
H
Thymine

*CH

.CH

^N
H
Uracil

Figure 9.8 Components of nucleic acids: phosphate, sugars,
and bases. Primes are used to number the ring positions in the
sugars to differentiate them from the ring positions in the bases.

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BOX 9.2

Without exception, the ge-
netic material is either
DNA or RNA; it is RNA
only in a few viruses. Since virtually
all transmissible diseases are of bacte-
rial or viral origin, this means that
transmissible diseases are also caused
by organisms with DNA or RNA as
their genetic material. However, in
one interesting situation, a transmissi-
ble disease appears to be caused by an
agent without genetic material. Four
human neurological diseases and six
similar animal diseases are caused, we
believe, by proteins without DNA or
RNA. (Two conditions in yeast are
probably caused in a similar way.) The
human diseases are kuru, Creutzfeldt-
Jakob disease, Gerstmann-Straussler-
Scheinker syndrome, and a recently
discovered fatal familial insomnia. The
animal diseases are scrapie (sheep
and goats), four encephalopathies
(bovine, feline, ungulate, and mink),
and chronic wasting disease (deer
and elk). All of these diseases are ex-
tremely slow to develop, all are fatal,
and all are believed to be caused ei-
ther by the ingestion of a protein
from an infected individual or from a
mutation of the normal gene. None of
the diseases as yet has a cure, and the
mechanism of action is not completely
understood.

The diseases appear to be caused
by a protein, similar to one normally
produced in the brain of healthy indi-
viduals. The term prion (taken from
/)roteinaceous infectious particle) has
been given to these agents by Stanley
Prusiner at the University of Califor-
nia in San Francisco, a 1997 Nobel
laureate. He, along with colleagues,

Biomedical
Applications

Prions: The Biological
Equivalent of Ice-Nine

isolated the prion protein (PrP) and
recently located the gene that codes
for the protein on the short arm of
chromosome 20. In addition to the
infective form, a familial (inherited)
form of these diseases can result from
a mutation of the gene that codes for
the prion protein active in normal in-
dividuals (probably at least all mam-
mals). The normal protein is termed
PrP c , and the mutated form is re-
ferred to as PrP Sc . Normally, PrP c is a
glycoprotein found on the membrane
surface of the cells of the brain and
some other tissues.

Although no cures exist for these
diseases, kuru, at least, seems to be al-
most eradicated. It was found only
among people in part of New Guinea
who practiced cannibalism. Once the
people stopped this practice, the
spread of the disease also ceased;
kuru does not seem to be generated
to any major extent by mutation. By
controlling feeding practices, it is
believed, bovine spongiform en-
cephalopathy will also disappear. In
the past, cows were fed protein sup-
plements contaminated by material
from infected animals.

In England, a recent epidemic of
bovine spongiform encephalopathy

(BSE, or mad cow disease) peaked in
1992-1993, affecting over 160,000
cattle. At least fourteen cases of a
variant of Creutzfeldt-Jakob disease in
people in England and France were
attributed to eating affected beef, cre-
ating a panic in England. With a
change away from using animal mat-
ter in cattle feed and a culling of cat-
tle herds, the epidemic has ended.
However, new human cases may
show up in the future owing to the
long incubation period of this prion
disease.

The obvious question is, how does
a protein that does not appear to con-
tain genetic material cause a trans-
missible disease when ingested?
Prusiner has suggested several mech-
anisms that would allow an infective
protein to induce copies of the nor-
mal protein to become infective. One
of these mechanisms involves a cas-
cade in which an infective PrP Sc
binds with a normal PrP c , resulting in
two infective PrP Sc proteins. From
this, one produces two, two produce
four, four produce eight, and so on.
As Nancy Touchette, writing in The
Journal of NIH Research, pointed
out, this is the way Kurt Vonnegut de-
scribed the behavior of the mythical
ice-nine in his 1963 book, Cat's Cra-
dle. In this fictional account, a single
seed caused all of the water on earth,
by a chain reaction cascade, to form
into a novel type of ice. We have not
yet resorted to science fiction to an-
swer the mystery of prion function;
however, it seems reasonable to guess
that an eventual understanding of the
mechanism of prion function will pro-
vide us with a biological novelty.

The sugars differ only in the presence (ribose in RNA)
or absence (deoxyribose in DNA) of an oxygen in the 2'
position. (The carbons of the sugars are numbered Y to
5 '.The primes are used to avoid confusion with the num-
bering system of the bases; see fig. 9.8.) DNA and RNA
both have four bases (two purines and two pyrim-
idines) in their nucleotide chains. Both molecules have

the purines adenine and guanine and the pyrimidine
cytosine. DNA has the pyrimidine thymine; RNA has

the pyrimidine uracil. Thus, three of the nitrogenous
bases are found in both DNA and RNA, whereas thymine
is unique to DNA, and uracil is unique to RNA.

A nucleotide is formed in the cell when a base attaches
to the 1 ' carbon of the sugar and a phosphate attaches to

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Chapter Nine Chemistry of the Gene

O P — o —

-o

O CH

OH H

Nucleoside

Nucleoside monophosphate

Nucleoside triphosphate

Figure 9.9 The structure of a nucleoside and two nucleotides:
a nucleoside monophosphate and a nucleoside triphosphate.

the 5' carbon of the same sugar (fig. 9.10); the nucleotide
takes its name from the base (table 9.2). Nucleotides are
linked together (polymerized) by the formation of a
bond between the phosphate at the 5' carbon of one nu-
cleotide and the hydroxyl (OH) group at the 3' carbon of
an adjacent molecule. Very long strings of nucleotides
can be polymerized by this phosphodiester bonding
(fig. 9.11).

Biologically Active Structure ^C*

Although the identities of the nucleotides that polymer-
ized to form a strand of DNA or RNA were known, the ac-
tual structures of these nucleic acids when they function
as the genetic material remained unknown until 1953.
The general feeling was that the biologically active struc-
ture of DNA was more complex than a single string of nu-
cleotides linked together by phosphodiester bonds, and
that several interacting strands were involved. In 1953,
Linus Pauling, a Nobel laureate who had discovered the

3' 2'
OH H
Adenosine monophosphate

Purine nucleotides

Adenine

O CH

2'
OH H
Guanosine monophosphate

Cytosine

^N
5 3

O CH

Pyrimidine nucleotides

OH H
Cytidine monophosphate

O CH

OH H
Thymidine monophosphate

Figure 9.10 Structure of the four deoxyribose nucleotides.

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Table 9.2

Nucleotide Nomenclature

Base

Nucleotide (nucleoside
monophosphate)

Abbreviation

Monophosphate

Diphosphate

Triphosphate

Ribose

Deoxyribose

Ribose

Deoxyribose

Ribose

Deoxyribose

Guanine

Guanosine monophosphate
Deoxyguanosine monophosphate

GMP

dGMP

GDP

dGDP

GTP

dGTP

Adenine

Adenosine monophosphate
Deoxyadenosine monophosphate

AMP

dAMP

ADP

dADP

ATP

dATP

Cytosine

Cytidine monophosphate
Deoxycytidine monophosphate

CMP

dCMP

CDP

dCDP

CTP

dCTP

Thymine

Deoxythymidine monophosphate

dTMP

dTDP

dTTP

Uracil

Uridine monophosphate

UMP

UDP

UTP

5 , -P0 4 end

H C Base

,0.

HX Base

.0.

Figure 9.11

Polymerization of adjacent
nucleotides to form a
sugar-phosphate strand.
There is no limit to the
length the strand can be
or on the type of base
attached to each
nucleotide residue.

Nucleotide
residue

HX Base

.0.

3'-OH end

helical structure of proteins, was investigating a three-
stranded structure for the genetic material, whereas Wat-
son and Crick had decided that a two-stranded structure
was more consistent with available evidence. Three lines
of evidence directed Watson and Crick: the chemical na-
ture of the components of DNA, X-ray crystallography,
and Chargaff's ratios.

DNA X-Ray Crystallography

All the time Watson and Crick were studying DNA struc-
ture, Maurice Wilkins, Rosalind Franklin, and their col-
leagues were using X-ray crystallography to analyze the
structure of DNA. The molecules in a crystal are arranged in
an orderly way, so that when a beam of X rays is aimed at
the crystal, the beam scatteres in an orderly fashion. The
scatter pattern can be recorded on photographic film or
computer-controlled devices. The nature of this pattern de-
pends on the structure of the crystal. The cross in the cen-
ter of the photograph in figure 9.12 indicates that the mol-
ecule is a helix; the dark areas at the top and bottom come
from the bases, stacked perpendicularly to the main axis of

Rosalind E. Franklin

(1 920-1 958). (Courtesy of Cold

Spring Harbor Laboratory.)

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Chapter Nine Chemistry of the Gene

the molecule. This image of the DNA molecule stimulated
Watson and Crick's understanding of its structure.

Chargaff y s Ratios

Until Erwin Chargaff 's work, scientists had labored under
the erroneous tetranucleotide hypothesis. This hy-
pothesis proposed that DNA was made up of equal quan-
tities of the four bases; therefore, a subunit of this DNA
consisted of one copy of each base. Chargaff carefully an-
alyzed the base composition of DNA in various species
(table 9.3). He found that although the relative amount of
a given nucleotide differs among species, the amount of
adenine equaled that of thymine and the amount of gua-
nine equaled that of cytosine. That is, in the DNA of all
the organisms studied, a 1:1 correspondence exists be-
tween the purine and pyrimidine bases. This is known as
Chargaff s rule. Chargaff's observations disproved the
tetranucleotide hypothesis; the four bases of DNA did not
occur in a 1 : 1 : 1 : 1 ratio. His results gave insight to Watson
and Crick in the development of their model.

The Watson-Crick Model

With the information available, Watson and Crick began
constructing molecular models. They found that a possible
structure for DNA was one in which two helices coiled
around one another (a double helix), with the sugar-

Erwin Chargaff (1905- ).
(Courtesy of Dr. Erwin Chargaff.)

^^JH*

Figure 9.12 Scatter pattern of a beam of X rays passed
through crystalline DNA. (Source: Reprinted by permission from R. E.
Franklin and R. Gosling, "Molecular configuration in sodium thymonucleate,"
Nature 171:740-41. Copyright 1953 by Macmillan Journals Limited.)

phosphate backbones on the outside and the bases on the
inside. This structure would fit the dimensions X-ray crys-
tallography had established for DNA if the bases from the
two strands were opposite each other and formed "rungs"
in a helical "ladder" (fig. 9.13). The diameter of the helix
could only be kept constant at about 20 A (10 angstrom
units = 1 nanometer) if one purine and one pyrimidine
base made up each rung. Two purines per rung would be
too big, and two pyrimidines would be too small.

After further experimentation with models, Watson and
Crick found that the hydrogen bonding necessary to form
the rungs of their helical ladder could occur readily be-
tween certain base pairs, the pairs that Chargaff found in
equal frequencies. (Hydrogen bonds are very weak bonds
in which two electronegative atoms, such as O and N, share
a hydrogen atom between them. They have 3 to 5% of the
strength of a covalent bond.) Thermodynamically stable

TablG 9.3 Percentage Base Composition of Some DNAs

Species Adenine

Thymine

Guanine

Cytosine

Human being (liver) 30.3

30.3

19.5

19.9

Mycobacterium tuberculosis 15.1

14.6

34.9

35.4

Sea urchin 32.8

32.1

17.7

18.4

Source: From E. Chargaff and J. Davidson, The Nucleic Acids, Academic Press, 1955.

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(a)

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate
(b)

Base

Sugar'

Phosphate

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate

Sugar

Phosphate

(c)

Figure 9.13 Double helical structure of DNA. (a) DNA magnified twenty-five million times by scanning tunneling microscopy.
(b) Component parts, (c) Line drawing, ([a] © John D. Baldeschweiler.)

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Chapter Nine Chemistry of the Gene

hydrogen bonding occurs between thymine and adenine
and between cytosine and guanine (fig. 9. 14). The relation-
ship is one of complementarity. There are two hydrogen
bonds between adenine and thymine and three between
cytosine and guanine.

Another point about DNA structure relates to the po-
larity that exists in each strand. That is, one end of a DNA
strand has a 5' phosphate and the other end has a 3'
hydroxyl group. Watson and Crick found that hydrogen
bonding would occur if the polarity of the two strands
ran in opposite directions; that is, if the two strands were
antiparallel (fig. 915).

base pair has three hydrogen bonds to every two in an A-T
(adenine-thymine) base pair, the higher the G-C content in
a given molecule of DNA, the higher the temperature re-
quired to denature it. This relationship exists (fig. 9.16).

Requirements of Genetic Material

Let us now return briefly to the requirements we have said
a genetic material needs to meet: (1) control of protein syn-
thesis, (2) self-replication, and (3) location on the chromo-
somes in the nucleus (in organisms with nuclei). Does DNA
(or when DNA is absent, RNA) meet these requirements?

DNA Denaturation

Denaturation studies indicated that the hydrogen bonding
in DNA occurs in the way Watson and Crick suggested. Hy-
drogen bonds, although individually very weak, give struc-
tural stability to a molecule in large enough numbers.
However, the hydrogen bonds can be broken and the DNA
strands separated when the DNA molecule is heated in wa-
ter. At a certain point, the thermal agitation overcomes the
hydrogen bonding, and the molecule becomes denatured
(or "melts"). It is logical that the more hydrogen bonds
DNA contains the higher the temperature needed to dena-
ture it. It thus follows that since a G-C (guanine- cytosine)

-N

Hydrogen bond

H |
N-H O

CH,

/
To deoxyribose

H N- -N

N^CH

\ /

,C N

/ \

O To deoxyribose

Adenine

Thymine

: ISL

/
To deoxyribose

\ /

H
-H-N

N-H N

\
C-

N :

\ /

N-H-
H

C N

/ \

To deoxyribose

Guanine

Cytosine

Figure 9.14 Hydrogen bonding between the nitrogenous
bases in DNA.

Control of Enzymes

In the next several chapters, we examine the details of
protein synthesis. We will see that DNA does possess the
complexity required to direct protein synthesis. Al-
though complementarity restricts the base opposite a
given base in a double helix, there are no restrictions on
the sequence of bases on a given strand. Later, we will
show that each sequence of three bases in DNA specifies
a particular amino acid during protein synthesis. The ge-

Base- - -Base-

Base- - -Base

Y
3'

Figure 9.15 Polarity of the DNA strands. Polarity is established
by the 3' and 5' carbons of a given sugar. For example, moving
down the left strand, the polarity is 5' -» 3' (read as five-prime
to three-prime). Moving down the right strand, the polarity is
3' -> 5'.

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100

•Jf

+ 60

— jik

■i— >
c

CD
O

S. 40

70 80 90 100

110

Melting (denatu ration) temperature (°C)

Figure 9.16 Relationship of the number of hydrogen bonds
(G-C content) and the thermal stability of DNA from different
sources. (From J. Marmur and P. Doty, Jr., "Relationship of the Number
of Hydrogen Bonds and the Thermal Bonds and the Thermal Stability of
DNA from Different Sources," Journal of Molecular Biology, 5:109-112.
Copyright © 1962 Academic Press LTD.)

netic code gives the relationship of DNA bases to the
amino acids in proteins.

Alternative Forms of DNA

The form of DNA we have described so far is called B
DNA. It is a right-handed helix: it turns in a clockwise
manner when viewed down its axis. The bases are
stacked almost exactly perpendicular to the main axis,
with about ten base pairs per turn (34 A; see fig. 9.13c).
However, DNA can exist in other forms. If the water con-
tent increases to about 75%, the A form of DNA (A DNA)
occurs. In this form, the bases tilt in regard to the axis,
and there are more base pairs per turn. However, this and
other known forms of DNA are relatively minor varia-
tions on the right-handed B form.

In 1979, Alexander Rich and his colleagues at MIT
discovered a left-handed helix that they called 2 DNA
because its backbone formed a zigzag structure (fig.
918). 2 DNA was found by X-ray crystallographic analy-
sis of very small DNA molecules composed of repeating
G-C sequences on one strand with the complementary
C-G sequences on the other (alternating purines and
pyrimidines). 2 DNA looks like B DNA with each base ro-
tated 180 degrees, resulting in a zigzag, left-handed
structure (fig. 9.19). (The original configuration of the
bases is referred to as the anti configuration; the rotated
configuration is called the syn configuration.)

Originally, it was thought that 2 DNA would not
prove of interest to biologists because it required very
high salt concentrations to become stable. However, it
was found that 2 DNA can be stabilized in physiologi-
cally normal conditions if methyl groups are added to
the cytosines. 2 DNA may be involved in regulating gene
expression in eukaryotes. We return to this topic in
chapter 16 (box 9.3).

Replication

Watson and Crick hinted in their 1953 paper how DNA
might replicate. Their observation stemmed from the
property of complementarity. Since the base sequence
on one strand is complementary to the base sequence on
the opposite strand, each strand could act as a template
for a new double helix if the molecule simply "unzipped,"
allowing each strand to specify the sequence of bases on
a new strand by complementarity (fig. 9.17). Mutability
would occur due to mispairings, other errors in replica-
tion, or damage to the DNA.

Location

DNA must reside in the nucleus of eukaryotes, where the
genes occur on chromosomes, or in the chromosomes of
prokaryotes and viruses. In both prokaryotes and eukary-
otes, the majority of the cell's DNA is in the chromosomes.
And all viruses contain either DNA or RNA. Thus, DNA ful-
fills all the requirements of a genetic material. RNA can ful-
fill the same requirements in RNA viruses and viroids.

Template
strand _

-T

A-

-T

A-

-c

G-

-c

G-

-G

Replication
fork (Y-junction)

Template
_ strand

Figure 9.17 Complementarity provides a possible mechanism
for accurate DNA replication. The parent duplex opens, and
each strand becomes a template for a new duplex.

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Chapter Nine Chemistry of the Gene

Groove

Major
groove

Minor
groove

ZDNA

BDNA

Figure 9.18 Z (left) and B (right) DNA. The lines connect
phosphate groups. (Reproduced with permission from the Annual
Review of Biochemistry, Volume 53, © 1 984 by Annual Reviews, Inc.)

^^:-:-&

/=&■-- -<=&

^^:-:-&

&tY:^3-

fi=^yym

4&-:-rf

^:-:-a

&tY:^3-

^^:-:-ffl

^t::^3-

-^^::Sr

-~&T

^:Y^^_

- B

- Z

- B

t
3'

Figure 9.19 B DNA converts to Z DNA by the rotation of
bases as indicated by the curved arrows. (Reproduced with

permission from Annual Review of Biochemistry, Volume 53, © 1984 by
Annual Reviews, Inc.)

DNA REPLICATION —
THE PROCESS

In their 1953 paper, Watson and Crick hinted that the repli-
cation of the double helix could take place as the DNA un-
winds, so that each strand would form a new double helix
by acting as a template for a newly synthesized strand (see
fig. 9. 17). For example, when a double helix is unwound at
an adenine-thymine (A-T) base pair, one unwound strand
would carry A and the other would carry T. During replica-
tion, the A in the template DNA would pair with a T in a
newly replicated DNA strand, giving rise to another A-T
base pair. Similarly, the T in the other template strand
would pair with an A in the other newly replicated strand,
giving rise to another A-T base pair. Thus, one A-T base pair
in one double helix would result in two A-T base pairs in
two double helices. This process would repeat at every
base pair in the double helix of the DNA molecule.

This mechanism is called semiconservative replica-
tion because, although the entire double helix is not con-
served in replication, each strand is. Every daughter DNA
molecule has an intact template strand and a newly rep-
licated strand. This is not the only way that replication
could occur. The alternative methods are conservative
and dispersive. In conservative replication, in which the
whole original double helix acts as a template for a new
one, one daughter molecule would consist of the original
parental DNA, and the other daughter would be totally new
DNA. In dispersive replication, some parts of the original
double helix are conserved, and some parts are not. Daugh-
ter molecules would consist of part template and part
newly synthesized DNA. In reality the dispersive category
is the all-inclusive "other" category, including any possibility
other than conservative and semiconservative replication.

The Meselson and Stahl Experiment

In 1958, M. Meselson and F. Stahl reported the results of
an experiment designed to determine the mode of DNA

Matthew Meselson
(1930- ). (Courtesy of
Dr. Matthew Meselson.
Photograph by Bud Gruce.)

Franklin W. Stahl
(1929- ). (Courtesy of
Dr. Franklin W. Stahl.)

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DNA Replication — The Process

221

BOX 9.3

Under natural conditions, single-
stranded RNA and double-
stranded DNA are the rule.
However, under laboratory condi-
tions, it is possible to induce a third
strand of DNA to interdigitate itself
into the major groove of the double
helix of normal DNA in a sequence-
specific fashion. That is, the third
strand of DNA will not just interdigi-
tate anywhere, but will form a stable
triplex at a specific sequence (fig. 1).
The rules of binding are a little less
precise than normal; not all sequences
are recognized, and recognition can
depend on surrounding sequences.
However, a thymine in the third strand
will usually recognize an adenine in an
adenine-thymine base pair (T*A-T)
and a cytosine in the third strand will
recognize a guanine in a guanine-
cytosine base pair (C+G-C).

Triple-stranded nucleotide chains
were first created in 1957 by three
scientists at the National Institutes
of Health — Alexander Rich, David
Davies, and Gary Felsenfeld — while
they were creating artificial nucleic
acids. At the time, triple-stranded
DNA seemed like a laboratory curios-
ity. Now it seems of interest because
it may have valuable uses both exper-
imentally and clinically. (Rich appar-

Biomedical
Applications

Multiple-Stranded DNA

ently had the same experience in his
codiscovery of Z DNA, which at first
seemed like an oddity but now is the
focus of some attention — see chap-
ter 16.) Now, researchers are able to
form triplexes in naturally occurring
DNA. Two applications of this tech-
nology are actively being pursued.

Both applications arise because a
single strand of DNA is capable of rec-
ognizing a relatively long sequence of
the double-stranded DNA in a chro-
mosome. Thus, it is possible to selec-
tively locate a particular genie se-
quence. Once the third strand locates
a particular sequence on a chromo-
some, two things can happen. First,
triplex DNA formation can prevent a
particular gene from expressing itself.
By the same technique, triplex DNA
can also be an abortifacient, a safe
method for preventing the implanta-
tion of a fetus by preventing the
expression of genes under the control
of the hormone progesterone.

The second use of triplex DNA is
to cut DNA at a specific place by
adding a cleaving compound to both
ends of the third strand of DNA. Once
the third strand has interdigitated, it
can then break the original double
helix. For example, S. Strobel and P.
Dervan at the California Institute of
Technology have used a chemical
complex containing iron attached to
both 3' and 5' ends of the third
strand of DNA. The addition of a third
chemical then initiates the cleavage
reaction. The cleavage of the original
duplex can be of value in modern re-
combinant DNA technology (see
chapter 13). Whether triplex DNA
will ever be of value is not certain at
this time. However, it seems to have
good potential for therapeutic use
and to help in studying and mapping
the human genome.

More recently, four-stranded DNA
molecules have been found, in which
double helices of certain sequences
interdigitate to form four-stranded
structures. These may be of impor-
tance in the formation of crossover
sites or in the structures at the ends
of eukaryotic chromosomes (see
chapter 15).

continued

replication. Some historians and philosophers of science
consider this the most elegant scientific experiment ever
designed. Meselson and Stahl grew E. coli in a medium
containing a heavy isotope of nitrogen, 15 N. (The normal
form of nitrogen is 14 N.) After growing for several gener-
ations on the 15 N medium, the DNA of E. coli was denser.
The researchers determined the density of the strands us-
ing a technique known as density-gradient centrifu-
gation. In this technique, a cesium chloride (CsCl) solu-
tion is spun in an ultracentrifuge at high speed for several
hours. Eventually an equilibrium arises between centrifu-
gal force and diffusion, so that a density gradient is estab-
lished in the tube with an increasing concentration of
CsCl from top to bottom. If DNA (or any other substance)
is added, it concentrates and forms a band in the tube at
the point where its density is the same as that of the
CsCl. If several types of DNA with different densities are
added, they form several bands. The bands are detectable

under ultraviolet light at a wavelength of 260 nm
(nanometers), which nucleic acids absorb strongly.

Meselson and Stahl transferred the bacteria with heavy

-15

14,

( N) DNA to a medium containing only N. The new

DNA, replicated in the N medium, was intermediate in

14 T

density between light C*N) and heavy ( 15 N) DNA, because
the replication was semiconservative (fig. 9.20). If replica-
tion had been conservative, two bands would have ap-
peared at the first generation of replication — an original
15 N DNA and a new 14 N double helix. And, throughout the
experiment, if the method of replication had been conser-
vative, the original DNA would have continued to show
up as a 15 N band. (This, of course, did not happen.) If
the method of replication had been dispersive, various
multiple-banded patterns would have appeared, depend-
ing on the degree of dispersiveness.The results figure 9.20
shows are completely consistent with semiconservative
replication and only with semiconservative replication.

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222

Chapter Nine Chemistry of the Gene

BOX 9-3 CONTINUED

~ZL

(a)

(b)

T
T
T

T
T

T
T
T
T

3'
I

T
T
T
T
T
C + G -

A
A
A

C + G -

A
A

C + G -

A
A
A
A

C + G -

A -

G -

C -

A -

G -

(c)

5'
I

A
C
G
G
A
T
C
C
T
T
T
T
T
C
T
T
T
C
T
T
C
T
T
T
T
C
T
T
C
C
G
G
G
T
C

Figure 1 A triple helix of DNA. (a) Space-filling model of (£>). (c) Sequence of third strand on left and double helix on
right. The asterisks in (c) and the Fe symbols in (b) refer to the attached iron compound. (© Courtesy Dr. Michael E. Hogan,
Triplex Pharmaceutical Corporation.)

Autoradiographic Demonstration
of DNA Replication

In 1963, J. Cairns used autoradiography to verify the
semiconservative method of replication photographi-
cally This technique makes use of the fact that radioac-
tive atoms expose photographic film. The visible silver
grains on the film can then be counted to provide an es-

timate of the quantity of radioactive material present.
Cairns grew E. colt bacteria in a medium containing ra-
dioactive thymine, a component of one of the DNA nu-
cleotides. The radioactivity was in tritium ( 3 H). Cairns
then carefully extracted the DNA from the bacteria and
placed it on photographic emulsion for a period of time.
He developed the emulsion to produce autoradiographs

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DNA Replication — The Process

223

Density-gradient
centrifuge tube

- 14 N

- 14 N/ 15 N

- 15 N

■ 14 N
- 14 N/ 15 N

■ 15 N

■ 14 N
- 14 N/ 15 N

■ 15 N

■ 14 N

- 14 N/ 15 N

■ 15 N

Original
15 N DNA

Generation
1 in 14 N
medium

Generation
2

Generation
3

DNA

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/ \ / \ / \

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i ii 11 II 1

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•vr

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:-'■ - : '-V -/- ' : -

■ ■"■/■ ; ". J T '■ j* ■

'■:»,

Figure 9.20 The Meselson and Stahl experiment to determine
the mode of DNA replication. The bands in the centrifuge tube
are visible under ultraviolet light. The pattern of bands (left)
comes about from semiconservative DNA replication (right) of
15 N DNA (blue) replicating in a 14 N medium (red).

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that he then examined under the electron microscope
(fig. 921). Each grain of silver represents a radioactive de-
cay. Interpretation of this autoradiograph reveals several
points. The first, known at the time, is that the E. coli
DNA is a circle. The second point is that the DNA is repli-
cated while maintaining the integrity of the circle. That
is, the circle does not appear to break during the process
of DNA replication; an intermediate theta structure
forms (topologically similar in shape to the Greek letter
theta, 6). Third, replication of the DNA seems to be oc-
curring at one or two moving Y- junctions in the circle,
which further supports the semiconservative mode of
replication. The DNA is unwound at a given point, and
replication proceeds at a Yjunction, in a semiconserva-
tive manner, in one or both directions (see fig. 9.17).

Figure 9.22 diagrams the way in which the two
Yj unctions move along the circle to the final step, form-
ing two new circles. The steps by themselves do not sup-
port either a unidirectional or a bidirectional mode of
replication. That is, a theta structure will develop if either
one or both Yj unctions is active in replication. But with
autoradiography, it is possible to determine whether new
growth is occurring in only one or in both directions.

In some cases, radioactivity was not applied to the cell
until DNA replication had already begun. In these cases,
the radioactive label appeared after the theta structure

(b)

Figure 9.21 (a) Autoradiograph of E. coli DNA during
replication, (b) Diagram has labels on the three segments, A, B,
and C, created by the existence of two forks, X and Y, in the
DNA. Forks are created when the circle opens for replication.
Length of the chromosome is about 1 ,300 |xm. ([a] From J.
Cairns, "The chromosome of E coli", Cold Spring Harbor Symposia on
Quantitative Biology, 28. Copyright © 1 963 by Cold Spring Harbor
Laboratory Press, Cold Spring Harbor, NY. Reprinted by permission.)

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Chapter Nine Chemistry of the Gene

Origin

Figure 9.22 Observable stages in the DNA replication of a circular chromosome,
assuming bidirectional DNA synthesis. The intermediate figures are called theta structures.

Origin

Unidirectional

Bidirectional

Radioactive label at
only one Y-junction

Radioactive label at
both Y-junctions

Figure 9.23 Radioactive labels distinguish unidirectional from
bidirectional DNA replication. In these hypothetical experiments,
DNA replication was allowed to begin, and then a radioactive
label was added. After a short period of time, the process was
stopped and the autoradiographs prepared. In bidirectional
replication (the actual case), the label appears at both Y-
junctions.

had already begun forming. Figure 9. 2 3 illustrates hypo-
thetical outcomes for either unidirectional or bidirec-
tional replication. By counting silver grains in auto-
radiographs, Cairns found growth to be bidirectional.
Both autoradiographic and genetic analysis have subse-
quently verified this finding.

In eukaryotes, the DNA molecules (chromosomes)
are larger than in prokaryotes and are not circular; there
are also usually multiple sites for the initiation of replica-
tion. Thus, each eukaryotic chromosome is composed of
many replicating units, or replicons — stretches of DNA
with a single origin of replication. In comparison, the
E. colt chromosome is composed of only one replicon. In
eukaryotes, these replicating units form "bubbles" (or
"eyes") in the DNA during replication (fig. 9.24).

Origin

(a)

(b)

Figure 9.24 Replication bubbles, (a) Formation of bubbles
(eyes) in eukaryotic DNA because of multiple DNA synthesis
sites of origin, (b) Electron micrograph (and explanatory line
drawing) of replicating Drosophila DNA showing these bubbles.
([£>] H. Kreigstein and D. Hogness, "Mechanism of DNA replication in
Drosophila chromosomes: Structure of replication forks and evidence for
bidirectionality," Proceeding of the National Academy of Sciences USA, 71
(1974):1 35-39. Reproduced by permission.)

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DNA Replication — The Enzymology

225

DNA REPLICATION
THE ENZYMOLOGY

* %

Let us turn now to the details of the processes that take
place during DNA replication. Like virtually all metabolic
processes, DNA replication is under the control of en-
zymes. The evidence for the details we describe comes
from physical, chemical, and biochemical studies of en-
zymes and nucleic acids and from the analysis of muta-
tions that influence the replication processes. More re-
cent techniques of recombinant DNA technology and
nucleotide sequencing have allowed us to determine the
nucleotide sequences of many of these key regions in
DNA and RNA. We will look first at E. coll

There are three major enzymes that will polymerize
nucleotides into a growing strand of DNA in E. colt.
These enzymes are DNA polymerase I, II, and III.
DNA polymerase I, discovered by Arthur Kornberg,
who subsequently won the Nobel Prize for his work, is
primarily utilized in filling in small DNA segments dur-
ing replication and repair processes. DNA polymerase
II can serve as an alternative repair polymerase; it can
also replicate DNA if the template is damaged. DNA
polymerase III is the primary polymerase during nor-
mal DNA replication.

Arthur Kornberg (1918- ). (Courtesy of Dr. Arthur
Kornberg. Photograph by Karsh.)

In the simplest model of DNA replication, new nu-
cleotides would be simultaneously added, according to
the rules of complementarity, on both strands of newly
synthesized DNA at the replication fork as the DNA
opens up. But a problem exists, created by DNAs an-
tiparallel nature; the two strands of a DNA double helix
run in opposite directions. Going in one direction on
the duplex, for example, one strand is a 5' — > 3' strand,
whereas the other is a 3' — > 5' strand. These directions

refer to the numbering of carbon atoms across the
sugar. In figure 9.25, going from the bottom of the fig-
ure to the top, the left-hand strand is a 3' — > 5' strand,
and the right-hand strand is a 5' —> 3' strand. Since
DNA replication involves the formation of two new an-
tiparallel strands with the old single strands as tem-
plates, one new strand would have to be replicated in
the 5' — > 3' direction and the other in the 3' — » 5' di-
rection.

However, all the known polymerase enzymes add nu-
cleotides in only the 5' — > 3' direction. That is, the poly-
merase catalyzes a bond between the first 5 / -P0 4 group
of a new nucleotide and the 3 -OH carbon of the last nu-
cleotide in the newly synthesized strand (fig. 9.25). The
polymerases cannot create the same bond with the 5'
phosphate of a nucleotide already in the DNA and the 3'
end of a new nucleotide. Thus, the simple model needs
some revision.

Continuous and Discontinuous ^%
DNA Replication ^*&

Autoradiographic evidence leads us to believe that repli-
cation occurs simultaneously on both strands. Continu-
ous replication is, of course, possible on the 3' — > 5'
template strand, which begins with the necessary 3 '-OH
primer. (Primer is double-stranded DNA — or, as we
shall see, a DNA-RNA hybrid — continuing as single-
stranded DNA template. The strand being synthesized
has a 3 -OH available; fig. 9.26.) A discontinuous form
of replication takes place on the complementary strand,
where it occurs in short segments, moving backward,
away from the Yjunction (fig. 9.27). These short segments,
called Okazaki fragments after R. Okazaki, who first
saw them, average about 1,500 nucleotides in prokary-
otes and 150 in eukaryotes. The strand synthesized con-
tinuously is referred to as the leading strand, and the
strand synthesized discontinuously is referred to as the
lagging strand.

Once initiated, continuous DNA replication can
proceed indefinitely. DNA polymerase III on the lead-
ing-strand template has what is called high processiv-
ity: once it attaches, it doesn't release until the entire
strand is replicated. Discontinuous replication, how-
ever, requires the repetition of four steps: primer syn-
thesis, elongation, primer removal with gap filling, and
ligation.

Primer Synthesis and Elongation

To synthesize Okazaki fragments, a primer must be cre-
ated de novo (Latin: from the beginning). None of the
DNA polymerases can create that primer. Instead, pri-
mase, an RNA polymerase coded for by the dnaG gene,
creates the primer, ten to twelve nucleotides, at the site

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Chapter Nine Chemistry of the Gene

Base- - -Base

Polymerase can act

Base- - -Base

Synthesis

U V 7

Polymerase cannot act

Base- - -Base

C 5'

Figure 9.25 New nucleotides can be added to DNA only during replication in the 5' -> 3' direction.

of Okazaki fragment initiation (fig. 9.28). The result is a
short RNA primer that provides the free 3 '-OH group
that DNA polymerase III needs in order to synthesize the
Okazaki fragment. DNA polymerase III continues until it
reaches the primer RNA of the previously synthesized
Okazaki fragment. At that point, it stops and releases
from the DNA.

All three prokaryotic polymerases not only can add
new nucleotides to a growing strand in the 5' — > 3' di-

rection, but also can remove nucleotides in the opposite
3' — > 5' direction. This property is referred to as 3' — > 5'
exonuclease activity. Enzymes that degrade nucleic
acids are nucleases. They are classified as exonucleases
if they remove nucleotides from the end of a nucleotide
strand or as endonucleases if they can break the sugar-
phosphate backbone in the middle of a nucleotide
strand. At first glance, exonuclease activity seems like an
extremely curious property for a polymerase to have —

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DNA Replication — The Enzymology

227

5' 3'

Base

Template
strand

Base

3' -OH

Base- - -Base-

Base- - -Base

Growing
progeny
strand

5' C

Base- - -Base

Figure 9.26 Primer configuration for DNA replication. A 3'-OH
group must be available on the nascent progeny strand
opposite a continuing single-stranded template.

Okazaki
fragment 3

Continuous

Discontinuous

replication

(leading

(lagging

strand)

Figure 9.27 Discontinuous model of DNA replication.
Lagging-strand replication requires Okazaki fragments to form
going backward, away from the Y-junction.

the proofreading function of DNA polymerase. In addi-
tion, exonuclease activity can remove the RNA primers
of Okazaki fragments.

Primer Removal with Gap Filling

DNA polymerase I is a polymerase when it adds nu-
cleotides, one at a time, and an exonuclease when it re-
moves nucleotides one at a time. To complete the
Okazaki fragment, DNA polymerase I acts in both capac-
ities. (DNA polymerase I mutants cannot properly con-
nect Okazaki fragments.) DNA polymerase I completes
the Okazaki fragment by removing the previous RNA
primer and replacing it with DNA nucleotides (fig. 9.29).
When DNA polymerase I has completed its nuclease and
polymerase activity, the two previous Okazaki fragments
are almost complete. All that remains is for a single phos-
phodiester bond to form.

curious unless we think about its ability to check com-
plementarity. If the complementarity is improper, mean-
ing that the wrong nucleotide has been inserted, the
polymerase can remove the incorrect nucleotide, put in
the proper one, and continue on its way. This is known as

Ligation

DNA polymerase I cannot make the final bond to join
two Okazaki fragments. The configuration needing com-
pletion is shown in figure 9. 30. An enzyme, DNA ligase,

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Chapter Nine Chemistry of the Gene

Continuous
replication

RNA primer

Last Okazaki
fragment

Continuous replication
on leading strand

RNA primer formation
on lagging strand

Continuous replication
on leading strand

DNA polymerase III results
in Okazaki fragment
synthesis on lagging strand

Figure 9.28 Primer formation and elongation create an Okazaki fragment during discontinuous DNA replication.

completes the task by making the final phosphodiester
bond in an energy-requiring reaction.

A question of evolutionary interest is why RNA is
used to prime DNA synthesis. Why not use DNA directly
and avoid the exonuclease and resynthesis activity seen
in figure 9. 29? Probably, making use of RNA primers low-
ers the error rate of DNA replication. That is, priming is
an inherently error-prone process since nucleotides are
initially added without a stable primer configuration. To
prevent long-term errors in the DNA, an RNA primer is
put in that can later be recognized and removed. Resyn-
thesis by polymerase I is in a much more stable primer
configuration (a long primer) and thus makes very few
errors.

Another question of evolutionary interest is why
DNA synthesis cannot take place in the 3' — > 5' direc-

tion. Probably, the answer has to do with proofreading
and the exonuclease removal of mismatched nu-
cleotides. When an incorrect nucleotide is found and re-
moved, the next nucleotide brought in, in the 5' — > 3' di-
rection, has a triphosphate end available to provide the
energy for its own incorporation (see fig. 9.25). Con-
sider what would happen if the polymerase were capa-
ble of adding nucleotides in the opposite direction. The
energy for the phosphodiester bond would be coming
from the triphosphate already attached in the growing
3' — » 5' strand (see fig. 9.25). Then, if an error in com-
plementarity were detected and the polymerase re-
moved the most recently added nucleotide from the
3' — > 5' strand, the last nucleotide in the double helix
would no longer have a triphosphate available to pro-
vide energy for the diester bond with the next nu-

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DNA Replication — The Enzymology

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Leading strand

Okazaki

fragment

3'*

Primer

Okazaki fragment

Primer

5'
3'

Removed primer
fragments

DNA polymerase I

5'
3'

Nick

Figure 9.29 The completion of an Okazaki fragment requires that DNA polymerase I replace the RNA primer base
by base with DNA nucleotides. A final nick in the DNA backbone remains {arrow).

cleotide. Continued polymerization would thus require
additional enzymatic steps to provide the energy
needed for the process to continue. This could stop or
slow the process down considerably As it is, the process
incorporates about four hundred nucleotides per sec-
ond with an error rate of about one incorrect pairing per
10 bases. (Other repair systems further improve this er-
ror rate — see chapter 12.)

The Origin of DNA Replication

Each replicon (e.g., the E. coli chromosome, or a segment
of a eukaryotic chromosome with an origin of replica-
tion) must have a region where DNA replication initiates.
In E. coli, this region is referred to as the genetic locus
oriC; it occurs at map location 84 minutes (see fig. 7.27).
For DNA replication to begin, several steps must occur.

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Chapter Nine Chemistry of the Gene

First, the appropriate initiation proteins must recognize
the specific origin site. Then the site must be opened and
stabilized. And, finally, a replication fork must be initiated
in both directions, involving continuous and discontinu-
ous DNA replication. Although most of the proteins in-
volved are known, there are still a few gaps in our knowl-
edge.

OriC, the origin of replication in E. coli, is about 245
base pairs long and is recognized by initiator proteins.
These proteins, the product of the dnaA locus, open up

the double helix. (Other DNA-binding proteins are also
involved here.) The initiator proteins then take part in
the attachment of DNA helicase, the product of the
dnaB gene, which unwinds DNA at the Yj unction.
Helicase is then responsible for recruiting (binding) the
rest of the proteins that form the replication initiation
complex. First is primase, which creates RNA primers.
Together, the helicase and primase comprise a primo-
some, attached to the lagging-strand template. As the
primosomes move along, they create RNA primers that

Okazaki

fragment

New Okazaki
fragment

Base

DNA ligase

Okazaki

fragment

New Okazaki
fragment

Base

Figure 9.30 After DNA polymerase I removes the RNA primer to complete an Okazaki fragment, a final gap
remains. DNA ligase closes it.

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DNA Replication — The Enzymology

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DNA polymerase III uses to initiate leading-strand syn-
thesis. As primers are being laid down on the lagging-
strand template, Okazaki fragment synthesis begins, and
Yj unction activity then proceeds as outlined earlier (see
figs. 9.28, 9.29, and 9.30).

DNA polymerase III holoenzyme is a very large pro-
tein composed of ten subunits (table 9.4). Three of the
subunits, a, s, and 6, form the polymerization core, with
both 5' — > 3' polymerase activity and 3' — > 5' exonucle-
ase activity. One subunit, the p subunit, is a "processivity
clamp." As a dimer (two identical copies attached head to
tail), the protein forms a "doughnut" around the DNA so
it can move freely on the DNA. When it is attached to the
core enzyme, the polymerase is held tightly to the DNA
and shows high processivity (fig. 93 1): the leading
strand is usually synthesized entirely without the enzyme
leaving the template (fig. 9.32). The remaining subunits

are involved in processivity control and replisome forma-
tion. They allow the polymerase to move off and on the
DNA of the lagging-strand template as Okazaki fragments
are completed (a process known as polymerase cy-
cling).

Eukaryotes have evolved at least nine DNA poly-
merases, named DNA polymerase a,p,7,8,e,£,iq,6, and
i. DNA polymerase 8 seems to be the major replicating
enzyme in eukaryotes, forming replisomes as in E. colt In
eukaryotes, the polymerase a-primase complex adds the
Okazaki fragment primers, first adding an RNA primer
and then a short length of DNA nucleotides. Polymerase
8 may be involved in repair or in normal DNA replication,
as is polymerase 8. DNA polymerase 7 appears to repli-
cate mitochondrial DNA. The remaining polymerases are
probably involved in DNA repair, with polymerase (3 be-
ing the major repair polymerase, as polymerase I is in

Table 9.4 Summary of the Enzymes

Involved in DNA Replication in E.

coli

Eniyme or Protein

Genetic Locus

Function

DNA polymerase I

polA

Gap filling and primer removal

DNA polymerase II

polB

Replicating damaged templates

DNA polymerase III

a subunit

dnaE

Polymerization core; 5' — > 3' polymerase

e subunit

dnaQ

Polymerization core; 3' — > 5' exonuclease

subunit

holE

Polymerization core

(3 subunit

dnaN

Processivity clamp (as a dimer)

t subunit

dnaX

Preinitiation complex; dimerization of core

7 subunit

dnaX

Preinitiation complex; loads clamp

8 subunit

holA

Processivity core

8' subunit

holB

Processivity core

X subunit

holC

Processivity core

iji subunit

holD

Processivity core

Helicase

dnaB

Primosome; unwinds DNA

Primase

dnaG

Primosome; creates Okazaki fragment
primers

Initiator protein

dnaA

Binds at origin of replication

DNA ligase

lig

Closes Okazaki fragments

Ssb protein

ssb

Binds single-stranded DNA

DNA topoisomerase I

topA

Relaxes supercoiled DNA

DNA topoisomerase type II

(DNA Gyrase)

a subunit

gyrA

Relaxes supercoiled DNA; ATPase

p subunit

gyrB

Relaxes supercoiled DNA

Topoisomerase IV

parE

Unconcatenates DNA circles

Termination protein

tus

Binds at termination sites

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Chapter Nine Chemistry of the Gene

Figure 9.31 Stereo view of sliding clamp, DNA polymerase, and DNA from bacteriophage
RB69. The clamp {red, blue, green) surrounds the DNA {brown) like a doughnut. The clamp is
attached to the proximal segment of the DNA polymerase (gray). (From Yousif Shamoo and
Thomas A. Steitz, "Building a replisome from interacting pieces" in Cell, 99:155-166, October 15. Reprinted by
permission of Cell.)

E. colt. Several of the polymerases most likely both repli-
cate and repair DNA.

Eukaryotes also have a clamp-loader complex, called
replication factor C, and a six-unit clamp called the pro-
liferating cell nuclear antigen. The RNA primers are re-
moved during Okazaki fragment completion (matura-
tion) by mechanisms similar to those in prokaryotes. In
eukaryotes, RNAase enzymes remove the RNA primers
in Okazaki fragments; a repair polymerase fills gaps; and
a DNA ligase forms the final seal. Helicases, topo-
isomerases, and single-strand binding proteins play roles
similar to those they play in prokaryotes. The completion
of the replication of linear eukaryotic chromosomes in-
volves the formation of specialized structures at the tips
of the chromosomes, which we discuss in chapter 15.
Thus, all of the enzymatic processes are generally the
same in prokaryotes and eukaryotes. DNA replication de-
veloped in prokaryotes and was refined as prokaryotes
evolved into eukaryotes.

T. Steitz and his colleagues have done much X-ray
crystallography work that has given us an excellent
look at the structure of a polymerase. (Most work has
actually been done on a fragment of DNA polymerase I
called the Klenow fragment.) The enzyme is shaped
like a cupped right hand with enzymatic activity taking
place in two places, separated by a distance of about

two to three nucleotides (fig. 9.33). It is proposed that
when the polymerization site senses a mismatch, the
DNA is moved so that the 3' end enters the exonucle-
ase site, where the incorrect nucleotide residue is then
cleaved. Polymerization then continues. There may be a
general mode of polymerase action among diverse
polymerases.

The replication of the E. colt chromosome may be
controlled by the methylation state of several sequences
within oriC. As we discuss in chapter 13, certain en-
zymes add methyl groups to specific DNA bases, and the
presence or absence of these methyl groups can serve as
signals to other enzymes.

Events at the Y-Junction

We now have the image of DNA replication proceeding as
a primosome moves along the lagging-strand template,
opening up the DNA (helicase activity), and creating RNA
primers (primase activity) for Okazaki fragments. One
DNA polymerase III moves along the leading-strand tem-
plate, generating the leading strand by continuous DNA
replication, whereas a second DNA polymerase III moves
backward, away from the Yj unction, creating Okazaki
fragments. Single-strand binding proteins (ssb pro-
teins) keep single-stranded DNA stabilized (open) during

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DNA Replication — The Enzymology

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oriC

Initiator proteins

Initiator proteins attach

Primosome

Helicase
attaches

Primosomes form

Primosomes move down DNA and initiate primers (5'— >*3')

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 i i'

Primer

1 1 1 1 1 1 1 1 1 1 1 1 1

Leading-strand synthesis begins

DNA
polymerase III

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Clamp

DNA
polymerase

I - ! I I I ■i- i j

iii

Primers for Okazaki fragments created

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

JT 71

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Okazaki primer

■ mil iii

Okazaki primer

Figure 9.32 Events at the origin of DNA replication in E. coll. The DNA opens up at oriC to create two moving Y-junctions.
Initiator proteins attach and then bind helicase. The helicase then binds primase, forming a primosome. After the primer forms
and two copies of DNA polymerase III are bound, the polymerization process begins.

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Chapter Nine Chemistry of the Gene

Clamp

Editing mode

Polymerizing mode

Figure 9.33 DNA polymerase (P) and exonuclease (E) activities
of the Klenow fragment of DNA polymerase I in E. coli. On the
right, 5' -» 3' polymerization is occurring. On the left, the 3'
end of the nascent strand has been backed up into the
exonuclease site, presumably when a mismatch was detected.

(With permission from the Annual Review of Biochemistry, Volume 63.
©1994 by Annual Reviews. www.AnnualReviews.org.)

this process, and DNA polymerase I and ligase connect
Okazaki fragments (fig. 9.34).

This simple picture is slightly complicated by the fact
that the lagging- and leading-strand synthesis is coordi-
nated. B. Alberts suggested an explanation: the repli-
some model, in which both copies of DNA polymerase
III are attached to each other and work in concert with
the primosome at the Y-junction (fig. 9.35). According to
this model, a single replisome, consisting of two copies
of DNA polymerase III, a helicase, and a primase, moves
along the DNA. The leading-strand template is immedi-
ately fed to a polymerase, whereas the lagging-strand
template is not acted on by the polymerase until an RNA
primer has been placed on the strand, meaning that a
long (fifteen-hundred base) single strand has been
opened up (fig. 9.35^).

As the replisome moves along, another single-
stranded length of the lagging-strand template forms. At
about the time that the Okazaki fragment is completed, a
new RNA primer has been created (fig. 935fe>).The Okazaki
fragment is released (fig. 9. 3 5c), and a new Okazaki frag-
ment is begun (polymerase cycling), starting with the lat-
est primer (fig. 935d). This takes the replisome back to
the same configuration as in figure 935a, but one
Okazaki fragment farther along.

Figure 9. 36 gives us a closer look at the details of the
Y-junction at the moment of polymerase cycling. Pri-
mase, which is not highly processive, must be in touch
with an ssb protein to stay attached to the DNA when
forming a primer. At the appropriate moment, after the
primer is formed, the clamp loader contacts the ssb, dis-

— DNA polymerase III

Leading strand

Primase

Lagging strand

ssb proteins

Helicase

DNA polymerase III

Figure 9.34 Schematic drawing of DNA replication at a
Y-junction. Two copies of DNA polymerase III, ssb proteins, and
a primosome (helicase + primase) are present.

lodging the primase. The clamp loader also loads a sliding
clamp, which then recruits (attaches to) the polymerase
that is creating the lagging strand. The polymerase then
continues, creating the Okazaki fragment. The primase
can later attach at a new point on the lagging-strand
template to create the next primer.

Supercoiling

The simplicity and elegance of the DNA molecule
masks an inevitable problem: coiling. Since the DNA
molecule is made from two strands that wrap about
each other, certain operations, such as DNA replication
and its termination, face topological difficulties. Up to
this point, we have seen the circular E. coli chromo-
some in its "relaxed" state (e.g., figs. 9.21 and 9.22).
However, certain enzymes in the cell cause DNA to be-
come overcoiled (positively supercoiled) or under-
coiled (negatively supercoiled). Positive supercoiling
comes about in two ways: either the DNA takes too
many turns in a given length, or the molecule wraps
around itself (fig. 9.37).

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DNA Replication — The Enzymology

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Replisome

(a)

(b)

(c)

ssb proteins

Okazaki fragment A

(part of completed lagging strand)

Okazaki fragment B synthesized as
unreplicated loop elongates

New C primer formed as

Okazaki fragment B completed and

released; A to B gap to

be completed by DNA polymerase I

and ligase

Completion by DNA polymerase I
and ligase

C primer positioned to begin
formation of next Okazaki fragment

(d)

Figure 9.35 The replisome, which consists of two DNA polymerase III holoenzymes and a primosome (helicase + primase),
coordinates replication at the Y-junction. Parts b-d show "polymerase cycling," in which the polymerase on the lagging-strand
template releases a completed Okazaki fragment and then begins the next one.

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Chapter Nine Chemistry of the Gene

DnaB Helicase
Primase

Pol III core

Clamp loader
(7 complex)

Figure 9.36 A close-up view of the Y-junction during
polymerase cycling. The two polymerases (pol III core) are held
together by t subunits. Also pictured are the sliding clamp
((3 Clamp), clamp loader (7 complex), primase, helicase, and
ssb proteins. In (a), the primase has just finished creating a
primer. The x subunit of the clamp loader contacts the ssb
protein that is touching the primase; the primase is then
dislodged (£>). The clamp is loaded at the new primer and the
polymerase on the lagging strand is cycled to the clamp to
begin the next Okazaki segment (c). (Reprinted from Cell, Vol. 96,

Yuzhakov et al., "Trading Places on DNA-a Three Point Switch Underlies
Primer Handoff from Primase to the Replicative DNA Polymers," pp. 153-163,
Copyright © 1999, with permission from Elsevier Science.)

Positive supercoiling occurs when the circular du-
plex winds about itself in the same direction as the helix
twists (right-handed), whereas negative supercoiling
comes about when the duplex winds about itself in the
opposite direction as the helix twists (left-handed). The
former increases the number of turns of one helix
around the other (the linkage number, L), whereas the
latter decreases it. The three forms of DNA in figure 9. 37
all have the same sequence, yet they differ in linkage
number. Accordingly, they are referred to as topological
isomers (topoisomers). The enzymes that create or al-
leviate these states are called topoisomerases.

Topoisomerases affect supercoiling by either of two
methods. Type I topoisomerases break one strand of a
double helix and, while binding the broken ends, pass
the other strand through the break. The break is then
sealed (fig. 9.38). Type II topoisomerases (e.g., DNA gy-
rase in E. colt) do the same sort of thing, only instead of
breaking one strand of a double helix, they break both
and pass another double helix through the temporary
gap. Four topoisomerases are active in E. colt, with some-
what confusing nomenclature: topoisomerases I and III
are type I; topoisomerases II and IV are type II.

As DNA replication proceeds, positive supercoiling
builds up ahead of the Yjunction. This is eliminated by
topoisomerases that either create negative supercoil-

ing ahead of the Yjunction in preparation for replica-
tion or alleviate positive supercoiling after it has been
created.

Termination of Replication

The termination of the replication of a circular chromo-
some presents no major topological problems. At the end
of the theta-structure replication (see fig. 9.22), both
Yjunctions have proceeded around the molecule. The re-
gion of termination on the E. colt chromosome, the ter-
minus region, is 180 degrees from oriC on the circular
chromosome, between minutes 28 and 36. There are six
terminator sites (Ter)\ three arrest the Yjunction from the
left, and three arrest the one from the right when bound
by a termination protein, the protein product of the tus
gene. (Tus stands for terminus utilization substance; each
Ter site is about twenty base pairs.) One interesting as-
pect of the termination of E. colt DNA replication is that
the cells are viable even if the whole terminator region is
deleted. There are fewer viable cells and some growth
problems, but in general, E. coli can successfully termi-
nate DNA replication even without formal termination
sites. A topoisomerase, topoisomerase IV, then releases
the two circles, and DNA polymerase I and ligase close
them up (fig. 9.39).

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Negatively supercoiled

Relaxed

Positively supercoiled

L = 48

L=50

L = 52

Figure 9.37 Positive and negative supercoils. Enzymes called topoisomerases can take relaxed DNA {center) and add negative (left)
or positive (right) supercoils. L is the linkage number.

E. coll
Topoisomerase I

Bind strand I
(red)

Open strand I

►

Pass strand II

(blue) through

strand I

Close strand I

y%TA

Release
DNA

DNA is
less coiled

Figure 9.38 Topoisomerase I can reduce DNA coiling by breaking one strand of the double helix and passing the other strand through it.

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Chapter Nine Chemistry of the Gene

DNA replication at
Y-junctions

Topoisomerase
separates circles

Single-strand
gap

DNA polymerase I and
ligase close the circles

Figure 9.39 The replication of circular DNA terminates when
topoisomerase separates the circles and DNA polymerase I
and ligase close the gaps in each circle.

DNA Partitioning in E. coli

In chapter 3, we discussed processes that partition eu-
karyotic chromosomes between daughter cells during
mitosis and meiosis. Until very recently geneticists be-
lieved that the partitioning of the E. coli chromosome
was a passive process, unlike that in eukaryotes. Now,
however, we know that more complexity is involved in
E. coli DNA partitioning. When DNA replication begins,
the newly replicated origins of replication are segregated
to opposite ends of the bacterial cell, acting as cen-
tromeres do. A ring of proteins, the products of the FtsZ
gene, form a ring at the middle of the cell and begin to

create the septum that will divide the cell into two. The
full complexity involved in E. coli chromosomal parti-
tioning should be uncovered in the near future.

REPLICATION STRUCTURES

The E. coli model of DNA replication that we have pre-
sented here is by way of the intermediate theta-structure
(see fig. 922). Two other modes of replication occur in
circular chromosomes: rolling-circle and D-loop.

Rolling-Circle Model

In the rolling-circle mode of replication, a nick (a break
in one of the phosphodiester bonds) is made in one of
the strands of the circular DNA, resulting in replication of
a circle and a tail (fig. 9.40). This form of replication oc-
curs in the F plasmid or E. coli Hfr chromosome during
conjugation (see chapter 7). The F + or Hfr cell retains the
circular daughter while passing the linear tail into the F~
cell, where replication of the tail takes place. Several
phages also use this method, filling their heads (protein
coats) with linear DNA replicated from a circular parent
molecule.

D-Loop Model

Chloroplasts and mitochondria (in eukaryotic cells) have
their own circular DNA molecules (see chapter 17) that
appear to replicate by a slightly different mechanism. The
origin of replication is at a different point on each of the
two parental template strands. Replication begins on one
strand, displacing the other while forming a displace-
ment loop or D-loop structure (fig. 9.41). Replication
continues until the process passes the origin of replication
on the other strand. Replication then initiates on the sec-
ond strand, in the opposite direction. Normal Y-junction
replication, as described earlier, also occurs in mitochon-
drial DNA under some growth conditions.

EUKARYOTIC DNA
REPLICATION

As we saw earlier, linear eukaryotic chromosomes usu-
ally have multiple origins of replication, resulting in fig-
ures referred to as "bubbles" or "eyes" (see fig. 9. 24).
Multiple origins allow eukaryotes to replicate their
larger quantities of DNA in a relatively short time, even
though eukaryotic DNA replication is considerably
slowed by the presence of histone proteins associated

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Eukaryotic DNA Replication

239

Single-strand
nick occurs

>■

Synthesis continues

Discontinuous

in 5'-**3' direction

synthesis now

on circle, displacing ^

also occurs on

the 5' tail /^

the tail

I/&

A\\

3' //
^ II

•w

5' ' VVc

E )

' E

-*v

When the tail is complete
it can be removed by
a nuclease

C D E

The linear DNA can then be
rolled into a closed
double-stranded
circle

5' A B C D E 3'

Figure 9.40 Rolling-circle model of DNA replication. The letters A-E provide landmarks on the chromosomes.

D-loop

with the DNA to form chromatin (see chapter 15). For
example, the E. colt replication fork moves through
about twenty-five thousand base pairs per minute,
whereas the eukaryotic Yjunction moves through only
about two thousand base pairs per minute. The number
of replicons in eukaryotes varies from about five hundred
in yeast to as many as sixty thousand in a diploid mam-
malian cell.

In budding yeast, a lower eukaryote that is often used
as a model organism, DNA replication initiates at sites
called autonomously replicating sequences (ARS).
Each consists of a specific 11-base-pair sequence plus
two or three additional short DNA sequences encom-
passing 100-200 base pairs. Six proteins form a complex
that binds to this sequence, referred to as the origin
recognition complex (ORC). These proteins seem to
be bound all the time, and thus additional proteins are
needed to initiate DNA replication. Some of these addi-
tional proteins are cyclin-dependent kinases, proteins in-
volved in the control of the cell cycle (chapter 3). This
makes sense because in eukaryotes, DNA replication can
take place only once during the cell cycle, during the S
phase. Thus, the initiation of DNA replication must be
tightly controlled to avoid multiple replication of some
or all replicons.

Figure 9.41 D-loops form during mitochondrial and chloroplast
DNA replication because the origins of replication are at
different places on the two strands of the double helix.

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Chapter Nine Chemistry of the Gene

SUMMARY

STUDY OBJECTIVE 1: To understand the properties that
a genetic material must have 205-211.

A genetic material must be able to control the phenotype of
a cell or organism (i.e., to direct protein synthesis), it must
be able to replicate, and it must be located in the chromo-
somes. Avery and his colleagues demonstrated that DNA
was the genetic material when they showed that the trans-
forming agent was DNA. Griffith had originally demon-
strated transformation of Streptococcus bacteria in mice.
Hershey and Chase demonstrated that the DNA of bacterio-
phage T2 entered the bacterial cell. Fraenkel-Conrat
demonstrated that in viruses without DNA (RNA viruses),
such as tobacco mosaic virus, the RNA acted as the genetic
material. Thus, by 1953, the evidence strongly suggested
nucleic acids (DNA or, in its absence, RNA) as the genetic
material.

STUDY OBJECTIVE 2: To examine the structure of DNA,
the genetic material 211-224.

Chargaff showed a 1:1 relationship of adenine (A) to
thymine (T) and cytosine (C) to guanine (G) in DNA.
Wilkins, Franklin, and their colleagues showed, by X-ray
crystallography, that DNA was a helix of specific dimen-
sions. Following these lines of evidence, Watson and Crick
in 1953 suggested the double-helical model of the structure
of DNA. In their model, DNA is made up of two strands, run-
ning in opposite directions, with sugar-phosphate back-
bones and bases facing inward. Bases from the two strands
form hydrogen bonds with each other with the restriction
that only A and T or G and C can pair. This explains the
quantitative relationships that Chargaff found among the
bases. Melting temperatures of DNA also support this struc-

tural hypothesis because DNAs with higher G-C contents
have higher melting, or denaturation, temperatures; G-C
base pairs have three hydrogen bonds versus only two in an
A-T base pair. The Watson-Crick DNA model represents the
B form. DNA can exist in other forms, including the Z form,
a left-handed double helix that may be important in con-
trolling eukaryotic gene expression.

STUDY OBJECTIVE 3: To investigate the way in which
DNA replicates 220-239.

DNA replicates by unwinding of the double helix, with
each strand subsequently acting as a template for a new
strand. This works because of complementarity — only A-T,
T-A, G-C, or C-G base pairs form stable hydrogen bonds
within the structural constraints of the model. This model
of replication is semiconservative. Meselson and Stahl
confirmed it in an experiment with heavy nitrogen. Autora-
diographs of replicating DNA showed that replication
proceeds bidirectionally from a point of origin. Prokaryotic
chromosomes are circular, with a single initiation point of
replication. Eukaryotic DNA is linear, with multiple initia-
tion points of replication.

DNA polymerase enzymes add nucleotides only in the
5' — » 3' direction. Replication proceeds in small segments,
working backward from the Yj unction on the 5' — » 3' tem-
plate strand. Presumably, the 5' — > 3' restriction has to do
with the proofreading DNA polymerases do to correct er-
rors in complementarity. Polymerase III is the active repli-
cating enzyme, and polymerase I is involved in DNA repair.
Many other enzymes help create the Okazaki fragments, un-
wind DNA, and release the DNA from supercoiling. Prokary-
otic and eukaryotic systems follow similar steps.

SOLVED PROBLEMS

PROBLEM 1: What evidence led to the idea that DNA was
the genetic material?

Answer: Avery and his colleagues (MacLeod and McCarty)
performed experiments showing that DNA was the trans-
forming agent, and they are thus generally given credit for
formalizing the notion that DNA, not protein, is the genetic
material. Chargaff, Hershey and Chase, Fraenkel-Conrat,
and several others also helped shape the general view. At
the time that Watson and Crick published their model, the
scientific community knew that DNA was the genetic ma-
terial but didn't know its structure.

PROBLEM 2: How does DNA fulfill the requirements of a
genetic material?

Answer: DNA is located in chromosomes, has a structure
that is easily and accurately replicated, and has the se-
quence complexity to code for the fifty thousand or
more genes that a eukaryotic organism has.

PROBLEM 3: What enzymes are involved in DNA replica-
tion in E. coli?

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Exercises and Problems

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Answer: A replisome, consisting of a primosome (a pri-
mase and a helicase) and two polymerase III nonen-
zymes, forms at a Yjunction on DNA. One polymerase
acts processively synthesizing the leading strand, while
the other forms Okazaki fragments initiated by primers
created by the primase. DNA polymerase I completes the
Okazaki fragments, eliminating the RNA primer of the
previous Okazaki fragment and replacing it with DNA. Fi-
nally, DNA ligase connects the fragments. Also involved
in the process are single-strand binding proteins and
topoisomerases that relieve the DNAs supercoiling. Initi-
ation involves initiation proteins at oriC, and termination
requires termination proteins bound to the termination
sites and a topoisomerase.

PROBLEM 4: What can be concluded about the nucleic
acids in the following table?

Nucleic Acid

Molecule

°/oT

Answer: We must first look to see if U or T is present, for
this will indicate whether the molecule is RNA or DNA,
respectively. Molecule b is RNA; a and c are DNA. Now we
look at base composition. In double-stranded molecules,
A pairs evenly with T (or U) and G pairs with C. This rela-
tionship holds for molecules a and c, so they are double-
stranded; molecule b is single-stranded. Finally, the melt-
ing temperature increases with the amount of G-C, so the
melting temperature of c is greater than that of a.

EXERCISES AND PROBLEMS

CHEMISTRY OF NUCLEIC ACIDS

1. If the tetranucleotide hypothesis were correct re-
garding the simplicity of DNA structure, under what
circumstances could DNA be the genetic material?

2. Nucleic acids, proteins, carbohydrates, and fatty
acids could have been mentioned as potential ge-
netic material. What other molecular moieties
(units) in the cell could possibly have functioned as
the genetic material?

3. In what component parts do DNA and RNA differ?

4. Draw the structure of a short segment of DNA (three
base pairs) at the molecular level. Indicate the polar-
ity of the strands.

5. Roughly sketch the shape of B and 2 DNA, remem-
bering that B DNA is a right-handed helix and Z DNA
is a left-handed helix.

6. Deduce whether each of the nucleic acid molecules
in the following table is DNA or RNA and single-
stranded or double-stranded.

Nucleic Acid

Molecule

7. A double-stranded DNA molecule is 28% guanosine
(G).

a. What is the complete base composition of this
molecule?

b. Answer the same question, but assume the
molecule is double-stranded RNA.

8. The following are melting temperatures for five DNA
molecules: 73° C, 69° C, 84° C, 78° C, 82° C. Arrange
these DNAs in increasing order of percentage of G-C
pairs.

9. We normally think that single-stranded nucleic acids
should not melt, but many, in fact, do have a T m . How
can you explain this apparent mystery?

10. In a single-stranded DNA molecule, the amount of G
is twice the amount of A, the amount of T is three
times the amount of C, and the ratio of pyrimidines
to purines is 1.5:1. What is the base composition of
the DNA?

11. A double-stranded DNA measures 6.5 m in length.
Approximately how many base pairs does it contain?

DNA REPLICATION— THE PROCESS

12. Diagram the results that Meselson and Stahl would
have obtained (a) if DNA replication were conserva-
tive and (b) if it were dispersive.

13. What type of photo would J. Cairns have obtained if
DNA replication were conservative? Dispersive?

*Answers to selected exercises and problems are on page A-10.

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Chapter Nine Chemistry of the Gene

DNA REPLICATION— THE EN2YMOLOGY

14. Following is a section of a single strand of DNA. Sup-
ply a strand, by the rules of complementarity that
would turn this into a double helix. What RNA bases
would primase use if this segment initiated an
Okazaki fragment? In which direction would repli-
cation proceed?

5 '-ATTCTTGGCATTCGC-3 '

15. What is a primosome in E. colt? a replisome? What
enzymes make up each? What is the relationship be-
tween these structures?

16. What are the differences between continuous and
discontinuous DNA replication? Why do both exist?

17. Describe the synthesis of an Okazaki fragment.

18. Describe the enzymology of the origin, continua-
tion, and termination of DNA replication in E. coll

19. Can you think of any other mechanisms besides
topoisomerase activity that could release supercoil-
ing in replicating DNA?

20. Draw a diagram showing how topoisomerase II (gy-
rase) might work.

21. Retroviruses are single-stranded RNA viruses that in-
sert their genomes into the host DNA during their
life cycle. But only double-stranded DNA can be in-
serted into double-stranded DNA.

a. Propose a mechanism that retroviruses could use
to insert their genomes.

b. What novel enzymes might such viruses require?

22. Propose a mechanism by which a single strand of
DNA can make multiple copies of itself.

23. Progeria is a human disorder that causes affected in-
dividuals to age prematurely; a nine-year-old often re-
sembles a sixty- to seventy-year-old individual in ap-
pearance and physiology. Suppose you extract DNA
from a progeric patient and find mostly small DNA
fragments rather than the expected long DNA mole-
cules. What enzyme(s) might be defective in patients
with progeria?

REPLICATION STRUCTURES

24. Under what circumstances would you expect to see
a DNA theta structure? D-loop? rolling-circle? bub-
bles? What function does each structure serve?

EUKARYOTIC DNA REPLICATION

25. In developing sea urchins, just after fertilization, the
cells divide every thirty to forty minutes. In the
adult, the cells divide once every ten to fifteen
hours. The amount of DNA per cell is the same in
each case, but the DNA obviously replicates much
faster in developing cells. Propose an explanation to
account for the difference in replication time.

CRITICAL THINKING QUESTIONS

1. Mutants are used to study various aspects of the phe-
notype and genotype. How can we study genes that are
critically important in the functioning of an organism?
For example, how do we study mutations in the gene
for DNA polymerase III in E. coli, when changes in this
gene are usually lethal? Remember, to study the genes
in bacteria, we need the bacteria to grow and form
colonies in order to be scored for their phenotypes.

2. DNA and RNA differ in two major ways: DNA has de-
oxyribose sugar, whereas RNA has ribose, and DNA has
thymine, whereas RNA has uracil. Why might those dif-
ferences exist other than accidents of evolution?

Suggested Readings for chapter 9 are on page B-5.

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GENE EXPRESSION

Transcription

STUDY OBJECTIVES

1. To examine the types of RNA and their roles in gene
expression 245, 256

2. To look at the process of transcription, including start and stop
signals, in both prokaryotes and eukaryotes 246

3. To investigate posttranscriptional changes in eukaryotic
messenger RNAs, including an analysis of intron removal 260

A computer model of the serine transfer RNA. The

amino acid binding site is yellow; the anticodon

is red. (© Ken Eward/SPL/Photo Researchers.)

STUDY OUTLINE

Types of RNA 245

Prokaryotic DNA Transcription 246

DNA-RNA Complementarity 246

Prokaryotic RNA Polymerase 247

Prokaryotic Initiation and Termination Signals for
Transcription 248
Ribosomes and Ribosomal RNA 256
Transfer RNA 256

Similarities of All Transfer RNAs 257

Transfer RNA Loops 258
Eukaryotic DNA Transcription 260

The Nucleolus in Eukaryotes 260

Differences Between Eukaryotic and Prokaryotic
Transcription 261

Promoters 262

Caps and Tails 265

Introns 265

RNA Editing 275
Updated Information About the Flow of Genetic
Information 275

Reverse Transcription 276

RNA Self-Replication 276

DNA Involvement in Translation 276
Summary 277
Solved Problems 277
Exercises and Problems 278
Critical Thinking Questions 279

Box 10.1 Observing Transcription in Real Time 250
Box 10.2 Polymerase Collisions:

What Can a Cell Do? 252
Box 10.3 Arc Viroids Escaped Introns? 272

243

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Chapter Ten Gene Expression: Transcription

In this chapter, we continue our study of genetics at
the molecular level. After discussing the structure
of DNA and the way in which it replicates in the
last chapter, we turn our attention here and in the
next chapter to the way in which the genetic
material — primarily DNA — expresses itself. In this chap-
ter, we concentrate on the conversion of DNA informa-
tion into RNA information, the first step in gene expres-
sion. In the next chapter, we look at the conversion of
RNA information into proteins. Later chapters discuss the
control of these processes. We begin with prokaryotes
and later in the chapter discuss the conceptually similar
but functionally more complex process in eukaryotes.

All living things synthesize proteins. In fact, the types
of proteins that a cell synthesizes determine the kind of
cell it is. Hence, the genetic material must determine the
types and quantities of proteins a cell synthesizes. Pro-
teins (polypeptides) are made up of strings of amino acids
(three hundred to five hundred, on average) joined to-
gether by peptide bonds. (We cover protein structure and
synthesis in chapter 11.) Each protein contains a unique
combination of only twenty amino acids. The amino acid
sequence is specified by the sequence of nucleotides in
DNA or RNA. In all prokaryotes, eukaryotes, and DNA
viruses, the gene is a sequence of nucleotides in DNA that
codes for the sequence of RNA. That RNA then deter-
mines which amino acids are included in a polypeptide.
RNA usually serves as an intermediary between DNA and
proteins. (In RNA viruses, the RNA may serve as a tem-
plate for the eventual synthesis of DNA, or the RNA may
serve as genetic material without DNA ever being formed.
We will consider these cases at the end of the chapter.)

In 1958, Francis Crick originally described the flow of
genetic information as the central dogma: DNA trans-
fers information to RNA, which then directly controls
protein synthesis (fig. 10.1). DNA also controls its own
replication. Transcription is the process of synthesizing
RNA from a DNA template using the rules of comple-
mentarity — the DNA information is rewritten, but in the
same nucleotide language. RNA controls the synthesis of
proteins in a process called translation because the in-
formation in the language of nucleotides is translated
into information in the language of amino acids.

In the previous chapter, we introduced the idea of
proteins that recognize specific DNA sequences and bind
to those sequences. Specifically, we introduced the initia-
tor proteins that bind to oriC and the proteins that bind
to the terminator sequences. DNA polymerases and some
of the other proteins involved in DNA replication bind to
DNA, but they do not necessarily bind to any specific se-
quences. Proteins that recognize specific DNA sequences
are critically important to the transcriptional process. In
the next chapter, we spend more time on proteins, dis-
cussing their structures and how they are synthesized. It
is sufficient to say here that specific proteins recognize

Self-replication loop

Transcription

RNA

Translation

Protein

Figure 10.1 Crick's original central dogma depicted the flow
of genetic information. Dashed red lines indicate the possible
information transfers unconfirmed in 1958, when Crick
proposed the central dogma.

specific DNA sequences. They do so by interdigitating
the amino acid side chains of the proteins into the
grooves of the DNA, thereby recognizing specific se-
quences by hydrogen bonding and other electrostatic in-
teractions between the side chains of the amino acids of
the proteins and the bases of the DNA (fig. 10.2). Proteins
can have parts that recognize DNA sequences and parts
that recognize other proteins or that perform other en-
zymatic activities such as hydrolyzing ATP.

(a)

(b)

Figure 10.2 Computer model of the interaction of a yeast
transcriptional factor, GAL4 {blue), and a seventeen -base- pair
region of DNA {red). Zinc ions are in yellow. The protein is a
dimer; only the DNA recognition region and associated part are
shown. Part {b) is a space-filling model of part (a). (Reprinted
with permission from Nature, 2 April 1992, Vol. 356, p. 411, fig. 3b, c.
Copyright 1992 Macmillan Magazines Limited.)

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Types ofRNA

245

TYPES OF RNA

In the protein synthesis process, three different kinds of
RNA serve in three different roles. The first type is mes-
senger RNA (mRNA), which carries the DNA sequence
information to particles in the cytoplasm known as ribo-
somes, where the messenger RNA is translated. The sec-

ond type is transfer RNA (tRNA), which brings the
amino acids to the ribosomes, where protein synthesis
takes place. The third type of RNA is a structural and
functional part of the ribosome called ribosomal RNA
(rRNA). The general relationship of the roles of these
three types of RNA is diagrammed in figure 10.3. In addi-
tion, small RNAs play other roles in cellular metabolism,
some of which are described later in the chapter.

^O^G^O^O^

Transcription

T^O^O^O^

DNA

Ribosomal
RNA

Transfer
RNA

Messenger
RNA

+ Ribosomal
proteins

+ Amino acid

Modification
in eukaryotes

c=^

Q<\/\y\/\/N^\/\yN/\y\/\y^|yVyl/

Ribosome

Translation

Growing
polypeptide

Next amino acid

Figure 10.3 Relationship among the three types of RNA — ribosomal, transfer, and messenger — during protein
synthesis. All three types are found together at the ribosome during protein synthesis.

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Chapter Ten Gene Expression: Transcription

We know that DNA does not take part directly in pro-
tein synthesis because, in eukaryotes, translation occurs
in the cytoplasm, whereas DNA remains in the nucleus.
We suspected for a long time that the genetic intermedi-
ate in prokaryotes and eukaryotes was RNA because the
cytoplasmic RNA concentration increases with increas-
ing protein synthesis, and the cytoplasmic RNAs carry
nucleotide sequences complementary to the cell's DNA.
Proof of an RNA intermediate came when it was shown
that messenger RNA directs protein synthesis.

PROKARYOTIC DNA
TRANSCRIPTION

DNA-RNA Complementarity

What proof do we have that a messenger RNA exists? That
is, what proof convinced geneticists that gene-sized RNAs
(not transfer RNAs or ribosomal RNAs) were found in the
cytoplasm that were complementary to the DNA in the
nucleus? At least two lines of evidence exist. First, it was
shown that the RNAs produced by various organisms
have base ratios very similar to the base ratios in the same
organisms' DNA (table 10.1). The second line of evidence
comes from experiments by B. Hall, S. Spiegelman, and
others using DNA-RNA hybridization. This technique
denatures DNA by heating, which causes the two strands
of the double helix to separate. When the solution cools,
a certain proportion of the DNA strands rejoin and
rewind — that is, complementary strands "find" each other
and re-form double helices. When RNA is added to the de-
natured DNA solution and the solution is cooled slowly,
some of the RNA forms double helices with the DNA if
the RNA fragments are complementary to a section of the
DNA (fig. 10.4). The existence of extensive complemen-
tarity between DNA and RNA is a persuasive indication
that DNA acts as a template for complementary RNA.

In another experiment, DNA-RNA hybridization
showed that bacteriophage infection led to the produc-
tion of phage-specific messenger RNA. Gene-sized pieces
of RNA extracted from Escherichia coli before and after

Table 10.1 Correspondence of Base Ratios

Between DNA and RNA of the
Same Species

RNA

% G + C

DNA

% G + C

E. coli

T2 phage

Calf thymus gland

52
35
40

— A

— T

— A

— A
— G

— C

— C
— G

— T

— A

— T

— A

— A
— G

— C

— C
— G

— T

DNA
5' 3'

— A

T —

— T

A —

— A

T —

— A

T —

— G

c—

— C

G —

— C

G —

— G

c—

— T

A —

Heat (denature) and
add RNA

T —
A —

T —
T —

c —

G —
G —

c —

A —

RNA

— A

— U

— A

— A
— G

— C

— C
— G

— U

Cool (renature)

5' 3'

—A T —

— U A —

—A T —

— G C —

-C G —

-C G —
— G C —
— U A —

Figure 10.4 DNA-RNA hybridization occurs between DNA and
complementary RNA.

bacteriophage T2 infection were tested to see if they hy-
bridized with the DNA of the T2 phage or with the DNA
of the E. coli cell. The RNA in the E. coli cell was found to
hybridize with the E. coli DNA before infection but with
the T2 DNA after infection. Thus it is apparent that when
the phage attacks the E. coli cell, it starts to manufacture
RNA complementary to its own DNA and stops the
E. coli DNA from serving as a template.

Having reached the conclusion that RNA is transcribed
(synthesized) from a DNA template and then directs pro-
tein synthesis, we look at two questions. First, is this RNA
single- or double-stranded? Second, is it synthesized (tran-
scribed) from one or both strands of the parental DNA?

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Prokaryotic DNA Transcription

247

For the most part, cellular RNA does not exist as a double
helix. It can form double helical sections when comple-
mentary parts come into apposition (e.g., see fig. 10.16),
but its general form is not a double helix. The simplest, and
most convincing, evidence for this is that complementary
RNA bases do not occur in corresponding proportions
(Chargaff 's ratios). That is, in RNA, uracil does not usually
occur in the same quantity as adenine, nor does cytosine
occur in the same quantity as guanine (table 10.2).

The answer to the second question is that RNA is not
usually copied from both strands of any given segment
of the DNA double helix, although rare exceptions do
occur. Consider a sequence of nucleotides on one strand
of a DNA duplex that specifies a sequence of amino
acids for a protein, with the complementary nucleotide
sequence also specifying the amino acid sequence for
another functional protein. Since most enzymes are three
hundred to five hundred amino acids long, the virtual
impossibility of this task is obvious. It was, therefore,
assumed a priori that, for any particular gene — that is,
in any particular segment of DNA — the sequence on
only one strand is transcribed and its complementary
sequence is not. There is now considerable evidence to
support this assumption.

The most impressive evidence that only one DNA
strand transcribes RNA comes from work done with bac-
teriophage SP8, which attacks Bacillus subtilis. This
phage has an interesting property — a great disparity in
the purine-pyrimidine ratio of the two strands of its DNA.
The disparity is significant enough that the two strands
can be separated by density using density-gradient cen-
trifugation. After denaturation and separation of the two
strands, DNA-RNA hybridization can be carried out sepa-
rately on each of the two strands with the RNA produced
after the virus infects the bacterium. J. Marmur and his
colleagues found that hybridization occurred only be-
tween the RNA and the heavier of the two DNA strands.
Thus, only the heavy strand acted as a template for the
production of RNA during the infection process.

The idea that only one strand of DNA serves as a tran-
scription template for RNA has also been verified for sev-
eral other small phages. However, when we get to larger
viruses and cells, we find that either of the strands may
be transcribed, but only one strand is used as a template
in any one region. This was clearly shown in phage T4 of
E. coli, where certain RNAs hybridize with one DNA

Table 1 0.2 Base Composition in RNA (percentage)

Adenine

Uracil

Guanine

Cytosine

E. coli

Euglena

Poliovirus

strand, and other RNAs hybridize with the other. Let us
now look at the transcription process in prokaryotes,
then proceed to examine the three types of RNA in de-
tail, and finally look at transcription in eukaryotes.

Prokaryotic RNA Polymerase

In prokaryotes, transcription of RNA is controlled by
RNA polymerase. Using DNA as a template, this en-
zyme polymerizes ribonucleoside triphosphates (RNA
nucleotides). The complete RNA polymerase enzyme of
E. coli — the holoenzyme — is composed of a core en-
zyme and a sigma factor. The core enzyme is composed
of four subunits: a (two copies), p, and p'; this core is the
component of the holoenzyme that actually carries out
polymerization. The sigma factor is involved in recogniz-
ing transcription start signals on the DNA. Following the
initiation of transcription, the sigma factor disassociates
from the core enzyme.

Logically, transcription should not be a continuous
process like DNA replication. If there were no control of
protein synthesis, all the cells of a higher organism would
be identical, and a bacterial cell would be producing all
of its proteins all of the time. Since some enzymes de-
pend on substrates not present all of the time, and since
some reactions in a cell occur less frequently than oth-
ers, the cell — be it a bacterium or a human liver cell —
needs to regulate its protein synthesis. One of the most
efficient ways for a cell to exert the necessary control
over protein synthesis is to perform transcription selec-
tively. Transcription of nongenic regions or of genes
coding for unneeded enzymes is wasteful. Therefore,
RNA polymerase should be selective. It should use as
transcription templates only those DNA segments (genes
or small groups of genes) whose products the cell needs
at that particular time.

The mechanisms of transcriptional control need to
be examined in two ways. First, we need to understand
how the beginnings and ends of transcribable sections
(a single gene or a series of adjacent genes) are demar-
cated. Second, we need to understand how the cell can
selectively repress or enhance transcription of certain
of these transcribable sections. The latter issues — the
keys to bacterial efficiency and eukaryotic growth and
development — are covered in chapters 14 and 16,
respectively.

RNA polymerase must be able to recognize both the
beginnings and the ends of genes (or gene groups) on
the DNA double helix in order to initiate and terminate
transcription. It must also be able to recognize the cor-
rect DNA strand to avoid transcribing the DNA strand
that is not informational. RNA polymerase accomplishes
those tasks by recognizing certain start and stop signals
in DNA, called initiation and termination sequences,
respectively.

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Chapter Ten Gene Expression: Transcription

Prokaryotic Initiation and Termination ^%
Signals for Transcription ^

The DNA region that RNA polymerase associates with im-
mediately before beginning transcription is known as the
promoter. The promoter is an important part of gene
expression in both prokaryotes and eukaryotes. Promot-
ers contain the information for transcription initiation
and are the major sites in which gene expression is
controlled.

Without the sigma factor, the core enzyme of RNA
polymerase binds randomly along the DNA. Formation of
the holoenzyme brings about high affinity of RNA poly-
merase for DNA sequences in the promoter region. Ter-
mination of transcription comes about when the poly-
merase enzyme recognizes a DNA region known as a
terminator sequence. Let us elaborate on the various
stages of transcription (in this section and in boxes 10.1
and 10.2).

Promoters

The RNA polymerase molecule covers a region of about
sixty base pairs of DNA. This was determined by causing
the polymerase to bind to DNA and then digesting the
mixture with nucleases, in a technique known as foot-
printing (fig. 10.5). The polymerase "protects" or pre-
vents degradation of the region it covers. The undigested
DNA is then isolated and its size determined. Geneticists
have gained much new information about the nature of
recognition regions within promoters through recombi-
nant DNA technology and nucleotide sequencing tech-
niques (see chapter 13). Sequencing of numerous pro-
moters has shown that they contain common sequences.
If the promoter nucleotide sequences align with each
other, and each has exactly the same series of nucleotides
in a given segment, we say that the sequence of that seg-
ment comprises an invariant or conserved sequence.
If, however, there is some variation in the sequence,
but certain nucleotides occur at a high frequency (sig-
nificantly greater than by chance), we refer to those
nucleotides as making up a consensus sequence. Sur-
rounding a point in prokaryotic promoters about ten nu-
cleotides before the first transcribed base is just such a
consensus sequence — TATAAT. This sequence is known
as a Pribnow box after one of its discoverers (fig. 10.6).
The nucleotides in the Pribnow box are mostly
adenines and thymines, so the region is primarily held to-
gether by only two hydrogen bonds per base pair. Since lo-
cal DNA denaturation occurs during transcription by RNA
polymerase (the DNA is opened to allow transcription),
fewer hydrogen bonds make this process easier energeti-
cally. When the polymerase is bound at the promoter re-
gion (fig. 10.6), it is in position to begin polymerization six
to eight nucleotides down from the Pribnow box.

Protein

DNA

Nucleases

Isolate and characterize
1 remaining DNA

Figure 10.5 Footprinting technique. DNA in contact with a
protein (e.g., RNA polymerase) is protected from nuclease
degradation. The protected DNA is then isolated and
characterized.

Promoter
region

First base
transcribed

^P R

T7A1

T7A2

0XI74A

lac

SV40

• TGGCGGTGATAATGGTTGCATGT • • • 3'

CCTATAGGATACTTACAGGCAT- • •

CATGCAGTAAGATACAAATCGCTA- • •

• TGTATGTTTTCATGCCTCCAAAT* • •

• CGGCTCGTATGTTGTGTGGAAT • • •

TGCAGCTTATAATGGTTACAAATA • • •

Pribnow box

Figure 10.6 Nucleotide sequences of the promoter region and
the first base transcribed from several different genes. Lambda
(X), T7, and (}>X174 are bacteriophages. Lac is an E. coli gene,
and SV40 is an animal virus. Only the SV40 promoter has the
actual consensus sequence of TATAAT. Even when other
sequenced promoters not shown here are Included, no base is
found 100% of the time (conserved).

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Prokaryotic DNA Transcription

249

The sequences shown in figure 10.6 are those of the
coding strand of DNA. It is a general convention to
show the coding strand because both that strand and
messenger RNA have the same sequences, substituting U
for T in RNA; they are both complementary to the same
template strand (also referred to as the anticoding
strand or noncoding strand; fig. 10.7). Another con-
vention is to indicate the first base transcribed by number
+ 1 and to use positive numbers to count farther down
the DNA in the downstream direction of transcription. If
transcription is proceeding to the right, the direction to
the left is called upstream, with bases indicated by neg-
ative numbers (fig. 10.8). Under this convention, the Prib-
now box is often referred to as the — 10 sequence.

Figure 10.8 also indicates another region with similar
sequences among many promoters centered near —35
and referred to as the —35 sequence. The consensus
sequence at —35 is TTGTCA. Mutation studies have
attempted to determine the relative roles of the — 10 and
— 35 sequences in transcription. In other words, muta-
tions of bases in the —10 and —35 regions were exam-
ined to determine how they affected transcription initia-
tion. The conclusions from these studies are that both

RNA
5'AUGU

DNA

Coding strand
TGGCGGTGATAATGGTTGCATGT • • • 3'

ACCGCCACTATTACCAACGTACA • • •
Pribnow box

Template strand
(anticoding)

Figure 10.7 The template (anticoding) strand of DNA is
complementary to both the coding strand and the transcribed
RNA. The sequences are from the promoter of the XP R region
(see fig. 10.6).

regions contribute to the efficiency of polymerase bind-
ing. In other words, the more each sequence differs from
the consensus sequence, the less frequently that pro-
moter initiates transcription. The sigma factor recognizes
both the —35 and the — 10 sequences. The sigma factor is
also sensitive to the spacing between these sequences,
preferring (being most efficient at) seventeen base pairs.

Farther upstream from the —35 sequence is a recogni-
tion element in bacterial promoters that are very strongly
expressed, such as the ribosomal RNA genes (fig. 10.8).
This upstream element, or UP element, is about twenty
base pairs long, is centered at — 50, and is rich in A and T
By mutational studies, it has been shown that adding this
element to promoters that don't normally have it greatly
increases the rate of transcription. There are other recog-
nition sites in prokaryotes, both upstream and down-
stream, at which various proteins attach that can enhance
or inhibit transcription by direct contact with the poly-
merase (the a and a subunits). We discuss these in chapter
14 under control of transcription. They are not part of
what we think of as the core promoter, the DNA sequence
needed for efficient binding of RNA polymerase.

Since the holoenzyme recognizes consensus se-
quences in a promoter, it is not surprising that some
promoters are bound more efficiently than others or that
different sigma factors exist within a cell. In E. colt, the ma-
jor sigma factor is a protein of 70,000 daltons, referred to
as a 70 . (One dalton is an atomic mass of 1 .0000, approxi-
mately equal to the mass of a hydrogen atom.) The exis-
tence of about five less common sigma factors provides
the cell with a mechanism for transcribing different genes
under different circumstances. For example, in an E. colt
cell subjected to elevated temperatures, a group of new
proteins, referred to as heat shock proteins, appear, act-
ing to protect the cell to some extent against the elevated
temperatures. These proteins all appear at once because
they have promoters that a different sigma factor recog-
nizes, one with a molecular weight of 32,000 daltons
(a 32 ); this new sigma factor is produced by the cell after

-60

-50

-40

-30

Upstream Downstream
-< ►

-20

-10

-1+1

5' ••• TCAGAAAATTATTTTAAATTTCCTCTTGTCAGGCCGGAATAACTCCCTATAATGCGCCACCACT •••3'

Upstream element

-35 sequence

-10 sequence

First

base

transcribed

Transcription

Figure 10.8 Promoter of the Escherichia coli ribosomal RNA gene, rrnB. Note the -10 and -35 sequences and the
upstream element. The first base transcribed (the transcriptional start site) is noted (+1), as well as the upstream,
downstream, and transcription directions. (Data from W. Ross, et al., 1993. Science 262:1407.)

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Chapter Ten Gene Expression: Transcription

BOX 10.1

The overwhelming evidence
that molecular events, such
as transcription, take place
comes from genetic and biochemical
analyses and occasionally an electron
micrograph of one type or another
(fig. 1). Thus, it is refreshing and illu-
minating to be able to observe some
of the processes we know are taking
place in real time; that is, to sit at a mi-
croscope and actually see these events
happen. Such a study on transcrip-
tion was published in 1991 in Nature
by four scientists at Washington
University in St. Louis.

Figure 1 Visualizing transcription.
Image of DNA before (a) and after
(b) E. coli RNA polymerase (bright
oblong object in b) binds to a
promoter. Pictures are by scanning
force microscopy, a new laser tech-
nique that images molecules in
water. Image sizes are 300 by
300 nm. Dark brown represents
substrate level; the highest point is
white at about 10 nm high. Interme-
diate colors represent intermediate
heights. (Courtesy of Martin Guthold and
Carlos Bustamante, Institute of Molecular
Biology and HHMI, University of Oregon.)

Experimental
Methods

Observing Transcription in
Real Time

Although new methods of mi-
croscopy are being developed, nor-
mally we cannot see these molecular
events taking place; the components
are too small. Making them visible
in electron microscopes usually re-

(a)

quires fixation that destroys the abil-
ity of the components to actually
continue their tasks. The Washington
University group overcame this by at-
taching a gold particle to DNA, thus
rendering the motion of that DNA vis-
ible under the light microscope (fig.
2). The scientists immobilized the
RNA polymerase to a glass coverslip;
thus, as transcription took place, the
DNA moved and the length of the
tether of the gold particle increased.
At first they stopped the process by
limiting the concentration of nucleo-
side triphosphates (NTPs). They

(b)

heat shock. We discuss heat shock proteins and other sys-
tems of transcriptional control in chapters 14 and 16.

From mutational studies of promoters and the pro-
teins in the RNA polymerase holoenzyme, we now have
a picture of a holoenzyme that sets down on a DNA
promoter because the sigma factor recognizes the —10
and — 35 elements, the a proteins recognize the UP
element, and the a and cr subunits recognize proteins
bound to various other upstream elements, when pres-
ent (fig. 10.9a). This initiation complex is initially re-
ferred to as a closed complex because the DNA has not
melted, which is the next step in transcription initiation
(fig. 10.9&). After the transcription of 5-10 bases, the
sigma factor is released (fig. 10.9c and d).

About seventeen base pairs of DNA are opened, and
as transcription proceeds, about twelve bases of RNA

form a DNA-RNA duplex at the point of transcription.
Some of this information comes from studies with potas-
sium permanganate (KMn0 4 ), which modifies DNA
bases that are single-stranded but not double-stranded.
Thus, the lengths of melted DNA can be determined
experimentally. Also used is the technique of pho-
tocrosslinking, in which two moieties such as DNA and
one or two proteins are caused to be permanently
crosslinked, verifying their close contact. This is done by
attaching a chemical crosslinking element to one of the
moieties and then causing crosslinking to occur by
shining light, usually ultraviolet, on the mixture.

Transcription, like DNA replication, always proceeds
in the 5' — » 3' direction. That is, a single base is added de
novo and then new RNA nucleotides are added to the
3'-OH free end, as in DNA replication. However, unlike

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251

could then observe the motion of the
gold ball when no transcription was
taking place. The scientists predicted
that an immobilized gold ball would
not move, and a tethered gold ball
would show a limited amount of

Brownian motion. That is, it would
show a limited amount of blur in light
microscope video images averaged
over time. However, as soon as NTPs
were added, any tethered gold ball
would show an increased blur as it

moved out of the field of vision and
eventually would be released when
transcription was completed. That is
exactly what they saw (fig. 3). Thus,
they succeeded in watching tran-
scription take place in real time.

40 nm gold particle

Glass
coverslip

DNA

RNA
polymerase

Figure 2 An experimental design in visualizing
transcription in real time under the light
microscope. Here, an RNA polymerase is immobi-
lized on a coverslip, waiting for nucleoside triphos-
phates (NTPs) to be added. The gold particle is
tethered by the DNA, allowing us to keep track
visually of the end of the DNA.

Figure 3 Enhanced light microscope images of
the gold particles. In 3 and 4, presumably immobi-
lized particles show no change in focus over time.
In 1 and 2, Brownian motion — and hence blur —
increases through time consistent with the length-
ening of the tether (transcription). The particle in 1
was released 87 seconds (and in 2, 135 seconds)
after NTPs were added. Scale bar is 1 |xm. (Tran-
scription by single molecules of RNA polymerase observed by
light microscopy. Robert Landick, Department of Biology,
Washington University, St. Louis, MO.)

DNA polymerase, prokaryotic RNA polymerase does not
seem to proofread as it proceeds. That is, RNA polymerase
evidently does not verify the complementarity of the new
bases added to the growing RNA strand. This deficiency is
not serious; since many messenger RNAs are short-lived
and many copies are made from actively transcribed
genes, an occasional mistake will probably not produce
permanent or overwhelming damage. If a particular RNA
is not functional, a new one will be made soon. Evolu-
tionarily speaking, it seems that it is more important to
make RNA quickly than to proofread each RNA made.

Terminators

Transcription continues as RNA polymerase adds nu-
cleotides to the growing RNA strand according to the

rules of complementarity (C, G, A, and U of RNA pairing
with G, C, T, and A of DNA, respectively). The polymerase
moves down the DNA until the RNA polymerase reaches
a stop signal, or terminator sequence. Two types of termi-
nators, rho-dependent and rho-independent, differ in
their dependency on the rho protein (Greek letter p).
The functional form of rho is a hexamer, six identical
copies of the protein. Rho-independent terminators
cause termination of transcription even if rho is not pres-
ent. Rho-dependent terminators require the rho pro-
tein; without it, RNA polymerase continues to transcribe
past the terminator in a process known as read-through.
Both types of terminators sequenced so far have one
thing in common: They include a sequence and its in-
verted form separated by another short sequence, all
together forming an inverted-repeat sequence. The

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Chapter Ten Gene Expression: Transcription

BOX 10.2

Both RNA polymerase and DNA
polymerase move along the
DNA of a cell during a cell
cycle. The DNA polymerase moves
along at about ten times the speed of
the transcribing enzyme. Since many
genes are usually active in a cell, the
interaction (collision) of the two en-
zymes is inevitable. What happens
when this collision takes place? What
does the cell do? Although we cannot
directly observe these interactions,
various bits of data suggest that a
head-on collision could be fatal to the
cell, and certain patterns of gene
placement minimize the chance of a
head-on collision.

B. Brewer first analyzed the prob-
lem of the coexistence of these two
enzymes in a paper published in
1988 in the journal Cell In evolution-
ary terms, the cell could obviate the
problems of a head-on collision by
either avoiding them or resolving
them. Resolution would entail some
sort of right-of-way settlement when
the two enzymes met; for example,
the RNA polymerase could drop
off the DNA when a confrontation
takes place. The cell might avoid con-
frontations if the genes are oriented
so that transcription occurs for the
most part in the same direction as
DNA replication. That is, DNA replica-
tion begins at oriC, with Yjunctions
proceeding to the left and the right
until they meet 180 degrees later.
Thus, to avoid head-on collisions,
genes on the left and right arcs of the

Experimental
Methods

Polymerase Collisions:
What Can a Cell Do?

bacterial chromosome could be tran-
scribed away from the origin of repli-
cation (fig. 1).

Brewer analyzed the orientation
of genes on the E. coli chromosome;
more recently, D. Zeigler and D. Dean
did the same for the chromosome of
Bacillus subtilis. In B. subtilis, 95%
(91 of 96) of the genes analyzed were
in the proper orientation to avoid
a head-on collision of polymerases.
Among the exceptions were sporula-
tion genes, genes that would not be
transcribed during DNA synthesis
and whose orientation is thus not
relevant to DNA polymerase activity.
In E. coli, Brewer found that, overall,
74% (375 of 501) of the genes she
looked at were oriented to avoid
head-on collisions. Brewer's data
were more impressive when she
broke them down according to tran-
scription function and activity.

For genes that transcribe very
actively most of the time, the orienta-
tion is about 90% in the "safe" direc-
tion. For regulatory genes that are
transcribed only very rarely, the ori-
entation is random (50% safe). For

other genes, the orientation was
72% in the safe direction. Thus, an
organization clearly exists within the
bacterial chromosome that helps to
avoid head-on collisions of the two
polymerases.

Brewer also provided evidence
that a head-on collision between
polymerases could be fatal to the cell.
Studies selected inversions of the
E. coli chromosome to see the effects
of collision. (Inversions are regions
that have been cut out and put back
in the opposite orientation.) It was
impossible to isolate inversion muta-
tions that changed the orientation of
genes in respect to oriC Thus, it
appears that a cell may not be able to
resolve a head-on collision of poly-
merases and that evolution has solved
the problem by having gene tran-
scription generally oriented in the
same direction as DNA replication.

More amazingly, Alberts and his
colleagues recently studied what
happens when a replication fork
catches up to a stalled RNA poly-
merase. Not only does the replication
fork pass the transcription apparatus,
but the RNA polymerase can resume
transcription after the replication
fork passes without loss of the tran-
script. Although there are contrary
observations in other systems, it ap-
pears that gene orientation and the
behavior of polymerases allow cells
to survive with both replication
and transcription occurring on the
same DNA.

terminator in figure 10.10 has the sequence AAAG-
GCTCC, 5' — > 3', from both the left on the coding strand
and from the right on the template strand. A four-base-
pair sequence separates the inverted repeats. Inverted re-
peats can form a stem-loop structure by pairing com-
plementary bases within the transcribed messenger RNA.
Both rho-dependent and rho-independent termina-
tors have the stem-loop structure in RNA just before the
last base transcribed. Rho-independent terminators, as
figure 10.10 shows, also have a sequence of thymine-
containing nucleotides after the inverted repeat, whereas
rho-dependent terminators do not. Although the exact

sequence of events at the terminator is not fully known,
it appears that the RNA stem-loop structure forms and
causes the RNA polymerase to pause just after complet-
ing it. This pause may then allow termination under two
different circumstances.

In rho-independent terminators, the pause may occur
just after the sequence of uracils is transcribed
(fig. 10.11). Uracil-adenine base pairs have two hydrogen
bonds and are thus less stable thermodynamically than
guanine-cytosine base pairs. Perhaps during the pause,
the uracil-adenine base pairs spontaneously denature, re-
leasing the transcribed RNA and the RNA polymerase,

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Prokaryotic DNA Transcription

253

■<- oriC -*■

~~^

3: ^ ^

■ ' ^ ^ § § £

>*.

Figure 1 Location and orientation of gene transcription on the chromosome of Bacillus subtilis {arrows). DNA replication
begins at oriC and terminates approximately 180 degrees from the origin of replication. Note that the overwhelming number of
arrows point away from the origin of replication toward the termination point. (From D. R. Zeigier and D. H. Dean, "Orientation of genes
in the Bacillus subtilis chromosome," Genetics, 125:703-8. Copyright © 1990 Genetics Society of America.)

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254

Chapter Ten Gene Expression: Transcription

RNA polymerase holoenzyme

/7^0<^OR>0<^0^^^

DNA

Closed
complex

UP
element

(a)

$^£>f rrrr \^ Open

V \jxLUiy ^^f complex

(b)

\^0%0^0^

Initiation of
transcription

(c)

/T^CSXs^

<^O^y7^7^ Elongation

Core enzyme

(d)

RNA

Figure 10.9 Transcription begins after RNA polymerase
attaches to the promoter, with specificity imparted by the
sigma factor. The DNA opens to form the open complex,
transcription begins, the sigma factor leaves, and elongation
commences.

terminating the process, and making the polymerase
available for further transcription of other promoters.

Rho-dependent terminators do not have the uracil se-
quence after the stem-loop structure. Here, termination
depends on the action of rho, which appears to bind to

HighA-T

DNA

5' ' ' 3'

TT AAAGGCTCCTTT T GGAGCCTTTTTTTT

AATTTCCGAGGAAA A CCTCGG A A AA AAA A

Template
strand

Inverted repeat

Transcription

Last base
transcribed

Stem-loop

structure in

RNA

UUUUUU 3'

• •• UUA

Figure 10.10 An inverted -repeat base sequence characterizes
terminator regions of DNA. Stem-loop structures can occur as
the RNA forms because of complementary sequences. The 3'
poly-U tail indicates a rho-independent terminator.

the newly forming RNA. In an ATP-dependent process,
rho travels along the RNA at a speed comparable to the
transcription process itself (fig. 10.11). Possibly, when
RNA polymerase pauses at the stem-loop structure, rho
catches up to the polymerase and unwinds the DNA-RNA
hybrid, leting the DNA, RNA, and polymerase fall free.
Rho can do this because it has DNA-RNA helicase (un-
winding) properties.

The process of transcription termination is probably
more complex than described. Significant interactions
may take place with other proteins, and particular se-
quences surrounding the termination sequence may also
be significant in the termination process. This is an area
of active research.

Figure 10.12 shows an overview of transcription. The
information of a gene, coded in the sequence of nu-
cleotides in the DNA, has been transcribed into a com-
plementary sequence of nucleotides in the RNA. This
RNA transcript contains a complement of the template
strand of the gene's DNA and thus acts as a messenger
from the gene to the cell's protein-synthesizing complex.
The transcript contains nucleotide sequences that will
be translated into amino acids — coding segments — as
well as noncoding segments before and after. The trans-
latable segment, or gene, almost always begins with a
three-base sequence, AUG, which is known as an initiator
codon, and ends with one of the three-base sequences,
UAA, UAG, or UGA, known as nonsense codons. (We dis-
cuss these signals in chapter 11.)

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Prokaryotic DNA Transcription

255

W^G^y7^C,\

•••

Transcript

Figure 10.11 Rho-independent {top) and rho-dependent {bottom) termination of transcription are preceded
by a pause of the RNA polymerase at a terminator sequence. Presumably, the stem-loop structure in the
nascent RNA causes the pause in both cases.

Promoter

Sigma

Terminator

RNA polymerase

mRNA

Figure 10.12 Transcription overview and RNA polymerase molecules. RNA polymerase is transcribing
near the terminator. The rho factor — actually made up of six subunits — is shown on the newly formed
RNA. The sigma factor is shown nearby, detached from the core polymerase.

The portion of the RNA transcript that begins at the
start of transcription and goes to the translation initia-
tor codon (AUG) is referred to as a leader, or 5' un-
translated sequence. The length of RNA from the non-
sense codon (UAA, UAG, or UGA) to the last nucleotide
transcribed is the trailer, or 3' untranslated sequence.
These sequences play a role in recognizing messenger
RNA and ensuring its structural stability at the ribosome
during the process of translation; the leader region can

also have regulatory functions (see chapter 14). Figure
10.13 diagrams a complete prokaryotic RNA transcript.
In this simplified drawing, the transcript has only one
gene (AUG — > UAA). However, the average prokaryotic
transcript contains the information for several genes.
We will say more about the parts of a transcript later in
this chapter and the next. Now we turn our attention to
the types of transcripts: ribosomal, transfer, and mes-
senger RNA.

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Chapter Ten Gene Expression: Transcription

Promoter

Terminator

DNA

. . . yT^o^o^o^c^o^^

Transcription

RNA

AUG

Leader

UAA

Trailer

Figure 10.13 Transcribed piece of prokaryotic RNA and its DNA template region. Note the promoter and
terminator regions on the DNA and the leader and trailer regions on the RNA. The initiation (AUG) and nonsense
(UAA) codons for protein synthesis are shown. These signals are read at the ribosome at the time of translation.

RIBOSOMES AND
RIBOSOMAL RNA

Ribosomes are organelles in the cell, composed of pro-
teins and RNA (ribosomal RNA, or rRNA), where protein
synthesis occurs. In a rapidly growing E. colt cell, ribo-
somes can make up as much as 25% of the mass of the
cell. Ribosomes, as well as other small particles and mol-
ecules, are measured in units that describe their rate of
sedimentation during density-gradient centrifugation in
sucrose. This technique gives information on size and
shape (due to the speed of sedimentation) while simulta-
neously isolating the molecules. Isolation by centrifuga-
tion in sucrose is a relatively gentle isolation technique;
the molecules still retain their biological properties and
can be used for further experimentation. In the 1920s,
physical chemist T. Svedberg developed ultracentrifuga-
tion, giving his name to the unit of sedimentation: the
Svedberg unit, S.

In sucrose density-gradient centrifugation, the gradi-
ent is formed by layering on decreasingly concentrated
sucrose solutions. In a related technique, cesium chloride
density-gradient centrifugation, mentioned in chapter 9,
the gradient develops during centrifugation. The sucrose
centrifugation is stopped after a fixed time, whereas in
the cesium chloride technique, the system spins until it
reaches equilibrium. The sucrose method tends to be
more rapid. Samples can be isolated from a sucrose gradi-
ent by punching a hole in the bottom of the tube and col-
lecting the drops in sequentially numbered containers.
The first (lowest-numbered) containers will contain the
heaviest molecules (with the highest S values).

Ribosomes in all organisms are made of two subunits
of unequal size. The sedimentation value is 50S (Svedberg
units) for the large one in E. colt and 3 OS for the smaller
one. Together they sediment at about 70S. Eukaryotic ri-
bosomes vary from 55S to 66S in animals and 70S to 80S
in fungi and higher plants. Most of our discussion will be
confined to the well-studied ribosomes of E. colt.

Each ribosomal subunit comprises one or two pieces
of ribosomal RNA and a fixed number of proteins.
The 30S subunit of E. colt has twenty-one proteins and a
16S molecule of ribosomal RNA, and the 50S subunit has
thirty-four proteins and two pieces of ribosomal RNA —
one 23S and one 5S section (fig. 10.14). Advances in
understanding ribosomal structure have come about af-
ter protein chemists isolated and purified all the proteins
of the ribosome. This allowed researchers to experiment
on the proper sequence needed to assemble the subunits
and also allowed them to develop immunological tech-
niques to show the positions of many proteins in the
completed ribosomal subunits.

In E. colt, all three ribosomal RNA segments are tran-
scribed as a single long piece of RNA that is then cleaved
and modified to form the final three pieces of RNA (16S,
23S, and 5S). The region of DNA that contains the three
ribosomal RNA molecules also contains genes for four
transfer RNAs (fig. 10.15). There appear to be about five
to ten copies of this region in each chromosome of
E. colt The occurrence of the three ribosomal RNA seg-
ments on the same piece of RNA ensures a final ratio of
1:1:1, the ratio needed for ribosomal construction.

TRANSFER RNA

During protein synthesis (see fig. 10.3), a messenger
RNA, carrying the information transcribed from the
gene (DNA), is bound to the ribosome. Amino acids are
brought to the ribosome attached to transfer RNAs. The
code is read in sequences of three nucleotides, called
codons. The nucleotides of the codon on messenger
RNA are complementary to and pair with a sequence of
three bases — the anticodon — on a transfer RNA. Each
different transfer RNA carries a specific amino acid. Thus,
the transfer RNA recognizes the specificity of the genetic
code (fig. 10.16).

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Transfer RNA

257

(a) 30S

Twenty-one proteins
One16SrRNA

(b) 50S

Thirty-four proteins
One 23S rRNA
One 5S rRNA

Figure 10.14 The E. coll ribosome. (a) and (b) show models of the 70S ribosome of E. coll, revealing the relationship of the small
(yellow) and large {red) subunits at the time of translation. The 30S ribosomal subunit is composed of twenty-one proteins and one
16S piece of ribosomal RNA. The 50S subunit is composed of thirty-four proteins and two pieces of ribosomal RNA, 23S and 5S.
([a and b] James A. Lake, Journal of Molecular Biology 105 (1976):131-59. Reproduced by permission of Academic Press.)

RNA transcript

tRNAs

16S

23S

tRNAs

/ \

Figure 10.15 The E. coll transcript that contains the three ribosomal RNA segments also contains four tRNAs and some spacer
RNA (red) that separates the tRNA and rRNA genes.

The correct amino acid is attached to its transfer RNA
by one of a group of enzymes called aminoacyl-tRNA
synthetases. One specific aminoacyl synthetase exists
for every amino acid, but the synthetase may recognize
more than one transfer RNA because there are more
transfer RNAs (and codons) than there are amino acids.
(In chapter 1 1 we discuss the genetic code in more de-
tail.) R. W. Holley a Nobel laureate, and his colleagues
were the first to discover the nucleotide sequence of a
transfer RNA; in 1964, they published the structure of the
alanine transfer RNA in yeast (fig. 10.17). The average
transfer RNA is about eighty nucleotides long.

Similarities of All Transfer RNAs

Transfer RNAs have several unusual properties. For one,
all the different transfer RNAs of a cell have the same gen-
eral shape; when purified, the heterogeneous mixture of
all of a cell's transfer RNAs can form very regular crystals.

The regularity of the shape of transfer RNAs makes sense.
During the process of protein synthesis, two transfer
RNAs attach next to each other on a ribosome, and a
peptide bond forms between their amino acids. Thus, any
two transfer RNAs must have the same general dimen-
sions as well as similar structures so that they can be rec-
ognized and positioned correctly at the ribosome.

An obvious feature of the transfer RNA in figure 10. 17 is
that it has unusual bases. When this transfer RNA is origi-
nally transcribed from DNA, it is about 50% longer than the
final eighty nucleotides. In fact, some transcripts contain
two copies of the same transfer RNA, or sometimes several
different transfer RNA genes are part of the same transcript
(see fig. 10. 15). The original transcription of transfer RNAs
is completely regular: It does not involve unusual bases.
The transcript is then processed down to the final size of
a transfer RNA by various nucleases that remove trailing
and leading pieces of RNA. In eukaryotes, a CCA sequence
of nucleotides is added at the 3' end by a nucleotidyl

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Chapter Ten Gene Expression: Transcription

"O.

CH,

ch!

Glutamic acid
(amino acid)

NH 3 +

V J

r -s

Glutamic acid
tRNA

Anticodon

CGGGAAACCCGU

mRNA

L -r J
Codon

Figure 10.16 Specificity of the genetic code manifests itself
in the transfer RNA, in which a particular anticodon is
associated with a particular amino acid. In this case, glutamic
acid is attached to its proper transfer RNA, which has the
anticodon CUU.

transferase enzyme. Then the transfer RNA is further modi-
fied, frequently by the addition of methyl groups to the
bases already in the RNA (fig. 10.18). Presumably these un-
usual bases disrupt normal base pairing and are in part re-
sponsible for the loops the unpaired bases form (see fig.
10.17).

Alanine

O — A

T-loop

G
\

C — G

i i

C — G
/ i

U G

\ i

G — C

i i

u y

C — G -U-MG

C-C-G-G-A

G-C-G-C

D-loop

■ y

¥"

G-G-C-C-U

/
D.

Variable loop ^g

, G — C-MG
A A — U

C-G-C-G

■D-G

G —

\
U

Anticodon loop
Anticodon

5'-

3' mRNA

¥
I

D
T

MG
Ml

Codon

Unusual bases

Pseudouridine

Inosine

Dihydrouridine

Ribothymidine

Methylguanosine

Methylinosine

Figure 10.17 Structure and sequences of alanine transfer RNA
in yeast. Note the modified bases in the loops. The anticodon
of the transfer RNA is shown paired with its complementary
COdon in the DNA. (Source: Data from R. W. Holley, et al., "Structure
of a ribonucleic acid," Science, 147:1462-65, 1965.)

Transfer RNA Loops

It is believed that the first loop on the 3' side (the T- or
T-i|;-C-loop) is involved in making the transfer RNA recog-
nizable to the ribosome. The ribosome must hold each
transfer RNA in the proper orientation to check the com-
plementarity of the anticodon of the transfer RNA and
the codon of the messenger RNA. The center loop of
transfer RNA is the anticodon loop. The aminoacyl-tRNA
synthetases seem to recognize many points all over the
transfer RNA molecule (see chapter 11).

The amino acid is attached to the ACC sequence on
the 3' end of the transfer RNA. The ribosome-binding

loop on all transfer RNAs has the T-i|/-C-G sequence. The
anticodon on all is bounded by uracil on the 5 ' side and a
purine on the 3' side. Thus, there is a good deal of general
similarity among all the transfer RNAs, consistent with the
fact that they all enter protein synthesis in the same way.
The actual shape of the functional transfer RNA in the cell
is not an open cloverleaf, as shown in figure 10.17; rather,
the whole molecule exhibits helical twisting due to pair-
ing of complementary regions (fig. 10.19).

Earlier we considered a rough definition of a gene as a
length of DNA that codes for one protein. But we have
just encountered an inconsistency — genes code for both
transfer RNAs and ribosomal RNAs, yet neither is eventu-

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Transfer RNA

259

l-N

N-

C
C

Inosine (I)

•N

ISl'

,^ C \ N / C

Ribose

,CH,

C
C

•N

Ribose

1-Methylinosine (Ml)

H-

S N'

C H

,^ c \ n ^ c \

Ribose

^3 ^°-

"N
H

N-

•N

Ribose

1-Methylguanosine (MG)

*N'

C OH

■H

OH
Ribose

Ribothymidine (T)

Pseudouridine (\j/)

Dihydrouridine (D)

Figure 10.18 Structures of the modified bases found in alanine transfer RNA of yeast. The various modifications of
normal bases are shown in red.

T-loop

D-loop

T-stem

3' Acceptor
end

Anticodon
stem

Anticodon
loop

Variable loop

Anticodon

(a)

(b)

Figure 10.19 Structure of yeast phenylalanine transfer RNA. (a) A diagram showing coiling of the sugar-phosphate backbone. (£>) A
molecular model with bases in yellow and backbone in blue. The two parts of the figure (a and b) are in the same orientation.
{[b] Courtesy of Alexander Rich.)

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Chapter Ten Gene Expression: Transcription

ally translated into a protein. Their transcripts function as fi-
nal products without ever being translated. Thus, transfer
RNA and ribosomal RNA are the major exceptions to the
general rule that a gene codes for a protein.

EUKARYOTIC DNA
TRANSCRIPTION

The Nucleolus in Eukaryotes

Eukaryotes have four segments of ribosomal RNA in the ri-
bosome, compared with three in prokaryotes. The smaller
ribosomal subunit has an 18S piece of RNA, and the larger
subunit has 5S, 5.8S, and 28S segments. All but the 5S ribo-
somal RNA section are transcribed as part of the same
piece of RNA. However, eukaryotic cells have many copies
of these ribosomal RNA genes, depending on the species.
For example, the fruit fly, Drosophila melanogaster, has
about 130 copies of the DNA region that the larger seg-
ments of ribosomal RNA are transcribed from. These re-
gions occur in tandem on the sex (X and Y) chromosomes
and are known collectively as the nucleolar organizer (see
chapter 3). The smallest ribosomal RNA subunit is also pro-
duced from a duplicated gene, but at a different point in
the genome. For example, in D. melanogaster, the 5S sub-
unit is produced on chromosome 2.

Eukaryotes — unlike prokaryotes, which have only
one RNA polymerase — have three RNA polymerases. Eu-
karyotic RNA polymerase I (or polymerase A) transcribes
only the nucleolar organizer DNA. RNA polymerase II (or
polymerase B) transcribes most genes. RNA polymerase
III (or polymerase C) transcribes small genes, primarily
the 5S ribosomal RNA gene and transfer RNA genes
(table 10.3). In addition, mitochondria, chloroplasts, and
some phages have other RNA polymerases.

Table 1 0.3 Prokaryotic and Eukaryotic

RNA Polymerases

Enzyme

Function

Prokaryotic

RNA polymerase

Transcribes DNA template

Primase

Primer synthesis during DNA

replication

Eukaryotic

RNA polymerase I

Transcribes nucleolar organizer

RNA polymerase II

Transcribes most genes

RNA polymerase III

Transcribes 5S rRNA and tRNA

genes

Primase

Primer synthesis during DNA

replication

■ ■ ''■^^■W^Sw - ...

Figure 10.20 Transcription in the nucleolus of the newt,
Triturus. Tandem repeats of the large ribosomal RNA genes are
being transcribed. The polarity of the process (progressing from
small to large transcripts), as well as the spacer DNA (thin lines
between transcribing areas), is clearly visible. Magnification
18,000X. (© 0. L. Miller, B. R. Beatty, D. W. Fawcett/Visuals Unlimited.)

At the nucleolar organizer, the nucleolus forms the
familiar dark blob found in eukaryotic nuclei. The nucle-
olus is the place where ribosomes are assembled. The
various ribosomal proteins that have been manufactured
in the cytoplasm migrate to the nucleus and eventually
to the nucleolus, where, with the final forms of the
ribosomal RNAs, they are assembled into ribosomes.

In the nucleolar organizer, an untranscribed region of
spacer DNA separates each repeat of the large ribosomal
RNA gene. This is shown in figure 10.20 and diagrammed
in figure 10.21. In the electron micrograph in figure
10.20, the polarity of transcription is evident from the
short RNA at one end of the transcribing segment and
the long RNA at the other end, with a uniform gradation
between. Notice that many RNA polymerases are tran-
scribing each region at the same time. The regions be-
tween the transcribed DNA segments are the spacer
DNA regions.

Like transfer RNAs, ribosomal RNAs are also modified:
some uridines are converted to pseudouridines, and some

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ribose sugars are methylated. These conversions take
place in the nucleolus, orchestrated by particles com-
posed of small RNA segments and protein. The RNA seg-
ments are referred to as small nucleolar RNAs
(snoRNAs) and, when combined with protein, are re-
ferred to as small nucleolar ribonucleoprotein parti-
cles (snoRNPs). Each different snoRNP has a snoRNA
that is complementary to the regions surrounding the nu-
cleotide to be modified. Thus, sites for modification are
chosen based on complementarity to a snoRNA, which
then somehow directs the modification to take place.

Differences Between Eukaryotic and
Prokaryotic Transcription

Although all aspects of transcription differ to some extent
between prokaryotes and eukaryotes, we will look at two
major differences here: the coupling of transcription and
translation that is possible in prokaryotes, and the exten-
sive posttranscriptional modifications that occur in eu-
karyotic messenger RNA. In E. coli, translation of the newly
transcribed messenger RNA into a protein can take place
before transcription is complete (fig. 10.22). The messen-
ger RNA is synthesized in the 5' — » 3' direction, and it is

Double helix DNA

- rRNA at start
of transcription

rRNA near final size

i
Spacer
DNA

Figure 10.21 Details of the
transcription of the large
ribosomal RNA genes shown
in figure 10.20. Note the
polarity of the process and
the spacer DNA, as seen in
figure 10.20.

RNA polymerase I

Transcription

RNA polymerase^

• • •

(a)

• • •

Messenger RNA

Growing polypeptide

■--.;''.'?■-■>- ;•-'■■ -•' ■ '■.•.','-.- ■-':

Figure 10.22 (a) In

prokaryotes, translation of
messenger RNA by ribosomes
begins before transcription is
complete. Ribosomes attach
to the growing mRNA strand
when the 5' end becomes
accessible. They then move
along the RNA as it
elongates. When the first
ribosome moves from the 5'
end, a second ribosome can
attach, and so on. (b) Electron
micrograph of events
diagrammed in (a). The
growing polypeptides cannot
be seen in this preparation.
Magnification 44,000x.
([b] Courtesy of O. L. Miller, Jr.)

(b)

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Chapter Ten Gene Expression: Transcription

near the 5' end that translation begins. As soon as the 5'
end of the RNA is available, a ribosome can attach to the
messenger RNA and move along it in the 5' —> 3' direction,
lengthening the growing polypeptide as it moves. When
the first ribosome moves away from the 5' end of the tran-
script, a second ribosome can attach and begin translation.
These processes are repetitive, as electron micrographs
(fig. 10.22&) clearly show. In eukaryotes, however, messen-
ger RNA is synthesized in the nucleus, but protein synthe-
sis takes place in the cytoplasm. (This regional division of
labor is not present in E. colt because, among other rea-
sons, the bacterium has no nucleus.) Before a eukaryotic
messenger RNA leaves the nucleus, it is highly modified by
processes that generally do not occur in prokaryotes.

Promoters

Eukaryotic promoters are somewhat similar to prokary-
otic promoters; both are regions of DNA at the beginnings
of genes with signals that allow RNA polymerase to attach
and begin transcription. In eukaryotes, however, more
proteins are involved in promoter recognition, and more
proteins are involved in the control of transcription, many
recognizing signals thousands of base pairs away. We dis-
cuss these control processes in eukaryotes in chapter 16.

All three eukaryotic RNA polymerases (I, II, and III)
recognize a seven-base sequence, TATAAAA, located at
about —25 on the promoter DNA. It is similar to the — 10
sequence in prokaryotes and is called the TATA box (or
Hogness box after its discoverer, D. Hogness). Since
RNA polymerase II transcribes most of the genes in eu-
karyotes, we turn our attention specifically to it.

Among the large number of promoters that have
been sequenced, a few lack the TATA box, yet are still
transcribed. Transcription initiation in these promoters
appears to be controlled by a CT-rich area, called the ini-
tiator element (Inr), at + 1 of the transcript (close to
the transcription start site), coupled with a down-
stream promoter element (DPE) at about + 28 to + 34
of the transcript. InTATA-less promoters, a protein called
TFIID requires both these elements to bind. The initiator
element has a consensus sequence of TCA(G orT)T(T or
C), and the downstream promoter element has the con-
sensus sequence of (A or G)G(A or T)CGTG. We will con-
centrate on RNA polymerase II genes with TATA boxes.

Yeast RNA polymerase II is a protein of twelve subunits.
This enzyme cannot locate promoters or attach to DNA in
a stable fashion. To attach at the beginnings of genes, RNA
polymerase II must interact with several proteins called
general transcription factors. In eukaryotes, general
transcription factors are named after the polymerase they
work with. Thus, the transcription factor that recognizes
the TATA box for polymerase II genes is called TFIID (D be-
ing the fourth letter of the alphabet for the fourth tran-
scription factor so named). TFIID is composed of one

Figure 10.23 Molecular space-filling model of a yeast TATA-
binding protein attached to a TATA box on the DNA. The DNA
sugar-phosphate backbone is green and the bases are yellow.
The protein has twofold symmetry {red and blue). Note the
bending of the DNA through 80 degrees, which also opens up
the minor groove of the DNA. The upper white atoms are the
N-terminus of the TATA-binding protein; the lower white atoms
are the first base pair at which transcription begins. (Courtesy of

J. L Kim and S. K. Burley. From J. L Kim, J. H. Geiger, S. Hahn, and
P. B. Sigler, "Crystal Structure of a Yeast TBP TATA-box Complex." Nature
365 (6446): 520-27, Oct. 7, 1993. © Macmillan Magazines, Ltd. Figure
adapted from the work of S. K. Burley.)

subunit that recognizes the TATA sequence, called TATA-
binding protein (TBP), and up to a dozen other proteins
called TBP-associated factors (TAFs), which recognize
the initiator element, when present, and aid in regulating
transcription. TFIID is, in essence, similar to the sigma fac-
tors of prokaryotic RNA polymerase. One interesting as-
pect of the binding of TBP is that it causes a significant
bending and opening of the DNA (fig. 10.23). This bending
may be an important signal for other binding proteins.

Once TFIID binds to the TATA box, a cascade of re-
cruitment (binding) of other transcription factors takes
place. Transcription factors ILA, IIB, and IIF bind, as does
RNA polymerase II in an unphosphorylated state. Then
transcription factors HE and IIH bind, forming a pre-
initiation complex (PIC), equivalent to the E. colt
holoenzyme (fig. 10.24a). The RNA polymerase II is then
phosphorylated, presumably by TFIIH, which is a kinase;
at this point, most of the transcription factors drop off,
leaving the elongation complex, which carries out a

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Transcription

• • •

DNA

Promoter

(a)

Enhancers

Activator

Upstream

Transcription

• •

Promoter

(b)

Figure 10.24 (a) An RNA polymerase II pre-
initiation complex at a promoter. TFIID binds
to the TATA box (red). The other transcription
factors are then recruited with the
polymerase, (b) Two activators (yellow) are
shown bound at one end (their DNA
domains) to enhancers {blue and green)
upstream on the DNA. The activators are
bound at their other ends (their
transcriptional activation domains) to other
proteins associated with the polymerase
machinery. Phosphorylation of the
polymerase initiates activated transcription.

Table 1 0.4 Putative Roles of the General Transcription Factors of RNA Polymerase II

General Transcription

Factor

Function

TFIID, TBP

Recognizes TATA box

TFIID, TAFs

Recognizes initiator element and regulatory proteins

TFIIA

Stabilizes TFIID

TFIIB

Aids in start-site selection by RNA polymerase II

TFIIE

Controls TFIIH functions; enhances promoter melting

TFIIF

Destabilizes nonspecific interactions of RNA polymerase II and DNA

TFIIH

Melts promoter with helicase activity; activates RNA polymerase II with kinase activity

Source: Data from R. G. Roeder, "The Role of General Initiation Factors in Transcription by RNA Polymerase II" in Trends in Biochemical Sciences, 21:327-35, 1996.

basal rate of transcription (fig. 10.25). TFIIH also has a
role here, since it is also a helicase. Table 10.4 summarizes
the postulated roles of the general transcription factors.
For activated transcription, a high level of transcription,
to take place, other factors are needed that are involved in
controlling which promoters are actively transcribed.
These other factors are activators or specific transcrip-
tion factors that bind to DNA sequences called en-

hancers. Enhancers are often hundreds or thousands of
base pairs upstream from the promoter (fig. 10.24&).

Note that much of this information has been gathered
by footprinting, mutational studies, cloning and isolating
the genes and proteins involved, and then reconstituting
various purified combinations in the test tube. These
studies are combined with kinetic research to determine
which arrangements are stable, immunological research

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Chapter Ten Gene Expression: Transcription

to isolate various components with antibodies, and pho-
tocrosslinking studies to determine which moities are in
contact with each other.

These specific transcriptional activators have domains
(regions) that recognize their specific enhancer se-
quences, regions that recognize proteins associated with
the polymerase (general transcription factors), and re-
gions that allow the joint attachment of other transcrip-
tion factors (fig. 10.24&). Similar to activators and en-
hancers, repressors can bind to silencer regions of DNA,
often far upstream of the promoters, to repress transcrip-
tion. Thus, many genes are associated with numerous and
complex arrangements of transcription factors, providing
elaborate control of transcription (see chapter 16).

For specific transcription factors to attach to both en-
hancers and the polymerase machinery, possibly thou-
sands of base pairs apart, the DNA must bend to allow
them to come into the range of the polymerase. Electron
micrographs clearly show this DNA bending and looping
(fig. 10.26).

Although RNA polymerases I and III seem to have ter-
mination signals similar to rho-independent promoters in
prokaryotes, termination of transcription of RNA poly-
merase II genes is more complex, coupled with further
processing of the mRNA.

Before we move on, several other points merit dis-
cussion. First, unlike prokaryotic RNA polymerases, eu-
karyotic RNA polymerases do proofread (showing 3' — » 5'
exonuclease activity). Second, as we will discuss in chap-
ter 15, eukaryotic DNA is complexed with histone pro-
teins that can interfere with transcription. In turn, part of
the RNA polymerase II complex is made up of proteins
that can disrupt the histones bound to the DNA.

In addition, the RNA polymerase II complex contains
proteins that act as mediators between activators and the
polymerase holoenzyme. This complex coordination of
the initiation of transcription in eukaryotes has been
termed combinatorial control; the huge initiation
complex may contain 85 or more different polypeptides.

^TA'vT

Figure 10.25 The RNA polymerase II elongation complex with
part of the protein structure removed to show the DNA and
RNA within the cleft of the protein. The DNA is blue (template
strand) and green (nontemplate strand) with the RNA red. The
majority of protein is shown as gray; the part in yellow is a
domain that appears to open for DNA loading and is in a
closed state during elongation, thus acting as a clamp on the
DNA and RNA. Closure of the clamp allows for the high
stability of transcribing complexes and thus for processivity of
the polymerase. The purple part is a helix that crosses the
major cleft of the enzyme. The DNA template strand is led over
this helix towards the active site. The pink sphere is a
magnesium ion in the active site, where RNA synthesis occurs.
(P. Cramer, D. A. Bushnell and R. D. Kornberg. RNA polymerase II at 2.8A
resolution and A. L. Gnatt, P. Cramer, J. Fu, D. A. Bushnell and R. D.
Kornberg, Structure of an RNA polymerase II transcribing complex. Reprinted
by permission of the authors.)

(a)

(b)

*V !! * *WT **ii '

Figure 10.26 The interaction between an activator and RNA
polymerase (in this case, in prokaryotes). (a) In this system, the
RNA polymerase of E. coli (the more heavily stained sphere) is
controlled by an activator called NtrC (the more lightly stained
sphere). The activator is bound to an enhancer, and the
polymerase is bound to the promoter, (b) The activator has
bound to the polymerase, causing a looping of the DNA.
Compare with figure 10.24. (Courtesy of Sydney Kustu.)

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Finally, transcription in the archaea, although under
much simpler control than in the eukaryotes, resembles
transcription in eukaryotes rather than prokaryotes.

The study of the details of the transcription process —
its initiation, control, and termination — is one of the
most active and exciting areas in modern genetics.

Caps and Tails

Eukaryotic transcription results in a primary tran-
script. In contrast to most prokaryotic transcripts that
contain information from several genes, virtually all tran-
scripts from higher eukaryotes contain the information
from just one gene. (Transcripts from several genes are
found in some lower eukaryotes, such as nematode
worms.) Three major changes occur in primary tran-
scripts of RNA polymerase II before transport into the cy-
toplasm: modifications to the 5' and 3' ends and removal
of intervening sequences. We refer to these changes as
posttranscriptional modifications .

At the 5' end of polymerase II transcripts, 7-methyl
guanosine is added in the "wrong" direction, 5' — » 5'
(fig. 10.27). This cap allows the ribosome to recognize
the beginning of a messenger RNA. At the other end,
the 3' end of polymerase II transcripts, a sequence of
twenty to two hundred adenine-containing nucleotides,
known as a poly-A tail, is added by the enzyme poly-A
polymerase. Polyadenylation takes place after the 3' end
of the transcript is removed by a nuclease that cuts
about twenty nucleotides downstream from the signal
5'-AAUAAA-3'.The tail adds stability to the molecule and
aids in its transportation from the nucleus.

When messenger RNAs were first studied in eukary-
otes, the messenger RNAs in the nucleus were found to
be much larger than those in the cytoplasm and were
called heterogeneous nuclear mRNAs, or hnRNAs. It
now turns out that these were primary transcripts, RNAs

that had not had any of the major posttranscriptional
modifications. In essence, they were premessenger RNAs.

Introns

Eukaryotes have segments of DNA within genes that are
transcribed into RNA but never translated into protein se-
quences. These intervening sequences, or introns,
are removed from the RNA in the nucleus before its trans-
port into the cytoplasm (fig. 10.28). P. Sharp and his col-
leagues at MIT and R. Roberts, T. Broker, L. Chow, and
their colleagues at the Cold Spring Harbor Laboratory
first discovered introns in 1977. (Sharp and Roberts were
awarded 1993 Nobel prizes for their work.) An example
of a gene with introns appears in figure 10.29. The seg-
ments of the gene between introns, which are tran-
scribed and translated — and hence exported to the cyto-
plasm and expressed — are termed exons. The results of
intron removal are clear when a messenger RNA with its
introns removed is hybridized with the original gene (fig.
10.30). The DNA forms double-stranded structures with
the exons in RNA. The introns in DNA have nothing to
pair with in the RNA, so they form single-stranded loops.
Introns also occur in eukaryotic transfer RNA and ribo-
somal RNA genes.

For introns to be removed, the ends of the exons must
be brought together and connected in a process called
splicing. At least two types of splicing occur, although they
are related: self-splicing and protein-mediated splicing.

Self-Splicing

In 1982, Thomas Cech and his colleagues, building on the
work of others, including Sidney Altman, who showed
that RNA can have catalytic properties, discovered self-
splicing by RNA. (Cech and Altman were awarded 1989

mRNA

Figure 10.27 A cap of 7-methyl guanosine is added in the "wrong" direction (5' -> 5'), to the 5' end of eukaryotic
mRNAs. In some cases, the 2' -OH groups on the second or second and third riboses {red) are methylated.

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Chapter Ten Gene Expression: Transcription

DNA □□□!

■ ■

IDDD

Richard J. Roberts (1943- ).
(Courtesy of Richard J. Roberts.)

Philip A. Sharp (1944- ).
(Courtesy of Dr. Philip A. Sharp.)

*b*m

Thomas Broker (1944- ).
(Courtesy of Dr. Thomas Broker.)

Louise T. Chow (1943- ).
(Courtesy of Dr. Louise Chow.)

Nobel prizes in chemistry.) Working with an intron in the
35S ribosomal RNA precursor in the ciliated protozoan,
Tetrahymena, Cech and his colleagues found that they
could induce intron removal in vitro with no proteins
present. A guanine-containing nucleotide (GMP, GDP, or
GTP) had to be present. Figure 10.31 diagrams how self-
splicing occurs. The intron is acting as an enzyme; we call
an RNA with enzymatic properties a ribozyme.

During self-splicing, the U-A bond at the left (5') side
of the intron is transferred to the GTP. The U that is now
unbonded displaces the G at the right (3) side of the in-
tron, reconnecting the RNA with a U-U connection and
releasing the intron (fig. 10.31). Since all bonds are re-
versible transfers (transesterifications) rather than new
bonds, no external energy source is required. Self-
splicing introns of this type are called group I introns.
An extensive secondary structure (RNA stem-loops) that
forms is also important in intron removal (box 10.3).

Although the first enzymatic activity of the ribozyme
is its own removal, its secondary structure after removal
gives it the ability to further catalyze reactions (fig. 10.32).
The reactions that ribozymes catalyze are transesterifica-

Intervening
sequence I

Transcription

Intervening
sequence II

hnRNA

Gene
segment A

Gene
segment B

Gene
segment C

Modification

mRNA

5' cap

AAA/

3' tail

Nucleus

Cytoplasm

Transfer to
cytoplasm

Functional mRNA
at ribosome

mRNA

AAA/

Figure 10.28 In eukaryotic DNA, intervening sequences, or
introns, are removed from the RNA in the nucleus before the
mRNA is transported into the cytoplasm and translated. Other
modifications consist of splicing, 5' capping, and 3'
polyadenylation.

tions and the hydrolysis reaction of splitting an RNA mol-
ecule into two parts. Ribozymes can also perform other
functions, including peptide bond formation, covered in
chapter 1 1 . Currently, at least seven different classes of ri-
bozymes are known, based on their enzymatic properties.
A ribozyme that can split other RNAs and that occurs in
small plant pathogens is called a hammerhead ri-
bozyme (fig. 10.33) because of its shape. Because these
RNA molecules are small, they have the potential to be
modified in the laboratory for specific purposes related to
clinical treatment and further study of RNA processing.

Self-splicing has also been found in genes in the mito-
chondria of yeast. These introns are referred to as group
II introns because they use a different mechanism of
splicing that does not require an external nucleotide. In-
stead, the first bond is transferred within the intron to an
adenosine, forming a lariat structure (fig. 10.34). In order
for the lariat to form, the ribose of the adenosine must
make three phosphodiester bonds (fig. 10.35).

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cAp

GGCCAATCTGCTCACACAGGATAGAGAGGGCAGGAGCCAGGCAGAGCATATAAGGTGAGGTAGGATCAGTTGCTCCTCACATTTGCTTCTGACATAGTTG

100 TGTTGACTCACAACCCCAGAAACAGACATCATGGTGCACCTGACTGATGCTGAGAAGGCTGCTGTCTCTTGCCTGTGGGGAAAGGTGAACTCCATGAAG

MetValHisLeuThrAspAlaGluLysAlaAlaValSerCysLeuTrpGlyLysValAsnSerAspGluV

200 TTGGTGGTGAGGCCCTGGGCAGGTTGGTATCCAGGTTACAAGGCAGCTCAAGAAGAAGTTGGGTGCTTGGAGACAGAGGTCTGCTTTCCAGCAGACAC
alGlyGlyGluAlaLeuGlyArg 30

300 TAACTTTCAGTGTCCCCTGTCTATGTTTCCCTTTTTAGGCTGCTGGTTGTCTACCCTTGGACCCAGCGGTACTTTGATAGCTTTGGAGACCTATCCTCTG

31 LeuLeuValValTyrProTrpThrGlnArgTyrPheAspSerLeuLysGlyTh

400 CCTCTGCTATCATGGGTAATGCCAAAGTGAAGGCCCATGGCAAGAAGGTGATAACTGCCTTTAACGATGGCCTGAATCACTTGGACAGCCTCAAGGGCAC
laSerAlalleMetGlyAsnAlaLysValLysAlaHisGlyLysLysVallleThrAlaPheAsnAspGlyLeuAsnHisLeuAspSerLeuLysGlyTh

500 CTTTGCCAGCCTCAGTGAGCTCCACTGTGACAAGCTGCATGTGGATCCTGAGAACTTCAGGGTGAGTCTGATGGGCACCTCCTGGGTTTCCTTCCCCTGC
rPheAlaSerLeuSerGluLeuHisCysAspLysLeuHisValAspProGluAsnPheArg 104

600 TATTCTGCTCAACCTTCCTATCAGAAAAAAAGGGGAAGCGATTCTAGGGAGCAGTCTCCATGACTGTGTGTGGAGTGTTGACAAGAGTTCGGATATTTTA

700 TTCTCTACTCAGAATTGCTGCTCCCCCTCACTCTGTTCTGTGTTGTCATTTCCTCTTTCTTTGGTAAGCTTTTTAATTTCCAGTTGCATTTTACTAAATT

800 AATTAAGCTGGTTATTTACTTCCCATCCTGATATCAGCTTCCCCTCCTCCTTTCCTCCCAGTCCTTCTCTCTCTCCTCTCTCTTTCTCTAATCCTTTCCT

900 TTCCCTCAGTTCATTCTCTCTTGATCTACGTTTGTTTGTCTTTTTAAATATTGCCTTGTAACTTGCTCAGAGGACAAGGAAGATATGTCCCTGTTTCTTC

1000 TCATAGCTCAAGAATAGTAGCATAATTGGCTTTTATGCAGGGTGACAGGGGAAGAATATATTTTACATATAAATTCTGTTTGACATAGGATTCTTGTGGT

1100 GGTTTGTCCAGTTTAAGGTTGCAAACAAATGTCTTTGTAAATAAGCCTGCAGGTATCTGGTATTTTTGCTCTACAGTTATGTTGATGGTTCTTCCATATT

1200 CCCACAGCTCCTGGGCAATATGATCGTGATTGTGCTGGGCCACCTTGGCAAGGATTTCACCCCCGCTGCACAGGCTGCCTTCCAGAAGGTGGTGGCT
105 LeuLeuGlyAsnMetlleVallleValLeuGlyHisHisLeuGlyLysAspPheThrProAlaAlaPheGlnLysValValAla

1300 GGAGTGGCCACTGCCTTGGCTCACAAGTACCACTAAACCCCCTTTCCTGCTCTTGCCTGTGAACAATGGTTAATTGTTCCCAAGAGAGCATCTGTCAGTT
GlyValAlaThrAlaLeuAlaHisLysTyrHisTer A

1400 GTTGGCAAAATGATAGACATTTGAAAATCTGTCTTCTGACAAATAAAAAGCATTTATGTTCACTGCAATGATGTTTTAAATTATTTGTCTGTGTCATAGA
1500 AGGGTTTATGCTAAGTTTTCAAGATACAAAGAAGTGAGGGTTCAGGTCTCGACCTTGGGGAAATAAA

Gene Gene

segment segment
A ' B V

Amino acids

DNA □□□[
Nucleotides

cAp 1 30 31

t r

104

ZIZ

Gene
segment

105

ZIZ

Intervening
sequence I

Intervening
sequence II

Ter
ZIZ

PA
HZ

]DDD

1467 1567

hnRNA

mRNA

Transcription

Modification

1 30/31 104/105

Ter

5' cap

3' tail

Figure 10.29 Nucleotide sequence of the mouse (3-globin major gene. The coding DNA strand is shown; cAp (position 79)
indicates the start of the capped mRNA; pA indicates the start of the poly-A tail (position 1467); numbers inside the
sequence are adjacent amino acid positions; Ter is the termination codon (position 1334). The three-letter abbreviations
(e.g., Met, Val, His) refer to amino acids (see chapter 11). The TATA box begins at position 49. (Source: National institutes of
Health Research by David A. Konkel, et al., "The sequence of the chromosomal mouse p-globin major gene: Homologies in capping, splicing and poly
(A) Sites," Cell, 15:1125-32, 1978.)

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Chapter Ten Gene Expression: Transcription

Exon I

Intron

(a)

3'-poly-A tail

Single-stranded DNA

(b)

DNA-RNA
hybrid

Intron 3

Intron 2

Figure 10.30 The mRNA of adenovirus hybridized with its
DNA. Three introns are visible as single-stranded DNA loops.
They form single-stranded loops because they have nothing in
the RNA molecule to hybridize with. Also visible is the poly-A
tail of the mRNA. (a) Electron micrograph, (b) explanatory
diagram, ([a] Courtesy of Louise T. Chow and Thomas Broker.)

Thomas Cech (1947- ).
(Courtesy of Dr. Thomas Cech.
Photo by Ken Abbott.)

Sidney Altman (1939- ).
(Courtesy of Dr. Sidney Altman.
Photo: Michael Marsland, Yale
University Office of Public Affairs.)

UCUA

Exon II
G |U 3'

GTP

ucu

UCU

Exon I

Exon II

Intron + G

UCUU

Figure 10.31 Self-splicing of a ribosomal RNA precursor in
Tetrahymena. An external GTP is required. Two bond transfers
produce a shortened RNA and a free intron.

Protein-Mediated Splicing (the Spliceosome)

Eukaryotic nuclear messenger RNAs also have their in-
trons removed by way of a lariat structure, just as in type
II introns, but with the help of RNA-protein particles.
Figure 10.36 shows consensus sequences in nuclear mes-
senger RNA for the majority of introns. At the left (5')
side of the intron, the GU sequence is invariant, as is
the AG at the right (3') side. The right-most A of the
UACUAAC sequence is the branch point of the lariat
and is also invariant. (In DNA nucleotides, UACUAAC is
TACTAAC; therefore, that region is sometimes referred to
as the TACTAAC box.)

Unlike the mitochondrial group II introns, however,
nuclear messenger RNAs have their introns removed
with the help of a protein-RNA complex called a
spliceosome, named by J. Abelson and E. Brody The
splicing apparatus in eukaryotic messenger RNAs con-
sists of several components called small nuclear ri-
bonucleoproteins (discovered and named by J. Steitz
and colleagues), abbreviated as snRNPs and pronounced
"snurps." Five of these particles take part in splicing, each
composed of one or more proteins and a small RNA
molecule; they are designated Ul, U2, U4, U5, and U6.
The RNA molecules range in size from 100 to 215 bases.
The snRNPs and their associated proteins are located
in twenty to forty small regions in the nucleus called
speckles because of their appearance in the fluorescent
microscope.

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The RNAs of these particles have been sequenced, and
sequencing shows they have regions of complementarity
to either sites in the exons, sites in the introns, or sites in
the other snRNP RNAs (table 10.5). These sequences, to-
gether with the experimental techniques of pho-
tocrosslinking and the creation of selective mutations (us-
ing techniques of site-directed mutagenesis described in
chapter 13) have given us insight into the splicing mecha-
nism. Photocrosslinking tells us which components are in
contact. Mutations change pairings of components and
may disrupt the structure. The change can be rescued
(the pairing restored) by making a second change in the
complementary RNA. When this happens successfully, the

Joan A. Steitz (1941- ).
(Courtesy of Dr. Joan A. Steitz.)

Ribozyme

ggcccucuQH

(a)

'OH

ggcccucua 5

U G
A C

C A
G-C
G-C
A-U
C-G
U • G
A-U
U-A
U • G
G-C

5'-

Substrate

U • G

C-G

U-A

C-G

C — G — A — A— A— A— U--A--G — CAAGACCGUCAAAUU— A

A U U
G G
A C
U-A
A-U
A C
C-G
C-G
A-U
A-U
A-U A
U— A M
U-A
U-A
C-G
U • G

Figure 10.32 The intron removed from the ribosomal RNA of
Tetrahymena can catalyze the removal of the 3' end of an
RNA, diagrammed here as five AMP residues (a 5 ) from the
sequence 5'-GGCCCUCUA 5 -3'. The intron is called the
Tetrahymena ribozyme. Any sequence can be removed from an
RNA as long as there is a sequence complementary to the
GGGAGG-5' of the ribozyme to bring the RNA into position. In
(a), the reaction needs an external guanine-containing
nucleotide (Gqh); substrate nucleotides are in lowercase letters.
This transesterification requires no external energy. In (£>), the
secondary structure of the ribozyme is shown. Goh is the site
of cleavage, and the position of the G-binding site is shown.
Further structure must develop to bring the G-site to the
substrate. Wavy lines represent additional structure not
shown. (Reprinted with permission from Ann Marie Pyle, et al., "RNA
substrate binding site in the catalytic core of the Tetrahymena ribozyme,"
Nature, Volume 358, 1992. Copyright © 1992 Macmillan Magazines, Ltd.)

G-C

C-G

G A

A C

C L

A A

A-U
C-G
U • G
G • U
G-C
G-C

G U

A A

G
G
G
C
G

C
C
C
G
C

GC U M GUC

CUGGCUGU

A C

A A

AUCAG

UUG

G U AGAAGGG
U I I I I I -I MINI

A jjUCUUCUCAUAAGAUAUAGUCGGACC -

G-site

'A C G

• I I

UAG

(b)

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Chapter Ten Gene Expression: Transcription

Table 1 0.5 The Five Small Nuclear Ribonucleoproteins (snRNPs) Involved in

Nuclear Messenger RNA Intron Removal and Their RNAs

snRNP RNA

Partial Sequence

Complementarity

Role

3'-UCCAUUCAUA

5 ' end of intron

Recognizes and binds 5' site of intron

3'-AUGAUGU

Branch point of intron

Binds branch point of intron

3'-UUGGUCGU . . .
AAGGGCACGUAUUCCUU

Binds to (inactivates) U6

3'-CAUUUUCCG

Exon 1 and exon 2

Binds to both exons

3 -CGACUAGU . . . ACA

U2, 5' site

Displaces Ul and binds 5' site and
U2 at branch point

Stem III

L2.4

L2.3A

A13
11.4 11.1 G12

G G C C

L2.2A

L2.1

.CCGG

b 10 - 4 10.1^9

HO OH

15.5U A16.5

15.4 G C16.4

15.3C G16.3

1 5.2 C G 1 6.2 Cleavage site

15.1 A U16.1 /

A14 C17^

1.1 1.2 1.31.4 1.5 1.6 1.7

G G U C G C C

Stem II

G 8

G5
U7 Ae

C C A G C G G

2.1 2.2 2.3 2.4 2.5 2.6 2.7

OH
PPP

Stem I

Figure 10.33 The hammerhead ribozyme, first seen in the
RNAs of certain viruses (stems I, II, and III), (a) The cleavage
point of the substrate {red) is shown using original sequence
numbering, relating to the three stems of the hammerhead-
shaped structure, (b) The cleavage, a transesterification, creates
a cyclic 2',3'-phosphodiester bond and a free 5'-OH.
(Reprinted with permission from Nature, Vol. 372, Heinz W. Pley et al.,
"Three Dimensional Structure of a Hammerhead Ribozyme." Copyright
© 1994 Macmillan Magazines Limited.)

(a)

Cyclic 2 , ,3'-phosphodiester bond

Transesterification

O OH

(b)

Free 5-OH

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Exon I

Intron Exon II
A — GET

Lariat

Intron

Exon I

Exon II

Figure 10.34 Self-splicing of a group II intron results in a lariat
configuration of the released intron. No external GTP is
required since the first bond transfer takes place with an
internal nucleotide, forming the loop of the lariat. A second
bond transfer releases the lariat-shaped intron.

presumed pairing is then confirmed. For example, if an A-U
base pair occurs between two pieces of RNA, changing the
A to a C disrupts the pairing. However, if the U is converted
to a G, the pairing is restored (complementary A-U bases
are converted to complementary C-G bases via a noncom-
plementary C-U intermediate). From these techniques, we
believe that the following sequence of events takes place.

First, the Ul snRNP binds at the 5' site of the intron
and the U2 snRNP binds at the branch point (fig. 10.37).
The U4, U5, and U6 snRNPs form a single particle. The U4
snRNP releases, freeing the U6 snRNP to bind to the 5'
site, displacing the Ul snRNP. (The Ul snRNP, with the
help of other proteins, may bind at the 5' site simply to
mark it and initiate the process.) The U6 snRNP then also
binds the U2 snRNP, allowing the lariat to form in the in-
tron. The U5 snRNP binds the two exon ends together, al-
lowing the splice to be completed as the lariat is removed.

The splicing machinery for the majority of introns also
includes numerous other polypeptides called auxiliary
and splicing factors; the entire splicing process requires
about 50 polypeptides. A second, less common, intron,
called the U12-dependent intron, with different consensus
sequences, also exists. It is removed by a similar splicing
process involving different snRNPs (Ull, U12) as well as
many components shared with the major spliceosome.

Currently, we believe the splicing out of the intron may
be autocatalyzed, just as in the type II self-splicing introns.
The spliceosome may have evolved to ensure control over
the process, allowing different introns to splice with differ-
ing efficiencies and allowing alternative splicing to take
place. In many eukaryotic genes, alternative paths of splic-

Guanine

Figure 10.35 The lariat branch point (see fig. 10.34), formed
during removal of a group II intron, occurs as three
phosphodiester bonds form at the same ribose sugar. (The
lariat loop is formed by the 2'-phosphodiester bond.)

Exon I

AGPUAUGU •
5' site

Intron
••UACUAAO««§AG

Exon II

3' site

Branch point

Figure 10.36 Consensus sequences of nuclear introns,
showing the 5' and 3' sites and the branch point. Letters in
blue (GU, A, AG) represent invariant bases. The last A (in blue)
of the UACUAAC sequence is the lariat branch point.

ing can take place — different splice sites may be chosen or
splices may be avoided entirely. Thus, a single gene can pro-
duce several different proteins, depending on splicing
choice. For example, in yeast, the gene RPL32 codes for a
ribosomal protein. When this protein accumulates in ex-
cess, it somehow causes intron removal to fail. The result
is a nonfunctional messenger RNA and no further RPL32

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Chapter Ten Gene Expression: Transcription

BOX 10.3

Viroids are small (less than
four hundred nucleotides),
single-stranded RNA circles
that act as plant pathogens. They do
not have protein coats. In addition,
they do not seem to code for any pro-
tein. The nature of their pathogenic-
ity is not well understood. In 1986,
Gail Dinter-Gottlieb, at the University
of Colorado, pointed out numerous
regions of homology between viroids

Experimental
Methods

Are Viroids Escaped
Introns?

and group I introns, supporting pro-
posals by Francis Crick and Theodor

Diener that viroids are escaped
introns.

Many group I introns form circles
after they are released. The self-
splicing group I intron of Tetrahy-
mena thermophila is 399 bases in
circular form, whereas the potato
spindle tuber viroid (PSTV) is 359
bases. These similarities of size and
shape prompted the search for base
homologies. In figure 1, we compare

Group 1 consensus

UCUGUUGAU

A U

GC A/ _
GGAU (G_U_U

U I I I Ml

c I I I I I I I I

uagacaacug"ccug a ,cga

A U " /\ U

c cU
« or\ \ V G
a cAU N \ c \ gG

r G A°

A r U G A X pC C GAC

U A GA

A A — U

U — A
G— C
G— C

A _ i

Box 2

C U

GGUUUAAAGGC

I I I I I I I I I I I

CCAAAUUUCUG

AU A AG AU A

C
U

U A GU C

GGAc R u xV

\_A G

U C

C U A GC GG

I I I I I I I
GGU CG CC

G. G

16N

G A

AU A

A C

U
G

GUCUCAGGG G
I I I I I I I I

G— CG G
G— C
A — U
A — U
A G
C

i llAGAGUU UCA

\ X G A
G u

A c"

qGUUlZ/Cu.

A G

D-stem

A G— C

G— U

G— C

A — U

G— C

G— U

U — A

U— A

c c u

A
C
U
C

Au u[/> U ,

A G

//An

u u A '/ u

A U/ A
A A

A
U
U

A
G

Tetrahymena RNA

U A A

A U A A

Figure 1 Self-splicing group I Tetrahymena intron (left page) and potato spindle tuber viroid (PSTV, right page); 16N in the left
figure refers to sixteen nucleotides not shown. Note the similarities around the group 1 consensus area. (From G. Dinter-Gottlieb,
Proceedings of the National Academy of Sciences, page 6251 , 1986.)

protein produced. In human beings, the gene RBP-MS can
produce at least twelve different transcripts, depending on
alternative splicing.

One other mode of protein-mediated intron removal
is known. Nuclear transfer RNAs have introns that are not
self-splicing but are removed by an endonuclease; the ex-
ons are subsequently joined by a ligase. Archaean bacte-
ria seem to have this type of intron.

Intron Function and Evolution

Since the discovery of introns, geneticists have been try-
ing to figure out why they exist. Several views have

arisen. Walter Gilbert suggested that introns separate ex-
ons (coding regions) into functional domains — that is,
they separate different exons that presumably have spe-
cific tasks. In a given protein, one exon might code for a
membrane-binding region, one might code for the active
site of the enzyme, and one might code for ATPase activ-
ity. By recombinational mechanisms, or by excluding an
exon during intron removal, exon shuffling would al-
low the rapid evolution of new proteins whose structures
would be conglomerates of various functional domains.
In a 1990 article in Science, Gilbert, with two colleagues,
calculated that all proteins in eukaryotes can be ac-
counted for by as few as one thousand to seven thousand

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the Tetrahymena intron with PSTV
Note that both have an extensive
secondary structure (stem-loops)
and similarities of some sequences.
Most notable is the box 9L similarity.
This box lies within a 16-base con-
sensus sequence of all group I in-

trons and has similarities to the cor-
responding sequence in PSTV Note
the general shape around the group I
consensus region: two stems to the
left and one to the right with some
homologies. Note also the D-stem
similarities.

These similarities strongly indi-
cate that viroids and group I introns
are related. Whether viroids are es-
caped introns or both evolved from
a common ancestor has, as yet, not
been resolved.

CUy

c
u
c

u
u

c
u

Group 1 consensus

U— G

C— G

G — C

A C C— G

c
c

u P s puc

U GGG U
I I I I I

° u cc c

Box 2

U— GCCCUUGGAACCGCA

I I I I I I I I I I

UCCACCUUGGUGU

G
A
C

aCa p gH C

A \\ X c CG U r / / / C C A

AAAAQa
A G A A,

c u

GGAACUA A A

I I I I I I I
CCUUGGUq

G C C\

U \ \ /

r c c

A G C GG

A
G

A \

A U G '

A
A
A

G
A
A
C

PSTV RNA

U C Q //

c c G // /G

G A A / G

G G A

Figure 1 (continued)

exons; all proteins may be conglomerates of this primor-
dial number. However, this view is controversial.

J. Darnell and W. F. Doolittle have expanded Gilbert's
idea of exon shuffling into the introns-early view. They
suggest that introns arose before the first cells evolved. Af-
ter eukaryotes evolved from prokaryotes, the prokaryotes
lost their introns. This is supported by the evidence that,
generally, prokaryotes lack introns. This view is also consis-
tent with the opinion that the original genetic material was
RNA. In this "RNA world," introns arose as part of the ge-
netic apparatus; they were the first enzymes (ribozymes).

An alternative view is that introns arose later in evo-
lution, after the eukaryotes split from the prokaryotes.

At first, the justification for this introns-late view was
that introns evolved late to give the organism the ability
to evolve quickly to new environments by an exon-
shuf fling type of mechanism. However, evolutionary
biologists don't accept the rationale of evolution based
on future needs. An alternative explanation is that
introns are actually invading "selfish DNA," DNA that
can move from place to place in the genome without
necessarily providing any advantage to the host organ-
ism. We call these "jumping genes" transposons and
discuss them at length in chapters 14 and 16. Thus, both
time frames for the development of introns — late or
early — have conceptual support.

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Chapter Ten Gene Expression: Transcription

5' site

Branch point

3' site

EXON I

EXON

EXON I EXON II

Figure 10.37 Sequence of steps, explained in the text, in which U1, U2, U4, U5, and U6 snRNPs take
part in intron removal in a nuclear RNA.

Evidence exists to support both views. Gilbert's exon-
shuffling view is supported by the analysis of some genes
that do indeed fit the pattern of exons coding for func-
tional domains of a protein. (Analysis consists of DNA
sequencing, RNA sequencing, and protein structural
analysis.) For example, the second of three exons of the
globin gene binds heme. Similarly, the human low-density
lipoprotein receptor is a mosaic of exon-encoded mod-
ules shared with several other proteins. Autocatalytic
properties of introns lend credence to the view that RNA
was the original genetic material and that introns can
move within a genome.

Additional evidence for the introns-early hypothesis
includes the discovery of several introns in phage genes
and introns in transfer RNA and ribosomal RNA genes in
ancient bacteria (archaebacteria). Until recently, how-
ever, no introns were known in the true bacteria (eu-
bacteria). That changed with recent work from the labs

of D. Shub and J. Palmer, who independently discovered
an intron in a transfer RNA gene in seven species of
cyanobacteria (blue-green algae of the eubacteria).This
intron was suspected to exist because it occurred in the
equivalent chloroplast gene; the chloroplast evolved
from an invading cyanobacterium. However, this dis-
covery has been viewed as supporting both the introns-
early and introns-late view. The introns-early supporters
say this evidence confirms that introns arose before the
eukaryotes-prokaryotes split. Introns-late supporters
say they expect to see some introns in prokaryotes be-
cause of the mobility these bits of genetic material
have.

Both the introns-early and the introns-late views may
be correct. It is possible that introns arose early, were lost
by the prokaryotes, which prioritized small genomes and
rapid, efficient DNA replication, and later evolved to pro-
duce exon shuffling in eukaryotes.

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Updated Information About the Flow of Genetic Information

275

RNA Editing

In the last few years, several examples have arisen
in which DNA sequence does not predict protein se-
quence. In several cases, changes in the protein occur
that could have only come about by inserting or deleting
nucleotides in the messenger RNA before it is translated.
This insertion or deletion is almost exclusively of
uridines. The process is termed RNA editing.

RNA editing was particularly evident in the mito-
chondrial proteins of a group of parasites, the try-
panosomes (some of which cause African sleeping sick-
ness); in one case, more than 50% of the nucleotides in
the messenger RNA were added uridines. Uridines were
also deleted from the original sequence. These parasites
had another mysterious trait — the existence of minicir-
cles and maxicircles of DNA in specialized mitochondria
called kinetoplasts. In the average kinetoplast, there are
about fifty maxicircles and about five thousand minicir-
cles, concatenated like chain links (fig. 10.38a). The
maxicircles contain genes for mitochondrial function
(see chapter 17); as L. Simpson and his colleagues
showed in 1990, both maxicircles and minicircles are
templates for guide RNA (gRNA), RNA that guides the
process of messenger editing.

The guide RNA forms a complement with the mes-
senger RNA to be edited; however, the guide RNA has the
sequence complementary to that of the final messenger
RNA, the one with bases added. Since the bases have not
yet been added, a bulge occurs in the guide RNA where
the complement to be added is (fig. 10.38&). The mes-
senger RNA is then cleaved opposite the bulge by an ed-
iting endonuclease. A uridylate (U) is brought into the
messenger RNA as a complement to the adenine (A) with
the enzyme terminal-U-transf erase. An RNA ligase then
closes the nick in the messenger RNA, which now has a
uridylate added.

An exciting outcome of this research, aside from
learning about a novel mechanism of messenger RNA
processing, is the possibility of clinical rewards. Anytime
there is a specialized pathway in a parasite not found in
its host, it is possible to use that pathway to attack the
parasite. Thus, this research might lead to new ways of
combating these trypanosome parasites.

RNA editing also occurs in other species and by dif-
ferent mechanisms. For example, in the apolipoprotein-B
(apoJS) gene in mammals, one gene produces two forms
of the protein. In one case, nucleotide 6666, a cytosine, is
modified by deamination to a uracil in the messenger
RNA, resulting in the termination of translation and a pro-
tein about half the normal size. RNA editing also occurs
in plant mitochondria and chloroplasts in which the
usual change is also a cytosine to a uracil. RNA editing is
thus routinely seen in specific examples of posttran-
scriptional RNA modification in both animals and plants.

(a)

5'-A-A-G-G -G -A-A-A-3'

I I I I I I I I

3-U-U-C-C C-U-U-U-5'
\ /

mRNA
Guide RNA

Cleavage by an
editing endonuclease

A-A-G-G G-A-A-A

I I I I I I I I

U-U-C-C-A-C-U-U-U

Addition of U from UTP
by terminal-U-transferase

A-A-G-G -U G-A-A-A
I I I I I I I I I
U-U-C-C-A-C-U-U-U

RNA ligase

A-A-G-G -U-G -A-A-A
I I I I I I I I I
U-U-C-C-A-C-U-U-U

(b)

Figure 10.38 RNA editing, (a) Eight hundred seventy base pair
minicircles of DNA from Leishmania tarentolae. (b) Mechanism
by which a guide RNA is involved in the editing of a
messenger RNA. After the cycle shown, a uridine-containing
nucleotide has been added to the messenger RNA. The guide
RNA has the sequence complementary to the messenger RNA
with the base already added, {[a] Courtesy of Larry Simpson.)

UPDATED INFORMATION
ABOUT THE FLOW OF
GENETIC INFORMATION

The original description of the central dogma included
three information transfers that were presumed to occur
even though they had not been observed (see fig. 10.1).
Since then, researchers have documented these three
transfers: reverse transcription, RNA self-replication, and
the direct involvement of DNA in translation (fig. 10.39).

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Chapter Ten Gene Expression: Transcription

Reverse Transcription

First, the return arrow from RNA to DNA in figure 10.39
indicates that RNA can be a template for DNA synthesis.
All RNA tumor viruses, such as Rous sarcoma virus, as
well as the AIDS virus, can make an RNA-dependent DNA
polymerase (often referred to as reverse transcriptase)
that synthesizes a DNA strand complementary to the
viral RNA. (H. Temin and D. Baltimore received Nobel
prizes for their discovery of this polymerase enzyme.)
This enzyme is involved in a tumor virus's infection of a
normal cell and the transformation of that cell into a can-
cerous cell. When the viral RNA enters a cell, it brings re-
verse transcriptase with it. The enzyme synthesizes a
DNA-RNA double helix, which then is enzymatically con-
verted into a DNA-DNA double helix that can integrate
into the host chromosome. After integration, the DNA is
transcribed into copies of the viral RNA, which are both
translated and packaged into new viral particles that are
released from the cell to repeat the infection process.
(We cover this material in more detail in chapters 13 and
16.)

Howard Temin David Baltimore (1938-

(1934-1994). (Courtesy of (Courtesy of Kucerea and

Dr. Howard Temin. UW photo Company/Laxenburger

media.) Strasse 58.)

RNA Self-Replication

The second modification to the original central dogma is
the verification that RNA can act as a template for its own
replication. This process has been observed in a small
class of phages. These RNA phages, such as R17, f2,
MS 2, and Qp, are the simplest phages known. MS 2 con-
tains about thirty-five hundred nucleotides and codes for
only three proteins: a coat protein, an attachment protein
(responsible for attachment to and subsequent penetra-
tion of the host), and a subunit of the enzyme RNA repli-
case. The RNA replicase subunit combines with three of
the cell's proteins to form RNA replicase, allowing the
single-stranded RNA of the phage to replicate itself.

Since the new protein needed to construct the RNA
replicase enzyme must be synthesized before the phage
can replicate its own RNA, the phage RNA must first act
as a messenger when it infects the cell. Thus, protein syn-
thesis is taking place without a preceding transcription
process. The viral genetic material, its RNA, is first used as
a messenger in the process of translation and then used
as a template for RNA replication.

DNA Involvement in Translation

In the mid-1960s, B.J. McCarthy and J. J. Holland showed
that under certain experimental conditions, denatured
(single-stranded) DNA could bind to ribosomes and be
translated into proteins. The experimental conditions
usually involved the addition of antibiotics that inter-
acted with the DNA or the ribosome. Direct translation
of DNA is not known to occur naturally.

Even in our updated central dogma (fig. 10.39), no ar-
rows originate at the protein. In other words, protein
cannot self-replicate, nor can it use amino acid sequence
information to reconstruct RNA or DNA. Crick has called
these arrows "forbidden transfers." We know of no
cellular machinery that can produce these forbidden
processes. In the next chapter, we continue this discus-
sion of protein synthesis by describing the process of
translation, in which the information in messenger RNA
is used to form the sequences of amino acids in proteins.

Self-replication
loop

Transcription/'

' Reverse
r / transcription

\ Lab

x conditions

\
\

RNA

Translation

Protein

Self-replication
loop

Figure 10.39 An updated version of Crick's central dogma,
showing all known paths of genetic information transfer. Paths
confirmed since Crick proposed the original central dogma
appear as dashed red lines (reverse transcription, RNA self-
replication, and direct DNA translation). Direct DNA translation
is known only under laboratory conditions: the process
apparently does not occur naturally. There is no known
information flow beginning with protein.

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Solved Problems

277

SUMMARY

The central dogma is a description of how genetic infor-
mation is transferred among DNA, RNA, and protein.
In chapter 9, we described the DNA self-replication loop. In
this chapter, we described the transcriptional process, in
which DNA acts as a template for the production of RNA.

STUDY OBJECTIVE 1: To examine the types of RNA and
their roles in gene expression 245-246, 256-260

Messenger RNA (mRNA) is a complementary copy of the
DNA of a gene that carries the information of the gene to
the ribosomes, where protein synthesis actually takes
place. Transfer RNAs (tRNAs) transport the amino acid
building blocks of proteins to the ribosome. Complemen-
tarity between the messenger RNA codon and the transfer
RNA anticodon establishes the amino acid sequence in the
synthesized protein ultimately specified by the gene. Ribo-
somal RNA (rRNA) is also involved in this process of gene-
directed protein synthesis.

STUDY OBJECTIVE 2: To look at the process of transcrip-
tion, including start and stop signals, in both prokaryotes
and eukaryotes 246-256

Intracellular RNA is single-stranded, although extensive in-
tramolecular stem-loop structures may form. At any one
gene, RNA is transcribed from only one strand of the DNA
double helix. The transcribing enzyme is RNA polymerase.
In E. coli, the core enzyme, when associated with a sigma
factor, becomes the holoenzyme that recognizes the tran-
scription start signals in the promoter. Several consensus
sequences define a promoter. In prokaryotes, termination
of transcription requires a sequence on the DNA, called the
terminator, that causes a stem-loop structure to form in the
RNA. Sometimes the rho protein is required for termination
(in rho-dependent, as compared with rho-independent, ter-
mination). In eukaryotes, there are three RNA polymerases.

Eukaryotic genes have promoters with sequences analo-
gous to those in prokaryotic promoters as well as en-
hancers that work at a distance.

The ribosome is made of two subunits, each with pro-
tein and RNA components. Transfer RNAs are charged with
their particular amino acids by enzymes called aminoacyl-
tRNA synthetases. Each transfer RNA has about eighty
nucleotides, including several unusual bases. All transfer
RNAs have similar structures and dimensions. Transfer
RNAs and ribosomal RNAs are modified from their primary
transcripts.

STUDY OBJECTIVE 3: To investigate posttranscriptional
changes in eukaryotic messenger RNAs, including an
analysis of intron removal 260-276

Prokaryotic messenger RNAs are transcribed with a leader
before, and a trailer after, the translatable part of the gene.
In prokaryotes, translation begins before transcription is
completed. In eukaryotes, these processes are completely
uncoupled — transcription is nuclear and translation is cy-
toplasmic. Eukaryotic messenger RNA is modified after
transcription: a cap and tail are added, and intervening se-
quences (introns) are removed, before transport into the
cytoplasm. Introns can be removed by self-splicing or with
the aid of the spliceosome, composed of small nuclear ri-
bonucleoproteins (snRNPs). It is not known whether in-
trons arose early or late in evolution or what their functions
are. In some organisms, such as trypanosomes, RNAs can be
edited further by the addition or deletion of nucleotides un-
der the direction of guide RNA.

The study of several RNA viruses has shown that RNA
can act as a template to replicate itself and to synthesize
DNA; under laboratory conditions, DNA can be translated
directly into protein. These discoveries add new directions
of information transfer to the central dogma.

SOLVED PROBLEMS

PROBLEM 1: What would be the sequence of segments
on a prokaryotic messenger RNA with more than one
gene present?

Answer: The transcript would have unmodified 5'
(leader) and 3' (trailer) ends. Reading the sequence of nu-
cleotides on the RNA, you would come across an initia-
tion codon (AUG) and then, after perhaps nine
hundred more nucleotides, a termination codon (UAA,
UAG, or UGA). The nine hundred nucleotides would
be those translated into the protein. Then there would

be a spacer region of nucleotides, followed by another ini-
tiation codon, intervening nucleotides that are translated
into amino acids, and a termination codon. This sequence
of initiation codon, codons to be translated, a termination
codon, and spacer RNA would be repeated for as many
genes as are present in the messenger RNA.

PROBLEM 2: Can one nucleotide be a conserved sequence?

Answer: Conserved sequences are invariant sequences
of DNA or RNA recognizable to either a protein or a

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Chapter Ten Gene Expression: Transcription

complementary sequence of DNA or RNA. However, in
group II introns, an adenine is needed near the 3' end of
the intron for lariat formation. Thus, this single nu-
cleotide, given its relative position in the intron and pos-
sible surrounding bases, is a conserved sequence of one.

PROBLEM 3: Why might E. colt not have a nucleolus?

Answer: The nucleolus is the site of ribosomal construc-
tion in eukaryotes. It is centered at the nucleolus orga-
nizer, the tandemly repeated gene coding for the three
larger pieces of ribosomal RNA. In E. colt, there are only
five to ten copies of the ribosomal RNA gene, whereas
there is usually an order of magnitude or more copies in
eukaryotes. Thus, the simplest reason that a nucleolus is
not visible in E. colt is because there are too few copies
of the gene around which a nucleolus forms.

PROBLEM 4: If this sequence of bases represents the start
of a gene on double-stranded DNA, what is the sequence
of the transcribed RNA, what is its polarity, and what is
the polarity of the DNA?

GCTACGGATTGCTG
CGATGCCTAACGAC

Answer: Begin by writing the complementary strand to
each DNA strand: CGAUGCCUAACGACforthe top,
and G C U A C G G AU U G C U Gfor the bottom. Now look
for the start codon, AUG. It is present only in the RNA made
from the top strand, so the top strand must have been tran-
scribed. The polarity of the start codon is 5' -A U G-3'. Since
transcription occurs 5' — > 3', and since nucleic acids are
antiparallel, the left end of the top DNA strand is the 3' end.

EXERCISES AND PROBLEMS

TYPES OF RNA

1. Diagram the relationships of the three types of
RNA at a ribosome. Which relationships make use
of complementarity?

PROKARYOTIC DNA TRANSCRIPTION

2. How could DNA-DNA or DNA-RNA hybridization be
used as a tool to construct a phylogenetic (evolu-
tionary) tree of organisms?

3. Assume that prokaryotic RNA polymerase does not
proofread. Do you expect high or low levels of error
in transcription as compared with DNA replication?
Why is it more important for DNA polymerase than
RNA polymerase to proofread?

4. What are the transcription start and stop signals in
eukaryotes and prokaryotes? How are they recog-
nized? Can a transcriptional unit include more than
one translational unit (gene)? (See also EUKARYOTIC
DNA TRANSCRIPTION)

5. What is a consensus sequence? a conserved
sequence?

6. What would the effect be on transcription if
a prokaryotic cell had no sigma factors? no rho
protein?

7. Draw a double helical section of prokaryotic DNA
containing transcription start and stop information.
Give the base sequence of the messenger RNA
transcript.

8. In what ways does the transcriptional process
differ in eukaryotes and prokaryotes? (See also
EUKARYOTIC DNA TRANSCRIPTION)

9. What is a stem-loop structure? an inverted repeat? a
tandem repeat? Draw a section of a DNA double he-
lix with an inverted repeat of seven base pairs.

10. What is the function of each of the following se-
quences: TATAAT, TTGACA, TATA, TACTAAC? What is a
Pribnow box? a Hogness box? (See also EUKARYOTIC
DNA TRANSCRIPTION)

11. What is footprinting? How did it help define pro-
moter sequences?

12. What are the differences between rho-dependent
and rho-independent termination of transcription?

.32-

* Answers to selected exercises and problems are on page A-ll.

13. What are the differences between a a and a cr ?

14. Draw a typical mature messenger RNA molecule of a
prokaryote and a eukaryote. Label all regions. (See
also EUKARYOTIC DNA TRANSCRIPTION)

15. Determine the sequence of both strands of the DNA
from which this RNA was transcribed. Indicate the
5' and 3' ends of the DNA and, with an arrow, which
strand was transcribed.

5'-C C A U C A U G A C A G A C C C U U G C U A A C G C-3'

16. The following DNA fragment was isolated from the
beginning of a gene. Determine which strand is
transcribed, indicate the polarity of the two DNA
strands, and then give the sequence of bases in the
resultant messenger RNA and its polarity.

CCCTACGCCTTTCAGGTT
GGGATGCGGAAAGTCCAA

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Transcription

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Critical Thinking Questions

279

17. The following DNA fragment represents the begin-
ning of a gene. Determine which strand is transcribed,
and indicate polarity of both strands in the DNA.

ATGATTTACATCTACATTTACATT
TACTAAATGTAGATGTAAATGTAA

18. The following sequence of bases in a DNA molecule
is transcribed into RNA:

CCAGGTATAATGCTCCAGTATGGCATGGTACTTCCGG

If the T (arrow) is the first base transcribed, deter-
mine the sequence and polarity of bases in the RNA,
and identify the Pribnow box and the initiator
codon.

19. You have isolated a mutant that makes a tempera-
ture-sensitive rho molecule; rho functions normally
at 30° C, but not at 40° C. If you grow this strain at
both temperatures for a short period of time and iso-
late the newly synthesized RNA, what relative size
RNA do you expect to find in each case?

20. Suppose you repeat the experiment in problem 19
and find the same size RNA made at both tempera-
tures. Provide two possible explanations for this un-
expected finding.

21. Why do you think most promoter regions are A-T
rich? (See also EUKARYOTIC DNA TRANSCRIPTION)

EUKARYOTIC DNA TRANSCRIPTION

22. Would introns be more or less likely than exons to
accumulate mutations through evolutionary time?

23. What would be the effect on the final protein prod-
uct if an intervening sequence were removed with
an extra base? one base too few?

24. What is heterogeneous nuclear messenger RNA?
What are small nuclear ribonucleoproteins?

25. What product would DNA-RNA hybridization pro-
duce in a gene with five introns? no introns? Draw
these hybrid molecules.

26. What are the recognition signals within the majority
of introns?

27. What are the differences between group I and group
II introns?

28. Diagram ribozyme functioning in a group I intron.

29. How does a spliceosome work? What are its compo-
nent parts?

30. What is a transcriptional factor? an enhancer?

31. In the following drawing of a eukaryotic gene, solid
red lines represent coding regions, and dashed blue
lines represent introns. Draw the RNA-DNA hybrid
that would result if cytoplasmic messenger RNA is
hybridized to nuclear DNA.

32. RNA-DNA hybrids are formed by using messenger
RNA for a given gene that is expressed in the pituitary
and the adrenal glands. The DNA used in each case is
the full-length gene. Based on the figure, provide an
explanation for the different hybrid molecules. DNA
is a dashed red line; RNA is a solid blue line.

*
■

1
■
■
■

1
1
1
1

2 : : 3 *,/ 4

2 *.

4 :

Pituitary

Adrenal

33. Enhancers can often exert their effect from a dis-
tance; some enhancers are located thousands of
bases upstream from the promoter. Propose an ex-
planation to account for this.

UPDATED INFORMATION ABOUT THE FLOW OF GENETIC
INFORMATION

34. How do prions relate to the central dogma of fig-
ure 10.39?

CRITICAL THINKING QUESTIONS

1. What are the upper limits to the size of a gene in eu- 2. Present a scenario of an activator controlling transcrip-
karyotes? tion in eukaryotes.

Suggested Readings for chapter 10 are on page B-7.

Tamarin: Principles of
Genetics, Seventh Edition

Molecular Genetics

11. Gene Expression:
Translation

©TheMcGraw-Hil
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GENE
EXPRESSION

Translation

Computer-generated model of the enzyme

seryl tRNA synthetase, the enzyme that charges

tRNA's with the amino acid serine. (© Dr. Stephen

Cusack/EMBL/SPL/Photo Researchers, Inc.)

STUDY OBJECTIVES

1. To study the mechanism of protein biosynthesis, in which
organisms, using the information in DNA, string together amino
acids to form proteins 281

2. To examine the genetic code 304

STUDY OUTLINE

Information Transfer 281

Transfer RNA 281

Initiation Complex 288

Elongation 292

Termination 296

More on the Ribosome

In the last chapter, we briefly discussed the shape and
composition of the ribosomal subunits. All of the protein
and RNA components have been isolated. Assembly path-
ways are known. We know approximately where the mes-
senger RNA, initiation factors, and EF-Tu are located on the
30S subunit during translation (fig. 1 1 .24; cf. fig. 1 1 . 18).We
also know where peptidyl transferase activity and EF-G
reside on the 5 OS subunit, which has a cleft leading into
a tunnel that passes through the structure. At present, it
seems that the nascent peptide passes through this

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Translation

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300

Chapter Eleven Gene Expression: Translation

•••mRNA

30S

Figure 11.20 Chain termination at the ribosome. One of two
release factors recognizes a nonsense codon in the A site. In
this case, RF1 recognizes UAG. The complex then falls apart,
releasing the peptide.

iiftr^' ■■' - ^ - ^ ^DNA ^ -'V

HESE;"'-' :.■•■"■■■ -.■ ■■St. *•■■•■ ■-.-#- - ■■.■.,■■■-".,-■■• >v r ■-' '

■ Ribb^ome ^

W;V^Su

&is

. '. > :-

',;■■':/ v,. - v : .-.■?:i,..iV.:v- •'■• ■■■,■■<'."■'.'•■■'■ : /■ .■■■

50S^

<■■*■- ■.■■;-<■-::> T ■':■■;: -vv' fr*&vT?tV£< V'-- tfW&v ■

s; = .-: /^' r :V'lv' 'mRNA

t* -V

■

Figure 11.21 A polysome (i.e., multiple ribosomes on the
same strand of mRNA). Each ribosome is approximately 250 A
units across. Also visible in this illustration are DNA and RNA
polymerase. (Reproduced courtesy of Dr. Barbara Hamkalo, International
Review of Cytology, (1972) 33:7, fig. 5. Copyright by Academic Press, Inc.,
Orlando, Florida.)

Ribosomes

Nascent protein

mRNA
5'

Nascent protein

(a)

Figure 11.22 (a) Protein synthesis at a polysome. Nascent
proteins exit from a tunnel in the 50S subunit. Messenger RNA
is being translated by the ribosomes while the DNA is being
transcribed, (b) A messenger RNA from the midge, Chironomus
tentans, showing attached ribosomes and nascent polypeptides
emerging from the ribosomes. Note the 5' end of the
messenger RNA at the upper right (small peptides).
Magnification 165,000x. ([£>] Courtesy of S. L McKnight and
O. L Miller, Jr.)

■ .

I."**

' * ' - n

* >$"■ '-■:

(b)

Nascent polypeptide
(early)

mRNA
Ribosome

Nascent polypeptide
(late)

Gene 1

Gene 2

fMet •••Terminator

Gene 3

AGGAGGU...AUG...UGA...AGGAGGU...AUG...UGA ... AGGAGGU... AUG...UGA

Shine-Dalgarno
sequences

Figure 11.23 A prokaryotic polycistronic mRNA. Note the several Shine-Dalgarno sequences for ribosomal
attachment and the initiation and termination codons marking each gene.

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Translation

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Information Transfer

301

tunnel, emerging close to a membrane-binding site
(fig. 1 1.24).The tunnel can hold a peptide length of about
forty amino acids. Note that although every ribosome has
a membrane-binding site, not all active ribosomes are
bound to membranes.

The Signal Hypothesis

Ribosomes are either free in the cytoplasm or associated
with membranes, depending on the type of protein being
synthesized. Membrane-bound ribosomes, indistinguish-
able from free ribosomes, synthesize proteins that enter
membranes. These proteins either become a part of the
membrane or, in eukaryotes, either pass into membrane-
bound organelles (e.g., the Golgi apparatus, mitochon-
dria, chloroplasts, vacuoles) or are transported outside

Gunter Blobel (1936- ).
(Courtesy of Dr. Gunter
Blobel, Dept. of Cell Biology,
Rockefeller University.)

mRNA
5'

3' end

16S

rRNA

30S

Peptidyl
transferase

EF-Tu

Membrane-
binding site

EF-G

50S

Tunnel

Exit hole for
polypeptide

Messenger RNA EF-G

70S

Nascent
protein

Figure 11.24 Functional sites on the prokaryotic ribosome.
The ribosome is synthesizing a protein involved in membrane
passage. Note the position of the messenger RNA on the 30S
subunit and the cleft, tunnel, and membrane-binding site on
the 50S subunit. (From C. Bemabeu and J. A. Lake, Proceedings of
the The National Academy of Sciences; 79:3111-15, 1982. Reprinted by
permission.)

the cell membrane. The signal hypothesis of G. Blobel,
a 1999 Nobel laureate, and his colleagues, explains the
mechanism for membrane attachment. The mechanism
applies to both prokaryotes and eukaryotes. Here, we de-
scribe it in mammals.

The signal for membrane insertion is coded into the
first one to three dozen amino acids of membrane-
bound proteins. This signal peptide takes part in a
chain of events that leads the ribosome to attach to the
membrane and to the insertion of the protein. The first
step occurs when the signal peptide becomes accessi-
ble outside of the ribosome. A ribonucleoprotein parti-
cle called the signal recognition particle (SRP),
which consists of six different proteins and a 7S RNA
about three hundred nucleotides long, recognizes the
signal peptide. The complex of signal recognition
particle, ribosome, and signal peptide then passes, or
diffuses, to a membrane, where the SRP binds to a re-
ceptor called a docking protein (DP) or signal recog-
nition particle receptor (fig. 11.25). During this time,
protein synthesis halts. The ribosome is brought into
direct contact with the membrane, and other proteins
of the membrane help anchor the ribosome. Protein
synthesis then resumes, with the nascent protein
usually passing directly into a translocation channel
(translocon). Once through the membrane, the signal
peptide is cleaved from the protein by an enzyme
called signal peptidase. A striking verification of this
hypothesis came about through recombinant DNA
techniques (chapter 13). A signal sequence was placed
in front of the a-globin gene, whose protein product is
normally not transported through a membrane. When
this gene was translated, the ribosome became mem-
brane bound, and the protein passed through the
membrane.

Since different proteins enter different membrane-
bound compartments (e.g., the Golgi apparatus), some
mechanism must direct a nascent protein to its proper

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Chapter Eleven Gene Expression: Translation

Ribosome

GTP

Signal recognition
particle

Signal
peptide

(a)

(b)

GDP + P :

Signal peptidase

Translocon

(c)

Membrane

Docking
protein

(f)

Figure 11.25 The signal hypothesis. A signal recognition particle recognizes a ribosome with a signal peptide, then
draws the ribosome to a docking protein located near a translocon in the membrane. With the addition of GTP, the
signal recognition particle releases the signal peptide; hydrolysis of the GTP to GDP + P, causes the signal
recognition particle to leave the docking protein. Peptide synthesis then resumes, with the newly synthesized peptide
passing through the translocon in the membrane. A signal peptidase on the other side of the membrane removes the
signal peptide. When translation is completed, the ribosome dissociates and drops free of the translocon.

membrane. This specificity seems to depend on the ex-
act signal sequence and membrane-bound glycopro-
teins called signal-sequence receptors. Apparently after
the ribosome binds to the docking protein, the signal
peptide interacts with a signal-sequence receptor,
which presumably determines whether that protein is
specific for that membrane. If it is, the remaining
processes continue. If not, the ribosome may be re-
leased from the membrane.

The signal peptide does not seem to have a consen-
sus sequence like the transcription or translation recog-
nition boxes. Rather, similarities (at least for the endo-
plasmic reticulum and bacterial membrane-bound
proteins) include a positively charged (basic) amino acid
(commonly lysine or arginine) near the beginning

(N-terminal end), followed by about a dozen hydrophobic
(nonpolar) amino acids, commonly alanine, isoleucine,
leucine, phenylalanine, and valine (table 11.2).

Table 1 1 ,2 The Signal Peptide of the Bovine

Prolactin Protein*

NH 2 - Met Asp Ser Lys Gly Ser Ser Gin Lys Gly Ser Arg Leu
Leu Leu Leu Leu Val Val Ser Asn Leu Leu Leu Cys Gin Gly Val
Val Ser | Thr Pro Val.. Asn Asn Cys - COOH

Source: From Sasavage et al., Journal of Biological Chemistry, 257:678-81,
1982. Reprinted with permission.

* The vertical line separates the signal peptide from the rest of the protein,
which consists of 199 residues.

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Information Transfer

303

Figure 11.26 Electron micrograph of a chaperone protein
(GroEL) from E. coli. Note the hollow, barrel shape of the
protein. (Courtesy of Dr. R. W. Hendrix.)

The mitochondrion, which needs to import upwards
of one thousand proteins through both inner and outer
membranes, poses a specific problem. Recent research
has revealed a family of translocation proteins (called Tom
proteins) in the outer membrane and a different set of
translocation proteins (called Tim proteins) in the inner
membrane .These proteins control the passage of proteins
synthesized in the cytoplasm into the mitochondrion.

The Protein-Folding Problem

Since biochemist Christian Anfinsen won a 1972 Nobel
Prize for showing that the enzyme ribonuclease refolds
to its original shape after denaturation in vitro, scientists
have believed that the final protein shape (secondary and
tertiary structure) forms spontaneously. Recently it has
been shown, however, that many proteins do not nor-
mally form their final active shape in vivo without the
help of proteins called chaperones or molecular
chaperones. The chaperones do not provide the three-
dimensional structure of the proteins they help, but
rather bind to a protein in the early stages of folding to
prevent unproductive folding or to allow denatured pro-
teins to refold correctly. Like human chaperones, they
prevent or undo "incorrect interactions," according to
J. Ellis. That is, many proteins have a large number of
different structures they could fold into. Many of these
structures would have no enzymatic activity or would
form functionless aggregates with other proteins. Molec-
ular chaperones allow proteins to fold into a thermody-
namically stable and functional configuration. Each cycle
of refolding requires ATP energy.

A well-studied class of chaperones is known as the
chaperonins, or Hsp60 proteins, because they are heat
shock proteins about 60 kilodaltons (60,000 daltons) in
size. They occur in bacteria, chloroplasts, and mitochon-
dria. One of the best studied of these chaperonins is the
protein GroE of E. coli. This protein in its active form is
composed of two components, GroEL and GroES. GroEL
(Hsp60) is made up of two disks, each composed of
seven copies of a polypeptide. GroES (HsplO) is a smaller
component composed of seven copies of a small sub-
unit. GroEL forms a barrel in which protein folding takes
place (fig. 11. 26). The barrel is shaped in such a way that
entering proteins of a certain size make contact at inte-
rior points in either the upper or lower ring of GroEL
(upper ring shown in fig. 11.27). The attachment of
GroES, the cap, causes the ring to open outward at the
top, stretching the protein inside. This stretching takes
energy from the hydrolysis of ATP molecules located in-
side the rings. When GroES dissociates, the protein can
fold into a new, more functional, configuration. If it
doesn't, the cycle repeats.

There are several classes of molecular chaperones,
proteins of different sizes and shapes that recognize dif-
ferent groups of proteins or protein conformations.
GroEL recognizes about 300 different proteins, small
enough to fit into the barrel (20-60 kilodaltons) and hav-
ing hydrophobic surfaces. These include many proteins
in the transcription and translation machinery of the cell.
Hsp90, another heat shock protein, recognizes proteins
involved in signal transduction, discussed in chapter 16.
Hsp70 recognizes hydrophobic regions in polypeptide
side chains, many of which extend across membranes.

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Chapter Eleven Gene Expression: Translation

(a)

Folded
polypeptide

Unfolded
polypeptide

(c)

(b)

Figure 11.27 The change in structure of GroEL with GroES
attached explains how the chaperonin can unfold a partially
folded polypeptide to allow it to refold in a different way. (a) A
space-filling model of GroEL is shown without (left) and with
(right) GroES. GroEL's rings are blue and magenta, and GroES
is green, (b) The same structures are seen in a cutaway view,
(c) This diagram shows how the attachment of GroES causes
the top part of the top ring of GroEL to pull apart an
improperly folded polypeptide, ([a & b]\ Reprinted from Bernd Bakau
and Arthur L. Horwich, "The Hsp70 and Hsp60 Chaperone Machines" in Cell,
vol. 92, 351-366. Copyright 1988, with permission from Elsevier Science.)

THE GENETIC CODE

Researchers in the mid-1950s assumed that the genetic
code consisted of simple sequences of nucleotides speci-
fying particular amino acids .They sought answers to ques-
tions such as: Is the code overlapping? Are there nu-
cleotides between code words (punctuation)? How many
letters make up a code word (codon)? Logic, along with
genetic experiments, supplied some of the answers, but
only with the rapidly improving techniques of biochem-
istry did they eventually decode the genetic language.

Triplet Nature of the Code

Several lines of evidence seemed to indicate that the na-
ture of the code was triplet (three bases in messenger
RNA specifying one amino acid). If codons contained only
one base, they would only be able to specify four amino
acids since there are only four different bases in DNA (or
messenger RNA). A couplet code would have 4X4 = 16
two-base words, or codons, which is still not enough to
specify uniquely twenty different amino acids. A triplet

code would allow for 4 X 4 X 4 = 64 codons, which are
more than enough to specify twenty amino acids.

Evidence for the Triplet Nature of the Code

The experimental manipulation of mutant genes, pri-
marily by Francis Crick and his colleagues, reinforced the
triplet code concept. In these experiments, a chemical
mutagen, the acridine dye proflavin, was used to cause in-
activation of the rapid lysis (rllB) gene of the bacterio-
phage T4. Proflavin inactivates the gene by either adding
or deleting a nucleotide from the DNA (see chapter 12).
The rll gene controls the plaque morphology of this bac-
teriophage growing on E. colt cells. Rapid-lysis mutants
produce large plaques; the wild-type form of the gene,
rll + , results in normal plaque morphology.

Figure 11.28 shows the consequences of adding or
deleting a nucleotide. From the point of addition or dele-
tion onward, a frameshift causes codons to be read in
different groups of three. If a deletion is combined with
an addition to produce a double-mutant gene, the
frameshift occurs only in the region between the two mu-

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Translation

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The Genetic Code

305

tants. If this region is small enough or does not contain
coding for vital amino acids, the function of the gene may
be restored. Two deletions or two insertions combined
will not restore the reading frame. However, Crick and his
colleagues found that the combination of three additions
or three deletions did restore gene function. This finding
led to the conclusion that the genetic code was triplet, be-

Normal
(CAG repeat)

CAGCAGCAGCAGCAG

First A deleted
(AGC repeat)

C*G CAGCAGCAGCAGC

mRNA

• • •

Frameshift
A inserted after

third A of normal

(GCA repeat)

CAGCAGCAAGCAGCA

mRNA

Frameshift
Deletion and

insertion combined

(return to CAG repeat)

C*G CAGCAAGCAGCAG

mRNA

Frameshift

Restoration

Figure 11.28 Frameshift mutations in a gene result from the
addition or deletion of one or several nucleotides (any number
other than a multiple of three) in the DNA. The messenger RNA
shown here normally has a CAG repeat. A single-base deletion
shifts the three-base reading frame to a series of AGC repeats.
A later insertion restores the reading frame. Asterisks (*)
indicate points of deletion or insertion.

cause a triplet code would be put back into the reading
frame by three additions or three deletions (fig. 1 1.29).

Overlap and Punctuation in the Code

Questions still remained: was the code overlapping? Did it
have punctuation? Several logical arguments favored a no-
punctuation, nonoverlapping model (fig. 11.30). An over-
lapping code would be subject to two restrictions. First, a
change in one base (a mutation) could affect more than
one codon and thus affect more than one amino acid. But
studies of amino acid sequences almost always showed
that only one amino acid was changed, which argued
against codon overlap. Second, certain restrictions affected
which amino acids occurred next to each other in pro-
teins. For example, the amino acid UUU coded could never
be adjacent to the amino acid coded by AAA because one
or both (depending on the number of bases overlapped) of
the overlap codons UUA and UAA would always insert
other amino acids between them. Overlap, then, seemed to
be ruled out since every amino acid appears next to every
other amino acid in one protein or another.

Punctuation between codons was also tentatively ruled
out. The messenger RNA in the tobacco necrosis satellite
virus has just about enough codons to specify its coat pro-
tein with no room left for a punctuating base or bases be-
tween each codon.

Breaking the Code

Once geneticists had figured out that the genetic code is
in nonoverlapping triplets, they turned their attention to
the sixty-four codons. They wondered which amino acid,
for example, does ACU specify? The work was done in
two stages. In the first stage, M.W. Nirenberg, S. Ochoa,
and their colleagues made long artificial messenger RNAs
and determined which amino acids these messenger
RNAs incorporated into protein. In the second stage, spe-
cific triplet RNA sequences were synthesized. The re-
searchers then determined the amino acid-transfer RNA
complex that was bound by each sequence.

Original mRNA
(CAG repeat)

CAGCAGCAGCAGCAGCAG

mRNA

Three inserted
Gs restore CAG
repeat in mRNA

Frameshift

Restoration

mRNA

Figure 11.29 The coding frame of CAG repeats is first shifted and then restored by three additions
(insertions). Asterisks (*) indicate insertions.

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Chapter Eleven Gene Expression: Translation

Severo Ochoa (1905-1993).
(Courtesy of Dr. Severo Ochoa.)

Marshall W. Nirenberg
(1934- ). (Courtesy of Dr.
Marshall W. Nirenberg.)

Synthetic Messenger RNAs

The ability to synthesize long-chain messenger RNAs re-
sulted from the 1955 discovery of M. Grunberg-Manago
and Ochoa of the enzyme polynucleotide phosphory-
lase, which joins diphosphate nucleotides into long-
chain, single-stranded polynucleotides. Unlike a poly-
merase, polynucleotide phosphorylase does not need a
primer on which to act. This enzyme is found in all bacte-
ria. (Its main function in the cell is probably the reverse of
its use here. It most likely serves as an exonuclease, de-
grading messenger RNA.) In 1961, Nirenberg and J. H.
Matthei added artificially formed RNA polynucleotides of
known composition to an E. colt ribosomal system and
looked for the incorporation of amino acids into proteins.

The system just described is called a cell-free system,
a mixture primarily of the cytoplasmic components of
cells, such as E. colt, but missing nucleic acids and mem-
brane components.These systems are relatively easy to cre-
ate by disrupting and then fractionating whole cells. The
systems hold the advantage of containing virtually all the
components needed for protein synthesis except the mes-
senger RNAs .Their disadvantages are that they are relatively
short-lived (several hours) and are relatively inefficient in
translation. However, an added benefit to the E. colt cell-free
system is that it will translate, albeit inefficiently, RNAs that
normally are not translated in vivo because they lack trans-
lation initiation signals .This feature allowed these scientists
to use artificial messenger RNAs that contained no Shine-
Dalgarno sequence for ribosomal binding.

Nirenberg and Matthei found that when the enzyme
polynucleotide phosphorylase in the E. colt cell-free sys-
tem made uridine diphosphates into a poly-U messenger
RNA, phenylalanine residues were incorporated into a
polypeptide. Thus, the first code word established was
UUU for phenylalanine. Nirenberg and Ochoa and their
associates continued the work. They found that AAA was
the code word for lysine, CCC was the code word for pro-
line, and GGG was the code word for glycine.

They then made synthetic messenger RNAs by using
mixtures of the various diphosphate nucleotides in
known proportions. Table 11.3 gives an example. From
their experiments, it was possible to determine the bases
used in many of the code words, but not their specific or-
der. For example, cysteine, leucine, and valine are all
coded by two Us and a G, but the experiment could not
sort out the order of these bases (5'-UUG-3' 3 5'-UGU-3',
or 5-GUU-3) for any one of them. Determining the order
required an extra step in sophistication — that is, being
able to synthesize known trinucleotides.

Synthetic Codons

Once trinucleotides of known composition could be
manufactured, Nirenberg and P. Leder in 1964 developed
a "binding assay." They found that isolated E. colt ribo-
somes, in the presence of high-molarity magnesium chlo-

Nonoverlapping,
no punctuation

codon

CAGCAGCAGCAG

• • •

Overlapping, one
base no punctuation

codon 1

codon 2

codon 3

codon 4

codon 5

codon 6

G C

Overlapping,
two bases no
punctuation

codon 1

codon 2

codon 3

codon 4

codon 5

codon 6

codon 7

codon 8

codon 9

codon 10

• • •

Nonoverlapping,
punctuation

codon

Punctuation
CAGCAGCAGCAG

• • •

Figure 11.30 The genetic code is read as a nonoverlapping
code with no punctuation {top). Before that was proven, it was
suggested that the code could overlap by one or two bases
{middle) or have noncoded bases (punctuation) between code
words {bottom).

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Table 1 1 .3

Structure of Artificial mRNA Made by
Randomly Assembling Uracil- and
Guanine-Containing Ribose
Diphosphate Nucleotides with a
Ratio of 5U:1G

Codon

Frequency of Occurrence

uuu

(5/6) 3 = 0.58

UUG

(5/6) 2 (l/6) - 0.12

UGU

(5/6) 2 (l/6) = 0.12

GUU

(5/6) 2 (l/6) = 0.12

UGG

(5/6)(l/6) 2 = 0.02

GUG

(5/6)(l/6) 2 = 0.02

GGU

(5/6)(l/6) 2 = 0.02

GGG

(1/6) 3 = 0.005

ride, could bind trinucleotides as if they were messenger
RNAs.Also bound was the transfer RNA that carried the
anticodon complementary to the trinucleotide. It was
thus possible, using radioactive amino acids, to deter-
mine which messenger RNA trinucleotide coded for a
particular amino acid. A given synthetic trinucleotide was
mixed with ribosomes and aminoacyl-tRNAs, including
one radioactively labeled amino acid. The reaction mix-
ture was passed over a filter that would allow everything
except the large trinucleotide + ribosome + aminoacyl-
tRNA complex to pass through. If the radioactivity passed
through the filter, it meant that the radioactive amino
acid was not associated with the ribosome. The experi-
ment was then repeated with another labeled amino
acid. When the radioactivity appeared on the filter, the in-
vestigators knew that the amino acid was affiliated with
the ribosome. Thus, that amino acid was coded by the se-
lected trinucleotide codon. In other words, the radioac-
tive amino acid was attached to a transfer RNA whose an-
ticodon was complementary to the trinucleotide codon
and thus bound at the ribosome.

Figure 11.31 shows an example. In the figure, the tri-
nucleotide is 5'-CUG-3'The transfer RNA with the anti-

Phillip Leder (1934- ).
(Courtesy of Dr. Phillip Leder.)

codon 3 -GAC-5' is charged with leucine. The mixture is
passed through a filter. If threonine, or any other amino
acid except leucine, is radioactive, the radioactivity
passes through the filter. When the experiment is re-
peated with radioactive leucine, the leucine, and hence
the radioactivity, is trapped by the filter. In a short period
of time, all of the codons were deciphered (table 11.4).

Wobble Hypothesis

The genetic code is a degenerate code, meaning that a
given amino acid may have more than one codon. As you
can see from table 11.4, eight of the sixteen boxes con-
tain just one amino acid per box. (A box is determined by
the first and second positions; e.g., the UUX box, in
which X is any of the four bases.) Therefore, for these
eight amino acids, the codon need only be read in the
first two positions because the same amino acid will be
represented regardless of the third base of the codon.
These eight groups of codons are termed unmixed fam-
ilies of codons. An unmixed family is the four codons be-
ginning with the same two bases that specify a single
amino acid. For example, the codon family GUX codes for
valine. Mixed families code for two amino acids or for
stop signals and one or two amino acids.

Six of the mixed-family boxes are split in half so that
the codons are differentiated by the presence of a purine
or a pyrimidine in the third base. For example, CAU and

UGU

o ^

■C?.^

C3-

Ribosome

Trinucleotide

Passes
through
i filter v

I Blocked by filter

Cellulose

nitrate

filter

^ ^

Figure 11.31 The binding assay determines the amino acid
associated with a given trinucleotide codon. Transfer RNAs with
noncomplementary codons pass through the membrane.
Transfer RNAs with anticodons complementary to the
trinucleotide bind to the ribosome and do not pass through the
filter. When the transfer RNA is charged with a radioactive
amino acid, the radioactivity is trapped on the filter.

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Chapter Eleven Gene Expression: Translation

Table 1 1 .4 The Genetic Code

First Position (5' End)

Second Position

Third Position (3'
U

End)

C A

Phe

Ser

Tyr

Cys

Phe

Ser

Tyr

Cys

Leu

Ser

stop

Leu

Ser

stop

Trp

G
U

Leu

Pro

His

Arg

Leu

Pro

His

Arg

Leu

Pro

Gin

Arg

Leu

Pro

Gin

Arg

G
U

lie

Thr

Asn

Ser

Thr

Asn

Ser

Thr

Lys

Arg

Met (start)

Thr

Lys

Arg

G
U

Val

Ala

Asp

Gly

Val

Ala

Asp

Gly

Val

Ala

Glu

Gly

Val

Ala

Glu

Gly

CAC both code for histidine; in both, the third base, U
(uracil) or C (cytosine), is a pyrimidine. Only two of the
families of codons are split differently.

The lesser importance of the third position in the ge-
netic code ties in with two facts about transfer RNAs. First,
although there would seem to be a need for sixty-two trans-
fer RNAs — since there are sixty-one codons specifying
amino acids and an additional codon for initiation — there
are actually only about fifty different transfer RNAs in an E.
colt cell. Second, a rare base such as inosine can appear in
the anticodon, usually in the position that is complementary
to the third position of the codon. These two facts lead re-
searchers to believe that some kind of conservation of trans-
fer RNAs is occurring and that rare bases may be involved.

We should mention, to avoid confusion, that both
messenger RNA and transfer RNA bases are usually num-
bered from the 5' side. Thus, the number-one base of the
codon is complementary to the number-three base of the
anticodon (fig. 11. 32). Thus, the codon base of lesser im-
portance is the number-three base, whereas its comple-
ment in the anticodon is the number-one base.

Since the first position of the anticodon (5') is not as
constrained as the other two positions, a given base at that
position may be able to pair with any of several bases in the

third position of the codon. Crick characterized this ability
as wobble (fig. 11.33)Table 11.5 shows the possible pair-
ings that would produce a transfer RNA system compatible
with the known code. For example, if an isoleucine transfer
RNA has the anticodon 3-UAI-5', it is compatible with the
three codons for that amino acid (see table 11.4): 5'-AUU-
3', 5'-AUC-3', and 5'-AUA-3'.That is, inosine in the first (5')
position of the anticodon can recognize U, C, or A in the
third (3') position of the codon, and thus one transfer RNA
complements all three codons for isoleucine.

Universality of the Genetic Code

Until 1979, scientists concluded that the genetic code
was universal. That is, the codon dictionary (see
table 11.4) was the same for E. colt, human beings, and
oak trees, as well as all other species studied up to that
time. The universality of the code was demonstrated, for
example, by taking the ribosomes and messenger RNA
from rabbit reticulocytes and mixing them with the
aminoacyl-tRNAs and other translational components of
E. coll Rabbit hemoglobin was synthesized.

In 1979 and 1980, however, researchers noted discrep-
ancies when sequencing mitochondrial genes for struc-

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(Met J
3'

tRNA

Anticodon

1 2 3

3' mRNA

Codo

Figure 11.32 Codon and anticodon base positions are
numbered from the 5' end. The 3' position in the codon (5' in
the anticodon) is the wobble base.

tural proteins (see chapters 13 and 17). It was discovered
that there were two kinds of deviations from universality in
the way mitochondrial transfer RNAs read the code. First,
fewer transfer RNAs were needed to read the code. Second,
there were several instances in which the mitochondrial
and cellular systems interpreted a codon differently.

According to Crick's wobble rules (see table 11.5),
thirty-two transfer RNAs (including one for initiation)
can complement all sixty-one nonterminating codons.
Unmixed families require two transfer RNAs, and mixed
families require one, two, or three transfer RNAs, depend-
ing on the family. The yeast mitochondrial coding system
apparently needs only twenty-four transfer RNAs. The re-
duction in numbers is accomplished primarily by having
only one transfer RNA recognize each unmixed family

Table 1 1 .5 Pairing Combinations at the Third

Codon Position

Number-one Base

Number-three Base in

tRNA (5' End)

mRNA (3' End)

U or C

A or G

A, U, or C

(table 1 1 .6; cf. table 11.4). Because mitochondrial transfer
RNAs for unmixed families of codons have a U in the first
(wobble) position of the anticodon, apparently, given the
structure of the mitochondrial transfer RNAs, the U can
pair with U, C,A, or G. Presumably, evolutionary pressure
has minimized the number of transfer RNA genes in the
DNA of the mitochondrion, in keeping with its small size.
Reduction from thirty-two to twenty-four is a 25% sav-
ings. (Recent evidence suggests that mammalian mito-
chondria may need only twenty-two transfer RNAs.)

It has also been found that yeast mitochondria read
the CUX family as threonine rather than as leucine (ta-
bles 11.4 and 11.6) and the terminator UGA (opal) as
tryptophan rather than as termination. However, there
appear to be differences among different groups of or-
ganisms reading the CUX family. Human and Neurospora
mitochondria appear to read the CUX codons as leucine,
just as cellular systems do. Of the groups so far analyzed,
only yeast reads the CUX family as threonine. Similarly,
human and Drosophila mitochondria read AGA and AGG
as stop signals rather than as arginine (table 11.7).

In 1985, it was discovered that Paramecium species
read the UAA and UAG stop codons as glutamine within
the cell. In addition, a prokaryote {Mycoplasma capri-
colum) reads UGA as tryptophan. We do not yet know
how general this finding is: scientists have scrutinized the
genetic code of very few species. We can thus conclude
that the genetic code seems to have universal tendencies
among prokaryotes, eukaryotes, and viruses. Mitochon-
dria, however, read the code slightly differently: different
wobble rules apply, and mitochondria and cells read at
least one terminator and one unmixed family of codons
differently. Also, the mitochondrial discrepancies are not
universal among all types of mitochondria. Further work,
involving the sequencing of more mitochondrial DNAs,
should elucidate the pattern of discrepancies among the
mitochondria of diverse species. We also now know that
not every organism reads all codons in the same way. Cil-
iated protozoa and a mycoplasma read some stop signals
as coding for amino acids. Nuclear variants are known in
the following codons: CUG, AUA, UAA, UAG, UGA, CGG,
and AGA. Mitochondrial variants are known in CUX, AUA,
UAA, UAG, AAA, UGA, CGX,AGA, and AGG.

One other type of variation of codon reading occurs:
site-speciflc variation, in which the interpretation of a
codon depends on its specific location. We are already fa-
miliar with the fact that GUG and, rarely, UUG can serve as
prokaryotic initiation codons. This means that they are rec-
ognized by tRNA^ et . However, they are not recognized by
tRNA^ et (i.e., GUG and UUG are not misread internally in
messenger RNAs). In some cases, two of the termination
codons (UGA and UAG, but not UAA) are misinterpreted as
codons for amino acids. That is, termination will not occur
at the normal place, resulting in a longer-than-usual protein.
In some cases, these "read-through" proteins are vital — the

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Chapter Eleven Gene Expression: Translation

Guanine

H /-\ C ^^ CH

To ribose sugar

•C N

N N

C
O

/N\

Cytosine

To ribose
sugar

H
X ^CH

Uracil

Guanine

To ribose sugar

Figure 11.33 Base-pairing possibilities for guanine and inosine
in the third (3') position of a codon. In the wobble position,
guanine can form base pairs with both cytosine and uracil.
Inosine, in the wobble position, can pair with cytosine, adenine,
and uracil.

H
,C

Inosine

H ^\ c ^ ^ CH

-C.

r
Z

-N'

Cytosine

C
O

To ribose
sugar

To ribose sugar

H H

Inosine

// C N

-N

>N Adenine

To ribose
sugar

N-

V N

To ribose sugar

H
X ^CH

Uracil

Inosine

y
y

To ribose
sugar

J
\

^C 1ST

^-C ^CH
1ST

y(j

To ribose

sugar

organism depends on their existence. For example, in the
phage QP, the coat-protein gene is read through about 2%
of the time. Without this small number of read-through pro-
teins, the phage coat cannot be constructed properly.

One last example of site-specific variation involves
the amino acid selenocysteine (cysteine with a selenium
atom replacing the sulfur; see fig. 11.1). Although many
proteins have unusual amino acids, almost all are due to
posttranslational modifications of normal amino acids.

However, the amino acid selenocysteine is inserted di-
rectly into some proteins, such as formate dehydrogenase
in E. colt, which has selenium in its active site. Selenocys-
teine is inserted into the protein by a novel transfer RNA
that recognizes the termination codon, UGA, if that
codon is involved in a particular stem-loop secondary
structure in the messenger RNA. The selenocysteine
transfer RNA is originally charged with a serine that is
then modified to a selenocysteine. In addition to the

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The Genetic Code

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Table 1 1 .6 The Genetic Code Dictionary of Yeast Mitochondria*

First Position (5'

End)
U

Second Position

Third Position (3'

U
C
A
G
U
C
A
G
U
C
A
G
U
C
A
G

End)

C A

PheAAG

Ser AGU

Tyr AUG

Cys ACG

LeuAAU

stop

Trp ACU

Thr GAU

Pro GGU

His GUG

Arg GCA

Gin GUU

Ik UAG

Thr UGU

Asn UUG

Ser UCG

Met UAC

Lys UUU

Arg UCU

Val CAU

Ala CGU

Asp CUG

Gly CCU

Glu CUU

Source: Data from S. Bonitz, et al., "Codon recognition rules in yeast mitochondria," Proceedings of the National Academy of Sciences 77:3167-70, 1980.
* Anticodons (3' — > 5') are given within boxes. (The ACU Trp anticodon is predicted.)

Table 1 1 .7 Common and Alternative Meanings of Codons

Codon

General Meaning

Alternative Meaning

cux

Leu

Thr in yeast mitochondria

AUA

Met in mitochondria of yeast, Drosophila, and vertebrates

UGA

Stop

Trp in mycoplasmas and mitochondria other than higher plants

AGA/AGG

Arg

Stop in mitochondria of yeast and vertebrates
Ser in mitochondria of Drosophila

CGG

Arg

Trp in mitochondria of higher plants

UAA/UAG

Stop

Gin in ciliated protozoa

UAG

Stop

Ala or Leu in mitochondria of some higher plants

stem-loop structure 3' (downstream) from the amber
codon (UAG), a selenocysteine elongation factor (SELB) is
also needed at the ribosome.This same mechanism may
occur in eukaryotes, but not all of the components have
yet been identified.

Evolution of the Genetic Code

It has been theorized that the genetic code has wobble in
it because it originally arose from a code in which only
the first two bases were needed for the small number of
amino acids in use several billion years ago. As new
amino acids with useful properties became available,
they were incorporated into proteins by a code modified
by the third base, albeit with less specificity.This view has
support from the fact that codons starting with the same

nucleotide come from the same biosynthetic pathway.
This indicates that in early evolution, as biosynthetic
pathways were extended to new amino acids, the new-
comers were incorporated by use of the second and
third bases of the code.

However, the question remains as to whether the ge-
netic code is highly evolved or just a "frozen accident." In
other words, is there a relationship between the codons
and the amino acids they code for, or is the code just one
of many random possibilities? Recent computer simula-
tions of random codes indicate that the current genetic
code is far outside the range of random in its ability to
protect the organism from mutation. This suggests that
the genetic code is not a frozen accident, but rather is
highly evolved. Numerous examples in the current code
support this view.

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Chapter Eleven Gene Expression: Translation

For example, in the unmixed codon family 5'-CUX-3' ,
any mutation in the third position produces another
codon for the same amino acid. Wobble in the third
position and codon arrangement ensures that less than
half of the mutations in the third codon position result in
the specification of a different amino acid.

There are also patterns in the genetic code in which
the mutation of one codon to another results in an amino
acid of similar properties. A high probability exists that
such a mutation will produce a functional protein.All the
codons with U as the middle base, for example, are for
amino acids that are hydrophobic (phenylalanine,
leucine, isoleucine, methionine, and valine). Mutation in

the first or third positions for any of these codons still
codes a hydrophobic amino acid. Both of the two nega-
tively charged amino acids, aspartic acid and glutamic
acid, have codons that start with GA. All of the aromatic
amino acids — phenylalanine, tyrosine, and tryptophan
(see fig. 11.1) — have codons that begin with uracil. Such
patterns minimize the negative effects of mutation.

This chapter completes the discussion of the me-
chanics of gene expression. The next chapter deals with
recombinant DNA technology, followed by several chap-
ters concerned with the control of gene expression in
both prokaryotes and eukaryotes.

SUMMARY

STUDY OBJECTIVE l:To study the mechanism of protein
biosynthesis, in which organisms, using the information
in DNA, string together amino acids to form proteins
281-303

A charged transfer RNA has an anticodon at one end and a
specific amino acid at the other end. The transfer RNAs are
charged with the proper amino acid by aminoacyl-tRNA
synthetase enzymes that incorporate the energy of ATP into
amino acid-tRNA bonds. Hence, no additional source of en-
ergy is needed during peptide bond formation. During pro-
tein synthesis, the translation apparatus at the ribosome
recognizes the transfer RNA. Through complementarity, the
anticodon pairs with a messenger RNA codon.

An initiation complex forms at the start of translation. In
prokaryotes, this complex consists of the messenger RNA, the
3 OS subunit of the ribosome, the initiator transfer RNA with
N-formyl methionine (fMet-tRNA f Met ), and the initiation fac-
tors IF1 , IF2, and IF3The 50S ribosomal subunit is then added
and A and P sites form in the resulting 70S ribosome. The
charged N-formyl methionine transfer RNA is in the P site.
A GTP is hydrolyzed, and the initiation factors are released.

A transfer RNA enters the A site, which requires the in-
volvement of elongation factors EF-Ts and EF-Tu (in E. colt).
At least one GTP hydrolysis releases the elongation factor, EF-
Tu, which had originally brought the charged transfer RNA to
the ribosome. Peptidyl transferase, which appears to be a ri-
bozymic component of the 50S ribosomal subunit, transfers
the amino acid from the transfer RNA in the P site to the
amino end of the amino acid on the transfer RNA in the A site.

With the help of elongation factor G (EF-G), the ribo-
some translocates in relation to the messenger RNA.The de-
pleted transfer RNA is moved from the P site to the E site,
where it is released; the transfer RNA with the growing pep-
tide is moved into the P site. EF-G is then released. Elonga-
tion and translocation continue until a nonsense codon en-
ters the A site. With the aid of the release factors RF1 and

RF2, the protein is released, and the messenger RNA-
ribosome complex dissociates. Eukaryotes have slightly
more complex processes involving several more proteins.

Proteins pass through membranes with the help of a sig-
nal peptide synthesized at their N-terminal ends. Proteins fold
into their final, functional configurations with the help of mo-
lecular chaperones, proteins that aid the folding process.

Many antibiotics interfere with translation in prokary-
otes. Puromycin, streptomycin, tetracycline, and chloram-
phenicol all act at the ribosome. Studying the mode of ac-
tion of these antibiotics has provided insights into the
mechanism of the translation process.

STUDY OBJECTIVE 2: To examine the genetic code
304-312

The genetic code was first assumed to be triplet because of
logical arguments regarding the minimum size of codons.
With his work on deletion and insertion mutants, Crick pro-
vided evidence that the code was triplet. Part of the code
was worked out initially with the synthesis of long, artificial
messenger RNAs and then the synthesis of specific trinu-
cleotide codons. Crick's wobble hypothesis accounts for
the fact that fewer than sixty-one transfer RNAs can read
the entire genetic code. Fewer transfer RNAs are needed be-
cause additional complementary base pairings occur in the
third position (3') of the codon.

The rule of universality of the genetic code has to be
modified in light of findings regarding mitochondrial trans-
fer RNAs; only twenty-four are needed to read the code. In
addition, some sense codons are interpreted differently in
mitochondrial systems; some nonmitochondrial systems
read stop codons differently (a mycoplasma and ciliated
protozoan); and some site-specific variation in codon read-
ing also occurs. The structure of the code in both cells and
mitochondria seems to protect the cell against a good deal
of potential mutation.

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Exercises and Problems

313

SOLVED PROBLEMS

PROBLEM 1: What is the energy requirement of protein
biosynthesis?

Answer: The cost of adding one amino acid to a growing
polypeptide is four or five high-energy bonds: two from an
ATP during the charging of the transfer RNA, and two or
three from the hydrolysis of GTPs during transfer RNA
binding to the A site of the ribosome and during transloca-
tion. Thus, for an average protein of three hundred amino
acids, there is a cost of 1,200 to 1,500 high-energy bonds.

PROBLEM 2: What are the start and stop signals of trans-
lation?

Answer: Once a messenger RNA is attached at the ribo-
some, the start signal is the methionine initiation codon
(usually AUG), whereas the stop signal is one of the three
nonsense codons (UAA, UAG, and UGA). Binding to the ri-
bosome in order to position the messenger RNA in relation

to the A and P sites differs in prokaryotes and eukaryotes. In
prokaryotes, the Shine-Dalgarno sequence allows the 16S
ribosomal RNA and the messenger RNA to form hydrogen
bonds, locating the beginning of the messenger RNA at the
ribosome. In eukaryotes, the 5 ' cap is usually recognized by
the ribosome, and the ribosome then proceeds to scan the
messenger RNA for the initiation codon.

PROBLEM 3: What amino acids could replace methio-
nine if a one-base mutation occurred?

Answer: The codon for methionine (internal as well as
initiation) is AUG. If the A is replaced, we would get UUG
(Leu), CUG (Leu), or GUG (Val); if the U is replaced, we
would get AAG (Lys),ACG (Thr), or AGG (Arg); and if the
G is replaced, we would get AUA (Ile),AUU (lie), or AUC
(lie). Hence, a one-base change in the codon for methio-
nine could result in any of six different amino acids.

EXERCISES AND PROBLEMS

INFORMATION TRANSFER

1. Given the following end part of a gene, which will be
transcribed and then translated into a pentapeptide,
provide the base sequence for its messenger RNA.
Give the anticodons on the transfer RNAs by making
use of wobble rules. What amino acids are incorpo-
rated? Draw the actual structure of the pentapeptide.

3 '-TACAATGGCCCTTTTATC-5 '
5 '-ATGTTACCGGGAAAATAG-3 '

2. Give an alternative translation mechanism that would
require only one transfer RNA site on the ribosome.

3. Draw the details of a moment in time at the ribo-
some during the translation of the messenger RNA
produced in problem 1 . Include in the diagram the
ribosomal sites, the transfer RNAs, and the various
nonribosomal proteins involved.

4. How do prokaryotic and eukaryotic ribosomes rec-
ognize the 5' end of messenger RNAs? Could eu-
karyotic messenger RNAs be polycistronic?

5. How many aminoacyl-tRNA synthetases are there?
What do they use for recognition signals?

6. What are the similarities and differences among the
three nonsense codons? Using the wobble rules,
what are their theoretical anticodons?

* Answers to selected exercises and problems are on page A-12.

7. Describe an experiment that demonstrates that the
transfer RNA, and not its amino acid, is recognized at
the ribosome during translation.

8. Other than the antibiotics named in the chapter, sug-
gest five "theoretical" antibiotics that could interfere
with the prokaryotic translation process.

9. How many single-base deletions are required to re-
store the reading frame of a messenger RNA? Give an
example.

10. A "nonsense mutation" is one in which a codon for
an amino acid changes to one for chain termination.
Give an example. What are its consequences?

11. The reverse situation to problem 10 is a mutation
from a nonsense codon to a codon for an amino
acid. Give an example. What are its consequences?

12. What are the consequences when an internal me-
thionine codon recognizes a prokaryotic initiation
transfer RNA?

13. What role does EF-Ts play in elongation? EF-Tu? What
are their eukaryotic equivalents?

14. What roles do RF1 and RF2 play in chain termina-
tion? What are their eukaryotic equivalents?

15. What is a signal peptide? What role does it play in eu-
karyotes? What is its fate?

16. Why doesn't puromycin disrupt eukaryotic transla-
tion?

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314

Chapter Eleven Gene Expression: Translation

Y7. A peptide, fifteen amino acids long, is digested by
two methods, and each segment is sequenced ac-
cording to the Edman degradation technique (see
box 11.1). The fifteen amino acids are denoted by
the letters A through O, with F as the N-terminal
amino acid. If the segments are as follows, what is
the sequence of the original peptide?

Method 1: CABHLN; FGKI; OEDJM
Method 2: KICAB; JM; FG; HLNOED

18. In human hemoglobin, the p chain is 146 amino
acids long. What is the minimum length of RNA
needed to make this protein?

19. Part of a DNA strand to be transcribed has the fol-
lowing sequence:

3 '-TACTAACTTACGCTCGCCTCA-5 '

a. What is the sequence of RNA transcribed from
this part of the strand?

b. What sequence of amino acids does the RNA pro-
duce?

THE GENETIC CODE

20. If DNA contained only the bases cytosine and gua-
nine, how long would a code word have to be? How
could we tell if this DNA were double-stranded?

21. If an artificial messenger RNA contains two parts
uracil to one of cytosine, name the amino acids and
the proportions in which they should be incorpo-
rated into protein.

22. What would be proved or disproved if an organism
were discovered that did not follow any of the rules
of the codon dictionary? Would we expect organ-
isms from another galaxy (if they exist) to use our
codon dictionary?

23. What would the genetic code dictionary (see
table 11.4) look like if wobble occurred in the sec-
ond position rather than the third (i.e. , if an unmixed
family of codons were of the form GXU)?

24. In experiments using repeating polymers, (GCGC) n in-
corporates alanine and arginine into polypeptides, and
(CGGCGG) n incorporates arginine, glycine, and ala-
nine. What codon can probably be assigned to glycine?

25. If poly-G is used as a messenger RNA in an incorpo-
ration experiment, glycine is incorporated into a
polypeptide. If poly-C is used, proline is incorpo-
rated. If both poly-G and poly-C are used, no amino
acids are incorporated into protein. Why?

26. A protein has leucine at a particular position. If the
codon for leucine is CUC, how many different amino
acids might appear as the result of a single-base sub-
stitution?

27. Polymers of (GUA) n result in the incorporation of
only two different amino acids rather three, as for
most other three-base polymers. Why?

28. The sixth amino acid in the (3 chain of normal hu-
man hemoglobin is glutamate. Two different muta-
tions of this codon substitute valine and lysine. What
is the likely codon for glutamate?

29. A normal protein has the following C-terminal
amino acid sequence: ser-tbr-lys-leu-COOH. A mutant
is isolated with the following sequence: ser-thr-lys-
leu-leu-phe-arg-COOH.What has probably happened
to produce the mutant protein?

30. A segment of a normal protein and three different
mutants appears as follows:

normal
mutant 1
mutant 2
mutant 3

gly-ala-ser-his-cys-leu-phe_
gly-ala-ser-his
gly-ala-ser-leu-cys-leu-phe.
gly-val-ala-ile-ala-ser

What is the probable sequence of bases in the nor-
mal RNA?

31. A normal protein has histidine in a given position.
Four mutants are isolated and determined to have ei-
ther tyrosine, glutamine, proline, or leucine in place of
histidine. What are the possible codon assignments,
and what codon is probably used for histidine?

CRITICAL THINKING QUESTIONS

Suggested Readings for chapter 11 are on page B-8.

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12. DNA: Its Mutation,
Repair, and Recombination

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DNA

Its Mutatiotiy Repair,
and Recombination

STUDY OBJECTIVES

1. To look at the nature of mutation in prokaryotes 316

2. To analyze functional and structural allelism and examine the
mapping of mutant sites within a gene 317

3. To verify the colinearity of gene and protein 324

4. To study mutagenesis 325

5. To investigate the processes of DNA repair
and recombination 339

STUDY OUTLINE

Mutation 316

Fluctuation Test 316

Genetic Fine Structure 317

Colinearity 324

Spontaneous Versus Induced Mutation 325

Mutation Rates 326

Point Mutations 326

Spontaneous Mutagenesis 328

Chemical Mutagenesis 330

Misalignment Mutagenesis 337

Intergenic Suppression 337

Mutator and Antimutator Mutations 338
DNA Repair 339

Damage Reversal 340

Excision Repair 340

Double-Strand Break Repair 344

Postreplicative Repair 344
Recombination 347

Double-Strand Break Model of Recombination 347

Bacterial Recombination 349

Hybrid DNA 351
Summary 352
Solved Problems 354
Exercises a nd Problems 354
Critical Thinking Questions 356
Box 12.1 The Ames Test for Carcinogens 332
Box 12.2 In Vitro Site-Directed Mutagenesis 333
Box 12.3 Adaptive Mutation 339

Computer-generated space-filling model of a DNA
enzyme repairing damaged DNA.

(©James King-Holmes/SPL/Photo Researchers, Inc.)

315

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316

Chapter Twelve DNA: Its Mutation, Repair, and Recombination

The mutation, repair, and recombination of
DNA are treated together in this chapter be-
cause the three processes have much in com-
mon. The physical alteration of DNA is in-
volved in each; repair and recombination
share some of the same enzymes. We progress from mu-
tation — the change in DNA — to repair of damaged DNA,
and, finally, to recombination, the new arrangement of
pieces of DNA.

MUTATION

The concept of mutation (a term coined by de Vries, a re-
discoverer of Mendel) is pervasive in genetics. Mutation
is both the process by which a gene (or chromosome)
changes structurally and the end result of that process.
Without alternative forms of genes, the biological diver-
sity that exists today could not have evolved. Without al-
ternative forms of genes, it would have been virtually im-
possible for geneticists to determine which of an
organism's characteristics are genetically controlled.
Studies of mutation provided the background for our cur-
rent knowledge in genetics.

fluctuation Test

In 1943, Salvador Luria and Max Delbriick published a pa-
per entitled "Mutations of Bacteria from Virus Sensitivity
to Virus Resistance." This paper ushered in the era of bac-
terial genetics by demonstrating that the phenotypic
variants found in bacteria are actually attributable to mu-
tations rather than to induced physiological changes.
Very little work had previously been done in bacterial ge-
netics because of the feeling that bacteria did not have
"normal" genetic systems like the systems of fruit flies
and corn. Rather, bacteria were believed to respond to
environmental change by physiological adaptation, a

Salvador E. Luria
(1912-1991). (Courtesy of
Dr. S. E. Luria.)

Max Delbruck (1906-1981).
(Courtesy of Dr. Max Delbruck.)

non-Darwinian view. As Luria said, bacteriology remained
"the last stronghold of Lamarckism" (the belief that ac-
quired characteristics are inherited).

What Causes Genetic Variation?

Luria and Delbruck studied the Ton r (phage Tl -resistant)
mutants of a normal Ton s (phage Tl -sensitive) Escherichia
coli strain. They used an enrichment experiment, as de-
scribed in chapter 7, wherein a petri plate is spread with
E. coli bacteria and Tl phages. Normally, no bacterial
colonies grow on the plate: all the bacteria are lysed.
However, if one of the bacterial cells is resistant to Tl
phages, it produces a bacterial colony, and all descendants
of this colony are Tl resistant. There are two possible
explanations for the appearance of Tl -resistant colonies:

1. Any E. coli cell may be induced to be resistant to
phage Tl, but only a very small number actually are.
That is, all cells are genetically identical, each with a
very low probability of exhibiting resistance in the
presence of Tl phages. When resistance is induced,
the cell and its progeny remain resistant.

2. In the culture, a small number of E. coli cells exist that
are already resistant to phage Tl; in the presence of
phage Tl, only these cells survive.

If the presumed rates of physiological induction and
mutation are the same, determining which of the two
mechanisms is operating is difficult. Luria and Delbruck,
however, developed a means of distinguishing between
these mechanisms. They reasoned as follows: If Tl resis-
tance was physiologically induced, the relative frequency
of resistant E. coli cells in a culture of the normal (Ton s )
strain should be a constant, independent of the number of
cells in the culture or the length of time that the culture has
been growing. If resistance was due to random mutation,
the frequency of mutant (Ton r ) cells would depend on
when the mutations occurred. In other words, the appear-
ance of a mutant cell would be a random event. If a muta-
tion occurs early in the growth of the culture, then many
cells descend from the mutant cell, and therefore many re-
sistant colonies develop. If the mutation does not occur un-
til late in the growth of the culture, then the subsequent
number of mutant cells is few. Thus, if the mutation hy-
pothesis is correct, there should be considerable fluctua-
tion from culture to culture in the number of resistant cells
present (fig. 12.1).

Results of the Fluctuation Test

To distinguish between these hypotheses, Luria and Del-
briick developed what is known as the fluctuation test.
They counted the mutants both in small ("individual") cul-
tures and in subsamples from a single large ("bulk") cul-
ture. All subsamples from a bulk culture should have the
same number of resistant cells, differing only because of

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Repair, and Recombination

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Mutation

317

f=^

<£=?

$=?

f=^

<F=?

f=?>

«=?

<^>

f=^

(a) Physiological induction

f=^>

<F^>

«=?>

f=?

<F=^>

f=^>

f=?

f=? f=^

(b) Random mutation

Figure 12.1 Occurrence of E co// Ton r colonies in Ton s cultures. Ten cultures of E. coli cells were grown from
a standard inoculum in separate test tubes in the absence of phage T1 , then spread on petri plates in the
presence of phage T1 . The resistant cells grow into colonies on the plates. We expect a uniform distribution
of resistant cells if the physiological induction hypothesis is correct (a) or a great fluctuation in the number of
resistant cells if the random mutation hypothesis is correct (£>).

random sampling error. If, however, mutation occurs, the
number of resistant cells among the individual cultures
should vary considerably from culture to culture; the num-
ber would be related to the time that the mutation oc-
curred during the growth of each culture. If mutation arose
early, there would be many resistant cells. If it arose late,
there would be relatively few resistant cells. Under physio-
logical induction, the distribution of resistant colonies
should not differ between the individual and bulk cultures.
Luria and Delbruck inoculated twenty individual cul-
tures and one bulk culture with E. coli cells and incubated
them in the absence of phage Tl. Each individual culture
was then spread out on a petri plate containing a very high
concentration of Tl phages; ten subsamples from the bulk
culture were plated in the same way. We can see from the
results (table 12.1) that there was minimal variation in the
number of resistant cells among the bulk culture subsam-
ples but a very large amount of variation, as predicted for
random mutation, among the individual cultures.

If bacteria have "normal" genetic systems that un-
dergo mutation, bacteria could then be used, along with
higher organisms, to answer genetic questions. As we
have pointed out, the modern era of molecular genetics
began with the use of prokaryotic and viral systems in ge-
netic research. In the next section, we turn our attention
to several basic questions about the gene, questions
whose answers were found in several instances only be-
cause prokaryotic systems were available.

Genetic Fine Structure

How do we determine the relationship among several mu-
tations that cause the same phenotypic change? What are
the smallest units of DNA capable of mutation and recom-
bination? Are the gene and its protein product colinear?
The answers to the latter two questions are important
from a historical perspective. The answer to the first ques-
tion is relevant to our current understanding of genetics.

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318

Chapter Twelve DNA: Its Mutation, Repair, and Recombination

TablG 12.1 Results from the Luria and Delbruck

Fluctuation Test

Samples

from Bulk

Individual Cultures*
Ton r

Culture*

Ton r

Culture

Colonies

Sample

Colonies

Number

Found

Number

Found

107

20
Mean (w)

16.7

11.4

Standard deviation 27.4

4.3

Source: From E. Luria and M. Delbruck, Genetics, 28: 491. Copyright © 1943
Genetics Society of America.

* Each culture and sample was 0.2 ml and contained about 2 X 10 7 E. colt cells.

Complementation

If two recessive mutations arise independently and both
have the same phenotype, how do we know whether they
are both mutations of the same gene? That is, how do we
know whether they are alleles? To answer this question, we
must construct a heterozygote and determine the com-
plementation between the two mutations. A heterozy-
gote with two mutations of the same gene will produce
only mutant messenger RNAs, which result in mutant en-
zymes (fig. 12.2a). If, however, the mutations are not allelic,
the gamete from the a 1 parent will also contain an a 2 al-
lele, and the gamete from the a 2 parent will also contain
the a^~ allele (fig. 12.2£>). If the two mutant genes are truly
alleles, then the phenotype of the heterozygote should be

mutant. If, however, the two mutant genes are nonallelic,
then the a x mutant will have contributed the wild-type al-
lele at the A 2 locus, and the a 2 mutant will have con-
tributed the wild-type allele at the A 1 locus to the het-
erozygote. Thus, the two mutations will complement each
other and produce the wild-type. Mutations that fail to
complement each other are termed functional alleles.
The test for defining alleles strictly on this basis of func-
tionality is termed the cis-trans complementation test.

There are two different configurations in which a het-
erozygous double mutant of functional alleles can form
(fig. 12.3). In the cis-trans complementation test, only
the trans configuration is used to determine whether the
two mutations were allelic. In reality, the cis configura-
tion is not tested; it is the conceptual control, in which
wild-type activity (with recessive mutations) is always ex-
pected. The test is thus sometimes simply called a trans
test. Functional alleles produce a wild-type phenotype in
the cis configuration but a mutant phenotype in the
trans configuration. This difference in phenotypes is
called a cis-trans position effect.

From the terms cis and trans, Seymour Benzer coined
the term cistron for the smallest genetic unit (length of
genetic material) that exhibits a cis-trans position effect.
We thus have a new word for the gene, one in which
function is more explicit. We have, in essence, refined
Beadle andTatum's one-gene-one-enzyme hypothesis to a
more accurate one-cistron-one-polypeptide concept. The
cistron is the smallest unit that codes for a messenger
RNA that is then translated into a single polypeptide or
expressed directly (transfer RNA or ribosomal RNA).

From functional alleles, we can go one step further in re-
combinational analysis by determining whether two allelic
mutations occur at exactly the same place in the cistron. In
other words, when two mutations prove to be functional
alleles, are they also structural alleles? The methods used
to analyze complementation can be used here also. Crosses
are carried out to form a mutant heterozygote (trans con-
figuration) whose offspring are then tested for recombina-
tion between the two mutational sites. If no recombination
occurs, then the two alleles probably contain the same

Seymour Benzer (1921- ).
(Courtesy of Dr. Seymour Benzer,
1970.)

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(a) If allelic

(A locus
a 1 and a 2
alleles)

a i /a i

3q/ 3ry

Heterozygote

Mutant phenotype

(b) If nonallelic

(/A 1 and A 2 loci
a 1 and a^ alleles
a 2 and a 2 + alleles)

a^ (a 2 + /a 2 + )

« 2

i i

(a 1 + /a 1 + ) a 2 /a 2

Heterozygote

Wild-type
phenotype

Figure 12.2 The complementation test defines allelism. Are two mutations (a-i, a 2 ) allelic if they affect the
same trait? To find out, mutant homozygotes are crossed to form a heterozygote. (a) If the mutations are
allelic, then both copies of the gene in the heterozygote are mutant, resulting in the mutant phenotype.
(b) If the mutations are nonallelic, then there is a wild-type allele of each gene present in the heterozygote,
resulting in the wild-type phenotype. (The two loci need not be on the same chromosome.)

structural change (involving the same base pairs) and are
thus structural alleles. If a small amount of recombination
occurs that generates wild-type offspring, then the two al-
leles are not mutations at the same point (fig. 12.4). Alleles
that were functional but not structural were first termed
pseudoalleles because it was believed that loci were

trans

CIS

Allelic
arrangement

(A locus)

Phenotype:

a A /a 2

Mutant

Wild-type

Figure 12.3 A heterozygote of two recessive mutations can
have either the trans or cis arrangement. In the trans position,
functional alleles produce a mutant phenotype. (Red marks
represent mutant lesions.) In the cis position, functional alleles
produce a wild-type phenotype. The cis-trans position effect
thus reveals functional alleles.

made up of subloci. Fine-structure analysis led to the un-
derstanding that a locus is a length of genetic material di-
visible by recombination rather than a "bead on a string."

Eye-color mutants of Drosophila melanogaster can
be studied by complementational analysis. The white-eye
locus has a series of alleles producing varying shades of
red. This locus is sex linked, at about map position 30 on
the X chromosome. (Several other eye-color loci on the X
chromosome are not relevant to this cross — e.g., prune
and ruby) If an apricot-eyed female is mated with a
white-eyed male, the female offspring are all heterozy-
gous and have mutant light-colored eyes (fig. 12.5). Thus,
apricot and white are functional alleles: they do not com-
plement (table 12.2). To determine whether apricot and
white are structural alleles, light-eyed females are crossed
with white-eyed males, and the offspring are observed
for the presence of wild-type or light-eyed males. Though
their rate of appearance is less than 0.001%, this is
significantly above the background mutation rate. The
conclusion is that apricot and white are functional, but
not structural, alleles.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

Functional but not
structural alleles

Functional and
structural alleles

A locus

Heterozygous

F-| in

trans position

=DC

i — ** — h

Double
mutant

Recombination

Single
mutant

Recombination

Gametes

Wild-
type

Single
mutant

Figure 12.4 Functional alleles may or may not be structurally
allelic. (Red marks represent mutant sites.) Functional alleles that
are not also structural alleles can recombine between the mutant
sites, resulting in occasional wild-type (and double mutant)
offspring. Structural alleles (which are also always functional
alleles) are defective at the same base pairs and cannot form
either wild-type or double mutant offspring by recombination.

Fine- Structure Mapping

After Beadle andTatum established in 1941 that a gene con-
trols the production of an enzyme that then controls a step
in a biochemical pathway, Benzer used analytical tech-
niques to dissect the fine structure of the gene. Fine-
structure mapping means examining the size and number
of sites within a gene that are capable of mutation and re-
combination. In the late 1950s, when biochemical tech-
niques were not yet available for DNA sequencing, Benzer
used classical recombinational and mutational techniques
with bacterial viruses to provide reasonable estimates on
the details of fine structure and to give insight into the na-
ture of the gene. He coined the terms muton for the small-
est mutable site and recon for the smallest unit of recom-
bination. It is now known that both muton and recon are a
single base pair.

Before Benzer's work, genes were thought of as beads
on a string. The very low rate of recombination between

Apricot eye

White eye

X W *X W — Light eye 9
— Apricot eye o*

yw a y

Testcross

9
Light eye

White eye

Rarely —

X H

\fW a VLV

\rw a \j

Light eye 9

Apricot eye 8

\fW \fW

X W Y

White eye 9

White eye o

\(W?\fW

X W? Y

Light eye 9

Light eye 8

X+X w

X + Y

Wild-type 9

Wild-type o*

Figure 12.5 Crosses demonstrating that apricot and white
eyes are functional, but not structural, alleles in Drosophila.
Light-eyed females are heterozygous for both alleles. When
testcrossed, they produce occasional offspring that are wild-
type (X + allele) or light-eyed (X w? allele). This indicates a
crossover between the two mutant sites (white and apricot) in
the heterozygous females, producing, reciprocally, an allele with
both mutational sites and the wild-type.

Table 12.2 Complementation Matrix of X-Linked Drosophila Eye-Color Mutants

white

prune

apricot

buff

cherry

eosin

ruby

white (w)

—

prune (pn)

—

apricot (w a )

—

buff O bf )

—

cherry (w ch )

—

eosin (w e )

—

ruby (rb)

—

Note: Plus sign indicates that female offspring are wild-type; minus sign indicates that they are mutant.

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Mutation

321

sites within a gene hampered the analysis of mutational
sites within a gene by means of recombination. If two
mutant genes are functional alleles (involving different
sites on the same gene), a distinct probability exists that
we will get both mutant sites (and both wild-type sites) on
the same chromosome by recombination (see fig. 12.4);
but, in view of the very short distances within a gene, this
probability is very low. Although it certainly seemed de-
sirable to map sites within the gene, the problem of find-
ing an organism that would allow fine-structure analysis
remained until Benzer decided to use phage T4.

r II Screening Techniques. Benzer used the T4 bacterio-
phage because of the growth potential of phages, in
which a generation takes about an hour and the increase
in numbers per generation is about a hundredfold. Actu-
ally, any prokaryote or virus should suffice, but Benzer
made use of other unique screening properties of the
phage that made it possible to recognize one particular
mutant in about a billion phages. Benzer used rll mutants
of T4. These mutants produce large, smooth-edged plaques
on E. colt, whereas the wild-type produces smaller
plaques whose edges are not as smooth (see fig. 7.7).

The screening system that Benzer employed made use
of the fact that rll mutants do not grow on E. coli strain
K12, whereas the wild-type can. The normal host strain, E.
coli B, allows growth of both the wild-type and rll mu-
tants. Thus, various mutants can be crossed by mixed in-
fection of E. coli B cells, and Benzer could screen for wild-
type recombinants by plating the resultant progeny
phages on E. coli K12 (fig. 12.6), on which only a wild-
type recombinant produces a plaque. It is possible to de-
tect about one recombinant in a billion phages, all in an
afternoon's work. This ability to detect recombinants oc-
curring at such a low level of frequency allowed Benzer
to see recombinational events occurring very close to-
gether on the DNA, events that would normally occur at a
frequency too low to detect in fruit flies or corn.

Benzer sought to map the number of sites subject to
recombination and mutation within the rll region of T4.
He began by isolating independently derived rll mutants
and crossing them among themselves. The first thing he
found was that the rll region was composed of two
cistrons; almost all of the mutations belonged to one of
two complementation groups. The ^4 -cistron muta-
tions would not complement each other but would com-
plement the mutations of the B cistron. The exceptions
were mutations that seemed to belong to both cistrons.
These mutations were soon found to be deletions in
which part of each cistron was missing (table 12.3).

Deletion Mapping. As the number of independently iso-
lated mutations of the A and B cistrons increased, it be-
came obvious that to make every possible pairwise cross
would entail millions of crosses. To overcome this prob-

E. coli B

Wild-type
recombinant

Progeny phages
are plated on
E. coli K12

Only wild-type phages
produce plaques

Figure 12.6 Using E. coli K12 and B strains to screen for
recombination at the rll locus of phage T4. Two rll mutants are
crossed by infecting the same B-strain bacteria with both
phages. The offspring are plated on a lawn of K12 bacteria in
which only wild-type phages can grow. The technique thus
selects only wild-type recombinants.

lem, Benzer isolated mutants that had partial or complete
deletions of each cistron. Deletion mutations were easy to
discover because they acted like structural alleles to alleles
that were not themselves structurally allelic. In other
words, if mutations a, b, and c are functional — but not

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

structural — alleles of each other, and mutation d is a struc-
tural allele to a, b, and c, then d must contain a deletion of
the bases mutated in a, b, and c. Once a sequence of dele-
tion mutations covering the A and B cistrons was isolated,
a minimal number of crosses was required to localize a
new mutation to a portion of one of the cistrons. A second
series of smaller deletions within each region was then iso-
lated, further localizing the mutation (fig. 12.7).

Next, each new mutant was crossed with each of the
other mutants isolated in its subregion to localize the rela-
tive position of the new mutation. If the mutation was
structurally allelic to a previously isolated mutation, it was
scored as an independent isolation of the same mutation. If
it was not a structural allele to any of the known mutations
of the subregion, it was added as a new mutation point. The
exact position of each new mutation within the region was

determined by the relative frequency of recombination be-
tween it and the known mutations of this region (see chap-
ter 7). Benzer eventually isolated about 350 mutations from
eighty different subregions defined by deletion mutations.
An abbreviated map is shown in figure 12.8.

What conclusions did Benzer draw from his work?
First, he concluded that since all of the mutations in both
rll cistrons can be ordered in a linear fashion, the original
Watson-Crick model of DNA as a linear molecule was cor-
rect. Second, he concluded that reasonable inroads had
been made toward saturating the map, localizing at least
one mutation at every mutable site. Benzer reasoned that
since many sites were represented by only one mutation,
some sites must occur that were represented by zero mu-
tations (i.e., not yet represented by a mutation). Since he
had mapped about 350 sites, he calculated that there

Figure 12.7 Localization of an rll mutation by deletion
mapping. Newly isolated mutants are crossed with mutants
with selected deletions to localize the new mutation to a small
region of the cistron. If the new mutant (e.g., r960) is located
in the A5c2a2 region, it would not produce the occasional
wild-type by recombination with r1272, r1241, rJ3, rPT1, or
rPB242 (the solid part of the bar indicates deleted segments).
It would produce the wild-type by recombination with rA1 05
and r638, and thus the mutation would be localized to the A5
region. When crossed with r1605, r1589, and rPB230, it
would produce only the rare recombinant with rPB230,
indicating the mutation is in the A5c region. When crossed with
r1993, r1695, and r1168, the mutant would produce the wild-
type by recombination with r1993 and r1168, and the mutation
would be localized to the A5c2a2 region. Finally, the mutant
would be crossed pairwise with all the known mutants of this
region to determine relative arrangement and distance. (Source:
Data from Seymour Benzer, "The fine structure of the gene," Scientific
American, 206: 70-84, January 1962.)

/-1272

/"1 241

rJ3

rPT1

rPB242

rA105

r638

Site

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Mutation

323

TablG 12.3 Complementation Matrix of Ten rll

Mutants

—

7
8
9

—

Note: Plus sign indicates complementation; minus sign indicates no comple-
mentation. The two cistrons are arbitrarily designated A and B. Mutants 4
and 9 must be deletions that cover parts of both cistrons. Alleles: A cistron:
1, 2, 4, 5, 6, 8, 9, 10; B cistron: 3, 4, 7, 9-

of amino acids times three nucleotides per codon). Thus,
although Benzer had not saturated the map with muta-
tions, he certainly had made respectable progress in dis-
secting the gene and demonstrating that it was not an in-
divisible unit, a "bead on a string."

Hot Spots. Benzer also looked into the lack of unifor-
mity in the occurrence of mutations (note two major "hot
spots" at B4 and A6c of fig. 12.8). Presuming that all base
pairs are either AT or GC, this lack of uniformity was un-
expected. Benzer suggested that spontaneous mutation is
not just a function of the base pair itself, but is affected by
the surrounding bases as well. This concept still holds.

To recapitulate, Benzer s work supports the model of
the gene as a linear arrangement of DNA whose nu-
cleotides are the smallest units of mutation. The link be-
tween any adjacent nucleotides can break in the recombi-
national process. The smallest functional unit, determined
by a complementation test, is the cistron. Mutagenesis is
not uniform throughout the cistron, but may depend on
the particular arrangement of bases in a given region.

were at least another 100 sites still undetected by muta-
tion. We now know that 450 sites is an underestimate.
However, since the protein products of these cistrons
were not isolated, there were no independent estimates
of the number of nucleotides in these cistrons (number

Intra-Allelic Complementation

Benzer warned that certainty is elusive in the complemen-
tation test because sometimes two mutations of the same
functional unit (cistron) can result in partial activity. The

7 f ?T i T f T V i T f T T T " i ? " 7 " ? i i f t i i

A1a

A1b1

f-rr i i i i i Tt — "f- — m i " V-
A1b2 A2a A2b A2d A2f

A2c A2e

A4d A4c A4a A3h A3g A3f A3e

i m i l ,
A2g X,

A2h1^
A2h2^,

U> U4 Jr ' ' flH i U -4-Jr

A4b

A3i

^4 H

A3a-d A2h3 >£»>

Tn m

fr4 Y H

-f-rr T * M? i W i f ? f 1 1 TT? i f f i i TTJT???H? rfl

A5a A5b A5c1 A5c2

A5d

A6a1

A6a2 A6b

B6 B5

B3 B2

L i j^m^ A-U A nU4i ' "U i UU ' U U i-Ui U i Ufl

|MM|

c
o

A6d

c
o

T-Mr

rH frM ? t ? 7 7 1 ?f i J

B8 B9a

n t4 t

B9bB10

Figure 12.8 Abbreviated map of spontaneous mutations of the A and B cistrons of the rll region of T4. Each
square represents one independently isolated mutation. Note the "hot spots" at A6c and B4. (From Seymour
Benzer, "On the topography of the genetic fine structure", Proceedings of the National Academy of Sciences USA 47:403-15, 1961.
Reprinted by permission.)

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

Normal enzyme
(two subunits)

Active site

Mutant a 1

nonfunctional

homozygote

Mutant a 2

nonfunctional

homozygote

Functional
heterozygote
a 1 and a 2
subunits
together

Active site restored

Figure 12.9 Intra-allelic complementation. With certain mutations, it is possible to get enzymatic activity
in a heterozygote for two nonfunctional alleles, if the two polypeptides form a functional enzyme. (Active
site is shown in color.)

problem can be traced to the interactions of subunits at
the polypeptide level. Some proteins are made up of sub-
units, and it is possible that certain mutant combinations
produce subunits that interact to restore the enzymatic
function of the protein (fig. 12.9). This phenomenon is
known as intra-allelic complementation. With this in
mind, geneticists routinely use the complementation test
to determine functional relationships among mutations.

Colinearity

Next we look at the colinearity of the gene and the
polypeptide. Benzer's work established that the gene was a
linear entity, as Watson and Crick had proposed. However,
Benzer could not demonstrate the colinearity of the gene
and its protein product. To do this, it is necessary to show
that for every mutational change in the DNA, a correspond-
ing change takes place in the protein product of the gene.
Colinearity would be established by showing that nu-
cleotide and amino acid changes occurred in a linear fash-
ion and in the same order in the protein and in the cistron.

Ideally, Benzer himself might have solved the
colinearity issue. He was halfway there, with his 350 or
so isolated mutations of phage T4. However, Benzer did
not have a protein product to analyze; no mutant protein
had been isolated from rll mutants. In the midst of com-
petition to find just the right system, Charles Yanofsky of
Stanford University and his colleagues emerged in the
mid-1960s with the required proof, showing that the or-
der of a polypeptide's amino acids corresponded to the
nucleotide sequence in the gene that specified it. Yanof-
sky's success rested with his choice of an amenable sys-
tem, one using the enzymes from a biochemical pathway.

Yanofsky did his research on the tryptophan biosyn-
thetic pathway in E. colt. The last enzyme in the pathway,
tryptophan synthetase, catalyzes the reaction of indole-
3-glycerol-phosphate plus serine to tryptophan and

3-phosphoglyceraldehyde. The enzyme itself is made of
four subunits specified by two separate cistrons, with
each polypeptide present twice.

Yanofsky and his colleagues concentrated on the A
subunit. They mapped ^4-cistron mutations with transduc-
tion (see chapter 7) using the transducing phage PI. They
first tested each new mutant against a series of deletion
mutants to establish the region where the mutation was.
Then they crossed mutants for a particular region among
themselves to establish relative positions and distances.

The protein products of the bacterial genes were iso-
lated using electrophoresis and chromatography to estab-
lish the fingerprint patterns of the proteins (see chapter
1 1). Assuming a single mutation, a comparison of the mu-
tant and the wild-type fingerprints would show a difference
of just one polypeptide spot (fig. 12.10), avoiding the need
to sequence the entire protein. The mutant amino acid was

Ala-Arg-Trp-Ser-Ser spot

Ala-Lys-Trp-Ser-Ser spot

°o

o o

o O

Wild-type

Mutant

Figure 12.10 Difference in "fingerprints" between mutant and
wild-type polypeptide digests. The single spot that differs in the
mutant can be isolated and sequenced, eliminating the need to
sequence the whole protein.

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Figure 12.11 Amino acid sequence of the carboxyl terminal end of the tryptophan synthetase A protein
and its DNA. Mutations are shown on the DNA (e.g., A446), as are the changed amino acids of those
mutations (in color). DNA and protein changes are colinear.

identified by analysis of just this one spot. Figure 12.11
shows the details of nucleotide and amino acid changes for
nine of the mutations in this 267-amino acid protein.

We can see from this figure that nine mutations in the
linear A cistron of tryptophan synthetase are colinear
with nine amino acid changes in the protein itself. In two
cases, two mutations mapped so close as to be almost in-
distinguishable. In both cases, the two mutations proved
to be in the same codon: the same amino acid position
was altered in each (A23-A46, A58-A78).Thus, exactly as
predicted and expected, colinearity exists between gene
and protein. Brenner and his colleagues, using head-

protein mutants of phage T4, independently confirmed
this work at the same time.

Spontaneous Versus Induced Mutation

H. J. Muller won the Nobel Prize for demonstrating that
X rays can cause mutations. This work was published
in 1927 in a paper entitled "Artificial Transmutation of the
Gene." At about the same time, L. J. Stadler induced mu-
tations in barley with X rays. The basic impetus for their
work was the fact that mutations occur so infrequently
that genetic research was hampered by the inability to

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

Hermann J. Muller
(1890-1967). (Courtesy of
National Academy of Sciences.)

Lewis J. Stadler
(1896-1954). {Genetics, 41,
1956: frontispiece.)

As discussed in chapter 1 1 , one of the outcomes of redun-
dancy in the genetic code is partial protection of the cell
from the effects of mutation; common amino acids have
the most codons, similar amino acids have similar codons,
and the wobble position of the codon is the least important
position in translation. However, when base changes result
in new amino acids, new proteins appear. These new pro-
teins can alter the morphology or physiology of the organ-
ism and result in phenotypic novelty or lethality.

Frameshift Mutation

A point mutation may consist of replacement, addition,
or deletion of a base (fig. 12.12). Point mutations that add

obtain mutants. Muller exposed flies to varying doses of
X rays and then observed their progeny. He came to sev-
eral conclusions. First, X rays greatly increased the occur-
rence of mutations. Second, the inheritance patterns of
X-ray-induced mutations and the resulting phenotypes of
organisms were similar to those that resulted from natu-
ral, or "spontaneous," mutations.

Mutation Rates

The mutation rate is the number of mutations that arise
per cell division in bacteria and single-celled organisms,
or the number of mutations that arise per gamete in
higher organisms. Mutation rates vary tremendously de-
pending upon the length of genetic material, the kind of
mutation, and other factors. Luria and Delbriick, for ex-
ample, found that in E. colt the mutation rate per cell di-
vision of Ton s to Ton r was 3 X 10~ 8 , whereas the mutation
rate of the wild-type to the histidine-requiring phenotype
(His + to His - ) was 2 X 10~ 6 . The rate of reversion (re-
turn of the mutant to the wild-type) was 7.5 X 10~ 9 . The
mutation and reversion rates differ because many differ-
ent mutations can cause the His phenotype, whereas re-
version requires specific, and hence less probable,
changes to correct the His phenotype back to the wild-
type. The lethal mutation rate in Drosophila is about 1 X
10~ 2 per gamete for the total genome. This number is rel-
atively large because, as with His, many different muta-
tions produce the same phenotype (lethality, in this case).

Point Mutations

The mutations of primary concern in this chapter are
point mutations, which consist of single changes in the
nucleotide sequence. (In chapter 8 we discussed chromo-
somal mutations, changes in the number and visible struc-
tures of chromosomes.) If the change is a replacement of
some kind, then a new codon is created. In many cases, this
new codon, upon translation, results in a new amino acid.

Original sequence

Single-step

Replacement

Addition

Deletion

Double-step

Second independent change
(single-step replacement, then
second single-step
replacement)

Back mutation

(of single-step replacement)

Intragenic suppressor
(single-step addition, then
single-step deletion)

AAACCCGGG
TTTGGGCCC

AAACTCGGG

TTTGAGCCC

I I I I I I I I I I
AAACCCTGGG

TTTGGGACCC

I I I I I I I I
AAACCGGG

TTTGGCCC

\ I

AAACTCGAG

TTTGAGCTC

I I I I

AAACCCGGG

TTTGGGCCC

I I I I

TTT

AACCCTGGG
TTGGGACCC

Figure 12.12 Types of DNA point mutations. Single-step
changes are replacements, additions, or deletions. A second
point mutation in the same gene can result either in a double
mutation, reversion to the original, or intragenic suppression. In
this case, intragenic suppression is illustrated by the addition of
one base followed by the nearby deletion of a different base.

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or subtract a base are, potentially, the most devastating in
their effects on the cell or organism because they change
the reading frame of a gene from the site of mutation on-
ward (fig. 12.13). A frameshift mutation causes two prob-
lems. First, all the codons from the frameshift on will be
different and thus yield (most probably) a useless pro-
tein. Second, stop-signal information will be misread.
One of the new codons may be a nonsense codon, which
causes translation to stop prematurely. Or, if the transla-
tion apparatus reaches the original nonsense codon, it is
no longer recognized as such because it is in a different
reading frame, and therefore, the translation process con-
tinues beyond the end of the gene.

Back Mutation and Suppression

A second point mutation in the same gene can have one
of three possible effects (see fig. 12.12). First, the muta-
tion can result in either another mutant codon or in one
codon that has experienced two changes. Second, if the
change is at the same site, the original sequence can
be returned, an effect known as back mutation: the
gene then becomes a revertant, with its original function
restored. Third, intragenic suppression can take
place. Intragenic suppression occurs when a second mu-
tation in the same gene masks the occurrence of the
original mutation without actually restoring the original
sequence. The new sequence is a double mutation that
appears to have the original (unmutated) phenotype.
In figure 12.12, a T addition is followed by an A deletion

that substitutes the AACCCT sequence for the original
AAACCC. These sequences, when transcribed
(UUGGGA, UUUGGG), are codons for leucine-glycine
and phenylalanine-glycine, respectively. Intragenic sup-
pression occurs whether the new codons are for differ-
ent amino acids or the same amino acids, as long as the
phenotype of the organism is reverted approximately to
the original. Suppressed mutations can be distinguished
from true back mutations either by subtle differences in
phenotype, by genetic crosses, by changes in the amino
acid sequence of a protein, or by DNA sequencing.

Conditional Lethality

A class of mutants that has been very useful to geneti-
cists is the conditional-lethal mutant, a mutant that is
lethal under one set of circumstances but not under an-
other set. Nutritional-requirement mutants are good
examples (see chapter 7). Temperature-sensitive mu-
tants are conditional-lethal mutants that have made it
possible for geneticists to work with genes that control
vital functions of the cell, such as DNA synthesis. Many
temperature-sensitive mutants are completely normal at
25° C but cannot synthesize DNA at 42° C. Presumably,
temperature-sensitive mutations result in enzymes with
amino acid substitutions that cause protein denaturation
to occur at temperatures above normal. Thus, the en-
zyme has normal function at 25° C, the permissive
temperature, but is nonfunctional at 42° C, the re-
strictive temperature.

Original short mRNA

( Met Y Leu Y GI Y Y Ph e Y Ser Y Pr ° Y Pr ° Y Ser 1

(Amber)

AUGCUCGGGUUUAGUCCACCGUCAUAG

Stop

New mRNA

G inserted

( Met Y Leu Y Gly Y Leu J (Amber)

AUGCUCGGGUUGUAGUCCACCGUCAUAG

New
stop

Figure 12.13 Possible effects of a frameshift mutation. The insertion of a single base
results in the creation of a new stop sequence {amber). The result will be premature
termination of translation.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

The interesting thing about most conditional-lethal
mutants of E. colt that cannot synthesize DNA at the re-
strictive temperature is that they have a completely nor-
mal DNA polymerase I. From this information, we infer
that polymerase I is not the enzyme E. colt normally uses
for DNA replication. When an organism with a condi-
tional mutation of polymerase I was isolated, it was able
to replicate its DNA normally, but unable to repair dam-
age to the DNA. This led to the conclusion that poly-
merase I is primarily involved in repair rather than repli-
cation of DNA. Conditional-lethal mutants thus allow
genetic analysis on genes otherwise impossible to study.

Spontaneous Mutagenesis

Watson and Crick originally suggested that mutation
could occur spontaneously during DNA replication if
pairing errors occurred. If a base of the DNA underwent
a proton shift into one of its rare tautomeric forms (tau-
tomeric shift) during the replication process, an inap-
propriate pairing of bases would occur. Normally, ade-
nine and cytosine are in the amino (NH 2 ) form. Their
tautomeric shifts are to the imino (NH) form. Similarly,
guanine and thymine go from a keto (C=0) form to an
enol (COH) form (fig. 12.14). Table 12.4 shows the new

Normal form

Tautomeric form

Adenine

Amino

To deoxyribose

- /

To deoxyribose

■ N

.C — C

H
N

\ /'

Imino

Cytosine

Amino

H
C-

N-

To deoxyribose

H
N

H //

c — c

/ \

\ /

N-

To deoxyribose

Imino

Guanine

-N

HC'

Keto

,N-

To deoxyribose

C — C

N^C

HC^"

— N-

To deoxyribose

-N OH

■ w

/ ^

\ N

\ /

N^C

NH
H

Enol

Thymine

Keto

CH,

HC NH

\ /

n — a

To deoxyribose

CH„ OH

\ /

C C

/ w

HC N

\ /

n — a

/ \

To deoxyribose

Enol

Figure 12.14 Normal and tautomeric forms of DNA bases. Adenine and cytosine can exist in the
amino, or the rare imino, forms; guanine and thymine can exist in the keto, or rare enol, forms.

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Table 12.4 Pairing Relationships of DNA Bases in

the Normal and Tautomeric Forms

In Normal State

In Tautomeric State

Base

Pairs with

base pairings that would occur following tautomeric
shifts of the DNA bases. Figure 12.15 illustrates the
molecular structure of one of these tautomeric pairings.

During DNA replication, a tautomeric shift in either
the incoming base (substrate transition) or the base
already in the strand (template transition) results in mis-
pairing. The mispairing will be permanent and result in a
new base pair after an additional round of DNA replica-
tion. The original strand is unchanged (fig. 12.16).

In the example in figure 12.16, the replacement of
one base pair maintains the same purine-pyrimidine rela-
tionship: AT is replaced by GC and GC by AT. In both
examples, a purine-pyrimidine combination is replaced
by a purine-pyrimidine combination. (Or, more specifi-
cally, a purine replaces another purine: guanine replaces

Cytosine

H
C-

N-

/
To deoxyribose

H
NH

-c

To deoxyribose

Adenine (amino)

Cytosine

H H

NH N

/ x

C—

-C

^CH

// \

\ /

N HN

/ \
v /

N-

To deoxyribose

C :
H

: N

■N
\

To deoxyribose

Adenine (amino)

Figure 12.15 Tautomeric forms of adenine. In the common
amino form, adenine does not base-pair with cytosine; in the
tautomeric imino form, it can.

DNA
replication

T
_L

Template transition — tautomerization of adenine in the template

DNA
replication

G
C

Substrate transition — tautomerization of incoming adenine

Figure 12.16 Tautomeric shifts result in transition mutations.
The tautomerization can occur in the template base or in the
substrate base. Tautomeric shifts are shown in red; the
resulting transition in blue. The transition shows up after a
second generation of DNA replication.

adenine in the first example and adenine replaces gua-
nine in the second.) The mutation is referred to as a tran-
sition mutation: a purine (or pyrimidine) replaces an-
other purine (or pyrimidine) through a transitional state
involving a tautomeric shift. When a purine replaces a
pyrimidine or vice versa, it is referred to as a transver-
sion mutation.

Transversions may arise by a combination of two
events, a tautomerization and a base rotation. (We saw
base rotations in the formation of 2 DNA in chapter 9.)
For example, an AT base pair can be converted to a TA
base pair (a transversion) by an intermediate AA pairing

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

(fig. 12.17). Adenine can pair with adenine if one of the
bases undergoes a tautomeric shift while the other ro-
tates about its base-sugar (glycosidic) bond (fig. 12.18).
The normal configuration of the base is referred to as the
anti configuration; the rotated form is the syn configura-
tion. Since we now believe that as many as 10% of bases

T
A

T
_L

->■ etc.

Figure 12.17 A model for transversion mutagenesis. An AT
base pair can be converted to a TA base pair (a transversion)
by way of an intermediate AA base pair. One of the red bases
is in the rare tautomeric form, while the other is in the syn
configuration. After a second round of DNA replication, one
DNA duplex will have a transversion at that point (blue).

Adenine
(a) To deoxyribose

Tautomeric shift

H H

N HN

NH N

180

Adenine /\

To deoxyribose

anti configuration
syn configuration

:N.

(b) To deoxyribose

To deoxyribose

Figure 12.18 Transversion mutagenesis. An AA base pair can
form if one base undergoes a tautomeric shift while the other
rotates about its glycosidic (sugar) bond. In (a), both bases are
in their normal configurations; no hydrogen bonding occurs. In
(£>), hydrogen bonds are possible.

may be in the syn configuration at any moment, the trans-
version mutagenesis rate should be about 10% of the
transition mutagenesis rate, a value not inconsistent with
current information.

Some base-pair mutations can have serious results.
If guanine undergoes an oxidation to 8-oxoguanine
(fig. 12.19), it pairs with adenine. A GC base pair is con-
verted to a TA base pair through an 8-oxoguanine-
adenine intermediate. This transversion has been found
to be common in cancers.

Since 1953, when Watson and Crick first described
the structure of DNA, tautomerization has been ac-
cepted as the obvious source of most transition muta-
tions. However, recent structural data has cast some
doubt on this assumption. X-ray crystallography and nu-
clear magnetic resonance (NMR) studies indicate that
both bases in transition mismatches may be in their nor-
mal forms. Other mechanisms, similar to wobble base
pairing (see chapter 11), may be responsible for most
transition mutations. These studies also indicate that
some transversions result from direct purine-purine or
pyrimidine-pyrimidine base pairing during DNA synthe-
sis. Much work needs to be done to clarify the nature of
spontaneous mutagenesis.

Chemical Mutagenesis

Muller demonstrated that X rays can cause mutation.
Certain chemical and temperature treatments can also
cause mutation. Determining the mode of action of vari-
ous chemical mutagens has provided insight into the mu-
tational process as well as the process of carcinogenesis
(box 12.1). In addition, knowing how chemical muta-
gens act has allowed geneticists to produce large
numbers of specific mutations at will (box 12.2).

8-Oxoguanine

Adenine

To deoxyribose

Figure 12.19 The structure of 8-oxoguanine, which base pairs
with adenine, converting a GC base pair to a TA base pair — a
transversion. (After C. Mol, et al., "DNA repair mechanisms for the
recognition and removal of damaged DNA bases," Annual Review of
Biophysics and Biomolecular Structure, 28:101-28,1999, figure 6.)

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Mutation

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Transitions

Transitions are routinely produced by base analogues.
Two of the most widely used base analogues are the
pyrimidine analogue 5-bromouracil (5BU) and the
purine analogue 2-aminopurine (2AP; fig. 12.20). The
mutagenic mechanisms of the two are similar. The
5-bromouracil is incorporated into DNA in place of
thymine; it acts just like thymine in DNA replication and,
since it doesn't alter the hydrogen bonding, should in-
duce no mutation. However, it seems that the bromine
atom causes 5-bromouracil to tautomerize more readily
than thymine does. Thus, 5-bromouracil goes from the
keto form (fig. 12.20) to the enol form more readily than
thymine. Transitions frequently result when the enol
form of 5-bromouracil pairs with guanine.

5-Bromouracil

2-Aminopurine

To deoxyribose q

To deoxyribose

Figure 12.20 Structure of the base analogues 5-bromouracil
(5BU) and 2-aminopurine (2AP).

HC ;

To deoxyribose \

C-r— CH
^C ) N H

(a)

2-Aminopurine

HC :

To deoxyribose

C^CH

-/ x

N-
N G

H—

N-

(b)

2-Aminopurine

N CH

\ /

C N

/ \

To deoxyribose
Thymine

NH 2

C CH

/ \

-N CH

\ /

C N

/ \

To deoxyribose

Cytosine

Figure 12.21 Two possible base pairs with 2-aminopurine.
(a) In the normal state, 2-aminopurine acts like adenine and
pairs with thymine, {b) In the rare state, 2-aminopurine acts like
guanine and forms complementary base pairs with cytosine.

The 2-aminopurine is mutagenic by virtue of the fact
that it can, like adenine, form two hydrogen bonds with
thymine. When in the rare state, it can pair with cytosine
(fig. 12.21). Thus, at times it replaces adenine, and at
other times guanine. It promotes transition mutations.

Nitrous acid (HN0 2 ) also readily produces transi-
tions by replacing amino groups on nucleotides with
keto groups ( — NH 2 to =0).The result is that cytosine
is converted to uracil, adenine to hypoxan thine, and
guanine to xanthine. As figure 12.22 shows, transition

H-

H
N

\ /v

C C

N-

H-

N-

HC :

To deoxyribose

To deoxyribose O
Uracil

„c^\

■H-

Adenine

H
— N

N-

-H-

To deoxyribose \i q 1 1

Hypoxanthine

■N

O-

H-

O To deoxyribose
Cytosine

H
-N

N-

To deoxyribose

N-

■H-

N-
H

Xanthine

^ To deoxyribose
Cytosine

Figure 12.22 Nitrous acid converts cytosine to uracil, adenine
to hypoxanthine, and guanine to xanthine. Uracil pairs with
adenine, whereas its progenitor, cytosine, normally pairs with
thymine; hypoxanthine pairs with cytosine, whereas its
progenitor, adenine, normally pairs with thymine; and xanthine
pairs with cytosine — the same base that guanine, its
progenitor, pairs with. Thus, only the first two pairings result in
transition mutations.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

BOX 12.1

Which chemicals cause
cancer in human beings?
It is difficult to determine
whether any substance is a carcino-
gen. Tests for carcinogenicity usually
involve administering the substance
in question to laboratory rats or mice
to determine whether the substance
actually causes cancer in these ani-
mals. Even these tests, however, are
not absolute predictors of cancer in
people. Tests of this nature are very
expensive ($1 million to $2 million
each) and time-consuming (three to
four years). Since more than fifty
thousand different chemical com-
pounds are used in industry, with
thousands more added each year, the
challenge of making the working
environment as well as the general
environment safe seems overwhelm-
ing. We can, however, make a prelim-
inary determination about the
cancer-causing properties of any sub-
stance very quickly because of the
relationship between mutagenicity
and carcinogenicity. Many sub-
stances in the environment that can
cause cancer also cause mutations.
Both kinds of effects are related to
DNA damage.

Bruce Ames, at the University of
California at Berkeley, developed a
routine screening test for mutagenic-
ity Substances that prove positive in
this test are suspected of being car-
cinogens and would have to be tested
further to determine their potential
to cause cancer in mammals.

Ames worked with a strain of
Salmonella typhimurium that re-
quires histidine to grow. This strain
will not grow on minimal medium.

Biomedical
Applications

The Ames Test for
Carcinogens

However, the strain will grow if a mu-
tagen is added to the medium, caus-
ing the defective gene in the histidine
pathway to revert to the wild-type.
(Mutagens inducing gross chromoso-
mal damage, such as deletions or
inversions, will not be detected.) Un-
der normal circumstances, there is a
background mutation rate; a certain
number of Salmonella cells revert
spontaneously, and therefore a cer-
tain number of colonies will grow on
the minimal medium. A mutagen,
however, increases the number of
colonies that can grow on minimal
medium. This procedure is, therefore,
a rapid, inexpensive, and easy test for
mutagenicity.

To improve this test's ability to de-
tect carcinogens, Ames added a sup-
plement of rat liver extract to the
medium. It is known that, although
many substances are themselves not
carcinogens, the breakdown of these
substances in the liver creates sub-
stances that are carcinogenic. Rat
liver enzymes act on a substance the
same way human livers do, convert-
ing a noncarcinogenic primary sub-
stance into a possible carcinogen.
The liver enzymes can also make a
mutagen nonmutagenic.

Other short-term tests are in use
that effectively duplicate the Ames
test. These include tests for muta-
genicity in mouse lymphoma cells
and two tests in Chinese hamster
ovary cells: a test for chromosomal
aberrations and a test for sister-
chromatid exchanges. None of these
tests surpasses the Ames test, which
has scored better than 90% correct
when tested with hundreds of
known carcinogens. Thousands of
other substances have been sub-
jected to this test; many have proven
to be mutagenic. These substances
are usually withdrawn from the
workplace or home environment.
From time to time, we read that a cer-
tain substance is believed to be car-
cinogenic and is being removed from
grocery store shelves. Examples have
included hair dyes, food preserva-
tives, food-coloring agents, and artifi-
cial sweeteners. Many of these first
were suspected after they failed the
Ames test.

Bruce Ames (1928- ).
(Courtesy Dr. Bruce Ames.)

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Mutation

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BOX 12.2

One way of studying the way
that proteins work is to
change the sequence of
amino acids in the protein. For exam-
ple, if a scientist were working on
the active site of a particular enzyme,
he or she could learn how the en-
zyme modifies its substrates by
changing one or a few amino acids.
Changes could be made, for exam-
ple, in order to study the role of
shape or charge on the functioning
of the enzyme. Advances in recom-
binant DNA techniques have made it
possible for a research scientist to
create exactly the changes he or she
wants in a protein.

To begin with, the gene for the
protein or enzyme must be cloned so
that it can be manipulated (see chap-
ter 13). Once cloned, deletions are
easy to create with restriction en-

Experimental
Methods

In Vitro Site-Directed
Mutagenesis

donucleases (described in chapter
13). If a particular endonuclease cuts
the gene in two places, the interven-
ing segment can be spliced out (fig.
la; see chapter 13). If the endonucle-
ase cuts only once, exonucleases can
digest the ends of the cut, extending
the deletion away from the cut in
both directions (fig. lb). Insertions
can be created by either cutting the
gene and repairing the single-
stranded ends (fig. lc) or by creating

Restriction
sites

Endonuclease

Cloned
gene

.Restriction
>/site

Endonuclease

DNA
plasmid

(b)

Cloned
gene

an oligonucleotide (a linker) with the
desired sequence and inserting the
linker at the site of an endonuclease
cut (fig. Id).

Far more impressive, however, is
the ability to change a single specific
codon in order to replace any amino
acid in the protein with any other
amino acid. The process involves di-
rected mutagenesis using artificially
created oligonucleotides.

Basically, a short sequence of
DNA (an oligonucleotide) is synthe-
sized complementary to a region of
the cloned gene, but with a change
in one or more bases of a codon to
specify a different amino acid. That
oligonucleotide is then hybridized
with the single-stranded form of the
clone (fig. 2). Although one or more

continued

Recircularize

Limited
\ t digestion

Recircularize

>■

I J Limited
' digestion

Figure 1 A cloned gene can be mutated in several ways, (a) If a restriction endonuclease has two sites in the gene, the
intermediate piece can be spliced out. (b) If the endonuclease has only one site, the gene can be opened at that site,
and limited digestion by exonucleases will delete part of the gene. continued

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Repair, and Recombination

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

BOX 12.2 CONTINUED

GATTC
CTAAG

DNA
Plasmid

G ATTC

CTAA G

Hint

endonuclease

Cloned
gene

DNA repair
enzymes

GATT
CTAA

ATTC
TAAG

GATTATTC
CTAATAAG

Recircularize

Restriction
site

/ /

Linker

Endonuclease

DNA
Plasmid

Cloned
gene

Insert and
recircularize

(d)

Figure 1 — continued (c) If an endonuclease has an offset region between its splice points of three or six nucleotides
(one or two codons), that length can be inserted by repairing the single-stranded ends after cutting by the
endonuclease. The resulting blunt ends can be spliced together. (Note that actually an ATT region has been converted to
an ATTATT region. If reading codons along the DNA, the actual insertion is of a TAT codon.) (c/) A linker of any length
(usually the length of a specific number of codons) can be inserted at a restriction site.

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Repair, and Recombination

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Mutation

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bases will not match, hybridization
can usually be facilitated by adjusting
the pH or ionic strength of the solu-
tion. The hybridized oligonucleotide
is then used as a primer for DNA
replication; the whole plasmid is
replicated, resulting in hybrid DNA.
In subsequent DNA replications of
the hybrid, both the original gene
and the mutated DNA will be pro-
duced. The latter can be isolated by
appropriate selection methods; it is a
plasmid with a cloned gene that has
the exact mutation the researcher
wanted. Using techniques of this
type, geneticists have made many ad-
vancements in understanding ex-
actly how various components of an
enzyme contribute to its function.

Artificial
oligonucleotide
with base
changes
hybridized to:

Plasmid with
gene cloned

DNA
replication

Original plasmid

Plasmid with site-directed mutation

Figure 2 Site-directed mutagenesis can involve any nucleotide(s). In this case,
an inserted gene with an lle-Gly sequence is converted, at the direction of the
investigator, to an lie-Ala sequence. A single-stranded form of the plasmid is
isolated. A synthetically prepared oligonucleotide (twenty-three bases in this
example) is added. It can be made to hybridize at the complementary site
despite differing by three bases. Then DNA replication is carried out using the
oligonucleotide configuration as a primer. After the strands of the duplex are
separated, the investigators can isolate the original plasmid as well as the
mutated plasmid. (Note that the investigators changed two codons, although
they changed only one amino acid, because they also wanted to introduce an
Alu site at that point for future Studies.) (From J. E. Villafranca, et al., "Directed muta-
genesis of dihydrofolate reductase," Science 222:782-88. Copyright 1983 by the AAAS.)

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

mutation results from two of the changes. Uracil pairs
with adenine instead of guanine, thus leading to a UA
base pair in place of a CG base pair; hypoxanthine (H)
pairs with cytosine instead of thymine, the original base
paired with adenine. Thus, in this case, an HC base pair
replaces an AT base pair. Both of these base pairs (UA
and HC) are transition mutations. Xanthine, however,
pairs with cytosine just as guanine does. Thus, the re-
placement of guanine with xanthine does not cause
changes in base pairing.

Like nitrous acid, heat can also deaminate cytosine to
form uracil and thus bring about transitions (CG to TA).
Apparently, heat can also bring about transversions by an
unknown mechanism.

These sites where this happens are referred to as AP
(apurinic-apyrimidinic) sites. If the AP site is not re-
paired, any of the four DNA bases could be inserted into
the new strand opposite the gap (fig. 12.23). If thymine
is placed in the newly formed strand, then the original
base pair is restored; insertion of cytosine results in a
transition mutation; insertion of either adenine or gua-
nine results in a transversion mutation. Of course, the
gap is still there, and it continues to generate new muta-
tions each generation until it is repaired. During DNA
replication in E. colt, the polymerase tends to place ade-
nine opposite the gap more frequently than it places
other bases.

Transversions

Ethyl methane sulfonate (CH3SO3CH2CH3) and ethyl
ethane sulfonate (CH3CH2SO3CH2CH3) are agents that
cause the removal of purine rings from DNA. The
multistep process begins with the ethylation of a purine
ring and ends with the hydrolysis of the glycosidic
(purine-deoxyribose) bond, causing the loss of the base.

Insertions and Deletions

The molecules of the acridine dyes, such as proflavin and
acridine orange (fig. 12.24), are flat. Presumably, they ini-
tiate mutation by inserting into the DNA double helix,
causing the helix to buckle in the region of insertion,
possibly leading to base additions and deletions during
DNA replication. Crick and Brenner used acridine-
induced mutations to demonstrate both that the genetic

1 1 r

GAG

C T C
J I L

Ethyl

methane

sulfonate

1 I

G
C

T C
J L

Normal
replication

1 1
G

* C
1

T C

1 1

1
G

1 1
G

C
1

C C

1 1

1
G

1 1
G

C
1

A C

1 1

1
G

1 1
G

C
1

G C

1 1

Continued
mutagenesis

1 1
A G

C
1

T C

1 1

1
G

1 1
G G

C
1

C C

1 1

1
G

1 1
T G

C
1

A C

1 1

1
G

1 1
C G

C
1

G C

1 1

Restoration

Transition

Transversion

Figure 12.23 Four possible outcomes after treatment of DNA with an alkylating agent, which removes the
purine — adenine in this example. The bases shown in red are the four bases that DNA polymerase may insert
opposite the gap. After another round of DNA replication, the gap remains to generate further mutations. The
inserted base forms a base pair (blue), which can be a restoration or a transition or transversion mutation.

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Mutation

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code was read from a fixed point and that it was triplet
(chapter 11).

Misalignment Mutagenesis

Additions and deletions in DNA can also come about by
misalignment of a template strand and the newly formed
(progeny) strand in a region containing a repeated se-

Proflavin

Acridine orange

Figure 12.24 Structure of two acridine dyes: proflavin and
acridine orange.

quence. For example, in figure 12.25 we expect the prog-
eny strand to contain six adjacent adenines because the
template strand contains six adjacent thymines. Misalign-
ment of the progeny strand results in seven consecutive
adenines: six thymines replicated, plus one already repli-
cated but misaligned. Misalignment of the template
strand results in five consecutive adenines because one
thymine is not available in the template. Regions with
long runs of a particular base may be very mutation
prone. They may explain the "hot spots" observed by
Benzer (see fig. 12.8) and others.

Intergenic Suppression

When a critical mutation occurs in a codon, several
routes can still lead to survival of the individual; simple
reversion and intragenic suppression are two that we
have already considered. A third route is through inter-
genic suppression — restoration of the function of a
mutated gene by changes in a different gene, called a
suppressor gene. Suppressor genes are usually transfer
RNA genes. When mutated, intergenic suppressors
change the way in which a codon is read.

Suppressor genes can restore proper reading to
nonsense, missense, and frameshift mutations. Non-
sense mutations convert a codon that originally speci-
fied an amino acid into one of the three nonsense

Progeny strand 5' • • • T-C-A-A-A-A

Template strand 3' • • • A-G-T-T-T-T-T-T-G-T

Progeny-strand
misalignment

Template-strand
misalignment

T-C A-A-A

A-G-T-T-T-T-T-T-G-T

A V DNA synthesis

T-C A-A-A-A-A-A-C-A---

A-G-T-T-T-T-T-T-G-T

Addition of an A

T-C-A-A-A-A

A-G T-T-T-T-T-G-T

T-C-A-A-A-A-A-C-A

A-G T-T-T-T-T-G-T
\/

Deletion of an A

Figure 12.25 Misalignment of a template or progeny strand during DNA synthesis. If
the progeny strand is misaligned after DNA replication has begun, the resulting progeny
strand will have an additional base. If the template strand is misaligned during DNA
replication, the resulting progeny strand will have a deleted base. These changes will
show up after another round of DNA replication. (From J.W. Drake, B.W. Glickman, and LS.
Ripley, "Updating the Theory of Mutation," American Scientist, 71:621-630, 1983. Reprinted with
permission of American Scientist, magazine of Sigma Xi, The Scientific Research Society.)

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Repair, and Recombination

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

codons. Missense mutations change a codon so that it
specifies a different amino acid. Frameshift mutations, by
additions or deletions of nucleotides, cause an alteration
in the reading frame of codons. A frameshift mutation,
caused by the insertion of a single base, can be sup-
pressed by a transfer RNA that has an added base in its an-
ticodon (fig. 12.26a). It reads four bases as a codon and
thus restores the original reading frame.

The transfer RNA produced by a nonsense suppres-
sor gene reads the nonsense codon as if it were a
codon for an amino acid; an amino acid is placed into

tRNA

C U U U
G A A A

C T T T

5' • • • — ' — ' —

o • • • ' ' — ' — ' ' —

(a) Frameshift suppression

• • • 3' mRNA
•••5' DNA

A U C
U A G

A T C

• • • 3' mRNA
•••5' DNA

(b) Nonsense suppression

Figure 12.26 Frameshift and nonsense suppression by mutant
transfer RNAs. In (a), a thymine has been inserted into DNA,
resulting in a frameshift. However, a transfer RNA with four
bases in the codon region reads the inserted base as part of
the previous codon in the messenger RNA. The frameshift thus
does not occur. In (b), an amber mutation (UAG), which
normally results in chain termination, is read as tyrosine by a
mutant tyrosine transfer RNA that has the anticodon sequence
complementary to the amber codon.

the protein, and reading of the messenger RNA contin-
ues. At least three suppressors of the mutant amber
codon (UAG) are known in E. colt. One suppressor puts
tyrosine, one puts glutamine, and one puts serine into
the protein chain at the point of an amber codon. Nor-
mally, tyrosine transfer RNA has the anticodon 3 -AUG-
5 ' . The suppressor transfer RNA that reads amber as a
tyrosine codon has the anticodon 3-AUC-5', which is
complementary to amber. Hence, a mutated tyrosine
transfer RNA reads amber as a tyrosine codon (fig.
12.266).

If the amber nonsense codon is no longer read as a
stop signal, then won't all the genes terminating in the
amber codon continue to be translated beyond their
ends, causing the cell to die? In the tyrosine case, two
genes for tyrosine transfer RNA were found; one con-
tributes the major fraction of the transfer RNAs, and
the other, the minor fraction. It is the minor-fraction
gene that mutates to act as the suppressor. Thus, most
messenger RNAs are translated normally, and most am-
ber mutations result in premature termination, al-
though a sufficient number are translated (suppressed)
to ensure the viability of the mutant cell. In general, in-
tergenic suppressor mutants would be eliminated
quickly in nature because they are inefficient — the
cells are not healthy. In the laboratory, we can provide
special conditions that allow them to be grown and
studied.

Mutator and Antimutator Mutations

Whereas intergenic suppressors represent mutations
that "restore" the normal phenotype, mostly through mu-
tation of transfer RNA loci, mutator and antimutator
mutations cause an increase or decrease in the overall
mutation rate of the cell. They are frequently mutations
of DNA polymerase, which, as you remember, not only
polymerizes DNA nucleotides 5 ' — ► 3' complementary to
the template strand, but also checks to be sure that the
correct base was put in (they proofread). If, in the proof-
reading process, the polymerase discovers an error, it can
correct this error with its 3' — ► 5' exonuclease activity.
Mutator and antimutator mutations sometimes involve
changes in the polymerase's proofreading ability (exonu-
clease activity).

Phage T4 has its own DNA polymerase with known
mutator and antimutator mutants. Mutator mutants are
very poor proofreaders (they have low exonuclease-to-
polymerase ratios), and thus they introduce mutations
throughout the phage genome. Antimutator mutants,
however, have exceptionally efficient proofreading abil-
ity (high exonuclease-to-polymerase ratios) and, there-
fore, result in a very low mutation rate for the whole
genome (box 12.3).

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DNA Repair

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BOX 12.3

Although we discuss evolution
in detail at the end of the
book, note here that we view
mutation as a random process, not
one that occurs because a cell
"needs" a particular mutation. For ex-
ample, Luria and Delbriick's work
showed that the mutation in E. coli
for resistance to phage Tl occurred
randomly before exposure to the
phage, not because the cells would
benefit from the mutation.

Because of the entrenched dogma
that mutations occur through ran-
dom processes, the scientific commu-
nity was startled when, in 1988, John
Cairns — a highly respected senior
scientist — and colleagues reported a
new observation — adaptive, or
directed, mutation, occurring when
the cell needed it. Their system was
the lacZ gene in E. coli. Cells that
could not use lactose as an energy
source QacZ~} were plated on a
medium in which lactose was the sole
energy source. Some mutants already
there of course produced colonies
(lacZ + }. The expectation was that no
new mutations would occur over time
because the lacZ~ cells would have
either died or stopped metabolizing.
Unexpectedly, Cairns and colleagues
found that more and more colonies
appeared over time, coming from
cells that had mutated to lacZ + . As a
control, they looked for revertants of
other genes not involved with lactose
metabolism. These mutations did not
occur in a directed manner.

Scientists were extremely skepti-
cal of this work for two reasons. First,

Experimental
Methods

Adaptive Mutation

it seemed to fly in the face of our
common understanding of the muta-
tional process. Second, there were no
obvious explanations for how this
could occur. Numerous articles were
published refuting the notion of
adaptive mutations and suggesting
other explanations for the results.
These explanations included artifacts
of miscounting cells to mutants that
were extremely slow growing, but
there all the time.

In the past several years, other sci-
entists have found at least a half
dozen similar results in other organ-
isms and other genes. The debate is
ongoing, but work published in mid-
1994 seems to have recast it into the
realm of methods of mutagene-
sis rather than non-Darwinian pro-
cesses. Several scientists found that
the "directed" mutations seem to be
of a certain type, mainly single nu-
cleotide deletions within runs of the
same nucleotide: for example, a dele-
tion of a C in a CCCC sequence. This
type of error happens during DNA
replication and could be the result of
a repair deficiency. That is, under ex-
treme duress, the cells may be going
into a "hypermutational" mode, or

selection may favor hypermutable
genotypes in which repair mecha-
nisms are shut down in order to in-
tentionally create lots of errors in the
DNA. Any errors that do not alleviate
the problem result in cell death, a
death that was inevitable anyway;
however, some errors will correct
the problem QacZ~ to lacZ + }. Those
cells will survive. One recent study
indicated that a subpopulation of
about 0.06% of the population was
hypermutable, with a mutation rate
about 200 times that of the normal
cells. That group of cells could ac-
count for the adaptive mutations.
Other scientists found that when the
locus of importance was on a plas-
mid, the adaptive mutation could oc-
cur by increased replication of the
plasmid. (Currently, the favored term
is adaptive mutation, rather than
directed mutation, indicating that
the mutations are useful to the cell,
not that some unknown process di-
rects them.)

This controversy, although not
necessarily resolved, has actually
brought out the best in the scientific
method: A skeptical scientific com-
munity tried its best to refute an un-
reasonable observation. Other scien-
tists then repeated the observation
and extended it, making it more wor-
thy of further study. Finally, further
work has given us reasonable mecha-
nisms that not only require no rejec-
tion of the original concept of ran-
dom mutation, but actually give us
hypotheses to test further.

DNA REPAIR

Radiation, chemical mutagens, heat, enzymatic errors,
and spontaneous decay constantly damage DNA. For ex-
ample, it is estimated that several thousand DNA bases
are lost each day in every mammalian cell due to sponta-
neous decay. Some types of DNA damage interfere with

DNA replication and transcription. In the long evolution-
ary challenge to minimize mutation, cells have evolved
numerous mechanisms to repair damaged or incorrectly
replicated DNA. Many enzymes, acting alone or in con-
cert with other enzymes, repair DNA. Repair systems are
generally placed in four broad categories: damage rever-
sal, excision repair, double-strand break repair, and

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

postreplicative repair. Enzymes that process repair steps
have been conserved during evolution. That is, enzymes
found in E. colt have homologues in yeast, fruit flies, and
human beings. However, eukaryotic systems are almost
always more complex.

Damage Reversal

Ultraviolet (UV) light causes linkage, or dimerization, of
adjacent pyrimidines in DNA (fig. 12.27). Although
cytosine-cytosine and cytosine-thymine dimers are occa-
sionally produced, the principal products of UV irradia-
tion are thymine-thymine dimers. These can be repaired
in several different ways. The simplest is to reverse the
dimerization process and restore the original unlinked
thymines.

In E. colt, an enzyme called DNA photolyase, the
product of the^r gene (for photoreactivation), binds
to dimerized thymines. When light shines on the cell,

Bulge

Thymine

the enzyme breaks the dimer bonds with light energy.
The enzyme then falls free of the DNA. This enzyme thus
reverses the UV-induced dimerization. Another example
of an enzyme that performs direct DNA repair is O 6 -
mGua DNA methyltransferase, which removes the
methyl groups from 6 -methylguanine, the major prod-
uct of DNA-methylating agents (fig. 12.28). Although
other repair mechanisms seem to be present in all or-
ganisms, photoreactivation is not; it is apparently absent
in human beings.

Excision Repair

Excision repair refers to the general mechanism of
DNA repair that works by removing the damaged portion
of a DNA molecule. Various enzymes can sense damage
or distortion in the DNA double helix. During excision
repair, bases and nucleotides are removed from the dam-
aged strand. The gap is then patched using complemen-
tarity with the remaining strand. We can broadly catego-
rize these systems as base excision repair and nucleotide
excision repair, which includes mismatch repair. We will
discuss only the major repair pathways; others exist. Pre-
sumably, redundancy in repair has been selected for be-
cause of the critical need to keep DNA intact and rela-
tively mutation free.

Base Excision Repair

A base can be removed from a nucleotide within DNA
in several ways: by direct action of an agent such as ra-
diation, by spontaneous hydrolysis, by an attack of oxy-
gen free radicals, or by DNA glycosylases, enzymes
that sense damaged bases and remove them. Currently,
at least five DNA glycosylases are known. For example,
uracil-DNA glycosylase, the product of the ung gene in

O CH,

Figure 12.27 UV-induced dimerization of adjacent thymines in
DNA. The red lines represent the dimer bonds in the adjacent
thymines.

OH H

Figure 12.28 The structure of 6 -methylguanine. The red color
shows the modification of guanine, in which the normal
configuration is a double-bonded oxygen (keto form).

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341

E. colt, recognizes uracil within DNA and cleaves it out
at the base-sugar (glycosidic) bond. The resulting site is
called an AP (apurinic-apyrimidinic) site, because of
the lack of a purine or pyrimidine at the site (see fig.
12.23). An AP endonuclease then senses the minor
distortion of the DNA double helix and initiaties exci-
sion of the single AP nucleotide in a process known as
base excision repair. The AP endonuclease nicks the
DNA at the 5' side of the base-free AP site. A DNA
polymerase then inserts a nucleotide at the AP site; an
exonuclease, lyase, or phosphodiesterase enzyme then
removes the base-free nucleotide. (Lyases are enzymes
that can break C-C, C-O, and C-N bonds.) DNA ligase
then closes the nick (fig. 12.29). The replacement of
just one base occurs 80-90% of the time. In the re-
maining 10-20% of cases, several nucleotides may be
removed, depending probably on which DNA poly-
merase (I or III) first repairs the site (fig. 12.29). In
mammals, DNA polymerase p performs two roles in
base excision repair: It both inserts a new base where
the AP site was and also eliminates the AP nucleotide
residue by exonuclease activity.

One question that concerned scientists was how the
glycosylases gain access to the inappropriate or damaged
bases within the double helix. Recently, it has been
demonstrated that these enzymes remove the inappro-
priate or damaged bases by first flipping them out of the
interior of the double helix in a process called base
flipping. For example, the enzyme in human beings that
recognizes 8-oxoguanine in DNA (see fig. 12.19),
8-oxoguanine DNA glycosylase, flips the base out to ex-
cise it. Base flipping seems to be a common mechanism
in repair enzymes that need access to bases within the
double helix (fig. 12.30).

Nucleotide Excision Repair

Whereas base excision repair is initiated by glycosylases
and usually involves the replacement of only one nu-
cleotide residue, nucleotide excision repair is initi-
ated by enzymes that sense distortions in the DNA back-
bone and replace a short stretch of nucleotides. For
example, six enzymes in E. colt excise a short stretch of
DNA containing thymine dimers if the dimerization is
not reversed by photoreactivation. Two copies of the
protein product of the uvrA gene (for ultraviolet light —
UV — repair) combine with one copy of the product of
the uv rB gene to form a UvrA 2 UvrB complex that moves
along the DNA, looking for damage (fig. 12.31). (The
complex has 5 ' to 3 ' helicase activity.) When the com-
plex finds damage such as a thymine dimer, with moder-
ate to large distortion of the DNA double helix, the
UvrA 2 dimer dissociates, leaving the UvrB subunit alone.
This causes the DNA to bend and attracts the protein
product of the uv rC gene, UvrC. The UvrB subunit first

1
U

1
T

1
G

1
C

1
T

1
A

T
1

Uracil-DNA glycosylase

1
T

1
G

1
C

1
T

1
A

T
1

AP endonuclease

1
T

1
G

1
C

1
T

1
A

T
1

G A

1 1

T
1

80-90%

DNA
polymerase

C T

1
G

1
C

1
T

1
A

T
1

G A

1 1

T
1

DNA
polymerase

1
C

1
T

1
G

1
C

1
T

1
A

T
1

Exonuclease,
lyase, or
phosphodiesterase

Exonuclease

1
C

1
T

1
G

1
C

1
T

1
A

T
1

1
A

1
C

1
T

1
G

1
C

1
T

1
A

T
1

DNA ligase

1
C

1
T

1
G

1
C

1
T

1
A

T
1

1
A

1
C

1
T

1
G

1
C

1
T

1
A

T
1

Figure 12.29 Mechanism of base excision repair. In this case,
a uracil-DNA glycosylase enzyme removes a uracil {red) from
DNA. An AP (apurinic-apyrimidinic) endonuclease nicks the
DNA on the 5' side of the base-free site. Between 80 and
90% of the time, a DNA polymerase will replace the single
nucleotide {green); an exonuclease, lyase, or phosphodiesterase
will remove the base-free nucleotide. The final nick is sealed
with DNA ligase. Between 1 and 20% of the time, the DNA
polymerase will extend polymerization beyond the single
nucleotide. In those cases, an exonuclease and DNA ligase

finish the repair. (From T. Lindahl, "The Croonian Lecture, 1996:
Endogenous Damage to DNA," Philosophical Transactions of the Royal
Society of London, B351, pp. 1529-1538, figure 6, 1996.)

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

UvrA 2 UvrB
locates lesion

(a)

(b)

Figure 12.30 Two views of the human enzyme, uracil-DNA
glycosylase, bound to DNA that has a uracil present. In (a), the
uracil-containing nucleotide residue has been flipped out and
the uracil cleaved; in (b), the uracil-containing residue has been
flipped out, but the uracil has not yet been cleaved. In both,
DNA is a green stick figure with red oxygen, yellow
phosphorus, and blue nitrogen atoms. In (a), the enzyme is
shown as a ribbon diagram; note the cleaved uracil bound to
the ribbon. In (b), the enzyme is shown as a molecular surface;
the uracil-containing nucleotide is flipped out by the purple
knob just left of and below center of the structure with the
flipped-out residue to the right. (From s. Parikh, c. Moi, and

J. Tainer, "Base excision repair enzyme family portrait" in Structure, 1997,
5:1543-1550, fig 1 a&b, p. 1544. Courtesy of J.A. Tainer, The Scripps
Research Institute.)

I
G

I
C

I
G

I
C

I
G

I
A

I
C

I
T =

I
= T

I
C

I
G

I
A

T
I

UvrC

UvrA 2 dissociates
UvrC attaches

i i i i i i i i i r

GCGCGAAC T=T

UvrB

i i i i r

C G G A A

ACGCGCTTGAAGCCTT

J I I I I I I I I I I I I I I L

UvrB then UvrC

nick DNA

UvrD helicase unwinds

oligonucleotide with dimer

DNA polymerase I

+
DNA ligase

1
G

1
C

1
G

1
C

1
G

1
A

1
C

1
T =

1
= T

1
C

1
G

1
A

T
1

1
T

1
G

1
C

1
G

1
C

1
G

1
A

1
C

1
T

1
C

1
G

1
A

T
1

Figure 12.31 Nucleotide excision repair. A lesion in DNA (a
thymine dimer) is located by a protein made of two copies of
UvrA and one of UvrB. Then, the UvrA subunits detach, and
UvrC attaches on the 5' side of the lesion. UvrB nicks the
DNA on the 3' side and UvrC on the 5' side of the lesion;
UvrD helicase unwinds the oligonucleotide containing the
lesion {red). DNA polymerase I and DNA ligase then repair the
patch {green).

nicks (hydrolyzes) the DNA four to five nucleotides on
the 3' side of the lesion; next, the UvrC subunit nicks the
DNA eight nucleotides on the 5' side of the lesion. (The
three components, UvrA, UvrB, and UvrC, are together
called the ABC excinuclease, for excision endonucle-
ase.) The enzyme helicase II, the product of the uvrD
gene, then removes the twelve- to thirteen-base oligonu-
cleotide as well as UvrC. DNA polymerase I fills in the
gap and, in the process, evicts the UvrB, and DNA ligase
closes the remaining nick (fig. 12.31). This is another rel-
atively simple system designed to detect helix distor-
tions and repair them.

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DNA Repair

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Like base excision repair, nucleotide excision repair
is present in all organisms. In yeast, approximately
twelve genes are involved, many in what is called the
RAD 3 group. In human beings, twenty-five proteins are
involved; they remove twenty-seven to twenty-nine nu-
cleotides, as compared to twelve to thirteen in E. colt.

Transcription and nucleotide excision repair are
linked in eukaryotes. Transcription factor TFIIH (see chap-
ter 1 1) is involved in repair of UV damage; it has helicase
activity and is found in both processes. Since it has been
shown that genes that are actively being transcribed are
preferentially repaired, we can now envision a model in
which transcription, when blocked by a DNA lesion like a
thymine dimer, signals the formation of a repair complex,
using TFIIH in both processes. In prokaryotes, RNA poly-
merase dissociates from the DNA in this circumstance,
losing the nascent transcript. This would be inefficient in
eukaryotes, whose genes are much longer and more ex-
pensive to transcribe; for example, the human dystrophin
gene, defective in the disease Duchenne muscular dystro-
phy, is 2.4 million bases long and takes almost eight hours
to transcribe. We believe that eukaryotic RNA polymerase
II backs up when stalled at a DNA lesion and continues af-
ter the lesion is repaired, without losing the transcript.
Much active research is going on in this area.

In human beings, the autosomal recessive trait
xeroderma pigmentosum is caused by an inability to re-
pair thymine dimerization induced by UV light. Persons
with this trait freckle heavily when exposed to the UV
rays of the sun, and they have a high incidence of skin
cancer. There are seven complementation groups (loci
XPA-XPG) whose protein products are involved in the
first steps of nucleotide excision repair and whose de-
fects cause xeroderma pigmentosum in human beings.
One of them, XPD, is a component of TFIIH.

Excision repair triggered by mismatches is referred to
as mismatch repair, which encompasses about 99% of
all DNA repairs. As DNA polymerase replicates DNA,
some errors are made that the proofreading polymerase
does not correct. For example, a template G can be
paired with a T rather than a C in the progeny strand. The
GT base pair does not fit correctly in the DNA duplex.
The mismatch repair system, which follows behind the
replicating fork, recognizes this problem. This system,
whose members in E. coli are specified by the tnutH,
mutt, mutS, and mutU genes, is responsible for the re-
moval of the incorrect base by an excision repair
process. (The genes are called mut for mutator because
mutations of these genes cause high levels of sponta-
neous mutation in the cells. The mutU gene is also
known as uvrD.) The mismatch repair enzymes initiate
the removal of the incorrect base by nicking the DNA
strand on one side of the mismatch.

You might wonder how the mismatch repair system
recognizes the progeny, rather than the template, base as

the wrong one. After all, in a mismatch, there are no de-
fective bases — theoretically, either partner could be the
"wrong" base. In E. coli, the answer lies in the methyla-
tion state of the DNA. DNA methylase, the product of the
dam locus, methylates 5-GATC-3' sequences, which are
relatively common in the DNA of E. colt, at the adenine
residue. Since the mismatch repair enzymes follow the
replication fork of the DNA, they usually reach the site of
mismatch before the methylase does. Template strands
will be methylated, whereas progeny strands, being
newly synthesized, will not be. Thus, the methylation
state of the DNA cues the mismatch repair enzymes to
eliminate the progeny-strand base for repair. After the
methylase passes by, both strands of the DNA are methyl-
ated, and the methylation cue is gone.

In figure 12.32, we present one model of mismatch re-
pair. The MutS protein, in the form of a homodimer — two

Template 5'
Progeny 3'

Discovery by
MutS

1 ¥ 1

GAT

C T A

i i i

i
C

MutL binds: Methylation
signal located

MutH endonuclease nicks
▼ DNA at GATC site

i i i
GAT

C T A

i i i

i
C

MutU binds and unwinds
nicked progeny strand

i i i
GAT

C T A

i i i

i
C

i
G

\jS^

Exonuclease
digestion

i i i
GAT

C T A

i i i

i
C

i
G

* >

DNA polymerase III
plus DNA ligase

-I- <-
-< I--

-o o-

i
C

i
G

Figure 12.32 Mismatch repair. The MutS protein discovers
mismatches; MutL binds and the MutH endonuclease nicks
the progeny strand at the 3'-CTAG-5' sequence. MutU
helicase unwinds the nicked oligonucleotide with the
mismatch {red). Exonuclease digestion, followed by DNA
polymerase III and DNA ligase repair, completes the
operation.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

copies of the same protein — finds the mismatch. MutL,
also in the form of a homodimer, then binds, and together
they find the methylation signal. They also activate the en-
donuclease MutH, which then nicks the unmethylated
strand at the 3 -CTAG-5' recognition site, which can be
one thousand to two thousand bases away from the mis-
match. At the recognition site, the MutS-MutL tetramer
loads the helicase MutU (UvrD), which then unwinds the
nicked strand. Any one of at least four different exonucle-
ases then attacks the unwound oligonucleotide. DNA
polymerase III then repairs the gap, and DNA ligase seals
it. This sequence of events highlights a common theme in
DNA repair: Once a lesion is found, the damaged DNA has
some protein bound to it until the repair is finished.

Our understanding of DNA damage and repair helps
provide an answer to an evolutionary question — Why
does DNA have thymine while RNA has uracil? If we live
in an RNA world, in which RNA evolved first, why don't
DNA and RNA both contain uracil? One answer is that a
common damage to cytosine, spontaneous deamina-
tion, results in uracil. If uracil were a normal base in
DNA, the conversion of cytosine to uracil by deamina-
tion would not leave any clue to a mismatch repair sys-
tem that a mutation had occurred. Thus, thymine re-
places uracil in DNA, since thymine is not confused
with any other normal base in DNA by common spon-
taneous changes. In fact, cytosine, guanine, adenine,
and thymine are not converted simply to any other of
the bases in DNA. Hence, changes of these bases leave
clues for the repair systems.

Double-Strand Break Repair

Some damage to DNA, such as that caused by ionizing ra-
diation, is capable of breaking both strands of the double
helix. When that happens, the cell uses one of two
mechanisms to repair the broken ends: It can simply
bring the ends back together (a process called nonho-
mologous end joining), or it can use a mechanism that
relies on the nucleotide sequences of a homologous
piece of DNA, such as a sister chromatid or a homolo-
gous chromosome. That method is called homology-
directed recombination.

In nonhomologous end joining, a protein called Ku, a
heterodimer of Ku70 and Ku80, binds to broken chro-
mosomal ends. It then recruits a protein kinase (PK CS );
their interaction and the interaction with other proteins
is stabilized by a scaffold protein called XRCC4 (for X-ray
cross complementation group 4). The complex directs
the annealing of the broken ends by DNA ligase IV No
particular sequence information is used, and if more than
two broken ends are present, incorrect attachments can
take place (e.g., translocations). The second method,
homology-directed recombination, involves a second

piece of DNA homologous to the broken piece. The
method is very similar to our current model of DNA re-
combination and is discussed in the section entitled "Re-
combination" later in the chapter.

Postreplicative Repair

When DNA polymerase III encounters certain damage in
E. coli, such as thymine dimers, it cannot proceed. In-
stead, the polymerase stops DNA synthesis and, leaving a
gap, skips down the DNA to resume replication as far as
eight hundred or more bases away. If allowed to remain,
this gap will result in deficient and broken DNA. Since
part of one strand is absent and the other has damage,
there appears to be no viable template for replicating
new DNA. However, the cell has two mechanisms to re-
pair this gap: one uses polymerases that can replicate
these lesions, and the other is a repair process that uses
homologous DNA.

Originally, several proteins were known to facilitate
the replication of DNA with lesions; they were believed
to interact with the polymerase to make it capable of us-
ing damaged DNA as a template. We now know that
these proteins are, in fact, polymerases that have the
ability to replicate damaged DNA. In E. coli, polymerase
V can copy damaged DNA. In yeast, polymerases r\ and £,
also called REV3/7 and RAD30 polymerases, respec-
tively, can also copy damaged DNA. Some of these poly-
merases are relatively error free; polymerase V and poly-
merases r| put adenine-containing nucleotides opposite
dimerized thymines. However, polymerases £ and the E.
coli polymerase IV, which also appears during times of
damage, are error prone in their replicative roles. One
possible reason for this is that the error-prone poly-
merases developed by evolutionary processes: They cre-
ate mutations at a time when the cell might need vari-
ability. That is, DNA damage can occur when the
environment is stressful for the cell; variability might
help the cell survive. As we will see later, the cell can
sense DNA damage and act appropriately.

In addition to using repair polymerases, the cell can
use a second repair mechanism to replicate damaged
DNA when the polymerase leaves a gap. A replication
fork creates two DNA duplexes. Thus, an undamaged
copy of the region with the lesion exists on the other
daughter duplex. A group of enzymes, with one speci-
fied by the recA locus having central importance, re-
pairs the gap. Since the repair takes place at a gap cre-
ated by the failure of DNA replication, the process is
called postreplicative repair. The recA locus was
originally discovered and named in the recombination
process. In fact, postreplicative repair is sometimes
called recombinational repair, and it shares many en-
zymes with recombination.

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The RecA Protein

The RecA protein has two major properties. First, it coats
single-stranded DNA (fig. 12.33) and causes that coated,
single-stranded DNA to invade double-stranded DNA
(fig. 12.34). By invasion, we mean that the single-stranded
DNA attempts to form complementary base pairs with
the antiparallel strand of the double-stranded DNA while
displacing the other strand of that double helix. A mech-
anism for this activity, assuming two sites on the enzyme,
appears in figure 12.35. RecA continues to move the
single-stranded DNA along the double-stranded DNA un-

Figure 12.33 Scanning tunneling microscope picture of single-
stranded DNA coated with RecA protein {large arrow). The small
arrow indicates uncoated, double-stranded DNA. (In fig. 12.35,
we show how the very large, coated DNA can invade the very
small, uncoated DNA.) (© Science VU-IBMRLA/isuals Unlimited.)

til a region of homology is found. The second major prop-
erty of the RecA protein is that, when stimulated by the
presence of single-stranded DNA, it causes autocatalysis
of another repressor, called LexA, and thus initiates sev-
eral sequences of reactions.

The RecA protein is responsible for filling a
postreplicative gap in newly replicated DNA with a
strand from the undamaged sister duplex. Gap-filling

RecA

(a)

frtT^

\ I I 1 1 I I I I I I I I I I
\\ \ 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Figure 12.34 One property of the RecA protein causes single-
stranded DNA to invade double-stranded DNA and to move
along it until a region of complementarity is found,
(a) Diagrammatic representation of the invasion of the RecA-
coated single-stranded DNA. (b) A more realistic diagram of the
same event. (Reproduced, with permission, from the Annual Review of
Biochemistry, Volume 61 , © 1 992 by Annual Reviews, Inc.)

Figure 12.35 A model of how the RecA protein can cause single-stranded DNA to invade a
double-stranded molecule, (a) Axial view of one nucleotide (with two phosphate groups) of
single-stranded DNA attached at site I in this cross-sectional diagram of the RecA protein.
The protein is about 60% larger than actually shown, (b) Duplex DNA is bound at site II of
RecA. (c) RecA protein rotates the bases so that the single-stranded DNA forms a
complementary base pair with one strand of the duplex, leaving the other strand of the duplex
unpaired (see fig. 12.34). (Reprinted with permission from P. Howard-Flanders, et al., "Role of RecA protein
spiral filaments in genetic recombination," Nature, 309:215-20. Copyright © 1984 Macmillan Magazines, Limited.)

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

■Thymine dimer

I I I I I I I I I T=T I I I I I I I I

New DNA
synthesis

RecA

T=T

_l L

Endonuclease

nicks

T=T

_l L

II II II II II II

TTT

DNA polymerase I
+DNAIigase

T=T

J L

Figure 12.36 RecA-dependent postreplicative DNA repair. DNA
polymerase III skips past a thymine dimer during DNA
replication (a). With the help of RecA, the single strand with the
thymine dimer invades the normal sister duplex (b). An
endonuclease nicks the new duplex at either side of the
thymine dimer site, freeing the new duplex with the thymine
dimer and leaving the sister duplex single-stranded (c). Repair
enzymes then create two intact daughter duplexes (c/).

processes then complete both strands. In figure
12.36a, we see a replication fork with a gap in the
progeny strand in the region of a thymine dimer. The
RecA protein is responsible for the damaged single
strand invading the sister duplex (fig. 12.36&). Endonu-
clease activity then frees the double helix containing
the thymine dimer (fig. 12.36c). DNA polymerase I and
DNA ligase return both daughter helices to the intact
state (fig. 12.36^). The thymine dimer still exists, but
now its duplex is intact, and another cell cycle is avail-

able for photoreactivation or excision repair to remove
the dimer.

The SOS Response

Postreplicative repair is part of a cell reaction called the
SOS response. When an E. colt cell is exposed to exces-
sive quantities of UV light, other mutagens, or agents that
damage DNA (such as alkylating or cross-linking agents),
or when DNA replication is inhibited, gaps are created in
the DNA. In the presence of this single-stranded DNA, the
RecA protein interacts with the LexA protein, the prod-
uct of the lex A gene. The LexA protein normally re-
presses about eighteen genes, including itself. The other
genes include recA, uvrA, uvrB, and uvrD; two genes
that inhibit cell division, sulA and sulB; and several oth-
ers. Each of these genes has a consensus sequence in its
promoter called the SOS box: 5'-CTGX 10 CAG-3' (where
X 10 refers to any ten bases). The LexA protein normally
binds at the SOS box, limiting the transcription of these
genes. When single-stranded DNA activates RecA, RecA
interacts with the LexA protein to trigger the autocat-
alytic properties of LexA (fig. 12.37). Transcription then
follows from all the genes having an SOS box. The two
inhibitors of cell division, the products of the sulA and
sulB genes, presumably increase the amount of time the
cell has to repair the damage before the next round of
DNA replication.

Eventually, the DNA damage is repaired. There is no
single-stranded DNA to activate RecA, and, therefore,
LexA is no longer destroyed. LexA again represses the
suite of proteins involved in the SOS response, and
the SOS response is over. Table 12.5 summarizes some
of the enzymes and proteins involved in DNA repair.

As we will mention in chapter 14, X prophage can be
induced into vegetative growth by UV light. This is an-
other effect of the SOS response. RecA not only causes
the LexA protein to be inactivated, but also directly inac-
tivates the X repressor, the product of the X c/gene. From
an evolutionary point of view, it makes sense for phage X
to have evolved a repressor protein that the RecA protein
inactivates. As a prophage, X is dependent on the survival
of the host cell. When that survival might be in jeopardy,
the prophage would be at an advantage if it could sense
the danger and make copies of itself that could leave the
host. One of these times might be when the host has suf-
fered a lot of DNA damage. The SOS response is a signal
to a prophage that the cell has received that damage.
Hence, the prophage is induced when RecA acts as a pro-
tease; the X repressor is destroyed, the cro protein be-
comes dominant, and vegetative growth follows. From an
evolutionary perspective, the E. colt cell has not created
an enzyme (RecA) that seeks out the X repressor for the
benefit of X. Rather, the X repressor has evolved for its
own advantage to be sensitive to RecA.

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Recombination

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Figure 12.37 The LexA protein represses its own gene, recA,
and several other loci [uvrA, uvrB, uvrD, sulA, and sulB) by
binding at the SOS box in each of the loci. Activated RecA
protein causes autocatalysis of LexA, eliminating the repression
of all these loci, which are then transcribed and translated.

LexA protein

00 Repression

No
repression

LexA autocatalysis

f QDO

J es o oo

RecA activated?

Catalyzed
LexA proteins

LexA proteins ry^S-^^

Translation

Transcription

OO,

SOS uvrA
box uvrB
uvrD
sulA
sulB
+ others

SOS recA
box

SOS
box

lexA

RECOMBINATION

Although recombination, the nonparental arrangement
of alleles in progeny, can come about both by indepen-
dent assortment and crossing over, we are concerned
here with recombination due to crossing over between
homologous pieces of DNA (homologous recombina-
tion). We briefly discuss transpositional recombination
in chapter 14 and site-specific recombination (e.g., X in-
tegration) in chapters 7, 14, and 16.

Recombination is a breakage-and-reunion process.
Homologous parts of chromosomes come into apposi-
tion and are then reconnected in a crosswise fashion (see
fig. 6.4). This general model fits what we know about the
concordance of recombination and repair: Both involve
breakage of the DNA and a small amount of repair syn-
thesis, and both involve some of the same enzymes.

Double-Strand Break Model
of Recombination

In 1964, R. Holliday suggested a model of homologous re-
combination that involved simultaneous breaks in one
strand each of the two double helices that were to cross
over. In 1983, J. Szostak and colleagues put forth a different
model, initiated by a double-strand break in one of the
double helices. At first, this model was not considered seri-

ously because a double-strand break was thought too dan-
gerous a DNA lesion for cellular enzymes to create. How-
ever, we now know that the double-strand break model is
generally correct, and we refer to the Holliday junction
for an intermediate stage in the process. The model de-
pends on DNA complementarity between the recombin-
ing molecules and is thus a model of great precision.

We begin with two double helices lined up as they
would be, for example, in a meiotic tetrad, ready to un-
dergo recombination (fig. 12.38a). The first step of the
process is a double-stranded break in one of the double
helices. In eukaryotes, the protein Spoil accomplishes

R. Holliday (1932- ). (Courtesy
of James L German, III, M.D.)

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

TablG 12.5 Some of the Enzymes and Proteins Involved in DNA Repair in E. colt, Not Including DNA

Polymerase I and III, DNA Ligase, and Single-Strand Binding Proteins

Enzyme

Gene

Action

Damage Reversal

DNA photolyase

phr

Undimerizes thymine dimers

DNA methyltransferase

ada

Demethylates guanines in DNA

Base Excision Repair

Uracil-DNA glycosylase

ung

Removes uracils from DNA

Endonuclease IV

nfo

Nicks AP sites on the 5 ' side

Exonuclease, lyase, or phosphodiesterase

several

Removes base-free nucleotide

Nucleotide Excision Repair

UvrA

uvrA

With UvrB, locates thymine dimers and other distortions

UvrB

uvrB

Nicks DNA on the 3 ' side of the lesion

UvrC

uvrC

Nicks DNA on the 5 ' side of the lesion

UvrD (helicase II)

uvrD

Unwinds oligonucleotide

Mismatch Repair

MutH

mutH

Nicks DNA at recognition sequence

MutL

mutL

Recognizes mismatch

MutS

mutS

Binds at mismatch

MutU (UvrD)

mutU

Unwinds oligonucleotide

Exonucleases

recj, xseA, sbcB

Degrades unwound oligonucleotide

DNA methylase

dam

Methylates 5'-GATC-3' DNA sequences

Double-Strand Break Repair

Ku70, Ku80

Binds to broken chromosomal ends

PK CS

Protein kinase

DNA ligase IV

LIG4

Ligates broken ends of DNA

XRCC4

Stabilization protein

Postreplicative Repair

Polymerase IV

DinB

DNA polymerase

Polymerase V

UmuC, UmuD

DNA polymerase

Polymerase m,

RAD30

DNA polymerase

Polymerase £

REV3, REV7

DNA polymerase

RecA

recA

Single-stranded DNA invades double-stranded DNA;
causes LexA to autocatalyze; protease

LexA

lexA

Represses SOS proteins

SulA, SulB

sulA, sulB

Inhibit cell division

this. The break is followed by 5' — > 3' exonuclease ac-
tivity to widen the gaps formed in the double helix and
create 3' single-stranded tails (fig. 12.38&, c). These tails
are coated with RecA protein that then catalyzes the in-
vasion of one of the single strands into the intact double
helix in direct apposition (fig. 12.38d). Repair of single-
stranded DNA by DNA polymerase I and DNA ligase then
replaces sections of previously digested DNA (fig.

12.38e). At this point, there is no "lost" genetic material;
however the two double helices are interlocked and
need to be freed of each other. Before that happens,
however, branch migration can take place, a process
in which the crossover point can slide down the du-
plexes (fig. 12.38/). In E. coli, the RuvAB complex, the
product of the ruvA and ruvB genes that together form
an ATP-dependent motor, moves the junction point

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Recombination

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5'-
3'-

3'-
5'-

(a)

3'
5'

5'
3'

Endonuclease

(b)

5'^> 3' Exonuclease
3'

3'.

(c)

RecA- mediated
invasion

(d)

Repair synthesis

(e)

Branch
migration o

i^V<-i

(f)

Resolution

oo, oo Patch

(g)

ii, ii Patch

(h)

oo, ii Splice

(i)

ii, oo Splice

Figure 12.38 The double-strand break model of genetic
recombination. Two homologous duplexes (a: red, blue) of the
four present in a meiotic tetrad are shown. An endonuclease
creates a double-stranded break in one of the duplexes (b). A
5'-> 3' exonuclease then digests away from the break in both
directions, creating 3' tails (c). RecA-mediated invasion occurs
in the second duplex (d), followed by repair synthesis to close
all gaps (e). Branch migration then takes place (f). Each of the
Holliday junctions is then resolved independently, either by
nicks in the two outer strands (o) or the two inner strands (/).
Therefore, four resolution structures are possible ig-f). In
patches, the ends of each duplex are the same as the original,
indicating that there may not be recombination for loci flanking
the point of crossover. In splices, the ends of each duplex have
recombined, indicating that flanking loci may have crossed

over. (Reprinted from Cell, Vol. 87, Frank Stahl, "Meiotic Recombination in

Yeast: Coronation of the Double-Stranded-Break Repair Model,"

(fig. 12.39). As the junction points move, they create het-
eroduplex DNA, places where the two strands of each
double helix come from different original helices. These
stretches have the potential to produce mismatches
where the two chromatids differed originally. To resolve
the cross-linked duplexes, a second cut at each junction
is required.

Each of the two crossover points is a Holliday junc-
tion. If we open these junctions, we can see that each
can be resolved in two different ways. (RuvC endonucle-
ase, the protein product of the ruvC gene, resolves the
Holliday junctions in E. colt; see fig. 12.39. RuvC cuts the
Holliday junction at the consensus sequence 5 '[A or
T]TT[G or C]-3'. The cut is on the 3' side of the two
thymines.) Since there are two Holliday junctions per
crossover, there are four potential combinations, as
shown in figure 12 3Sg-j. Some of these combinations
produce patches, where no recombination takes place
among loci to the sides of the hybrid piece. Other com-
binations produce splices, where reciprocal recombina-
tion of loci takes place at the ends. The Holliday junc-
tions can be seen in the electron microscope (fig. 12.40).
Note that homology-directed recombination to repair
double-strand breaks is similar to the process shown in
figure 12.38.

Bacterial Recombination

In bacterial recombination, a linear molecule recombines
with a circular molecule (see fig. 7.15). Usually, invading
DNA originates in the linear molecule. The RecBCD pro-
tein, whose subunits are the products of the recB, recC,
and recD loci, initiates the first steps in forming an invad-
ing linear DNA molecule. RecBCD is a helicase, an exo-
nuclease, and an endonuclease.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

(a)

Rotate to
open

Open arms
(b) into cross

(d)

t
Branch
migration

* — 1 — ~

^xy?^Y?$d£KyAy?^y^

Heteroduplex

(e)

(c)

Structure with
RuvA and RuvB

Figure 12.39 Branch migration at a Holliday junction, (a) Two double helices {red, blue) are connected by a
crossover, (b) The structure opens when one of the double helices rotates, (c) We further clarify the structure by
separating the arms into an open cross, showing the direction in which the arms move during branch migration
{arrows: the equivalent of pulling out on the left and right arms, drawing in the top and bottom arms), (d) A more
realistic drawing with the RuvA and RuvB proteins, indicating one of the RuvA tetramers behind the center of the
cross. A second tetramer (not shown) is located above the cross center, forming a RuvA sandwich of the cross
center. The RuvB hexamers are shown on either side of the cross, (e) RuvC can resolve the cross to form either a
splice or a patch, depending on which cut is made. (Reprinted with permission from Nature, Vol. 374, C. Parsons, et al.,
"Structure of a Multisubunit Complex that Promotes DNA Branch Migration." Copyright © 1995 Macmillan Magazines Limited.)

The RecBCD protein enters a DNA double helix from
one end and travels along it in an ATP-dependent
process. As it travels along the DNA, it acts as a 3' — > 5'
exonuclease, degrading one strand of the linear double
helix (fig. 12.41). This process continues until RecBCD
comes to a chi site, the sequence 5 -GCTGGTGG-3',

which appears about a thousand times on the E. colt
chromosome. RecBCD s recognition of that sequence at-
tenuates its 3' — > 5' exonuclease activity and enhances
its 5' — » 3' exonuclease activity, begun after an endonu-
cleolytic cleavage. From that point on, RecBCD creates a
3' overhang or tail. That tail is coated by RecA and then

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Recombination

351

RecBCD

Figure 12.40 A Holliday intermediate structure, equivalent
to the structure seen in figure 12.39c. Each arm is about
1 micron long. (H. Potter and D. Dressier. "DNA recombination: In
vivo and in vitro studies," Cold Spring Harbor Symposium on Quantitative
Biology, Volume XLIII, 1979, pp. 969-85.)

invades the circular bacterial chromosome to initiate a
crossover event (fig. 12.41). After this pairing, the un-
paired segments of the double helix of the bacteria and
the exogenote are both degraded. Finally, DNA ligase
seals the circular double helix. The resulting hybrid DNA
will then be open to mismatch repair that can restore
either original base pairs or base pairs from the invading
DNA (see below).

Hybrid DNA

The result of bacterial recombination or meiotic recom-
bination with branch migration is a length of hybrid
DNA. This hybrid DNA, also called heterozygous DNA
or heteroduplex DNA, has one of two fates, if we as-
sume a difference in base sequences in the two strands.
Either the heteroduplex can separate unchanged at the
next cell division, or the cell's mismatch repair system
can repair it (fig. 12.42). Without appropriate methyla-
tion cues, the mismatch repair system can convert the
CA base pair to either a CG or a TA base pair. If TA were
the original bacterial base pair, conversion to CG would
be a successful recombination, whereas return of the CA
to TA would be restoration rather than recombination.

Recombination in yeast, or any other eukaryote, gen-
erates two heteroduplexes.The repair process can cause
gene conversion (fig. 12.43), the alteration of progeny
ratios indicating that one allele was converted to another,
a phenomenon seen in up to 10% of yeast asci. The
mismatched AC will be changed to an AT or a GC base
pair; the mismatched TG base pair will be changed to TA
or CG. The result of the repair, as shown at the bottom
of figure 12.43, can be gene conversion in which an ex-

3'.

5':

chi

(a)

3'.

5''

(b)

3'.
5''

(c)

3'.
5'.

3'.

5':

■ 5'

RecA- mediated
invasion

(d)

3'.

5'
3'

■ 5'

Branch
migration

(e)

Repair

3'.
5'-

3'i

(f)

Degradation of
linear DNA

5'.

3'i

5'
3'

3' Bacterial
5' chromosome

3'
5'

(9)

Figure 12.41 RecBCD enters a linear DNA double helix {red)
at one end and travels along it, digesting the 3' strand. When
the protein encounters a chi site {green), it cuts the other
strand and begins acting as a 3' -> 5' exonuclease, creating a
3' overhang (b, c). The 3' overhang can then invade a double
helix mediated by RecA. Repair and degradation of the linear
DNA results in hybrid DNA in the bacterial chromosome, which
can be fixed by the mismatch repair system.

a + a + ) is converted to a 3:1 ra-
a + ) or a 1:3 ratio (a~ a + a + a + ). If the het-

pected ratio of 2:2 (a
tio (a~ a~ a

eroduplexes are not repaired, then a single cell generates
both kinds of offspring after one round of DNA replica-
tion. Thus, the colony from the cell will be half wild-type
(a + ) and half mutant (cC). We see this phenomenon only
in an Ascomycete fungus, such as yeast, in which all the
products of a single meiosis remain together.

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

Replication

without

repair

i i i r

A C C G

T G G C

I I I I

and

"i — i — i — r

A C T G

T G A C
J I I L

i — i — i — r

A C C G

T G G C
J I I L

1 I I r
T G A C

A C T G
J I I L

Recombination

Y
T

i — r

A C C G

T G A C
J I I L

3' -i

3' - 1

— Parental duplexes

— Heteroduplex

Repair

1 I I T

A C C G

T G G C

J I I I

i — r

A C

i — r

T G

T G A C
J I I L

Figure 12.42 Fate of a heteroduplex DNA. Recombination results in heteroduplex DNA with
mismatched bases. Replication without repair produces two different daughter molecules.
Repair converts the mismatched base pair to one or the other normal base pair.

SUMMARY

STUDY OBJECTIVE 1: To look at the nature of mutation in
prokaryotes 316-317

In 1943, Luria and Delbriick demonstrated that bacterial
changes are true mutations similar to mutations in higher
organisms. They showed that a high variability occurs in
the number of mutants in small cultures as compared with
the number of mutants in repeated subsamples of a large
culture. Mutations thus occur spontaneously; they are
caused by mutagens, which include chemicals and radia-
tion. This chapter is concerned primarily with point muta-
tions rather than changes in whole chromosomes or chro-
mosomal parts.

STUDY OBJECTIVE 2: To analyze functional and struc-
tural allelism and examine the mapping of mutant sites
within a gene 317-324

Allelism is defined by the cis-trans complementation test.
Complementation implies independent loci, or nonallelic
genes. Lack of complementation implies allelism. Func-
tional alleles that differ from each other at the same nu-
cleotides are also called structural alleles. Benzer used T4
phages to do fine-structure studies using complementation
testing and deletion mapping.

STUDY OBJECTIVE 3: To verify the colinearity of gene
and protein 324-325

Yanofsky demonstrated colinearity of the gene and protein.
He had the advantage of working with a gene whose pro-
tein product was known.

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Summary

353

Tetrad

Homologous
chromosomes

Can be repaired to

(a - ) A or G (a + )
T C

(a") A or G (a + )

a
a"

T
I

I
A

T
I

I
G

C
I

I
G

C
I

Sister
chromatids

Recombination between
middle two, producing
two heteroduplexes

T
1

1
A

C
1

1
G

T
1

1
G

C
1

Heteroduplex

a + a a +

or or or

Gene conversion

Figure 12.43 Recombination and repair can cause gene
conversion. During recombination, heteroduplex DNA is formed,
containing mismatched base pairs. Without methylation cues,
repair enzymes convert the mismatch to a complementary base
pair, in a random fashion — that is, an AC base pair can be
converted to either an AT (a~ allele) or a GC (a + allele) base
pair. Two of the four possible repair choices create 3:1 ratios of
alleles rather than the expected 2:2 ratios in the offspring. The
3:1 ratio represents gene conversion.

STUDY OBJECTIVE 4: To study mutagenesis 325-338

After a mutation, the normal phenotype, or an approxima-
tion of it, can be restored either by back mutation or sup-
pression. Intragenic suppression occurs when a second
mutation within the same gene causes a return of normal or
nearly normal function. Intergenic suppression occurs
when a second mutation happens, usually in a transfer RNA
gene, that counteracts the original mutation. Nonsense,
missense, and frameshift mutations can all be suppressed.

Spontaneous mutation probably occurs primarily be-
cause of tautomerization of the bases of DNA. If a base is in
the rare form during DNA replication, it can form unusual
or mutant base pairings. We describe the mechanisms of
action of the most common mutagens.

STUDY OBJECTIVE 5: To investigate the processes of
DNA repair and recombination 339-352

DNA repair processes can be divided into four categories:
damage reversal, excision repair, double-strand break re-
pair, and postreplicative repair. Photoreactivation is an ex-
ample of damage reversal. Thymine dimers are undimerized
by a photolyase enzyme in the presence of light energy. Ex-
cision repair removes a damaged section of a DNA strand.
Repair enzymes fill in the gap. Excision repair can be di-
vided into three types. In base excision repair, bases are re-
moved by environmental causes or by glycosylases that
sense damaged bases. AP endonuclease and an exonucle-
ase, phosphodiesterase, or lyase then removes the base-free
nucleotide. Some enzymes use base flipping to gain access
to nucleotides in the double helix.

In nucleotide excision repair, enzymes in the Uvr sys-
tem remove a patch containing the lesion, usually a
thymine dimer. In mismatch repair, enzymes of the Mut sys-
tem use methylation cues to remove a progeny patch con-
taining the mismatch.

Double-strand break repair relies on one of two mecha-
nisms. In nonhomologous end joining, the cell simply
brings the broken ends back together. In homology-
directed recombination, the cell repairs the broken ends us-
ing a recombinational mechanism.

Postreplicative repair fills in gaps left by DNA poly-
merase III. Some polymerase enzymes can use lesions as
templates. Otherwise, the RecA protein is central to the
process. A single strand from the undamaged duplex is

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

used to fill the gap in the damaged duplex. Single-stranded
DNA induces the SOS response, which temporarily elimi-
nates LexA-mediated repression.

Recombination in eukaryotes begins with a double-
stranded break in one double helix, followed by invasion of
one of the ends into the other double helix. Repair, ligation,
and branch migration follow. The crossover points, called

Holliday junctions, need to be resolved, resulting in patches
and splices. In E. coli, the RecBCD protein invades linear
DNA, creating tails for invasion of the circular bacterial
chromosome. Recombination results in heteroduplex DNA,
which, if repaired, can lead to gene conversion. Thus, a bat-
tery of enzymes within the cell can modify DNA. These en-
zymes serve in DNA replication, repair, and recombination.

SOLVED PROBLEMS

PROBLEM 1: An investigator isolates two recessive wing
mutants of Drosophila melanogaster. The flies differ in
wing vein pattern. Are the mutations that cause these
variants allelic?

Answer: To verify allelism, the investigator must create a
heterozygote of the two mutations by either mating the
flies, if they are of opposite sexes, or breeding each mutant
into a separate stock for new matings. If the heterozygotes
are of the wild-type, then the mutations are not allelic. If
the heterozygote has a mutant phenotype, then we pre-
sume the mutations are functional alleles. (Allelism should
be verified in females to be sure that the locus is not on the
X chromosome, since males have only one.) If the muta-
tions are functional alleles, it is possible to determine
whether they are also structural alleles by looking for wild-
type offspring of the heterozygote. If they occur at a rate
higher than the background mutation rate, the alleles are
not structural alleles. If wild-type offspring occur at the
mutation rate, the alleles are presumably structural.

PROBLEM 2: What is the difference between mismatch
repair and AP repair?

Answer: Both processes are similar in that they entail
removal of an incorrect base in a DNA double helix by an
excision process folio wed by a repair process. The
processes differ in the event that triggers them.
Mismatch repair is triggered by a base pair that does not
occupy the correct space in the double helix — that is, by
a non-Watson and Crick pairing (not AT or GC). AP re-
pair is triggered by enzymes that recognize a missing
base.

PROBLEM 3: What role does the RecBCD protein play in
recombination?

Answer: For recombination to take place in E. coli, a sin-
gle strand of DNA from the exogenote must insinuate it-
self into the chromosomal double helix with the help of
the RecA protein. It is the RecBCD protein that creates
the single-stranded DNA. It does so by traveling down the
double helix, creating a single-strand tail in its wake. At
chi sites, it switches the activity of the enzyme from a
3' — » 5' exonuclease to a 5' — > 3' exonuclease, creating a
3' tail. RecA can then act on this single-stranded tail to
initiate recombination.

EXERCISES AND PROBLEMS

MUTATION

1. Construct a data set that Luria and Delbriick might
have obtained that would prove the mutation theory
wrong.

2. What types of enzymatic functions are best studied
using temperature-sensitive mutations?

3. Seven arginine-requiring mutants of E. coli were inde-
pendently isolated. All pairwise matings were done
(by transduction) to determine the number of loci
(complementation groups) involved. If a (+) in the fol-
lowing figure indicates growth and a (— ) no growth
on minimal medium, how many complementation
groups are involved? Why is only "half" a table given?
Must the upper left to lower right diagonal be all (— )?

5
6

+
+

+
+
+
+

* Answers to selected exercises and problems are on page A-13-

4. Several rll mutations (M to S) have been localized to
the A cistron because of their failure to complement
with a known deletion in that cistron. The phages
carrying these mutations are then mated pairwise
with the following series of subregion deletions. The
mating is done on E. coli B and plated out on E. coli

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Exercises and Problems

355

K12 (see the figure). A (+) shows the presence of
plaques on K12, whereas a (— ) shows an absence of
growth. Look at the deletion map of the area and lo-
calize each of the rll A mutations on this map.

Deletion

Mutant

—

A cistron

I I

1 I

'> I

Relative positions of deletions

5. A Drosophila worker isolates four eye-color forms of
the fly: wild-type, white, carmine, and ruby. (The
worker does not know that white, carmine, and ruby
are three separate loci on the X chromosome.) What
crosses could the researcher make to determine al-
lelic relations of the genes? What results would be
expected? A new mutant, eosin, is isolated. What
crosses should be carried out to determine that
eosin is an allele of white?

6. Define structural and functional alleles. What is the
cis part of a cis-trans complementation test?

7. Did Benzer and Yanofsky work with genes that had
intervening sequences? What relevance might in-
trons have to their work?

8. How can intra-allelic complementation result in in-
correct conclusions about allelism?

9. E. colt bacteria of strain K12 are lysogenic for phage
A. Why won't rll mutants of phage T4 grow in these
bacteria?

10. Diagram the tautomeric base pairings in DNA. What
base pair replacements occur because of the shifts?

11. What is the difference between a substrate and a
template transition mutation?

12. Describe two mechanisms for transversion muta-
genesis.

13. 5-bromouracil, 2-aminopurine, proflavin, ethyl
ethane sulfonate, and nitrous acid are chemical mu-
tagens. What does each do?

14. A point mutation occurs in a particular gene. Describe
the types of mutational events that can restore a func-
tional protein, including intergenic events. Consider
missense, nonsense, and frameshift mutations.

15. Why does misalignment result in addition or dele-
tion of bases?

16. What are the differences and similarities between in-
tergenic and intragenic suppression?

17. Eight independent mutants of E. colt requiring trypto-
phan (trp) are isolated. Complementation tests are
performed on all pairwise combinations. Based on the
results shown, determine how many genes you have
identified and which mutants are in which genes
(+ = complementation; — = no complementation).

1
2

5
6

7
8

+
+

+
+
+

18. Complementation tests are usually done with reces-
sive mutations for if the mutations were dominant,
all progeny would be mutant, regardless of whether
the genes are allelic. Suppose you have isolated in a
diploid species two independent dominant muta-
tions that each confer resistance to the drug cyclo-
heximide. Call these mutations Chx-1 and Chx-2.
What crosses can you perform to determine
whether the mutations are allelic? Your crosses
should allow you to determine whether the muta-
tions are allelic, nonallelic and unlinked, or nonal-
lelic and linked.

19. A series of overlapping deletions in phage T4 are iso-
lated. All pairwise crosses are performed, and the
progeny scored for wild-type recombinants. In the
following table, + = wild-type progeny recovered;
— = no wild-type progeny recovered.

1
2

+
+

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Chapter Twelve DNA: Its Mutation, Repair, and Recombination

a. Draw a deletion map of these mutations.

b. A point mutation, 6, is isolated and crossed with
all of the deletion strains. Wild-type recombinants
are recovered only with strains 2 and 3. What is
the location of the point mutation?

20. Hydroxylamine is a chemical that causes exclusively
C — > T transition mutations. Can hydroxylamine
reverse nonsense mutations? Explain.

21. A nonsense suppressor is isolated and shown to in-
volve a tyrosine transfer RNA. When this mutant
transfer RNA is sequenced, the anticodon turns out
to be normal, but a mutation is found in the dihy-
drouridine loop. What does this finding suggest
about how a transfer RNA interacts with the mes-
senger RNA?

22. Devise selection-enrichment procedures for isolat-
ing the following kinds of mutants:

a. extra-large bacterial cells

b. nonmotile ciliated protozoans

23. Two chemically induced mutants, x and y are
treated with the following mutagens to see if rever-
tants can be produced: 2-amino purine (2AP),
5-bromouracil (5BU), acridine dye (AC), hydroxyl-
amine (HA), and ethylmethanesulfonate (EMS). In
the following table, + = revertants and — = no
revertants. For each mutation, determine the proba-
ble base change that occurred to change the wild-
type to the mutant.

Chemical

Mutant

2AP

5BU

EMS

—

25. What situation will lead to false negatives in a com-
plementation test — or, in other words, indicate mu-
tations are in the same gene when, in fact, they are in
different genes?

26. Suppose you repeat the Luria-Delbriick fluctuation
test, but this time you look for lac colonies. Your "in-
dividual cultures" produce the following numbers of
lac colonies: 20, 25, 22, 18, 24, 19, 17, 25, 26, and 18.
Subsamples from the bulk culture give results identi-
cal to these. What can you conclude?

27. You have isolated a new histidine auxotroph, and,
despite all efforts, you cannot produce any rever-
tants. What probably happened to produce the orig-
inal mutant?

DNA REPAIR

28. UV light causes thymine dimerization. Describe the
mechanisms, in order of efficiency, that can repair
the damage. Name the enzymes involved.

29. What types of damage do excision repair endonucle-
ases recognize?

30. What are the functions of the RecA protein? How is
it involved in phage X induction? (See also RECOM-
BINATION)

RECOMBINATION

31- Diagram, in careful detail, a recombination by way of
the double-strand break model. What enzymes are re-
quired at each step?

32. What are the different enzymes involved in recipro-
cal and nonreciprocal recombination?

24. What situation will lead to a false positive in a com-
plementation test — or, in other words, indicate two
genes when, in fact, the mutations are in the same
gene?

CRITICAL THINKING QUESTIONS

1. Charles Yanofsky demonstrated colinearity of the gene 2. Comment on the statement that DNA is a molecule de-
and its protein product. What are the alternatives to signed for replication and repair,

colinearity?

Suggested Readings for chapter 12 are on page B-10.

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GENOMICS,
BIOTECHNOLOGY,

AND
RECOMBINANT

DNA

STUDY OBJECTIVES

1. To look at the techniques of gene cloning 358

2. To examine the techniques of creating restriction maps 377

3. To study the methods of DNA sequencing 383

4. To look at the goals and methods of the Human Genome
Project 390

5. To look at the practical benefits and human issues of genetic
engineering 397

STUDY OUTLINE

Genomic Tools 359

Restriction Endonucleases 359

Prokaryotic Vectors 360

Cloning a Particular Gene 364

Southern Blotting 366

Probing for a Cloned Gene 368

Eukaryotic Vectors 369

Expression of Foreign DNA in Eukaryotic Cells 372
Restriction Mapping 377

Constructing a Restriction Map 379

Double Digests 379

Restriction Fragment Length Polymorphisms 380
Polymerase Chain Reaction 381
DNA Sequencing 383

The Dideoxy Method 383

Creating a General-Purpose Primer 386
Mapping and Sequencing the Human Genome 390

Locating a Gene of Interest 390

The Human Genome Project 391

Ethics 397
Practical Benefits from Gene Cloning 397

Medicine 397

Agriculture 398

Industry 398
Summary 398
Solved Problems 399
Exercises and Problems 400
Critical Thinking Questions 404
Box 13.1 The Recombinant DNA Dispute 370
Box 13.2 Cloning Dolly 374
Box 13.3 Genes Within Genes 388

Artificially colored transmission electron micrograph of

DNA plasmids from the bacterium Escherichia coli.

These plasmids are used in genetic engineering.

357

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

In the spring of 2000, J. Craig Venter, CEO of Celera
Genomics, and Francis Collins, director of the
National Institutes of Health's Human Genome
Research Institute, jointly announced that they and
their colleagues had completed the sequence of
the human genome. Although there is still work ahead to
finish the project, the accomplishment was enormous. To
some, it was working out the very secret of life. This
accomplishment firmly established the science of
genomics, the study of the mapping and sequencing of
genomes.

Vector

Foreign DNA

J. Craig Venter (1946- ).
(Courtesy of Celera Genomics.)

Francis S. Collins (1950- ).
(Courtesy of Francis Collins.)

Since the mid-1970s, the field of molecular genetics
has undergone explosive growth, noticeable not only to
geneticists, but also to medical practitioners and re-
searchers, agronomists, animal scientists, venture capital-
ists, and the public in general. Medical practitioners and
researchers have new treatments for diseases available.
Agronomists see the possibility of greatly improved crop
yields, and animal scientists have gained the possibility of
greatly improving food production from domesticated
animals. Geneticists and molecular biologists are gaining
major new insights into understanding gene expression
and its control.

The new DNA manipulation techniques, centered on
the isolation, amplification, sequencing, and expression
of genes, are based on the insertion of a particular piece
of foreign DNA into a vector — a plasmid or phage. A plas-
mid is placed into a host cell, either prokaryotic or eu-
karyotic, which then divides repeatedly, producing nu-
merous copies of the vector with its foreign piece of
DNA. A phage simply multiplies in host cells (fig. 13. 1). In
both cases, the foreign piece of DNA is amplified in num-
ber; it can be expressed (transcribed and translated into
a protein) when in a plasmid in a host cell. A commonly
used host cell is E. coli. Following its amplification, the

Insertion

Hybrid vector

Uptake
by E. coli

Vector

Host chromosome

Growth of E. coli cells

"Many copies of
foreign DNA

> Expression of
foreign DNA
(presence of
protein product)

Figure 13.1 Overview of recombinant DNA techniques. A
hybrid vector is created, containing an insert of foreign DNA.
The vector is then inserted into a host organism. Replication of
the host results in many copies of the foreign DNA and, if the
gene is expressed, quantities of the gene product. (All DNA
shown is double-stranded.)

foreign DNA can be purified and its nucleotide sequence
determined. When it is expressed, large quantities of the
gene product of the foreign DNA can be obtained. The
new technology is variously referred to as gene cloning,
recombinant DNA technology, or genetic engineer-
ing. In this chapter, we look in detail at the methods and
procedures of recombinant DNA technology, including
DNA sequencing.

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Genomic Tools

359

GENOMIC TOOLS Q
Restriction Endonucleases

In 1978, Nobel prizes in physiology and medicine were
awarded to W. Arber, H. Smith, and D. Nathans for their pi-
oneering work in the study of restriction endonucle-
ases. These are enzymes that bacteria use to destroy for-
eign DNA, presumably, the DNA of invading viruses. The
enzymes recognize certain nucleotide sequences
(restriction sites) found on foreign DNA, usually from
four to eight base-pairs long, and then cleave that DNA at
or near those sites. (Restriction endonucleases were orig-
inally so named because they restricted phage infection
among strains of bacteria. Phages that could survive in
one strain could not survive in other strains with differ-
ent restriction enzymes.)

Three types of restriction endonucleases are
known. Their groupings are based on the types of se-
quences they recognize, the nature of the cut made in
the DNA, and the enzyme structure. Types I and III re-
striction endonucleases are not useful for gene cloning
because they cleave DNA at sites other than the recog-
nition sites and thus cause random, unpredictable
cleavage patterns. Type II endonucleases, however,
cleave at the specific sites they recognize, leading to
predictable cleavage patterns. The sites type II endonu-
cleases recognize are inverted repeats; they have
twofold symmetry. To see the symmetry, you must read
outward from a central axis on opposite strands of the
DNA. For example, the type II restriction endonuclease
BamHI recognizes

5'-GGA|TCC-3'
3'-CCT|AGG-5'

Reading out from the center (vertical line) is AGG on the
top strand and AGG on the bottom strand. The sequence
is, in a sense, a palindrome, a sequence that reads
the same from either direction. (Palindrome is from the
Greek palindromos, which means "to run back."
The name Hannah and the numerical sequence 1238321
are palindromes.) In figure 13.2 are some palindromic
sequences that type II restriction endonucleases recog-
nize; well over one hundred type II enzymes are known.
The host cell protects its DNA not by being free of
these restriction sites, but usually by methylating its DNA
in these regions (fig. 13. 3). The same sequences that the

®-

H
H

H N H

Methylation

OCK

N-^O

,0.

®-

OCH

OH H

Cytosine

OH H

5-Methylcytosine

Figure 13.3 A methylase enzyme adds a methyl group to
cytosine, converting it to 5-methylcytosine.

Endonuclease

HindU

EcoRI

BamH\

Pst\

3'
5'

Sequence recognized

I I G T Py Pu A C I I
| | C A Pu Py T G | |

| 1

1 G A A T T C 1 1
| C T T A A G | |

| 1

1 ' G G A T C C ' '
| | C C T A G G | |

1 ' C T G C A G ' '
| | G A C G T C | |

5'
3'

J L

i r

3'
5'

i r

G T Py

3' 5'

Pu A C

C A Pu f p y T G

5' 3

J L

■3'

5'-

, , C T T A A

3' J 1 5'

A A T T C

G
3' —

i r

■3'

5'-

G A T C C

C C T A G

5'
3'

C T G C A

A C G T C

-L-L5'
— r- 3'

5'
3'

Blunt ends

5' Overhang

3' Overhang

Figure 13.2 Sequences cleaved by various type II restriction endonucleases. Py is any pyrimidine and Pu is any purine. Arrows
denote places where endonucleases cleave the DNA. In 1971, K. Danna and D. Nathans showed that a restriction endonuclease
would consistently cut DNA into pieces of the same size. This precision and repeatability of enzyme action made enzymes useful for
further research. Not all restriction endonucleases make staggered cuts with 3' and 5' overhangs; some produce blunt ends.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

endonucleases attack in the unmethylated condition are
protected when methylated. After host DNA replication,
new double helices are hemimethylated; that is, the old
strand is methylated but the new one is not. In this con-
figuration, the new strand is quickly methylated
(fig. 13.4). Foreign DNA, without methyl groups on either
strand, is not methylated.

Restriction endonucleases are named after the bacte-
ria from which they were isolated: BamHl from Bacillus
amyloliquefaciens, strain H; EcoTH from E. coli, strain
RY13; Hindll from Haemophilus influenzae, strain Rd;
and Bgll from Bacillus globigii. From here on, we will re-
fer to type II restriction endonucleases simply as restric-
tion enzymes.

Restriction enzymes cut the DNA in two different
ways. For example, HindW cuts the recognition sequence
down the middle, leaving "blunt" ends on the DNA (see
fig. 132). We will discuss how pieces of DNA with blunt
ends can be used in cloning. The staggered cuts made, for
example, by BamYU leave "sticky" ends (a 5' overhang)
that can reanneal spontaneously as hydrogen bonds form
between the complementary bases (see fig. 13.2). The
ability to reanneal these sticky ends, first demonstrated
by S. Cohen, H. Boyer, and colleagues in 1971, opened up
the field of gene cloning.

Prokaryotic Vectors

With current technology it is routine to join together, in
vitro, DNAs from widely different sources. In figure 135,

5' - CCGG - 3'

3'-GGCC-5'
*

Methylated DNA

DNA replication

5' -CCGG -3'
3'-GGCC-5'

5' - CCGG - 3'
3'-GGCC-5'

Methylation

Hemimethylated DNA

5' -CCGG -3'
3'-GGCC-5'

Methylated DNA

Figure 13.4 Host DNA is methylated in the /-/pall restriction
site. Asterisks indicate methyl groups on cytosines. After DNA
replication, the DNA is hemimethylated; the new strands have
no methyl groups. Hemimethylated DNA is then fully methylated
by cellular enzymes.

we see how a circular DNA molecule cleaved by a spe-
cific restriction enzyme can recircularize if it is cleaved in
only one place, or how different molecules with the
same free ends can anneal to form hybrid molecules.
Only the action of a DNA ligase is needed to make the
molecules complete (see chapter 9).

One of the pieces of DNA involved in the annealing
can be a plasmid, a piece of DNA that can replicate in
a cell independently of the cellular chromosome. The
recombinant plasmid (fig. 13.6) can be transferred
into a cell. (A recombinant plasmid is also known as a
hybrid plasmid, hybrid vehicle, hybrid vector, or
chimeric plasmid. The latter is after the chimera, a
mythological monster with a lion's head, a goat's body,
and a serpent's tail.) Many procedures exist that can
introduce this recombinant plasmid into a host cell.
For example, a bacterial cell can be made permeable
to this, or any, plasmid by the addition of a dilute solu-
tion of calcium chloride. Once inside the cell, the for-
eign DNA is replicated each time the plasmid DNA
replicates.

Note that in the process of inserting a piece of for-
eign DNA, the restriction site is duplicated, with one
copy at either end of the insert. This property makes it
easy to remove the cloned insert at some future time, if
needed, since restriction sites enclose it (fig. 13.6).

Cloning with Restriction Enzymes

A few conditions must be met in order to succeed in
cloning DNAs from different sources. A plasmid vehicle
should be cleaved at only one point by the endonucle-
ase. If it is cleaved at more than one point, it will frag-
ment during the experiment. However, some phage ve-
hicles must be cleaved at two points so that the foreign
DNA can replace a length of the phage DNA rather than
simply being inserted. Common vehicles, derivatives of
phage X, have been named Charon phages (pro-
nounced "karon") after the mythical boatman of the
River Styx. (See chapter 14 for a detailed discussion of
phage X.)

During normal phage infection (see chapter 7), only
DNA the size of a phage genome is packaged into X
heads. Thus, for X to be a useful vector, the foreign DNA
must replace part of its DNA. We note that X can func-
tion quite well as a hybrid vehicle with a 15,000 base-
pair (15 kilobases, or 15 kb) section replaced by for-
eign DNA because that section of phage DNA is used
for integration into the E. coli chromosome, a
nonessential phage function. That is, the phage can in-
fect a bacterium, replicate inside the bacterium, and
burst out without the integration region. Genetic engi-
neers have created a X DNA molecule with the
nonessential region missing and an EcoRl cleavage site
in its place. Only hybrid DNA can thus be incorporated

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First plasmid

Second plasmid

EcoRI

Recircularize

EcoRI

Recircularize

A A T T C
G

I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I

C T T A A

A A T T C
G

I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I

C T T A A

3' 5'

A A T T C
G

I I I

I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 1 I I I I I I I I

G A A T T C

I I I I I I I I I

I I I I

i i i

C T T A A G
I I I I I I I I

5'3'

i i i i i

C T T A A

Ligase is needed
to join the plasmids

Figure 13.5 Circular plasmid DNA with a palindrome recognized by EcoRI. After the DNA is cleaved by the endonuclease, it has two
exposed ends that can join to recircularize the molecule or unite two or more linear molecules of DNA cleaved by the same
restriction endonuclease. The final nicks are closed with DNA ligase. S. Cohen, H. Boyer, and their colleagues first joined plasmids
with this technique in 1971.

into phage heads because the diminished phage DNA,
without an insert, is too small to be properly packaged.

One disadvantage of cloning with normal E. colt plas-
mids is that they are unstable if the foreign DNA is very
large, greater than about 15 kb.That is, if a large chromo-
somal segment is cloned, the plasmid tends to lose parts
of the clone as the plasmid replicates. Primarily for this
reason, geneticists began using phage X as a vector (see
fig. 7.21) because these phages could successfully main-
tain foreign DNA as large as 24 kb.

The phage chromosome is about 50 kb of DNA;
within the phage head it is linear, and within the cell it is
circular. The DNA to fill the phage head is recognized
during infection because it has a small segment of single-
stranded DNA called a cos site (twelve bases; derived
from the term "cohesive ends") at either end. Reanneal-
ing the cos sites allows X chromosomes to circularize
when they enter a host cell; cutting the DNA at the cos
site opens the circle into a linear molecule (fig. 13.7).
Geneticists have taken advantage of these cos sites to
clone even larger segments of foreign DNA because it
turns out that even 24 kb is not adequate to study some
eukaryotic genes or gene groups. Many eukaryotic genes
are very large because of their introns and transcrip-

tional control segments. DNA up to 50 kb can be cloned
if cos segments are attached to either end with a plasmid
origin of DNA replication and a selectable antibiotic
gene. These cos-site-containing plasmids are called cos-
mids (fig. 137). Cosmids not only allow the cloning of
very large pieces of DNA, they actually select for large
segments of foreign DNA because small cosmids are not
incorporated into phage heads. Thus, foreign DNA rang-
ing from 2.5 to 50 kb in size can be cloned using plas-
mids, Charon phages, or cosmids. (Much larger pieces of
DNA, about a million bases, can be cloned in yeast, as we
will describe later.)

Selecting for Hybrid Vectors

In the methods we have described, restriction enzymes
separately cut both vector and foreign DNA. The two are
then mixed in the presence of ligase. The many products
that are created can be divided generally into three cate-
gories: vectors with foreign DNA, vectors without for-
eign DNA, and fragments. In a later section, we will dis-
cuss methods of finding a particular piece of foreign
DNA in a vector. Here, we point out how vectors with in-
serts of any kind are selected.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Plasmid

Foreign DNA

CCAGG
GGTCC

Treat with a
restriction
endonuclease EcoRW

CCAGG
GGTCC

Treat with ligase

Recombinant plasmid

Figure 13.6 Formation of a recombinant plasmid. The same restriction endonuclease, in this case EcoRII, is used to cleave both host and
foreign DNA. Some of the time, cleaved ends will come together to form a plasmid with an insert of the foreign DNA. Ligase seals the
nicks. R Berg was the first scientist to clone a piece of foreign DNA when he inserted the genome of the SV40 virus into phage X in 1973.

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Part of plasmid

Cloned insert

Part of

plasmid

„ GGGCGGCGACCT
5

1 i 1 1 1 1 1 1 1 1 1 1 5'
CCCGCCGCTGGA

cos

Package in X heads

Linear DNA

Infect E. coli

Bacterial
chromosome

Cosmid

Circular DNA

E. coli

Figure 13.7 A cosmid is a plasmid with cos sites that can be transferred into bacteria within phage lambda heads, a
very efficient method of infection. The cos sites are single-stranded; they reanneal to a circle when inside the host. (The
heavy lines of the linear DNA, bacterial chromosome, and cosmid are double-stranded DNA.)

Charon phages are selected simply by their ability to
infect E. coli cells. As we mentioned, after manipulation,
only \ DNA with a foreign insert is packaged because of
the size requirement. Plasmids that contain foreign DNA
can be selected through screening for antibiotic resis-
tance. For example, a widely used cloning plasmid is
named pBR322. (Plasmids are often named with the ini-
tials of their developers. The vector pBR322 was first de-
scribed in a paper published in 1977 by authors F. Bolivar
and R. Rodriguez, hence pBR.) Plasmid pBR322 contains
genes for tetracycline and ampicillin resistance and vari-
ous restriction sites. There is, for example, a BamHI site
in the tetracycline-resistance gene (fig. 138). After the
ligating procedure, plasmids with and without foreign
DNA will be present. E. coli cells are then exposed to this
DNA mixture in the presence of calcium chloride; after
taking up the DNA, the E. coli cells are plated on a
medium without antibiotics. Replica-plating is done onto
plates with one or both antibiotics. Colonies resistant to
both antibiotics are composed of cells with plasmids hav-
ing no inserts; those resistant only to ampicillin have a
plasmid with an insert. Colonies resistant to neither an-
tibiotic have cells with no plasmids.

Blunt-End Ligation

Restriction endonuclease treatment may not suffice for
cloning; an endonuclease may cut in the wrong place, say
in the middle of a desired gene, or the foreign DNA may
have been isolated by other methods, such as physical
shearing. In these cases, several other methods of cloning
can be used.

The most common method of joining foreign and ve-
hicle molecules that do not have sticky ends is called
blunt-end ligation; the phage enzyme, T4 DNA ligase,
can join blunt-ended DNA. Blunt ends can be generated
when segments of DNA to be cloned are created by phys-
ically breaking the DNA or by using certain restriction
endonucleases, such as Hindll (see fig. 13.2), that form
blunt ends. Since the ligase is nonspecific about which
blunt ends it joins, many different, unwanted products re-
sult from its action. Restriction enzymes that produce
sticky ends are preferred for cloning.

A variation of blunt-end ligation uses linkers —
short, artificially synthesized pieces of DNA containing
a restriction endonuclease recognition site. When these
linkers are attached to blunt pieces of DNA and then

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

amp r

BamYW site

Yeast DNA

BamYW sites

pBR322

amp r

BamYW
treatment

Yeast DNA

Recombinant plasmids lose tetracycline resistance

Hybrid plasmid

Figure 13.8 E. coli plasmid pBR322. This plasmid carries two genes, amp r and tef, that confer resistance to
ampicillin and tetracycline, respectively. A BamYW restriction site occurs within the tef gene. A cloned fragment
within the tef gene therefore destroys the tetracycline resistance. (Heavy black, blue, and red lines represent
double-stranded DNA.)

treated with the appropriate restriction endonuclease,
sticky ends are created. In figure 139, the linkers are
twelve base-pair (bp) segments of DNA with an EcoRl
site in the middle. They are attached to the DNA to be
cloned with T4 DNA ligase. Subsequent treatment with
Eco RI will result in DNA with Eco RI sticky ends.

DNA for cloning can be obtained generally in two
ways: (1) a desired gene or DNA segment can be synthe-
sized or isolated or (2) the genome of an organism can be
broken into small pieces and the small pieces can be ran-
domly cloned (shotgun cloning). Then the desired DNA
segment must be "fished" out from among the various
clones created. Let us look first at synthesizing or isolating
a desirable gene before cloning it, and then look at the
process of locating a desired gene after it has been cloned.

Cloning a Particular Gene

Creating DNA to Clone

To clone a particular gene (or DNA segment), a scientist
must have a purified double-stranded piece of DNA con-
taining that gene. There are numerous ways to obtain
that DNA; several entail creating or isolating a single-
stranded messenger RNA that is then enzymatically con-
verted into double-stranded DNA. The problem is then re-
duced to obtaining the desired messenger RNA.

The messenger RNA for a particular gene can be ob-
tained in several different ways, depending on the partic-
ular gene. If large quantities of the RNA from a particular
cell are available, the RNA can be isolated directly. For ex-
ample, mammalian erythrocytes have abundant quanti-

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Foreign DNA

T4 DNA Ligase

EcoRI

CTTAA

DNA with
EcoRI ends

Figure 13.9 Linkers: small segments of DNA with an internal restriction site. Linkers can be added to
blunt-ended DNA by T4 DNA ligase. The restriction enzymes create DNA with ends that are
compatible with any DNA cut by the same restriction enzyme (in this case, EcoRI).

ties of a- and p-globin messenger RNAs. Also, ribosomal
RNA and many transfer RNAs are relatively easy to isolate
in quantities adequate for cloning.

Double-stranded DNA for cloning is made from the pu-
rified RNA with the aid of the enzyme reverse transcrip-
tase, isolated from RNA tumor viruses (see chapter 10). We
describe here the conversion of RNA to DNA using a eu-
karyotic messenger RNA with a 3' poly-A tail (fig. 13.10).
In the first step, a poly-T primer is added, which base-pairs
with the poly-A tail of the messenger RNA. This short,
double-stranded region is now a primer for polymerase
activity — a free 3 -OH exists. The primed RNA is then
treated with the enzyme reverse transcriptase, which will
polymerize DNA nucleotides using the RNA as a template.
The result is a DNA-RNA hybrid molecule (fig. 13.10c).

The hybrid is now treated with the enzyme RNaseH,
which creates random nicks in the RNA part of the RNA-
DNA hybrid. These nicks provide the primer configura-
tion for repair synthesis, the same repair done on
Okazaki fragments when RNA primer is removed and re-
placed by DNA. Thus, the hybrid is treated with DNA
polymerase I, which replaces each small RNA segment
with DNA, base by base. Finally, the short DNA segments
of the second DNA strand are united with DNA ligase
(fig. I3.IQ/). The resulting double-stranded DNA is re-
ferred to as complementary DNA (cDNA). Hence,

starting with a piece of single-stranded messenger RNA,
we have generated a piece of double-stranded DNA. This
piece can now be cloned using the blunt-end methods
we have described.

If the RNA is not available in large enough quantities,
it is possible to synthesize DNA in vitro if the amino acid
sequence of its expressed protein is known. A possible
nucleotide sequence can be obtained from the genetic
code dictionary (see table 1 1 .4) if the sequence of amino
acids is known from the protein product of the gene.
This method will probably not re-create the original DNA
because of the redundancy in the genetic code. In other
words, any one of six different codons could have coded
a particular leucine in a protein. Despite an element of
guesswork, it is possible to synthesize a piece of DNA
that will code for a particular protein. Currently, auto-
mated machines that add one base at a time in ten-minute
cycles can synthesize DNA sequences of over one hun-
dred bases.

You will notice that the methods we have described,
making cDNA or synthetic DNA using the genetic code
dictionary, produce DNA missing the gene's promoter
and its transcriptional control sequences as well as other
untranslated areas of the DNA (introns). If it is desirable
to clone an intact gene with its promoter and introns,
then cloning can be done by creating random pieces of

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

(a)

IIIIIIIIIIIIIIIIIIIIIIAAAAAA

3' mRNA

Primer

(b)

IIIIIIIIIIIIIIIIIIIIIIAAAAAA

3' mRNA

T T T T T T 5 ' DNA

Reverse transcription

(c)

(d)

(e)

(f)

IIIIIIIIIIIIIIIIIIIIIIAAAAAA

3' mRNA

3 > TTTTTT g , DNA

(first strand)

FT I I I I

RNase H

IIIIIIIIIIIIIIIIAAAAAA

3' mRNA

TTTTTT 5 , DNA

Polymerase I

FT" I I I I

I I I I I
J I I I L

IIIIIIIIIIIAAAAAA 3 ' P NA

(second strand)

TTTTTT

5' DNA

DNA ligase

3' DNA

IIIIIIIIIIIIIIIIIIIIIIAAAAAA

I TTTTTT 5 , DNA

Figure 13.10 (a) A messenger RNA, shown in black, begins as a single strand, (b) A poly-T DNA segment (red) is
added as primer; it complements the 3' poly-A tail of the eukaryotic messenger RNA. (c) Reverse transcriptase acts
on this primed configuration to synthesize a single strand of DNA from the RNA template, (d) The RNA is then nicked
randomly by RNase H. (e) The RNA segments are then replaced by DNA (blue) by the action of DNA polymerase I.
(/) After DNA ligase treatment, the final result is double-stranded complementary DNA (cDNA).

the genome. The gene of interest can be found either be-
fore or after cloning it, although it is usually done after
cloning.

Creating a Genomic Library

When cDNA or synthetic DNA cannot be used for
cloning, the total DNA of an organism can be broken into
small pieces to isolate the desired gene or DNA fragment.
The desired DNA can be isolated either before or after
cloning. This DNA is referred to as genomic DNA to dif-
ferentiate it from cDNA.

If the original DNA is isolated before cloning, then only
that DNA need be cloned. Alternatively, a "shotgun" ap-

proach can be used to clone a sample of the entire genome
of an organism (in small pieces, of course), creating a ge-
nomic library, a set of cloned fragments of the original
genome of a species (fig. 13.1 1). In a genomic library, a de-
sired gene can be located after it is already cloned.

Southern Blotting ^*

When DNA segments are generated randomly, usually by
endonuclease digestion, a desired gene must be located.
As mentioned, we can look for the gene either before or
after it is cloned. We consider first the procedure for lo-
cating a specific gene in a DNA digest, before the DNA
has been cloned.

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To locate a specific gene in the midst of a DNA digest,
one must have a specific probe. Probes are generally nu-
cleic acids with sequences that precisely locate a comple-
mentary DNA sequence by hybridization. The probes are
labeled so they can be identified later with autoradiogra-
phy or chemiluminescent techniques (techniques
in which tags are used that fluoresce under ultraviolet
or laser light). Thus, if we wish to locate the gene for
|3-globin, we could use radioactively labeled p-globin mes-
senger RNA or radioactively labeled cDNA. RNA-DNA or
DNA-DNA hybrids would form between the specific gene

and the radioactive probe. Autoradiography or chemilu-
minescence would then locate the radioactive probe.

Let us assume that we wanted to clone the rabbit
P-globin gene. First, we would create a restriction digest
of rabbit DNA (fig. 13.12). We would then subject this di-
gest to electrophoresis on agarose to separate the various
fragments according to size. Agarose is a good medium
for separating DNA fragments of a wide variety of sizes.

(a)

Rabbit DNA

Begin with genome
of organism

Create fragments with blunt
ends or restriction-generated
sticky ends

Clone in
plasmids or phage

Amplify and isolate
vector in E. coli

Plate out

Produce plaques
containing
genomic library

Figure 13.11 Creating a genomic library using the shotgun
approach in creating inserts. First, the genome is fragmented.
The fragments are then cloned randomly in vectors. The
collection of these vectors is referred to as a genomic library.

Endonuclease

(b)

Small segments
of double-stranded
DNA

Agarose
gel

Electrophoresis

(c)

Longer segments

Nitrocellulose
filter

Shorter segments

Denature DNA to single strands and
Southern blot

(d)

Radioactive probe (single-stranded)

Autoradiograph

(e)

Figure 13.12 Locating the rabbit p-globin gene within a DNA
digest using the Southern blotting technique. The rabbit DNA
(a) is segmented with a restriction endonuclease (b) and then
electrophoresed on agarose gels (c). Southern blotting transfers
the DNA to nitrocellulose filters (d). Finally, a radioactive probe
(p-globin messenger RNA) locates the DNA fragment with the
p-globin gene after autoradiography (e).

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Edwin M. Southern (1938- ).
(Courtesy of Edwin Southern.)

In a digest of this kind, however, there are usually so
many fragments that the result is simply a smear of
oligonucleotides, from very small to very large. To pro-
ceed further, we have to transfer the electrophoresed
fragments to another medium for probing, or the DNA
fragments would diffuse out of the agarose gel. Nitrocel-
lulose filters or nylon membranes are excellent for hy-
bridization because the DNA fragments bind to these
membranes and will not diffuse out. The transfer proce-
dure, first devised by E. M. Southern, is called Southern
blotting. In this technique, the double-stranded DNA on
the agarose gel is first denatured to single-stranded DNA,
usually with NaOH.Then the agarose gel is placed directly
against a piece of nitrocellulose filter, and the resulting
sandwich is placed agarose-side-down on a wet sponge.
Dry filter paper placed against the nitrocellulose side
wicks fluid from the sponge, through the gel, and past the
nitrocellulose filter, carrying the DNA segments from the
agarose to the nitrocellulose (fig. 13.13). NaOH is used as
the transfer solution in the tray. The DNA digest fragments
are then permanently bound to the nitrocellulose filter by
heating. DNA-DNA hybridization takes place on the filter.
(A similar technique can be performed on RNA, which is
called, tongue-in-cheek, northern blotting. Immunolog-
ical techniques, not involving nucleotide complementar-
ity, can be used to probe for proteins in an analogous
technique called western blotting.)

A labeled probe can be obtained in several different
ways. In this example, the easiest way to obtain a ra-
dioactive probe would be to isolate p-globin messenger
RNA from rabbit reticulocytes and construct cDNA using
the reverse-transcriptase method described. The deoxyri-
bonucleotides used during reverse transcription are then
synthesized to contain radioactive phosphorus, 32 P. As
figure 1312 shows, after hybridization, a single radioac-
tive band locates a DNA segment with the p-globin gene.
Note that the probe, originating from messenger RNA,
will lack the introns present in the gene. However, prob-
ing is successful as long as there are complementary re-
gions in the two nucleotide strands.

Movement of
sodium hydroxide
solution and
DNA fragments

Weight

Filter paper
(dry)

Nitrocellulose
filter

Agarose
gel

Kitchen
sponge

Tray with
sodium
hydroxide
solution

Figure 13.13 Arrangement of gel and filters in the Southern
blotting technique. The NaOH buffer is drawn upwards by the
dry filter paper, transferring the DNA from the agarose gel to
the nitrocellulose filter.

To clone the p-globin gene, a second agarose gel
would be run with a sample of the digest used in figure
13.12. That gel, not subject to DNA-DNA hybridization,
would have the p-globin segment in the same place. The
band, whose location is known from the autoradiograph,
could be cut out of the agarose gel to isolate the DNA. We
could then clone the DNA by methods discussed earlier
in the chapter.

Probing for a Cloned Gene

Dot Blotting

The methods we have described are also useful in locat-
ing genes already cloned within plasmids, for example,
after a genomic library has been constructed. In this
case, electrophoresis and Southern blotting are not
needed since we will be probing for a particular se-
quence of DNA already cloned rather than DNA seg-
ments within a digest.

For example, the DNA of a human-mouse hybrid cell
line was cloned in order to locate human DNA. In this
case, a hybrid cell line had only one human chromosome,
chromosome 20. In order to locate DNA from that chro-
mosome, probes were used that were isolated from hu-
man chromosomes. The probes were radioactively la-
beled. Meanwhile, 288 E. colt colonies, each containing a
hybrid plasmid, were grown and transferred directly to a
nitrocellulose filter. In preparation for probing, the cells
were lysed and their DNA denatured. The plasmid DNA
within the cells of each clone was then hybridized with
the radioactive probes. Figure 13.14 is an autoradiograph
of the 288 clones. The two dark spots indicate clones car-

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rying DNA from human chromosome 20 (those clones
"light up" autoradiographically). This technique, hy-
bridization of cloned DNA without an electrophoretic-
separation step, is referred to as dot blotting.

These techniques can also be carried out without the
grid arrangement of colonies by using replica plating as
described in chapter 7. Thus, a specific gene can be lo-
cated after shotgun cloning.

Western Blotting

An entirely different method used to locate particular
cloned genes utilizes the actual expression of the cloned
genes in the plasmid-containing cells. If a eukaryotic
gene is cloned in an E. colt plasmid downstream from an
active promoter, that gene may be expressed (tran-
scribed and translated into protein). Plasmids that allow
the expression of their foreign DNA are termed expres-
sion vectors. There are many problems with this tech-
nique because bacteria normally would not express eu-
karyotic genes. However, special vectors have been
developed in which the cloning site is just downstream
from a promoter. The eukaryotic gene thus becomes part
of the prokaryotic gene, producing a fusion protein, usu-
ally with only a few amino acids from the prokaryote. Of
course, the eukaryotic gene must be in the correct orien-
tation and in the correct codon reading frame for appro-
priate translation, meaning that the success rate of this
technique is relatively low.

A particular protein product can be located by west-
ern blotting, a method completely analogous to either

Southern blotting or dot blotting. In this technique, prob-
ing is done with antibodies specific for a particular pro-
tein, rather than using a radioactive oligonucleotide
probe. A second antibody, specific to the first and labeled
with a marker, usually fluorescent, locates the first anti-
body. For example, assume we are looking for the ex-
pression of a particular protein in the clones of figure
1314. The clones would be transferred to a nitrocellu-
lose membrane, where they would be lysed (e.g., with
chloroform vapor). Then an antibody, specific for the par-
ticular protein, would be applied. A second antibody,
specific for the first antibody and labeled with a fluores-
cent marker, would be applied to the filters. Fluores-
cence of the second antibody would locate the presence
of the first antibodies and thus indicate which of the
clones is expressing the particular gene (fig. 13.15).

Eukaryotic Vectors

The work we have described so far involves introducing
chimeric plasmids into bacteria, primarily E. colt How-
ever, there are several reasons why we want to extend
these techniques to eukaryotic cells (box 13.1). First, a
prokaryote like E. colt is not capable of fully expressing
some eukaryotic genes since it lacks the enzyme systems
necessary for some posttranscriptional and posttransla-
tional modifications such as intron removal and some
protein modification. Second, we also wish to study the
organization and expression of the eukaryotic genome in

Figure 13.14 Dot blot autoradiograph of 288 clones of DNA
from a mouse-human hybrid cell line. After lysing samples of
each clone on a nitrocellulose filter, the investigator hybridized
the clones with radioactive probes for human-specific
sequences. The two dark spots indicate clones carrying human
DNA. The slight background radiation in most other spots
provides the spot pattern needed to orient the investigator.
(Source: Courtesy of Nick 0. Bukanov.)

• • •

• • •
o o •

o o o
o o o
• o o
o o o

Nitrocellulose filter with
twelve clones carrying pBR313
plasmids with inserts

Lyse cells, apply antibodies

Ultraviolet light

reveals clone expressing

the probed-for gene

Figure 13.15 Western blot technique is used to locate an
expressed protein from among many clones. Clones that may
carry the expressed protein are lysed. Tagged antibodies are
applied to locate the protein; a second antibody locates the
first antibody either through a fluorescent color marker, as
shown here, or with autoradiography.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

BOX 13.1

Paul Berg (1926- ).
(Courtesy of Dr. Paul Berg.)

Paul Berg shared the 1980 No-
bel Prize in chemistry for cre-
ating the first cloned DNA
molecule, a hybrid A phage that con-
tained the genome of the simian tu-
mor virus, SV40. The fact that he
could do this work was worrisome
to many people, himself included.
The recombinant DNA dispute was
underway. Berg voluntarily stopped
inserting tumor virus genes into

Ethics and Genetics

The Recombinant DNA
Dispute

phages that attack the common in-
testinal virus E. coll

People continue to worry about
the dangers of working with recom-
binant DNA. One immediate and ob-
vious concern is that cancer or toxin
genes will "escape" from the labora-
tory. In other words, recombinant
DNA technology could create a bac-
terium or plasmid that contained
toxin or tumor genes. The modified
bacterium or plasmid could then ac-
cidentally infect people. A 1974 re-
port by the National Academy of Sci-
ences led to a February 1975
meeting, which took place at the
Asilomar Conference Center south of
San Francisco. Berg convened this
meeting, which over one hundred
molecular biologists attended. The
recommendations of the Asilomar
Committee later formed the basis for

official guidelines developed by the
National Institutes of Health (NIH). In
essence, NIH established guidelines
of containment.

Containment means erecting
physical and biological barriers to the
escape of dangerous organisms. The
NIH guidelines defined four levels of
risk, from minimal to high, and four
levels of physical containment for
them (called PI through P4). The
most hazardous experiments, dealing
with the manipulation of tumor
viruses and toxin genes, require ex-
treme care, which included negative-
pressure air locks to the laboratory
and experiments done in laminar-
flow hoods, with filtered or inciner-
ated exhaust air.

Biological containment means de-
veloping host cells and manipulated
vectors that are incapable of success-
ful reproduction outside the lab, even
if they escape. High-risk work was
done with host cells or vectors that
were modified. For example, a bac-
terium of the E. coli strain EK2 can-
not survive in the human gut be-
cause it has mutations that do not
permit it to synthesize thymine or
diaminopimelate.The lack of thymine-

vivo (in the living system), something we can only ac-
complish by working directly with eukaryotic cells. Fi-
nally, we wish to learn how to manipulate the genomes
of eukaryotes for medical as well as economic reasons. To
these ends, we discuss eukaryotic plasmids and the di-
rect manipulation of eukaryotic genomes in vivo.

Yeast Vectors

Yeasts, small eukaryotes that can be manipulated in the
lab, like prokaryotes, have been studied extensively.
Baker's yeast, Saccharomyces cerevisiae, has a naturally
occurring plasmid. In addition, bacterial plasmids have
been introduced into yeast. Unfortunately, the cells tend to
lose these plasmids. This tendency has been overcome,
however, by constructing bacterial plasmids that contain a
yeast centromere (CEN) and the origin of yeast DNA repli-
cation (ARS for autonomously replicating sequence; fig.
13.16). The yeast then carries the plasmids from one gen-

eration to the next. The plasmids can have telomeric se-
quences inserted, and they can then be made linear by cut-
ting the telomeric sequences with endonuclease. Alterna-
tively, the plasmids can be linearized first, and then have
telomeric sequences added to their ends. The plasmids are
then called yeast artificial chromosomes (YACs). The
particular advantage YACs have is that they are capable of
accepting very large pieces of inserted DNA. Remember
that a cosmid can hold about 50 kb; a YAC can hold as
much as 800 kb or more. The ability to clone this much
DNA is valuable when working with large eukaryotic
genes and in the Human Genome Project (see the section
with this title later in the chapter).

Recombinant DNA studies in yeast have increased
our knowledge about gene regulation in eukaryotes,
about how the centromere works, and about the way in
which the tips of eukaryotic linear chromosomes are
replicated. In addition, YACs have allowed us to analyze
and sequence very large segments of eukaryotic DNA.

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synthesizing ability is lethal because
the cell cannot replicate its DNA. The
diaminopimelate is a cell-wall con-
stituent; without it, the cells burst.
These bacteria also carry mutations
that make them extremely sensitive
to bile salts. Thus, if by accident the
cells were to escape, they would
pose virtually no threat. The plasmids
used for recombinant research were
modified so that they could not be
transferred from one cell to the next.
Again, if containment failed, neither
the host cells nor their plasmids
would survive.

In 1979, the guidelines were re-
laxed. Although it was wise to be cau-
tious, it appears that initial fears were
unwarranted. Recombinant DNA
work now seems to pose little dan-
ger: Containment works very well,
and engineered bacteria do very
poorly under natural conditions.
E. colt has been living in mammalian
guts for millions of years, so it has had
numerous opportunities to incorpo-
rate mammalian DNA into its genome
(intestinal cells are dying and slough-
ing off into the gut all the time). No
"Andromeda strain" has arisen, nor do
we foresee one in the future.

Current concern is focused on the
acceptability of genetically modified
crops (GM crops). As we will discuss
later, one fourth of American crop-
land is planted with genetically modi-
fied crops, modified mainly for insect
resistance. These modifications have
curtailed our use of insecticides. (For
cotton and corn, for example, liquid
insecticide use dropped by 3.6 mil-
lion liters and powdered insecticide
by 300,000 kilograms in 1999.) How-
ever, people are concerned with the
effects these modifications might
have on natural ecosystems: How
many valuable insects will be killed
by mistake? Although Third World
countries are desperate for these
technologies, the United States, Euro-
pean and Asian trading partners are
demanding that the crops we export
be genetically unmodified. Farmers
are also concerned that genetically
modified crops have been modified
to be sterile (so called "terminator
technology") so that farmers would
need to buy new seeds each year.

More recently, the recombinant
DNA dispute has taken a whole new
twist. It now has surfaced as a con-
flict between academic freedom and

industrial secrecy. It seems that re-
combinant DNA technology is very
lucrative. Numerous academic scien-
tists have either begun genetic
engineering companies or become
affiliated with pharmaceutical com-
panies. However, the philosophies of
private enterprise and academia are
often in conflict. Academic endeav-
ors are presumably open, with free
exchange of information among col-
leagues, whereas private enterprise
entails some degree of secrecy, at
least until patents are obtained to
protect the investments of the com-
panies. Thus, a basic conflict can arise
for scientists trained in gene cloning.
The conflict has been prevalent since
late 1980, when the first patent for re-
combinant DNA techniques was
awarded to Stanford University and
the University of California. When, in
April 2000, United States President
Bill Clinton and British Prime Minis-
ter Tony Blair issued a joint statement
asking that human genome data not
be patented, the American stock mar-
ket took a major downturn. This is a
tumultuous time for biotechnology.

Animal Vectors

The vehicle most commonly used in higher animals is the
DNA tumor virus SV40. (SV, or simian vacuolating virus,
was first isolated in monkeys; however, it can transform
normal mouse, rabbit, and hamster cells. Unlike the use
of the word transformation in bacteria, transformation
in eukaryotes refers to the changing of a normal cell into
a rapidly growing, cancerous one.) SV40 is an icosahedral
particle with a small (5,224 base pairs) chromosome,
which is a circular, double-stranded DNA molecule.

Like X vectors, SV40 virions allow foreign DNA to re-
place part of their DNA. The viruses can then be used in
recombinant DNA studies in one of two ways (fig. 13. 17).
They can replicate and complete their life cycle with the
help of nonrecombinant viruses, or they can replicate in
the host without making active virus particles by existing
as circular plasmids in the cytoplasm or by integrating
into the host's chromosomes. SV40 has become a valu-
able tool in mammalian genomic studies. For example,

the rabbit (3-globin gene was cloned in SV40, and en-
hancer sequences (see chapter 10) were discovered in
SV40. DNA tumor viruses have also deepened our under-
standing of transformation in eukaryotes (oncogenesis).

Plant Vectors

The best-studied system for introducing foreign genes into
plants is the naturally occurring crown gall tumor system.
The soil bacterium Agrobacterium tumefaciens causes tu-
mors, known as crown galls, in many dicotyledonous
plants (fig. 1318). In essence, the crown gall is made of
transformed plant cells. These cells have been transformed
by a plasmid within the bacterium called the tumor-
inducing, or Ti, plasmid. Transformation occurs when a
piece of the plasmid called T-DNA (for transferred DNA) is
integrated into the chromosome of the plant host. Crown
gall cells produce amino acid derivatives, termed opines,
that the A. tumefaciens cells use. By manipulating this

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Antibiotic
resistance
(e.g., amp")

Bacterial -
origin of
replication
(on)

Yeast origin
of DNA
replication (ARS)

Centromeric

region

(CEN)

Make linear cut (with endonuclease)
and add telomeric ends

Telomere

amp r oh CEN ARS

Telomere

YAC

Figure 13.16 Escherichia coii plasmid pBR322 modified for
use in yeast. This plasmid survives and replicates in both yeast
and E. coii because it contains the origin of replication for both,
as well as a yeast centromeric region (CEN). When it is made
linear and telomeres are added, the yeast artificial chromosome
(YAC) becomes suitable for cloning large pieces of DNA.

system, geneticists have begun to understand the transfor-
mation process in plants as well as to develop a manipu-
latable system for introducing foreign genes into plants.

The study of genetics in plants has been boosted a
great deal by the availability of model organisms similar
to E. coii, yeast, and fruit flies. Recently, much attention
has focused on the meadow weed,Arabidopsis thaliana
(fig. 13. 19). This small plant is ideal for studying plant ge-
netics because its genome is small, approximately 100
million base pairs located in only five chromosomes
(2n = 10). This is only about five times the genome of
yeast or twenty times the genome of E. coii. Thus, in
terms of genome size, it is quite manageable. A. thaliana
has joined the ranks of organisms whose genomes have
been sequenced. The plants are easy to grow in very
large numbers, and each plant produces as many as ten
thousand seeds. Hence, this organism compares very fa-
vorably with fruit flies and yeast for studying questions of
gene control in a eukaryote, in this case a plant.

Expression of Foreign DNA in Eukaryotic Cells

Foreign DNA can be introduced into eukaryotic cells in
methods similar to bacterial transformation. However,
the process in eukaryotes is called transfection be-
cause, as we described, the term transformation in eu-
karyotes is used to mean cancerous growth. Eukaryotic
organisms that take up foreign DNA are referred to as
transgenic. Most of the techniques described here tran-
scend taxonomic lines.

SV40© PrOteinC ° at
^DNA

Remove DNA from
protein coat

Origin of DNA replication

Early genes

- Late genes
SV40 DNA
Replace part of SV with foreign DNA

o ° r o

Infect
cell

After infection of host
three choices for
continued life cycle

Add normal helper
SV40 virus

Add no
helper SV40

SV40
vehicles

Normal
SV40

Figure 13.17 SV40 virus can be used as a gene cloning
vehicle. Although part of the virus is replaced by inserted DNA
during cloning, it can still replicate with the aid of normal helper
viruses (nonrecombinant SV40). Without the aid of helper
viruses, it can either replicate as a plasmid or integrate into the
host chromosome.

Figure 13.18 Crown gall on tobacco plant (Nicotiana
tabacum) produced by Agrobacterium tumefaciens containing
Ti plasmids. (Courtesy of Robert Turgeon and B. Gillian Turgeon,
Cornell University.)

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Animal cells, or plant cells with their walls removed
(protoplasts), can take up foreign chromosomes or DNA
directly from the environment with a very low efficiency
(in the presence of calcium phosphate). Directly injecting
the DNA greatly improves the efficiency. For example,
transgenic mice are now routinely prepared by injecting
DNA into either oocytes or one- or two-celled embryos
obtained from female mice after appropriate hormonal
treatment (fig. 13.20). After injection of about 2 picoliters
(2 X 10~ 12 liters) of cloned DNA, the cells are reim-
planted into the uteruses of receptive female hosts. In
about 15% of these injections, the foreign DNA incorpo-
rates into the embryo. Transgenic animals are used to
study the expression and control of foreign eukaryotic
genes. In 1988, a transgenic mouse prone to cancer was
the first genetically engineered animal to be patented.
This mouse provides an excellent model for studying can-
cer (see chapter 16). (A controversy arose as to whether
engineered higher organisms should be patentable; cur-
rently they are.) Mice have already been successfully
transfected with a rat growth-hormone gene (fig. 13.21),
and transgenic sheep have been produced that express
the gene for a human clotting factor. The latest recombi-

Figure 13.19 A dwarf form of the plant Arabidopsis thaliana.
(Source: Science, Vol. 243, March 10, 1989, cover. ©1989 AAAS,
Washington, D.C. Photo by DeVere Patton. Courtesy of E. I. DuPont de
Nemours and Company.)

nant DNA dispute arises from the cloning of sheep in
1997 (box 13.2).

Transfection can also be mediated by retroviruses
(RNA viruses containing the gene for reverse transcrip-
tase). For example, a retroviral vector infected and re-
paired human white blood cells lacking the enzyme

» :m

1 < ■ J 1 - ■-»■

*^~*"" , - U *'- •-■*;.

Injecting

Oocyte

Suction pipette

needle

Figure 13.20 Injection of DNA into the nucleus (germinal
vesicle) of a mouse oocyte. The oocyte is held by suction from
a pipette. (© John Gardon/Phototake.)

Figure 13.21 Mouse littermates. The larger one is a transgenic
mouse containing the rat growth-hormone gene. (Source:
Richard D. Palmiter, Ralph L. Brinster, et al., "Dramatic growth of mice that
develop from eggs microinjected with metallothionein growth hormone fusion
genes," Nature 300, 16 December 1982, cover. Photograph by Dr. Ralph L
Brinster. Copyright © 1982, Macmillan Magazines Ltd.)

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

BOX 13.2

Fiction writers in the past have
created stories in which scien-
tists cloned one person to cre-
ate numerous copies. The themes of
these stories have varied from the
cloning of Adolph Hitler to the
cloning of a very busy man to help
him fulfill his day-to-day obligations.
The possibility of those scenarios
came a bit closer to reality in Febru-
ary 1997 when a group of scientists
from the Roslin Institute and PPL
Therapeutics, both in Edinburgh,
Scotland, reported in Nature maga-
zine that they had successfully
cloned a sheep from a cell taken from
the udder of a six-year-old ewe. The
cloned lamb was named Dolly (fig. 1).
In the past, genetically identical ani-
mal embryos had been created only
with amphibian cells, and those cre-
ated from adult nuclei had never suc-
cessfully reached adulthood. Cloning
in which the nuclei came from fetal
cells or cells from cell lines had been
successful before in mammals.

To clone an animal, it is necessary
to begin with an egg, the only cell
known to initiate and support devel-
opment. In order to clone an individ-
ual, using the word clone to mean
create a genetically identical copy, it
is necessary to get an egg without a
nucleus and then to transplant a nu-
cleus of known origin. Techniques for

Ethics and Genetics

Cloning Dolly

nuclear transplantation had been
worked out with frogs and toads in
the 1950s. The Scottish scientists suc-
ceeded in obtaining sheep eggs, enu-
cleating them (removing their nu-
clei), and then transferring in donor
nuclei by fusing the donor cells and
the enucleated eggs with an electri-
cal pulse. The electrical pulse also ini-
tiated development of the egg.
Although only one pregnancy of the
twenty-nine initiated was successful,
the lamb that was born seems normal
in every way; it has since produced
offspring.

Others had tried this type of ex-
periment with many types of ani-
mals, including mice. They were not
successful for numerous reasons. The
most likely explanation for the recent
success, according to the scientists, is
that the donor cells were kept in a
nongrowth phase for several days,
which may have synchronized them
with the oocyte. Thus, the nucleus
and the oocyte were at the same

stage of the cell cycle and thus com-
patible. Other reorganizations that
had to take place in the donor chro-
mosomes are not really known for
certain, but one thing is clear: the nu-
cleus of an adult cell in the sheep has
all of the genetic material needed to
support normal growth and develop-
ment of an egg. (The work has since
been repeated with goats, cattle, and
mice.)

There are numerous ramifications
to the success of this work. First,
mammal cloning could become a rou-
tine procedure. This would allow us
to study mammalian development
and to replicate genetically identical
individuals, particularly transgenic
animals that would have particular
genomes of value. We can also use
these techniques to study aging,
since an "old" nucleus is initiating the
development of a new organism. Also
of interest is the interaction of a par-
ticular genome with a particular cyto-
plasm, since the cytoplasm contains
not only the materials needed for
early development, but also cell or-
ganelles, including mitochondria that
have their own genetic material.

Finally, ethical issues must be con-
sidered if this technique is successful
with human beings. Parents might
wish to clone a deceased child or to
obtain an immunologically compati-

adenosine deaminase. A retrovirus responsible for a form
of leukemia in rodents, the Moloney murine leukemia
virus, was engineered so that all the viral genes were re-
moved and replaced with an antibiotic marker (neomycin
resistance) and the human adenosine deaminase gene. The
virus binds to the cell surface and is taken into the cell, its
RNA is converted to DNA by reverse transcription, and the
DNA is incorporated into one of the cell's chromosomes.
It is not possible for this highly modified virus to attack
and damage the cells unless a helper virus is added. Unlike
the SV40 viruses in figure 13. 17, the modified Moloney
viruses cannot initiate a successful infection without the
helpers because vital genes have been removed.

Three other recent techniques deliver recombinant
DNA to eukaryotic cells: electroporation, liposome-

mediated transfer, and "biolistic" transfer. In electro-
poration, exogenous DNA is taken up by cells subjected
to a brief exposure of high-voltage electricity. Presum-
ably, this electric field creates transient micropores in the
cell membrane, allowing exogenous DNA to enter.

Liposome-mediated transfection is a technique that
encapsulates foreign DNA in artificial membrane-bound
vesicles called liposomes. The liposomes are then used
to deliver their DNA to target cells. In one experiment,
50% of mice injected with these DNA-containing lipo-
somes were successfully transfected — they expressed
the proteins the transfecting DNA encoded.

Last are techniques developed to deliver foreign
DNA into mitochondria and chloroplasts. These have
proven difficult targets for genetic engineering because,

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ble sibling for a child who needs an
organ or bone marrow transplant.
Others might oppose cloning based
on their religious and moral convic-
tions. In response to these latter con-
siderations, President Bill Clinton
urged Congress to ban the cloning of
human beings in the United States in
1997 for at least five years.

Figure 1 Dolly is the first cloned sheep produced by the
transfer of a nucleus from the cell of an adult sheep. (AP/Wide

World Photos.)

among other reasons, they have double-membrane walls
that have not proven amenable to delivery of recombi-
nant DNA. Recently, transfection has been successful in
both mitochondria and chloroplasts using a biolistic
(biological ballistic) process, literally shooting recombi-
nant DNA coated on tungsten microprojectiles into
these organelles.

Knockout Mice

Normally, a gene used to transfect mice is incorporated
randomly in the mouse genome. However, in about one
in one thousand experiments, the gene replaces the nor-
mal gene by a process similar to meiotic recombination
(homologous recombination; see chapter 12). With this

process in mind, geneticists have been able to select for
homologous recombination; by transfecting with defec-
tive genes, they have created mice without working
copies of a particular gene. The mice produced are called
knockout mice, and they give geneticists the opportu-
nity to study the phenotype of an animal that lacks a par-
ticular gene.

The geneticist first creates a vector with the modified
gene in question. In addition, flanking regions to that
gene are added so that homologous recombination can
occur. Finally, two antibiotic genes are introduced so that
selection for successful transfection takes place. Within
the flanking regions, the gene for neomycin resistance
(neo r ) is inserted; its product inactivates the antibiotic
neomycin (fig. 13. 22). Outside of the flanking regions,

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

the gene for thymidine kinase (tk) is inserted. This gene
phosphorylates the drug gancyclovir; the phosphory-
lated gancyclovir is a nucleotide analogue that is incor-
porated during DNA synthesis, killing the cell. Thus, the
combination of the tk gene and gancyclovir is lethal;
without the tk gene, gancyclovir is harmless. If cells are
exposed to both drugs, neomycin and gancyclovir, nor-
mal cells will be killed by neomycin, cells with the tk
gene will be killed by gancyclovir, and only the cells with
the neo r gene but lacking the tk gene will survive. These
alternative outcomes allow us to select the cells in which
homologous recombination took place (fig. 13.22).

Cells that did not incorporate the vector will die from
the effects of neomycin (they are neomycin sensitive).
Cells that randomly took up the vector DNA by nonho-
mologous recombination will contain the neo r and tk
genes and will be killed by gancyclovir. However, cells
that underwent homologous recombination will contain
the neo r gene but lack the tk gene; these cells will there-
fore survive in the presence of both antibiotics (fig.
13.22). Geneticists can isolate embryonic stem cells from
mice, cells that can produce any mouse tissue. The cells
are transfected and then grown in tissue culture in the

presence of neomycin and gancyclovir, and only cells
that undergo homologous recombination will survive.
These cells are then injected into early-stage mouse em-
bryos to become part of the developing mouse. The mice
that develop will be chimeric; some will have incorpo-
rated the transfected cells into the germ line and become
heterozygotes for the disabled gene. Finally, when mice
like this are mated, one fourth of their offspring will be
homozygous for the disabled gene. Thus, knockout mice,
have been created through this ingenious technique.

Knockout mice are especially useful for studying de-
velopment and immunology. For example, if the gene for
the Mullerian-inhibiting substance is knocked out, males
are infertile because they develop female reproductive
organs. This experiment led to insight into the genetic
path for sex determination (see chapter 5). Hundreds of
knockout experiments are published each year.

Reporter Systems

We conclude this section by discussing two reporter
systems, systems used to indicate that a transfection ex-
periment was successful. Plants can be transfected with

Zi$

(a) Homologous
recombination

(b) Nonhomologous
recombination

Host

chromosome
(homologous
region)

Vector

Host

chromosome
(nonhomologous
region)

Figure 13.22 Creating a knockout mouse. A vector is created that has a disabled (nonfunctional) form
of the gene in question (X*; red). Next to the gene in question is the neomycin resistance gene (neo r ;
green); both genes are surrounded by regions (blue) that flank the normal gene on its chromosome.
Finally, outside the flanking regions in the vector is the gene for thymidine kinase (tk; yellow). In
homologous recombination (a), involving crossovers in the homologous flanking regions, the disabled
gene and the neomycin resistance gene replace the normal gene on the cell's chromosome. In
nonhomologous recombination (b), almost the entire vector is incorporated into the host chromosome,
including the thymidine kinase gene. Techniques then allow for the selective growth of cells with the
rare homologous recombination event. One fourth of the offspring of heterozygous chimeric mice will be
knockouts for the gene in question.

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Restriction Mapping

377

Figure 13.23 Luminescent transgenic tobacco plant containing
the firefly luciferase gene. The plant was watered with luciferin,
resulting in a firefly glow. (© Science VU/Keith V. Wood/Visuals
Unlimited.)

the Ti plasmids of Agrobacterium tumefaciens, as al-
luded to earlier. When a plant is infected with A. tume-
faciens containing the Ti plasmid, a crown gall tumor is
induced when the Ti plasmid transfects the host plant,
transferring the T-DNA region. Those cells transfected
with the T-DNA are induced to grow as well as to pro-
duce opines that the bacteria feed on. Much recent re-
search has concentrated on engineering Ti plasmids to
contain other genes that are also transferred to the host
plants during infection, creating transgenic plants. One
series of experiments has been especially fascinating.

Tobacco plants have been transfected by Ti plasmids
containing the luciferase gene from fireflies. The product
of this gene catalyzes the ATP-dependent oxidation of lu-
ciferin, which emits light. When a transfected plant is wa-
tered with luciferin, it glows like a firefly (fig. 13. 2 3). The
value of these experiments is not the production of glow-

Figure 13.24 Expression of green fluorescent protein in root
cells of the plant Arabidopsis thaliana under fluorescent light.
Chromosomes are visible in a cell undergoing mitosis and
chromatin is visible as circles of green in interphase nuclei.
The green fluorescent protein gene was fused to the carboxy
terminus of the gene for the transcription factor Cry2, controlling
genes for the phototrophic response (bending toward light).
(Copyright Sean Cutler, Stanford University Plant Biology Department.)

ing plants, but rather the use of the glow to "report" the
action of specific genes. In further experiments, the pro-
moters and enhancers of certain genes were attached to
the luciferase gene. As a result, luciferase would only be
produced when these promoters were activated; thus,
the glowing areas of the plant show where the trans-
fected gene is active.

One of the more recent reporter systems developed
uses a gene from jellyfish that produces a green fluo-
rescent protein. The value of this system is that it "re-
ports" when ultraviolet light shines on it, rather than re-
quiring an addition, as in the luciferase system. The gene
for the green fluorescent protein is recombined with a
gene in question, and then the transfection is performed.
If the gene in question has transferred successfully, car-
rying the gene for the green fluorescent protein, the fluo-
rescent protein will report it when activated by ultravio-
let light (fig. 13.24).

RESTRICTION MAPPING

The number of cuts that a restriction enzyme makes in a
segment of double-stranded DNA depends on the size of
that DNA, its sequence, and the number of base pairs in
the recognition sequence of the particular enzyme. That
is, a restriction enzyme with only three base pairs in its
recognition sequence will cut more times than one with
six base pairs in its sequence, since the probability of a
sequence occurring by chance is a function of the length
of that sequence. A sequence of three bases occurs more
often by chance (1/4 3 = 1/64 base pairs) than a se-
quence of six bases (1/4 6 = 1/4,096 base pairs). Hindu,

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

for example, cuts the circular DNA of the tumor virus
SV40 into eleven pieces; some restriction enzymes can
cut E. colt DNA into hundreds of pieces. The product of
the action of a restriction enzyme on a DNA sample is
called a restriction digest.

Using electrophoresis, we can separate the fragments
of a restriction digest by size. With techniques to be de-
scribed later, we can locate the restriction sites on the
original gene or piece of DNA. That is, we can construct
a map of the restriction recognition sites that will give us
the physical distance between sites, in base pairs (fig.
1325). This restriction map is extremely valuable for
several reasons. For example, when the radioactive nu-
cleotide tritiated thymidine was added for a very short
period of time during the beginning of DNA replication
in SV40 viruses, the radioactivity always appeared in only
one restriction fragment. This demonstrated that SV40
replication started from a single, unique point; that point

was localized to a particular segment of the SV40 chro-
mosome.

In addition, a restriction map often allows re-
searchers to correlate the genetic map and the physical
map of a chromosome. Certain physical changes in the
DNA, such as deletions, insertions, or nucleotide changes
at restriction sites, can be localized on the genetic map.
These changes can be seen as changes in size, or in the
total absence, of certain restriction fragments when com-
pared with wild-type DNA. This information allows us to
see changes in the DNA; it also gives us information
about the evolution of species (see chapter 21). The dif-
ferences in fragment sizes are called restriction frag-
ment length polymorphisms (RFLPs) and have
proven valuable in pinpointing the exact location of
genes and determining the identity or relatedness of in-
dividuals. A restriction digest is also useful for isolating
short segments of DNA that can be easily sequenced.

200

400

100

(a)

Gel of

total

digest

800-

700-

600-

500-

400-

400

300-

200-

200

100-

100

Gel of
partial
digest

750

650

550

500

450
400

250
200

100
50

(b)

Figure 13.25 Restriction map from electrophoresis of a restriction endonuclease
digest, (a) Original piece of DNA, showing restriction sites marked by A.
(b) Agarose gels showing bands of total and partial restriction digests. Asterisks
mark radioactive bands produced by end-labeled segments. At the left is the
scale of molecular weight markers in base pairs (e.g., 800 bp, 700 bp). The total
digest produces fragments that are 400, 200, 100, and 50 bp — and the 200 and
100 bp fragments are end labeled. The partial digest yields six additional bands.

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Restriction Mapping

379

Constructing a Restriction Map

How do we construct a restriction map? Figure 13. 2 5
shows a hypothetical piece of DNA cut by restriction
enzyme A. Below this map is a diagram of the elec-
trophoresed digest on agarose gels, which are usually
used because their porosity allows DNA fragments of rel-
atively large size to move. The restriction enzyme makes
three cuts in the DNA, generating four fragments that are
200, 50, 400, and 100 base-pairs long. The banding pat-
tern on the gel at the left in figure 13.25& is the result of
the electrophoresing of that digest. (Note that smaller
segments move faster than larger segments.) The sizes of
the segments are determined by comparison with stan-
dards of known size (not shown, although the scale is in-
dicated on the left). The gel does not reveal the order of
these segments on the chromosome. Several methods
can be used to determine the exact order of the restric-
tion segments on the original piece of DNA.

Before restriction enzyme digestion, the 5' ends of
the DNA can be labeled radioactively with 32 P using the
enzyme polynucleotide kinase. Since the enzyme is act-
ing on double-stranded DNA, both ends will be labeled.
Upon electrophoresis after digestion of the DNA in figure
13.25, the 200-base-pair and 100-base-pair (bp) bands
will be labeled radioactively, indicating that these seg-
ments are the termini of that piece of DNA. However, we
still don't know the order of the middle pieces.

The order of the other segments can be determined
by slowing down the digestion process to produce a
partial digest. If the reaction is cooled or allowed to
proceed for only a short time, not all restriction sites will
be cut. Some pieces of DNA will not be cut at all, some
will be cut once, some twice, and some cut at all three re-
striction sites. The result of electrophoresis of this partial
digest is seen at the right in figure 13.25&. From this gel,
we can reconstruct the segment order. This gel contains
the four original segments plus six new segments, each
containing at least one uncut restriction site.

From the total digest gel, we know that the 200 and
100 bp segments are on the outside because they were
labeled radioactively. This means the 50 and 400 bp seg-
ments are on the inside. In the partial digest, we find a
250 bp segment but not a 150 bp segment, which tells us
that the 50 bp segment lies just inside and next to the
200 bp terminus (fig. 13. 26b). There is a 500 bp segment
but not a 600 bp segment, which tells us that the 400 bp
segment lies adjacent to the 100 bp terminus (fig.
1326c). An unlabeled 450 bp segment confirms that the
400 and 50 bp segments are adjacent and internal in the
DNA. We thus unequivocally reconstruct the original
DNA (compare fig. 13.26e with fig. 13. 2 5a), creating a
map of sites of restriction enzyme recognition regions
separated by known lengths of DNA.

Double Digests

In practice, restriction mapping is usually done with sev-
eral different restriction enzymes. Figure 1 3.27 is a map of
the DNA of figure 1325, with the recognition sites of a
second endonuclease, B, included. Using the same
methodology just outlined, we can show that the order of
the B segments is 350, 250, and 1 50 base pairs arising from
two cuts by endonuclease B. What we do not know is how
to overlay the two maps. Do the B segments run left to
right or right to left with respect to the A segments (fig.
1327a and by. We can determine the unequivocal order
by digesting a sample of the original DNA with both en-
zymes simultaneously, thus producing a double digest.

The two orders shown in figure 13. 21a and b are
used to make different predictions about the double di-
gest. From the first order (a), we predict a 200 bp end
segment, radioactively labeled. From the second order
(b), we predict that the labeled 200 bp segment will be
cut back to 150 base pairs: there should not be a labeled
200 bp segment. The double digest shows a labeled 200
bp segment, indicating order (a). All other aspects of or-
der (a) are consistent with the double digest.

Restriction mapping thus provides us with a physical
map of a piece of DNA, showing restriction endonuclease
sites separated by known lengths of DNA. This technique

(a)

200

100

200

(b)*r

(c)

250

(d)

400

I 100

500

50,

400

/icn

(e)*r

200

I 50 i

400

100

750

Figure 13.26 Steps in the reconstruction of the DNA from
figure 13.25. Asterisks show 32 P end labels. From the total
digest, the 1 00 and 200 bp segments are established as the
end segments (a). Since there are also 50 and 400 bp
fragments within the DNA (established from the total digest),
only certain bands (fragments) are possible from the partial
digest, which establishes that the 50 bp fragment is adjacent
to the 200 bp fragment and the 400 bp fragment is adjacent
to the 100 bp end segment (steps b and c). The occurrence of
an unlabeled 450 bp fragment in the partial digest verifies the
existence of the 50 and 400 bp fragments (c/), yielding the final
structure (e). All the fragments in the partial digest are
consistent with this arrangement.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

200

100

_l_

400

1
A

Actual

100

350

(a)

250

150

' Hypothetical

A alternative

500-

Gelof

total

A digest

Gel of

total

B digest

Gel of

total

double

digest

400-

400

300-

200-

200

100-

100

350

250

150

250
200

100
50

(c)

Figure 13.27 Overlay of the recognition sites affected by two different restriction endonucleases
{A and B) on the same piece of DNA. (a) Actual arrangement, (b) Hypothetical alternative
arrangement, (c) Electrophoresis of the total restriction digests by A alone, B alone, and both.
Asterisks indicate radioactive end-labeled bands. Order (a) is consistent with all the bands found
in all the digests, whereas order (b) is not. For example, in order {b) an internal (unlabeled) 1 50
bp fragment is predicted, but this fragment is not found in the total digest.

gives us short DNA segments of known position that we
can sequence, as well as a physical map of the DNA that
can be compared with the genetic map and can locate
mutations and other particular markers.

Restriction Fragment Length Polymorphisms

Restriction fragment length polymorphisms (RFLPs), ob-
tainable from restriction digests, are proving to be very

valuable genetic markers in two areas of study: human gene
mapping and forensics. In a restriction digest of the whole
human genome, there might be thousands of fragments
from a single restriction enzyme. Unique probes have been
developed for Southern blotting these digests. Genetic vari-
ation usually comes in the form of a second allele that, due
to a mutation, lacks a restriction site and is therefore part of
a larger piece of DNA (fig. 13.28). Some probes have un-
covered hypervariable loci with many alleles (any one

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Polymerase Chain Reaction

381

Allele ... JL
A,

-650-

Probe

j "TnilllliTT ▼ ... 750 bp

-750-

K Ihh TS

jccns.

4ug &uq

shirt

\ 1hfc

Allele ... JL
A n

Probe

^Tniniirrf ▼ ... 1 ; 4oo bp

1,400-

1,400

750

Southern blot

Figure 13.28 Restriction fragment length polymorphism (RFLP)
analysis. Allele /\ 1 is a gene segment, 750 bp long, identified
by probe binding. In allele A 2 , the restriction site to the left of
allele A 1 has been changed. The probe thus recognizes a
1 ,400 bp fragment instead of the 750 bp fragment. During
Southern blotting, two different bands show up from the two
alleles. Arrows indicate restriction sites.

person has, of course, only two of the many possible al-
leles). A population's genetic variation is generated because
these hypervariable loci contain many tandem repeats of
short (10 to 60 bp) segments. Due presumably to unequal
crossing over (see chapter 8), just one of these loci, called
variable-number-of-tandem-repeats (VNTR) loci, can
generate much variation. As a result, probing for one of
these VNTR loci in a population reveals many alleles.

The Southern blots of such digests create a DNA
fingerprint of extreme value in forensics. DNA extracted
from blood or semen samples left by a criminal can be
compared with DNA patterns of suspects (fig. 13.29).
When a single probe recognizes a number of different loci,
each individual will have many bands on a Southern blot,
with most people producing unique patterns. In one sys-
tem, developed by A.Jeffreys, a single probe locates fifty or
more variable bands per person. If Jeffreys 's probes are

Alec Jeffreys (1950- ).
(Courtesy of Dr. Alec Jeffreys.)

Figure 13.29 Forensic use of DNA fingerprinting. Southern blot
of DNA from victim (V) and defendant (D) in a crime. Jeans
and shirt refer to blood samples taken from the clothing of the
defendant. The pattern on the shirt clearly matches the victim's
blood, not the defendant's own blood. All of the other lanes of
the blot contain controls and size standards. The probability
that the blood stains were not from the victim was estimated
at one in thirty-three billion, more than the number of people
on earth. However, these probabilities are controversial,
depending on statistical assumptions about variability within
racial and ethnic subpopulations. (Courtesy of Cellmark Diagnostics,
Germantown, MD.)

used to compare the patterns, the likelihood that the two
patterns would match randomly is infinitesimally small.
This technique thus has greater power to identify individ-
uals than using the prints from their fingertips.

POLYMERASE CHAIN
REACTION

In the past, in many instances (in museum specimens,
dried specimens, crime scene evidence, and fossils), a
DNA sample was available, but in such small quantity or
so old as to be considered useless for study. That situation
changed in 1983 when Kary Mullis, a biochemist working
for the Cetus Corporation, devised the technique we now
refer to as the polymerase chain reaction (PCR). PCR
can be used to amplify whatever DNA is present, however
small in quantity or poor in quality. The only requirement
is that the sequence of nucleotides on either side of the
sequence of interest be known. That information is
needed to construct primers on either side of the se-
quence of interest. Once that is done, the sequence be-
tween the primers can be amplified.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

In the PCR technique, the primers and the ingredi-
ents for DNA replication are added to the sample. Then,
the mixture is heated (e.g., 95° C for twenty seconds) to
denature the DNA. The temperature is then lowered
(e.g., 55° C for twenty seconds) so that primers can an-
neal to their complementary sequences. The tempera-
ture is then raised again (e.g., 72° C for twenty seconds)
for DNA replication. Then, a new cycle of replication is
initiated (fig. 13.30). The various stages in the cycle are
controlled by changes in temperature since the tempera-
tures for denaturation, primer annealing, and DNA repli-
cation are different. About twenty cycles of PCR pro-
duces a million copies of DNA; thirty cycles make a
billion copies. The technique is aided by using DNA
polymerase from a hot-springs bacterium, Thermus
aquaticus, that can withstand the denaturing tempera-
tures. Thus, after each cycle of replication, no new com-
ponents have to be added to the reaction mixture.
Rather, the cycling can be continued without interrup-
tion in PCR machines (simply programmable water baths
that accurately and rapidly change the water temperature
that surrounds the reaction mixture). Some machines
can process ninety-six samples at a time.

PCR has been used to create DNA fingerprints by am-
plifying microsatellite DNA. These are repeats of very
short sequences of DNA dispersed throughout the
genome. For example, cytosine-adenine (CA) repeats oc-
cur tens of thousands of times in eukaryotes, in repeats
of from twenty to sixty base pairs. As in the case ofVNTR
loci, there is tremendous variability among people in the
number of these repeats at a locus, due presumably to
crossover errors. Unlike the situation with VNTR loci,
however, PCR amplification of one of these loci can be
done without restriction cutting, Southern blotting, and
probing — PCR gives the results directly upon elec-
trophoresis. All we need are the surrounding primer se-
quences to any microsatellite locus. PCR is now a rou-
tinely used tool in the laboratories of molecular
geneticists. They use it to rapidly amplify the DNA re-
gions of interest for research or forensic uses.

Figure 13.30 Polymerase chain reaction. DNA is denatured,
(step 1), primer oligonucleotides that are complementary to end
sequences on the two strands anneal (step 2), and DNA
replication takes place (step 3). Each step in the cycle is
controlled by temperature changes. The targeted sequence is
shown as red on one stand and blue on the other. Primers are
shown as either green or yellow lollipops. A green primer
begins the copying of the red strand into a complementary
blue strand; a yellow primer begins the copying of a blue
strand into a complementary red stand. In three cycles, one
double-stranded region of DNA becomes eight. The process
requires the addition of primers, deoxynucleotide triphosphates,
and DNA polymerase, as well as changing temperature cycles.

Targeted
sequence

Cycle 1

Steps
1 and 2

5' ■ '3'

Step
3

Cycle 2

Steps
1 and 2

Step
3

Cycle 3

Steps
1 and 2

Step
3

T T T T

/u\

owio

tiff

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We now turn our attention to a major result of recom-
binant DNA technology, DNA sequencing. Recombinant
DNA technology, with its ability to isolate and amplify
small, well-defined regions of chromosomes, has allowed
the development of DNA sequencing techniques.

DNA SEQUENCING Q\

Paul Berg of Stanford University, Walter Gilbert of Har-
vard University, and Frederick Sanger of the Medical Re-
search Council in Cambridge, England, shared the 1980
Nobel Prize in chemistry. Berg won for creating the first
cloned DNA molecules when he spliced the SV40
genome into phage X. Gilbert and Sanger were awarded
the prize for independently developing methods of se-
quencing DNA. Gilbert, along with Allan Maxam, devel-
oped a method of DNA sequencing called the chemical
method. It involves chemically breaking down the DNA
at specific bases. Sanger, who won a Nobel Prize in 1959
for sequencing the insulin protein, later took part in de-
veloping methods for sequencing RNA. His sequencing
method, developed with Alan Coulson, involved DNA
synthesis and was called the plus-and-minus method.
The further development of the method by Sanger, Coul-
son, and S. Nicklen, using specific chain-terminating nu-
cleotides, led to a modification of the plus-and-minus
method known as the dideoxy method.

'rT

1 ]

•1

fyli

■

Walter Gilbert (1932- ). Frederick Sanger (1918- ).

(Photo: Rick Stafford.) (Courtesy of Dr. Frederick

Sanger.)

The Dideoxy Method

In the dideoxy method, manipulation of DNA synthesis
enables DNA sequencing. Remember from chapter 9 that
DNA synthesis occurs at a primer configuration, one in
which double-stranded DNA ends with a 3 -OH group on
one strand. The other strand continues as single-stranded
DNA (fig. 1331, middle). The dideoxy method creates a
primer configuration of the DNA to be sequenced and
enables replication to proceed. A trick, using chain-

terminating nucleotides, stops DNA synthesis at known
positions. These chain-terminating nucleotides are
formed of sugars lacking OH groups at both the 2' and 3'
carbons (hence the term dideoxy). Without a 3' -OH
group, a dideoxynucleotide cannot be used for further
DNA polymerization (fig. 1331).

Chain-terminating nucleotides permit synthesis to be
stopped at a known base. The sample to be sequenced is
elongated separately in four different reaction mixtures,
each having all four normal nucleotides but also having a
proportion of one of the chain-terminating dideoxy nu-
cleotides. For example, if the pool of thymine-containing
triphosphate nucleotides contains a portion of the
dideoxythymidine triphosphate molecules, then synthe-
sis of the growing strand is sometimes terminated when
adenine (the complement of thymine) appears on the
template, creating fragments that end in thymine. Similar

PHPHP

5'
C Thymine

A' V
W 27

OH H

Deoxythymidine triphosphate (dTTP)

5'
C Thymine

PMPMP

\3' 27

H H

Dideoxythymidine triphosphate (ddTTP)

Normal primer

Dideoxy
configuration
(not a primer)

Figure 13.31 Dideoxy nucleotides cause chain termination
during DNA replication. The dideoxy primer configuration lacks
the 3'-OH group needed for chain lengthening in a normal
primer configuration.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

reactions are carried out in separate test tubes for each of
the other nucleotides, producing fragments that termi-
nate when the respective complementary nucleotide is
present. The resulting fragments from each reaction are
electrophoresed, generating a pattern on the gel that re-
veals the sequence of the newly synthesized DNA. Let us
go through an example.

In figure 13. 32a, we show the DNA to be sequenced,
a small segment of nine base pairs. To sequence this seg-
ment, one must get one strand of this double-stranded
segment into the configuration shown in figure 1332&.
The DNA to be sequenced must be the template for new

DNA synthesis. (We will soon discuss how we obtain the
required configuration.) Having created the necessary
primer configuration, we take four subsamples of it, each
including all four nucleoside triphosphates plus DNA
polymerase I. At least one of the nucleoside triphosphates
is radioactively labeled, usually with 32 P. This label allows
us to identify newly synthesized DNA by autoradiography.
To each of the four subsamples, one of the
dideoxynucleotides (dd) is added — one subsample gets
ddTTP, one gets ddATP, one gets ddCTP, and one gets
ddGTP. These dideoxynucleotides are added in addition
to the regular deoxynucleotides to increase the probabil-

(a)

5'-ATACCGTAC-3'
3' -TATGGCATG -5'

Create primed
t configuration

(b)

3'- OH

TATGGCATG

ddTTP
(dideoxy)

AT*
TATGGCATG

Add DNA polymerase I plus

four deoxynucleotides

(dTTP, dATP, dCTP, dGTP), one

of which is radioactive, and subdivide

into four subsamples

ddATP
(dideoxy)

TATGGCATG

ATA*
TATGGCATG

ddCTP
(dideoxy)

ddGTP
(dideoxy)

ATAC*
TATGGCATG

ATACCG'
TATGGCATG

ATACC*
TATGGCATG

(c)

ATACCGT*
TATGGCATG

ATACCGTA*
TATGGCATG

ATACCGTAC*
TATGGCATG

Figure 13.32 Initial steps in the dideoxy method of DNA sequencing. The asterisks indicate the dideoxynucleotides. The
DNA to be sequenced is placed into a primer configuration (a, b). Four reaction mixtures are created, each with all four
normal nucleotides plus one of the dideoxynucleotides. Thus DNA synthesis in each reaction mixture is stopped a
percentage of the time when the complement to the dideoxynucleotide appears in the template (c).

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DNA Sequencing

385

ity that chain termination will occur at every appropriate
position. If the dideoxynucleotide were added in place of
the deoxynucleotide, then the chain would be termi-
nated the first time the complement of that base ap-
peared in the template strand. By mixing the dideoxynu-
cleotides and the deoxynucleotides, we are assured that
termination will occur in every appropriate position.

In figure 13. 3 2c, we see that the template has two
adenines. Therefore, in the ddTTP reaction mixture, ade-
nine's complement (thymine) is needed twice. There are

thus two possible points for ddTTP to incorporate, two pos-
sible chain terminations, and therefore two fragments that
could end in dideoxythymidine, of two and seven bases, re-
spectively. Similarly, there are three possible fragments end-
ing in adenine, of one, three, and eight bases; three ending
in cytosine, of four, five, and nine bases; and one ending in
guanine, of six bases (fig. 13. 32c and fig. 13. 33, top).

After DNA synthesis is completed, the old primer is
removed, leaving only newly synthesized DNA fragments
(fig. 13. 3 3). Newly replicated segments of various lengths

Isolate newly synthesized DNA

ddTTP

AT*
+
ATACCGr

ddATP

+
ATA*

+
ATACCGTA'

ddCTP

ATAC*

+
ATACC*

+
ATACCGTAC

ddGTP

ATACCG*

Electrophoresis
and autoradiography

ddTTP

ddATP

ddCTP

ddGTP

Sequence

9 —

8 —

7 —
6 —

T
G

5 —
4 —
3 —
2 —

C
C
A

1 —

—

Figure 13.33 Electrophoresis of segments produced by the dideoxy method of
DNA sequencing. This method allows direct reading of the sequence. The
asterisks indicate the dideoxynucleotides. The newly synthesized reaction
products seen in figure 13.32 are isolated by removal of the primer and
template. Each reaction mixture (e.g., ddTTP is the mixture containing
dideoxythymidine triphosphates) produces specific products of specific lengths
that can be determined by electrophoresis. In the case of the ddTTP mixture,
two fragments ending in thymine are possible; one is two bases long, the other
seven bases long. Thus, the complement of thymine, adenine, appears in
positions 2 and 7 of the original piece of DNA. However, either the original
strand or its complement (the new synthesis) yields the original sequence since
DNA is a double helix; the sequence in one strand is always defined by the
complementary sequence in the other strand.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

from each reaction mixture are placed in separate slots
and then electrophoresed on polyacrylamide gels to de-
termine the lengths of the segments. Since only newly
synthesized DNA segments are radioactive, autoradiogra-
phy lets us keep track of newly synthesized DNA. As you
can see from the autoradiograph of the gel in figure
13. 33, each subsample produces segments that begin at
the primer configuration (beginning of synthesis) and
end with the chain-terminating dideoxy base. By starting
at the bottom and reading up, back and forth across the
gel, we can directly determine the exact sequence of the
DNA segment. Because they have the appearance of
stepladders in each lane (fig. 13. 34), the gels are usually
referred to as stepladder gels or ladder gels.

This technique (in the form of the original plus-and-
minus method) was first used to sequence the genome of
the DNA phage c|)X174 (box 13. 3). That phage was used
because it lent itself to the sequencing method. It has
single-stranded DNA within the phage coat, yet its DNA
becomes double-stranded once it enters the bacterium.
Creating a primer configuration was thus relatively easy.
The double-stranded circle from within the host could be
treated with a restriction endonuclease to produce
double-stranded fragments (fig. 13.35). These fragments
could then be denatured. From this mixture, a particular
fragment could be isolated by electrophoresis. The iso-
lated strand would reanneal to the single-stranded DNA
taken from the phage heads, forming a primer for new
growth. The same restriction endonuclease would free
the new growth after it had taken place. Thus, the
dideoxy method was relatively easy to apply to the 5,387-
base chromosome of c))X174.

Creating a General-Purpose Primer

To make the dideoxy method efficient, researchers cre-
ated a general primer for routine sequencing work by re-
combinant DNA engineering of an E. colt vector, the
single-stranded DNA phage Ml 3. This phage is similar to
c))X174 in that both are packaged as single-stranded DNA,
and both are replicated to double helices within the host.
Therefore, the double-stranded form within the host,
called the replicating form, can be engineered by stan-
dard methods, and the single-stranded form can be used
for sequencing. The system works as follows.

By very clever engineering, J. Messing and his col-
leagues created cloning sites for a variety of restriction
enzymes in a bacterial gene QacZ) that had been inserted
into Ml 3 (fig. 1336). The gene is for the p-galactosidase
enzyme that normally breaks down lactose. It also breaks
down an artificial substrate of the enzyme, X-gal, which is
normally colorless. When cleaved by p-galactosidase,
X-gal becomes blue. Thus, in the presence of the func-
tional lacZ gene, Ml 3 plaques are blue. If the gene is dis-
rupted by a cloned insert, X-gal does not break down,

A
G
G
A
T
T
T
A
A
C
A
C
G
G
A
C
G
A
T
A
G
G
A
T
C
G
G
C
G
A
T
C
G
A
T
C
G
G
C
T
G
T
A
G
T
G
G
A
A
A
G
A
T
T

Figure 13.34 Autoradiograph of a dideoxy sequencing gel. The
letters G, A, T, and C along the bottom refer to the ddGTP,
ddATP, ddTTP, and ddCTP reaction mixtures, respectively.
Lanes are repeated for easier identification of the bands. The
sequencing is also verified by sequencing the complementary
strand and checking for agreement. (Courtesy of Richard J.
Roberts.)

GATGCATC

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DNA Sequencing

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Restriction
endonuclease
produces four
fragments

~r 2

Fragments are isolated
by electrophoresis
and denatured

i i i i

5' 3'

Double-stranded circle

One of the fragments isolated
by electrophoresis joins the
complementary region of the
single-stranded DNA circle
isolated from phage heads

Primer

DNA polymerase

Discard

Polymerization produces extension
products of various lengths

DNA circle is
treated with original
endonuclease

and denatured

>►

New growth

New growth is isolated

Figure 13.35 The genome of phage c|)X174 lent itself to the dideoxy method (originally, the plus-and-minus method) of
DNA sequencing. Because the phage occurs in both the single- and double-stranded forms, it can be manipulated for
sequencing. The double-stranded form is fragmented with an endonuclease. One fragment is isolated by
electrophoresis and hybridized to the single-stranded form, creating a primer for new DNA synthesis and thus for
dideoxy sequencing. Newly synthesized DNA can be isolated by treating it with the same restriction enzyme, which will
create the same cut made originally. The newly isolated pieces can then be electrophoresed as in figure 13.34.

and hence the plaques are colorless. (Ml 3 doesn't form
true plaques because it doesn't lyse the E. colt cells. It
does form turbid sites due to reduced bacterial growth.)

An oligonucleotide primer can be synthesized that is
complementary to a region of the phage DNA upstream
from the cloning sites. Single-stranded phage DNA con-
taining a cloned insert is isolated and hybridized with
the synthetic oligonucleotide. This operation creates the
primer configuration for dideoxy sequencing of the
cloned DNA. Virtually any clonable segment of DNA can
be sequenced using this very general method. Theoreti-
cally that segment could be any size.

Stepladder gels, however, are effective only up to
about four hundred base pairs. To sequence larger regions
requires sequencing overlapping segments and reconsti-
tuting the sequence by the overlap pattern, similar to the

methods we described for amino acid sequencing (chap-
ter 11, box 11.1). Overlapping segments of DNA are usu-
ally obtained by using two or more restriction enzymes.

The most recent innovation in DNA sequencing in-
volves using four fluorescent dyes, each fluorescing at a
different wavelength (505, 512, 519, and 526 nm); each
of the four dideoxy nucleotides has a different dye at-
tached. After the newly synthesized fragments are iso-
lated, the products from all four reactions are run to-
gether in the same lane of a polyacrylamide gel. The gel is
then scanned with an argon laser that excites the dye
molecules. An instrument records the color of the peaks,
reading the sequence directly and automatically (fig.
13. 37). This method greatly simplifies sequencing since it
is automated. It also alleviates the necessity for radioac-
tive tags.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

BOX 13.3

Complete sequencing of a DNA
genome using Sanger and Coul-
son's plus-and-minus method
(the forerunner to the dideoxy
method) was first accomplished with
<|>X174, a virus that contains a single-
stranded DNA circle of 5,387 bases
within its protein capsule. Once in-
jected into the host, the DNA is repli-
cated to form a double helix that
then proceeds in normal viral fashion
to replicate itself, manufacture its
own coat proteins, lyse the cell, and
escape. This virus has nine genes. The
virion is a small, twenty-faced polyhe-
dron with a small spike at each of its
twelve vertices. This spike attaches
4>X174 to E. coll The coat accounts
for one protein and the spike ac-
counts for two. Thus, three of the
virus's nine genes manufacture coat
proteins. Figure 1 illustrates the loca-
tion of the genes in 4>X174, obtained
through standard mapping methods.

From the information obtained
from the sequencing of MS2, an RNA
virus, geneticists believed that there
should always be a nontranslated se-
quence between genes, presumably
for the purpose of controlling expres-
sion of each gene. However, careful
perusal of the nucleotide sequence of
4>X174 provided several surprises.
First, the ends of three genes
overlapped the beginnings of the
next genes (A-C, C-D, and D-J); in the
first two cases, the initiation codon is
entirely within the end of the previ-
ous gene, but read in a different frame
of reference. In the sequence ATGA,
the ATG is the initiation of the next
gene, whereas the TGA is the termina-
tion of the previous gene. In the D-J

Experimental
Methods

Genes Within Genes

interface, one A is shared: TAATG
(UAAUG in ribose nucleotides; fig. 2).
It is the number 3 base of the termi-
nation codon and the number 1 base
of the initiation codon. The surprises
did not end there.

At first, with the sequence of nu-
cleotides spread out in front of them,
the researchers could not find the B
and the E genes; they appeared to be
missing. Upon careful analysis, how-
ever, the scientists found that the B
gene was entirely within the A gene
and the E gene was entirely within
the D gene (fig. 3). Their finding went
against theory. We were led to be-
lieve, from logical arguments, that
genes cannot substantially overlap.
There would be too much of a con-
straint on function: The functional se-
quence of one gene would also have
to be a functional sequence in the
other. Similarly, there would be an
evolutionary constraint involved. The
genes would have to evolve together.
But here we have two cases in which
genes do overlap. How could over-
lapping genes come about?

There are a large number of
thymine bases in the <|>X174 genome.
In the D gene particularly, many of
the codons end with thymine. The
imbedded E gene is read on a shifted
frame with D so that the terminal
bases of Ds codons are the middle

bases of Es. A look at the genetic
code (see table 11.4) shows that the
codons with U in the middle (iTs
codons) are mainly for hydrophobic
amino acids. Thus, E is a protein with
detergent properties. In fact, it is the
protein responsible for the dissolu-
tion of the outer cell wall of the host
bacterium, a process that a detergent
can accomplish in vitro. The proper-
ties of the E gene, then, are more the
properties of its individual amino
acids rather than their exact sequence.

In the A-B case, there is an indica-
tion that the two genes were once
autonomous. This indication is based
on the patterns of the codons; ^4's
codons tend to end in thymine before
the overlap, but thereafter, in the re-
gion of overlap, Bs codons end in
thymine, whereas ^4's codons do not.
Presumably, a mutational event
tagged the B material onto the end of
the earlier, shorter A gene and im-
proved its enzymatic ability. We can
only speculate, however.

The amazing arrangement of this vi-
ral DNA is one of extreme economy.
The protein package is small, yet a min-
imum of nine genes had to be packed
into it. We have seen this kind of econ-
omy before in the codon usage of mi-
tochondrial DNA (see chapter 11).

As more sequencing has taken
place, geneticists have discovered
other novel overlap situations. For ex-
ample, in one case, two genes were
transcribed from opposite strands of
the same region of DNA from a rat. On
one strand, the gonadotropin-releasing
hormone gene (GnRH) is located. On
the other is a gene (RH) that produces
a protein expressed in the heart.

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DNA Sequencing

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Gene overlap is known to occur in
bacteria as well. In E. colt, the pro-
moter for the ampC gene (coding for
the enzyme (3-lactamase) begins
within the last ten codons for the frdC
gene, which codes for a subunit of the
enzyme fumarate reductase. There is
evidence that in this arrangement the
frdC terminator can have some regu-
latory control of ampC transcription.
(See chapter 14 for a discussion of
regulatory processes in prokaryotes.)

With DNA sequence data, includ-
ing the complete sequences of other
chromosomes such as those of SV40
and mitochondria, we have accumu-
lated much information about gene
arrangements. Overlap to one degree
or another has been found in small
viruses (cf>X174, SV40), large viruses
(A), mitochondrial chromosomes,
bacterial DNA, and even eukaryotes,
in which several cases are now
known in which genes are located

within introns of other genes. In one
of the few examples known, three
genes are located in an intron of the
neurofibromatosis gene, a gene that
causes a disfiguring neurological dis-
ease. Although relatively uncommon,
overlap and embedding of genes may
have some regulatory role in tran-
scription in addition to minimizing
the length of the chromosome.

End transcription

Figure 1 Presumed location of the nine genes of
phage cj>X174 on its circular chromosome.
Transcription begins at three different places, each
marked p, for promoter. The function of each gene
appears within the circle.

J
5'
E
D

Met Ser
AAGGAGUGAUGUAAUGUCU

• • •

Lys Glu Stop

Glu Gly Val Met Stop

Figure 2 Sequence, shown as ribose nucleotides,
where genes E and D end and gene J begins. Each is
out of register with the other two. The A of AUG for
gene J, for example, is the second A of the UAA termi-
nator of gene D.

Figure 3 The actual map of the nine genes of phage
cj)X1 74. Note that B is entirely within A and E is entirely
within D.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Cloning site within /acZgene

M13

Double-stranded
replicating
form (RF)
isolated from
infected cell

Foreign DNA

Insert foreign DNA
into cloning site;
transform host
E. coli for phage
replication

Isolate single-stranded
DNA from phage heads

3' - OH Y

Primer

Add synthetic
oligonucleotide, which
hybridizes adjacent to
cloning site, forming
primer configuration

Figure 13.36 Phage M13, a useful vector for sequencing a
piece of cloned DNA by the dideoxy method since it exists in
both single- and double-stranded forms. In addition, it contains
restriction sites within a copy of the lacZ gene (blue). This
allows for the selection of clones with inserted pieces of foreign
DNA. An artificial oligonucleotide, which hybridizes adjacent to
the cloning site, provides the primer configuration needed for
new synthesis.

Either the dideoxy or the chemical method of se-
quencing (not discussed) allows us to read the sequence
of hundreds of nucleotides on a single gel. Whole viral,
prokaryotic, and eukaryotic genomes, and numerous re-
gions of interest in prokaryotes, eukaryotes, and viruses
have been sequenced. As W. Gilbert said in his Nobel
Prize acceptance speech in 1981, "When we work out
the structure of DNA molecules, we examine the funda-
mental level that underlies all processes in living cells."

MAPPING AND SEQUENCING £%
THE HUMAN GENOME V

Locating a Gene of Interest

Genes of importance can be searched for directly. A
breast cancer gene provides a good example. Other genes
that have been found this way include the genes for cystic
fibrosis and Huntington disease. The concept of finding a
gene is relatively simple; the methodology is tedious.
Searching for many genes, including medically important
genes such as one for breast cancer, means looking for a
gene only by its symptoms; that is, we don't know the
protein product of the gene or its location. Searching be-
gins by looking at pedigrees of families segregating the
disease and then trying to correlate the occurrence of the
disease with a particular RFLP or microsatellite marker.
When this is done, the gene has been localized to a par-
ticular region of a particular chromosome. Then, with a
genomic library, chromosome walking (see the next sec-
tion) is done until a gene in the neighborhood of the
marker is found that could be the target gene. With the
gene in hand, its sequence and protein product can be de-
termined, a first step in medical treatment.

Chromosome Walking

Despite the limited size of any one inserted piece of for-
eign DNA, it is possible to learn about longer stretches of
DNA by using a technique of overlapping clones called
chromosome walking. Let us say that a particular gene
(in region A) is located in clone 1 , as discovered through
probing. The cloned insert can be removed, using the
same restriction enzyme initially used to insert it in the
vector, and broken into small pieces that are used as
probes themselves. The idea is to locate another clone
with an inserted region that overlaps the first one (fig.
13. 38). The second clone is now treated the same way —
with segments used to probe for yet another overlap far-
ther down the chromosome. In this way, relatively long
segments of a chromosome can be available for study in
overlapping clones.

One obvious use of chromosome walking is to dis-
cover what genes lie next to each other on eukaryotic
chromosomes. The technique is very tedious and is halted
at certain areas not amenable to walking, such as repeated
sequences found in the DNA of eukaryotes (see chapter
15). Once an overlapping probe contains a commonly re-
peated sequence, it hybridizes to many clones that do not
contain adjacent segments. This "cross-referencing"
lessens the value of the technique. Currently, newer tech-
niques (termed chromosome jumping), designed to
bypass regions not amenable to walking, are being devel-
oped. These techniques depend on the ability to locate

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Figure 13.37 Processed data from automated DNA analysis using fluorescent dyes. The DNA is
sequenced by attaching a different fluorescent dye to each dideoxy base. Thus, the dideoxy bases
can be identified by their fluorescent color in a laser light rather than by which lane they occupy in a
gel. Only one lane need therefore be run. In this diagram, guanine is yellow, cytosine is blue, adenine
is green, and thymine is red. The sequence is read left to right, top to bottom. (From L Johnston-Dow,
et al., BioTechniques, 5:754-65, 1987, copyright © 1987. Eaton Publishing, Natick, MA. Reprinted with permission.)

the two ends of a segment without having to walk
through the middle. Ends of a segment can be located if
the region has been inverted or if a large region is cloned
and the middle part later removed, leaving just the ends.
A probe of the ends allows the investigator to locate
clones with first one end and then the other, effectively
jumping over the intervening region.

The Breast Cancer Gene

The initial location of the breast cancer gene BRCA1 was
determined by M. King in 1990 using a marker (Dl 7S74)
on the long arm of chromosome 17 (fig. 13.39); it was the
183rd marker that King had tried (fig. 13.40). The breast
cancer gene BRCA1 was particularly difficult to locate
because it accounts for only about 5% of all breast can-

Mary-Claire King
(1946- ). (Courtesy of
Office of Public Information,
Berkeley Campus, University
of California. Photograph ©
Jane Scherr.)

cers. However, it accounts for a much higher percentage
of inherited, early onset breast cancers, those in women
under fifty years of age. One woman in two hundred in-
herits this gene, and among those women, 80 to 90% risk
developing the disease. The actual locating and cloning
of this gene was done in 1994 by a team led by M. Skol-
nick.The gene codes for a protein of 1,863 amino acids;
it seems to act as a tumor suppressor protein (see chap-
ter 16). Its mechanism of action is as a transcription fac-
tor associated with RNA polymerase II (see chapter 10).

The Human Genome Project

The Standard Method

In chapter 6, we developed a human chromosome map.
Generally, a locus was located on a particular chromo-
some by tissue culture techniques (somatic-cell hy-
bridization). Loci could be pinpointed further using
aberrant chromosomes, such as those with deletions. If a
locus was present when the intact human chromosome
was present but absent if the deletion chromosome was
present, the gene could be localized to the deleted re-
gion. In addition, probes for specific genes can show us
roughly where that gene is located (fig. 13.41).

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Insert A in clone 1
fragment and subclone

A B

Use probe to detect clone
with adjacent region

Clone 2 detected; has A-B region
fragment, subclone, and probe

B C

Clone 3 detected; has B region
fragment, subclone, and probe

Clone 4 detected; has B-C region
fragment, subclone, and probe

Continue

ABC

Figure 13.38 Chromosome walking technique. This technique
allows one to study long chromosomal regions by locating
overlapping cloned inserts. We begin with a specific cloned
piece of DNA, referred to as insert A. This piece is fragmented
to create probes for other clones in a genomic library that
contains regions that overlap A (the next region down is referred
to as B). The A-B clone is itself then fragmented to create
probes to repeat the process, moving down the chromosome.

D17S74

Chromosome 17

Figure 13.39 A Giemsa-banded chromosome 17, showing the
numbering of the regions and the location in region 21 q in
which marker D17S74 is located. The terminology of the
marker is that of section 74 of chromosome 17. This marker
correlated to the position of the BRCA1 gene.

Two general methods were developed for mapping
the human genome, the standard method, supported in
large part by federal funding, and the whole-genome
shotgun method used by the Celera Genomics Com-
pany In the standard method, the project is reduced to
finding a segment of the genome and locating where it
belongs. The segment is then sequenced. By the overlap
of sequenced pieces, the whole genome is pieced to-
gether. Mapping is done chromosome by chromosome
since individual chromosomes can be isolated in large
numbers by the methods of flow cytometry, described in
chapter 15. In the initial stages of the Human Genome
Project, when the primary task was mapping, yeast artifi-
cial chromosomes (YACs) were the primary cloning
agent. However, as the emphasis of the project shifted

Figure 13.40 A pedigree of a family of individuals in which
early onset breast cancer is segregating. At the bottom of the
figure is a gel of the various bands produced, showing the
alleles of D17S74, marked A-E in decreasing size of fragment
probed. The individuals in the pedigree are shown directly over
their lanes in the gel. The original parents were dead {diagonal
line) and thus were not typed. The mother, two of her
daughters, and two of her granddaughters were diagnosed
with breast cancer in ages ranging from twenty-three to forty-
five years of age (yellow). Note that in every case of breast
cancer, the woman has the B allele of marker D17S74. It is
this correlation that localized the breast cancer gene to that
region of the chromosome. D17S74 was the 183rd marker M.
King and her colleagues studied; the other markers showed no
correlation with breast cancer. (Reprinted with permission from J. M.
Hall et al., "Linkage of Early-Onset Familial Breast Cancer to Chromosome
17q21," Science, 250:1684-89, 1990. Copyright © 1990 American
Association for the Advancement of Science.)

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Figure 13.41 The physical location of a gene or marker can
be found by probing chromosomes with a complementary DNA
sequence that has a specific fluorescent compound bound to
it. When activated, the probe is seen {bright yellow spots) in a
laser scanning confocal microscope. The chromosomes are
counterstained with propidium iodide, which makes them
fluoresce red. In this case, the probe has located a sequence
on human chromosome 11. (© Peter Menzel/Photographed at Yale
University Medical School.)

to sequencing, bacterial artificial chromosomes
(BACs) were used. The bacterial artificial chromosomes
are derivatives of the fertility factor (F factor, see chapter
7). They have properties of stability and homogeneity
that make them more compatible with automated se-
quence techniques.

To begin sequencing, each individual chromosome is
broken up into overlapping segments of about 150,000
bp in a BAC library Each BAC is then digested into
smaller pieces that are cloned in cosmids or PI phages di-
gested into smaller pieces for sequencing.

Before we define the techniques further, we should
mention that we are not dealing with just one map of the
genome, but several different kinds of maps. Although
the ultimate goal was the complete DNA sequence of the
genome, yielding the exact location of every gene, we
needed to go through several stages to get there —
remember, we are trying to keep track of 33 billion
bases. We are familiar with the genetic linkage map of
chromosomes described in chapter 6. These maps are
called classical linkage maps; they define distances in
recombination frequencies. A modern linkage map is
one that uses RFLP markers along its length instead of
genes. There is also a physical map, in which distances
are in physical units of base pairs. These maps can be of
microsatellite markers or of sequence-tagged sites
(STSs). Sequence-tagged sites are DNA lengths of 100-
500 base pairs that are unique in the genome. They are
created by polymerase chain reaction amplification of
primers obtained by sequencing segments of the
genome. The primers are then tested to be sure the se-
quence is unique. About 50% of attempts yield sequence-
tagged sites.

The physical map can also be marked off in differ-
ences among individuals that amount to changes in
single base pairs. These differences are called single-
nucleotide polymorphisms (SNPs — pronounced
"snips"). These are located about every one thousand
bases along the human genome. These single-nucleotide
polymorphisms are expected to be especially useful in
keeping track of differences among individuals in genes
responsible for diseases.

RFLPs, microsatellite markers, STSs, and SNPs allow
us to keep track of BACs and cloned pieces in cosmids
and PI phages. However, as we locate various DNA
pieces, we will be building up continuous regions of a
chromosome by overlapping these pieces. These over-
lapping, contiguous clones are referred to as contigs.
This process is repeated chromosome by chromosome.
In other words, we are creating a library of overlapping
clones that cover the complete length of each chromo-
some. In essence, we are putting together a linear jigsaw
puzzle. Contigs are created by comparing the segments
that clones have in common, if any (fig. 13.42). From
shared segments, we can infer which parts of the clones
overlap. Through this process, contigs of parts of the
chromosome can be built up (fig. 13.43). Later, contigs
comprising part of a chromosome can be ordered by tak-
ing an end clone of a completed contig and using it as a
probe to begin chromosome walking to find an end
clone of a nearby contig (fig. 13.44).

For example, let's begin with a BAC from chromo-
some 7 of 150,000 base pairs long with three sequence-
tagged sites located along its length. We can determine
neighboring BACs by shared sequence-tagged sites.
The BAC is then digested and cloned into cosmids. The

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Restriction-Fragment Fingerprints

(a) Clone 1 overlapping clone 2

r s -\-

Y Y Y

6.5

•|-

Restriciton sites
of EcoRI

Clone 1

Clone 2 | Y Y

6.5

~\-

(b) Fingerprints of clones 1 and 2

Clone 1

if)

H — '

C
CD

H —

c
o

-I— »

CO
CD

6.5
6.0

- 5.0

4.0
3.5

■= 2.0

Clone 2

7.0
6.5

5.0

4.0

3.0

2.0

Gel patterns

(c) Regions of overlap and nonoverlap inferred from fingerprint date in (b). Fragments
are arbitrarily ordered, from largest to smallest, within each region.

3.5

6.5

Clone 1

Clone 2

6.5

Nonoverlap

yy_

Overlap

^K_

Nonoverlap

Figure 13.42 To create contigs, researchers must find overlapping clones and determine
their region of overlap. In part (a), we have two overlapping pieces of DNA, found by
chromosome walking. The pieces are digested with EcoRI and electrophoresed,
producing the blots in part (d). From these gels, we see that fragments of 2.0, 4.0, 5.0,
and 6.5 kb pairs are in common, indicating that they are in the region of overlap in both
clones. We have thus isolated the overlap region and the unique end regions of both
clones (compare c with a). Restriction maps can then be made of each segment,

ordering the pieces. (Reprinted courtesy of Los Alamos Science, Volume Number 20, a publication of
Los Alamos National Library, Los Alamos, NM.)

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Mapping and Sequencing the Human Genome

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Assembly of a Contig

Overlap

Clone B

J I l_L

Overlap

Clone A

J I l_L

J l_LL

Clone C

J l_LL

Overlap

Figure 13.43 A contig is further built up by assembling
pairwise overlapping clones into longer sequences. Here we
see that clone A overlaps clone B to the left and clone C to
the right In this case, there is one fragment common to all
three clones. By comparing clones in this manner, we can
march down the chromosome, creating a larger and larger
contig. (Reprinted courtesy of Los Alamos Science, Volume Number 20,
a publication of Los Alamos National Library, Los Alamos, NM.)

overlap of cosmids can be determined by sequence-
tagged sites, RFLPs, SNPs, or microsatellites in common.
The cosmids are then digested and sequenced. From the
sequences we work back, finding overlap and thereby
constructing a contig of that BAC. The same process is
carried out on neighboring BACs, extending the contig
eventually to cover the entire chromosome.

At the initiation of the Human Genome Project, vari-
ous goals were set. A modern linkage map of microsatel-
lite markers of the human genome was targeted to be
complete when markers were spaced about 0.7 centi-
morgans (about 700,000 base pairs) apart. That goal was
reached in 1996 with 2,335 microsatellite markers lo-
cated on the genome. The physical map of sequence-
tagged sites would be considered complete with markers
every 100,000 bases, the equivalent of 30,000 sequence-
tagged sites in the genome cloned in BACs. That goal was
reached in 1997. The sequence of the complete genome
was targeted for 2001 and announced in 2000.

Closing the Gap Between Two Contigs

Only one walking step is needed to bridge the gap between two contigs

Contig X

Contig Y

Four walking steps are needed to bridge the gap between two contigs

Contig W

Contig Z

Clone in a contig

Probe used to find the next clone in a walk

Next clone in a walk

Figure 13.44 When contigs of large parts of a chromosome are built up, they need to
be connected. We can do this directly if there is an overlap at the end of one contig and
the beginning of the next. Barring that, we must do chromosome walking to find clones
that bridge the gap between two contigs. At the top of the figure, in typical chromosome
walking technique, the DNA of an end clone is fragmented and used to probe for an
overlap. In this case, one clone is found that overlaps two contigs and thus joins them
into one long contig. In the bottom portion of the figure, the walk requires finding four
overlapping clones that bridge the gap between the two contigs. In both cases, the
process is successful, joining two contigs into one longer one. (Reprinted courtesy of Los
Alamos Science, Volume Number 20, a publication of Los Alamos National Library, Los Alamos, NM.)

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

One of the reasons that goals were optimistic is that
methods of mass production have been developed as the
project has moved along. These methods include the au-
tomation of sequencing and cloning and the develop-
ment of some new technology. For example, scientists at
Affymetrix, Inc., have developed the equivalent of a DNA
probe computer chip. Thousands of known DNA se-
quences are synthesized on a glass substrate. The DNA to
be probed is introduced to this chip, where hybridization
will take place. Using fluorescent technology, successful
probing can be determined using a laser confocal scan-
ning system (fig. 13.45). These chips allow extremely
rapid analysis of DNA sequences. Several other manufac-
turers have developed similar technologies.

As mentioned at the beginning of this chapter, J. C.
Venter of Celera Genomics was a co-announcer of the
completion of the sequencing of the entire human
genome. Venter and his colleagues used a whole-genome
shotgun method in which the entire human genome was
broken into small segments, cloned, and sequenced. The
Celera group will then piece together the genome with a
massive computing effort. Previously, it had been
thought that this method could not work on a genome as

large as the human genome. Venter and his colleagues,
however, had sequenced the Drosophila genome (180
million base pairs) by March of 2000 by this method. Ven-
ter and his colleagues had also sequenced the first true
organism, the bacterium Haemophilus influenzae (1.8
million base pairs) in July of 1995. Since that time, the
yeast Saccharomyces cerevisiae (12 million base pairs)
was sequenced in 1996, and a significant genetic model
organism, the nematode worm, Caenorhabditis elegans
(97 million base pairs; see chapter 16), was sequenced in
1998. Since 1995, numerous other bacteria and eukary-
otes have had their genomes sequenced.

Bioinformatics and Proteomics

These incredible accomplishments in genomics have
given rise to two newly named sciences, bioinformat-
ics and proteomics. Bioinformatics is the science of
mining the data from the DNA sequences obtained from
sequencing. Mining refers to the storage, retrieval, and
analysis of the data. Proteomics is the study of the pro-
teome, from proteins of the genome, and refers to the
study of the complete set of proteins from a particular

GeneChip* Probe Array

Hybridized Probe Cell

Single-stranded fluoresce ntly
labeled DNA targel

Oligonucleotide probe

f f I '
Nv N\N\ Mr

50 um

Each probe cell or feature contains
millions of copies of a specific
oligonucleotide probe

Over 50,000 different probes
complementary to genetic
information of interest

Image of Hybridized Probe Array

Figure 13.45 The Gene Chip® DNA probe array is a glass wafer containing from sixty-five thousand to one million or
more different DNA sequences. The chips are created by photolithographic techniques, similar to those used in
computer chip manufacture. The DNA being probed must have fluorescent molecules attached to permit rapid
screening. (Courtesy of Affymetrix, Inc.)

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Practical Benefits from Gene Cloning

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genome. It is the protein analogue to genomics. It is esti-
mated that there are from 50,000 to 2 million different
proteins in biological systems, although the number of
distinct shapes — motifs — may only be about five thou-
sand. The role of proteomics is to characterize all pro-
teins and determine their structures and shapes. As clas-
sical genetics worked from phenotype to genotype,
modern molecular genetics is working from genotype
(genomics) to phenotype (proteomics).

Ethics

In addition to the expected scientific and medical infor-
mation that we will gain from sequencing of the human
genome are ethical problems that the project will create.
We will shortly have the ability to test people for various
genes that we cannot test for now, such as genes for la-
tent diseases like coronary artery disease and cancer. Can
insurance companies then demand to test individuals for
a whole battery of genes and decide afterward whether
that person is insurable or what that person's insurance
rates should be? Will many persons find themselves unin-
surable because they have genes that might predispose
them to cancer? Will individuals find themselves unem-
ployable because of similar problems? What should doc-
tors do about diagnosing a genetic disease (such as Hun-
tington disease) that has no cure? Should they tell the
patient? Another ethical issue is the extent to which ge-
netic intervention should be used to change the course
of a person's life. With the knowledge of the sequence
and location of our genes, and the technology to transfer
genes into people, will transgenic people become the
norm (see box 13.2)? Should we not only cure diseases
this way but tailor a person to some ideal? Will genetic in-
tervention into our basic genetic blueprint be routine? To
address these questions, an ethics panel has been set up
as part of the Human Genome Project.

PRACTICAL BENEFITS FROM
GENE CLONING

Throughout this chapter, we have mentioned applica-
tions of genetic engineering. Here we summarize some
of the accomplishments and future directions in the
medical, agricultural, and industrial arenas.

Medicine

In medicine, genetic engineering has had remarkable
successes in some areas. On the one hand, basic knowl-
edge about how genes work (and don't work) has ad-
vanced tremendously. On the other hand, recombinant
DNA methodology has made available large quantities of

substances previously in short supply. These include in-
sulin, interferon (an antiviral agent), growth hormone,
growth factors, blood-clotting factors, and vaccines for
diseases such as hepatitis B, herpes, and rabies. Advances
in AIDS and cancer research are discussed in chapter 16.
Genetic engineering is making it possible to manufacture
antibodies to diagnose and treat diseases. The sequenc-
ing of the human genome will further aid medicine by
identifying the genes for various diseases, a first step in
discovering cures. So far, several genes of great impor-
tance have been located, cloned, and sequenced. We also
pointed out the use of restriction fragment length poly-
morphisms and the polymerase chain reaction as tech-
niques of tremendous power in identifying individuals
for forensic purposes.

On another front, transgenic mice and cloned sheep
have shown that genetic engineering can be applied to
higher organisms (fig. 13.46). The use of this technology
to treat human diseases, however, is only just beginning.
In July 1990, the National Institutes of Health approved
gene therapy treatments on people: A child was infused
with cells to replace a gene for the enzyme adenosine
deaminase, an enzyme whose absence results in a dys-
functional immune system. Although the latter treatment
was successful, it had been augmented by other treat-
ments, rendering the conclusions equivocal. Mice and
dogs have had hemophilia B corrected by infusion of a
genetically modified adenovirus. AIDS, hemophilia, cystic
fibrosis, and diabetes are other diseases that should be
amenable to gene therapy in the near future.

Figure 13.46 The sow shown is transgenic, producing large
quantities of human protein C in her milk. The protein controls
blood clotting and is normally found only in trace quantities in
human blood. (Courtesy of William H. Velander, Virginia Tech.)

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Agriculture

Currently, in the United States, approximately one quar-
ter of farmland is planted with crops that are genetically
modified. Most are resistant to certain insect pests be-
cause they contain genes from Bacillus thuringiensis
(often referred to as Bt). These genes are for insecticidal
proteins called 8 endotoxins. For example, the proteins
CrylA and CrylC from Bacillus thuringiensis protect
the plants against larval forms of lepidopterans such as
the European corn borer. Cry3A protects against
coleopterans such as the Colorado potato beetle. In ex-
cess of fifty genetically altered crop plants have been ap-
proved for planting, including those protected against in-
sect pests, frost, and premature ripening. Rice is being
modified so that its vitamin A potential is maintained
even after the husks are removed, a procedure done to al-
low for storage since the husks become rancid. That
change alone will improve the health of millions of peo-
ple throughout the world. Box 13.1 discussed some of
the ethical concerns surrounding genetically modified
crop plants.

Industry

Industrial applications of biotechnology include engi-
neering bacteria to break down toxic wastes, modifying
yeast to use cellulose to produce glucose and alcohol for
fuel, using algae in mariculture (the cultivation of marine
organisms in their natural environments) to produce
both food and other useful substances, and developing
better food processing methods and waste conversion.
As an example, baker's yeast (Saccharomyces cerevisiae)
has been modified with a plasmid that contains two cel-
lulase genes, an endoglucanase and an exoglucanase, that
convert cellulose to glucose. The yeast can then convert
glucose to ethyl alcohol. These yeasts are now capable of
digesting wood (cellulose) and converting it directly to
alcohol. The potential exists to harvest the alcohol the
yeast produces as a fuel to replace fossil fuels that are in
dwindling supply and are polluting the planet.

As you can see, there is no one direction that
biotechnology is taking. Many advances are being made
that will probably affect every person's life in a beneficial
way. Cautious optimism is certainly in order.

SUMMARY

STUDY OBJECTIVE 1: To look at the techniques of gene
cloning 359-377

Recombinant DNA techniques revolve around the cloning
of foreign DNA in a plasmid or phage. Cloned DNA can be
amplified, expressed, and sequenced. Gene cloning tech-
niques came about with the discovery of restriction en-
donucleases. Type II restriction endonucleases cleave DNA
at palindromic regions, which have twofold symmetry.

Recombinant vectors can be constructed several differ-
ent ways. Foreign and vector DNA can be made compatible
by treating each with the same restriction endonuclease —
each will then have the same sticky ends. If that does not
work, T4 DNA ligase can join blunt ends. In a variation of
this method, linkers containing restriction sites are added
to vector and foreign DNA. These linkers are then treated
with a restriction endonuclease that gives the DNA sticky
ends.

DNA to be cloned can be synthesized from an RNA tem-
plate (cDNA) or isolated by various techniques. If messen-
ger RNA is available, it can be converted into a clonable
complementary DNA with the enzyme reverse transcrip-
tase. If DNA is to be isolated directly, it must be identified
among all the other DNA fragments created. Locating a de-
sirable piece of DNA is done with probes, complementary
nucleic acids labeled with radioactivity or chemilumines-
cence. Southern blotting, a transfer technique, is used first,

followed by DNA-DNA or DNA-RNA hybridization and au-
toradiography. If the DNA is cloned first, as in the creation
of a genomic library, probes can be created or expression of
the cloned gene can be determined.

Eukaryotic vectors have been developed, including
yeast plasmids, tumor virus vehicles in animals, and crown
gall tumor plasmids in plants. Eukaryotes can be transfected
by foreign DNA and express it in transgenic organisms.
DNA can be injected, shot in on projectiles, electroporated,
or introduced by viruses, plasmids, or liposomes. Knockout
mice, lacking a specific gene, can be created.

STUDY OBJECTIVE 2: To examine the techniques of cre-
ating restriction maps 377-383

Restriction digests can be separated by electrophoresis,
then used to construct a restriction map. This is a map of
the DNA showing the location of restriction enzyme recog-
nition sites. The genetic maps, generated by mating analy-
sis, can then be superimposed on the restriction maps, lo-
cating regions of interest on the physical map. Restriction
fragment length polymorphisms (RFLPs) provide a tool for
locating genes through linkage analysis and are also valu-
able in forensic science. The polymerase chain reaction
(PCR) is a technique used to rapidly amplify particular seg-
ments of DNA.

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Solved Problems

399

STUDY OBJECTIVE 3: To study the methods of DNA se-
quencing 383-390

DNA is usually sequenced by one of two methods. The
dideoxy method developed by Sanger and his colleagues re-
quires the synthesis of DNA in the presence of chain-
terminating (dideoxy) nucleotides. Electrophoresis fol-
lowed by autoradiography allows the sequence of
nucleotides synthesized to be determined directly. Fluores-
cent labeling allows computerized sequence determina-
tions. The phage cj)X174 was sequenced in its entirety
through the forerunner of this technique, the plus-and-
minus method. Gilbert and Maxam's chemical method also
is used widely.

STUDY OBJECTIVE 4: To look at the goals and methods of
the Human Genome Project 390-397

The Human Genome Project is a massive, international ef-
fort to map and sequence all 33 billion bases of the human

genome. Initial success was announced in the spring of
2000. Modern linkage maps are being created of restriction
sites, microsatellite markers, sequence-tagged sites, and
single-nucleotide polymorphisms. They are being coordi-
nated with physical maps created with overlapping con-
tiguous clones of chromosomes. These techniques cur-
rently allow us to find genes of interest. The project also
includes the sequencing of the genomes of other relevant
organisms.

STUDY OBJECTIVE 5: To look at the practical benefits and
human issues of genetic engineering 397-398

Genetic engineering is moving forward on a number of
fronts. Medical, agricultural, and industrial applications are
becoming widespread.

SOLVED PROBLEMS

PROBLEM 1: A piece of eukaryotic DNA is obtained by
using a restriction endonuclease that leaves blunt ends
(Haelll). How could we get this piece of DNA into a
BamHI site in plasmid pBR322, and how would we
know when the foreign DNA has been cloned?

Answer: Since the two pieces of DNA (the eukaryotic
piece and the plasmid) have different ends, they must be
made compatible before cloning. The simplest way
would be to attach blunt-ended linkers to the foreign
DNA with phage T4 DNA ligase (see fig. 13.9). The link-
ers, of course, would have a BamHI site within. After the
linkers are attached to the foreign DNA, it would be
treated with the BamHI restriction enzyme, giving the
foreign DNA BamHI ends. The plasmid is then also
treated with the restriction enzyme and the two (the for-
eign DNA and the cut plasmid) are now mixed together
in the presence of E. colt DNA ligase, which seals up the
plasmids, with or without cloned inserts (see fig. 13.6).
Since they have compatible ends, some of the time, a
piece of foreign DNA is inserted into a plasmid. The plas-
mids are then taken up by E. coli cells that are grown
overnight in an incubator. The bacterial colonies are then
replica-plated on media with the antibiotics ampicillin or
tetracycline. Colonies that are resistant to ampicillin but
sensitive to tetracycline are assumed to be bacteria con-
taining plasmids with cloned inserts (see fig. 13.8).

PROBLEM 2: How does a reporter system work?

Answer: A reporter system is a genetically manipulated
system that displays a particular phenotype or reaction

when a desired event has taken place. In this chapter, we
discussed the firefly luciferase reporter system in which
the desired result (transcription of a particular promoter)
causes a transgenic tobacco plant to glow. Let us say that
we are studying the control of transcription of a particu-
lar eukaryotic gene. We could attach the promoter of that
gene to the firefly luciferase gene in a Ti plasmid by
cloning techniques. The plasmid could then transfect
tobacco plants, and we could continue our experiment
to determine whether the promoter under study is active
under various conditions. We would know whether it
was active by watering the plants with luciferin. If the
plant glows, then the luciferase gene product is present,
which means that the promoter under question is active.
In other words, the glowing of the plant "reports" the ac-
tion of the promoter under question; the promoter is ac-
tive because it allowed the transcription of the luciferase
gene. We also discussed the green fluorescent protein re-
porter system.

PROBLEM 3: A piece of DNA has the sequence 3'-GGCG-
TATTC-5'. It is sequenced using the dideoxy method.
How many bands are found on the ladder gel? How many
bands and of what size are found for each reaction mix-
ture?

Answer: Since the piece of DNA is nine bases long, the
total number of bands in all four lanes of a sequencing gel
add up to nine (see fig. 13. 33). By each reaction mixture,
we mean the four reaction mixtures each with one of the
dideoxynucleotides. In the reaction mixture with ddTTP,

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

E
as

CD
N

chain termination occurs at the adenine in the piece of
DNA; that is, a DNA segment was synthesized that is six
bases long (see the following figure). In the reaction mix-
ture with ddATP, chain termination occurs opposite each
of the thymines, producing DNA segments of five, seven,
and eight nucleotides. In the reaction mixture with
ddGTP, chain termination occurs opposite the cytosines
in positions three and nine. And, in the reaction mixture
with ddCTP, chain termination occurs after synthesis of
segments one, two, and four bases long. Note that the gel
gives us the sequence of the complement strand of the
original piece of single-stranded DNA.

9
8
7
6
5
4
3
2
1

ddTTP

ddATP

ddGTP

ddCTP

G
A
A
T
A
C
G
C
C

PROBLEM 4: A linear DNA molecule 1,000 base pairs
long is digested with the following restriction enzymes,
producing the following results:

EcoW

Bglll

Ecom + Bglll

400 bp, 600 bp
250 bp, 750 bp
250, 350, 400 bp

Determine the restriction map.

Answer: Each enzyme alone produces two fragments, so
the molecule has one site for each enzyme. Since we get
different-sized fragments with each enzyme, the sites
must be located asymmetrically along the DNA. Draw
these sites:

EcoRl Bglll

400 | 600 and 250 | 750

The EcoRl fragment that lacks a Bglll site should appear
in the double digest. If Bglll cuts within the 400 base-pair
fragment, we would expect to see 150, 250, and 600
base-pair fragments. We don't see this, so the Bglll site is
not within the 400 base-pair EcoRl fragment. Thus, the
map looks like this:

EcoRl Bglll

400 i 350 | 250

EXERCISES AND PROBLEMS

GENOMIC TOOLS

1. What specific properties of type II endonucleases
make them useful in gene cloning?

2. The following is a double helix of DNA. What, if any,
are potential restriction enzyme recognition se-
quences?

5 '-TAGAATTCGACGGATCCGGGGCATGCAGATCA-3 '
3 '-ATCTTAAGCTGCCTAGGCCCCGTACGTCTAGT-5 '

3. Assuming a random arrangement of nucleotides on a
piece of DNA, what is the probability that a restric-
tion endonuclease whose recognition site consists
of four bases (a four-cutter) will cut the DNA? What
is the probability for a six-cutter? an eight-cutter?

4. Under what circumstances is a restriction endonu-
clease unsuitable for cloning a piece of foreign DNA?

5. What methods exist to create sticky ends or create
ends for joining two incompatible pieces of DNA?
When is each method favored?

6. Diagram a possible heteroduplex between two
phage A vectors, one with and one without a cloned
insert, created by DNA-DNA hybridization.

7. What are the differences among plasmid, cosmid,
expression vector, and YAC? Under what circum-
stances is each useful?

8. What are the steps by which messenger RNA can be
converted into cDNA? How would we obtain ra-
dioactive cDNA? radioactive messenger RNA?

9. What is chromosome walking? When is it used?

10. How would we isolate a human alanine transfer RNA
gene for cloning? How would we locate a clone with
a human alanine transfer RNA gene in a genomic
library?

11. What are the differences among Southern, western,
northern, and dot blotting?

* Answers selected exercises and problems are on page A-14.

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Exercises and Problems

401

12. How would you develop a probe for a gene whose
messenger RNA could not be isolated? How could an
expression vector be used to isolate a cloned gene?

13. How are E. colt plasmids manipulated to survive in
yeast? How can virus genomes, such as SV40 and
phage A , survive as functioning vectors when parts
of their genomes are replaced by cloned DNA?

14. What methods are used to get foreign DNA into eu-
karyotic cells? What is transfection? What is a trans-
genic mouse?

15. Exonuclease III is an enzyme that sequentially re-
moves bases from the 3' end of double-stranded
DNA. The following two molecules, each 100 bp
long, are digested with exonuclease III. Molecule 1 is
completely digested; molecule 2 is only partially di-
gested. Explain these results.

Molecule 1:

Molecule 2:

CGTTCAG...
GCAAGTC...
AAAAAAAAAA.

r | ir | ir ■ ir ■ ir ■ ir ■ ir ■ ir ■ ir ■ ir ■ i

16. A plasmid that contains an EcoRl site within a gene
for ampicillin resistance is cut with EcoRl, and then
religated. This plasmid is used to transform E. colt
cells, and the plasmid is reisolated from the ampi-
cillin-resistant colonies. The reisolated plasmids
from two different colonies are electrophoresed,
and the results appear in the following figure.

Undigested
plasmid

Colony 1
Digested,
religated

Colony 2
Digested,
religated

How do you account for the two bands in colony 2?

17. Most human genes contain one or more introns. Since
bacteria cannot excise introns from nuclear messen-
ger RNA (snRNPs are needed), how can bacteria be
used to make large quantities of a human protein?

RESTRICTION MAPPING

18. How are DNA fingerprints useful in forensic cases?
Could they be used in paternity exclusion?

19. The following segment of DNA is cut four times by the
restriction endonuclease EcoRl at the places shown.
Diagram the gel banding that would result from elec-
trophoresis of the total and partial digests. Note the
end-labeled segments and regions where several seg-
ments form bands at the same place on the gel.

100 I 300 1 50 1 250 I 150

20. The following figure shows a gel of a total and partial
digest of a DNA segment treated with Hindll. End-
labeled segments are noted by asterisks. Draw the
restriction map of the original segment.

Total
digest

Partial
digest

800
700
600
500
400
300
200
100

750
700

*
*

400

200

100
50

500

400
350
300
250
200

100
50

21. Several mutants of the DNA segment shown in prob-
lem 19 were isolated. They gave the following gel pat-
terns when the total digests were electrophoresed.
Asterisks denote the end-labeled segments. Can you
determine the nature of the mutations?

Mutants
B

400-

300-

200

100-

22. Restriction maps of a segment of DNA were worked
out separately for BamWl and Taq\. Two overlays of
the maps are possible. The double-digest gel is
shown in the following figure (asterisks denote end
labels). Which overlay is correct?

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

Bam HI
Taq\

BamH\
Taq\

100

50'

100

300

250

j£2(

200

150 ' 100 ' 100

300

"lOO 1 150

j!2l

200

250

Alternative

200-

150-

100-

50-

Double

digest

gel

23. A linear DNA molecule 1,000 bp long gives the fol-
lowing size fragments when treated with these re-
striction enzymes. Derive a restriction map.

EcoRl: 300 bp, 700 bp

BamHI: 150 bp, 200 bp, 250 bp,

400 bp
Ecom + BamHI: 50 bp, 100 bp, 200 bp,

250 bp, 400 bp

24. A linear DNA molecule cut with EcoRl yields frag-
ments of 3 kb, 4.2 kb, and 5 kb. What are the possi-
ble restriction maps?

25. You have double-stranded DNA that you radioac-
tively label at the 5' ends. Digestion of this molecule
with either EcoRl or BamHI yields the following
fragments. The numbers are in kilobases (kb), and an
asterisk indicates the fragments that are labeled.

Ecom: 2.8, 4.6, 6.2*, 7.4, 8.0*
BamHI: 6.0*, 10.0*, 13.0

If unlabeled DNA is digested with both enzymes si-
multaneously, the following fragments appear: 1.0,
2.0, 2.8, 3.6, 6.0, 6.2, 7.4. What is the restriction map
for the two enzymes?

26. A 12 kb DNA molecule cut with EcoRl yields one 12 kb
fragment. When the original molecule is cut with
BamHI, three fragments of 2 kb, 4.5 kb, and 5.5 kb
are produced. When the fragment from EcoRl is
treated with BamHI, four fragments of 2 kb, 2.5 kb,
3.0 kb, and 4.5 kb are produced. Draw a restriction
map.

27. A plasmid 3 kb in length contains a gene for ampi-
cillin resistance and a gene for tetracycline resist-
ance. The plasmid has a single site for each of the fol-
lowing enzymes: EcoRl, Bglll, Hindlll, Pstl, and Sail.
If DNA is cloned into the EcoRl site, resistance to ei-
ther antibiotic is not affected. DNA cloned into the
BgFLl, Hindlll, or Sail sites abolishes tetracycline re-
sistance, and DNA inserted into the Pstl site elimi-

nates ampicillin resistance. If the plasmid is digested
completely with enzyme mixes, the following frag-
ments result:

Mixture

Fragment Size (kb)

EcoRl + Pstl
EcoRl + Bglll
EcoRl + Hindlll
EcoRl + Sail
EcoRl + Bglll + Pstl

0.7,2.3

0.3,2.7

0.08,2.92

0.85,2.15

0.3,0.7,2.00

Draw a restriction map of the plasmid, and indicate
the locations of the resistance genes and the sites of
enzymatic cleavage.

28. A gene has the following EcoRl restriction map (in
kilobases):

1.0 I 0.7 | 2.0

Draw the gel pattern expected from

a. a mutant that has lost the site between the 1.0
and 0.7 kb fragments.

b. a mutant that has a new site within the 2.0 kb
fragment.

29. A DNA fragment 8 kb in size is labeled with 32 P at
the 5' ends. It is then digested with EcoRl, Bglll, or a
mixture of both enzymes. The size of the fragments
and the labeled fragments (*) appear as follows. Sizes
are in kilobases.

3.5
3.0
2.0
1.5
1.0
0.5

EcoRl

Bglll

Mix

Which of the following two maps is consistent with
the results?

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Exercises and Problems

403

0.5

1.0

3.0

3.5

1.0

1.5

2.0

0.5

3.0

0.5

1.0

3.0

3.5

3.0

0.5

2.0

1.5

1.0

30. You now take an unlabeled molecule from problem
29, digest it with Hindlll, and get two fragments, 5.5
and 2.5 kb in size. If Hindlll does not cut within the
3.5 kb EcoRl fragment, what size fragments do you
expect in a double digest of Hindlll and Eco RI?

31. Two normal individuals have a child with Down syn-
drome. RFLP analysis with a probe from chromo-
some 21 is performed on all three individuals, and
the results of the gels appear as follows. Based on
these results, what can you conclude about the ori-
gins of the number 2 1 chromosomes?

Mother

Father

Child

—

POLYMERASE CHAIN REACTION

32. What is PCR? When is it used?

DNA SEQUENCING

33. What are the steps in the dideoxy method of DNA
sequencing? How has the technique been improved
with fluorescent dyes?

34. The following diagram is of a dideoxy sequencing
gel. What is the sequence of the DNA under study?

21 -

20-

19-

18-

17-

16-

15-

14-

13-

12-

11 -

10-

1 -

ddCTP

ddGTP

ddATP

ddTTP

35. How can a particular piece of DNA be manipulated
to be in the appropriate configuration for dideoxy
sequencing?

36. Provide, if possible, DNA sequences that can mark
the termination of one gene and the initiation of an-
other, given that the genes overlap in one, two,
three, four, five, six, or seven bases.

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Chapter Thirteen Genomics, Biotechnology, and Recombinant DNA

37. Draw the expected gel pattern derived from the
dideoxy sequencing method for a template strand
with the following sequence:

5 '-CAGCGAATGCGGAA-3 '

38. A DNA strand with the sequence 3'-GACTATTCC-
GAAAC-5' is sequenced by the dideoxy method. If
the reaction mixture contains all four radioactive de-
oxynucleotide triphosphates plus dideoxythymi-
dine, what size labeled bands do you expect to see
on the gel?

MAPPING AND SEQUENCING THE HUMAN GENOME

39. What is hypervariable DNA? a RFLP? a VNTR locus?
microsatellite DNA? a sequence-tagged site? (See
also RESTRICTION MAPPING)

PRACTICAL BENEFITS FROM GENE CLONING

40. Describe some areas of practical benefit from ge-
netic engineering. Why might some people be con-
cerned about its widespread use?

CRITICAL THINKING QUESTIONS

1. In the past, geneticists have used several different meth-
ods to splice pieces of DNA that do not have compati-
ble "sticky ends." We mentioned blunt-end ligation and
the addition of linkers containing specific restriction
sites. Given that nucleotides can be added to the 3'
ends of double-stranded DNA with the enzyme de-
oxynucleoside terminal transferase, can you see an-

other way to create compatible ends on foreign and ve-
hicle DNA?

2. The motion picture Jurassic Park was based on the
premise that DNA of dinosaurs could be extracted from
the blood-meals of mosquitoes preserved in amber and
inserted into the genome of a frog, which would then
produce living dinosaurs. Is this premise reasonable?

Suggested Readings for chapter 13 are on page B-ll.

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Control in Prokaryotes and
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GENE
EXPRESSION

Control in
Prokaryotes and

Phages

STUDY OBJECTIVES

1. To study the way in which inducible and repressible operons
work 406

2. To examine attenuator control in bacteria 415

3. To analyze the control of the life cycle of phage X 418

4. To determine the way in which transposable genetic elements
transpose and control gene expression in bacteria 425

5. To look at other transcriptional and posttranscriptional
mechanisms of control of gene expression in bacteria and
phages 430

Artificially colored transmission electron micrograph of
a T4 bacteriophage attached to an Escherichia coli
bacterium. (© Biozentrum, University of Basel/SPL/Photo

Researchers, Inc.)

STUDY OUTLINE

The Operon Model 406

Lac Operon (Inducible System) 406

Lactose Metabolism 406

The Regulator Gene 406

The Operator 408

Induction of the Lac Operon 409

Lac Operon Mutants 409
Catabolite Repression 412
Trp Operon (Repressible System) 413

Tryptophan Synthe sis 413

Operator Control 414
Trp Operon (Attenuator-Controlled System) 415

Leader Transcript 415

Leader Peptide Gene 416

TRAP Control 417

Redundant Controls 418
Lytic and Lysogenic Cycles in Phage A. 418

Phage \ Operons 418

Early and Late Transcription 420

Repressor Transcription 421

Maintenance of Repression 421

Lysogenic Versus Lytic Response 423
Transposable Genetic Elements 425

IS Elements 425

Composite Transposons 427

Mechanism of Transposition 427

Phenotypic and Genotypic Effects of Transposition 429
Other Transcriptional Control Systems 430

Transcription Factors 430

Promotor Efficiency 430
Translational Control 430
Posttranslational Control 433

Feedback Inhibition 433

Protein Degradation 433
Summary 434
Solved Problems 435
Exercises and Problems 436
Critical Thinking Questions 438

405

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406

Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

Genes are transcribed into RNA, which, for
the most part, is then translated into pro-
tein. Control mechanisms are exercised
along the way. Without some control of
gene expression, an Escherichia coli cell, for
example, would produce all its proteins in large quanti-
ties all the time, and all the cells in a eukaryotic organism
would be identical. Although most control mechanisms
are negative (preventing something from happening),
controls can also be positive (causing some action to oc-
cur or enhancing some action). This chapter is devoted
to analyzing control processes in prokaryotes and
phages; in chapter 16, we examine control processes in
eukaryotes.

In the process leading from a sequence of nu-
cleotides in DNA to a protein, control is exerted in many
places. In general, control of gene expression can take
place at the levels of transcription, translation, or protein
functioning. The most efficient place to control gene ex-
pression is at the level of transcription.

One of the best-understood mechanisms exerts con-
trol of transcription, regulating the production of mes-
senger RNA according to need. E. coli messenger RNAs
are short-lived in vivo: They degrade enzymatically
within about two minutes. A complete turnover (degra-
dation and resynthesis) in the cell's messenger RNA oc-
curs rapidly and continually, and this rapid turnover is a
prerequisite for transcriptional control, a central feature
of the regulation of prokaryotic gene expression.

THE OPERON MODEL

Not all of the proteins prokaryotes can produce are
needed in all circumstances in the same quantities. For
example, some metabolites, such as sugars, which the
cell breaks down for energy and as a carbon source, may
not always be present in the cell's environment. If a given
metabolite is not present, enzymes for its breakdown are
not useful, and synthesizing these enzymes is wasteful. If
the cell produces enzymes for the degradation of a par-
ticular carbon source only when this carbon source is
present in the environment, the enzyme system is known
as an inducible system. Inducible enzymes are synthe-
sized when the environment includes a substrate for
those enzymes. The enzymes will then catabolize (break
down) the substrate.

On the other hand, the enzymes in many synthetic
pathways are in low concentration or absent when an
adequate quantity of the end product of the pathway is
already available to the cell. That is, if the cell encounters
an abundance of the amino acid tryptophan in the envi-

ronment or if it is overproducing tryptophan, the cell
stops the manufacture of tryptophan until a need arises
again. A repressible system is a system of enzymes
whose presence is repressed, stopping the production of
the end product when it is no longer needed. Repressible
systems are repressed by an excess of the end product of
their synthetic (anabolic) pathway.

The best-studied inducible system is the lac operon
in E. coli. Since the term operon refers to the control
mechanism, we will defer a definition until we describe
the mechanism.

LAC OPERON
(INDUCIBLE SYSTEM)

Lactose Metabolism

Lactose (milk sugar — a disaccharide) is a p-galactoside
that E. coli can use for energy and as a carbon source af-
ter it is broken down into glucose and galactose. The en-
zyme that performs the breakdown is P-galactosidase
(fig. 14.1). (The enzyme can additionally convert lactose
to allolactose, which, as we will see, is also important.)
There are very few molecules of p-galactosidase in a
wild-type E. coli cell grown in the absence of lactose.
Within minutes after adding lactose to the medium, how-
ever, this enzyme appears in quantity within the bacterial
cell. When the synthesis of p-galactosidase (encoded by
the lacZ, or z gene) is induced, the production of
two additional enzymes is also induced: P-galactoside
permease (encoded by the lacY, or y gene) and
P-galactoside acetyltransferase (encoded by the lacA,
or a gene). The permease is involved in transporting lac-
tose into the cell. The transferase is believed to protect
the cell from the buildup of toxic products created by
p-galactosidase acting on other galactosides. By acetylat-
ing galactosides other than lactose, the transferase pre-
vents p-galactosidase from cleaving them.

The Regulator Gene ^l*

Not only are the three lac genes (z, y, a) induced to-
gether, but they are adjacent to one another in the E. coli
chromosome; they are, in fact, transcribed on a single,
polycistronic messenger RNA (fig. 14.2). Induction in-
volves the protein product of another gene, called the
regulator gene, or / gene QacI). Although the regulator
gene is located adjacent to the three other lac genes, it is
a totally independent transcriptional entity. The regula-
tor specifies a protein, called a repressor, that inter-
feres with the transcription of the genes involved in lac-
tose metabolism.

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Lac Operon (Inducible System)

407

CH 2 OH

H OH

hV a

H OH

(a)

Lactose

CH 2 OH

p-Galactosidase

H 2

CH 2 OH
— O

\OH H/

hV/h

H OH
Galactose

CH 2 OH

— O

\OH H/
HOV_/ H

H OH
Glucose

CH 2 OH
— O

\OH H/

hV/h

H OH

(b)

H OH

OH
H

CH 2 OH

Lactose

(3-Galactosidase

CH 2 OH
— O

O — CH,

l\OH h/I H

hV/h ,

■o

H OH HO

\OH H/

H OH

Allolactose

Figure 14.1 The enzyme (3-galactosidase hydrolytically cleaves lactose into glucose and galactose
(a). The enzyme can also convert lactose to allolactose (£>).

□□□□:

P o

I I I

mRNA

Transcription

I I

5' | 3'

Translation

Transcription

Repressor
protein

z protein

Translation

y protein

DNA
]□□□

~] Polycistronic mRNA
3'

a protein

Figure 14.2 The lac operon is transcribed as a multigenic (polycistronic) mRNA. The z, y, and a indicate the
lacZ, lacY, and lacA loci. The mRNA transcript is then translated as individual proteins. The lac operon
regulator gene is denoted as /'; the o stands for operator and the p for promoter. Both the operon and the
regulator gene have their own promoters. (Source: Data from R. C. Dickson, et al., "Genetic regulation: The lac
control region," Science, 187:27-35, January 10, 1975.)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

The Operator

For the repressor protein to exert its influence over tran-
scription, there must be a control element (receptor site)
located near the beginning of the p-galactosidase (lacZ)
gene. This control element is a region referred to as the op-
erator, or operator site (fig. 14.2). The operator site is a se-
quence of DNA that the product of the regulator gene, the
repressor, recognizes. When the repressor is bound to the
operator, it either interferes with RNA polymerase binding
or prevents the RNA polymerase from achieving the open
complex (see chapter 10). In either case, transcription of
the operon is prevented (fig. 14.3). The repressor is re-
leased when it combines with an inducer, a derivative of
lactose called allolactose (see fig. 14.1).

Note that the promoter not only is recognized by
RNA polymerase but also has other controlling elements
in the immediate vicinity of the initiation site of tran-
scription. We can now define an operon as a sequence
of adjacent genes all under the transcriptional control of
the same promoter and operator.

The nucleotide sequence of the lac operator region is
shown in figure 14 A. The operator in figure 14.3 is referred
to as the primary operator, o 1 , centered at + 1 1. Two other
operator sequences have been found. One, o 2 , is centered at
+41 2. The third overlaps the C-terminal end of the / gene, is
centered at —82, and is referred to as o 3 . The structure of
the repressor and its interaction with the operator sites was
worked out recently with X-ray crystallography. The func-
tional repressor is a homotetramer of the protein product of

DDDI

RNA polymerase

Repressor

]DDD DNA with repressor

DDDI

DDD DNA without repressor

Polycistronic mRNA

Figure 14.3 The repressor. By binding to the operator, the repressor either prevents RNA polymerase from
binding to the promoter and transcribing the lac operon as shown, or prevents the polymerase from
achieving the open configuration. In either case, transcription of the lac operon is prevented. When the
repressor is not present, transcription takes place. The functional repressor is a tetramer.

E.coli □□□[
chromosome

] DDD

CAP

site

Cap

-35

-10

site

sequence

DNA
sequence

5' Glu Ser|Gly Gin Stop

Repressor
binding

Shine-
Dalgarno
sequence

fMetThrl

GGAAAGCGGGC/1Gre/1GCGC/l/ICGC^rrAATGTGAGTTAGCTCACTCATTAGGCACCCCAGGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGG/l/irrarG/1GCGG/lJ"/l/lC/l/irnCACACAGGAAACAGCTATGACCATG
CCTTTCGCCCGTC/ICTCGCGrreCGra/ITTACACTCAATCGAGTGAGTAATCCGTGGGGTC^^

I I I I III I I 1 U I b

-80 -70 -60 -50 -40 -30 -20 -10

+10 +20 +30

Figure 14.4 The lac operon promoter and operator regions. The CAP site is described later. The base
sequence corresponds to the diagram above it. The terminal amino acids of the / gene are shown, as well
as the initial amino acids of the lacZ gene. In addition, we picture the Shine-Dalgarno sequence of the DNA,
the repressor-binding region (centered at around +10 of the gene), the -10 and -35 sequences of the
promoter, and primary (o-i) and secondary (o 3 ) operator sites (see text). (Data from R. c. Dickson, et ai., "Genetic
regulation: The lac control region," Science, 187:27-35, January 10, 1975.)

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Lac Operon (Inducible System)

409

the / gene; that is, it is formed from four identical copies of
the repressor protein. Since each operator site has twofold
symmetry, two repressor monomer proteins bind to each
operator site. The monomer is shaped so that it fits into the
major groove of the DNA to locate the exact base sequence
of the operator; it then binds at that point through electro-
static forces. A tetramer can bind to two of the operator
sites at the same time, presumably o 1 and o 3 or o x and o 2 . In
the process, the DNA is formed into a loop (fig. 14.5).

Induction of the Lac Operon

Under conditions of repression, before the operon can
be "turned on" to produce lactose-utilizing enzymes, the
repressor will have to be removed from the operator. The
repressor is an allosteric protein; when it binds with
one particular molecule, it changes the shape of the pro-
tein, which changes its ability to react with a second par-
ticular molecule. Here the first molecule is the inducer al-
lolactose and the second molecule is the operator DNA.
When allolactose is bound to the repressor, it causes the
repressor to change shape and lose its affinity for opera-
tor sequences (fig. 14.5).

With allolactose bound to the repressor, the ability of
the repressor to bind to the operator is greatly reduced,
by a factor of 10 3 . Since no covalent bonds are involved,
the repressor simply dissociates from the operator. After
the repressor releases from the operator, RNA poly-
merase can now begin transcription. The three lac
operon genes are then transcribed and subsequently
translated into their respective proteins.

This system of control is very efficient. The presence
of the lactose molecule permits transcription of the
genes of the lac operon, which act to break down the lac-
tose. After all the lactose is metabolized, the repressor re-
turns to its original shape and can again bind to the op-
erator. The system is "turned off." Using very elegant
genetic analysis, details of this system were worked out
by Francois Jacob and Jacques Monod, who subsequently
won 1965 Nobel prizes for their efforts.

CAP
site

(a)

^J7

(b)

Figure 14.5 Because the lac operator DNA sequences are
palindromes, each half can bind one repressor subunit. (a) The
tetrameric repressor binds to o 1 and o 3 , causing the DNA in
between to form a loop. Each of the subunits is shown in a
different color. The round portion of the subunit in touch with
the DNA is the N-terminal end of the repressor subunit; the
C-terminal ends form tails that bind the subunits together. Also
indicated are the CAP site and the -10 and -35 sequences.
(b) When each of the subunits binds an allolactose molecule
[black circles), the shape of the middle portion of the subunit
changes, causing the subunit to fall free of the operators.

Frangois Jacob (1920- ).
(Courtesy of Dr. Frangois Jacob.)

Jacques Monod (1910-1976).
(Archives Photographiques, Musee
Pasteur.)

Lac Operon Mutants

Merozygote Formation

Discovery and verification of the lac operon system
came about through the use of mutants and partial
diploids of the lac operon well before DNA sequencing
techniques had been developed. The structural (enzyme-
specifying) genes of the lac operon, z, y, and a, all have
known mutant forms in which the particular enzyme
does not perform its function. These mutant forms are
designated z~, y , and a~ . The alleles for normal forms
of the enzymes are z + ,y + , and a + .

Partial diploids in E. colt can be created through sex-
duction (chapter 7) because some strains of E. colt have

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

the lac operon incorporated into an F' factor. Since F +
strains can pass the F' particle into F~ strains, lac operon
diploids (also called merozygotes, or partial diploids) can
be formed. By careful manipulation, various combina-
tions of mutations can be looked at in the diploid state.

Constitutive Mutants

Constitutive mutants are mutants in which the three
lac operon genes are transcribed at all times — that is,
they are not turned off even in the absence of lactose. In-
spection of figure 14.3 shows that constitutive produc-
tion of the enzymes can come about in several ways. A
defective repressor, produced by a mutant regulator
gene, will not turn the system off, nor will a mutant op-

erator that will no longer bind the normal repressor. The
regulator constitutive mutants are designated i~\ the op-
erator constitutive mutants are designated o c . Both types
of mutants produce the same phenotype: constitutive ex-
pression of the three lac operon genes.

When a new mutant is isolated, it is possible to deter-
mine whether it is caused by a regulator or operator
mutation. For example, we can determine the exact loca-
tion of a mutation on the bacterial chromosome by stan-
dard mapping techniques (see chapter 7) or, more
recently, by DNA sequencing (see chapter 13). Alterna-
tively, the Jacob and Monod model predicts different
modes of action for the two types of mutations. In
merozygotes, a constitutive operator mutation affects
only the operon it is physically a part of. Operator muta-

(a)

of regulator
gene

7 +

E. coli
chromosome

^^ Defective repressor: operon transcribed

(b)

♦

Normal repressor: both operons repressed

Figure 14.6 (a) A lac operon in E. coli with a mutation of the regulator gene (/"). Transcription and translation
of this gene yield a defective repressor; the cell thus has constitutive production of the lac operon. In (b), the
wild-type regulator gene is introduced in an F' factor; there is both a bacterial chromosome and an F' factor,
each containing a regulator gene. (The F' operon carries a mutant z allele, allowing us to keep track of the
transcriptional control of the chromosomal operon only.) In this case, the phenotype is now normal (inducible)
because enough repressor is produced by the F' allele (/ + ), by transcription and translation, to bind to both
operators. RNA polymerase is shown as solid spheres on the DNA; the wild-type repressor is shown as a green
square; the mutant repressor, which cannot bind to the operator, is shown as a red diamond.

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Lac Operon (Inducible System)

411

tions are therefore called cis -dominant. However, a
constitutive /-gene mutation, since it works through an al-
tered protein, is recessive to a wild-type regulator gene in
the same cell, regardless of which operon (chromosomal
or F' factor) the mutation is on. Constitutive regulator
mutations are, therefore, trans-acting. (If two mutations
are on the same piece of DNA, they are in the cis config-
uration. If they are on different pieces of DNA, they are in
the trans configuration.) Trans-acting mutations usually
work through a protein product that diffuses through the
cytoplasm. Cis-acting mutants are changes in recognition
sequences on the DNA.

In figure 14. 6a, the bacterium has a regulator consti-
tutive mutation (/~); the cell has constitutive production

of the operon. If the wild-type regulator is introduced in
an F' plasmid (fig. 14. 6b), the normal (inducible) pheno-
type is restored because the F' / + allele is dominant to
the chromosomal mutation — the i + regulates both the
chromosomal and F' operons. Hence, both operons are
inducible. We don't need to be concerned about the
other components of the F' plasmid because it carries a
z~ allele; only the activity of the chromosomal operon
will be observed. In figure 14.7 a, however, the chromo-
somal operon carries an operator constitutive mutation;
the cell also has constitutive production of the operon.
When a wild-type operator is introduced into the cell in
an F' plasmid (fig. 14. 7b), the cell still has the constitu-
tive phenotype because the operator allele on the F'

(a)

of regulator
gene

(b)

Translation

.,+

E. coli
chromosome

Mutant operator: operon transcribed

of regulator
gene

Mutant operator: chromosomal operon

still transcribed

♦

Translation

Defective
repressor

Figure 14.7 (a) A lac operon in E. coli with a mutation of the operator (o c ). The cell has a constitutive phenotype;
the operator cannot bind the wild-type repressor protein, and thus transcription is continuous, even in the absence
of lactose. The phenotype is unchanged even when a wild-type operator is introduced into the cell in an F factor
(b) m , there is both a bacterial chromosome and an F factor, each containing an operator. (The F operon carries
mutant regulator and z alleles, allowing us to keep track of the transcriptional control of the chromosomal operon
only.) The F operator does not change the phenotype of the cell because the wild-type operator exerts no control
over the chromosomal operator, which exerts a cis-dominant effect; another operator on another operon has no
effect. RNA polymerase is shown as solid spheres on the DNA; the wild-type repressor is shown as a green
square; the mutant repressor, which cannot bind to the operator, is shown as a red diamond.

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

plasmid does not control the bacterial operon; the lac
operon on the bacterial chromosome will be continually
transcribed. The chromosomal operon has a cis-dominant
operator mutation that has a constitutive phenotype.
Note, too, that only the bacterial chromosome deter-
mines the phenotype because the introduced F' plasmid
has a z~ allele.

Other Lac Operon Control Mutations

Other mutations have also been discovered that support
the Jacob and Monod operon model. A superrepressed
mutation, f, was located. This mutation represses the
operon even in the presence of large quantities of the in-
ducer. Thus, the repressor seems to have lost the ability
to recognize the inducer. Basically, the /-gene product is
acting as a constant repressor rather than as an allosteric
protein. In an f/i + merozygote, both operons are re-
pressed because the f repressor binds to both operators.
Another mutation, i Q , produces much more of the
repressor than normal and presumably represents a mu-
tation of the promoter region of the i gene.

In 1966, W. Gilbert and B. Muller-Hill isolated the lac
repressor and thereby provided the final proof of the va-
lidity of the model. At about the same time, M. Ptashne
and his colleagues isolated the repressor for phage X
operons. Control of gene expression in phage X is dis-
cussed later in this chapter.

Mark Ptashne (1940- ).
(Courtesy of Dr. Mark
Ptashne.)

OH OH

Adenylcyclase

Inhibited by glucose

® o

Phosphodiesterase

5' AMP

OH OH

Figure 14.8 Structure of cyclic AMP (cAMP). Glucose uptake
lowers the quantity of cyclic AMP in the cell by inhibiting the
enzyme adenylcyclase, which converts ATP to cAMP

CATABOLITE REPRESSION

An interesting property of the lac operon and other oper-
ons that code for enzymes that catabolize certain sugars
(e.g., arabinose, galactose) is that they are all repressed in
the presence of glucose. That is, glucose is catabolized in
preference to other sugars; the mechanism (catabolite
repression) involves cyclic AMP (cAMP; fig. 14.8). In

eukaryotes, cAMP acts as a second messenger, an intra-
cellular messenger regulated by certain extracellular hor-
mones. Geneticists were surprised to discover cAMP in
E. colt, where it works in conjunction with another regu-
latory protein, the catabolite activator protein (CAP),
to control the transcription of certain operons.

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Trp Operon (Repressible System)

413

In the absence of glucose, cAMP combines with CAP,
and the CAP-cAMP complex binds to a distal part of the
promoter of operons with CAP sites (e.g., the lac operon;
see fig. 14.4). This binding apparently enhances the affin-
ity of RNA polymerase for the promoter, because without
the binding of the CAP-cAMP complex to the promoter,
the transcription rate is very low. The uptake of glucose
by E. colt cells causes the loss of cAMP from the cell,
probably by inhibiting adenylcyclase (fig. 14.8), and thus
lowers the CAP-cAMP level. The transcription rate of
operons with CAP sites will, therefore, be reduced (fig.
14.9). The same reduction of transcription rates is no-
ticed in mutant strains of E. colt when this part of the dis-
tal end of the promoter is deleted. The binding of CAP-
cAMP to the CAP site causes the DNA to bend more than
90 degrees (fig. 14.10). This bending, by itself, may en-
hance transcription, making the DNA more available to
RNA polymerase.

In addition, at some point in the process of initiation
of transcription, the CAP is in direct contact with RNA
polymerase. This was shown by photo cross-linking stud-
ies in which the CAP was treated with a cross-linking
agent that bound the a subunit of RNA polymerase when
irradiated with UV light. For the two proteins to cross-
link, they must be in direct contact during the initiation
of transcription.

Catabolite repression is an example of positive regu-
lation: Binding of the CAP-cAMP complex at the CAP site
enhances the transcription rate of that transcriptional

CAP cAMP RNA No glucose;

protein / polymerase cAMP present;

active transcription

lac operon
DNA

(a)

Promoter

CAP
protein

Glucose present;
no cAMP present;
little transcription

(b)

CAP site

Figure 14.9 Catabolite repression. When cAMP is present in
the cell (no glucose is present), it binds with CAP protein,
and together they bind to the CAP site in various sugar-
metabolizing operons, such as the lac operon shown here.
The CAP-cAMP complex enhances the transcription of the
operon. When glucose is present, it inhibits the formation of
cAMP. Thus no CAP-cAMP complex forms, and transcription
of the same operons is reduced.

Figure 14.10 CAP-DNA interaction: model of cap protein and
DNA. The cap site has twofold symmetry, like the operator.
The cAMP-binding domain is dark blue, the DNA-binding
domain is purple, and the cyclic AMP molecules within the
protein are red. The DNA sugar-phosphate backbones are
shown in yellow, the bases in light blue. DNA phosphates in
red (on the double helix) are those whose modification
interfere with CAP binding. DNA phosphates in dark blue (also
on the double helix) are those especially prone to nuclease
attack because of the bending of the DNA. (Courtesy of

Thomas A. Steitz.)

unit. Thus, the lac operon is both positively and
negatively regulated; the repressor exerts negative con-
trol, and the CAP-cAMP complex exerts positive control
of transcription.

TRP OPERON (REPRESSIBLE
SYSTEM)

The inducible operons are activated when the substrate
that is to be catabolized enters the cell. Anabolic operons
function in the reverse manner: They are turned off (re-
pressed) when their end product accumulates beyond
the needs of the cell. Two entirely different, although not
mutually exclusive, mechanisms seem to control the tran-
scription of repressible operons. The first mechanism fol-
lows the basic scheme of inducible operons and involves
the end product of the pathway. The second mechanism
involves secondary structure in messenger RNA tran-
scribed from an attenuator region of the operon.

Tryptophan Synthesis

One of the best-studied repressible systems is the trypto-
phan, or trp, operon in E. colt. The trp operon contains

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

the five genes that code for the synthesis of the enzymes
that build tryptophan, starting with chorismic acid
(fig. 14.1 1). It has a promoter-operator sequence (p, o) as
well as its own regulator gene (trpK).

Operator Control

In this repressible system, the product of the trpR gene,
the repressor, is inactive by itself; it does not recognize
the operator sequence of the trp operon. The repressor
only becomes active when it combines with tryptophan.
Thus, when tryptophan builds up, enough is available to
bind with and activate the repressor. Tryptophan is thus
referred to as the corepressor. The corepressor-repressor
complex then recognizes the operator, binds to it, and
prevents transcription by RNA polymerase.

After the available tryptophan in the cell is used up,
the diffusion process causes tryptophan to leave the re-
pressor, which then detaches from the trp operator. The
transcription process no longer is blocked and can pro-
ceed normally (the operon is now derepressed). Tran-
scription continues until enough of the various enzymes
have been synthesized to again produce an excess of
tryptophan. Some becomes available to bind to the re-
pressor and make a functional complex, and the operon
is again shut off and the process repeated, ensuring that
tryptophan is being synthesized as needed (fig. 14.12).
This regulation is modified, however, by the existence
of the second mechanism for regulating repressible
operons — attenuation.

Repressed
state

□ □□I

trp structural genes

i — ' — i

p o e d c b a

nnn nnn

Chorismic acid

Glutamine

Gene

trpE

Anthranilate
synthase

Enzyme

Anthranilic acid

Phosphoribosyl pyrophosphate

trpD

Anthranilate
phosphoribosyltransferase

Phosphoribosyl anthranilic acid

trpC

lndole-3-glycerolphosphate
synthetase

Carboxyphenylamino-1 -deoxyribulose phosphate

trpC

lndole-3-glycerolphosphate
synthetase

lndole-3-glycerolphosphate

Serine

trpA

trpB

Tryptophan
synthetase

Tryptophan

Figure 14.11 Genes of the tryptophan operon in E coli. The
enzymes they produce control the conversion of chorismic acid
to tryptophan. The symbol o on the chromosome refers to the
trp operator, which has its own repressor, the product of the
trpR gene.

RNA polymerase

Repressor + corepressor (tryptophan)
e

]DDD DNA with repressor-corepressor complex

l-o^

Derepressed
state

DDDI

DDD DNA without repressor-corepressor complex

Inactive
repressor

Figure 14.12 The repressor-corepressor complex binds at the operator and prevents the transcription of
the trp operon in E coli. Without the corepressor, the repressor cannot bind, and therefore transcription is
not prevented. The blue wedge is the corepressor (two tryptophan molecules), and the partial red circle is
the repressor.

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Trp Operon (Attentuator-Controlled System)

415

TRP OPERON (ATTENUATOR-
CONTROLLED SYSTEM)

Details of the second control mechanism of repressible
operons have been elucidated primarily by C. Yanofsky
and his colleagues, who worked with the tryptophan
operon in E. colt. This type of operon control, control by
an attenuator region, has been demonstrated for at
least five other amino acid-synthesizing operons, includ-
ing the leucine and histidine operons. This regulatory
mechanism may be the same for most operons involved
in the synthesis of an amino acid.

Leader Transcript jg^

In the trp operon, an attenuator region lies between the
operator and the first structural gene (fig. 14.13). The
messenger RNA transcribed from the attenuator region,
termed the leader transcript, has been sequenced, re-

Charles Yanofsky (1925- ).
(Courtesy of Dr. Charles Yanofsky.)

vealing two surprising and interesting facts. First, four
subregions of the messenger RNA have base sequences
that are complementary to each other so that three dif-
ferent stem-loop structures can form in the messenger
RNA (fig. 14.14). Depending on circumstances, regions
1-2 and 3-4 can form two stem-loop structures, or re-
gion 2-3 can form a single stem-loop. When one stem-
loop structure is formed, the others are preempted. As

P o
DDDr~TT

Leader
peptide
gene

Attenuator region

Structural
genes (e, d, ...)

IDDD

Figure 14.13 Attenuator region of the trp operon,
which contains the leader peptide gene {red). This
region is transcribed into the leader transcript.

50
I
■A-G.

u* Y« G " u ' A v\

\ .100

C
*

■

C A — U
G • U —A— 110
C— C — C
A— U C
C— G— C

140

Stem-loops
1-2 and 3-4

U— A A — U-U-U-U-U-U-U-
U C — G — C

C— G— C — G
.C— G— C — G
JJ G— C— G— 130

70— G C— G— C

•
U

C— G«
C— G'

G N

Stem-loop
2-3

Figure 14.14 Nucleotide
sequence of part of the
leader transcript of the trp
attenuator region (bases 50
to 140). Stem-loops 1-2
and 3-4, or stem-loop
2-3, can form because of
complementarity of the
nucleotides. All possible
base pairings are shown in
the middle of the figure.
(From D. L Oxender, et al.,
"Attenuation in the Escherichia
coli tryptophan operon: Role of
RNA secondary structure
involving the tryptophan codon
region," Proceedings of the
National Academy of Sciences,
76:5524-28, 1979. Reprinted by
permission.)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

we will see, the particular combination of stem-loop
structures determines whether transcription continues.

Leader Peptide Gene *^

The second fact obtained by sequencing the leader tran-
script is that there is a small gene coding information for
a peptide from bases 27 to 68 (fig. 14.15). The gene for
this peptide is referred to as the leader peptide gene. It
codes for fourteen amino acids, including two adjacent
tryptophans. These adjacent tryptophan codons are criti-
cally important in attenuator regulation. The proposed
mechanism for this regulation follows.

Excess Tryptophan

Assuming that the operator site is available to RNA poly-
merase, transcription of the attenuator region will be-
gin. As soon as the 5' end of the messenger RNA for the
leader peptide gene has been transcribed, a ribosome
attaches and begins translating this messenger RNA.
Depending on the levels of amino acids in the cell,

three different outcomes can take place. If the concen-
tration of tryptophan in the cell is such that abundant
tryptophanyl-tRNAs exist, translation proceeds down
the leader peptide gene. The moving ribosome overlaps
regions 1 and 2 of the transcript and allows stem-loop
3-4 to form, as shown in the configuration at the far left
of figure 14.16. This stem-loop structure, referred to as
the terminator, or attenuator, stem, causes transcrip-
tion to be terminated. Note that stem-loop 3-4, the ter-
minator stem, followed by a series of uracil-containing
bases, is a rho-independent transcription terminator
(see chapter 10). Hence, when existing quantities of
tryptophan, in the form of tryptophanyl-tRNA, are ade-
quate for translation of the leader peptide gene, tran-
scription is terminated.

Tryptophan Starvation

If the quantity of tryptophanyl-tRNA is lowered, the ribo-
some must wait at the first tryptophan codon until it ac-
quires a Trp-tRNA Trp . This is shown in the configuration in
the middle part of figure 14.16. The stalled ribosome will

([MetY Lysjf Alajf lie Y PheY Valjf LeuY LysY GlylfTrp YTrpYArgY ThrJfSer) Stop
AAGUUCACGUAAAAAGGGUAUCGACAAUGAAAGCAAUUUUCGUACUGAAAGGUUGGUGGCGCACUUCCUGAAACGGGCAG • • •

Figure 14.15 Base sequence of the trp leader transcript and the amino acids these nucleotides code. Note the presence of
adjacent tryptophan COdons. (From D. L Oxender, et al., "Attenuation in the Escherichia coli tryptophan operon: Role of RNA secondary structure
involving the tryptophan codon region," Proceedings of the National Academy of Sciences, 76:5524-28, 1979. Reprinted by permission.)

Excess trp:
termination

ftp-starved:
no termination

No translation:
termination

Ribosome

Terminator
stem

Figure 14.16 Model for attenuation in the E. coli trp operon. The circle represents the ribosome attempting to translate the leader
transcript of figure 14.14. Under conditions of excess tryptophan, the 3-4 stem-loop forms (the terminator stem), terminating
transcription. Under conditions of tryptophan starvation, the ribosome is stalled, and stem-loop 2-3 forms, allowing continued
transcription. Under general starvation, there is no translation, resulting in the formation of stem-loops 1-2 and 3-4, which again
results in the termination of transcription. (From D. L Oxender, et al., "Attenuation in the Escherichia coli tryptophan operon: Role of RNA secondary
structure involving the tryptophan codon region," Proceedings of the National Academy of Sciences, 76:5524-28, 1979. Reprinted by permission.)

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Trp Operon (Attentuator-Controlled System)

417

permit stem-loop 2-3 to form, which precludes the forma-
tion of the terminator stem-loop (3-4). In this configuration,
transcription is not terminated, so that eventually, the whole
operon is transcribed and translated, raising the level of tryp-
tophan in the cell. The stem-loop 2-3 structure is referred to
as the preemptor stem. Note that the preemptor stem is
not a rho-independent transcription terminator and thus,
without the rho protein present, will not terminate tran-
scription (see chapter 10).

General Starvation

A final configuration is possible, as shown on the far right
in figure 14.16. Here, no ribosome interferes with stem
formation and, presumably, stem-loops 1-2 and 3-4 (ter-
minator) form. This configuration also terminates tran-
scription because of the terminator stem. It is believed
that this configuration occurs if the ribosome is stalled
on the 5' side of the trp codons, which happens when

the cell is starved for other amino acids. Presumably, it
makes no sense to manufacture tryptophan when other
amino acids are in short supply. Hence, the cell can care-
fully bring up the levels of the various amino acids in the
most efficient manner.

TRAP Control

The tryptophan operon in bacilli such as Bacillus subtilis
is also controlled by attenuation, but secondary structure
in the mRNA transcript is induced by binding not the ri-
bosome, but a trp RNA-binding attenuation protein
(TRAP). This protein attaches to the nascent messenger
RNA only after the protein binds tryptophan molecules;
the result is a terminator stem that forms in the messen-
ger RNA. In the absence of excess tryptophan, TRAP does
not bind to the messenger RNA, a preemptor (also called
an antiterminator) stem, not the terminator stem, forms,
and transcription continues (fig. 14. Hot). Recently, the

(a)

dna [ii

P T
*P 35

.T^AJ^nt^^^^ n trpEDCFBA

60 91 108 111 133 £04

Ia
u

0S3

'A J

\a\u g

,q)u

c u

^^AU^A^LIU^A^UUA^^WfLJA^H 1 | 1 tfpEDCFB^ -

transcription

- tryptophan

anti -termination

108 111
+ tryptophan

termination

UUUAUULJ

* ^t^

^?m&ft

(b)

Figure 14.17 The trp operon control by attenuation in Bacillus subtilis.
(a) The top of the figure shows the leader region of the DNA with the two
parts of the antitermination stem (A, B) and the termination stem (C, D).
The triplets (GAG and TAG in DNA, or GAG and UAG in the messenger
RNA) that the trp RNA-binding attenuation protein (TRAP) binds to are
circled. The label trpEDCFBA refers to the structural genes of the trp
operon. Nucleotides 108-133 (C, D) form the terminator stem, and
nucleotides 60-1 1 1 (A, B) form the antiterminator stem. The arrows below
the boxed letters A-D indicate the inverted repeat sequences forming the
stems. Without TRAP, the antiterminator stem forms; with TRAP, the
terminator stem forms as TRAP is bound by nucleotides 36-91 of the
messenger RNA. Part (b) is a close-up of the mRNA (ball-and-stick model)
wrapped around TRAP (ribbon diagram with subunits in different colors)
bound by tryptophan molecules (spheres). (From Alfred A. Antson, et ai.,
"Structure of the trp RNA-binding attenuation protein, TRAP, bound to RNA" in Nature,
Vol.40, September 16, 1999, fig. 1 p. 234 and fig. 2a p. 237. Reprinted by permission of
Macmillan Ltd.)

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418

Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

structure of the protein was worked out; it has eleven
symmetrical loops, each of which can bind a tryptophan
molecule. When TRAP is bound to tryptophan molecules,
it can attach to triplets in the messenger RNA transcript,
triplets of GAG or UAG. The TRAP wraps the mRNA
around itself, forming an elegant pinwheel (fig. 14.17&).

Redundant Controls

Some amino acid operons are controlled only by attenua-
tion, such as the his operon in E. coli, in which the leader
peptide gene contains seven histidine codons in a row, or
the trp operon in B. subtilis. Redundant control (repres-
sion and attenuation) of tryptophan biosynthesis in
E. coli allows the cell to test both the tryptophan levels
(tryptophan is the corepressor) and the tryptophanyl-
tRNA levels (in the attenuator control system). The atten-
uator system also allows the cell to regulate tryptophan
synthesis on the basis of the shortage of other amino
acids. For example, when there is a shortage of both
tryptophan and arginine, operator control allows tran-
scription to begin, but attenuator control terminates
transcription because stem-loops 1-2 and 3-4 form
(fig. 14.16).

LYTIC AND LYSOGENIC
CYCLES IN PHAGE X

When a bacteriophage infects a cell, it must express its
genes in an orderly fashion; some gene products are
needed early in infection, and other products are not
needed until late in infection. Early genes usually control
phage DNA replication; late genes usually determine
phage coat proteins and the lysis of the bacterial cell. A
phage is most efficient if it expresses the early genes first
and the late genes last in the infection process. Also, tem-
perate phages have the option of entering into lysogeny
with the cell; here, too, control processes determine
which path is taken. One generalization that holds true
for most phages is that their genes are clustered into
early and late operons, with separate transcriptional con-
trol mechanisms for each.

Phage X is perhaps the best-studied bacteriophage. It
has a chromosome of about 48,500 base pairs. Since it is
a temperate phage, it can exist either vegetatively or as a
prophage, integrated into the host chromosome. This
phage warrants our attention because of the interesting
and complex way that its life-cycle choice is determined.
It is a model system of operon controls. The complexity
results from having two conflicting life-cycle choices.

Briefly, the expression of one of the two life-cycle
alternatives, lysogenic or lytic cycles, depends on
whether two repressors, CI and Cro, have access to op-

erator sites. The CI repressor acts to favor lysogeny: it
represses the lytic cycle. The Cro repressor favors the
lytic cycle and represses lysogeny. The operator sites,
when bound by either CI or Cro, can either enhance or
repress transcription. Other control mechanisms are
also involved in determining aspects of the X life cycle,
including antitermination and multiple promoters for
the same genes.

Phage X Operons

Phage X (see fig. 7.21) exhibits a complex system of con-
trols of both early and late operons, as well as controls for
the decision of lytic infection versus lysogenic integra-
tion. The X genes are grouped into four operons: left,
right, late, and repressor (fig. 14.18). The left and right
operons contain the genes for DNA replication and re-
combination and phage integration. The late operon con-
tains the genes that determine phage head and tail pro-
teins and lysis of the host cell. The sequence of events
following phage infection is relatively well known.

The map of X (fig. 14.18) is a circle, but the X chro-
mosome has two linear stages in its life cycle (fig. 14.19).
It is packed within the phage head in one linear form,
and it integrates into the host chromosome to form a
prophage in another linear form (fig. 14.19). Those two
linear forms do not have the same ends (figs. 14.18 and
14.19&). The mature DNA, which is packed within the
phage heads before lysis of the cells, is flanked by cos
sites (chapter 13). It results from a break in the circular
map between the A and R loci. The prophage is inte-
grated at the att site, and the circular map is thus broken
there at integration.

The homologous integration sites on both X and the
E. coli chromosome consist of a 15 bp core sequence
(called "O" in both), flanked by different sequences on
both sides in both the bacterium and the phage (fig.
14.20). In the phage, the region is referred to as POP',
where P and P' (P for phage) are two different regions
flanking the O core on the phage DNA. In the bacterium,
the region is called BOB', where B and B' (B for bac-
terium) are two different regions flanking the O core on
the E. coli chromosome. Integration, which is a part of
the lysogenic life cycle, requires the product of the X int
gene, a protein known as integrase, and is referred to as
site-specific recombination. Later excision of the
prophage, during induction, when the phage leaves the
host chromosome to enter the lytic cycle, requires both
the integrase and the protein product of the neighboring
xis gene, excisionase.

After infection of the E. coli cell by a X phage, the
phage DNA circularizes, using the complementarity of
the cos sites. Transcription begins, and within a very
short time the phage is guided toward either entering
the lytic cycle and producing virus progeny or entering

Tamarin: Principles of
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Molecular Genetics

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Control in Prokaryotes and
Phages

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Lytic and Lysogenic Cycles in Phage A

419

^pressor oper 0n

exo

Prophage
ends

Nu1

Mature
DNA ends
(cos sites)

H T G

Figure 14.18 Genetic map of phage X. There are four operons present: the repressor, left, right, and late. The prophage, a linear
form integrated into the bacterial chromosome, begins and ends at att. The mature phage, another linear form found packed into the
phage heads, begins and ends at Nu1 (cos sites).

Mature DNA

cos A W B

Q S R cos

(a)

Prophage gal int xis exo

• • •

E. coli

(b)

K I J b

bio

• • •

E. coli

Figure 14.19 The two linear forms of X phage, (a) False color electron micrograph of the X chromosome, approximately 16 |xm in
length. This is the linear form of the phage chromosome found within the phage heads, (b) The mature linear DNA (found within
phage protein coats) is flanked by cos sites. The prophage is flanked by E. coli DNA (bio and gal loci), ([a] Courtesy of Martin Guthoid

and Carlos Bustamante, Institute of Molecular Biology and HHMI, University of Oregon.)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

aft site of X

att site of E. coli

I I I I I I I I I I I I I I I I I II I I
TCAGCTTTTTTATACTAAGTT

AGTCGAAAAAATATGATTCAA

I I I I I I I I I I I I I I I I I I I I I
CCTGCTTTTTTATACTAACTT

GGACGAAAAAATATGATTGAA

POP'

BOB'

CCTGCTTTTTTATACTAAGTT . . . T CAGCTT T TT T AT ACT A A CT T

GGACGAAAAAATATGATTCAA. . . AGT CG A A A A A AT ATGAT T GA A

(b)

BOP'

Prophage

POB'

Figure 14.20 Integration of the X phage into the E. coli chromosome requires a crossover between the two attach sites, called POP'
(phage) and BOB' (bacteria), (a) General pattern of this site-specific attachment, (b) Nucleotide sequences of the various components.

the lysogenic cycle and integrating into the host chro-
mosome. What events lead up to this "decision" on
which path to take?

Early and Late Transcription

When the phage first infects an E. coli cell, transcription
of the left and right operons begins at the left (p L ^ and
right (p#) promoters, respectively: The N (left) and cro
(right) genes are transcribed (fig. 14.21) and then trans-

lated into their respective proteins. Transcription then
stops on both operons at rho-dependent terminators
(Sri, t L j ). Transcription cannot continue until the protein
product of the N gene is produced. This protein is called
an antiterminator protein. When it binds at sites up-
stream from the terminators, called nutL and nutR (nut
stands for N utilization; L and R stand for left and right),
the polymerase reads through the terminators and con-
tinues on to transcribe the left and right operons.
(Although it is not completely clear why antitermination

L L1

rex cl

cro

> >

nutL p L

Transcription

p R nutR

Transcription —

> f >

l R1

Figure 14.21 Transcription begins at the left and
right promoters {p L , p R ) and proceeds to the left
and right terminators {t L1 , t m ). Transcription
continues through these terminators when the
protein product of the N gene binds to the nutL
and nutR sites.

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Lytic and Lysogenic Cycles in Phage A

421

has evolved here, it seems to give the phage better con-
trol over the timing of events.)

Transcription then continues along the left and right
operons through the ell and cIII genes (see fig. 14.18).
Later, if the lytic response is followed, the Q gene, which
codes for a second antiterminator protein in the right
operon, has the same effect on the late operon as the N
gene did on the two early operons: Without the <Q-gene
product, transcription of the late operon proceeds about
two hundred nucleotides and then terminates. With the
<Q-gene product, the late operon is transcribed. Hence, in
phage X, proteins that allow RNA polymerase to proceed
past termination signals mediate general control of tran-
scription. If only the previously described events were to
transpire, the lytic cycle pathway would always be fol-
lowed. However, a complex series of events can also take
place in the repressor region that may lead to a "decision"
to follow the lysogenic cycle instead.

Repressor Transcription

The c///-protein product inhibits a host cell protease,
called FtsH, that would break down the c/7-gene product.
The c//-gene product binds at two promoters, enhancing
their availability to RNA polymerase, just as the CAP-cAMP
product enhances the transcription of the lac operon.
The ell protein binds at the promoters for cl transcription
and for int transcription (fig. 14.22). At this point, the
phage can still "choose" between either the lytic or the
lysogenic cycles. Integrase (the product of the int gene)

and cl (repressor) proteins are now produced, favoring
lysogeny, as well as the cro-gene product, the antirepres-
sor, which is a repressor of cl and therefore favors the
lytic pathway. (Cro stands for control of repression and
other things; the c of cl, the repressor, stands for "clear,"
which is the appearance of X plaques that have cl muta-
tions. These mutants can only undergo lysis without the
possibility of lysogeny. Normal X infections produce tur-
bid plaques, accounted for by lysogenic bacterial growth
within the plaques.) We now focus further on the repres-
sor region with its operators and promoters.

Maintenance of Repression

The cl gene, with the aid of the cIT-gene product, is tran-
scribed from a promoter known as p RE , the RE standing
for repression establishment (fig. 14.23). Once cl is tran-
scribed, it is translated into a protein called the X repres-
sor, which interacts at the left and right operators, o L and
o R , of the left and right operons. When these operators
are bound by cl protein, transcription of the left and right
operons (and therefore also the late operon) ceases.
There are several ramifications of the repression. First,
lysogeny can be initiated because the int gene has been
transcribed at the early stage of infection. Second, since
CII and CIII are no longer being synthesized, cl transcrip-
tion from the^?££ promoter stops. However, cl can still be
transcribed because there is a second promoter, pj^ (RM
stands for repression maintenance), that allows low lev-
els of transcription of the cl gene.

cIII protein
breaks down
FtsH protease

int ... clll N rex cl cro ell

Pi Pre

Transcription A

Transcription ^

Figure 14.22 The c//-gene product of phage X
binds to the cl promoter {p RE ) and the int
promoter (p/), enhancing transcription of those
genes. The clll protein breaks down the FtsH
protease that would normally break down the
ell protein.

ell protein

□ □□[

rn:

Prm Pr

rex

Pre

cro

I II I

p L - Left promoter
o, - Left operator

p R - Right promoter
o R - Right operator

]□□□

Prm~ Maintenance-of-

repression promoter

- Establishment-of-
repression promoter

Figure 14.23 Early regulation region of phage
X. Two promoters, p RE and p RM , transcribe the
cl and rex genes. The left operator overlaps the
left promoter, and the right operator overlaps
both the right promoter and the maintenance-
of-repression promoter.

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

The cl gene can further control its own concentra-
tion in the cell. When the right and left operators were
sequenced, each was discovered to have three sites of re-
pressor recognition (fig. 14.24). On the right operator, for
example, the right-most site (p R1 ) was found to be most
efficient at binding repressor. When repressor was bound
at this site, the right operon was repressed, and tran-
scription of cl was enhanced (again, in a way similar to
enhancement of transcription by the binding of CAP-
cAMP at the CAP site in the lac operon). Excess repres-
sor, when present, however, was also bound by the other
two sites within o R . The foregoing process results in the
repression of the cl gene itself. Hence, maintenance lev-
els of CI can be kept within very narrow limits.

A third ramification of repression is the prevention of
superinfection. That is, bacteria lysogenic for X phage are
protected from further infection by other X phages
because repressor is already present in the cell. Thus, the
excess of repressor controls new invading X phages. (We
say that bacterial cells lysogenic for phage X are immune
from infection by additional X phage.) These bacteria
are also protected from infection by T4 phage with rll
mutants. The rex-gene product, the product of the other
gene in the repressor operon, controls this protection.

The promoters for maintenance and establishment of
repression differ markedly in their control of repressor
gene expression. When jD^ is active, a very high level of
repressor is present, whereas Prm produces only a low
level of repressor. The level of repressor is due to the
length of the leader RNA transcribed on the 5' side of the
cl gene. The^?^ promoter transcribes a very long leader
RNA and is very efficient at translation of the cl region. In
contrast, the Prm promoter begins transcription at the
initiation codon of the protein. This leaderless messenger
RNA is translated very inefficiently into CI.

The X repressor is a dimer of two identical subunits
(fig. 14.25). Each subunit is composed of two domains, or

Amino acids

Prm

DUUL

Tr-TT

°R3 °R2 °R1

]□□□

Preferential binding
of cro-gene product

Preferential binding
of c/-gene product

Figure 14.24 The right operator on the phage X chromosome
overlaps the p RM and p R promoters. There are three repressor
recognition sites within the operator: o R1 , o R2 , and o R3 .
Preferential binding by the Cro repressor to o R3 and the Cl
repressor to o R1 determines whether transcription occurs to the
left or the right.

COOH » COOH

132-236

(COOH end)

93-131

1-92

(NH 2 end)

/j^O^X

DNA

(a)

Axis of
symmetry

6
5

3
2

1
-1

o R1 /R1-69

(b)

Figure 14.25 The X repressor, (a) The X repressor is a dimer
with each subunit having helical amino- and carboxyl-terminal
ends. The helical structure of each amino-terminal end binds in
the major DNA groove, {b) Diagram of the interaction of amino
acid residues 1-69 (blue) with o R1 {red) in the closely related
phage 434. Dashed lines are hydrogen bonds. Numbers -1 to
6 and 4' and 5' are phosphate numbers. Amino acids are
designated by the single-letter code (fig. 11.1). Small red
Circles are water molecules. (From D. W. Rodgers and s. c.
Harrison, "The complex between phage 434 repressor DNA-binding domain
and operator site R3 : Structural differences between consensus and non-
consensus half-sites," Structure, 1:227-40, Dec. 15, 1993. © Current
Biology Ltd.)

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Phages

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Lytic and Lysogenic Cycles in Phage A

423

"ends." The carboxyl- and amino-terminal ends are sepa-
rated by a relatively open region, susceptible to protease
attack. The alpha-helical regions of the amino-terminal
ends interdigitate into the major groove of the DNA to lo-
cate the specific sequences making up the left and right
operator sequences. As described earlier for the lac op-
erator, o R1 , o R2 , and o R3 each have twofold symmetry.

The binding of the X repressor in o R1 enhances the
binding of another molecule of repressor into o R2 . To-
gether, they enhance Prm transcription, presumably
through contact with RNA polymerase. The repressors
also block p R transcription (see fig. 14.24).

Lysogenic Versus Lytic Response

We have described the mechanism by which X estab-
lishes lysogeny How then does X turn toward the lytic cy-
cle? Here, control is exerted by the cro-gene product, an-
other repressor molecule that works at the left and right
operators in a manner antagonistic to the way the CI
repressor works. In other words, using the right operator
as an example, cro-gene product binds preferentially to
the leftmost of the three sites within o R and represses cl
but enhances the transcription of cro (see fig. 14.24).

The cro-gene product can direct the cell toward a
lytic response if it occupies the o R and o L sites before the
X repressor, or if the X repressor is removed. From the
point of view of phage X, when would be a good time for
the CI repressor to be removed? Thinking in evolutionary
terms, we would expect that a prophage might be at an
advantage if it left a host's chromosome and began the
lytic cycle when it "sensed" damage to the host. In fact,
one of the best ways to induce a prophage to enter the
lytic cycle is to direct ultraviolet (UV) light at the host
bacterium. (Actually, this was how lysogeny was discov-
ered, by French geneticist Andre Lwoff.) UV light causes
damage to DNA and induces several repair systems. One,
called SOS repair (see chapter 12), makes use of the pro-
tein product of the recA gene. Among the activities of
this enzyme is to cleave the X repressor in the suscepti-
ble region between domains. The cleaved repressor falls
free of the DNA, making the operator sites available for
the cro-gene product. The lytic cycle then follows.

Initially, however, when the phage first infects an
E. colt cell, the "decision" for lytic versus lysogenic
growth is probably determined by the c//-gene product.
This protein, as we mentioned, is susceptible to a bacte-
rial protease, which, in turn, is an indicator of cell
growth. When E. colt growth is limited, its proteases tend
to be limited, a circumstance that would favor lysogeny
for the phage. It is the ell protein that, when active, fa-
vors lysogeny and when inactive favors the lytic cycle.
Thus, under active bacterial growth, the ell protein is

more readily destroyed, it thus fails to enhance cl tran-
scription, and lysis follows. When bacteria are not grow-
ing actively, the ell protein is not readily destroyed, it en-
hances cl transcription, and lysogeny results. Thus, under
initial infection, the choice between lysogeny or the lytic
cycle depends primarily on the ell protein, which gauges
the health and activity of the host. After lysogeny is es-
tablished, it can be reversed by processes that inactivate
the cl protein, indicating genetic damage to the bac-
terium (the SOS response) or an abundance of other
hosts in the environment (zygotic induction, see chapter
7). In zygotic induction, the lytic cycle is induced during
conjugation, presumably when an Hfr cell sends a copy
of the X prophage into an F~ cell. At that point, without
repressor present, the prophage can reassess whether to
continue lysogeny or enter the lytic cycle.

Not all the details regarding the Cl-Cro competition are
known, but an understanding of the relationship of lytic
and lysogenic life cycles and the nature of DNA-protein
recognition has emerged (fig. 14.26 and table 14.1).

Table 14.1 Elements in Phage A. Infection

Gene Products

cl Repressor protein whose function favors lysogeny

ell Enhances transcription at the p T and p RE promoters

clll Inhibits the FtsH protease

cro Antirepressor protein that favors lytic cycle

N Antiterminator acting at nutR and nutL

rex Protects bacterium from infection by T4 rll mutants

int Integrase for prophage integration

Q Antiterminator of late operon

FtsH Bacterial protease that degrades ell protein

Promoters of

p R Right operon

p L Left operon

p RE Establishment of repression at repressor region

Pxm Maintenance of repression at repressor region

p R Late operon

pj int gene

Terminators

t R1 Terminates after cro gene

t L1 Terminates after N gene

Antiterminators

nutR In cro gene

nutL In TV gene

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Molecular Genetics

14. Gene Expression:
Control in Prokaryotes and
Phages

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424

Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

(1 ) Initial infection. Transcription from p R and p L through cro and N.

Termination at t R1 and t Lr

int

cIII N rex cl

cro ell o ... Q S

t L1 p L /o L

Prm Pre

PR /0 R f R1

(2) N protein allows antitermination at t L1 and t Rr Transcription continues through ell and clll. Protein product
of ell allows transcription at p l and p RE .

Prm ^Pre
Pr /o r *ri

(3) Repressor and antirepressor {cl- and cro-gene products) compete for o R and o L
sites.

Lytic growth
Antirepressor {cro protein) gains access to o R3 ,

°R2' °L3> ar| d °L2- ^'9 nt > ' e ft> anc ' ' ate operons
transcribed. Repressor region {cl, rex) repressed.

rex

cro

p L /o L

Prm
Pr /o r

fffj Absorbed phage

.Cell's DNA

Lysogeny

Repressor {cl protein) gains access to o R1 ,

°R2> °LV ai1C ' °L2 m ^'9 nt ' ' e ^' anC ' ' ate OP^fOnS

repressed. Transcription at p RM enhanced.

rex

cro

Bacterium

p L /o L

Prm
Pr /o r

Lysis
of cell

Injection of phage
genetic material

Rarely, mutagenic chemicals or radiation
cause expression of lytic cycle

Assembly of
new phage

Manufacture of
phage proteins

0(";

Replication of phage genetic
material and breakdown of
host's genetic material

Cell division

Integration of phage DNA

Figure 14.26 Summary of regulation of phage X life cycles. {1) In the initial infection, transcription begins in cro and N but
terminates shortly thereafter at left and right terminators. (2) The product of the N gene allows transcription through the
initial terminators; in essence, all genes can now be transcribed. (3) Lysogeny will occur if the cl protein gains access to
the right and left operators; the lytic cycle will prevail if the cro-gene product gains access to those two operators.

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Control in Prokaryotes and
Phages

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Transposable Genetic Elements

425

TRANSPOSABLE GENETIC
ELEMENTS

Up until this point, we have thought of the genome in
fairly conservative terms. If we map a gene today, we ex-
pect to see it in the same place tomorrow. However, our
discovery of mobile genetic elements has modified that
view to some extent. We now know that some segments
of the genome can move readily from one place to an-
other. Their moving can have an effect on the phenotype
of the organism, primarily at the transcriptional level. We
thus begin our discussion of mobile genetic elements
here, and we conclude it in chapter 16, because mobile
elements also affect the phenotypes of eukaryotes.

IS Elements

Transposable genetic elements, transposons, or even
jumping genes, are regions of the genome that can move
from one place to another. In some cases, transposition is
conservative: the transposons move without copying them-
selves. They are liberated from the donor site by double-
strand breaks in the DNA. In other cases, transposition is
replicative: a copy of the transposon is inserted while the
original stays in place. This mechanism involves only single-
strand breaks of the DNA at the donor site.

Barbara McClintock first discovered transposable ele-
ments in corn in the 1940s (see chapter 16); they were dis-
covered in prokaryotes in 1967, where they first showed
up as polar mutants in the galactose operon of E. coll No

genes of the operon were expressed past the point of the
polar mutation. This effect was explained by assuming that
the transposon brought with it a transcription stop signal.
The presence of an inserted piece of DNA in these polar
mutants was verified by heteroduplex analysis (fig. 14.27).

The first transposable elements discovered in bacte-
ria were called insertion sequences or IS elements. It
turns out that these are the simplest transposons. The IS
elements consist of a central region of about 700 to 1,500
base pairs surrounded by an inverted repeat of about 10
to 30 base pairs, the numbers depending on the specific
IS element. Presumably, the inverted repeats signal the
transposing enzyme that it is at the ends of the IS ele-
ment. The central region of the IS element contains a
gene or genes for the transposing event (usually genes for
transposase and resolvase enzymes); the relatively small
IS elements carry no bacterial genes (fig. 14.28).

The target site that the transposable element moves to
is not a specific sequence, as with the att site of X. It be-
comes a direct repeat flanking the IS element only after
insertion, giving rise to a model of insertion (fig. 14.29).
The target site is cut in a staggered fashion, leaving single-
stranded ends. The IS element is then inserted between
the single-stranded ends. Repair processes convert the
two single-stranded tails to double-stranded segments
and, hence, to direct flanking repeats. When DNA is se-
quenced, the pattern of a direct flanking repeat surround-
ing an inverted repeat, with a segment in the middle, sig-
nals the existence of a transposable element. Currently,
we know of more than fifteen families, including a total of
over five hundred known members, of IS elements.

(a)

Duplex DNA

Extraneous
DNA

(b)

Figure 14.27 Heteroduplex analysis revealing a transposon. (a) Two plasmids were hybridized, one with and one without a
transposon. (b) The transposon is seen as a single-stranded loop {red); it has nothing to pair with in the heteroduplex. ([a] Courtesy of
Richard P. Novick, M.D.)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

Figure 14.28 An IS element
(IS5) inserted into a target site
in a bacterial chromosome
creates a direct repeat on
either side of the IS element.
The explanation is shown in
figure 14.29. An inverted
repeat {red) is seen as the
same sequence read inwards
from outside on the upper
strand (left) and the lower
Strand (right). (Reprinted with
permission from Nature, Vol. 297,
M. Kroger and G. Hoborn,
"Structural Analysis of Insertion
Sequence IS5." Copyright © 1982
Macmillan Magazines Limited.)

Chromosomal target site

— TTAG —

— AATC —

Transposon (IS5)

GGAAGGTGCGAATAAG
CCTTCCACGCTTATTC

* * •

CTTGTTCGCACCTTCC
GAACAAGCGTGGAAGG

Inverted repeats

Insertion

— TTAGGGAAGGTGCGAATAAG

— AATCCCTTCCACGCTTATTC

• • •

CTTGTTCGCACCTTCCTTAG —
GAACAAGCGTGGAAGGAATC —

Inverted repeats
- Direct repeats -

Figure 14.29 Insertion of an
IS element (IS5 of fig. 14.28)
results in a direct flanking
repeat surrounding the
transposon in the host
chromosome. This occurs
because the insertion takes
place at a point in which a
staggered cut is made in the
host DNA, leaving
complementary regions on
either side of the transposon.
Repair replication results in
two copies of the flanking
sequence.

Chromosome

— o o —

— < I —
_l_ < —

— h- < —

Transposon (IS)

Target site

Staggered cuts

TTAG

AATC

Insertion

TTAG

AATC

I I I I

Repair

— o o —

— < I

_l_ < —
_l_ < —

— o o —

— < I

_l_ < —

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Transposable Genetic Elements

421

Composite Transposons

After the discovery of IS elements, a more complex type
of transposable element, a composite transposon, was
discovered. A composite transposon consists of a central
region surrounded by two IS elements. The central region
usually contains bacterial genes, frequently antibiotic re-
sistance loci. For example the composite transposon
Tn 10 contains the genes for transposase and resolvase, as
well as the bacterial gene for p-lactamase, which confers
resistance to ampicillin (fig. 14.30). Arrangements of
composite transposons can vary quite a bit. The IS ele-
ments at the two ends can be identical or different; they
can be in the same or different orientations; they can be
similar to known IS elements or different from any freely
existing IS elements. In the latter case, they are called IS-
like elements.

Two IS elements can transpose virtually any region
between them. In fact, composite transposons most
likely came into being when two IS elements became lo-
cated near each other. We can see this very clearly in a
simple experiment. In figure 14.31, there is a small plas-
mid constructed with transposon Tn 10 in it. The "re-

verse" transposon, consisting of the two IS elements and
the plasmid genes, or the normal transposon, could each
transpose.

Mechanism of Transposition

Transposition can come about by several mechanisms;
however, it does not use the normal recombination ma-
chinery of the cell (see chapter 12). One model, by
J. Shapiro, explains the fact that many transposons in the

J. A. Shapiro (1943- ).
(Courtesy of Dr. J. A. Shapiro.)

Tn70-

Inverted repeats

(

Central

region

1 1

Transposase resolvase p-lactamase

Inverted repeats

Figure 14.30 A composite transposon consists of a central region flanked by two IS elements. Transposon Jn10 contains the
transposase and resolvase enzyme genes as well as the bacterial gene p-lactamase, which protects the cell from the antibiotic ampicillin.

Jn10

ISH

r is

Figure 14.31 Two IS elements in a
plasmid can transpose either of the two
regions between them. In the case
shown, either the Tn70 transposon or
the "reverse" transposon ("other genes")
is transposed.

Other genes
Transposition

Jn10

New composite
transposon

IS Other genes

Normal transposition

"Reverse" transposition

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

process of transposition go through a cointegrate state
(fig. 14.32), in which there is a fusion of two elements.
During the process of transposition (in this case from
one plasmid to another), an intermediate cointegrate
stage is formed, made up of both plasmids and two
copies of the transposon. Then, through a process called
resolution, the cointegrate is reduced back to the two
original plasmids, each now containing a copy of the
transposon.

A diagram of Shapiro's mechanism appears in figure
14.33. At first, staggered cuts are made in the donor and
recipient DNA molecules (fig. 1433a and £>). Then non-
homologous ends are joined so that only one strand of
the transposon connects them (fig. 14. 33c). This process
is presumably controlled by the transposon-coded trans-
posase enzyme. Repair-DNA replication now takes place
to fill in the single-stranded segments. The result is a coin-
tegrate of the two plasmids with two copies of the trans-
poson. The last step is a recombination event at a homol-
ogous site within the two transposons. This is catalyzed
by a resolvase enzyme, which resolves the cointegrate
into the original two plasmids, each with a copy of the
transposon (fig. I4.33e).

Transposon

Plasmid
II

Transposition

Transposon

Cointegrate

Transposon

Resolution

Transposon

Plasmid
II

Figure 14.32 Transposition frequently goes through an
intermediate cointegrate stage. In this case, the transposon is
copied from one plasmid to another, with an intermediate stage
consisting of a single large plasmid.

Transposon

(a)

Staggered
cuts

(b)

Joining of nonhomologous strands

(c)

DNA repair replication

Resolution

Transposons

(e)

Figure 14.33 The Shapiro mechanism of transposition.
Staggered cuts are made at the site of transposon insertion
and at either side of the transposon itself (a and b).
Nonhomologous single strands join, resulting in two single-
stranded copies of the transposon in the cointegrate (c). Repair
replication of these single strands produces two copies of the
transposon (d). A crossover at the transposon resolves the
cointegrate into two plasmids, each with a copy of the
transposon (e).

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Transposable Genetic Elements

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Phenotypic and Genotypic Effects
of Transposition

Transposition can have several effects on the phenotype
and genotype of an organism. If transposition takes place
into a gene or its promoter, it can disrupt the expression
of that gene. Depending on the orientation of a transpo-
son, it can prevent the expression of genes. A transposon
can also cause deletions and inversions.

Direct repeats on a chromosome can come about, for
example, by the sequential transposition of the same IS
or transposon, in the same orientation. Pairing followed
by recombination results in a deletion of the section
between the repeats (fig. 14.34). In the case of inverted
repeats, pairing followed by recombination results in an
inversion of the section between the repeats.

Direct repeats

— >■

(a)

— ►

A well-known case of transposon orientation control-
ling a phenotype in bacteria occurs in Salmonella ty-
phimurium. The flagella of this bacterium occur in two
types. Any particular bacterium has either type 1 or type
2 flagella (called phase 1 or phase 2 flagella). The differ-
ence is in the flagellin protein the flagella are composed
of. Phase 1 flagella are determined by the HI gene and
phase 2 flagella are determined by the H2 gene. The
change from one phase to another occurs at a rate of
about 10~ 4 per cell division. After extensive genetic
analysis, the following scheme was suggested and later
verified using recombinant DNA techniques.

The HI and H2 genes are at separate locations on the
bacterial chromosome (fig. 14.35). H2 is part of an operon
that also contains the rHl gene, the repressor of HI. The
promoter of this operon lies within a transposon upstream
of the operon. When the promoter is in the proper orien-
tation, the H2 operon is expressed, resulting in phase 2 fla-
gella. The rHl gene product represses the HI gene (fig.
14. 35a). If the inverted repeat ends of the transposon un-
dergo recombination, the transposon is inverted (see fig.
14.34), moving the promoter into an incorrect orientation
for the transcription of the H2 operon. No HI repressor is
made, so the HI gene is expressed (fig. 14.35&).

As N. Kleckner has summarized, transposons can
have marked effects on the phenotype by their actions in
transposition and by the fact that they may carry genes
valuable to the cell. However, they can also exist without
any noticeable consequences. This fact has led some
evolutionary geneticists to suggest that transposons are
an evolutionary accident that, once created, are self-
maintaining. Since they may exist without a noticeable
benefit to the host's phenotype, transposons have been

Inverted repeats

^ ^ ^.

♦0*

w ^

(b)

^ ^. -^.

Figure 14.34 Pairing and recombination in DNA repeats,
(a) Direct repeats can result in deletion (in the form of a circle)
due to a single crossover, (b) Inverted repeats can result in an
inversion of the region between the repeats due to a crossover.

Transposon

Repression

(a)

(b)

-*»<-

►

rH1

Phase 2

Transcription

Inversion

rH1

Phase 1

No transcription

Transcription

Figure 14.35 Arrangement of flagellin genes on the Salmonella
chromosome. The promoter (p) is within a transposon. In one
orientation (a), the H2 operon is transcribed, which results in
H2 flagellin and rH1 protein, the repressor of the H1 gene. In
the second orientation (b), the H2 operon is not transcribed,
resulting in uninhibited transcription of the H1 gene.

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

Nancy Kleckner (1947- ).
(Courtesy of Nancy Kleckner.
Photo by Stu Rosner.)

referred to as selfish DNA. In recent theoretical and ex-
perimental studies, however, some scientists have sug-
gested that transposons improve the evolutionary fitness
of the bacteria that have them (see chapter 20).

OTHER TRANSCRIPTIONAL
CONTROL SYSTEMS

Transcription Factors

Phage T4

Phage T4, a relatively large phage with seventy-three
genes, has transcription controlled by particular RNA
polymerase specificity factors. Like phage X, phage T4 has
early, middle, and late genes, genes that need to be
expressed in a particular order. Early T4 genes have
promoters whose specificity of recognition depends on
the sigma factor of the host (cr 70 of E. colt). Two products
of early phage genes are the AsiA and MotA proteins. AsiA
binds to the — 35 recognition region of cr 70 , preventing
transcription from both host genes and early phage genes.
AsiA is thus called an anti-sigma factor, a protein that in-
terferes with a sigma factor. Middle phage promoters do
not have — 35 recognition regions but do bind MotA. Host
RNA polymerase bound with the cr 70 -AsiA complex recog-
nizes these promoters. Finally, late phage genes have pro-
moters that depend on the phage-encoded sigma factor
cr 8p55 . Some proteins are needed both early and late in the
infection process; they are specified by genes that have
promoters recognized by different specificity factors.

Heat Shock Proteins

A response to elevated temperature, found in both prokary-
otes and eukaryotes, is the production of heat shock pro-
teins (see chapter 10). In E. coli, elevated temperatures
cause the general shutdown of protein synthesis concomi-
tant with the appearance of at least seventeen heat shock
proteins. These proteins help protect the cell against the
consequences of elevated temperature; some are molecu-
lar chaperones (see chapter 11). The production of these

proteins is the direct result of the gene product of the htpR
gene, which codes for cr 32 . The normal sigma factor is a 70 ,
the product of the rpoD gene; the heat shock genes have
promoters recognized by a 32 rather than a 70 . Heat causes
the htpR gene to become active, as well as stabilizing a 32 .
From DNA sequence data, the difference in promoters be-
tween normal genes and heat shock genes seems to lie in
the —10 consensus sequence (Pribnow box). In normal
genes, it is TATAAT; in heat shock protein genes, it is CCC-
CATXT, in which X is any base.

Promoter Efficiency

In addition to the mechanisms previously described, there
are other ways to regulate the transcription of messenger
RNA. One is to control the efficiency of various processes.
For example, we know that the promoter sequence of dif-
ferent genes in E. coli differs. Since the affinity for RNA
polymerase is different for the different sequences, the
rate of initiation of transcription for these genes also
varies. The more efficient promoters are transcribed at a
greater rate than the less efficient promoters. An example
is the promoter of the / gene of the lac operon. This pro-
moter is for a constitutive gene that usually produces only
about one messenger RNA per cell cycle. However, mu-
tants of the promoter sequence are known that produce
up to fifty messenger RNAs per cell cycle. Here, then, the
transcriptional rate is controlled by the efficiency of the
promoter in binding RNA polymerase. Efficiency can be
controlled by the direct sequence of nucleotides (i.e., dif-
ferences from the consensus sequence) or by the distance
between consensus regions. For example, promoters vary
in the number of bases between the —35 and —10 se-
quences. Seventeen seems to be the optimal number of
bases separating the two. Presumably, more or fewer than
seventeen reduces the efficiency of transcription.

TRANSLATIONAL CONTROL

When considering control of gene expression, it is im-
portant to remember that all control mechanisms are
aimed at exerting an influence on either the amount, or
the activity, of the gene product. Therefore, in addition to
transcriptional controls, which influence the amount of
messenger RNA produced, there are also translational
controls affecting how efficiently the messenger RNA is
translated. (Attenuator control — see fig. 14.16 — can also
be viewed as translational control because the environ-
ment is tested by translation even though attenuation
results in the cessation of transcription.) In prokaryotes,
translational control is of lesser importance than tran-
scriptional control for two reasons. First, messenger
RNAs are extremely unstable; with a lifetime of only

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Translational Control

431

about two minutes, there is little room for controlling the
rates of translation of existing messenger RNAs because
they simply do not last very long. Second, although there
are some indications of translational control in prokary-
otes, such control is inefficient — energy is wasted syn-
thesizing messenger RNAs that may never be used.

Translational control can be exerted on a gene if the
gene occurs distally from the promoter in a polycistronic
operon. The genes that are transcribed last appear to be
translated at a lower rate than the genes transcribed first.
The three lac operon genes, for instance, are translated
roughly in a ratio of 10:5:2. This ratio is due to the polar-
ity of the translation process. That is, in prokaryotes,
translation is directly tied to transcription — a messenger
RNA can have ribosomes attached to it well before tran-
scription ends. Thus, genes at the beginning of the
operon are available for translation before genes at the
end. In addition, exonucleases seem to degrade messen-
ger RNA more efficiently from the 3' end. Presumably,
natural selection has ordered the genes within operons
so that those producing enzymes needed in greater
quantities will be at the beginning of an operon.

Translation can also be regulated by RNA-RNA
hybridization. RNA complementary to the 5' end of a
messenger RNA can prevent the translation of that mes-
senger RNA. The regulating RNA is called antisense
RNA. In figure 14.36, the messenger RNA from the ompF
gene in E. colt is prevented from being translated by com-
plementary base pair binding with an antisense RNA
called micF RNA (mic stands for mRNA-mterfering com-
plementary RNA). The ompF gene codes for a membrane
component called ?l porin, which, as the name suggests,
provides pores in the cell membrane for transport of ma-
terials. Surprisingly, a second porin gene, ompC, seems to
be the source of the micF RNA. Transcription of the op-
posite DNA strand (the one not normally transcribed)
near the promoter of the ompC gene yields the antisense

RNA. One porin gene thus seems to regulate the expres-
sion of another porin gene, for reasons that are not com-
pletely understood. Antisense RNA has also been impli-
cated in such phenomena as the control of plasmid
number and the control of transposon TnlO transposi-
tion. Control by antisense RNA is a fertile field for gene
therapy because antisense RNA can be artificially synthe-
sized and then injected into eukaryotic cells.

A third translational control mechanism consists of
the efficiency with which the messenger RNA binds to
the ribosome. This is related to some extent to the se-
quence of nucleotides at the 5' end of the messenger RNA
that is complementary to the 3' end of the 16S ribosomal
RNA segment in the ribosome (the Shine-Dalgarno se-
quence). Variations from the consensus sequences
demonstrate different efficiencies of binding and, there-
fore, the initiation of translation occurs at different rates.

The redundancy in the genetic code can also play a
part in translational control of some proteins since differ-
ent transfer RNAs occur in the cell in different quantities.
Genes with abundant protein products may have codons
that specify the more common transfer RNAs, a concept
called codon preference. In other words, certain
codons are preferred; they specify transfer RNAs that are
abundant. Genes that code for proteins not needed in
abundance could have several codons specifying the rarer
transfer RNAs, which would slow down the rate of trans-
lation for these genes. The codon distribution of the
phage MS 2 in table 14.2 shows that every codon is used
except the UGA stop codon. (The numbers in the table re-
fer to the incidence of a particular codon in the phage
genome.) However, the distribution is not random for all
amino acids. For example, the amino acid glycine has two
common codons and two rarer codons. The same holds
for arginine but not, for example, valine.

Finally, translational control can be exerted at the ribo-
some. When an uncharged transfer RNA finds its way into

3' end
HO-UUUUUU

c c
W

C-G
A-U
U-A
U-G
C-G
G-C
G-C
U-A
U-A

micF RNA

u
u

A
A

c c

C A
/ \

^ C N

A C

AJ-CX
G A

A C

N,, .• C

G A-A

\j_a'

U-A

G-C

U-A U-A r

C-G A-U

A-U U-A

A-U r U-A

A-Us A °U-A

A-U' U-A

C-G A-Up

CUUUAUCC-C-CAUUUGUCUGUAAGUCUUUACUUACUGCCAUUAUUUA UUACUACUAUCGC

AUAAGAUC UCACUUCCAAAA

5' end

GACGGC-AGUGG-CAGGUG-UCAUAAA

AUGAGGGUAAUAAAU — AAUGAUGA-AGCGCAAUAUUCUGG-C-AGUGAUCG

— GACAGAACUU
5' end

/AC,
A

A A
^A^

c Shine-Dalgarno Initiation
sequence codon

ompFmRNA

CUGCUCUGUUA"
3' end

Figure 14.36 Complementarity between the RNA of the ompF gene and antisense RNA, or micF RNA. The region of overlap
includes the Shine-Dalgarno sequence and the initiation codon, effectively preventing ribosome binding and translation of the ompF
RNA. Notice the Stem-loops on each side Of the overlap. (Reproduced, with permission, from the Annual Review of Biochemistry, Volume 55,
©1986 by Annual Reviews, Inc.)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

the A site of the ribosome, a likely event under amino acid
starvation, it causes an idling reaction in the ribosome,
which entails the production of the nucleotide guanosine
tetraphosphate (5 / -ppGpp-3 / ; fig. 14.37). This is part of a
control mechanism called the stringent response. A pro-
tein called the stringent factor, the product of the relA
gene, produces guanosine tetraphosphate (ppGpp), origi-
nally called magic spot because of its sudden appearance
on chromatograms. (The gene is called rel from the re-
laxed mutant, which does not have the stringent re-
sponse.) The stringent factor is associated with the ribo-
some, where ppGpp is synthesized, although it is not one
of the structural proteins of the ribosome. The SpoT pro-
tein, the product of the spoTl gene, breaks down ppGpp;
thus, the concentration of ppGpp is regulated.

The ppGpp interacts with RNA polymerase, causing
an almost complete cessation of the transcription of ribo-
somal RNA; thus, no energy is wasted synthesizing ribo-
somes when translation is not possible. However, many
amino acid-synthesizing operons require ppGpp for tran-
scription; ppGpp thus inhibits ribosomal RNA production
and enhances the production of enzymes to synthesize
amino acids, all when the cell is starved for amino acids.

One other thing a ribosome can do when faced with
amino acid shortages is to slide past hungry codons,
codons for which a charged transfer RNA is not available.
At that point, the growing peptide chain will be attached
to the last charged transfer RNA to enter the ribosome,
the one now in the peptidyl (P) site. The complex of the
transfer RNA and the ribosome slides down the messen-
ger RNA until it encounters the next codon for the trans-
fer RNA. At this point, it is hoped, the next codon en-

5'-GDP

Stringent factor

ATP

AMP

0" O

P — o — P
o o

OCH

O OH

O^P 0"

— o-

ppGpp

Figure 14.37 The idling reaction. The stringent factor catalyzes
the conversion of GDP to 5'-ppGpp-3'. The added
pyrophosphate groups come from ATP.

Table 14.2

Codon Distribution in MS2, an

RNA Virus

First Positioi

Second Position

Third Position

ti U

Phe

Ser

Tyr

Cys

Phe

Ser

Tyr

Cys

Leu

Ser

stop

Leu

Ser

stop

Trp

Leu

Pro

His

Arg

Leu

Pro

His

Arg

Leu

Pro

Gin

Arg

Leu

Pro

Gin

Arg

Thr

Asn

Ser

Thr

Asn

Ser

Thr

Lys

Arg

Met

Thr

Lys

Arg

Val

Ala

Asp

Gly

Val

Ala

Asp

Gly

Val

Ala

Glu

Gly

Val

Ala

Glu

Gly

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Posttranslational Control

433

countered in the aminoacyl (A) site will code for a
charged transfer RNA present.

POSTTRANSLATIONAL
CONTROL

Feedback Inhibition

Even after a gene has been transcribed and the messenger
RNA translated, a cell can still exert some control over the
functioning of the enzymes produced if the enzymes are
allosteric proteins. We have discussed the activation and
deactivation of operon repressors (e.g., lac, trp) owing to
their allosteric properties. Similar effects occur with
other proteins. The need for posttranslational control is
apparent because of the relative longevity of proteins as
compared with RNA. When an operon is repressed, it no
longer transcribes messenger RNA; however, the messen-
ger RNA that was previously transcribed has been trans-
lated into protein, and this protein is still functioning.
Thus, during operon repression, it would also be efficient
for the cell to control the activity of existing proteins.

An example of posttranslational control occurs with
the enzyme aspartate transcarbamylase, which catalyzes
the first step in the pathway of pyrimidine biosynthesis in
E. colt (fig. 14.38). An excess of one of the end products
of the pathway, cytidine triphosphate (CTP), inhibits the
functioning of aspartate transcarbamylase. This method of
control is called feedback inhibition because a product
of the pathway is the agent that turns the pathway off.

Aspartate transcarbamylase is an allosteric enzyme. Its
active site is responsible for the condensation of carbamyl
phosphate and L-aspartate (fig. 14.38). However, it also has
regulatory sites that have an affinity for CTP. When CTP is
bound in a regulatory site, the conformation of the enzyme
changes, and the enzyme has a lowered affinity for its nor-
mal substrates; recognition of CTP inhibits the condensa-
tion reaction the enzyme normally carries out (fig. 14.39).
Thus, allosteric enzymes provide a mechanism for control
of protein function after the protein has been synthesized.

Protein Degradation

A final control affecting the amount of gene product in a
cell is control of the rate at which proteins degrade. The
normal life spans of proteins vary greatly. For example,
some proteins last longer than a cell cycle, whereas oth-
ers may be broken down in minutes. Several models have
been suggested for control of protein degradation, in-
cluding the N-end rule and the PEST hypothesis.

According to the N-end rule, the amino acid at the
amino-, or N-terminal, end of a protein is a signal to pro-
teases that control the average length of life of a protein.

In recent experiments, the life span of the P-galactosidase
protein was determined with almost complete predictabil-
ity based on its modified N-terminal amino acid. Protein life
spans range from two minutes for those with N-terminal
arginine to greater than twenty hours for those with N-
terminal methionine or five other amino acids (table 14.3).
According to the PEST hypothesis, protein degrada-
tion is determined by regions rich in one of four amino
acids: proline, glutamic acid, serine, and threonine. (The
one-letter abbreviations of these four amino acids are P, E,
S, and T, respectively.) Proteins that have these regions
tend to degrade in less than 2 hours. In one study of
thirty-five proteins with half-lives of between 20 and 220
hours, only three contained a PEST region. We see that
not only are different proteins programmed to survive for
varying lengths of time in the cell, but that programming
seems to be based on the N-terminal amino acid as well
as various regions rich in the PEST amino acids.

Pyrimidines,
TTP. CTP

Carbamyl phosphate + L-aspartate

Inhibits Aspartate
>► transcarbamylase

N-Carbamyl-L-aspartate

Inorganic phosphate

Figure 14.38 Aspartate transcarbamylase catalyzes the first
step in pyrimidine biosynthesis. An end product, cytidine
triphosphate (CTP), inhibits the enzyme.

TablG 14.3 Relationship Between N-Terminal

Amino Acid and Half-Life of E. colt
p-Galactosidase Proteins with Modified
N-Terminal Amino Acids

N-Terminal Amino Acid

Half-Life

Met, Ser, Ala, Thr, Val, Gly

>20 hours

lie, Glu

30 minutes

Tyr, Gin

10 minutes

Pro

7 minutes

Phe, Leu, Asp, Lys

3 minutes

Arg

2 minutes

Source: Data from Bachmair, et al., Science, 234:179-86, 1986.

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

Figure 14.39 Aspartate transcarbamylase. Left: the enzyme bound with cytidine triphosphate (CTP). Right: the enzyme without CTP.
Notice the difference in shape with and without CTP. With CTP, the enzyme literally closes up. (Courtesy of Evan R. Kantrowitz.)

SUMMARY

STUDY OBJECTIVE 1: To study the way in which in-
ducible and repressible operons work 406-414

Most bacterial genes are organized into operons, which can
either be repressed or induced. Transcription begins in
inducible operons, such as lac, when the metabolite that
the operon enzymes act upon appears in the environment.
The metabolite (or a derivative), the inducer, combines
with the repressor (the product of the independent regula-
tor gene) and renders the repressor nonfunctional. In the
absence of the inducer, the repressor binds to the operator,
a segment between the promoter and the first gene of the
operon. When in place, the repressor blocks transcription.
After combining with the inducer, the repressor diffuses
from the operator, and transcription proceeds.

All operons responsible for the breakdown of sugars in E.
coli are inducible. In the presence of glucose, other inducible
sugar operons (such as the arabinose and galactose operons)
are repressed, even if their sugars appear in the environ-
ment. This process is called catabolite repression. Cyclic
AMP and a catabolite activator protein (CAP) enhance the

transcription of the nonglucose sugar operons. Glucose low-
ers the level of cyclic AMP in the cell and thus prevents the
enhancement of transcription of these other operons.

Repressible operons, such as the trp operon in E. coli,
have the same basic traits as an inducible operon —
polycistronic transcription controlled by an operator site
between the promoter and the first structural gene. How-
ever, the repressor protein, controlled by an independent
regulator gene, is functional in blocking transcription only
after it has combined with the corepressor This corepres-
sor is the end product of the operon's pathway or some
form of the end product (tryptophan in the trp operon).

STUDY OBJECTIVE 2: To examine attenuator control in
bacteria 415-418

Amino acid-synthesizing operons often have an attenuator
region. The ability of a ribosome to translate a leader
peptide gene determines the secondary structure of the
messenger RNA transcript. If the ribosome can translate the
leader peptide gene, there must be adequate quantities of

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Solved Problems

435

the amino acid present, and a terminator stem and loop
form in the messenger RNA, causing termination of tran-
scription. In the trp operon of bacilli, a protein that binds
tryptophans serves the same purpose.

STUDY OBJECTIVE 3: To analyze the control of the life cy-
cle of phage X 418-424

Control of gene expression in X phage is complex. The "de-
cision" for lytic versus lysogenic response is determined by
competition between two repressors, CI and Cro.

STUDY OBJECTIVE 4: To determine the way in which
transposable genetic elements transpose and control
gene expression in bacteria 425-430

Transposons are mobile genetic elements; copies of them
can be inserted at other places in the genome. Their ends
are inverted repeats. Upon insertion, they are flanked by
short, direct repeats. They can be simple (IS elements) or

complex. Their presence can cause inversions or deletions.
The flagellar phase in Salmonella is controlled by the ori-
entation of a transposon.

STUDY OBJECTIVE 5: To look at other transcriptional and
posttranscriptional mechanisms of control of gene ex-
pression in bacteria and phages 430-434

Affinity of early and late operons in phages for different
sigma factors are another form of transcriptional control, as
seen in phage T4 transcription and the transcription of heat
shock proteins. Translational control can be exercised
through a gene's position in an operon (genes at the begin-
ning are transcribed most frequently), through redundancy
of the genetic code, or through a stringent response that
shuts down most transcription during starvation. Posttrans-
lational control is primarily regulated by feedback inhibi-
tion. The N-terminal amino acid or particular regions within
the proteins program the rate of protein degradation.

SOLVED PROBLEMS

PROBLEM 1: How could you determine whether the
genes for the breakdown of the sugar arabinose are un-
der inducible control in E. colt?

Answer: Inducible means that the genes to break down
the substrate — arabinose, a five-carbon sugar, in this
case — are not active in the absence of the inducer
(again, arabinose). Therefore, in the absence of arabinose
in the cells' environment, the arabinose utilization en-
zymes should not be active within the bacterial cells, but
after arabinose is added to the medium, the enzymes
should be present. We thus need to assay the contents of
the cells before and after arabinose is added to the
medium, performing the assay after the cells are broken
open and the DNA destroyed so as not to confound the
experiment. Using a standard biochemical analysis for
arabinose, we should find that the bacterial cell is inca-
pable of metabolizing arabinose before induction but ca-
pable of metabolizing it afterward. If the cells were capa-
ble of metabolizing arabinose in both cases, we would
say that arabinose utilization is constitutive. If the cells
were incapable of utilizing arabinose in both cases, we
would conclude that the bacterium is incapable of using
the sugar arabinose as an energy source. (In fact, arabi-
nose utilization is inducible.)

PROBLEM 2: Why would the RecA protein of E. coli
cleave the X repressor?

Answer: Since the cleaving of the X repressor is a signal
to begin the lytic phase of the life cycle of the phage, it
seems odd that the lysogenized bacterial cell would be

an accomplice to its own destruction. However, the phe-
nomenon makes much more sense if we realize that the
RecA protein has several other functions critically im-
portant to the bacterial cell (see chapter 12). The X
phage has evolved the ability to take advantage of the
existence of the RecA protein by evolving a repressor
sensitive to it. Evolutionary biologists view this as "co-
evolution," two interacting organisms evolving to take
advantage of or minimize properties of the other. The
bacterium, however, might be at a disadvantage. Since
RecA has many functions involving interactions with
other proteins, it may be highly limited in how it can
change. This is one plausible explanation as to why RecA
liberates phage X.

PROBLEM 3: What are the differences in action of the X
promoters p RE and^?^?

Answer: The promoters p^ and^? i?M are both promoters
of the repressor operon of phage X. Transcription from
these promoters allows production of the cl repressor
protein, the repressor that favors lysogeny. Initially, the
promoter pj^ is activated. For it to be a transcription site,
it must be activated by the product of the ell gene, which
lies in the right operon. This promoter, p^, produces a
messenger RNA with a long leader that is translated very
efficiently. Once the repressor binds at the operators of
the left and right operons, the ell gene is no longer tran-
scribed, and therefore p^ is no longer a site for transcrip-
tion. However, the repressor gene can still be transcribed
from the p km promoter, which does not need the product

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

of the ell gene. This promoter produces a transcript with
no leader and thus is translated very inefficiently. At that
point, however, only a very small quantity of repressor is
needed to maintain lysogeny.Thus, the two promoters are
the sites for the initiation of the repressor operon under
different circumstances: one early in the infection stage
and one after lysogeny is under way.

PROBLEM 4: What are the phenotypes of the following
partial diploids for the lac operon in E. colt in the pres-
ence and absence of lactose?

a. (F')z + o + p + z /i

o + p + z + (chromosome)

+ _.+ ,,+ +

b. (F') C o c p^ z^/i o p

z (chromosome)

Answer: Consider one DNA molecule at a time. If one
DNA molecule can never make the enzyme, it can be ig-
nored. In (a), the plasmid DNA (F') will never make en-
zyme (it is z), and the chromosomal DNA will never
make repressor (it is O- The functional repressor in the
plasmid (i + ) will bind to both DNAs, and hence the chro-
mosomal operon will not be transcribed in the absence
of lactose and will be induced to transcribe in the pres-
ence of lactose. In (£>), the plasmid DNA (F') will always
be transcribing (operator constitutive) because the re-
pressor can never bind the operator (o c ); hence, the
operon will be transcribed all the time. The chromo-
somal DNA can never make RNA (it isp~).

EXERCISES AND PROBLEMS

LAC OPERON (INDUCIBLE SYSTEM)

1. Are the following E. colt cells constitutive or in-
ducible for the z gene?

a. /

b. i~

c. i~

d. t

e. f

f. fi

+ +

o z
o c z

+ +

o z

2. Determine whether the following lac operon
merozygotes are inducible or constitutive for the z

gene.
a. i +

+ +

cT zV¥' i

b. /

c. i

d. i

e. i

o
o

z + /F
z + /F'

i + o

z
z
z

o z

z /V
o z~/V i o~ z

3. You have isolated a repressor for an inducible
operon and have determined that it has two differ-
ent binding sites, one for the inducer and one for the
operator. Mutants of the repressor result in three dif-
ferent phenotypes as far as binding is concerned.
What are these phenotypes?

4. An E. colt strain is isolated that produces (3-
galactosidase (lac z) and permease (lac y) constitu-
tively Provide two possible mutations that could
cause this phenotype, and then describe how each
mutation would behave in a partial diploid in which
the second operon is wild-type for the entire lac
system.

5. You have isolated two E. colt mutants that synthesize
p-galactosidase constitutively

* Answers to selected exercises and problems are on page A-16.

a. If these mutants affect different functions, in what
two functions could they be defective?

b. You can make a partial diploid of the mutants
with the wild-type. What result do you expect for
each mutant?

6. A hypothetical operon has a sequence of sites, Q R S
T U, in the promoter region, but the exact location of
the operator and promoter consensus sequences
have not been identified. Various deletions of this
operator region are isolated and mapped. Their loca-
tions appear as follows, with a "/" representing a
deleted region.

Deletion

////////

4
5

////////

/////////////////

mini

Deletions 3 and 4 are found to produce constitutive
levels of RNA of the operon, and deletion 1 is found
to never make RNA. Where are the operator and
promoter consensus sequences probably located?

CATABOLITE REPRESSION

7. Describe the role of cyclic AMP in transcriptional
control in E. colt.

8. Operon systems exert negative control by acting
through inhibition. The CAP system exerts positive
control because it acts through enhancement of
transcription. Describe how an operon could work if
it were dependent only upon positive control.

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Exercises and Problems

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9. J. Beckwith isolated point mutations that were si-
multaneously uninducible for the lac, ara, mal, and
gal operons, even in the absence of glucose. Provide
two different functions that could be missing in
these mutants.

TRP OPERON (REPRESSIBLE SYSTEM)

10. Construct a merozygote of the trp operon in E. colt
with two forms of the first gene (e gene: e 2 , e 2 ) in the
operon. Describe the types of cis and trans effects
that are possible, given mutants of any component of
the operon. Can this repressible system work for any
type of operon other than those that control amino
acid synthesis?

11. The tryptophan operon is under negative control; it
is on (transcribing) in the presence of low levels of
tryptophan and off in the presence of excess trypto-
phan. The symbols a, b, and c represent the gene for
tryptophan synthetase, the operator region, and the
repressor — but not necessarily in that order. From
the following data, in which superscripts denote
wild-type or defective, determine which letter is the
gene, the repressor, and the operator (+ is trypto-
phan synthetase activity; — is no activity).

Tryptophan

Strain

Genotype

Absent

Present

a~ b + c +

a b c

a + b~ c +

—

a + b~ c + /a~

b +

a b c la

—

a + b + c~/a~

c +

—

a b c la

12. The histidine operon is a repressible operon. The
corepressor is charged tRNA Hls , and its gene is not part
of the operon. For the following mutants, tell whether
the enzymes of the operon will be made; then tell
whether each mutant would be czs-dominant in a
partial diploid.

a. RNA polymerase cannot bind the promoter.

b. The repressor-corepressor complex cannot bind
operator DNA (the operator has the normal se-
quence).

c. The repressor cannot bind charged tRNA Hls .

TRP OPERON (ATTENUATOR-CONTROLLED SYSTEM)

13. Describe the interaction of the attenuator and the
operator control mechanisms in the trp operon of E.
colt under varying concentrations of tryptophan in
the cell. How does attenuator control react to short-
ages of other amino acids?

LYTIC AND LYSOGENIC CYCLES IN PHAGE X

14. What is the fate of a X phage entering an E. colt cell
that contains quantities of X repressor? What is the
fate of the same phage entering an E. colt cell that
contains quantities of the cro-gene product?

15. Describe the fate of X phages during the infection
process with mutants in the following genes: cl, ell,
cIII, N, cro, att, Q.

16. What is the fate of X phages during the infection
process with mutants in the following areas: o R1 ,
o R3 , Pl, Pre, Prm, Pr, t L1 , t R1 , nutl, nutR?

17. What are the three different physical forms that the
phage X chromosome can take?

18. How does ultraviolet light (UV) damage induce the
lytic life cycle in phage X?

19. The X prophage is sometimes induced into the lytic
life cycle when an Hfr lysogen (lysogenic cell) con-
jugates with a nonlysogenic F" cell. How might in-
duction come about in this instance?

20. A temperature-sensitive mutant of the X cl gene has
been isolated. At 30° C the cl repressor binds X DNA,
but it cannot bind DNA at 42° C (it denatures). What
is the consequence of incubating E. colt that are lyso-
genic for this X mutant at 42° C?

21. The mutant in problem 20 is heated to 42° C for five
minutes, cooled to 30° C, and grown for one hour so
that the cells divide several times. The temperature is
then raised to 42° C, and you wait for lysis. Many of
the cells are not lysed and are in fact able to form
colonies. Explain these results.

TRANSPOSABLE GENETIC ELEMENTS

22. Why are IS elements sometimes referred to as
"selfish DNA"?

23. What are the differences among an IS element, a
transposon, an intron, a plasmid, and a cointegrate?

24. Describe the Shapiro model of transposition. What
are the roles of transposase, DNA polymerase I, ligase,
and resolvase?

25. Why are transposons flanked by direct repeats?

26. How do transposons induce deletions? inversions?

27. Describe how a transposon controls the expression
of the flagellar phase in Salmonella.

28. What is a polar mutation? What can cause it?

OTHER TRANSCRIPTIONAL CONTROL SYSTEMS

29. List the steps from transcription through translation
to enzyme function, noting all the points at which
control could be exerted. {See also TRANSLA-
TIONAL CONTROL and POSTTRANSLATIONAL
CONTROL)

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Chapter Fourteen Gene Expression: Control in Prokaryotes and Phages

30. What are the advantages of transcriptional control
over translational control? {See also TRANSLA-
TIONAL CONTROL)

31- How are heat shock proteins induced?

32. In phage T4, the genes rllA and rllB lie adjacent to
each other on the T4 chromosome. During the early
phase of infection, rllA and rllB products are present
in equimolar amounts. In the late phase of infection,
the amount of rllB protein is ten to fifteen times
higher than that of rllA protein. Nonsense mutations
(mutations to a stop codon) in rllA eliminate early but
not late rllB transcription. In the mutants that contain
small deletions near the end of rllA, the amount of
rllA product is always equal to the amount of rllB
product, regardless of the time of infection. Based on
this information, devise a map of the rll region. In-
clude the location(s) of the promoter(s).

TRANSLATIONAL CONTROL

33. What is the stringent response? How does it work?
What is an idling reaction?

34. What is antisense RNA? How does it work? What is
the obvious source of this regulatory RNA? How
could this RNA be used to treat a disease clinically?

POSTTRANSLATIONAL CONTROL

35. What is feedback inhibition? What other roles do al-
losteric proteins play in regulating gene expression?

36. What controls the rate of degradation of proteins?

CRITICAL THINKING QUESTIONS

1. From an evolutionary perspective, why do you think Es-
cherichia coli evolved a CAP system of positive control
of gene expression? Why not just metabolize any and all
sugars in the environment as they appear?

2. Why might some proteins and messenger RNAs pro-
duced in Escherichia coli be degraded so quickly?

Suggested Readings for chapter 14 are on page B-13-

Tamarin: Principles of
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Molecular Genetics

15. The Eukaryotic
Chromosome

©TheMcGraw-Hil
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THE EUKARYOTIC
CHROMOSOME

STUDY OBJECTIVES

1. To examine the arrangement of DNA and proteins comprising
the eukaryotic chromosome 440

2. To look at the nature of centromeres and telomeres in
eukaryotic chromosomes 453

3. To analyze the nature of the DNA in eukaryotic
chromosomes 457

Artificially colored scanning electron micrograph

of part of a polytene salivary gland chromosome

from a fruit fly (Drosophila), revealing the underlying

banding pattern. (© Professors P. Motta and T. Naguro/

SPL/Photo Researchers, inc.)

STUDY OUTLINE

The Eukaryotic Cell 440

The Eukaryotic Chromosome 440

DNA Arrangement 440

Nucleoprotein Composition 442

Chromosomal Banding 451

Centromeres and Telomeres 453

The C-Value Paradox 457
Summary 461
Solved Problems 462
Exercises and Problems 462
Critical Thinking Questions 463
Box 15.1 How Big Is Big, How Small Is Small? 441
Box 15.2 High-Speed Chromosomal Sorting AW

439

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Molecular Genetics

15. The Eukaryotic
Chromosome

©TheMcGraw-Hil
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440

Chapter Fifteen The Eukaryotic Chromosome

In chapter 14, we looked at the control of gene
expression in prokaryotes and bacteriophages.
Compared to eukaryotes, bacteriophages and pro-
karyotes are relatively simple. Of fundamental
importance is that, in these lower forms, the operon
model of induction and repression of transcription is a uni-
fying theme for control of gene expression. Despite nu-
ances such as catabolite repression and attenuator control,
the operon model provides a relatively clear picture of how
genes are turned on and off in phages and prokaryotes. This
model does not exist for eukaryotes. In attempting to elu-
cidate models for control of gene expression in eukaryotes,
we must take one very important factor into account: the
complexity of the structure of the eukaryotic chromo-
some. In this chapter, we cover the current understanding
of how these very large structures are organized.

THE EUKARYOTIC CELL

Eukaryotes and prokaryotes are the two superkingdoms
of organisms. The following comparisons, using E. colt as
a general model for prokaryotes, show how much more
complex eukaryotes are:

1. An E. colt chromosome contains approximately 4.2 X
10 6 base pairs of DNA. The haploid human genome
contains nearly one thousand times as much DNA.

2. Eukaryotic DNA is in the form of nucleoprotein, a
DNA-histone protein complex. Although a few his-
tonelike proteins have been found in E. colt, its chro-
mosomal DNA is not complexed with protein to any-
where near the same extent.

3. An E. colt cell has very little internal structure. Eu-
karyotes have a number of internal organelles and an
extensive lipid membrane system, including the nu-
clear envelope itself.

4. An E. colt cell is small (0.5 to 5.0 jxm in length for bac-
teria). Eukaryotic cells are generally larger than
prokaryotes (10 to 50 jim in length for animal tissue
cells; box 15.1).

5. The messenger RNA of E. colt is translated while it is
being transcribed. Eukaryotic messenger RNA is mod-
ified within the nucleus before it is transported out
for translation in the cytoplasm.

6. Almost no messenger RNA isolated from eukaryotic
cells, including the messenger RNA of animal viruses,
has been found to be polycistronic (containing many
genes). Most prokaryotic messenger RNAs are poly-
cistronic.

7. Most E. colt genes are parts of inducible or repressible
operons; there are almost no operons in eukaryotes.

8. E. colt exists as a simple, single cell. Although some
prokaryotes do aggregate, sporulate, and show a few

other limited forms of differentiation, they are pri-
marily one-celled organisms. And, although some eu-
karyotes are single-celled (e.g., yeast), the essence of
eukaryotes is differentiation. In human beings, a zy-
gote gives rise to every other cell type in the body in
a relatively predictable manner.

To fully appreciate the complexity of eukaryotes, we
begin by looking at the eukaryotic chromosome. In the
next chapter, we look at the patterns of development in
eukaryotes and some mechanisms of control of gene
expression.

THE EUKARYOTIC
CHROMOSOME

DNA Arrangement

Evidence that the eukaryotic chromosome is uninemic —
that is, contains one double helix of DNA — comes from
several sources. The best data are provided by radioactive-
labeling studies, first done by J.Taylor and his colleagues
in 1957. If a eukaryote is allowed to undergo one DNA
replication in the presence of tritiated ( 3 H-) thymidine,
each of the daughter chromatids would be expected to
contain a double helix with one unlabeled DNA template
strand and one labeled strand of newly synthesized bases
(fig. 15.1). This configuration is expected on the basis of
semiconservative replication, with each chromatid con-
taining one double helix. A second round of DNA replica-
tion, in the absence of 3 H-thymidine, should produce
chromosomes in which one chromatid would have unla-
beled DNA and one would have labeled DNA. Figure 15.2
shows the chromosomes after this second replication in
nonlabeled media. As expected, one chromatid of every
pair is labeled and one is not.

In another kind of experiment, R. Kavenoff, L. Klotz,
and B. Zimm demonstrated that Drosophila nuclei con-
tained pieces of DNA of the size predicted from their
DNA content, based on the premise that each chromo-
some contains one DNA molecule. They isolated the
DNA and measured the size of the largest DNA molecules
using the viscoelastic property of DNA, the rate at which

Ruth Kavenoff (1944- ).
(Courtesy of Dr. Ruth
Kavenoff.)

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The Eukaryotic Chromosome

441

BOX 15.1

Generally, eukaryotic cells are
large, and prokaryotic cells
are small. For example, an av-
erage eukaryotic cell is about 50 |jim
in diameter, whereas an average bac-
terium is about 5 |xm in length. The
average virus is about 0.05 |xm in di-
ameter. These size differences occur
because eukaryotic cells have com-
plex substructures and internal archi-
tecture that prokaryotic cells lack.
Since we believe that prokaryotic
cells depend on diffusion to ex-
change materials with the environ-
ment, they would have to be small.
And viruses, intracellular parasites,
would of necessity be very small.
There are, of course, exceptions.

In 1999, a team of scientists from
Germany, Spain, and the United States
isolated large sulfur bacteria off the
Namibian coast of Africa and named
them Thiomargarita namibiensis,
the sulfur pearl of Namibia. These bac-
teria can be almost half a millimeter in
diameter, the size of the period at the
end of this sentence (fig. 1). Based on
the sequence of 16S ribosomal DNA,
these bacteria were shown to be
close relatives of other marine sulfur
bacteria. They are almost one hun-
dred times the volume of the bacteria
previously believed to be largest,
Epuliscium flshelsoni, known only
from the intestine of the brown sur-
geonfish.

The smallest prokaryotes are the
Mycoplasmas, at about 0.2 |jim in di-
ameter, rivaling the viruses in size.
They are animal pathogens and de-
composing organisms. The smallest
eukaryote, Ostreococcus tauri, a green
alga found in the plankton, was dis-
covered in 1994 from a water sample
in a French lagoon on the Mediter-
ranean Sea by a group of French sci-

Experimental
Methods

How Big Is Big, How
Small Is Small?

entists. These organisms are less than
1 ixm in diameter. Scientists believe
that the lower limit on the size of
a cell (not counting viruses) is about
200 nm (0.2 |Jim), based on the size
of DNA and ribosomes that a cell
must contain.

With T namibiensis as the largest
prokaryote, we note that the largest
eukaryotic cell with a single nucleus
is most likely the ostrich egg. The
largest organisms are the blue whale,
Balaenoptera musculus, weighing in

at 118,000 kilograms; giant redwood
trees, Sequoiadendron giganteum,
100 meters tall and weighing 5.5 mil-
lion kilograms; a quaking aspen
clone, Populus tremuloides, weigh-
ing 6 million kilograms; and Armil-
laria bulbosa, a fungus. In 1992,
three scientists from the University of
Toronto and Michigan Technological
University, using restriction fragment
length polymorphisms (RFLPs) and
polymerase chain reaction (PCR)
techniques, showed that the huge hy-
phal mass of this tree-root colonizing
fungus growing in a forest in north-
ern Michigan was a single organism.
It covered about eight hectares, prob-
ably weighed more than 10,000 kilo-
grams, and probably has existed for
more than 1,500 years.

Although we don't want to get dis-
tracted by the oddities and extremes of
nature, size differences are remarkable.

Figure 1 The bacterium Thiomargarita namibiensis shown with a fruit fly
(Drosophila virilis, 3 mm in length) for size comparison. The arrow points to a
single bacterial cell, 0.5 mm wide, bright with sulfur inclusions. Above the cell
are empty sheaths of dead bacteria. (From h.n. Schuiz, et ai., "Dense populations of a
giant sulfur bacterium in Namibian Shelf Sediments" in Science, Vol. 284, pp. 493-95, April 16,
1999. Reprinted by permission of the American Association for the Advancement of Science.)

stretched molecules relax. From other sources, primarily
UV absorbance studies, it was estimated that the largest
Drosophila chromosome had about 43 X 10 9 daltons of
DNA. Results from the viscoelastic measurements indi-
cated the presence of DNA molecules of between 38 and

44 X 10 daltons. Viscoelastic measurements of inver-
sions, which changed the ratio of the arms but not the
overall size of the chromosome, yielded similar results.
However, a translocation that radically changed the size
of the chromosome to 59 X 10 9 daltons resulted in an

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Chapter Fifteen The Eukaryotic Chromosome

DNA

Chromosome

Replication in
3 H-thymidine

Separation

Replication in
unlabeled medium

Figure 15.1 Radioactive labeling of a uninemic eukaryotic
chromosome following semiconservative replication. Replication
occurs first in the presence of 3 H -thymidine and then in its
absence. Red represents labeling. After the second round of
replication, one chromatid of each chromosome is labeled,
whereas the other is not, confirming that there is only one DNA
molecule per chromatid and that the chromosome is thus
uninemic.

equivalent change in the viscoelastic estimates to be-
tween 52 and 64 X 10 9 daltons.

The conclusion from these studies is that the largest
Drosophila chromosome, and by extension every eu-
karyotic chromosome, contains a single DNA molecule
running from end to end, encompassing both arms. The
viscoelastic values were corroborated by carefully isolat-
ing and measuring the lengths of long DNA molecules, an
especially difficult task given DNAs propensity to break.

%vi

•^i*

~ *•

■ H

v* * ■«** £

Figure 15.2 Second metaphase in hamster cells in culture
after one replication in the presence of 3 H-thymidine followed
by one in nonradioactive medium, verifying the uninemic
nature of the eukaryotic chromosome. Cases in which the
label apparently switches from one chromatid to the other are
caused by sister chromatid exchanges (at arrows). (Source:
G. Marin and D. M. Prescott, "The frequency of sister chromatid exchanges
following exposure to varying doses of 3 H-thymidine or X-ray," Journal of Cell
Biology, 21, (1964): 159-67, by copyright permission of the Rockefeller
University Press.)

The longest molecule that the investigators found was
1.2 cm long, equivalent to between 24 and 32 X 10 9 dal-
tons (fig. 15.3), close to the predicted size. Thus, the evi-
dence is in complete concordance with the simple
uninemic model of eukaryotic chromosomal structure
(box 15.2).

Nucleoprotein Composition

Nucleosome Structure

Since each eukaryotic chromosome consists of a single,
relatively long piece of duplex DNA, the average diploid
cell contains many of these long pieces of DNA. For chro-
mosomes to be properly distributed to each daughter
cell during mitosis and meiosis, they must be condensed
into structures that are more easily managed. Wrapping
the DNA around "spools" of protein constitutes the first
step in a series of coiling and folding processes that even-
tually result in the fully compacted chromosome we see
at metaphase.

Interphase nuclei can be disrupted by placing them
in a hypotonic liquid such as water. When this happens,
chromatin material is released. When this material is ob-
served under the electron microscope, small particles

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Chromosome

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The Eukaryotic Chromosome

443

Figure 15.3 Auto radiograph of a 1.2 cm radioactive DNA
molecule carefully isolated from Drosophila melanogaster
chromosomes. Drops of DNA solution were placed on
microscope slides, then tilted to allow the DNA to spread
slowly down the slide. A photographic emulsion was applied
and later developed after a five-month exposure period.

(From Ruth Kavenoff, Lynn C. Klotz, and Bruno H. Zimm, Symposia on
Quantitative Biology (Cold Spring Harbor), 38(1973):4.)

I 1 mm

called nucleosomes can be seen (fig. 15.4). These are
the spools that the DNA is wrapped around. They are
made of histone proteins and associated DNA (table
15.1). The histones, a group of arginine- and lysine-rich
basic proteins, have been well characterized. They are es-
pecially well suited to bind to the negatively charged
DNA (table 15.2).

When chromatin is treated with micrococcal nucle-
ase, individual nucleosomes can be isolated, indicating
that the DNA between nucleosomes is accessible to di-
gestion. The results of these studies indicate that a length
of 168 base pairs (bp) of DNA, the core DNA, is inti-
mately associated with the nucleosome, and another 50
to 75 base pairs, depending on species, connects the nu-
cleosomes (linker DNA; fig. 15.5). When the quantities of
the various histones were measured, there were two
each of histones H2A, H2B, H3, and H4 per nucleosome
and only one molecule of histone HI. Reconstitution and
degradation studies have indicated that histone HI is not
a necessary component in the formation of nucleo-
somes. We believe that histone HI is associated with the
linker DNA as it enters and emerges from the nucleo-
some (fig. 15.6), although its exact position is not known
with certainty. Histone HI may be more off center and in-
ternally located than illustrated. The term chromato-
some has been suggested for the core nucleosome plus
the HI protein, a unit that includes approximately 168
base pairs of DNA. Nucleosomes, then, are a first-order
packaging of DNA; they reduce its length and undoubt-
edly make the coiling and contraction required during
mitosis and meiosis more efficient (fig. 15.7).

When DNA is replicated, twice as many nucleosomes
are needed since one double helix becomes two. Recent
studies indicate that a parental nucleosome is partly dis-
assembled during DNA replication and reassembled on
one or the other daughter strand, apparently randomly.
The other DNA strand has a new nucleosome constructed

&-■ ®

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Figure 15.4 Electron micrograph of chromatin fibers. Photo
shows nucleosome structures {spheres) and connecting strands
of DNA called linkers. The bar is 100 nm long. (Source: D. E.
Olins and A. L. Olins, "Nucleosomes: The structural quantum in
chromosomes," American Scientist, 66: 704-11, November 1978.
Reproduced by permission.)

Table 15.1 The Constituency of Calf

Thymus Chromatin

Constituent

Relative Weight*

DNA

100

Histone proteins

114

Nonhistone proteins

RNA

Weight relative to 100 units of DNA.

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Chapter Fifteen The Eukaryotic Chromosome

BOX 15.2

To facilitate the creation of re-
combinant genomic libraries,
for mapping purposes, and for
other reasons, it is useful to be able to
isolate individual human chromo-
somes. To these ends, several methods
have been developed to isolate chro-
mosomes. Here we discuss a high-
speed sorting method based on fluo-
rescent staining and flow cytometry.

DNA can be treated with several
fluorescent dyes. Chromosomes can
then be recognized individually by
their relative fluorescent intensities.
The dyes Hoechst 33258 and chro-

Experimental
Methods

High-Speed Chromosomal
Sorting

momycin A3 are a valuable combina-
tion because they respond to differ-
ent wavelengths of light and they
bind DNA differently. Hoechst binds
preferentially to DNA rich in adenine
and thymine, whereas chromomycin

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Chromomycin A3

fluorescence intensity

Mini

mum Maximum

Figure 1 Flow karyotype of human chromosomes at very high resolution,
measured under low-speed sorting (fifteen to thirty-five chromosomes per
second). The ordinate is Hoechst 33258 fluorescence intensity, and the
abscissa is chromomycin A3 fluorescence intensity. All chromosomes are
resolved except numbers 9-12. (Reprinted with permission from J. W. Gray, et al.,
"High-Speed Chromosome Sorting," Science, 238:323-329, 1987. Copyright © 1987 American
Association for the Advancement of Science.)

binds preferentially to DNA rich in
guanine and cytosine. Thus, since
every human chromosome has a
unique ratio of bases, the relative in-
tensity of each chromosome is differ-
ent when fluoresced.

Chromomycin fluoresces in the
presence of a laser tuned to 458 nm,
and Hoechst fluoresces in the pres-
ence of a UV laser. The chromosomes
can be identified when their relative
fluorescence in the two lasers is plot-
ted, producing a flow karyotype
(fig. 1). Modern flow cytometry tech-
niques then allow the isolation of
these identified chromosomes.

In practice, chromosomes are iso-
lated in large numbers from cells that
have been arrested in metaphase by
treatment with colcemid, which in-
hibits spindle formation. These chro-
mosomes are then purified in buffer
and treated with the two dyes. The
chromosomes are separated at high
speed (two hundred chromosomes
per second) in a flow cytometry de-
vice (fig. 2). As the chromosome-
containing buffer passes through the
laser beams, identification is made.
The liquid is then forced to form
minute droplets (215,000 per sec-
ond) by passing through a vibrator.
Specific droplets carrying the identi-
fied chromosomes are then charged,
either positively or negatively, and
passed between deflection plates.
Positively charged droplets pass one
way, and negatively charged droplets
pass the other way, thus allowing the
simultaneous isolation of two differ-
ent chromosomes. At a rate of two
hundred chromosomes per second, it

of histones from the cellular pool with the help of pro-
teins called chromatin assembly factors; at least three
of these factors are known.

For example, in fruit flies, a protein complex called
the replication-coupling assembly factor assembles
new nucleosomes. In addition, a protein complex called
condensin is needed for the condensation of interphase

chromosomes to mitotic chromosomes. This complex in-
cludes two SMC proteins (for structural maintenance of
chromosomes) and two non-SMC proteins. SMC proteins
also aid other chromosomal activities, such as mitotic
segregation, sister-chromatid adhesion, dosage compen-
sation, and recombination. Thus, a diverse array of pro-
teins is involved in creating nucleosomes and chromato-

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445

is possible to isolate 0.1 g of DNA in
less than an hour; 0.1 g of DNA is ad-
equate for library construction and
represents about 5 X 10 5 average
chromosomes.

The technique is not perfect. Dur-
ing isolation, debris and clumps of
chromosomes are produced that

cause contamination problems. Then,
some chromosomes are so similar in
their fluorescence that they are hard
to separate. This is true, for example,
for chromosomes 9 to 12. Also, chro-
mosome 21 is hard to separate be-
cause its fluorescence tends to fall
into the debris area.

Some of these problems, however,
can be overcome by using hybrid cell
lines of hamsters, for example, con-
taining only one human chromosome.
It is much easier to isolate the human
chromosome from the hybrid line. Pu-
rity values of 90% are not unreason-
able, with some in excess of 95%.

< — Chromosomes

Sample

Negatively
charged
deflection
plate

Collection tube
(charged droplets)

Flask for

undeflected

droplets

Figure 2 The flow cytometry device used to separate chromosomes at high speed. A buffer with chromosomes enters the
device. Lasers cause fluorescence that is analyzed with the aid of the photomultiplier tubes. Droplet formation is induced by
vibration, and, based on a flow rate of 50 m/sec, appropriate drops are charged. Charged drops are then separated by
charged deflection plates and collected. Uncharged droplets pass through. (Reprinted with permission from J. w. Gray, et ai.,
"High-Speed Chromosome Sorting," Science, 238:323-329, 1987. Copyright © 1987 American Association for the Advancement of Science.)

somes, condensing interphase chromosomes, and per-
forming numerous other activities of chromosomes.

Dosage compensation has recently been associated
with a change in nucleosome structure. The inactivated X
chromosome appears to have a different type of histone
present. Histone H2A is replaced by a variant called
mH2A. The details of this mechanism are under study

Nucleosomes apparently play a major role in con-
trolling gene expression; DNA with nucleosomes has a
much lower transcription rate than DNA without nucle-
osomes. It makes sense that the positions of nucleo-
somes can provide or prevent access to promoters.
There are regions of the DNA, known as nuclease-
hypersensitive sites, that appear to be nucleosome

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15. The Eukaryotic
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Chapter Fifteen The Eukaryotic Chromosome

free. These sites, usually mutiples of a nucleosomal re-
gion of about two hundred base pairs, are particularly
sensitive to digestion by different nucleases. When these
regions are isolated, they usually have sequences that
control functions in replication, transcription, or other
activities of DNA. For example, numerous promoter
regions in Drosophila, mouse, and human DNA are in
nuclease-hypersensitive sites. Hence, some specific DNA
sequences are kept free of nucleosomes, and these se-
quences appear to be recognized by various enzymes
such as RNA polymerase. In many other cases, however,
nucleosomes do appear to cover promoters and repress
transcription. For transcription to occur in these cases,
some form of chromatin remodeling must take place.
Two general classes of proteins are involved in chro-
matin remodeling. First are proteins that acetylate the
N-terminal tails of the hist ones, a process that may cause
the nucleosomes to bind the DNA less tightly and thus
make it available for attachment of transcription factors.
These enzymes are called histone acetyl transferases
(HATs). Deacetylating enzymes have the reverse effect:
They act to repress transcription. Second, a class of ATP-
dependent proteins such as the SWI/SNF complex in
yeast also affect chromatin remodeling. (Some workers
called the proteins SWI because they were involved in
mating type snatching, and others called them SNF for
sucrose /^on/ermenting.) The SWI/SNF complex is a
group of eleven proteins involved in transcription activa-
tion in many genes, presumably allowing transcription

(a)

(b)

Histone core

60 A

Core DNA

— Linker DNA

- Nucleosome

110A

Figure 15.5 The eukaryotic chromosome is associated with
histone proteins to form nucleosomes. The protein core is
wrapped with 1 .7 loops of DNA and connected with a length
of DNA called a linker.

Histone
octamer

DNA

(c)

Figure 15.6 Nucleosome structure, (a) Schematic comparison
of the eight histones comprising the nucleosome in salt
solution. A dimer consists of one H2A and one H2B histone
molecule; a tetramer consists of two H3 and two H4 histones.
(b) DNA fits in surface grooves on the more compacted
structure found in physiological conditions, (c) The diagram
shows the presumed position of the H1 histone, encompassing
166 base pairs of DNA.

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447

Table 15.2

Composition

of Histones

Number of

Percentage of

Fraction

Class

Amino Acids

Basic Amino Acids

Very lysine rich

213

H2A

Lysine, arginine rich

129

H2B

Moderately lysine rich

125

Arginine rich

135

Arginine, glycine rich

102

Figure 15.7 Nucleosome core particle at 2.8 A resolution.
Shown are 1 46 base pairs of DNA {brown and turquoise) and the
eight histone protein chains {purple: H3; green: H4; yellow: H2A;
and red: H2B). Note the protein tails (N-terminal ends) of the
histone polypeptides extending out of the nucleosome. On the
left is the view down the DNA helix, and on the right is the
perpendicular view. (From Karolin Luger, et al., "Crystal structure of the
nucleosome core particle at 2.8A resolution" in Nature, 389:251-260,
September 18, 1997, fig. 1a p. 252. Reprinted by permission of Macmillan,
Ltd.)

factors to access promoters by remodeling chromatin.
These proteins are able to reposition a nucleosome on
DNA by sliding the nucleosome down the DNA.

We thus conclude that although nucleosomes serve
as a general, first-order packing mechanism in eukaryotic
DNA, they can be positioned precisely and can attenuate
transcription. It is interesting to note that once transcrip-
tion begins, RNA polymerase apparently moves along nu-
cleosomed DNA by translocation of the histones by 75 to
80 base pairs without disrupting the nucleosome itself.
This seems to be accomplished by the RNA polymerase
moving the DNA and then re-forming the nucleosome in
its wake (fig. 15.8).

Figure 15.8 RNA polymerase steps around a nucleosome
without disrupting it. {1) The RNA polymerase begins at a
promoter (P) and heads for the nucleosomed DNA (filled in),
whose border is noted with a line and the letter B. As the
polymerase encounters the nucleosome, it begins to unwrap
the DNA from the histones (2). The displaced DNA then
reencounters the histones (3), about seventy-five to eighty base
pairs upstream from the original point of nucleosome formation.
The polymerase continues on its way (4 and 5), and the
nucleosome re-forms in its displaced position without disrupting
the histones or ever fully losing contact with the core DNA.
(From Vasily M. Studitsky, et al., "A histone octamer can step around a
transcribing polymerase without leaving the template," Cell, 76: 371-82,
January 28, 1994. Copyright © 1994 by Cell Press. Reprinted by
permission.)

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Chapter Fifteen The Eukaryotic Chromosome

Higher-Order Structure of Chromatin

Since the nucleosome has a width of only 110 A, and
metaphase chromosomes appear to be constructed of a
fiber having a diameter of about 2,400 A (fig. 15.9), sev-
eral additional levels of chromatin compaction lead to
the metaphase chromosome. Various experiments, which
change the ionic strength the chromatin is subjected to,
indicate that the 110 A DNA spontaneously forms a
300 A, solenoidlike fiber with increased ionic strength. It
seems that this fiber results from the coiling of the nu-
cleosomal DNA (fig. 15.10). This 300 A fiber is not, how-
ever, the final form of the DNA. We can account for the
contraction of the 300 A fiber to the 2,400 A fiber found
in metaphase chromosomes by the formation of a second
solenoidlike structure from the winding of the 300 A
fiber (fig. 15.11).

If the histones are removed from a chromosome, the
DNA billows out, leaving a proteinaceous structure
termed a scaffold (fig. 15.12). This scaffold structure is
formed from nonhistone proteins; two of them pre-
dominate, namely SCI and SC2. SCI has been identified

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Figure 15.9 Chinese hamster chromosome. Note the fibers
making up the chromosome; they are approximately 2,400 A in
diameter. Magnification 1 1 ,800X. (Source: Courtesy of Dr. Hans Ris.)

as topoisomerase II. It would not be unreasonable to ex-
pect several hundred different proteins, many in minute
quantities, to be associated with the chromosome and in-
volved in replication, repair, and transcription.

Figure 15.10 Solenoid model for the formation of the 300 A
chromatin fiber. Nucleosomal DNA wraps in a helical fashion,
forming a hollow core. Although histone H1 is not shown, it is
known to be on the inside of the solenoid.

•2,400 A-

-H

-300 A

Figure 15.11 The 2,400 A fiber of the eukaryotic chromosome
is a hollow, solenoidlike structure. It is formed by the coiling of
the 300 A fiber, which itself is a solenoid.

Tamarin: Principles of
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15. The Eukaryotic
Chromosome

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The Eukaryotic Chromosome

449

? - >, ■ «■

Figure 15.12 Scaffold protein. When the histones are removed
from a eukaryotic chromosome, a fibrous scaffold remains.
The DNA loops out from this scaffold. The bar is 2 |xm long.

(J. Paulson and U. Laemmli, "The structure of histone-depleted metaphase
chromosomes," Cell, 12:817-28, 1977. Micrograph courtesy of James R.
Paulson.)

Polyteny, Puffs, and Balbiani Rings

Drosophila's salivary glands, as well as some other tis-
sues of Drosophila and other diptera, contain giant
banded chromosomes (see fig. 6.12) that result from the
replication of the chromosomes and the synapsis of ho-
mologues without cell division (endomitosis). These
chromosomes consist of more than one thousand copies
of the same chromatid and appear as alternating dark
bands and lighter interband regions. The dark bands are
referred to as cbromomeres. Also seen are diffuse areas
called chromosome puffs (fig. 15.13). Chromosome
puffs are also referred to as Balbiani rings. These rings
were originally defined as puffs in the midge, Chirono-
mus, whose polytene chromosomes were discovered by
E. G. Balbiani in 1881. Currently, the term applies to all
puffs, or at least the larger puffs, in all species with poly-
tene chromosomes.

The structure of the polytene chromosome can be ex-
plained by the diagram in figure 15.14. Dark bands (chro-
momeres) are due to tight coiling of the 300 A fiber; light
interband regions are due to looser coiling. The figure

:- . «**

■ \

Figure 15.13 A chromosome puff on the left arm of
chromosome 3 of the midge Chironomus pallidivittatus. (Jan-
Erik Edstrom, et al., Developmental Biology 91:131-37, 1982, Figure 1B,
Academic Press.)

Band

Puff

Band

Interband —

Band

-Puff

Band
Interband

300 A fibers

Figure 15.14 Polytene chromosome with bands and a puff.
Three of the approximately one thousand synapsed chromatids
are shown diagrammatically on the right.

also shows how chromosome puffs would come about as
fibers unfold in regions of active transcription.

Staining with reagents specific for RNA, such as tolui-
dine blue, or autoradiography with tritiated ( 3 H) uridine,
have been used to demonstrate that active transcription
is going on in the puffs but not in neighboring regions of
the polytene chromosomes. The messenger RNA isolated
from cells with puffs has also been shown to hybridize
only to the puffed regions of the chromosomes. Thus,
these regions of the DNA are complementary to the
messenger RNA (fig. 15.15) and represent areas of active
transcription. Modern recombinant DNA techniques
have also shown that many puffs probably represent
the transcription of only one gene, although there are
exceptions.

Puffs generally fall into four categories. Stage-specific
puffs appear during a certain stage of development, such

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Chromosome

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Chapter Fifteen The Eukaryotic Chromosome

as molting. Tissue-specific puffs are active in one tissue
but not another. (In dipteran larvae, tissues other than
the salivary glands, such as the midgut and Malpighian
tubules, have polytene chromosomes.) Constitutive
puffs are active almost all the time in a specific tissue.
And environmentally induced puffs appear after some
environmental change, such as heat shock (fig. 15.16). In
Drosophila, about 80% of the puffs are stage specific; in
Chironomus, only about 20% are. For example, at the
time of molt in insects, the hormone ecdysone is
secreted by the pro thoracic gland. At the same time,
many puff patterns change (fig. 15.17). Similar changes in
puff patterns can be induced by the injection of
ecdysone. Hence, molting, a stage-specific developmen-
tal sequence, is related to a sequential transcription se-
quence in the chromosomes.

Lampbrush Chromosomes

Lampbrush chromosomes, which occur in amphibian
oocytes, are so named because their looped-out configu-
ration has the appearance of a brush for cleaning lamps,
now a relatively uncommon household item (fig. 15.18).
The loops of the lampbrush chromosomes are covered
by an RNA matrix and are the sites of active transcrip-
tion. Presumably, the loops are unwindings of the single
chromosome, similar to the unwindings in the polytene
chromosome shown in figure 15.14. Thus, under certain
circumstances, such as in polytene chromosomal puffs

.jul .li

Figure 15.16 Puff 4-81 B of the salivary gland in Drosophila
hydei is induced by heat shock (37° C for one-half hour). At
the top, normal activity. At the bottom, temperature shock in
vitro, resulting in the puff. (Source: H. D. Berendes, et al.,
"Experimental puffs in salivary gland chromosomes of Drosophila hydei,"
Chromosoma [Berl.] 16:35-46, Fig. 4a-b, 1965. © Springer- Verlag.)

Figure 15.15 Hybridization at a Chironomus tentans salivary
gland chromosome puff. The chromosomal DNA is hybridized
with labeled RNA (black dots) transcribed from the locus. The
activity of the locus is forming the puff. (Reprinted by permission
from B. Lambert, "Repeated DNA sequences in a Balbiani ring," Journal of
Molecular Biology, 72:65-75, 1972. Copyright by Academic Press, Inc.
(London) Ltd.)

_o_

"XT

— U~

r^L_n

<^L_Q

Time

Figure 15.17 Puff patterns on a segment of a Chironomus
tentans salivary gland chromosome during molt. As time
proceeds, puffs appear and disappear and change in size.

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The Eukaryotic Chromosome

451

300 A fiber

? mm

Figure 15.18 Lampbrush chromosome of the newt,
Notophthalmus viridescens. Centromere is at the the left
{arrow); the two long homologues are held together by three
Chiasmata. Magnification 238 X. (Source: Joseph G. Gall, figure 2 in
D. M. Prescott, ed., Methods in Cell Physiology, vol. 2 [New York: Academic
Press, 1966], 39. Reproduced by permission.)

and in lampbrush chromosomes, active transcription can
be seen in the light microscope. Since only certain bands
puff at any one moment in polytene chromosomes, and
since the loops of lampbrush chromosomes are of vari-
ous sizes (with some regions not looped at all), we have
evidence of specific transcription. However, we have no
indication, so far, of the nature of the control of that tran-
scription.

Chromosomal Banding

Several chromosomal staining techniques reveal consis-
tent banding patterns. By means of these patterns, all of
the human chromosomes can be differentiated (see fig.
5.1). Of possibly greater importance is the fact that these
staining techniques have provided some insight into the
structure of the chromosome. The techniques for stain-
ing the C, G, and R chromosomal bands will serve as an
illustration.

G-bands are obtained with Giemsa stain, a complex
of stains specific for the phosphate groups of DNA. Treat-
ment of fixed chromatin with trypsin or hot salts brings
out the G-bands. Giemsa stain enhances banding that is
already visible in mitotic chromosomes. The banding pat-
tern is caused by the arrangement of chromomeres. Un-
der careful observation, the major G-bands prove to con-
sist of many smaller chromomeres. This banding
appearance has led D. Comings to suggest the mecha-
nism of chromosomal folding shown in figure 15.19.

C-bands are Giemsa-stained bands after the chromo-
somes are treated with NaOH. The C is for "centromere,"
because these bands represent constitutive heterochro-

Nuclear
matrix or
nuclear -
membrane
fragment

Small

chromomere
band \

Interband

Small

chromomere

band

>T\ Larger
L/ chromomere

Chromomeres and
interband chromatin

Clustering of
chromomeres

Interbands

( \ G-bands as
^ chromomere
yl clusters

Chromosomal
bands

Figure 15.19 Model of eukaryotic (mammalian) chromosomal
banding. G-bands are chromomere clusters, which result from
the contraction of smaller chromomeres. These, in turn, result
from looping of the 300 A fiber. (Reproduced with permission, from
the Annual Review of Genetics, Volume 12, © 1978 by Annual Reviews, Inc.)

matin surrounding the centromeres (fig. 15.20). The DNA
is also usually satellite rich. Satellite DNA differs in
buoyant density from the major portion of cellular
DNA. When eukaryotic DNA is isolated and centrifuged
in CsCl, forming a density gradient, the majority of the
DNA forms one band in the gradient at a single buoyant
density. The buoyancy is determined by the G-C content
of the DNA. However, smaller secondary bands are also
usually present, indicating regions of DNA having se-
quences different from the majority of the cell's DNA
(fig. 15.21). DNA isolated this way is referred to as satel-
lite DNA because of the secondary, or satellite, bands
formed in the density gradient. As we will see, this DNA
is found primarily around centromeres and consists of
numerous repetitions of a short sequence.

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Chapter Fifteen The Eukaryotic Chromosome

R-bands are visible with a technique that stains the
regions between G-bands. The chromosomes are fixed,
stained with Giemsa, and then viewed with a phase con-
trast microscope. Since the dark-light pattern is the op-
posite of the G-band pattern, these bands are called re-
verse bands.

(a)

^^^ m^^

dfe

m* mf *

^^j

m . mm

■^^BL-^^H

mm*

^^m

^^^H -mmm

\m\

\mm\m\ U^M

^^%\ *

^ mk

•

*. ^^

^k. M

•

\ml mmmmmi

(b)

Figure 15.20 (a) C banding of chromosomes from a cell in the
bone marrow of the house mouse, Mus musculus. The arrow
indicates that the Y chromatids have already separated into
two chromosomes, (b) Yellow fluorescence indicates a satellite
DNA probe in human chromosomes (centromeres), ([a] B. Vig,
"Sequence of centromere separation: Role of centromeric heterochromatin,"
Genetics, 102:795-806, 1982. [b] Photograph Courtesy of Oncor, Inc.
Gaithersburg, Maryland.)

From the information gleaned from these staining
techniques, D. Comings distinguished between three
basic chromatin types: euchromatin, constitutive
heterochromatin, and intercalary heterochromatin
(table 15.3). Presumably, the only chromatin involved in
transcription is euchromatin. Constitutive hetero-
chromatin surrounds the centromere and is rich in

David E. Comings (1935- ).

(Courtesy Dr. David E. Comings.)

Figure 15.21 Satellite DNA in Drosophila virilis. The quantity of
DNA is graphed against the buoyant density (g/cc), resulting in
four peaks. The large peak (at left) is the major DNA
component of the cell; the other three bands are satellite DNA.
The left-most of the satellite peaks (1 .692) is DNA with a
repeating sequence of ACAAACT; the middle satellite peak
(1 .688) is a sequence of ATAAACT; and the right-most satellite
peak (1 .671) has a sequence of ACAAATT. (From Joseph G. Gall,
et al., Cold Spring Harbor Laboratory Symposia on Quantitative Biology,
38:417-21. Copyright © 1974 Cold Spring Harbor Laboratory, Cold Spring
Harbor, NY. Reprinted by permission.)

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Table 1 5.3 The Three Major Types of Chromatin in Eukaryotic Chromosomes

Centromeric Constitutive

Intercalary

Euchromatin

Heterochromatin

Relation to bands

In R-bands

In C-bands

In G-bands

Location

Chromosome arms

Usually centromeric

Chromosome arms

Condition during interphase

Usually dispersed

Condensed

Genetic activity

Usually active

Inactive

Probably inactive

Relation to chromomeres

Interchromomeric

Centromeric chromomere

Intercalary chromomeres

satellite DNA. Intercalary heterochromatin is found
throughout the chromosome. Thus, it becomes appar-
ent that the eukaryotic chromosome is a relatively com-
plex structure.

Centromeres and Telomeres

Centromeres

Two regions of the eukaryotic chromosome have spe-
cific functions — the centromere and the telomeres. The
centromere is involved in chromosomal movement dur-
ing mitosis and meiosis, whereas the telomeres terminate
the chromosomes. As we pointed out in chapter 3, the
terms centromere and kinetochore, while occasionally
used interchangeably, are distinct. The kinetochore is the
interface between the visible constriction in the chro-
mosome (the centromere) and the microtubules of the
spindle. The kinetochore of higher organisms (e.g., mam-
mals) contains proteins and some RNA. Microscopically,
it is a trilaminar structure, attached to chromatin at the
inner layer and to microtubules at the outer layer
(fig. 15.22).

Most of our knowledge of the genetics of cen-
tromeres has come from work in yeast (Saccharomyces
cerevisiae). Cells did not maintain most artificially cre-
ated yeast plasmids because they were lost during mito-
sis. However, plasmids were isolated that did replicate
normally during cell division. Presumably, they contained
centromeres, allowing them to replicate and move in syn-
chrony with the host's chromosomes. Further genetic en-
gineering made it possible to isolate smaller and smaller
regions that could serve as centromeres. After se-
quencing the centromeres of fifteen of the sixteen yeast
chromosomes, it was possible to conclude that the cen-
tromere from yeast is about 250 base pairs long with
three consensus regions (fig. 15.23); we are defining a
centromere as a sequence of DNA called the CEN locus
or CEN region. Recent data indicate that this region may
contain a single, modified nucleosome associated with

jr.*** \ r> ' ' V

Figure 15.22 The kinetochore of a metaphase chromosome of
the rat kangaroo. IL, ML, and OL refer to inner, middle, and
outer layers, respectively, of the kinetochore. Note the
microtubules attached to the kinetochore and the large mass of
dark-staining chromatin making up most of the figure.
Magnification 30,800X. (From B. R. Brinkley and J. Cartwright, Jr.,
J. Cell Biology, 50:41 6-31 , 1 971 .)

region II. The 250 base-pair length of the CEN regions of
yeast chromosomes is about 200 A, the same as the di-
ameter of a microtubule, indicating that only one micro-
tubule attaches to each centromere during mitosis or
meiosis in a yeast cell. This region is called a point cen-
tromere (fig . 15.24).

Higher eukaryotes have larger centromeric regions
that attach more microtubules. These regions are re-
ferred to as regional centromeres (see figs. 15.22 and
also 312). Regional centromeres range from nineteen to
one hundred kilobases (kb; 19,000-100,000 bases) with
unique and satellite (repeated sequence) DNA that is
heterochromatic and may include expressed genes. We
know much less about regional centromeres than we do
about point centromeres.

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Chapter Fifteen The Eukaryotic Chromosome

PuTCACPuTG

- 78-86 bp
91-95% AT

III

TGTTTPyTGXTTTCCGAAAXXXXAAA

Figure 15.23 Consensus sequence for the three regions (l-lll) of fifteen
yeast centromeres. Pu represents any purine, Py represents any pyrimidine,
and X represents any base. The arrows appear over inverted repeat
sequences. (Source: Data from L. Clarke and J. Carbon, "The structure and function of
yeast centromeres," Annual Review of Genetics, 19:29-56, 1985.)

Figure 15.24 Schematic view of a yeast centromeric region. The arrows are the nuclease-
hypersensitive sites. A microtubule is about the same width as the centromeric region. (With permission,
from the Annual Review of Genetics, Volume 1 9 © 1 985 by Annual Reviews www.AnnualReviews.org)

Telomeres

Since eukaryotic chromosomes are linear, each has two
ends, referred to as telomeres, that not only mark the
termination of the linear chromosome but also have sev-
eral specific functions (fig. 15.25). Telomeres must pre-
vent the chromosomal ends from acting in a "sticky" fash-
ion, the way that broken chromosomal ends act (see
chapter 8). In other words, chromosomal ends must not
elicit a DNA repair response (see chapter 12). Telomeres
must also prevent the ends of chromosomes from being
degraded by exonucleases and must allow chromosomal
ends to be properly replicated.

Most telomeres isolated so far are repetitions of se-
quences of five to eight bases. In human beings, the
telomeric sequence is TTAGGG, repeated 300 to 5,000
times at the end of each chromosome. The human telo-
mere was discovered by R. Moyzis and his colleagues
when they probed the highly repetitive segment of hu-
man DNA. (Highly repetitive DNA, as its name implies,
consists of numerous copies of a single sequence and
usually comprises the satellite components of the cell's
DNA; see next section.) When a probe for this sequence
was applied to human chromosomes, the sequence was

found at the tip of each chromosome in roughly the same
quantity (fig. 15.26). This is a highly conserved sequence,
found in all vertebrates studied as well as in unicellular
trypanosomes. Similar sequences are found in various
other eukaryotes (table 15.4); the first sequence was iso-
lated by E. Blackburn and J. Gall in 1978.

When a linear DNA molecule is replicated, the 3' — > 5'
strand can be replicated to the end (see chapter 9).
The 5' — > 3' strand, however, is replicated with RNA
primers that are then degraded, leaving a short gap on
the progeny strand (fig. 15.27). It is always the G-rich
strand of telomeric DNA that ends up single-stranded,
forming a 3' overhang of twelve to sixteen nucleotides.
Thus, the normal replication process of a linear DNA mol-
ecule leaves an incomplete terminus. Hence, scientists
suspected that there would be a unique mechanism for
the replication of telomeres.

Telomeric sequences appear to be added de novo
without, DNA template assistance by an enzyme called
telomerase, discovered by E. Blackburn and her col-
leagues. This was seen when telomeres from another
species were engineered into yeast cells. After a cell cy-
cle, the yeast telomeric sequence had been added on at

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Figure 15.25 Polytene chromosome from the salivary gland of
a Drosophila larva showing bands, interbands, puffs, and
telomeres. (© David M. Phillips/Visuals Unlimited.)

Elizabeth H. Blackburn
(1948- ). (Courtesy of Dr.
Elizabeth H. Blackburn.)

Figure 15.26 The human genome probed for the telomeric
sequence, TTAGGG, using fluorescent staining techniques. The
yellow dots at the tips of the chromosomes are the probes.
(From Robert K. Moyzis, et al., Proceedings of the National Academy of
Science, USA, 85:6622-26, 1988. Figure 4, left.)

Table 1 5.4 Telomeric Sequences in Eukaryotes;

The G-Rich Strand of the Double
Helix Is Shown

Organism

Telomeric Repeat

Human beings, other mammals,

TTAGGG

birds, reptiles

Trypanosomes

TTAGGG

Holotrichous ciliates

GGGGTT

(Tetrahymend)

Hypotrichous ciliates

GGGGTTTT

(Sty lony chid)

Yeast

GT, GGT, and GGGT

Plants

TTTAGGG

the ends of the foreign chromosome, the result, presum-
ably, of the telomerase enzyme.

When Blackburn and her colleagues isolated telo-
merase, they discovered that a segment of RNA, about
160 base pairs, is an integral part of the enzyme. That
RNA has a region that is complementary to the G-rich re-
peat of the telomeric DNA sequence of the species. After
careful experimentation, including modifying the gene
for the telomerase RNA, Blackburn and her colleagues
concluded that telomerase uses its RNA as a template for
adding telomeric repeats to the ends of chromosomes.
Telomerase is thus a reverse transcriptase, using RNA nu-
cleotides as a template to polymerize DNA nucleotides.

Blackburn and her colleagues proposed that the first
step in telomere extension is hybridization of the 3' end
of the telomere with the RNA component of telomerase
(fig. 15.28a). Then, with the telomerase RNA as a
template, the 3' end of the telomere is extended
(fig. 15.28&). Finally, a translocation step takes place that
displaces the telomere in respect to the RNA, returning
to the configuration at the beginning of the process
(fig. 15.28c). The single-stranded C-rich strand is then
synthesized with DNA polymerase and DNA ligase.

Once telomeres have been added to the ends of eu-
karyotic chromosomes, different organisms use any of
three different methods known to protect the ends of

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Chapter Fifteen The Eukaryotic Chromosome

5'
3'-

Telomere

3'
5'

DNA replication, including
formation of final
Okazaki fragment

Primer

Removal of
final primers

Figure 15.27 Removal of final primers after the replication of
linear DNA creates single-stranded ends.

the chromosomes. First, the guanine-rich DNA can form
complex structures. Biochemists have discovered that
four guanines can form a planar G-tetraplex, with the
four bases hydrogen bonded to each other (fig. 15.29).
Several structures have been hypothesized to explain the
novel ends of these chromosomes (fig. 15.30). Second,
proteins have been discovered that bind to the 3' ends of
telomeres. In the ciliate Oxytricha nova, a protein called
the telomere end-binding protein (TEBP) attaches to the
3' ends of telomeres and protects them (fig. 15.31).
Finally, a novel structure called the t-loop has been dis-
covered at the end of mammalian telomeres. This loop
forms at the ends of chromosomes under the direction of
a protein called TRF2 (telomere repeat-binding /actor),
which causes the 3' end of the chromosome to loop
around and interdigitate into the double helix, forming
the loop (fig. 15.32).

How do cells keep track of the number of their telo-
meric repeats? Proteins have been isolated that bind to
telomeres (Rapl in Saccharomyces cerevisiae, TRF1 in
human beings). By mutating these proteins or the telo-
meric sequences, scientists have changed the equilib-
rium number of telomeric repeats. This led to the current
model that the cell counts the number of these proteins
bound to the telomeres, not the number of telomeres di-
rectly, to know whether telomeres should be added. This
is a very active area of research.

In yeast, protozoa, and other single-celled organisms,
telomerase is active, keeping the ends of the chromo-

• • •

(a)

Telomerase

Telomerase
RNA

Reverse transcription
(telomere extension)

(b)

New DNA

Translocation of
telomere and
gap filling by DNA
polymerase

(c)

Figure 15.28 Telomerase extends telomeres using telomerase
RNA {red) as a template. Gap filling (green) by DNA
polymerase and ligase complete the double helix. (Source: Data

from Shippen-Lentz and Blackburn, Science, 247:550, 1990.)

somes at the appropriate lengths. These cells can divide
potentially forever. However, in most cells of higher or-
ganisms, telomerase is not active, and the ends of the
chromosomes get shorter with each cell division. At a
certain telomeric length, the cells no longer divide. How-
ever, if telomerase becomes active, and the ends of the
chromosomes lengthen, a signal is conveyed to keep
cells dividing, which can lead to cancerous growth. In
fact, human telomerase was isolated from an immortal
cell line (HeLa) derived from cervical cancer cells. Thus,

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R--N N

Figure 15.29 A G-tetraplex can form from four guanines in a
plane, hydrogen bonded with each other. (Source: Data from
Yong Wang and Dinshaw J. Patel, "Solution structure of the human
telomeric repeat d[AG 3 (T 2 AG 3 ) 3 ] G-tetraplex," Structure, 1 :263-82,
December 15, 1993.)

subunit
C-term.

P
subunit

Figure 15.31 Telomere of the ciliate Oxytricha nova shown
bound by the dimeric protein called telomere end-binding
protein (TEBP). The a and (3 subunits of the protein form a
deep cleft in which the 3' end of the telomere lies. The folding
of the protein into its final form around the DNA may only
occur after the DNA has bound, explaining how the DNA could
be recognized and placed into such a deep cleft. (Reprinted

courtesy of Dr. Martin Horvath.)

Figure 15.30 Based on G-tetraplexes (fig. 15.29), the
illustrated structure can form at the very tip of a telomere. The
sequence d(GGTTGGTGTGGTTGG) is shown forming a
four-Stranded Structure. (Reproduced, with permission, from the
Annual Review of Biophysics and Biomolecular Structure, Volume 23, ©
1994 by Annual Reviews, Inc.)

The C-Value Paradox

Why do eukaryotes have so much DNA, and why is there
huge variation in the DNA content between species of
comparable complexity? These questions define the
C-value paradox, in which C refers to the quantity of
DNA in a cell. For an example of the paradox, although
human beings have 33 billion base pairs in the haploid
genome, an amoeba has more than 200 billion base pairs.
And although an average bony fish has over 300 billion
base pairs of DNA in its haploid genome, the Japanese
puffer fish has less than half a billion base pairs. If the
basic bony fish pattern can be created with less than
half a billion base pairs, why does the average bony fish
have over 600 times that much DNA? What is this excess
DNA doing? To explain the C-value paradox, researchers
examined the repetitiveness of DNA, and more re-
cently, probed and sequenced DNA to understand its
properties.

attention is now turning to the possible clinical applica-
tion of this knowledge: If telomerase can be deactivated
in tumor cells, the cells may stop dividing or die, thereby
eliminating the cancer. Further, studying normal telo-
mere shortening, which appears to act as a biological
clock, may help us understand the aging process and
senescence.

DNA-DNA Hybridization

R. Britten and his colleagues, using the technique of
DNA-DNA hybridization, first systematically analyzed
the repetitiveness of the DNA within eukaryotes. When
DNA is heated, it denatures or unwinds into single
strands; when it cools, it renatures. The rate of renatura-
tion depends on the DNA sequences. If the sample con-
tains DNA with repeated sequences, it will hybridize

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Chapter Fifteen The Eukaryotic Chromosome

(a)

(b)

Figure 15.32 The t-loop at the end of the mammalian
telomere, (a) A diagram of how the t-loop is formed by the
interdigitation of the 3' end of the telomere into the double
helix, (b) Electron micrograph of a t-loop from a mouse liver
cell. The loop is about 10,000 bases around, ([b] From Jack D.
Griffith, et al., "Mammalian telomeres end in a large duplex loop" in Cell,
97:503-14, May 14, 1999. Copyright © Cell Press.)

faster than DNA that does not have repeated sequences.
From these studies, Britten and his colleagues found that
eukaryotic chromosomes contain regions of unique,
moderately repetitive, and highly repetitive DNA.
Unique DNA is, as its name implies, DNA with unre-
pealed sequences. Repetitive DNA is DNA whose se-
quences are repeated in the genome.

Satellite DNA, found around centromeres (see fig.
15.20), is highly repetitive DNA with a unique repeat
length of about two hundred base pairs. Given the

Roy J. Britten (1919- ).
(Courtesy of Dr. Roy J. Britten.)

quantity of satellite DNA per cell, there must be more
than one million repetitions of this two-hundred-
nucleotide sequence in higher eukaryotes. At the other
end of the spectrum is unique DNA, which makes up
most of the transcribed genes of an organism. The rest
of the DNA is repetitive DNA in a few to several hun-
dred thousand copies. This repetitive DNA comprises at
least three categories. One is "junk" DNA, DNA that is
not useful to the organism, made up of un transcribed
and parasitic sequences (selfish DNA). Another cate-
gory is transcribed genes in many copies that have di-
verged from each other, such as antibody, collagen, and
globin genes. We use the term gene family to refer to
genes that have arisen by duplication, with or without
divergence, from an ancestral gene. And finally, tran-
scribed genes in many copies that are virtually identical,
such as ribosomal RNA and histone genes, make up a
third category of repetitive DNA.

Junk DNA

We saw in chapter 13 that transposons in prokaryotes are
generally viewed as selfish or parasitic: They serve no
purpose to the cell. The transposons replicate on their
own, increasing in number. Eukaryotic transposons are
mostly retrotransposons, transposable elements that
move by way of an RNA intermediate. That is, the retro-
transposon is transcribed into RNA and then, by reverse
transcription, converted to a cDNA that is then inserted
into the genome. These elements can make up 50% of the
eukaryotic genome, existing in hundreds of thousands of
copies. They generally fall into two categories: LINES and
SINES. Long interspersed elements (LINES), are up to
seven thousand base pairs each and contain genes for re-
verse transcription, RNA binding, and endonuclease ac-
tivity They thus have the ability to jump by way of an
RNA intermediate. Human DNA is believed to be com-
posed of about 15% LINES.

Short interspersed elements (SINES) are gener-
ally derivatives of transfer RNA genes and do not have the
ability to retrotranspose on their own. That is, in the past,
their transcripts were modified, converted to cDNA by
reverse transcription, and then reinserted into the host's

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genome. They rely on the reverse transcriptase provided
by the genes of LINES or retroviruses. One group of
SINEs not derived from transfer RNA is derived from the
RNA of the signal recognition particle (see chapter 1 1);
members of this group occur in human beings in about
five-hundred thousand copies of a three-hundred-base-
pair sequence. Because these sequences are cleaved by
the restriction endonuclease^4/z/I, they are called the Alu
family. The human genome is also permeated by rem-
nants of at least a dozen distinct families of ancient retro-
viruses scattered throughout our chromosomes.

At this point, we can see some potential explanations
for the C-value paradox. Much eukaryotic DNA is junk,
apparently doing no harm. In some cases, 97% of the host
genome is composed of junk DNA. Recent work seems to
indicate that gross differences in DNA content between
higher organisms may be due to the differing abilities of
different species to rid themselves of this parasitic DNA.
If it builds up without being removed, the DNA content
of the species can soar. Thus, the wide differences in
DNA content among higher eukaryotes mentioned at the
beginning of this section have little to do with the com-
plexity of the organism, but rather with the ability of the
organism to remove junk DNA as it forms.

Expressed Genes in Many Copies

Several types of genes create a product that is needed in
such large quantity that one copy of the gene could not
fulfill the cell's needs. We are familiar with the nucleolus,
the site of the ribosomal RNA genes (see fig. 10.20). Hu-
man beings have about two hundred copies of the major
ribosomal RNA gene and about two thousand copies of
the 5S ribosomal RNA gene. Fruit flies have about two
hundred and one hundred copies, respectively, of the
two genes.

In some cases, the normal number of multiple copies
of a gene is still not enough. The cell must then resort to
gene amplification, a process whereby the cell in-
creases the number of copies of the gene. For example,
during oogenesis, ribosomal RNA genes (rDNA) are often
amplified. In Xenopus, rDNA is amplified about one
thousand times, which allows an oocyte to accumulate
about 10 12 ribosomes. The amplified DNA is in the form
of small, circular, extrachromosomal molecules of DNA.
Several models have been proposed as to how cells actu-
ally amplify their DNA. One model relies on unequal
crossing over (as in Bar eye in Drosophila), whereas an-
other model is based on unscheduled extra DNA replica-
tion in a region, followed by recombinational events that
generate linear and circular forms of the excess DNA. It is
not presently clear which model is correct.

In addition to ribosomal RNA genes, other genes are
repeated, ensuring adequate gene products. The number
and location of repeated genes are usually discovered by

hybridization studies using probes, similar to the way
that telomeric DNA was shown to be at the tips of the
chromosomes (see fig. 15.26). Repeated genes include
the genes for transfer RNAs and hist ones. The average
transfer RNA is repeated about a dozen times in
Drosophila. Human beings have over thirteen hundred
copies of transfer RNA genes in the haploid genome. In
many species, the five histone genes form a repeated
cluster, although each gene is transcribed independently
(fig. 15.33), while prokaryotic operons are transcribed as
a unit. The arrangement of histone genes may be more
complex in higher forms. There are indications that in
mammals, histone genes may lie in small groups or even
as individual genes.

Several types of genes occur in similar but not identi-
cal forms — that is, an original gene was duplicated but,
unlike histone or ribosomal RNA genes, the copies di-
verged in function. These gene families include globin
genes, immunoglobulin genes (see chapter 16), chorion
protein (insect eggshell) genes, and Drosophila heat
shock genes.

The Globin Gene Family

Globins are oxygen-transporting and storage molecules
found in animals, some plants, and microorganisms. In
higher vertebrates, there are two types of globins: myo-
globin, which stores oxygen in muscles, and hemoglobin,
found in red blood cells. Myoglobins function as single
molecules, whereas hemoglobins occur as tetramers,
two each of two protein chains. Evolution in the globin
gene family can be traced by comparative studies of glo-
bins in different species as well as molecular studies of
globins within a species (see chapter 21). Studying he-
moglobins has provided a great deal of information on
gene expression and evolution. We turn our attention to
the globin gene family in human beings.

During human development, four major hemoglobins
appear: embryonic hemoglobin, Hb F, Hb A, and Hb A 2

Sea urchin

(Psammechinus

miliaris)

H2B

H2A

H2A H2B

Fruit fly

(Drosophila

melanogaster)

Figure 15.33 The arrangement of histone genes {red) within
the five-gene cluster in sea urchins and fruit flies. Arrows
indicate the direction of transcription. Spacer DNA {black)
separates the genes.

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Chapter Fifteen The Eukaryotic Chromosome

(table 15.5). Structurally, the £ (Greek, zeta) subunit (a
component of embryonic hemoglobin) is a-like, whereas
the rest are p-like (fig. 15.34; see also fig. 10.29). Fetal he-
moglobin has a higher affinity for oxygen than does adult
hemoglobin, thus allowing fetuses to draw oxygen from
their mother's blood. From a comparative study of the
DNA sequences, the evolution of the various hemoglobin
genes has been inferred (fig. 15.35).

The a genes are located in a cluster on chromosome
16; the p genes are located in a cluster on chromosome 1 1
(fig. 15.36). These two clusters provide a clear case his-
tory of gene duplication, presumably by unequal crossing

over, followed by divergence. Having a second or third
copy of a gene allows one of the duplicates to diverge
(and perhaps to become nonfunctional in the process),
whereas the original still performs the required function.
Many diseases of genetic interest involve the hemo-
globins. In fact, hemoglobinopathies, including sickle-
cell anemia and the thalassemias, are the most common
genetic disorders in the world population. The best-
known mutation of a hemoglobin gene itself is the one
that causes sickle-cell anemia, a mutation of the sixth
amino acid of the p chain. In the homozygous state, the
disease is usually fatal. However, heterozygotes show an

a-globin

Intron I
(95 bp)

Intron II
(125 bp)

— I
100

141

Figure 15.34 The structure of adult human a-
and p-globin genes. The numbers refer to
amino acids (or translated codons).

(3-globin

104

105 146

Intron I
(150 bp)

Intron II
(850 bp)

Ancestral
primitive
globin gene

Ancestral

gene

Ancestral

Hba

gene

Ancestral

Hb(3

gene

Myoglobin

Figure 15.35 The presumed evolution of the
various human globin genes from an ancestral
primitive gene. The diagram represents a
branching tree that begins on the left and
progresses to the right. Each branch point is
an evolutionary step in which the genes
presumably were duplicated and then either
diverged or simply endured as duplicates, as in
present-day genes (on the right).

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Summary

461

\|/<x1

— m—

i i i

Kb 30
5' — >3'

I I

-DD-
l I

— DD DQ-

I I I I I

-DH-
I I

¥P2

■DD-

y y

-on -on— [

VP1

5 p
-D-ED-

I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I

Kb 60 50 40 30 20 10

5' — ^3'

Figure 15.36 The a- and (3-globin gene clusters in human
beings. The i|j(31 and 2 and the i]/a1 refer to nontranscribing
genes (pseudogenes). Mutation has rendered the pseudogenes
inactive. Within each gene box, solid color refers to exons and
open regions refer to introns. (Reproduced, with permission, from the
Annual Review of Genetics, Volume 14, © 1980 by Annual Reviews, Inc.)

increased resistance to malaria. One of the ramifications
is that the sickle-cell allele is maintained at relatively high
frequencies in malarial regions (see chapter 21).

The thalassemias are a group of diseases that affect
the regulation of the a and p hemoglobin genes. (Thal-
assemia comes from the Greek for "sea blood," because
the disease is best known in individuals living around the
Mediterranean Sea.) In a and p thalassemias, the a or p
subunit, respectively, is present in very low quantities or
entirely absent. Many of the genetic defects are deletions,
possibly due to unequal crossing over within the globin
gene complexes. T. Maniatis showed that p thalassemia is
caused by a mutation in the p-globin gene that disrupts
RNA splicing. The body compensates by forming y 4 or p 4

Table 15.5 Types of Human Hemoglobin

Generally

Type

When Present

Composition

Embryonic

Up until eight weeks of
gestation and beyond

£ 2 e 2

Fetal (Hb F)

Eight weeks to birth

<* 2 72

Adult (Hb A)

Just before birth and
beyond

«2^2

Adult (Hb A 2 )

In immature cells

a 2 °2

Note: Subscripts refer to the numbers of subunits present.

Tom Maniatis (1943- ).
(Courtesy of Dr. Tom Maniatis.)

hemoglobin in a thalassemias, or a 2 72 or a 2°2 m P thal-
assemias. These are relatively unsuitable or inefficient re-
sponses; the diseases range from very mild to very severe
and frequently fatal. More information is needed regarding
the control of hemoglobin production in the thalassemias.

SUMMARY

STUDY OBJECTIVE 1: To examine the arrangement of
DNA and proteins compromising the eukarayotic chro-
mosome 440-452

To study developmental control in eukaryotes, we must un-
derstand the eukaryotic chromosome, which is uninemic:
It consists of one DNA double helix per chromosome.
Nucleoprotein is composed of DNA, histones, and nonhis-
tone proteins. The nucleosome, a uniform packaging of the
DNA, is made of histones. The majority of the nonhistone
proteins create the scaffold structure of the chromosome
and are not involved in gene regulation. Presumably, very
small quantities of the nonhistone proteins take part in the
regulation of transcription.

Core DNA, wrapped around nucleosomes, is separated
by linker DNA between nucleosomes. There are regions of
DNA, vulnerable to nucleases, that do not contain nucleo-
somes; these are referred to as nuclease-hypersensitive sites.
Nucleosomes generally inhibit transcription. The 110 A nu-
cleosomed DNA forms a 300 A fiber by coiling into a sole-
noidlike structure. Coiling of this fiber presumably forms
the thick, 2,400 A fiber seen in metaphase chromosomes.

STUDY OBJECTIVE 2: To look at the nature of centromeres
and telomeres in eukaroyotic chromosomes 453-457

The centromere and telomeres are specific functional re-
gions of a chromosome. Centromeres isolated from yeast

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Chapter Fifteen The Eukaryotic Chromosome

chromosomes have three consensus areas. Telomeres are
tandem repeats of a short (five-base-pair to eight-base-pair)
segment. Telomeric sequences are added to the ends of
chromosomes by the enzyme telomerase, which uses RNA
as a template for adding DNA nucleotides. The number of
telomeric repeats varies, declining as a cell ages. Telomeric
repeat number may control the ability of a cell to replicate
and may be implicated in cancerous growth.

Substructuring in the eukaryotic chromosome is demon-
strated by G-, C-, and R-banding techniques. C-bands (con-
stitutive heterochromatin) appear to be around the cen-
tromeres. These bands consist primarily of satellite DNA,
which seems to have a structural role in the chromosome.
G-bands (Giemsa bands) presumably represent intercalary
heterochromatin and, also presumably, do not have an ac-
tive transcriptional role. R-bands (reverse bands) appear be-

tween the G-bands and represent intercalary euchromatin,
the site of transcribed, structural genes.

STUDY OBJECTIVE 3: To analyze the nature of the DNA in
eukaryotic chromosomes 457-461

Eukaryotes have very large genomes with huge differences in
DNA content between organisms similar in complexity, lead-
ing to the C-value paradox. Eukaryotic chromosomes contain
both unique and repetitive DNA. Highly repetitive DNA is
structural (centromeres, telomeres). Junk DNA is mainly
short and long interspersed elements. These SINEs and LINEs
are often present in hundreds of thousands of copies and can
account for 50% of an organism's DNA. They are retrotrans-
posons, transposons that jump by way of an RNA intermedi-
ate. Some functional genes also occur in many copies, such
as ribosomal RNA genes, histone genes, and globin genes.

SOLVED PROBLEMS

PROBLEM 1: Why is higher-order chromosomal structure
expected in eukaryotes but not prokaryotes?

Answer: The simplest explanation is the difference in
amount of the genetic material in prokaryotes and eu-
karyotes. Since the average human chromosome has sev-
eral centimeters of DNA, that DNA must be contracted to
a si2e in which it can be moved during mitosis and meio-
sis without tangling and breaking. Nucleosomes provide
the first order of coiling, and then several levels of coiling
of the nucleosomed DNA bring it down to a manageable
size for nuclear divisional processes.

PROBLEM 2: Why might we expect to see chromosomal
puffs that are tissue- and stage-specific, constitutive, and
environmentally induced?

Answer: The various patterns of chromosomal puffing
are expected because puffing indicates transcription, the
activity of specific genes. Thus, since various tissues are
different because they have different proteins, each tis-
sue is expected to have a unique suite of active genes and
thus a unique suite of puffs. Similarly, different stages in
an insect's development would require different genes to
be active, and different puffs should therefore appear at
different stages of development. Some genes are active
all the time because they specify proteins, such as ribo-
somal protein genes, that are needed all the time. Finally,
environmental insults such as heat shock are known to
induce a group of genes that are needed to react to the
specific insult, resulting in a suite of puffs that respond
consistently to an environmental insult.

EXERCISES AND PROBLEMS

THE EUKARYOTIC CELL

1. Summarize the major differences between eukary-
otes and prokaryotes, including the structures of
their DNAs.

THE EUKARYOTIC CHROMOSOME

2. Summarize the evidence that the eukaryotic chro-
mosome is uninemic.

3. What results would you get in the experiment
shown in figure 15.1 if the eukaryotic chromosome
were not uninemic, but instead had some other
number of complete DNA molecules (e.g., binemic)?

4. What are the major protein components of the eu-
karyotic chromosome? What are their functions?

5. What evidence is used to determine the length of
DNA associated with a nucleosome? What is a
nuclease-hypersensitive site? What functions are
associated with these sites?

* Answers to selected exercises and problems are on page A-18.

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Critical Thinking Questions

463

6. What is the protein composition of a nucleosome?
What function does histone HI have?

7. What are the relationships among the 1 10 A, 300 A,
and 2,400 A fibers of the eukaryotic chromosome?

8. Draw a mitotic chromosome during metaphase. Dia-
gram the various kinds of bands that can be brought
out by various staining techniques. What information
is known about the DNA content of these bands?

9. Give a 300 A fiber model of the chromosome to
account for G-bands.

10. Give a 300 A fiber model of the chromosome to
account for polytene chromosomal puffs.

11. What are the differences among polytene chromo-
somes, lampbrush chromosomes, puffs, and Balbiani
rings? Draw an example of each.

12. Under what circumstances does a chromosomal puff
occur? What does it signify?

13. What is satellite DNA? What does it signify?

14. What is a centromere? a kinetochore? What do we
know about the sequences within a yeast cen-
tromere?

15. What is a telomere? What are its functions? What is
its structure?

16. Describe three ways in which cells protect their
telomeres.

17. What functions exist in unique, repetitive, and
highly repetitive DNAs?

18. How would you use recombinant DNA techniques
to locate the number and position of Alu members
in the human chromosomes?

19. How could you use modern recombinant DNA tech-
nology to determine the direction of transcription of
the histone genes in figure 15.33?

20. How many functional globin genes are there in mam-
mals?

21. How could you determine, using modern recombi-
nant DNA techniques, that the a- and p-globin
pseudogenes exist?

22. Kavenoff and colleagues determined the size of DNA
in Drosophila chromosomes in two ways: (1) Spec-
trophotometry measurements were made on the
largest intact chromosome. These measurements
were then used to calculate the amount of DNA in
each chromosome. (2) Nuclei were gently lysed and
chromosomes isolated. The lengths of the longest
DNA molecules were measured, and those lengths
were used to determine the amount of DNA in each
molecule. What results for each method would you
expect if

a. the chromosomes contain one DNA molecule?

b. the chromosomes contain more than one DNA
molecule?

23. What can be said about the base composition of the
satellite DNA with a density of 1.671 in figure 15.21?

24. When chromatin is partially digested with an en-
donuclease, the proteins removed, and the DNA sep-
arated in a sizing gel, DNA fragments in multiples of
two hundred base pairs are found. Provide an expla-
nation for this observation.

25. If chromatin is digested with an endonuclease to
produce two hundred base-pair fragments, and these
fragments are then used for transcription experi-
ments, very little RNA is made. Provide an explana-
tion for this observation.

26. Can nucleosomes contain the DNA for one gene? Ex-
plain.

27. If radioactive probes are made from highly repetitive
DNA, these probes hybridize in situ mainly to cen-
tromeric and telomeric regions. What does this re-
sult suggest about the organization of chromo-
somes?

28. Would you expect archaeal species to have nucleo-
somes?

29. What is the C-value paradox, and how is it explained?

30. What are the origins of SINEs and LINEs?

CRITICAL THINKING QUESTIONS

1. How could comparative DNA studies aid us in under-
standing the roles of the different kinds of DNA present
in the eukaryotic chromosome?

2. How could mutations involving telomeres lead to
cancer?

Suggested Readings for chapter 15 are on page B-14.

Tamarin: Principles of
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Molecular Genetics

16. Gene Expression:
Control in Eukaryotes

©TheMcGraw-Hil
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GENE
EXPRESSION

Control in
Eukaryotes

Artificially colored scanning electron micrograph of

T-lymphocytes, white blood cells involved in the

immune system. The cells are seen in the

thymus gland, where they mature.

STUDY OBJECTIVES

1. To examine the control of transcription in eukaryotes 465

2. To analyze the genetic control of development
in eukaryotes 469

3. To study the mechanisms causing cancer 484

4. To study the genetic mechanisms that generate antibody
diversity 492

STUDY OUTLINE

Control of Transcription in Eukaryotes 465

Chromatin Remodeling 465

Specific Transcription Factors 466

Methylation of DNA 466

Signal Transduction 467

Transposons 468
Patterns in Development 469

Drosophila Development 469

Developmental Genetics of Drosophila 471

Plants 479

Other Models of Development 483
Cancer 484

Mutational Nature of Cancer 484

Viral Nature of Cancer 487

Environmental Causes of Cancer 492
Immunogenetics 492

Immunoglobulins 493

Antibody Diversity 494

T-Cell Receptors and MHC Proteins 498
Summary 505
Solved Problems 505
Exercises and Problems 506
Critical Thinking Questions 507
Box 16.1 Protein Motifs of DNA Recognition 480
Box 16.2 Chromosomal Painting 486
Box 16.3 AIDS and Retroviruses 502

464

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Control of Transcription in Eukaryotes

465

In this chapter, we turn our attention to the control
of gene expression in eukaryotes. We concentrate
on the roles that chromatin remodeling, specific
transcription factors (transcription activating pro-
teins), and DNA methylation play in determining
which genes are expressed at a particular time in a par-
ticular cell. We also look at some other possible factors in
the control of gene expression: transposons and 2 DNA.
We then look at the control of gene expression during
development, using the fruit fly as a model system. A sin-
gle cell, the zygote, becomes a whole organism through
controlled cascades of gene expression, pathways that
are highly conserved in evolution and relatively few in
number. Finally, we look at cancer — cell growth out of
control — and immunogenetics, the way in which im-
munological diversity is generated.

CONTROL

OF TRANSCRIPTION

IN EUKARYOTES

In prokaryotes, an RNA polymerase holoenzyme with its
promoter-recognizing sigma factor is generally active,
transcribing at high levels; repressors are needed to pre-
vent transcription. In eukaryotes, an RNA polymerase
holoenzyme (e.g., RNA polymerase II), with its promoter-
recognizing TFIID, is generally not transcribing; it needs
access to the promoter, which is usually wrapped
around nucleosomes, and it needs specific transcription
factors to become active (fig. 16.1). Thus, although the
parts of the transcribing machinery of prokaryotes and
eukaryotes are generally similar, the essence of prokary-
otic transcription is activity, whereas the essence of eu-
karyotic transcription is inactivity. In addition, eukary-
otes generally do not have operons; however, groups of
eukaryotic genes involved in the same pathway or func-
tion can be induced simultaneously by having common
enhancers that respond to the same specific transcrip-
tion factors. Such a group of genes is called a syn-
expression group.

Prokaryotes

RNA Polymerase

DNA

mommmmmfm^

Promoter

►

Transcription

Nucleosomes

►

No transcription

Eukaryotes

Chromatin
remodeling

Specific

transcription

factors

wmo^mmmmiK

►

Transcription

Figure 16.1 In prokaryotes, the default condition is active
transcription. In eukaryotes, the default condition is no
transcription since the DNA of promoters is usually wrapped
around nucleosomes and specific transcription factors are
needed to recruit the polymerase holoenzyme. Transcription in
eukaryotes is generally initiated when specific transcription
factors bind to enhancer sequences near the promoter, and
chromatin is remodeled at the promoter.

Chromatin Remodeling

For transcription to take place in eukaryotes, the DNA
must be available for the preinitiation complex to form,
with its RNA polymerase and general transcription fac-
tors. It appears that DNA wrapped around nucleosomes is
often not accessible for the formation of the preinitiation
complex, but is available for recognition by transcription-
activating proteins, also called specific transcription fac-
tors (as compared to the general transcription factors of
the RNA polymerase machine; see chapter 10). One
model of initiation of transcription by genes whose pro-

moters are wrapped around nucleosomes is for specific
transcription factors to recruit chromatin-remodeling
proteins. As we discussed in chapter 15, there are two
general classes of proteins that remodel nucleosomes:
histone acetyl transferases and ATP-dependent chro-
matin remodeling proteins such as the SWI/SNF complex
in yeast. Thus, the presence of one or more specific tran-
scription factors can begin the process of transcription
by recruiting chromatin-remodeling proteins that allow
the RNA polymerase access to the promoter.

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Chapter Sixteen Gene Expression: Control in Eukaryotes

Specific Transcription Factors

As we discussed in chapter 10, eukaryotic transcription
begins with the formation of a preinitiation complex
formed by the amalgamation of a group of general tran-
scription factors (such as TFIID in RNA polymerase II
formation). Proteins that exert control over transcrip-
tion at specific promoters are the specific transcription
factors (see figure 10.24). These proteins generally
have two domains: a domain that recognizes a specific
DNA sequence, and a domain that recognizes another
protein, such as a protein in the preinitiation complex.
Thus, these proteins recognize signals in the vicinity of
the promoter of a gene, bind there, and initiate tran-
scription. Currently, we believe that the majority of
specific transcription factors act by recruiting the
components of the RNA polymerase holoenzyme.
Thus, the binding of a specific transcription factor at a
promoter is the first step in the formation of a pre-
initiation complex at the promoter of a gene. Some
transcription-activating proteins also recruit chromatin-
remodeling proteins.

An example of a specific transcription factor is Dor-
sal, the product of the dorsal gene in fruit flies, active in
development. Dorsal controls the transcription of several
genes and at several different levels of protein concen-
tration. The ability to have different effects at different
concentrations is extremely important, allowing gradi-
ents of the same protein to control the expression of dif-
ferent genes. One gene Dorsal controls is rhomboid,
which has three sites in its promoter that Dorsal binds to,
initiating transcription. Another gene, twist, also has
three sites in its promoter that bind Dorsal, also initiating
transcription. However, the rhomboid sites are more ef-
ficient in binding Dorsal; thus, rhomboid is transcribed
at lower concentrations of Dorsal than twist is (fig. 16.2).
One other signal in the control of transcription that is of
current interest is methylation.

Methylation of DNA

The importance of methylation in DNA-protein interac-
tions is well known. In chapter 13, we showed that a par-
ticular DNA sequence could be protected from restriction
endonucleases if it were methylated. A small percentage
of cytosine residues are methylated in many eukaryotic
organisms, mainly in CpG sequences (see fig. 13.3); 80%
of the cytosines in CpG sequences in human DNA are
methylated. (Often, when we refer to a sequence of two
bases on the same strand of DNA, we put a "p" between
them — CpG — to indicate that they are on the same strand
connected by a phosphodiester bond and not on two
different strands as a hydrogen-bonded base pair.)

The degree of methylation of DNA is related to the si-
lencing of a gene. Genes that are dormant in one cell type
but active in another, or genes that are dormant at one
stage of development but active in another, are usually less
methylated when active and more fully methylated when
inactive. For example, adenovirus, a cancer-causing virus,
has been observed in many eukaryotic cell lines. In most
lines in which the adenovirus DNA has integrated into the
host chromosome, late viral genes are turned off. These
genes are highly methylated at their CCGG or GCGC sites.

In addition, chemicals that prevent methylation fre-
quently activate previously dormant genes. For example,
5-azacytidine inhibits methylation; X chromosomal
genes, which are normally deactivated, can be reacti-
vated by treatment with 5-azacytidine. There are numer-
ous other examples of the activation of genes after treat-
ment with this chemical. The activated genes lack
methylated cytosines that were previously methylated.
Finally, the possibility exists that DNA methylation can af-
fect the pattern of chromatin structure.

Recent work has also indicated that the methylation it-
self may not prevent transcription, but rather may be a sig-
nal for transcriptional inactivity In the thale cress plant, Arab-
idopsis thaliana, a protein named Mom (for Morpheus

fmj^m^fmjmm

rhomboid

Dorsal

DNA

twist

Low concentration

High concentration

Figure 16.2 The rhomboid and
twist genes each have three
enhancer sequences that are
recognized by the Dorsal
transcription factor. However, the
recognition sequences of the
rhomboid gene are more efficient
at binding Dorsal than the
recognition sequences of the
twist gene. Thus, rhomboid is
induced at both low and high
concentrations of Dorsal,
whereas twist is induced only at
high concentrations.

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16. Gene Expression:
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Control of Transcription in Eukaryotes

467

molecule), has been discovered that, when mutated, results
in genes that have heavy methylation levels but are actively
transcribed. Thus, the methylation level can be separated
from the transcriptional activity of genes, although the two
usually occur together. Arabidopsis is proving to be a good
model in the study of the role of methylation in transcrip-
tional activation because other common model organisms,
namely fruit flies, yeast, and the nematode, Caenorhabditis
elegans, do not have methylation of their DNA.

Further interest has been generated in the role of
methylation in controlling gene expression by the discov-
ery of 2 DNA, and the fact that 2 DNA can be stabilized by
methylation (see chapter 9). This observation has led to a
model of transcriptional regulation based on alternative
DNA structures. Sequences (such as CpG repetitions) that
could exist as 2 DNA exist as B DNA when being tran-
scribed. If the gene is to be silenced (turned off), the CpG
sequences are converted to stable 2 DNA by methylation,
which then blocks transcription. This possibility has gained
some interest because of the recent discovery of an
enzyme, double-stranded RNA adenosine deaminase
(ADAR1), that binds to 2 DNA sequences.

Signal Transduction

We return to the question of how specific transcription
activation factors appear at specific times. As we will de-
scribe in the section on development, control of gene

expression requires that genes be expressed at specific
times and under specific circumstances. If transcription
is usually controlled by specific transcription factors,
what determines the appearance of these factors at the
appropriate times and places? One common mechanism
is a signal transduction pathway, in which signals
pass from the external environment through the cyto-
plasm, into the nucleus.

For example, in a signal transduction pathway in-
volved in development of the fruit fly, the Toll protein
spans the cell membrane (fig. 16.3). It acts as a receptor
for the Spatzle protein, which, when detected, causes a
change in the cytoplasmic end of Toll, activating it. Acti-
vated Toll activates Pelle, a protein kinase that phos-
phorylates the Cactus protein, causing it to dissociate
from Dorsal. Once Dorsal dissociates from Cactus, which
acts to repress Dorsal, Dorsal becomes an active specific
transcription factor that can cross the nuclear membrane
and activate its target gene (fig. 16.3). We thus see that
Spatzle attaching to its receptor protein (Toll) on the cell
surface results in the activation of the target gene of the
Dorsal protein in the nucleus. These pathways can be-
come very complex, with many protein elements. More
elements mean more sensitive control of various
processes, often requiring that several conditions be met
before a gene is activated. In addition, these pathways are
usually conserved in evolution. A similar pathway,
though more complex, occurs in mammals in which the

Outside

Cell membrane

Inside

Cactus

Dorsal

Cytoplasm

Nuclear membrane

Nucleus

Figure 16.3 The signal transduction pathway. In this mechanism, the Spatzle protein outside of a cell interacts with the Toll receptor
protein, freeing the Dorsal protein to act as a transcription factor in the nucleus. When Toll binds Spatzle, spanning the cell
membrane, it changes the configuration of the interior domain of Toll, which then interacts with Pelle, causing it to phosphorylate the
Cactus protein. Previously, Cactus had been bound to Dorsal, making Dorsal inactive; phosphorylation of Cactus releases it from
Dorsal. Dorsal is then free to cross the nuclear membrane and act as a transcription factor.

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Chapter Sixteen Gene Expression: Control in Eukaryotes

target gene is interleukin-1, a protein in the immune sys-
tem that induces fever. In the mammalian pathway, the
signal protein is called Toll-like receptor-4, and the spe-
cific transcription factor is called NF-kB.

Transposons

We have already shown that transposons can affect gene
expression in prokaryotes, as, for example, in controlling
the flagellar phase in Salmonella (see chapter 14). Here
we show how transposons can alter or regulate eukary-
otic gene expression.

Barbara McClintock discovered transposons in eukary-
otes in the 1940s, without the aid of the tools of modern
molecular genetics. She won the Nobel Prize for her work
in 1983. She observed corn kernels that were streaked or
spotted, indicating a high mutation rate. After careful ge-
netic analysis, she showed that the mutability was due to
transposons, which she called controlling elements.

The Ac-Ds System

The Ac-Ds system consists of two transposons. McClin-
tock referred to the Ac (activator) transposon as an au-
tonomous element and to the Ds (dissociation) transpo-
son as a nonautonomous element. Ds cannot transpose
until Ac enters the genome. At that time, Ds can trans-
pose, be excised, or cause the chromosome it occurs on
to break. Ds affects the phenotype by blocking expres-
sion of the genes it transposes into, as well as by causing
the loss of alleles in acentric chromosomal fragments lost
when Ds breaks its chromosome.

In figure 16.4, we see three kinds of corn kernels: pur-
ple, bronze (light-colored), and bronze with purple
spots. The purple kernels result from dominant function-

ing alleles that provide enzymes in the pathway for pur-
ple pigment. In the kernels that are bronze without
spots, Ds elements have transposed into both copies of
the Bz2 locus, disrupting the pigment pathway. Without
the Ac element, the Ds elements remain in place, and the
kernels are a uniform bronze color. In the bronze kernels
with purple spots, the Ac element has entered the
genome in the genetic cross. In the presence of Ac, Ds
leaves its site in some of the cells, restoring activity to the
Bz2 locus. This restored activity creates purple spots in
those cells and in their progeny with the functioning Bz2
allele (see fig. 14. 34a). Ds and Ac elements have been
cloned and sequenced. They are typical transposons that
are very similar to each other. As might be expected,
however, Ds has a deletion that prevents it from produc-
ing transposase. For Ds to transpose, Ac must provide the
transposase. Ds apparently arose from Ac by deletion.

It is interesting to note that one of Mendel's original
seven characteristics of pea plants, wrinkled peas (rr: see
fig. 2.3), is caused by a transposon that inserts in the gene
for Starch-branching enzyme I. When this gene is func-
tional, the cells produce both branch-chained amy-
lopectins and straight-chained amylose. If the gene fails to
produce this enzyme, more sugar is present in the seeds,
leading to greater osmotic pressure and, therefore, greater
water content. More water is lost from these seeds upon
maturation, resulting in greater shrinkage and wrinkling
than in the wild-type seeds (RR and Rr). The transposon
that disrupts this gene is about eight hundred base pairs
long and is very similar to the Ds transposon in maize.

The Ac-Ds system disrupts transcription through an
invasive element that seems harmful (or at best neutral)
to the organism. Mating-type control in yeast, by con-
trast, is a highly evolved system whose alternative ex-
pressions are advantageous to the organism.

Figure 16.4 The Ac-Ds mutability system in corn. Shown is an ear of corn with purple and bronze kernels. The purple kernels have
no transposons. The bronze kernels (light-colored) lack the purple pigment because they have a Ds element in both copies of the
Bz2 locus, disrupting pigment production. Without an Ac element present, the kernel remains bronze. In the presence of the Ac
element, the Ds element can leave its position, restoring the allele and producing a purple spot in a bronze kernel. Spots differ in size
based on when the Ds element was excised during the development of the kernel: early excision yields large spots; late yields small

spots. (Com ear courtesy of Dr. Neelima Sinha; Photo by the author.)

Tamarin: Principles of
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Control of Mating Type in Yeast

Transposons determine the mating type in yeast. Haploid
yeast cells exist in one of two mating types, a and a, de-
termined by the MATa and MATa alleles. Homothallic
strains of yeast switch mating types, as often as every
generation. (The term homothallic, a misnomer, means
that every cell is alike — each can mate with any other.
The term was applied before scientists realized that the
cells could change mating types.) Homothallism is deter-
mined by the dominant HO allele that codes for an en-
donuclease that initiates transposition. Strains that do not
change mating type are heterothallic, determined by
the recessive bo allele; no active endonuclease is present
to allow transposition, and thus they undergo no change
in mating type.

The ability to switch mating types in a single cell im-
plies that both forms of the mating-type gene are present
in each cell. In 1971, Y. Oshima and I. Takano proposed
that mating type was controlled by a transpositional
event, similar to the Ac-Ds system in corn or the flagellar
phase in Salmonella. Later genetic and recombinant
DNA studies revealed the exact mechanism.

The third chromosome in yeast contains the mating-
type locus (MAT). Silent (unexpressed) copies of the
mating-type alleles are found on the left and right arms of
the same chromosome (fig. 16.5). HML contains the silent
a allele and HMR contains the silent a allele. In transpo-
sition, a copy of one or the other (HMR or HML) moves
to the MAT site, replacing whatever allele was there to
begin with. This mechanism has been called a cassette

Centromere

(a)

HML

MAT

HMR

HML

MATa

HMR

(b)

-o-

HML

(c)

MATa

HMR

Figure 16.5 Role of transposition in controlling the mating type
in yeast, (a) Mating-type loci on the third chromosome. MAT is
the active mating-type locus. HML and HMR are silent loci,
carrying the two mating-type alleles, a and a, respectively.
(b) Transposition of HML to MAT results in the MATa allele at
the MAT site and the a mating type, (c) Transposition of HMR to
MAT results in an active MATa allele, yielding the a mating type.

mechanism. The MAT site is analogous to a cassette
player, with HMR and HML similar to cassette tapes.
Transposition brings a new "tape" to the "cassette player."

MATa and MATa each begin a genetic cascade that ac-
tivates certain genes and represses others. For example,
MATa codes for two proteins. The MATal protein acti-
vates the transcription of an a-factor (a pheromone)
gene and an a-factor (pheromone) receptor gene.
(Pheromones are chemical signals, analogous to hor-
mones, that convey information between individuals.)
The MATa2 protein represses the a-specific genes. Con-
jugation requires the emission of one type of pheromone
and the reception of the other type: An a cell emits a fac-
tor and is receptive to a factor; an a cell emits a factor
and is receptive to a factor.

In summary, then, transposons can affect eukaryotic
gene expression. However, with the exception of a few
systems such as mating-type determination in yeast,
transposons appear to have a random, disruptive effect
on developmental processes.

PATTERNS IN DEVELOPMENT

Development is the orderly sequence of change that
produces increasing complexity during the growth of an
organism; it is controlled by the differential expression of
genes. A central problem of development is explaining
genomic equivalence, how cells with identical genetic
material can give rise to different cell types. A favored
approach to understanding the genetic control of devel-
opment in higher organisms requires first learning the
details of the normal developmental process in an organ-
ism and then studying the disruption of this normal
process by mutation and experimental manipulation.

At one point, scientists believed that development
might take place through permanent changes in chro-
mosomes. The idea was that subtle changes might occur
in chromosomes during development; these changes
would not be observable by karyotyping a cell. Geneti-
cists have explored this hypothesis by several methods.
However, the cloning of a mammal, such as the sheep
Dolly (see chapter 13), from the cell of an adult demon-
strates that adult nuclei are totipotent: Any adult nu-
cleus can give rise to the whole organism and all its cel-
lular types, indicating the chromosomes are intact.

Drosophila Development

The fruit fly, Drosophila melanogaster, has emerged as
an excellent model organism for the study of develop-
ment. The zygote develops from the egg, in maternal
cytoplasm. Maternal messenger RNAs and proteins are
the first expressed in the embryo. These substances

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Chapter Sixteen Gene Expression: Control in Eukaryotes

first determine the broad pattern of the embryo. Then,
through signal pathways involving numerous specific
transcription factors, they initiate a cascade of gene ex-
pression that eventually determines the fate of each
cell. As we will see, many parallels exist between the
fruit fly and higher organisms.

We will concentrate on two overall patterns of devel-
opment here: the formation of the basic body plan
(anterior-posterior and dorsal-ventral polarity, which re-
sults in a segmented embryo that has a front, back, top,
and bottom) and the determination of gene expression
within segments.

Drosophila development begins within a follicle that
contains the oocyte surrounded by follicle and nurse
cells. The fifteen nurse cells, along with the oocyte, were
derived from four divisions of an earlier germ-line cell
(fig. 16.6). The nurse cells maintain connections to each
other and to the oocyte by cytoplasmic bridges, open-
ings in the membranes surrounding the cells. Thus, the
nurse cells can readily pass materials (messenger RNAs
and proteins) into the oocyte.

After fertilization, the diploid nuclei divide thirteen
times in the space of about 3.5 hours, forming a synci-
tium — a group of nuclei without cell membranes. Dur-
ing this time, most of the nuclei migrate to the inner sur-
face of the developing embryo, where cell membranes
eventually form, producing a cellular blastoderm. Dur-
ing the syncitial period, materials can move freely
through the cytoplasm. At the posterior end of the em-
bryo, several cells, called pole cells, that will eventually
form the germ cells of the developing fly are set aside
(fig. 16.7). Development then proceeds through gastru-
lation, in which cells grow inward, forming the basic
germ layers of the embryo (mesoderm, endoderm,
and ectoderm). From these layers, various adult struc-

Nucleus

Nurse cells

Follicle cells

Figure 16.6 The follicle from a fruit fly, Drosophila, consisting of
the oocyte, fifteen nurse cells arising from four divisions of a
germ-line cell that also gave rise to the oocyte, and follicle cells.

tures will arise. At about six hours of development, fur-
rows become visible in the embryo, delineating seg-
ments. The first segments visible are called paraseg-
ments. They do not give rise to the later segments of the
embryo, but rather overlap the later segments in a simple
fashion: Each later segment is made up of the anterior
end of one parasegment and the posterior end of the
next (fig. 16.8). This distinction is meaningful since, as
we shall see later, some genes express themselves within
the borders of parasegments rather than segments.

The fully segmented embryo has an anterior region,
destined to be the head; three thoracic segments, which
will give rise to the thorax (the middle region of the fly
containing wings and legs); and eight abdominal seg-
ments that will give rise to the abdomen. The embryo
also has an anterior tip, the acron, that will give rise to
structures at the very head end — eyes, and antennae; and

0.5 hours

1.5 hours

^ooo o o o o o O OOOOOOOOOO o~q

°°0 ~ - ~n OO

o o o o o oo qo£-H

2.5 hours

3.25 hours

Figure 16.7 Development of the fertilized Drosophila egg after
laying. Pole cells, which will be future germ cells, are set apart
at about 2 hours. A syncitial blastoderm forms at about 2.5
hours, followed by a cellular blastoderm, consisting of about
five thousand cells, at about 3.25 hours.

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a posterior tip, called the telson, that will give rise to the
internal structures at the very posterior end of the fly.
The fates of these segments have been determined by
treating them with various harmless dyes and tracing
where the dyes end up. A projection of adult structures
on embryonic tissue is called a fate map.

Developmental Genetics o/Drosophila

The General Body Plan

The role genes play in determining the general axes of
the body plan has been worked out at several levels.
First, mutations causing female sterility were isolated.
(C. Nusslein-Volhard and E. Wieschaus were instrumen-
tal in systematically isolating many of these mutants;
they were awarded Nobel prizes for this work.) For ex-
ample, among normal female flies that were sterile, some

Christiane Nusslein-Volhard
(1942- ). (Courtesy of
Christiane Nusslein-Volhard.)

Eric F. Wieschaus
(1947- ). (Courtesy of Dr.
Eric F. Wieschaus. Photograph
by Denise Applewhite.)

T2a

C1 C2 C3 T1 T2 T3 A1 A2 A3 A4 A5 A6 A7 A8

P a

12 3 4

P a

6 7 8 9 10 11 12 13 14

Figure 16.8 The relationship between
parasegments, segments, and the adult fruit fly.
The initial segments of the fly are called
parasegments; the nonsegmented parts of the
embryo are called the acron at the head end
(accounting for eyes and antennae) and the
telson at the tail end (accounting for the end of
the alimentary canal). Later segments are made
up of the posterior end of one parasegment and
the anterior portion of the next (p, a). The later
segments map directly on the adult body,
accounting for mouthparts (mandible, maxilla,
and labium), thoracic segments (1-3), and
abdominal segments (1-8). (H is for head.)
(From P. A. Lawrence, The Making of a Fly, Copyright ©
1992 Blackwell Science, Ltd., Oxford, England. Reprinted
by permission.)

Parasegments

7 8 9 10 11

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Chapter Sixteen Gene Expression: Control in Eukaryotes

produced embryos without heads or thoracic struc-
tures. The gene for this mutation, which has since been
cloned and sequenced, is called bicoid (fig. 16.9). It
codes for a specific transcription factor, the Bicoid pro-
tein. (Remember that gene names are italicized, using
the first letter, lowercase for recessive and uppercase for
dominant; the protein product of these genes is not ital-
icized, but the first letter is capitalized.)

Pricking the anterior end of a normal embryo, caus-
ing the loss of cytoplasm from that end (fig. 16.10), can
mimic these mutants. This experiment indicates there is
some cytoplasmic localization determining the develop-
ment of the anterior end of the fly. To support that idea
further, it was possible to get normal development from
a bicoid fly by injecting the anterior end with cytoplasm
from a normal embryo (fig. l6.10£>).This process of facil-
itating normal development by manipulating the embryo
is termed a rescue experiment. By probing with a com-
plementary oligonucleotide to the bicoid messenger
RNA, researchers found that the bicoid messenger RNA is
formed in the nurse cells and then passed into the
oocyte, where it becomes localized at the anterior tip
(fig. 16. 11a). After fertilization, this messenger RNA is
translated into Bicoid, which begins to diffuse from the
anterior end of the egg, until it reaches about 50% of the
length of the egg. The protein can be visibly located
by treating the eggs with antibodies to the protein;
these antibodies can then themselves be made visible
(fig. 16.11&).

The Bicoid protein is called a morphogen, a sub-
stance that diffuses through the egg and by its concen-
tration determines the developmental fate of that part of
the embryo. Although nurse cells are germ-line cells, they
are of maternal origin and not from the embryo. Since

Posterior

.... »

^ ■ "tr- *+■

T2 .l

i i

r3 A*

■ - , ' -

j i . A *

1 J # 4

wild-type

A3 A4 A5

bicoid

Figure 16.9 Two Drosophila larvae, with cuticular patterns
visible on the ventral surfaces. On the top is the wild-type with
the cuticular pattern coinciding with thoracic and abdominal
segments. On the bottom is a bicoid mutant, lacking head and
thoracic Structures. (Courtesy of Christiane Nusslein-Volhard.)

maternal cells, not the embryo itself, produce this mor-
phogen, the gene responsible for its production is called
a maternal-effect gene.

Other maternal-effect genes are involved in formation
of the anterior pattern that produces headless embryos.
However, they don't appear to produce a morphogen.
Rather, these genes seem to be involved in the transport,
stabilization, and modification of the morphogen. In mu-
tants of these other genes (swallow, exuperantia), Bi-
coid is found in the nurse cells but not in the embryo; cy-
toplasm from the nurse cells of these mutants can rescue
bicoid mutants, indicating that the morphogen is present
but not delivered to the oocyte. Only mutants of the bi-
coid gene itself cannot rescue the various headless mu-
tants because only in bicoid mutants is the morphogen it-
self missing.

Through experiments similar to the ones described
for bicoid, four independent signaling pathways of
maternal-effect genes have been isolated. These pathways
determine the general body plan of the developing em-
bryo: anterior, posterior, terminal, and dor so-ventral. The
posterior pattern is controlled by the gradient of a pro-
tein, Nanos. Before the nanos gene is active, producing
messenger RNA, the first posterior gene active is oskar;
the localization of oskar messenger RNA then defines the
localization of nanos messenger RNA. Mutant embryos
can be rescued by wild-type cytoplasm; the nanos mes-

Anterior Wild-type egg Posterior

bicoid larva

Remove anterior
cytoplasm

A1 A2A3A4A5 A6A7 A8

(a)

bicoid egg

Wild-type larva

^fil^AlA2A 3 A4 A5 A6 A7 A8

Inject wild-type
anterior cytoplasm

(b)

Figure 16.10 Experiments to demonstrate that a cytoplasmic
localization at the anterior end of the fruit fly egg determines
anterior structures, (a) A wild-type egg has anterior cytoplasm
removed, resulting in a larva lacking anterior structures, similar
to a bicoid mutant, (b) A bicoid mutant egg has anterior
cytoplasm from a wild-type egg injected into the anterior of the
egg, resulting in a larva indistinguishable from the wild-type.

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(a)

b/'co/'c/mRNA

(b)

Bicoid protein

Figure 16.11 The bicoid morphogen first appears in the fruit
fly egg as (a) messenger RNA at the anterior end of the egg.
After fertilization, the messenger is translated into [b) Bicoid
protein that diffuses toward the posterior end of the embryo.
(Courtesy of Daniel St. Johnston.)

senger RNA is localized at the posterior tip of the embryo
and produces a protein that diffuses from that tip.
Maternal-effect genes that act in a somewhat different
manner control the other two pattern systems in the de-
veloping embryo.

The terminal pattern controls development of both
ends of the embryo; a key gene is torso. This gene codes
for a membrane-bound tyrosine kinase receptor protein
that is found evenly distributed on the outer surface of

the developing embryo. (Tyrosine kinases phosphorylate
the amino acid tyrosine in specific proteins.) Apparently
other genes in follicle cells located only at the poles of
the egg produce a substance that activates the torso ty-
rosine kinase receptor, making it active in only the poles
of the egg (fig. 16.12). A maternal-effect gene, Toll, that
also produces a membrane receptor, controls the dorso-
ventral axis. Thus, we see that four pathways of maternal-
effect genes determine the major body plan of the egg.
Two of the pathways are determined by genes that result
in diffusion of a morphogen (bicoid and nanos), and two
are determined by genes for membrane receptors (torso
and Toll). About thirty maternal-effect genes are known
(table 16.1).

Follicle cells

Figure 16.12 The Drosophila follicle, showing follicular cells
(green) at the tip of the oocyte that secrete a substance that
activates the Torso {torso gene) tyrosine kinase at the areas
marked by red lines; the inactivated kinase is located around
the surface of the oocyte.

Table 16.1 Maternal-Effect Genes in Drosophila (Allelic Designations in Parentheses)

Posterior

Terminal

Dorso-Ventral

bicoid (bed)

nanos (nos)

torso (tor)

Toll (Tl)

swallow (swa)

oskar (osk)

trunk (trk)

nudel (ndl)

exuperantia (exu)

vasa (vas)

torsolike (tsl)

pipe (pip)

bicaudal (bic)

tudor (tud)

polehole lfs(l) ph]

windbeutel (wbl)

Bicaudal-D (BicD)

stauffen (stau)

Nasrat ffs(l) N]

snake (snk)

Bicaudal-C (BicC)

valois (val)

easter (ea)

pumilio (pum)

cactus (cact)
spdtzle (spz)
tube (tub)
pelle (pll)

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Activity of maternal-effect genes in the follicle cells is
controlled by an interaction between the oocyte itself
and the follicle cells. Follicle cells at the anterior of the
oocyte produce bicoid messenger RNA as a default con-
dition. At the posterior of the oocyte, the follicle cells
produce nanos messenger RNA, along with several other
gene products. These follicle cells are induced to action
by the product of the gurken gene in the oocyte; the
oocyte nucleus is located posteriorly at this point, and its
gene products can be directed to the posterior of the

oocyte, where they diffuse to adjacent follicle cells.
These cells have a receptor on their surfaces, the product
of the torpedo gene, that recognizes the gurken gene
product. Through signal transduction, these follicle cells
are induced to express the nanos gene (fig. 16.13).

At this point, some product of these follicle cells in-
duces a reorganization of the microtubules in the oocyte,
causing the oocyte nucleus to move anteriorly and dor-
sally Now, the same gurken-torpedo interaction takes
place, causing these follicle cells to induce the dorso-

Nurse cell

Follicle cell

Posterior

Dorsal

Ventral

(a)

(b)

Figure 16.13 The interaction of the oocyte nucleus and follicle cells early in development, (a) The oocyte nucleus, located posteriorly
in the oocyte, activates posterior follicle cells. These cells will later provide the nanos messenger RNA to control posterior
development of the embryo. After this interaction, a product of the follicle cells causes a rearrangement of microtubules in the oocyte,
moving the oocyte nucleus anteriorly and dorsally. There the same interaction takes place, in this case activating follicle cells to
control dorsal development, (b) The oocyte signal {red circles) is the product of the gurken gene; it interacts with a receptor (blue
Y-shapes) on the surface of follicle cells, the product of the torpedo gene.

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ventral axis. As of yet, we don't know all of the signaling
going on, nor why two similar cell types react differently
to the same oocyte signal (Torpedo), but we do know
that maternal-effect genes in the follicle cells are induced
by the oocyte itself.

Maternal-effect genes are the first in a series control-
ling a cascade of gene expression that eventually deter-
mines the fates of individual cells in the developing fly
embryo. The rest of the genes are zygotic genes, genes ac-
tive in the cells of the embryo itself. As we move down
this cascade of genes, we go from broad patterns to more
and more focused gene activity. We go from a single cell
with gradients of morphogens to stripes of cells with
different genes active.

Segmentation Genes

Once the general body plan of the fly is in place, devel-
opment continues in the formation of parasegments and
then segments. The various organs of the fly's body are
produced from these segments. Further development is
now under the control of the zygote's own genes, gener-
ally referred to as segmentation genes. These genes fall
into three general categories: gap genes, pair-rule genes,
and segment-polarity genes (table 16.2; fig. 16. 14). These
genes are activated sequentially, each by the genes acti-
vated before it; each group controls a smaller and more
focused domain of the fly's development. In this discus-

sion, we will concentrate on the anterior-posterior
system.

The maternal-effect genes of the anterior-posterior
system have created Bicoid and Nanos gradients. The seg-
mentation genes increment, narrow, and focus these gra-
dient signals until fourteen distinct bands form, corre-
sponding to the fourteen parasegments that develop,
creating compartments that the tissues of the fly arise
from (e.g., wings, legs, bristles).

The gap genes were first discovered as mutants that
resulted in missing segments in the embryo (fig. 16.14).
The Bicoid and Nanos gradients act on gap genes, specif-
ically hunchback. Although the Hunchback protein is
present in the egg from maternal production, the mater-
nally supplied quantity is apparently not significant. Bi-
coid and Nanos independently create a Hunchback gra-
dient that is maximal at the anterior end of the embryo,
due to activation by Bicoid, and absent at the posterior
end, due to Nanos repression. Bicoid is a specific tran-
scription factor that can bind to at least six sites in the
promoter region of the hunchback gene. Three of these
sites are strong binding sites and three are weak. Thus,
depending on the concentration of Bicoid in the gradi-
ent, different levels of Hunchback are produced, creat-
ing the Hunchback gradient. Experiments with extra
copies of the bicoid gene show that it is the actual quan-
tity of Bicoid present at a particular point, and not the
shape of the gradient, that actually determines the effect.

Table 16.2

Segmentation

Genes in Drosophila

Class

Locus

Allelic Designation

Chromosome

Gap

Kruppel

knirps

kni

hunchback

Pair-rule

paired

prd

even-skipped

eve

odd-skipped

odd

barrel*

brr

runt

run

engrailed

Segment

■polarity

cubitus interruptus

wingless

gooseberry

gsb

hedgehog

fused

patch

pat

Source: Reprinted with permission from Nature, Vol. 287, C. Nusslein-Volhard and E. Wieschaus. Copyright © 1980 Macmillan Magazines Limited.
* barrel is a synonym of the hairy gene.

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Chapter Sixteen Gene Expression: Control in Eukaryotes

Gap

Pair-rule

Segment polarity

Kruppel

Even-skipped

Odd-skipped

Gooseberry

Figure 16.14 Segmentation genes of Drosophila fall into three categories: gap, pair-rule, and segment polarity. To the left of
each pair is the wild-type larva with cuticular pattern that indicates segment position; to the right is the mutant larva. An
example of a gap mutation is Kruppel, which eliminates the three thoracic and five of the eight abdominal segments (shaded
in the wild-type larva). Pair-rule genes are shown that eliminate even {even-skipped) or odd {odd-skipped) segments (counting
from the abdominal segments). An example of a segment polarity gene is gooseberry, in which the posterior portion of each
segment behaves like a mirror image of the anterior portion of the segment. (Reprinted with permission from Christiane Nussiein-
Volhard and Eric Wieschaus, "Mutations affecting segment number and polarity in Drosophila," Nature, 287:795-80, 1980. Copyright © 1980 Macmillan
Magazines, Ltd., London, England.)

Presumably, as more Bicoid is present, it binds to more
of the hunchback promoter sites, resulting in greater
transcriptional activity.

At least three gap genes are controlled by the con-
centrations of the specific transcription factor hunch-
back: Kruppel, knirps, and giant. In response to the
Hunchback gradient, these three genes are expressed in
discrete stripes in the embryo (fig. 16.15). Both anterior
and posterior edges of the Kruppel stripe are controlled
by Hunchback concentration; Hunchback concentration
also controls the anterior edges of the Knirps and Giant
stripes. The posterior edges of the Knirps and Giant
stripes are controlled by the gradient of the Tailless pro-
tein, which is controlled in turn by the terminal
maternal-effect gene, torso (fig. 16.15). We know the dis-
tributions of these proteins by antibody studies, and we
know the limits of the protein distributions from studies
of various mutants that lack the clear edges of the stripes.
For example, the borders of the Kruppel stripe are
changed in hunchback mutants in accordance with the
number of copies of the genes. We have thus gone from
very broad and fuzzy regions of maternal-effect gene
products to more defined bands of gap gene products.

Interaction of the gap gene proteins then controls
transcription of the pair-rule genes (see fig. 16. 14). These
genes affect alternate sets of segments, even and odd. For
example, mutants of the even-skipped gene cause the
loss of the even-numbered segments, counting by the ab-
dominal segments (loss of two thoracic segments as well

Kruppel knirps giant

Posterior

■*— •

CD
O

o
O

Figure 16.15 Three discrete bands of gene expression
{Kruppel, knirps, and giant) in the developing Drosophila
embryo. These bands come about because of the gradients of
Hunchback and Tailless proteins. The Hunchback protein level
controls the anterior edge of gene expression of Kruppel,
knirps, and giant, as well as the posterior edge of the Kruppel
gene expression. The Tailless protein level controls the posterior
end of knirps and giant gene expression. The nature of these
border edges is verified in mutations of the hunchback and
tailless genes that result in different limits. The three genes
{Kruppel, knirps, and giant) are transcription factors, further
controlling gene expression in these regions of the embryo.

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as abdominal segments 2, 4, 6, and 8). Finally, the segment-
polarity genes are controlled by the pair-rule genes, re-
sulting in genes that affect all segments (see fig. 16.14).
For example, mutants of the gooseberry gene modify the
posterior half of each segment, making it the mirror im-
age of the anterior half.

As development continues, and different classes of
segmentation genes are activated, the borders of stripes
of activation for these various genes become sharper and
sharper, until cell-cell interactions focus the expression
of different genes to neighboring cells. For example, we
see in figure 16.16 the narrowing and sharpening of the
even-skipped and fushi tarazu bands in the developing
embryo. (The gene fushi tarazu, meaning "not enough
segments" in Japanese, is a pair-rule gene.)

Most segmentation genes are specific transcription
factors, genes that interact with DNA to activate or re-
press transcription. Thus, pattern formation in develop-

ment is a process of activating different genes in se-
quence, gradually narrowing the scope of which cells ex-
press a particular gene. There is one final group of genes
we will discuss in this developmental cascade in
Drosophila. At this early stage of development, these
genes, the homeotic genes, take control of the devel-
opment of the segments.

Homeotic Mutants

In homeotic mutants, one cell type follows the develop-
mental pathway other cell types normally follow. These
genes define the future development of segments based on
the pattern of expression of the segmentation genes before
them. When they mutate, they switch the development of
that segment to an adjacent segment, usually anterior to it.
Homeotic genes are also called memory genes because
they set the developmental fate of a segment, a fate that is

(a)

(b)

(d)

(e)

(c)

Figure 16.16 Photos a-e show how the margins of expression of two gap genes, fushi tarazu (brown) and even-skipped (gray),
narrow and sharpen as time goes on (between about hours 3 and 4 of embryonic development). The stripes appear from staining
with antibodies against the proteins. (From P. A. Lawrence, "The making of a fly," Blackwell Publications, 1992.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

"remembered" from one cell division to the next. They are
also called master-switch genes since they control the ac-
tivity of many other genes.

Two major homeotic gene complexes are known in
Drosophila melanogaster (fig. 16.17): the bithorax
complex (BX-C), analyzed extensively by E. Lewis,
D. Hogness, and their colleagues, and the Antennapedia
complex (ANT-C), worked on extensively by W. Gehring,T
Kaufman, and their colleagues. (The two regions together
are known as the Hom-C region.} Genes in the Antenna-
pedia complex control the fate of the anterior develop-

Edward B. Lewis
(1918- ). (Courtesy of
Dr. Edward B. Lewis.)

Drosophila

Walter J. Gehring
(1939- ). (Courtesy Dr.
Walter J. Gehring.)

Abd-B

abd-A

I I

Ubx Antp Scr Dfd Zen pb lab

l l

i i r

1 1 i

i i

BX-C

ANT-C

Figure 16.17 A map of the homeotic complexes ANT-C and
BX-C in Drosophila, and the regions of the body in which the
genes are expressed, mapped on a ten-hour embryo. Note
that the genes are expressed from right to left, or in an anterior
to posterior direction. Dotted lines indicate lack of detectable
function at this stage in development. Embryonic segments are
intercalary (int), maxillary (max), labial (lab), thoracic (T1-T3),
and abdominal (A1-A8). Genes are labial (lab), proboscipedia
(pb), Zerknullt (Zen), Deformed (Dfd), Sex combs reduced (Scr),
and Antennapedia (Antp) in ANT-C and Ultrabithorax (Ubx),
abdominal-A (abd-A), and Abdominal-B (Abd-B) in BX-C. Note
that Zen is unique in specifying information for the dorsal-
ventral axis rather than the anterior-posterior axis. (Reprinted
from Cell, Vol. 68, W. McGinnis and R. Krumlauf, pp. 283-302, Copyright ©
1992, with permission from Elsevier Science.)

ment of the fruit fly (head and anterior thorax), whereas
genes in the bithorax complex control the fate of poste-
rior development (posterior thorax and abdomen). Muta-
tions in the genes of these complexes can change the fate
of development of whole sections of the fly. For example,
Nasobemia, an Antennapedia-complex mutant, causes
legs to grow where antennae would normally be located
(fig. 16.18); and bithorax, a bithorax-complex mutant,
produces flies with two thoraxes (four-winged diptera;
fig. 16. 19). The genes in these complexes are arranged in
order of their progressive action from anterior to poste-
rior on the fly (see fig. 16.17). One model of action for
these genes suggests that they require the action of the
genes of the adjacent anterior segment plus the action of
that homeotic gene itself. Thus, loss of function of a par-
ticular gene by mutation would cause a segment to de-
velop like the previous section in the anterior direction.

The Homeo Box

Using recombinant DNA techniques, W. Gehring and his
colleagues found a consensus sequence of 180 base pairs
of DNA in genes of the Antennapedia and bithorax
complexes. Further probing localized this same segment
of 180 base pairs to about a dozen genes in Drosophila,
all with homeotic or segmentation properties. They thus
called this DNA sequence the homeo box. The nu-
cleotides of the homeo box are translated into a peptide
region of 60 amino acids called the homeo domain
(fig. 16.20, box 16.1).

Using a recombinant probe for the homeo box, or a
computer search for the consensus sequence, re-
searchers found it in the genes of plants, yeast, sea
urchins, frogs, and human beings. This high degree of se-

-X

O, ^

Figure 16.18 Nasobemia, a mutation that causes legs to grow
in the place of antennae on the head of a Drosophila.

(Courtesy of Dr. Walter J. Gehring.)

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Patterns in Development

479

(a)

(b)

Figure 16.19 A normal fruit fly (a) and a bithorax mutant (b).
The bithorax mutant is actually the product of a combination of
three mutations that produce a fly with an almost perfect
second thorax with its own set of wings. (Courtesy of E. B. Lewis,
California Institute of Technology.)

quence conservation across widely divergent groups of
organisms indicates that the sequence is crucial to the
functioning of homeotic genes and that the mechanism
arose early in evolutionary time.

The conservation of homeotic control throughout
evolutionary history is evident from the fact that the
homeotic genes that control the development of fruit
flies are also found in mammals. During evolutionary his-
tory, the mammalian genome has been duplicated four
times. Thus, there are four homeotic clusters in mice,
called Hox clusters, on four different chromosomes
(fig. 16.21). With multiple copies, the genes could be
modified by evolution while still maintaining one copy
functioning as originally intended. This duplication has
allowed increased complexity in higher eukaryotes. As the
Drosophila homeobox genes function from anterior to
posterior, so do the homeobox genes in other organisms.

Plants

Much work is being done in determining the genetic
control of development in plants. A favored model is the
thale cress, Arabidopsis thaliana, a member of the mus-
tard family (fig. 16.22). It is a dicotyledonous an-
giosperm, ideal for the study of flower development, a
current focus of attention. Flowers have an arrangement
of repeated units not unlike the segmentation found in
fruit flies.

Flower development takes place in two phases, floral
induction and pattern formation. In floral induction, the
shoot apical meristem sets aside a floral meristem.
The organ primordia are then generated. There are four
primordia, in the form of four whorls, that make up a
flower. Outermost is the sepal whorl, then the petal
whorl, then the stamen whorl, responsible for the male
parts of the flower, and finally the innermost carpel
whorl, responsible for the female parts of the flower (the
pistil; fig. 16.23). The genetics of development in plants is

Antennapedia
Mouse MO- 10
Frog MM3

1 20

Arg Lys Arg Gly Arg Gin Thr Tyr Thr Arg Tyr Gin Thr Leu Glu Leu Glu Lys Glu Phe

Ser Lys Arg Gly Arg Thr Ala Tyr Thr Arg Pro Gin Leu Val Glu Leu Glu Lys Glu Phe

Arg Lys Arg Gly Arg Gin Thr Tyr Thr Arg Tyr Gin Thr Leu Glu Leu Glu Lys Glu Phe

Antennapedia
Mouse MO- 10
Frog MM3

21 40

His Phe Asn Arg Tyr Leu Thr Arg Arg Arg Arg lie Glu lie Ala His Ala Leu Cys Leu

His Phe Asn Arg Tyr Leu Met Arg Pro Arg Arg Val Glu Met Ala Asn Leu Leu Asn Leu

His Phe Asn Arg Tyr Leu Thr Arg Arg Arg Arg lie Glu lie Ala His Val Leu Cys Leu

Antennapedia
Mouse MO- 10
Frog MM3

Thr Glu Arg Gin lie Lys He

Thr Glu Arg Gin lie Lys lie

60
Trp Phe Gin Asn Arg Arg Met Lys Trp Lys Lys Glu Asn
Trp Phe Gin Asn Arg Arg Met Lys Tyr Lys Lys Asp Gin
Trp Phe Gin Asn Arq Arg Met Lys Trp Lys Lys Glu Asn

Figure 16.20 The homeo domain of three genes: the MO-10 gene from the mouse (Mus), the MM3 gene from
the frog (Rana), and the Antennapedia gene from Drosophila, which is considered the consensus sequence;
amino acids in red differ from this sequence. (From W. J. Gehring, Scientific American, November 1985. Reprinted with
permission of Walter J. Gehring.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

BOX 16.1

Different motifs have been
found in specific transcrip-
tion factors and other pro-
teins that bind to DNA. In homeo do-
mains, amino acids 31 to 38 and 41 to
50 form a helices. The configuration
of two a helices in a protein, sepa-
rated by a short segment (called a
"turn"), has been found in many pro-
teins that bind to DNA (e.g., Cro, A. re-
pressor, CAP protein). It is called the
helix-turn-helix motif. One a he-
lix recognizes a DNA sequence by fit-
ting into the major groove, and the
other helix stabilizes the configura-
tion (fig. 1).

The helix-turn-helix (or helix-
loop-helix) motif appears in some
proteins that bind to DNA. However,
different motifs have also been found
in other proteins that bind to DNA.
These include the zinc finger, the
leucine zipper, and the basic/
helix-loop-helix/leucine zipper.
The zinc finger, a fingerlike projection
of amino acids, whose base consists
of cysteine and histidine residues
binding a zinc ion, was first discov-
ered in 1985 by A. Klug and his

Experimental
Methods

Protein Motifs of DNA
Recognition

colleagues in the transcription factor
TFIIIA in Xenopus (fig. 2). These fin-
gers are referred to as C 2 H 2 proteins
because two cysteines (C 2 ) and two
histidines (H 2 ) are involved. There are
also C x proteins in which x is either
4, 5, or 6, referring to the number of
cysteines involved in the chelation
of the zinc ion, and other variants of
protein structures formed around
zinc ions.

Another motif was discovered in
analyzing a DNA-binding protein
from rat liver nuclei. Scientists no-
ticed that in a-helical regions of the
protein, a repetition of leucines oc-
curred every seven residues for se-
quences as long as forty-two residues.
In a helical configuration, these
leucines would line up on one side of

the protein. When a computer search
for sequences of this type was done,
several other proteins, believed to
bind to DNA, showed up with this
configuration, including three cancer-
causing genes, c-myc, fos, and jun,
and a transcription-regulating protein
in yeast. Using the computer, the sci-
entists developed the leucine-zipper
model, in which two helices with
leucine repeats would interdigitate
the leucines, in zipper fashion, to
form a stable molecule (fig. 3). This
zipper could provide a scaffolding for
other amino acids that could then
recognize specific DNA sequences in
order to perform their functions.

A recently discovered DNA-binding
motif, the basic/helix-loop-helix/
leucine zipper, is a series of basic
amino acids followed by the helix-
loop-helix and then a leucine zipper
(fig. 4). This motif is found in the Myc
oncoprotein and in a transcription
factor, Max, that binds with Myc.
Knowing that specific motifs bind to
DNA gives us an idea of the function
of many proteins as soon as their
amino acid sequences are determined.

Figure 1 The helix-turn-helix motif of a DNA-binding protein. The
two helices are pictured as cylinders. The a-helix 1 recognizes the
DNA sequence in the major groove; the a-helix 2 stabilizes the
configuration.

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Figure 2 The zinc-finger configuration of the TFIIIA protein. Zinc chelates with
cysteines and histidines to form the base of the finger structure.

Figure 3 Three-dimensional model of the
leucine-zipper region of the Max transcription
factor. The leucine residues line up opposite
each other in the two strands. (From A. R. Ferre-
D'Amare, et al., "Recognition by Max of its cognate DNA
through a dimeric b/HLH/Z domain," Nature 363:38-45,
May 6, 1993. © Macmillian Magazines Limited.)

Figure 4 Diagram of a dimer of basic/helix-
loop-helix/leucine zipper interacting with DNA.
One basic region, interacting with DNA, is
shown in red, followed by the first helix in yel-
low, the loop in purple, the second helix in
blue, and the zipper portion in orange. The
second monomer is shown in gray. (From A. R.
Ferre-D'Amare, et al., "Recognition by Max of its cognate
DNA through a dimeric b/HLH/Z domain," Nature
363:38-45, May 6, 1993. © Macmillan Magazines Limited.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

Abd-B abd-A Ubx Antp Scr Dfd Zen pb lab

— Hom-C

A13

A11 A10 A9

A7 A6 A5 A4 A3 A2 A1

— Hox-A

B13

— Hox-B

C13 C12 C11 C10 C9 C8 C6 C5 C4

Hox-C

Figure 16.21 The

homeobox genes in
Drosophila (Hom-C) are
aligned with the four
homeobox clusters of the
mouse, labeled Hox-A,
Hox-B, Hox-C, and Hox-D.
Note that not all genes are
present in all four mouse
Hox clusters and that as
many as four additional
genes (10-13) are present
in each mouse region as
compared with the fly.

D13 D12 D11 D10 D9 D8

— Hox-D

Pistil (C)

Stamen (B, C)

Petal (A, B)

Sepal (A)

Figure 16.23 Cutaway view of a typical angiosperm flower.
The flower develops from four whorls: sepal, petal, stamen,
and carpel. Homeodomain genes in the A group are active in
sepal and petal whorls; homeodomain genes of the B group
are active in petal and stamen whorls; and homeodomain
genes in the C group are active in the stamen and carpel
whorls (the pistil develops from the carpel whorl).

Figure 16.22 The thale cress plant, Arabidopsis thaliana.
(Courtesy of Dr. John Celenza.)

studied by mutational analysis, selective ablation (re-
moval or killing) of cells during development, and other
techniques used in animal studies.

Many genes have been isolated that affect the se-
quence of steps of floral induction and pattern forma-
tion. The first stage to be controlled in floral induction is
its timing. That is, flower formation usually occurs at a
specific time in the life cycle of a plant, affected by envi-
ronmental cues (day length, temperature). In Arabidop-
sis, at least three dozen genes have been isolated that af-
fect the timing of flower formation. These genes include

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CONSTANS, a late-flowering gene, EARLY FLOWERING 1,
an early-flowering gene, and GIBBERELLIN INSENSI-
TIVE, a gene for late flowering only in short days (au-
tumn).

The next stage in floral induction is generating floral
meristem at the point where a flower will form. At least
five genes are known that impart identity on floral
meristem (floral-meristem identity genes); when mu-
tated, these genes result in either shoots instead of flow-
ers or in highly abnormal flowers. These genes include
LEAFY, UNUSUAL FLORAL ORGANS, APETALA1, and
APETALA2.

Floral development continues by the creation of or-
gan primordia. Although far removed from animals in
both taxonomy and DNA sequences, plants have
homeotic genes, some producing proteins homologous
to those produced by animal genes. Currently, floral
homeotic genes are classified into three categories, A, B,
and C. Genes from category A affect sepals and petals;
genes from category B affect petals and stamens; and
genes from category C affect stamens and carpels
(fig. 16.23). This is not unlike the model of action in
Drosophila's homeotic gene clusters, which acts sequen-
tially, controlling development along the head-to-tail axis
of the fly. It appears that genetic control of floral devel-
opment is highly conserved across angiosperms, the
dominant plant group.

An example of a homeotic gene is AGAMOUS, a gene
in the C group required for the development of stamens
and carpels. Expression of this gene takes place in the
third and fourth whorls of the flower, the stamen and
carpel whorls. After its expression in the appropriate
whorls, AGAMOUS is repressed. Its repressor is another
gene, CURLY LEAF. When the protein product of CURLY
LEAF was compared with protein sequences from
Drosophila, it proved to have similarities in amino acid
sequence with a gene in Drosophila called Enhancer of
zeste. This gene is also a repressor of a homeotic gene,
but in fruit flies.

Thus, several valuable conclusions come from this
study of Arabidopsis. Most important is the fact that

plants and animals seem to use similar mechanisms in de-
velopment. Both groups have repeated units (segments)
in development; both have homeotic genes that control
developmental pathways in these units; both have re-
pressors of homeotic genes that maintain the proper de-
velopmental fate in their segments; and, despite large tax-
onomic distances, there is some homology between the
proteins in plants and animals.

Other Models of Development

Although the study of development in animals has pro-
gressed markedly by using Drosophila as a model, other
organisms have been used as well. Historically, amphib-
ians were the focus of developmental research because
they have large eggs that can be easily observed and
manipulated. The same reasoning made the chick em-
bryo a classical model of development. The nematode
Caenorhabditis elegans has emerged as another model
organism for developmental studies because of its
simplicity (fig. 16.24). Each individual consists of only
about one thousand cells; its life cycle lasts only 35 days;
and with only 8 X 10 7 base pairs of DNA, it has the
smallest genome of any multicellular organism. In 1963,
S. Brenner proposed learning the lineage of every cell in
the adult. With the efforts of numerous colleagues, that
work was completed in about twenty years. From the fer-
tilized egg to the adult, the division and fate of every cell
of this nematode worm is known. The worm has been
especially useful in studying homeotic mutants and

Sydney Brenner (1927- ).
(Courtesy of Dr. Sydney Brenner.)

(a)

(b)

Figure 16.24 The roundworm Caenorhabditis elegans.
(a) Self-fertilizing hermaphrodite, (b) Male. The worms are
about 0.3 mm long. (J. E. Sulston and H. R. Horvitz, "Post-
Embryonic Cell Lineages of the Nematode Caenorhabditis elegans,"
Developmental Biology, 56:1101-56, 1977, Academic Press.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

apoptosis, programmed cell death. As we will see later,
apoptosis is important in development as well as in the
elimination of infected or cancerous cells. Also being
used as animal models are mice and zebra fish. In plants,
the snapdragon, Antirrhinum majus, is another model
organism.

Development is a growth process that is, among
other things, an orderly process of cell division. As we
discussed in chapter 3, the cell cycle is controlled by
checkpoints; the cycle is allowed to continue if the cell is
ready for the next stage. "Ready" means, among other
things, that the cell has successfully completed DNA
replication and repair of DNA damage. If the cell is not
ready, the cell cycle stops until the cell is ready. If the cell
is damaged beyond repair, including being cancerous,
programmed cell death (apoptosis) is initiated. It is clear
that numerous checks and balances are involved in as-
suring that only healthy, ready cells continue in the cell
cycle. Interference to these checks and balances can lead
to uncontrolled cell growth — cancer.

CANCER

Cancer is an informal term for a diverse class of more
than 100 distinct diseases marked by abnormal cell pro-
liferation; white blood cells proliferate at an inappropri-
ate rate, or other cell types form growths known as
tumors (neoplasms). Benign tumors grow in only one
place and do not invade other tissues. The cells of malig-
nant tumors not only continue to proliferate but also in-
vade nearby tissues and, by a process called metastasis,
spread to distant parts of the body through blood or
lymph vessels and start new centers of uncontrolled cell
growth wherever they go.

Cancers are generally divided into four groups, de-
pendent on the type of cells originally involved. Two
types of cancer cause overproduction of white blood
cells. Leukemias are diseases that cause excessive pro-
duction of leukocytes, which originate in the bone mar-
row. Lymphomas cause excessive production of lym-
phocytes, which originate in the lymph nodes and
spleen. Sarcomas are tumors of tissue such as muscle,
bone, and cartilage that arise from the embryological
mesoderm. About 85% of cancers are carcinomas, tu-
mors arising from epithelial tissue such as glands, breast,
skin, and the linings of the urogenital, digestive, and re-
spiratory systems.

All cancers are genetic: They come about from alter-
ations in genes that control cell growth. Most evidence
indicates that cancers are clonal — they arise from a sin-
gle aberrant cell that then proliferates. Therefore, analyz-
ing the causes of cancer comes down to trying to under-
stand how one cell is changed, or transformed, from a

normal cell to a cancerous one. As we will see, most can-
cers come about from a series of genetic changes, pro-
gressing from an aberrant cell to an aggressively cancer-
ous one. This view is called the clonal evolution
theory of cancer.

Historically, cancers were understood to be caused
by either mutation or by viruses. We now know that
viruses can bring cancer-causing genes into cells, where
their mutated form or inappropriate location can lead to
cancer. Thus, both the mutational and viral views of can-
cer are ultimately concerned with mutation. In essence,
cancers result from the inappropriate activity of certain
genes, whether those genes were changed by mutation
or were imported or activated by viruses.

Mutational Nature of Cancer

Mutations, both point and chromosomal, have been im-
plicated in carcinogenesis (table 16.3). For example, the
disease xeroderma pigmentosum in human beings is
caused by mutations in any of seven loci (XpA-XpG) that
inactivate the mutation repair system that corrects UV-
light damage (see chapter 12); exposure to the sun then
results in skin lesions that often become malignant. A
related disease, ataxia- telangiectasia, is caused by a
defect in the double-strand break repair mechanism, of-
ten the result of X-ray induced damage. (Ataxia refers to
difficulty in balance; telangiectasia refers to dilated
blood vessels in the eye membranes.) By binding to the
ends of DNA, as would happen when a double-strand
break occurs, ATM (the protein product of the ataxia-
telangiectasia locus, atm) begins a signaling pathway
that tells the cell there are broken ends of DNA. Persons
with this defect are at risk for acute and chronic
leukemia and lymphomas; women with it are also at risk
for ovarian cancers.

Most cancers are associated with chromosomal de-
fects; improved chromosomal banding techniques have
demonstrated that a specific chromosomal defect is often
associated with a specific cancer (table 16.3, fig. 16.25,
box 16.2). The implication is that when a gene is in a
new location (because of translocation or the deletion of
intervening material), that gene may fall under the con-
trol of more powerful promoters or promoters outside
the range of that gene's normal control. As we shall see,
genes that are known to be able to transform cells
(oncogenes) are often the ones that are relocated into
regions of new control. These oncogenes then become
more active, and transformation follows.

Cancer-Family Syndromes

In some cases, a predisposition for malignancies is inher-
ited. When four thousand clinic registrants were inter-
viewed, almost half reported virtually no family history of

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Cancer

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Table 1 6.3 A Small Sample of Chromosomal Rearrangements Associated

with Specific Cancers

Disease

Chromosomal Rearrangement

Genes Affected

Burkitt's lymphoma

t(8; 14)*

c-MYC

Non-Hodgkin's lymphoma

t(3; 4)

Laz3, BCL-6

B-cell chronic lymphocytic leukemia

t(ll; 14)

BCL-1, PRAD-1

Follicular lymphoma

t(l4; 18)

BCL-2

T/B-cell lymphoma

Inversion, chromosome 14

TCR-a

Chronic myelogenous leukemia/

t(9; 22)

CABL

Acute lymphocytic leukemia

Ewing's sarcoma

t(ll;22)

FLU, EWS

Melanoma of soft parts

t(12; 22)

ATF1, EWS

Liposarcoma

t(12; 16)

CHOP, FUS

Reprinted with permission from Nature, Vol. 372, T. H. Rabbits. Copyright © 1994 Macmillan Magazines Limited.
* The notation of the form t(8; 14) indicates a translocation between chromosomes 8 and 14.

U M n H If II I

M
II

'§*:*

Figure 16.25 G-banded chromosomes from a patient with
chronic myelogenous leukemia showing a translocation of
chromatin (arrows) from chromosome 22 to chromosome 9.

(Courtesy of Charles Rubin, M.D., University of Chicago, Department of
Pediatric Hematology/Oncology.)

cancer, whereas about 7% reported that many family
members had cancer. This 7% was considered cancer
prone because three or more close relatives of the inter-
viewed person had cancer. The interpretation of the study
is that some families are predisposed toward cancer, but
most are not, displacing the idea that everyone in the pop-
ulation has a uniform and low probability of developing
cancer. Lending support to this interpretation are the
cancer-family syndromes, in which family members
seem to inherit a nonspecific predisposition toward tu-
mors of various types. At least twenty cancer-family syn-
dromes are known. In figure 16.26, we see a pedigree for
a cancer-family syndrome in which the predisposition for
several different types of cancers, rather than a particular

irO

o Doeo

Breast cancer

Colon cancer

Prostate cancer

Qj Pancreatic cancer

Figure 16.26 Pedigree of a type I cancer-family syndrome. This
is interpreted as the inheritance of the propensity toward cancer
rather than the inheritance of any specific type of cancer.

type of cancer, seems to be inherited. Women in this fam-
ily get breast, colon, and pancreatic cancers, whereas men
get colon, prostate, and pancreatic cancers.

Tumor-Suppressor Genes

There is a class of cancer-related genes called the tumor-
suppressor genes (also called anti-oncogenes).

These genes act by suppressing malignant growth. Muta-
tions are recessive, and in the homozygous state, cancer

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Chapter Sixteen Gene Expression: Control in Eukaryotes

A technique has been devel-
oped that allows investigators
to differentiate all of our chro-
mosomes very quickly and accurately
by seeing them painted in different
fluorescent colors. This technique al-
lows a scientist or clinician to deter-
mine quickly whether any chromo-
somal anomalies exist, either in
number (aneuploidy) or structure
(deletions, translocations). The tech-
nique, chromosomal painting, is a

% <^V

(a)

(c)

K n u

1 2 3

SB SB

A 5

U u n to m n n

6 7 3 9 10 11 12

Ml 46 no

13 14 15

RH Aft Hi

16 17 ia

a% Kit

19 20

** AA Q

BOX 16.2

Experimental
Methods

Chromosomal Painting

variant of the technique known as
fluorescent in situ hybridization
(FISH), in which a fluorescent dye is
attached to a nucleotide probe that

then binds to a specific site on a chro-
mosome and makes itself visible by
fluorescence (see fig. 13.41). A whole
chromosome can be made visible by
this technique if enough probes are
available to mark enough of the chro-
mosome. However, there are not
enough fluorescent markers known
to paint all 24 of our chromosomes
(autosomes 1-22, X, Y) a different
color. Now, with as few as five differ-
ent fluorescent markers and enough
probes to coat each chromosome, it
is possible to make combinations of
the different marker dyes so that each
chromosome fluoresces a different
color. Because the colors are not gen-
erally distinguishable by the human
eye, they have to be separated by a
computer that then assigns each
chromosome its own color. As figure
1 shows, the technique works very
well. With it, we can rapidly deter-
mine any chromosomal anomaly in a
given cell. This technique is helpful in
clinical diagnosis of various syndromes
and diseases, including cancer.

hi! itt in

i i a

u m

4 n

in* ml is it

6 7 * »

IIM l*H ll«!

in 11 i? "

13 i* IS

III *I **

16 17 13

if utatt

■f* 20

34 at *

(b)

(d)

Figure 1 Chromosomal spreads after treating with
probes specific for all human chromosomes and
attached to flourescent tags. Colors are generated by
computer. Left (a, b) are the spread and karyotype of
a normal cell; right (c, d) are the same for an ovarian
cancer cell with complex chromosomal anomalies.

(Courtesy of Michael R. Speicher and David C. Ward, "The col-
oring of cytogenetics," Nature Genetics, 2:1046-48, 1996, figs.
2 and 3. Photos courtesy David C. Ward.)

ensues. The first tumor-suppressor gene to be isolated
was the gene for retinoblastoma, a tumor of retinoblast
cells, which are precursors to cone cells in the retina of
the eye. This is a disease young children contract, be-
cause after the retinoblast cells differentiate, they no
longer divide and apparently can no longer form tumors.
The disease occurs both in a hereditary and a sporadic
form. Both forms are presumably due to the recessive ho-
mozygous state of the locus. In the hereditary form, indi-
viduals inherit one mutant allele; a second mutation re-
sults in the disease. In the sporadic form, with identical

symptoms, both alleles have apparently mutated sponta-
neously in the somatic tissue of the retina. The retinoblas-
toma gene has also been implicated in other cancers, in-
cluding sarcomas and carcinomas of the lung, bladder,
and breast.

How do we know that retinoblastoma results from
the loss of suppression rather than simply the activity of
an oncogene? J. Yunis, who examined cells from several
retinoblastoma patients, found a frequently deleted part
of chromosome 13, specifically band ql 4. Yunis noticed
that the exact points of deletion varied from individual to

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individual, indicating that the phenomenon was due to
loss of gene action rather than enhancement of gene ac-
tivity due to the new placement of genes previously sep-
arated by the deleted material.

Under normal circumstances, the retinoblastoma pro-
tein, RB, inhibits the cell cycle from advancing. If the ap-
propriate checkpoint is passed, RB is phosphorylated by
cyclin-dependent kinase and cyclin complexes, and the
cell cycle progresses. As the protein product of a
homozygous recessive mutant, RB no longer inhibits the
cell cycle advance, even if the checkpoint has not been
cleared. Thus, DNA-damaged and cancerous cells are al-
lowed to continue to grow.

The retinoblastoma gene has been isolated and
cloned. The gene specifies a 105-kilodalton protein
(pi 05) found in the nucleus, as would be expected if it
were a suppressor of DNA transcription. It binds with at
least three known oncogenic proteins: the El A protein of
adenovirus, the SV40 (a simian virus) large T antigen, and
the 16E7 protein of human papillomavirus, a virus asso-
ciated with 50% of cervical carcinomas. The implications
are that these three viruses may use a similar mechanism
in transformation, and this mechanism involves inactiva-
tion of the retinoblastoma pi 05 protein.

Further support for the existence of tumor-
suppressor genes came from work by E. Stanbridge and
his colleagues with another childhood cancer, Wilm's
tumor. This is a kidney cancer that is also believed to be
caused by loss of action in a tumor-suppressor gene. It is
associated with the loss of band pl3 on chromosome 1 1 .
Researchers introduced a normal chromosome 11 into
Wilm's tumor cells growing in culture. The result was
normal cell growth, exactly what we would predict if the
introduced normal gene were a tumor-suppressor gene.

A third tumor-suppressor gene is the p53 gene,
named for its 53-kilodalton protein product and located
on chromosome 17. This gene is the most common mu-
tation in cancers, found in more than 50% of human tu-
mors. It achieved the status of Science magazine's 1993
"Molecule of the Year." Since the p53 protein is found in
so many cases, it is clear that its role as a tumor suppres-
sor was of great importance in the normal activity of
cells. Normally, p53 is highly unstable: the MDM2 protein
binds its amino terminal end and ubiquinates it, leading
to the rapid degradation of p53 in the proteasome within
several minutes. However, p53 is stabilized when it is
phosphorylated by cell-cycle checkpoint kinases. For ex-
ample, ATM binds to double-stranded DNA breaks. Bound
this way, it activates the protein CHK2, a checkpoint ki-
nase that then phosphorylates p53. In the active state,
p53 is a transcription factor that induces at least thirty-
four different genes, genes involved either in stopping
the cell cycle, inducing apoptosis, or regulating itself.

First, p53 stops the cell cycle to give the cell a chance
to repair its DNA. Cell growth is arrested by the induction

(also called upregulation) of cyclin-dependent kinase
inhibitors (proteins such as p21, WAF1, and CIP1). This
action stops the cell cycle. In fact, if DNA repair does not
take place, cells can be forced to remain permanently in
Gl phase. Alternatively, p53 can induce cell death by up-
regulating the bax gene. Its protein is involved in the
pathway to induce caspases, proteinases that destroy the
cell. (Caspases get their name from the fact that they are
cysteine-requiring aspwcXic acid proteinases. The bax
gene's name comes from bcl-2 associated-^ gene; bcl-2 is
from B-cdX /eukemia//ymphoma-2.)

Finally, the p53 protein is a transcription factor for
the gene for MDM2, the protein that regulates p53-
Thus, the p53 protein has a narrow window in which to
stop the cell cycle or induce apoptosis, giving the cell a
chance to repair its DNA damage or commit suicide.
After this, the p53 protein is itself repressed (fig. 16.27).

It is clear that the loss of p53 activity allows DNA
damage to build up in a cell. This is why more than 50%
of cancers involve loss of p53 activity. More than twenty
other tumor-suppressor genes are known.

Viral Nature of Cancer

Retroviruses

Animal viruses come in many different varieties, with
DNA or RNA as their genetic material (fig. 16.28). Several
classes of viruses, both DNA and RNA, can transform cells,
a process that may or may not be caused by an oncogene
the virus carries. Some DNA viruses do carry oncogenes,
such as the adenovirus that carries the gene for the El A
protein, which may act by binding to the retinoblastoma
repressor protein. Oncogenes, however, were originally
discovered in retroviruses, a group of very simple RNA
viruses that contain the enzyme reverse transcriptase. Af-
ter the virus enters the host cell, this enzyme converts the
viral RNA into DNA. In 1910, Peyton Rous, who much
later won the Nobel Prize for his work, discovered that a
sarcoma in chickens could be induced by a cell-free ex-
tract from a tumor in another chicken. The transmitted
agent was later found to be a retrovirus, named Rous sar-
coma virus, the first retrovirus to be discovered.

Peyton Rous (1879-1970).
(Courtesy of Rockefeller University
Archives.)

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NoDNA
damage

Free
MDM2

Degradation of p53
in proteasome

MDM2

www^o(^)

Activate

DNA

CHK2

DNA
damage

Phosphorylation of p53
(active transcription factor)

dna wwe mowx. ^m(R

Stop cell
cycle

Induce
apoptosis

Induce
MDM2

Figure 16.27 Normally, MDM2 ubiquinates the p53 protein, which leads to its degradation in the
proteasome. If there is DNA damage that produces broken DNA ends (double-strand breaks), MDM2
binds to these ends and activates the checkpoint kinase CHK2. CHK2 phosphorylates p53, making it
stable and an active transcription factor. Next, the active p53 binds to the promoters of synexpression
group genes for stopping the cell cycle and apoptosis. In addition, p53 induces the MDM2 gene,
which can then cause p53 degradation if the broken DNA has been repaired. (MDM2 is named for the
cell culture in which it was cloned — murine c/ouble minute chromosome clone number 2.)

The retrovirus, which often carries only three genes,
integrates into the host genome in a series of steps
(fig. 16.29). When the virus enters the host, it is in the form
of a plus (+) RNA strand (capable of acting as a messen-
ger RNA; the minus strand is the complement to the
strand). At either end is a repeated sequence (R) located
outside two unique sequences (U3 and U5). Through re-
verse transcription, using the reverse transcriptase the
virus brings in, the viral RNA is converted to a double-
stranded DNA. During that process, the ends of the DNA
take on the configuration of long terminal repeats (LTRs),
repetitions of U3-R-U5. The linear DNA then circularizes
and integrates into the host genome just as a transposon
does, generating short direct repeats at either end.

As we mentioned, retroviruses can cause cellular
transformation through direct integration or from the
oncogenes they carry Transformation from integration
comes about because the integrated provirus either inac-
tivates a tumor-suppressor gene or activates an oncogene

in a process called insertion mutagenesis. The U3
region of the retrovirus contains both an enhancer and a
promoter. Since a long terminal repeat lies at either end
of the provirus, cellular genes can be turned on when the
virus integrates.

Oncogenes

Genetic analysis and recombinant DNA studies showed
that Rous sarcoma virus transforms cells through the action
of a single gene. This gene, called src for sarcoma, was the
first viral oncogene discovered. Since then at least fifty have
been discovered, and each has been given a three-letter
designation (table 16.4). Unlike tumor suppressors, which
lead to cancer when in the homozygous mutant condition,
oncogenes act in a dominant fashion: Only one copy of the
activated gene need be present for transformation to occur.
With the viral oncogene in hand, researchers could
create a probe for the gene and look within the DNA of

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Enveloped

Nonenveloped

dsDNA

Herpesviridae
(herpes simplex virus)

Poxviridae, Choropoxvirinae
(vaccinia virus)

Hepadnaviridae
(hepatitis B virus)

dsDNA

Iridoviridae
(frog virus)

ssDNA

Adenoviridae
(human adenovirus 2)

Papovaviridae
(polyoma virus)

Parvoviridae
(feline parvovirus)

ssRNA

Coronaviridae Paramyxoviridae
(human coronavirus) (measles)

Bunyaviridae
(bunyamwera virus)

Orthomyxoviridae
(influenza virus)

Arenaviridae

(lymphocytic

choriomeningitis virus)

OOOOOOOOOOOOOOOOOOOOOOOOO

Togaviridae Flaviviridae Retroviridae

(rubella virus) (yellow fever virus) (human

immunodeficiency
virus)

oooooooooooooooooooooooooo

Rhabdoviridae (rabies virus)

_2_£_2_2_2_2_2_2_2_2_2__2__2__2_2_2_2_2_2

Filoviridae

(Marburg virus, Ebola virus) 100 nm

dsDNA

ssDNA

Reoviridae
(human rotavirus)

Birnaviridae

(infectious pancreatic

necrosis virus)

Picomaviridae

(human poliovirus 1,

human hepatitis A virus)

Caliciviridae

(vesicular exanthema

of swine virus)

Figure 16.28 Representatives of families of animal viruses. The abbreviations ss and ds refer to single-stranded and double-stranded,
respectively. (From R. I. B. Francki, et al., Classification and Nomenclature of Viruses, fifth report, 1991. Springer-Verlag, Vienna. Reprinted by permission.)

the host organism. To the surprise of virtually everyone,
these oncogenes were found in untransformed cells.
Since transforming viruses can function quite well as
viruses without their oncogenes, and since cellular onco-
genes have introns and viral oncogenes do not, geneti-
cists generally accept the theory that these oncogenes
originated in the host and were picked up, presumably as
messenger RNAs, by the retroviruses. We believe that
retroviruses pick up cellular genes by transcription read-
through, transcribing beyond the end of the integrated
virus and producing a messenger RNA that is then incor-
porated into a viral particle after intron removal.
Retroviruses can thus pick up genes adjacent to their
point of integration.

To distinguish oncogenes within viruses and hosts,
we prefix the name of a viral oncogene, such as src, with
a v (y-src) and a cellular oncogene with a c (c-src). Cellu-

lar oncogenes within a nontransformed cell are called
proto-oncogenes. How are proto-oncogenes induced
to become oncogenes, and what do proto-oncogenes
normally do in the cell?

Oncogene Induction

Proto-oncogenes can be induced in at least three differ-
ent ways. First, a mutation can cause a proto-oncogene to
transform its host cell. For example, a ras proto-oncogene
(see table 16.4) was converted to an oncogene when one
codon, GGC (glycine), was converted to GTC (valine).
Second, a proto-oncogene can be activated if it is moved
to a region with a strong promoter or enhancer. Burkitt's
lymphoma, for example, is associated with a transloca-
tion involving the proto-oncogene c-myc, which is nor-
mally located on chromosome 8. When translocated to

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Chapter Sixteen Gene Expression: Control in Eukaryotes

Table 16.4

Some Oncogenes, Their Origins, and Their Protein Products

Oncogene

Virus

Species

of Origin

Gene Function

abl

Abelson murine leukemia virus

Mouse

Tyrosine kinase

src

Rous sarcoma virus

Chicken

Tyrosine kinase

erbB

Avian erythroblastosis virus

Chicken

Tyrosine kinase

fms

McDonough feline sarcoma virus

Cat

Growth factor

mos

Avian myeloblastosis virus

Chicken

Protein kinase

sis

Simian sarcoma virus

Woolly monkey

Growth factor

Ha-ras

Harvey murine sarcoma virus

Rat

GTP-binding protein

Ki-ras

Kirsten murine sarcoma virus

Rat

GTP-binding protein

fas

FBJ osteosarcoma virus

Mouse

Binds DNA

myb

Avian myeloblastosis virus

Chicken

Binds DNA

erbA

Avian erythroblastosis virus

Chicken

Binds DNA

rel

Reticuloendotheliosis virus

Turkey

Binds DNA

jun

Avian sarcoma virus 17

Chicken

Binds DNA

Source: Reprinted with permission from J. Marx, "What Do Oncogenes Do?," Science, 223:673-76, 1984. Copyright © 1984 American Association for the Advance-
ment of Science.

R U5 PBS gag pol env U3 R 3' po |y-A
5 cap 1 — 1 —

20-80 bases

80-100 bases

tail
200-1 ,000 bases

Reverse
transcription

U3 R U5 PBS gag
— i — i — i

pol env U3 R U5
1 1 — i —

LTR

Circularization

and

integration into host

cell chromosome

RNA

DNA

Figure 16.29 A retroviral RNA genome. R is a
repeated sequence; U3 and U5 are unique sequences;
PBS is the primer-binding site, LTR is the long terminal
repeat; gag, pol, and env are viral genes. During the
process of reverse transcription, the LTRs are created at
the ends of the DNA. Direct host repeats are created
when the viral DNA integrates into the host
chromosome.

LTR

H h

Direct Inverted
host repeats
repeat

PBS gag pol env

LTR

-I h

Inverted
repeats

<-^

Proviral
> DNA

Direct
host
repeat

chromosome 14, c-myc is placed contiguous with the
immunoglobulin IgM constant gene. This gene is very
active in lymphocytes (as we will see later). Hence c-myc
is now transcribed at a much higher rate than normal,
resulting in cellular transformation. The c-myc gene
normally occurs near a fragile site, a region of a chro-
mosome that has a tendency to break. Many proto-
oncogenes occur near fragile sites on chromosomes.

The simple capture of a gene by a retrovirus might be
enough for transformation, since the gene is brought
under the influence of viral transcriptional control.
However, not all genes captured this way are onco-
genes. Third, a proto-oncogene can be activated if it is
amplified. Several cases are known in which amplified
genes (e.g., c-ras and c-abl) or genes on trisomic chro-
mosomes are related to transformation.

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Viral oncogenes can cause transformation by the
same mechanisms. Either a mutation of the oncogene it-
self or the placement of the gene next to an active viral
promoter can cause high levels of transcription of the
oncogene and, hence, transformation of the cell. What
are the gene products of these proto-oncogenes?

Oncogene Function

We know that proto-oncogenes are important to the
cell because they have been conserved evolutionarily
For example, c-src is found in fruit flies as well as in ver-
tebrates; c-ras is found in yeasts and in human beings.
They are all genes that can promote cell growth; they
are specific transcription factors or other components
of growth-stimulating signal pathways. The known pro-
tein products of oncogenes can be classified into at
least four categories: tyrosine kinases, growth factors,
GTP-binding proteins, and DNA-binding proteins (see
table 16.4). As proto-oncogenes, they normally func-
tion at low levels; in transformed cells, they function at
high levels. Can these proteins explain cancerous
growth?

Tyrosine kinases are enzymes that add a phosphate
group to tyrosine residues in proteins. Other kinases
phosphorylate serine and threonine. (These three amino
acids have OH groups available for phosphorylation.)
Proteins that are phosphorylated at their tyrosine
residues are involved in signal pathways, cytoskeleton
shape (transformed cells are shaped differently than nor-
mal cells), and glycolysis (cancer cells tend toward the
anaerobic glycolytic pathway). Overactivation of the cel-
lular oncogene can result in inappropriate kinase activity,
thereby changing many of the cellular activities and lead-
ing to cancer.

The c-sis oncogene encodes platelet-derived growth
factor, which stimulates cells to grow. Its potential in
transformation is obvious. GTP-binding proteins, the
product of v-ras, for example, play a role in transmitting
endocrine signals across membranes. Increased quanti-
ties of GTP-binding proteins can send continuous or
amplified signals to certain cells and thus enhance
growth. The v-myc gene product is a protein that binds to

DNA; specific transcription factors can signal inappropri-
ate transcription, also inducing transformation.

From the initial lesion in a gene to full-blown cancer
normally takes many steps (clonal evolution; fig. 16.30).
For example, in the colorectal cancer familial adenoma-
tous polyposis (FAP), at least seven genetic changes are
needed. Through these steps, a normal mass of cells
passes through hyperplasia, an increased growth with-
out any obvious change in cells; to dysplasia, in which
overgrowth continues with changes in cell and nuclear
structures (polyp formation); to the cancerous state,
with invasion of surrounding tissues and metastasis.
B.Vogelstein and colleagues have discovered many of the
genetic changes involved in the formation of this cancer.
First, the APC gene (adenomatous polyposis coli) mu-
tates, leading through hyperplasia to dysplasia, a condi-
tion referred to as aberrant cryptic foci. Although the ex-
act role of the APC protein is not known, it does bind
P-catenin, which is involved in cell adhesion and acti-
vates the cyclin Dl promoter, exerting a direct effect on
cellular proliferation. Thus, mutation of APC results in an
accumulation of p-catenin, which then has effects on cell
cycle progression and cell adhesion.

The next genetic change results in an early adenoma
(a benign growth). Mutation of the ras oncogene leads to
intermediate adenoma, due to the autonomous growth
signals sent by the Ras GTP-binding protein. This is fol-
lowed by late adenoma caused by the mutation of a gene
in region q21 of chromosome 18, a gene called deleted in
colorectal cancer (DCC). This gene codes for a trans-
membrane protein involved in the adhesion of cells to
each other.

At this point, mutations resulting in the loss of p53
protein result in full-blown cancer. Throughout this se-
ries of events, it is a cell from the previous state that
mutates into the next state, consistent with our concept
of clonal evolution. Although it may seem odd that so
many mutations appear consecutively in the same cells,
remember that mutations in some genes, such as p53, re-
sult in an overall higher mutation rate within cells. In one
study, when cancer cells were compared with non-
cancerous progenitor cells, the cancer cells showed
eleven thousand genetic changes.

Normal
cells

Mutation of
APC

*■ Hyperplasia,—

Dysplasia,

then

Early Adenoma

Mutation of
ras

Intermediate
Adenoma

Mutation of

DCC,

DPC4, or

JV18

■> Late —
Adenoma

Mutation of
p53

> Carcinoma

■>• Invasive
and
Metastatic

Figure 16.30 Some of the known steps in converting a normal colon cell into a cancerous one. At least four known genes are
involved, two oncogenes (APC and ras) and two tumor-suppressor genes (DCC — or DPC4 or JV18 — and p53). (Reprinted from Cell,
Vol. 87, K. W. Kinzler and B. Vogelstein, "Lessons from Hereditary Colorectal Cancer," pp. 159-170, Copyright © 1996, with permission from Elsevier Science.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

In summary, all cancers share the following traits.
First, they provide their own growth signals while ignor-
ing inhibitory signals; in essence, cancer cells can grow
without limit. Second, cancer cells avoid apoptosis.
Third, tumor cells create new blood supplies by a
process known as angiogenesis; new blood vessels grow
in the tumors, allowing them unlimited increase in size.
And finally, all malignant tumors have the capability of in-
vasive growth and metastasis.

Environmental Causes of Cancer

Environment plays a major role in carcinogenesis, and
many environmental carcinogenic agents are known
(table 16.5). Many of these agents are also mutagens (see
chapter 12). Avoidable substances in the environment
and the diet are estimated to cause 80 to 90% of all can-
cers, although the exact mechanisms by which these
agents induce transformation are generally unknown.
Perhaps the most effective cancer prevention strategy
would be to avoid as many carcinogens from the envi-
ronment as possible.

In the final section of the chapter, Immunogenetics,
we look at another genetic system of transcriptional con-
trol. We try to answer the question: How does a single or-
ganism produce such a vast array of immunological pro-
tection?

Table 16.5 Carcinogenic Substances in

the Environment

Carcinogen

Cancer Site(s)

Aromatic amines

Bladder

Arsenic

Liver, lung, skin

Asbestos

Lung

Benzine

Bone marrow

Chromium

Lung, nose, nasopharynx sinuses

Cigarettes

Lung

Coal products

Bladder, lung

Dusts

Lung

Ionizing radiation

Bone, bone marrow, lung

Iron oxide

Lung

Isopropyl oil

Nasopharynx sinuses, nose

Mustard gas

Lung

Nickel

Lung, nasopharynx sinuses, nose

Petroleum

Lung

Ultraviolet irradiation

Skin

Vinyl chloride

Liver

Wood and leather dust

Nasopharynx sinuses, nose

IMMUNOGENETICS

Vertebrates have evolved the ability to protect them-
selves against invading bacteria, viruses, and parasites
and against their own cancer cells by creating an enor-
mous amount of immune diversity with relatively few
genes. Here we concentrate on the genetic control of
immunity, the ability of an animal to resist infection.
The foreign substance from the bacterium, virus, para-
site, or cancer cell that evokes an immune response is
called an antigen. The immune response itself is a
complex interaction of various cell types, signaling
pathways, and other components. The immune system
of a mammal can destroy millions of different antigens
without harming its own cells — quite an amazing
accomplishment.

The two major components of the immune system
are the B and T lymphocytes, white blood cells that
originate in bone marrow and mature in either the
bone marrow (B cells) or the thymus gland (T cells).
The B cells are responsible for producing very specific
proteins called antibodies, or immunoglobulins
(Igs), which protect the organism from antigens in
three general ways. Immunoglobulins can coat anti-
gens so that they are more readily engulfed by phago-
cytes (white blood cells that engulf foreign material);
immunoglobulins can combine with the antigens — for
example, by covering the membrane-recognition sites
of a virus — and thereby directly prevent their ability to
function; or, in combination with complement, a blood
component, immunoglobulins can cause the cell to die
if the antigen is from an intact cell. B cells are the ma-
jor component of humoral immunity, immunity con-
trolled by antibodies in the serum and lymph; T cells
are the major component of cellular immunity, im-
munity against infected cells.

Whereas the B cells produce immunoglobulins, one
type of T cell is concerned with locating and destroying
infected cells to prevent invading organisms from escap-
ing detection within those infected cells. The cytotoxic
T lymphocytes attack host cells infected by a virus, bac-
terium, or parasite. Thus, infected cells are destroyed be-
fore new viruses, bacteria, or parasites can be produced,
helping to terminate the infection. Cytotoxic T lympho-
cytes recognize infected host cells by surface receptors
called T-cell receptors. These receptors recognize an
infected host cell by two aspects of the infected cell's
surface: major histocompatibility complex (MHC)
gene products, and antigens. All host cells have MHC
components on their surfaces; an infected cell has the
ability to cause part of the antigen to appear on its sur-
face with the MHC protein, as if the MHC protein were
"presenting" the antigen to the T-cell receptor
(fig. 16.31).

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Virus

f cells

Tcell

T-cel
receptor

MHC protein + antigen

Infected
cell

Figure 16.31 When a mammal (e.g., mouse) is infected by a
virus, part of the viral coat is recognized as an antigen,
triggering an immune response. B cells produce antibodies that
specifically attach to the viral antigen (humoral immunity).
Infected cells "present" the antigenic part of the viral coat to
the outside at the major histocompatibility complex protein on
the cell surface. T-cell receptors recognize this MHC-antigen
complex and then trigger the destruction of the infected cell
(cellular immunity).

Antigen-
binding
region

Heavy

C H 3

COOH

Figure 16.32 Schematic view of an immunoglobulin protein
(IgG). V = variable region; C = constant region; L = light
chain; H = heavy chain. The S-S bonds are sulfhydryl bridges
across two cysteines. The NH 2 ends of the molecule form the
antigen-recognition parts. The internal sulfhydryl bonds roughly
mark areas called domains, two each on the light chains and
four each on the heavy chains. The heavy chain also has a
hinge domain. Similar domains are found in the T-cell receptors
and the MHC proteins. These domains indicate the evolutionary
relatedness of these three types of molecules.

The dual attack by B and T cells has three main com-
ponents of genetic interest: antibodies (immunoglobu-
lins), T-cell receptors, and products of the major histo-
compatibility complex. These three protein families are
evolutionarily related to each other, and each provides a
diversity of protein products.

Immunoglobulins

Immunoglobulins, produced by the B cells, are large pro-
tein molecules composed of two identical light polypep-
tide chains (about 214 amino acids) and two identical
heavy chains (about 440 amino acids), held together by
sulfhydryl bonds (fig. 16.32). Each polypeptide chain has
a variable and a constant region of amino acid sequences.
The variable regions recognize the antigens and thereby
give specificity to the immunoglobulins (fig. 16.33).
There are five major types of heavy chains (7, a, jx, 8, and
e), giving rise to five types of immunoglobulins: IgG, IgA,
IgM, IgD, and IgE. Each has slightly different properties;

for example, only IgG can cross the placenta, giving im-
munity to the fetus. In addition, every immunoglobulin
has one of two types of light chains, k or X (kappa or
lambda).

Mutations of the constant region of the chains, called
allotypes, follow the rules of Mendelian inheritance. In
the variable region, however, called the idiotypic varia-
tion, diversity is much greater than two alleles per indi-
vidual. The average individual has the potential to ex-
press between 10 6 and 10 9 different immunoglobulins,
each with a different amino acid sequence. The lower
limit, 10 , is arrived at through the study of persons with
multiple myeloma, a malignancy in which one lymphatic
cell divides over and over until it makes up a substantial
portion of that person's lymphocytes. From these per-
sons, we can isolate a relatively purified immunoglobulin
that is the product of a single clone of cells and is re-
ferred to as a monoclonal antibody. A very low pro-
portion of a normal person's lymphocytes produces any
one specific immunoglobulin.

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Chapter Sixteen Gene Expression: Control in Eukaryotes

Multiple myeloma cells can be fused to spleen cells.
The resulting cells, called hybridomas, which produce
monoclonal antibodies, can be perpetuated in tissue cul-
ture indefinitely, thus providing a ready supply of specific
monoclonal antibodies. Recent work with hybridomas
has allowed us to locate, isolate, and sequence im-
munoglobulin genes. How can one genome produce 10 9
different antibody molecules?

Antibody Diversity

Since the mammalian genome does not have 10 9 genes
(10 5 genes is a better estimate), different models were
suggested to explain antibody diversity. In 1965, W. J.
Dreyer and J. C. Bennett suggested that a given chain of
an immunoglobulin was not the result of one gene, but of
a combination of genes, one for the constant region and
one for the variable region. In addition, they suggested
that a particular organism, in its haploid genome, had only
one constant gene but several hundred or thousand vari-
able genes. The final product would be the result of the
action of a combination of one of the variable genes and

(a)

(b)

Figure 16.33 A computer-generated view of the interaction of
an antigen-binding region of an immunoglobulin (green) with an
antigen (purple: in this case, the hormone angiotensin II,
composed of only eight amino acid residues). Note how the
antigen fits into the variable end of the immunoglobulin, as an
apple fits into a cupped hand, (a) Top view; (b) side view.

(Courtesy of L. Mario Amzel.)

the constant gene. Modern recombinant DNA technol-
ogy has verified the essence of that model.

In reality, several genes contribute to form the vari-
able regions of the heavy and light chains, given that we
are using the word gene for DNA segments that code for
a part of the final heavy or light chain of the im-
munoglobulin. Genes for the k, X, and heavy chains are
located on chromosomes 2, 22, and 14, respectively, in
human beings. Each is a multigene complex. Let us ex-
amine the k light-chain gene complex as an example of
how the DNA must be modified to produce the final pro-
tein product (fig. 16.34).

The first step in DNA rearrangement is the joining of
aV (variable) and a J (joining) gene in a B cell (fig. 16.34),
a process called V-J joining. Since any one of eighty
V genes can combine with any one of five J genes, four
hundred different combinations are possible (80 X 5).
Since we expect this to be another example of site-
specific recombination, as we saw with phage X integra-
tion in chapter 14, recombinational signal sequences must
be flanking all genes so that any two can be moved next
to each other. Through DNA sequencing, these signals,
termed recombination signal sequences, have been de-
termined to be a heptamer (seven bases) and a nonamer
(nine bases), separated by twelve bases on one side and
twenty-three bases on the other (known as the 12-23
rule; fig. 16.35).

V K i V

k80

Jk1 ^k2 Jk3 ^k4 Jk5

-DO D-CKHKh-C^

Figure 16.34 The complex for the human k light chain is
composed of about eighty variable genes (V k1 -V k80 ), five
joining genes (J k1 — J k5 ), and one constant gene, C K , in the
undifferentiated cell (germ line). The final k light chain will be
composed of the products of one variable gene, one joining
gene, and the constant gene.

nGGTTTTTGT 10h CACTGTG
CCAAAAACA^ Dp GTGACAC

'=]'

Nonamer

Heptamer

CACAGTG 9 o hn ACAAAAACC

GTGTCAC J P TGTTTTTGG

Heptamer

Nonamer

Figure 16.35 In order for V-J joining to take place (site-
specific recombination), there must be signals at the V side
and at the J side. One signal is a heptamer (seven base pairs)
and a nonamer (nine base pairs) separated by twenty-three
base pairs, and the other signal is the same heptamer and
nonamer separated by twelve base pairs, in reverse orientation.

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Two proteins, RAG1 and RAG2 (from recombination
activating genes 1 and 2), form a recombinase enzyme
capable of recognizing 12-23 signals and producing
double-strand breaks in DNA at the junction of coding
(V and J) regions and signal sequences (12-23 regions;
fig. 16.36). Recognition of the 12-23 signals is done with
the help of HMG1 or HMG2 (Mgh mobility group pro-
teins 1 and 2), proteins that bind DNA and bend it, en-
hancing the activity of the RAG recombinase. The recom-
binase brings together one of the variable regions and
one of the joining regions and links them in a process
that also generates a circle of DNA with the intervening
segments, which is then eliminated (fig. 16. 36).

The RAG recombinase nicks each DNA strand at its
recombination signal sequence. Then by a transesterifica-
tion (shifting a phosphodiester bond), the recombinase
forms a hairpin and a blunt end, a double-strand break of
the DNA (fig. 16.37). The remainder of the DNA processing
is done with double-strand break repair enzymes that repair
radiation-induced DNA damage (fig. 16.37; see chapter 12).
The result is an eliminated circle of intervening genes and
a chromosome with aV region adjacent to a new J region.

The point of crossover at theV-J junction is itself vari-
able, generating junctional diversity. Not only are any
two V and J genes capable of coming together, but also
the sequence at the junction of the two genes can vary.
For example, we see in figure 16.38 that the junction in
the protein at amino acids 95 and 96 can be Pro-Trp, Pro-
Arg, or Pro-Pro, depending on exactly where the
crossover occurred.

In figure 16.39, we see the DNA after V k50 and J k4
join. This gene is now transcribed. The region between
J k4 and C K (the constant gene) is then removed by RNA
splicing, leaving the final messenger RNA product, which
is then translated into a k light chain. In this cell, the ho-
mologous k region is repressed as well as both X regions,
a phenomenon known as allelic exclusion. Thus,
this cell produces only one light chain, the V k50 -J k4 -C k
protein.

Similar types of events take place in the heavy-chain
gene and the X light-chain gene if it has been activated.
There are some differences, however (fig. 16.40). The X
complex in human beings has only two variable genes,
with four J genes and one C gene. The heavy-chain

V k2 ». V k3

DNA

>-C=^-C=^--<t=l-<J=-0

CZI

RAG1 + RAG 2
HMG1 orHMG2

■

v k2 J k2
* ■ >■

Figure 16.36 V-J joining. The variable genes are
shown as blue boxes with blue arrowheads as the
recombination signal sequences (the 12 spacer
signal); the joining genes with their 23 spacer signals
are in red. With the RAG1-RAG2 recombinase and
either HMG1 or HMG2, a variable and a joining gene
are brought together. Recombination results in
double-strand breaks that are then repaired so that a
variable and a joining gene are spliced together and
the intermediate material is released. (Source: Diagram

modeled from S. D. Fugmann, et al., "The RAG proteins and
V(D)J recombination: Complexes, ends, and transposition,"
Annual Review of Immunology, 18:495-527, 2000.)

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Chapter Sixteen Gene Expression: Control in Eukaryotes

DNA

Recombination
signal sequence

k41

'k5

RAG1 + RAG2
HMG1 orHMG2

■
■
■

3'OH

=4>

■
■
■

I Transesterification

■
■
■

:> c

1 2 34

1 2

34 5

1 *

M \

TCT
AGA

CCT CCC ACA
GGA GGG TGT

■ ■ ■ ^^ ■

■ ■ ^fl ■ ■ ■

TGG
ACC

TGG ACG
ACC TGC

Ser

Pro

Trp Thr

96 97

Pro

CCG

TGG

Trp

Pro

CCT

TGG

Trp

Pro

CCT

CGG

Arg

Pro

CCT

CCG

Pro

CCT

CCC

Pro

Figure 16.38 Variability in crossing over during V-J joining
generates junctional diversity. In this case, V k41 and J k5 are
shown. Amino acid codons 94 and 97 are always the same,
TCT and ACG, respectively, as are the first two bases of codon
95, CC. Depending on the exact point of crossover, five
different codon pairs can be generated. Codons for proline
(Pro) are always the first in each pair (95), but codons for
tryptophan (Trp), arginine (Arg), or Pro are all possible second
codons (96). Matching numbered arrows indicate crossover
points for the five possibilities. (Source: Data from E. E. Max, et al.,
"Sequences of five potential recombination sites encoded close to an
immunoglobulin k constant region gene," Proceedings of the National
Academy of Sciences, 76:3450-54, 1979.)

Processing

Repair

Figure 16.37 The mechanism of site-specific recombination
between a variable and joining gene. The RAG recombinase,
with the help of HMG1 or HMG2, recognizes the recombination
signal sequences (red and blue arrowheads). A nick at each
signal is made, producing 3'OH ends; transesterification then
forms hairpin loops and recombinational signals with double-
strand breaks. The hairpin loops are brought together, opened,
and repaired, with some processing taking place. This creates
junctional diversity, including crossover point variability and N
segments (to be discussed later). The enzymes responsible for
the processing are repair enzymes.

complex has about one hundred to three hundred V genes,
nine J genes, and the five C genes of the five major types (7,
a, |jl, 8, e). In addition, heavy-chain regions have another set
of genes, called diversity (D) genes. At least five such genes

are in the human heavy-chain complex, and they add still
another variable region to the final protein. In the heavy
chain, D-J joining first takes place, then V-DJ joining; lastly,
splicing creates the final heavy-chain product (fig. 16.41).

As we pointed out earlier, the final form of the heavy-
chain protein in human beings has five regions or do-
mains — C H 3, C H 2, hinge, C H 1, and variable region (see
fig. 16.32). Each of the constant regions, as well as the
hinge region, comes from its own exon (fig. 16.42). (The
variable region, of course, comes from the extensive re-
combination just described: fig. 16.41.) The heavy chain
is thus another example of the relationship between
exon structure and domain function, a topic we dis-
cussed in chapter 10. Heavy-chain structure would sup-
port the exon shuffling view (introns early).

V-J, D-J, and V-DJ joining, collectively called V(D)J
joining, are the only known examples of site-specific re-
combination in vertebrates. The genes responsible are ac-
tive only in pre-B and pre-T cells.

In addition to V(D)J joining, junctional diversity is
also added during heavy-chain recombination by the ad-
dition of nucleotides in a template-free fashion. In other
words, added nucleotides, called N segments, appear at
the joining junctions; they are not specified in the DNA.
For example, in one case, the sequence GTGGGGGCC
(three codons long) was found at a D-J junction, but not

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V K 1 v k80

Jk1 ^k2 ^k3 ^k4 ^k5

-D-D — LHHHHh-dh

V-J joining v v

v k51 v k

DNA

J v-1 J v-O J

80 u k1 u k2 u k3

Excise

C D-D--EHH^

V K 49 V k50 J k 4 J k5

DNA

Transcription

V k 50^k4 ^k5 Ck

□HHZI

RNA

1 1 '

Splice out

V K 50 ^k4 ^k

RNA

Translation

V K J K ^K

Protein

Figure 16.39 V-J joining in the human k region. In this
example, V k50 is joined to J k4 and then to C K . First, V-J joining
takes place using the heptamer-nonamer signals shown in
figure 16.35. Then the region from the V k50 to the C K genes is
transcribed. Splicing the RNA removes the region containing
the extra J gene, J k5 . The final RNA, containing V k50 -J k4 -C k , is
then translated into the k light chain.

seen in the undifferentiated (germ-line) genome. Mice
that lack the gene for terminal deoxynucleotide trans-
ferase lack these N segments, implicating that gene in the
process of N-segment formation. The enzyme adds nu-
cleotides at the 3' ends of the DNA strands; these free
ends are created during V(D)J joining. The enzyme is
found in high levels in immature lymphocytes.

There is a final way in which variability is generated.
Sequencing studies indicate that mutation occurs in vari-
able regions after recombination has taken place. The
mechanism of this specific mutagenesis, called somatic
hyper mutation, is not known. Given the number of
variable, constant, joining, and diversity genes, as well as
the variation at the joining junctions, it is easy to see how
10 9 different immunoglobulin combinations could be
generated (table 16.6).

TablG 16.6 Three Hundred Immunoglobulin

Genes Can Generate 1.8 Billion
Different Antibodies

Source

Factor

Light Chains

V genes

40 X

J genes

V-J recombination*

10X - 2,000X

Heavy Chains

V genes

200 X

D genes

J genes

V-D, D-J recombination*

100X = 900,000X

Total

2,000 X 900,000 = 1.8 billion

Source: Data modified from P. Leder, "The genetics of antibody diversity,"
Scientific American, 102-15, May 1982.

* Junctional diversity, N-segment formation, and hypermutability.

Light

V K 1 Vk80 Jk1 Jk2 Jk3 Jk4 ^k5 ^k

* -— -n — EHHHHr- LZh-

Vm V^ 2 J^1 J^2 \z \a C?c

* -L>CKHHKhLZr-

Heavy

Vm V H20 o D-, D 5 J H1 J H9 C^ C 6

-D-O--D-0-K

Figure 16.40 Arrangement of the
genes in the light and heavy chains of
the human immunoglobulin complexes.

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Vi V 20 o Di D 5 J-| J 9 C^ C 6 C y C e

DNA

-D-D^}-£K-D-£h>

D 4 J 2

D-J
joining

Excise

€3

^1 V 2 qo D-| D 3 J 3 Jc

CM}

C ^

c 8

C t

c e

c«

V-DJ
joining

Excise

V51 D 2

^1 V 50 D3J3 Jg C^ C § C y C e

-d-othph:

Transcription

V DJ J 9 C

RNA

RNA processing

V DJ C

Figure 16.41 Formation of an immunoglobulin heavy chain (IgM).
First, D-J joining takes place, followed by V-DJ joining. In each
case, intervening DNA is spliced out by site-specific
recombination. Then, as in light-chain formation (fig. 16.39), the
modified region is transcribed; RNA processing (splicing) then
brings the final regions together, which, when translated, form the
V-D-J-C heavy chain. (Source: Data from F. W. Alt, et al., "Development
of the primary antibody repertoire," Science, 238:1079-87, 1987.)

T-Cell Receptors and MHC Proteins

As we mentioned earlier, genetic diversity also exists in
the T-cell receptors and the major histocompatibility
complex (MHC). From its function (recognizing both
the antigen and the MHC "self" gene product), it seems
evident that the T-cell receptor must show the same
type of diversity that immunoglobulins have. In fact,
the T-cell receptor genes are very similar to the im-
munoglobulin genes. T-cell receptors are composed of
a and p subunits; there are V, J, and C components of
the a subunit and V, J, D, and C components of the p
subunits (fig. 16.43). In a sampling of T-cell receptors
from one individual, approximately one million differ-
ent p chains and 25 different a chains were found, yield-
ing approximately 25 million (1 million X 25) different
T-cell receptors.

V-D-J region

Exon 1 Hinge Exon 2 Exon 3

DNA

CTCT

Heavy-chain
protein

C H 2

C H 3

Figure 16.42 The constant portion of heavy-chain genes is
made of four domains, each transcribed from its own exon.

The major histocompatibility complex (MHC) region
(also known as the human leukocyte antigen or HLA re-
gion in people) comprises a region of 3.6 million base
pairs with 224 identified genetic loci. The genes are
generally referred to as class I, II, and III genes. Class III
genes code for proteins in the complement system,
which is involved in the destruction of foreign cells. Class
I and II genes code, in part, for proteins that present anti-
gens to T cells. That is, class I and II proteins form struc-
tures with grooves on their surfaces that are shaped to
hold small polypeptides. These polypeptides can be nor-
mal breakdown products of cellular metabolism in
healthy cells ("self" proteins) or parts of foreign invaders
or their gene products in infected cells. Although similar,
the two types of MHC proteins are found in different
places and serve somewhat different functions.

Class I MHC proteins consist of a membrane-bound
a chain and a second chain called p 2 macroglobulin
(fig. 16. 44a). Class II MHC proteins consist of an a and
p chain (fig. \6.44b). These two proteins present anti-
gens somewhat differently. In the class I molecules, the

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'a1

'an

-DO

D-D

V P1 V (364 D (31.1 J (31.1 J (31.6

p -D-CHH-0-

(a)

a (3
NH 2

V a K § § JJ V p

Cp1 Dp2.1 Jp2.1 Jp2.7 Cp 2 ^P 65

{H}-{H

s - s

Membrane

(b)

COOH

Figure 16.43 The T-cell receptor is made up of two protein
chains, a and (3, anchored in the cell membrane. Each protein
is similar to an immunoglobulin chain and is created the same
way, with V-J types of joining taking place, (a) The a gene
complex is composed of numerous V and J genes and one
constant gene. The (3 gene complex has V, D, J, and C
genes, (b) Each protein chain has two domains, similar to the
domains of the immunoglobulins, indicating a common
evolutionary ancestry.

groove is bound on both sides so that the presented
polypeptide is small and defined (fig. 16.45). In the
class II protein, the groove is unbound, allowing for a
longer polypeptide (fig. 16.46).

Class I MHC proteins are found on almost all cells.
The polypeptides that an infected cell presents with the
MHC I proteins come from the breakdown of proteins
within the cytoplasm of the cell. Peptides are targeted for
breakdown when a ubiquitin molecule binds to the pro-
tein, sending a cellular signal that the protein is to be de-
graded. (Ubiquitin is a small polypeptide of 76 amino acid
residues, highly conserved in eukaryotes.) The ubiquitin-
tagged protein is unfolded, in an ATP-dependent process,
and then fed into a proteasome, a barrel-shaped cellular
organelle for protein breakdown (fig. 16.47). Then the
peptide fragments associate with two proteins, together
called TAP (transporter for antigen processing), that pre-
vent further degradation of the peptide as well as trans-
port the peptide into the endoplasmic reticulum, where
the peptide binds to the class I MHC proteins. The MHC
I proteins with antigen are then transported to the cell
surface. Passing T cells, called killer T cells or CD8T cells
because of their CD8 receptor protein, recognize the for-
eign antigen presented by the MHC I protein and release
substances that kill the infected cells (fig. 1 6.48a).
(White blood cells are classified by their surface anti-
gens; the CD designation comes from cluster of differen-
tiation antigens used for this purpose.)

Groove in which
antigen is presented

COOH

— (3 2 macroglobulin

Membrane

COOH

HOOC

COOH

(a) Class I

(b) Class II

Figure 16.44 The major histocompatibility complex (MHC) class I protein (a) is composed of two protein chains.
The a chain is composed of three domains similar to the immunoglobulins and T-cell receptors. The second
chain is p 2 macroglobulin. The MHC class II protein (b) is composed of an a and a (3 chain. The MHC proteins
present antigens to the T-cell receptors to signal that a foreign agent has invaded the cell.

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Figure 16.45 A three-dimensional computer model of the
antigen-presenting site of an MHC class I protein. The
presented peptide is nine amino acid residues long and
internally bound at each end. (Courtesy of Don c. Wiley.)

Class II MHC proteins are found only on cells in-
volved in the immune system, such as macrophages and
B lymphocytes. These cells are most likely to have en-
countered foreign objects like bacteria or parasites by
having engulfed them. The MHC II proteins present for-
eign antigens not from the cytoplasm, but from endo-

somic vesicles within the cells. These vesicles form by
budding from the cell surface and often contain foreign
proteins and protease enzymes. The MHC II proteins mi-
grate into the vesicles, where they pick up foreign
polypeptides and then migrate to the cell surface. The re-
sponse of passing T cells to the presentation by MHC II
proteins is different from the response to MHC I pro-
teins. The MHC II proteins with antigens are recognized
by helper T cells, also called CD4 cells because of their
surface receptor protein. Rather than kill cells such as in-
fected macrophages that are useful in the immune sys-
tem, the helper T cells stimulate the macrophages to de-
stroy the foreign bacteria in their endosomic vesicles.
The helper T cells also activate antibody-producing B
cells. CD4 cells are the prime targets of the HIV virus,
making individuals with AIDS very prone to bacterial in-
fections and other immune problems (box 16.3).

One last point is worth mentioning about the MHC
system. The loci for MHC proteins do not have V(D)J join-
ing to produce the high levels of variability found in B
and T cells. The MHC loci are, however, very variable,
with many alleles. (That variability is one reason why or-
gan transplants are usually rejected without immunosup-
pressive therapy.) Each individual can have only two al-
leles at each locus, but hundreds of alleles exist in any
population. Presumably, the different alleles allow for
somewhat different affinities for different antibodies.
They may have been selected over evolutionary time to
give certain individuals in a population more chances to
be able to identify and eliminate foreign substances.

Figure 16.46 A three-dimensional computer model of the antigen-presenting site of an MHC class II protein. The presented
peptide is fifteen amino acid residues long and is not internally bound at each end. In (a), the view as seen by the T cell; (£>) is a
side view. The presented peptide is shown in red; the electron surface of the MHC protein is blue. (From: J. H. Brown, et ai., Nature

364:33-39, 1993, fig. 4, p.35.)

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Virus

Outside

Cell membrane

Viral
protein

Inside

Ubiquitin

/WW

Ubiquitin

WvV

TAP

A/W

MHC I

Proteasome

Endoplasmic
reticulum

Figure 16.47 How the MHC I protein obtains foreign peptide to display at the cell surface. In this example, a virus attacks a cell.
The viral protein is recognized as foreign and is tagged with ubiquitin. The tagged protein is then unfolded and fed into a
proteasome. With the aid of TAP, a piece of the degraded protein enters the endoplasmic reticulum, where it combines with the
MHC I protein, which is then transported to the cell surface.

Killer T cell

T cell receptor

MHC I with antigen /

Releases toxins
to kill body cell

Body ce

(a)

Figure 16.48 MHC class I and II proteins are found on
different types of cells and recognized by different types of T
cells. In (a), a normal body cell presents a foreign antigen in
an MHC class I protein that a passing killer T cell recognizes.
The T cell then releases toxins that kill the infected cell. In (b),
a macrophage presents an antigen in an MHC class II protein
that a helper T cell recognizes. The T cell stimulates the
macrophage to destroy its invaders and also stimulates a
B-cell reaction.

(b)

Activates
B cells

T cell receptor

Stimulates
macrophage

Macrophage

MHC II with antigen/ ?o destroy

r invaders within

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Chapter Sixteen Gene Expression: Control in Eukaryotes

BOX 16.3

In 1983, Robert C. Gallo of the
National Cancer Institute and Luc
Montagnier of the Pasteur Insti-
tute of Paris co-discovered HIV — the
human immunodeficiency virus, caus-
ative agent of acquired immune defi-
ciency syndrome, or AIDS (fig. 1). HIV
is a retrovirus causing a disease first
diagnosed in 1981 among young male
homosexuals in the United States.

The AIDS virus attacks helper T
cells; a particular protein on the sur-
face of these T cells, called CD4, is a
receptor for the HIV virus coat pro-
tein, gpl20 (fig. 2). A secondary re-
ceptor, the protein CCR5, is also
needed for the virus to gain entry into
the cell. (CCR5 refers to cysteine-
cysteine linked cytokine receptor 5.)
HIV also attacks macrophages. With
destruction of the T cells, a person's
immune system loses the ability to
fight off common diseases. Persons
who develop the disease frequently
fall victim to opportunistic diseases

Biomedical
Applications

AIDS and Retroviruses

such as pneumonia caused by the pro-
tozoan Pneumocystis carinii; Kaposi's
sarcoma, a rare cancer found in people
taking immunosuppressive drugs; and
several other conditions, normally rare
except in people with suppressed im-
mune systems. These conditions collec-
tively became known as the acquired
immune deficiency syndrome.

EPIDEMIOLOGY

AIDS has spread throughout the
world. A 1959 blood sample from cen-
tral Africa contained the first known
human infection. By sequencing simi-
lar viruses in primates (simian im-
mune deficient viruses, SIVs), re-

searchers discovered that the com-
mon form of AIDS, caused by HIV-1,
jumped from chimpanzees to human
beings in the region of Gabon in west-
ern Africa. HIV-2, causing the less
common form of AIDS, came from
sooty mangabeys; SIVs have jumped
to human beings at least seven times.
There seem to be two worldwide
patterns in the spread of AIDS, which
is not contracted by casual contact.
In the New World, Australia, and
Western Europe, homosexual men
and intravenous drug users primarily
spread the disease and are the groups
at highest risk. In Africa and the
Caribbean, the disease is spread pri-
marily through heterosexual sex.
Parts of southern Africa have infec-
tion rates between 16 and 32%; East-
ern Europe, Asia, and North Africa
have relatively low infection rates. In
the United States, over 750,000 per-
sons have the AIDS virus, with
350,000 deaths reported. Worldwide,

Robert C. Gallo (1937- ).
(Courtesy of Dr. Robert Gallo.)

Luc Montagnier (1932- ).
(Courtesy of Dr. Luc Montagnier.)

Figure 1 Scanning electron micrograph of a T-lymphocyte (green) infected with
the AIDS virus. Small spherical structures (red) on the surface of the cell are
new virus particles budding Off. (© NIBSC, Science Source/Photo Researchers, Inc.)

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Immunogenetics

503

Lipid
membrane

Reverse
transcriptase

RNA

HLA antigens from
host cell membrane

Surface

glycoprotein

gp120

Transmembrane

glycoprotein

gp41

Matrix
p17

Capsid
p25

Figure 2 AIDS virion structure. Numbers associated with proteins are kilodalton masses (e.g., gp120 is a 120-kilodalton
protein). (From Nester et al., Microbiology: A Human Perspective, 3rd edition. Copyright © 2001 The McGraw-Hill Companies, Inc. Reprinted
with permission.)

over 33 million people are affected.
Most of those who got the disease be-
fore 1990 have died. However, the in-
fection rate seems to have peaked in
the United States in 1985; the only
area in which infections are increas-
ing is through heterosexual sex.

HIV GENES

As mentioned, a retrovirus minimally
contains only the gag (group antigen
gene), pol (polymerase), and env (en-
velope) genes. The viral messenger
RNA is translated starting with gag
(fig. 3). There is a translation termina-
tion signal at the end of the gag gene
that is occasionally read through, re-
sulting in a gag-pol protein. The env
gene is translated only after the viral
RNA is spliced to remove the gag-pol
region. The protein products of all
three genes are further modified by
cleavage and other changes (phos-
phorylation and glycosylation), result-
ing in core virion proteins from gag,
reverse transcriptase, protease, and in-

tegrase from pol, and envelope glyco-
proteins from env.

The HIV retrovirus is especially
complicated. Not only does it have
the gag, pol, and env genes, but it

also has six other genes (fig. 4), in-
cluding two main regulatory genes,
tat and rev. One, tat (for ^rans-^cti-
vating transcription factor), has a pro-
tein product that binds at a sequence

R U5

5' cap h — i \-

gag pol env

Mostly gag
Rarely gag-pol

Translatio

Cap

\ Splice out

R 3'poly-A
tail

gag-pol

env

Cleave

/ J

Core Reverse
virion transcriptase and
proteins protease and
integrase

— i — Tail

Translate

and

cleave

and
modify

Envelope
glycoproteins

Figure 3 Expression of a retroviral mRNA. Translation begins with the gag
gene and occasionally, due to read-through, proceeds through the gag-pol
genes. The results are core virion proteins and the enzymes reverse transcrip-
tase, protease, and integrase. Splicing must take place before env can be
translated. Cleavage of the primary transcript and some modification
produces envelope glycoproteins. continued

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Chapter Sixteen Gene Expression: Control in Eukaryotes

BOX 16.3 CONTINUED

in the long terminal repeat named
TAR, for frans-<2ctivating response
element. Tat enhances the processiv-
ity of transcription of the proviral
DNA and also recruits chromatin-
remodeling proteins to the promoter.
The product of the other regulatory
gene, rev (regulation of expression of
virion proteins), binds at a region in
the env gene called RRE (for rev re-
sponse element) and enhances the
transport of viral messenger RNAs
into the cytoplasm. Together, tat and
rev are responsible for the major ex-
pression of viral structural genes
(gag, pol, and env).

The four remaining genes — vif,
vpr, nef, and vpu — are called the ac-
cessory genes because it first seemed
that their action was not necessary
for viral functioning. We now know
that each gene produces a protein
that has a role in viral replication and
infectivity The Vpr protein (tiral pro-
tein R) is involved in transporting the
viral RNA to the nucleus. Vpr can also
induce cell cycle arrest at G2, which
may have a role in protecting in-

fected cells from cytotoxic T-cell ac-
tivities. Vpu (Viral protein U) de-
grades CD4; this action frees viral sur-
face protein precursors from the
endoplasmic reticulum. In addition,
degradation of CD4 helps prevent su-
perinfection of cells, keeping them
alive longer. The main function of Vif
(Viral infectivity /actor) is to stabilize
the virion. Nef (negative /actor) was
originally thought to be a negative
regulator of viral activity, hence its
name. However, it is now known that
Nef can reduce production of cellular
CD4 protein and enhance infection
by viruses free in the blood.

TESTING AND TREATMENT

AIDS testing is done by various tech-
niques, such as western blots, looking
for antibodies to the AIDS proteins,
usually gpl20, gp4l, and reverse tran-
scriptase. Initially, dideoxy nucleo-
tides, such as the drug 3'-azido-2',
3'-dideoxythymidine (AZT, fig. 5) and
dideoxyinosine were used to treat
AIDS. AZT is a thymidine analogue
without a 3'-OH group, meaning that

it causes chain termination during
DNA replication. It seems that during
the reverse transcription process,
reverse transcriptase preferentially
chooses AZT over normal thymidine-
containing nucleotides, whereas
mammalian DNA polymerases prefer
the opposite. Thus, AZT preferentially
prevents the reverse transcription of
the HIV RNA, keeping it at levels that
are not toxic to the cell.

Dideoxyinosine has the same ef-
fect and has also been licensed as an
AIDS treatment. Unfortunately, the
AIDS virus mutates at a high rate, ren-
dering these single-substance treat-
ments ultimately ineffective. In 1996,
treatment success improved remark-
ably when new therapies involving
combinations of drugs, including pro-
tease inhibitors, were developed.
(Dr. David Ho of the Aaron Diamond
AIDS Research Center in New York
City was named Time Magazine's
Man of the Year for his role in this
therapy.) Thus, at the moment, opti-
mism is rising that AIDS may be con-
trollable and eventually curable.

LTR

pol

5'
TAR

p17 p24

9^9

>p6
\ P 7

vif

CH,

.0.

LTR

env

gp120 gp41

vpu
-vpr
<rtat+\

nef

TAR

<rrev+»

Thymine

N 3 H

Figure 4 The genome of HIV-1 . Boxes represent
different genes. The gag gene is responsible for
four proteins, p17 (matrix), p24 (capsid), p7 (nu-
cleocapsid), and p6. The pol gene is responsible for
protease (PR), reverse transcriptase (RT), and inte-
grase (IN). The env gene is responsible for two
envelope proteins, gp120 and gp41. Intervening
DNA separates the tat and rev genes into two
parts each. TAR is in the long terminal repeat (LTR),
and RRE is in env. (From R. H. Miller and N. Sarvar,
"HIV Accessory Proteins as Therapeutic Targets," Nature
Medicine 3:389-91, 1987. Copyright © 1987 Nature Pub-
lishing Group.)

Figure 5 AZT; it differs from deoxythymidine
monophosphate at the 3' position of the sugar.

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16. Gene Expression:
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Solved Problems

505

SUMMARY

STUDY OBJECTIVE 1: To examine the control of tran-
scription in eukaryotes 465-469

In eukaryotes, specific transcription factors that can gain
access to promoters generally control transcription. In ad-
dition, nucleosome structure can obstruct the RNA poly-
merase holoenzyme's access to the promoter. Signal trans-
duction pathways provide environmental cues that lead to
remodeled nucleosomes and the appearance of specific
transcription factors at gene promoters that are scheduled
to begin transcription. Methylation is involved in the con-
trol of gene expression in some higher eukaryotes; the
methylation level is high in nontranscribed genes in the
cells of these organisms. Transposons can also control gene
expression. Mutable loci in corn and mating type in yeast
are both determined by transposition. Z DNA may also play
a role in the eukaryotic control of transcription.

STUDY OBJECTIVE 2: To analyze the genetic control of
development in eukaryotes 469-484

The ultimate goal of the developmental geneticist is to un-
derstand the role of genes in controlling development, the
orderly sequence of changes that give rise to a complex or-
ganism. Development does not have to proceed by perma-
nently changing the chromosomes; cloning has shown that
differentiated nuclei can be totipotent. The Drosophila em-
bryo begins development with morphogens, diff usable
messenger RNAs and proteins secreted by maternal-effect
genes. These genes provide anterior, posterior, dorso-
ventral, and terminal patterns of transcription for zygotic
segmentation genes; these segmentation genes fall into the
gap, pair-rule, and segment-polarity classes. They eventually

determine differential gene expression in neighboring
cells. Finally, homeotic genes determine the fates of entire
regions of the body. Flowering in angiosperms also involves
repeated units (whorls) and homeobox genes.

STUDY OBJECTIVE 3: To study the mechanisms causing
cancer 484-492

Cancer is a generic term for genetic diseases in which cells
proliferate inappropriately. Mutations in oncogenes (cancer-
causing genes) or tumor-suppressor genes, such as p53, can
lead to cancer. Mutation can take place by base pair changes,
chromosomal rearrangements, and amplification. Viruses
can also bring oncogenes into cells, causing the activation
of the oncogenes. For the full development of cancer, sev-
eral genes usually must mutate.

STUDY OBJECTIVE 4: To study the genetic mechanisms
that generate antibody diversity 492-504

Immunoglobulins (antibodies) have tremendous diversity;
about 10 9 different antibodies can be generated by the hu-
man genome. This number comes about by V(D)J joining
between several genes among hundreds, as well as by junc-
tional diversity caused by the location of the crossover
points during site-specific recombination, template-free ad-
dition of codons (N segments), and somatic hypermutation.
Similar diversity exists in T-cell receptors that recognize the
major histocompatibility complex (MHC) proteins. These
proteins present foreign polypeptides for destruction by
the immune system.

SOLVED PROBLEMS

PROBLEM 1: Relate the homeo box, homeo domain, and
master-switch concepts.

Answer: A master-switch gene is a gene in a eukaryote
that controls many other genes. In a prokaryote, this con-
trol is achieved with operon organization; that is, many
genes controlling the same function are transcribed as a
unit. Thus, a gene that represses transcription of an
operon represses all of the genes in that operon. A
master-switch gene is viewed in a similar manner, given
that polygenic transcripts are very rare in eukaryotes. A

master-switch gene would translate to a specific tran-
scription factor, a protein that might control transcrip-
tion of many genes (a synexpression group). For this to
happen, the master-switch gene would need to interact
with DNA. Thus, the finding of a homeo box that tran-
scribes a homeo domain in genes that control large phe-
notypic changes is consistent with this view. The homeo
domain is the part of the transcription factor that binds
to DNA.

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Chapter Sixteen Gene Expression: Control in Eukaryotes

PROBLEM 2: What stages in the formation of an im-
munoglobulin molecule generate diversity?

Answer: Variability is generated through four general
processes: choice of which subunit genes to combine,
choice of how to combine these subunit genes, de novo
generation of diversity at junctions, and unusually high mu-
tation rates. Thus, in our description of the formation of a k
chain, diversity is added by (1) the choice of which variable
and joining genes to combine; (2) recombinational variabil-
ity at the point of recombination; (3) the creation of N seg-
ments at the junctions; and (4) somatic hypermutation.

PROBLEM 3: How can you reconcile the viral and muta-
tional natures of cancer?

Answer: The two theories are reconciled because both
define cancer as a disease caused by the inappropriate
actions of genes. In the mutational view, inappropriate
activity is generated by a gene mutation. In the viral
view, a gene brought into the cell by a virus generates the
inappropriate activity.

EXERCISES AND PROBLEMS

CONTROL OF TRANSCRIPTION IN EUKARYOTES

1. Diagram the sequence on the yeast third chromo-
some as the mating type changes from a to a and
back again.

2. What are the differences between a general tran-
scription factor and a specific transcription factor?

3. Tissue culture cells are exposed for five minutes to
radioactive dUTP in the presence or absence of
5-azacytidine. Radioactivity in RNA is determined to
be 1,500 counts per minute without azacytidine and
27,300 in the presence of azacytidine. Propose an
explanation to account for these results.

4. A retrovirus, lacking a cellular oncogene, is shown to
be integrated 3 kilobases from a proto-oncogene.
When the RNA for this oncogene is quantified, in-
fected cells are found to have ten times more
oncogene-specific messenger RNA than uninfected
cells. How can you account for this increase in RNA
synthesis?

PATTERNS IN DEVELOPMENT

5. What is genomic equivalence, and why is explaining
it a central problem in developmental genetics?

6. What is the relationship between parasegments and
segments in the developing Drosophila embryo?

7. What are the three classes of segmentation genes in
Drosophila embryos? What are the effects of muta-
tions of genes in each class?

8. How does the hunchback gene function in
Drosophila development?

9. What are the differences between a syncitial and a
cellular blastoderm in a Drosophila embryo?

* Answers to selected exercises and problems are on page A-19-

10. What is meant by the statement that homeotic genes
have been conserved evolutionarily?

11. What are the four regions of the body plan of the de-
veloping Drosophila embryo laid out by maternal-
effect genes? What are the four major maternal-effect
genes?

12. What is a morphogen? How does the Bicoid protein
of Drosophila function as one?

13. What is the helix-turn-helix motif of DNA binding?
What other motifs are known for DNA-binding pro-
teins?

14. If drugs that inhibit transcription are injected into
fertilized eggs, early cell division and protein synthe-
sis still occur. Why?

15. Why do you suppose so much early research on de-
velopmental genetics was done with amphibians?

CANCER

16. What gross chromosomal abnormalities are associ-
ated with cancers?

17. From the pedigree of figure 16.26, what modes of in-
heritance would be consistent with each type of
cancer, assuming that a single gene controlled each?

18. What chromosomal abnormality is associated with
retinoblastoma? with Wilm's tumor?

19. What is the proposed mechanism of action of the
retinoblastoma gene? What evidence supports this
mechanism? Why is it called an anti-oncogene?

20. Retinoblastoma has been called a recessive onco-
gene. Explain.

21. What are the general forms of animal viruses? What
types of genetic material do they have?

22. What is the minimal genetic complement of a retro-
virus? What does each of the genes code for?

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Critical Thinking Questions

507

23. What translation mechanisms exist for the expres-
sion of the genes of a retrovirus?

24. Assume that a particular oncogene produces a
growth factor.

a. How could a retrovirus affect the oncogene so
that the cell becomes cancerous?

b. How could you test your hypothesis?

25. What are the differences among v-src, c-src, and
proto-src genes?

26. How can the proto-src gene be activated?

27. What is the evidence that the c-src gene came before
the v-src gene?

28. How does translocation activate the c-myc gene in
Burkitt's lymphoma?

29. A cDNA probe for a proto-oncogene is constructed.
Cellular DNA from normal cells and a clone of cells
infected with a retrovirus that lacks the oncogene
(clone 1) is digested with a particular restriction en-
zyme. The DNA is separated in a gel and hybridized
with the radioactive probe. The results appear in the
following figure.

Normal

Clone 1

Interpret these results by describing where the
retrovirus has inserted.

IMMUNOGENETICS

30. What is the general mechanism that allows an anti-
body to "recognize" an antigen?

31. What components go into making an Ig light chain?
a heavy chain?

32. How many different antibodies does a B lymphocyte
produce? How many can it potentially produce be-
fore it differentiates?

33. What are the nucleotide recognition signals in V-J
joining?

34. What are B and T lymphocytes? What roles do they
play in the immune response?

35. What is a T-cell receptor?

36. What is the major histocompatibility complex?

37. A disorder of the immune system is characterized by
a complete lack of antibody production. Provide two
possible molecular defects that would result in such
a condition.

38. Many alleles for the genes for the constant region of
antibodies have been found. Suppose that two such
alleles for the X light chain are called Cj and c 2 . In a
heterozygote, CjC 2 , some cells are found to make
only Cj and others only c 2 . Propose an explanation.

39. Complementary DNA is made from messenger RNA
for the light chain of an antibody molecule. DNA
from embryonic cells and from mature B lympho-
cytes is isolated and digested with a restriction en-
zyme, and the fragments are separated in a gel. Ra-
dioactive cDNA is used to probe this gel, and the
results appear in the figure that follows. Provide an
explanation for these results.

Embryonic

Lymphocytes

CRITICAL THINKING QUESTIONS

1. The E1B gene of adenovirus produces a protein that
binds with p53, allowing the virus to multiply in the
cell. Given that more than 50% of cancer cells lack p53
activity, how might you engineer the adenovirus to at-
tack only cells without p53 activity? That is, can you en-
gineer adenovirus to attack a large proportion of can-
cerous cells?

2. Given what you know about flower development in
plants, what might be the simplest mechanism plants
could use to produce male-only or female-only flowers?

Suggested Readings for chapter 16 are on page B-15.

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Molecular Genetics

17. Non-Mendelian
Inheritance

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NON-MENDELIAN
INHERITANCE

STUDY OBJECTIVES

1. To analyze the inheritance patterns of maternal effects
509

2. To analyze the patterns of cytoplasmic inheritance 511

3. To analyze the patterns of imprinting 524

Artificially colored scanning electron micrograph of a

mitochondrion in the cytoplasm of an intestinal

epithelial cell. (© Professors P. Motta & T. Naguro/

SPL/Photo Researchers, Inc.)

STUDY OUTLINE

Determining Non-Mendelian Inheritance 509
Maternal Effects 509

Snail Coiling 509
Moth Pigmentation 510
Cytoplasmic Inheritance 511

Mitochondria 511

Chloroplasts 515

Infective Particles 518

Prokaryotic Plasmids 522
Imprinting 524
Summary 524
Solved Problems 525
Exercises and Problems 525
Critical Thinking Questions 527

508

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Maternal Effects

509

The phenotype can be controlled by chromo-
somal genes behaving according to Mendelian
rules and the environment. In this chapter, we
deal with another mode of inheritance, non-
Mendelian inheritance (also called extrachro-
mosomal, cytoplasmic, and nonchromosomal inheri-
tance; maternal effects; or imprinting). Maternal effects
are the influences of a mother's genotype on the pheno-
type of her offspring; examples include snail coiling and
moth pigmentation (we started a discussion of maternal
effects in chapter 16, when we looked at development in
Drosophila). Cytoplasmic inheritance is controlled by
nonnuclear genomes found in chloroplasts, mitochon-
dria, infective agents, and plasmids. And imprinting is a
process in which gene expression depends on the parent
from which the gene came. None of these modes of in-
heritance follow the usual Mendelian rules and ratios.

Maternal effects result from the asymmetric contribu-
tion of the female parent to the development of zygotes.
Although both male and female parents contribute
equally to the zygote in terms of chromosomal genes
(with the exception of sex chromosomes), the sperm
rarely contributes anything to development other than
chromosomes. The female parent usually contributes
the zygote's initial cytoplasm and organelles. Zygotic de-
velopment, therefore, usually begins within a maternal
milieu, so that the maternal cytoplasm directly affects
zygotic development (see chapter 16).

Cytoplasmic inheritance refers to the inheritance pat-
tern of organelles and parasitic or symbiotic particles
that have their own genetic material. Chloroplasts, mito-
chondria, bacteria, viruses, and, of course, plasmids all
have their own genetic material. These genomes are
open to mutation. As we shall see, their inheritance pat-
tern does not follow Mendel's rules for chromosomal
genes.

Imprinting occurs in more than twenty genes and is
responsible for several human diseases.

DETERMINING NON-
MENDELIAN INHERITANCE

How does one determine that a trait is inherited? The
question does not have as obvious an answer as we might
expect. Environmentally induced traits can mimic inher-
ited pheno types, as with the phenocopies we discussed
in chapter 5. For example, the inheritance of vitamin
D-resistant rickets is mimicked by lack of vitamin D in
the diet. It is possible to determine that the rickets is not
inherited by simply administering adequate quantities of
vitamin D. Inherited rickets does not respond to vitamin
D until about 150 times the normally adequate amount is
administered.

Some environmentally induced traits persist for sev-
eral generations. For example, a particular Drosophila
strain that normally grows at 21° C was exposed to 36° C
for twenty-two hours. Dwarf progeny were produced.
When they were mated among themselves, fewer and
fewer dwarfs appeared in each generation, but smaller-
than-normal flies were produced as late as the fifth gen-
eration. The appearance of an environmentally induced
trait that persists for several generations has been termed
dauermodification.

Extrachromosomal inheritance is usually identified
by the odd results of reciprocal crosses. If the progeny of
reciprocal crosses are not followed for several genera-
tions, the results can be misleading when extrachromo-
somal inheritance is involved. Where feasible, nuclear
transplantation has proved useful in identifying extra-
chromosomal inheritance. In this technique, the nucleus
of a cell, such as an amoeba or frog egg, is removed by mi-
crosurgery or destroyed by radiation, and another nucleus
substituted. Thus, not only can a nucleus be isolated
from its cytoplasm, but various nuclei can be implanted
in the same cytoplasm.

A similar experiment, called a heterokaryon test, can
be done with various fungi such as Neurospora and
Aspergillus: Mycelia can fuse and form a heterokaryon, a
cell containing nuclei from different strains. Thus, nu-
clei of both strains exist in the mixed cytoplasm. Subse-
quently, spores (conidia) that have one or the other nu-
cleus in the mixed cytoplasm can be isolated. The
phenotype of the colonies produced from these isolated
conidia show whether the trait under observation is con-
trolled by the nucleus or the cytoplasm.

Chromosomal genes in a particular cytoplasm can
also be isolated by repeated backcrossing of offspring
with the male-parent type. In each cross, the content of
the female chromosomal genes is halved, but, presum-
ably, the cytoplasm remains similar to the female line.
Thus, after several generations, male genes can be iso-
lated in female cytoplasm. The phenotypic results of
the final cross will indicate whether inheritance was
chromosomal or extrachromosomal.

MATERNAL EFFECTS

Snail Coiling

Snails are coiled either to the right (dextrally) or to the
left (sinistrally) as determined by holding the snail with
the apex up and looking at the opening. The snail is dex-
trally coiled if the opening comes from the right-hand
side and sinistrally coiled if it comes from the left-hand
side (fig. 17.1). The inheritance pattern of the coiling is at
first perplexing.

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Chapter Seventeen Non-Mendelian Inheritance

Sperm

P 1 Dextral \^ Sinistral

Self

Egg Sperm

Self

Figure 17.1 Inheritance of coiling in the pond snail, Limnaea peregra. Reciprocal crosses {D, dominant
dextral; d, recessive sinistral coiling) are shown {DD mated with dd in each case). The F 1 individuals in both
crosses have the Dd genotype but reflect the mother's genotype in respect to coiling; DD mothers produce
dextrally coiled offspring, whereas dd mothers produce sinistrally coiled offspring. The F 2 individuals in both
cases are identical because the genotypes of the F 1 mothers are identical {Dd). The coiling of a snail's shell is
determined by its mother's genotype, not phenotype.

In the left half of figure 17.1, a dextral snail provides
the eggs, and a sinistral snail provides the sperm. The off-
spring are all dextral; presumably, therefore, dextral coil-
ing is dominant. When the V 1 are self-fertilized (snails are
hermaphroditic), all the offspring are dextrally coiled.
The result is unexpected. Nevertheless, when the F 2 are
self-fertilized, one-fourth produce only sinistral offspring,
and three-fourths produce only dextral offspring. If self-
fertilization is continued through ensuing generations,
this 3:1 phenotypic ratio will be revealed as a Mendelian
1:2:1 genotypic ratio, thereby reaffirming the notion of a
single locus with two alleles, and dextral dominant. How-
ever, something interfered with the expected pheno-
typic pattern.

When the reciprocal cross is made (fig. 17.1, right),
the F : have the same genotype as just described but are
coiled sinistrally, as is the female parent. From here on,
the results are exactly the same for both crosses. In both
cases, the F : are pheno typically similar to the female par-
ent even though the offspring in both crosses have the
same genotype (Dd). The explanation is that the geno-
type of the maternal parent determines the phenotype of
the offspring, with dextral dominant. Thus, the DD
mother in figure 17.1 produces V 1 progeny that are dex-
tral with a Dd genotype, and the dd mother produces
progeny that are sinistral with the same Dd genotype.
Why does this pattern occur?

A process of spiral cleavage takes place in the zy-
gotes of mollusks and some other invertebrates. The spin-
dle at mitosis is tipped in relation to the axis of the egg.
If the spindle is tipped one way, a snail will be coiled
sinistrally; if it is tipped the other way, the snail will be
coiled dextrally. The direction of tipping is determined
by the maternal cytoplasm, which is under the control of
the maternal genotype. Obviously, maternal control af-
fects only one generation — in each generation, the coil-
ing is dependent on the maternal genotype.

Moth Pigmentation

There are other examples of maternal effects in which
the cytoplasm of the mother, under the control of chro-
mosomal genes, controls the phenotype of her offspring.
In the flour moth, Ephestia kiihniella, kynurenin, which
is a precursor for pigment, accumulates in the eggs. The
recessive allele, a, when homozygous, results in a lack of
kynurenin. Reciprocal crosses give different results for
larvae and adults. When a nonpigmented female is
crossed with a pigmented male, the results are strictly
Mendelian; but when the mother is pigmented (a + a), all
the larvae are pigmented regardless of genotype (fig.
17.2). The initial larval pigmentation comes from residual
kynurenin in the eggs, which is then diluted out so that
an adult's pigmentation conforms to its own genotype.

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Cytoplasmic Inheritance

511

Parents

Larvae

Adults

aa X a + a

a + a x aa

Nonpigmented

Pigmented

Nonpigmented

aa a + a

Nonpigmented

Pigmented

a + a

Nonpigmented

Pigmented

Nonpigmented

Pigmented

Figure 17.2 Inheritance pattern of larval
and adult pigmentation in the flour moth,
Ephestia kuhniella. A single locus controls
the presence (a + ) or absence (a) of
kynurenin. In the cross on the left, the
mother is aa (nonpigmented). Her aa
offspring, in both the larval and adult
stages, are also nonpigmented. In the
reciprocal cross (right), the mother has
the a + a genotype and is pigmented. Her
aa offspring are nonpigmented as adults
but are pigmented as larvae because of
residual kynurenin from the egg, which
eventually dilutes out.

CYTOPLASMIC INHERITANCE

Mitochondria

The mitochondrion is an organelle in eukaryotic cells
in which the electron transport chain takes place. The ac-
tual number of mitochondria per cell can be determined
by serial sectioning of whole cells and examination un-
der the electron microscope. This is a tedious and diffi-
cult procedure. Estimates range between ten and ten
thousand per cell, depending on the organism and cell
type. As far as we are concerned, the most interesting
aspect of the mitochondrion is that it has its own DNA.
In most animal cells, the mitochondrial DNA (mtDNA) is
a circle of about sixteen thousand base pairs (fig. 17.3).
However, some organisms (yeast, higher plants) have mi-
tochondrial DNAs five to twenty-five or more times larger
than in animals. And some organisms have linear mito-
chondrial chromosomes.

Two general patterns are found in mitochondrial in-
heritance in animals. First, the mitochondria are generally
inherited in a maternal fashion; that is, the male gamete
usually does not contribute mitochondria to the zygote.
However, a small amount of "leakiness" occurs in this
process. For example, it has recently been shown that
about one mitochondrion per thousand is of paternal ori-
gin in mice. In some species, such as mussels, it appears
that mitochondrial inheritance is biparental. That is, the
population of mitochondria in an offspring derives al-
most equally from the male and female parent. In some
gymnosperm plants, such as coastal redwoods, mito-
chondria are inherited paternally — only paternal mito-
chondria are passed into the zygote. However, these are
all exceptions to the general rule of maternal inheritance
of mitochondria.

The second general pattern of mitochondrial inheri-
tance is homoplasmy, the existence of a uniform popu-
lation of mitochondria within an organism. In general, all

the mitochondria within an individual are genetically
identical. Certainly, biparental inheritance and leakiness
of paternal mitochondria violate that principle, resulting
in heteroplasmy, a heterogeneity of mitochondria
within a cell or organism.

Figure 17.3 Electron micrograph of the circular DNA from
within a mouse cell mitochondrion. Magnification 48,000x.
(M. M. K. Nass, "The circularity of mitochondrial DNA," Proceedings of the
National Academy of Sciences, USA, 56 (1966):1215-22. Reproduced by
permission of the author.)

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Chapter Seventeen Non-Mendelian Inheritance

Mitochondrial Genomes

Numerous mitochondrial DNAs have been sequenced, in-
cluding the human mitochondrial DNA, which is 16,569
base pairs long. It is a model of economy, with very few
noncoding regions and no introns (fig. 17.4). Each strand
of the duplex is transcribed into a single RNA product
that is then cut into smaller pieces, primarily by freeing
the twenty-two transfer RNAs interspersed throughout
the genome. Also formed are a 16S and a 12S ribosomal
RNA. Although proteins and small molecules such as ATP
and tRNAs can move in and out of the mitochondrion,
large RNAs cannot. Thus, the mitochondrion must be rel-
atively self-sufficient in terms of the RNAs needed for
protein synthesis. We previously discussed mitochondrial
protein synthesis when we looked at unique attributes of
the mitochondrial genetic code in chapter 1 1 .

Oxidative phosphorylation, the process that occurs
within the mitochondrion, requires at least sixty-nine
polypeptides. The human mitochondrion has the genes
for thirteen of these: cytochrome b, two subunits of
ATPase, three subunits of cytochrome-c oxidase, and
seven subunits of NADH dehydrogenase. The remaining
polypeptides needed for oxidative phosphorylation are
transported into the mitochondrion; they are synthesized
in the cytoplasm under the control of nuclear genes. Pro-
teins targeted for entry into the mitochondrion have spe-
cial signal sequences (see chapter 1 1).

The signal sequences range up to eighty-five amino
acids long. Signal sequences examined so far do not have
consensus amino acids but do have certain attributes
(fig. 17.5), including a somewhat regular alternation of
basic (positively charged) and hydrophobic (negatively
charged) residues. In addition, they form a helices with
opposite hydrophobic and hydrophilic faces that must
somehow be important in the protein's ability to enter
the mitochondrion. When a signal sequence (such as that
in fig. 17.5) is attached to nonmitochondrial proteins by
DNA manipulations, those proteins are transported into
the mitochondrion.

The mitochondrial ribosomal RNA is more similar to
prokaryotic ribosomal RNA than to eukaryotic ribosomal
RNA. The mitochondrial ribosome, although constructed
of imported cellular proteins, is sensitive to prokaryotic
antibiotics; for example, streptomycin and chlorampheni-
col inhibit their function. This affinity (close resem-
blance) between mitochondria and prokaryotes is strong
support for the symbiotic origin of mitochondria. That is,
we now accept the model advocated by L. Margulis that
organelles such as mitochondria and chloroplasts were
originally free-living bacteria and cyanobacteria, respec-
tively. These prokaryotes invaded or were eaten by early
cells and, over evolutionary time, became the organelles
we see today. Since they arose as prokaryotes, these or-
ganelles retain certain evolutionary similarities to other
prokaryotes.

Transcription

p hp 1/16,569
tRNA Mne

tRNA Thr

Cytochrome
b

tRNA Trp

tRNA Asp

-"*""*"' ATPase 6

CCO II ATPase 8

tRNA Lys

NADH 5

tRNA Leu
tRNA Ser
tRNA Hls

NADH 4

NADH 4L

tRNA Arg
NADH 3

tRNA Gly
CCO III

Figure 17.4 Gene map of the
human mitochondrial
chromosome. All but nine loci
are on the heavy (H) strand. The
light-strand (L) loci are labeled
inside the circle; the H-strand
loci are labeled on the outside.
Also shown are the origins of H-
and L-strand replication and the
directions of transcription. The
twenty-two tRNA genes are
colored red. NADH refers to
NADH dehydrogenase (subunits
1-4, 4L, 5, and 6); CCO refers
to cytochrome-c oxidase

(subunits l-lll). (Source: Data from
V. McKusick, Mendelian Inheritance in
Man, 7th edition, 1986.)

Tamarin: Principles of
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Molecular Genetics

17. Non-Mendelian
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Cytoplasmic Inheritance

513

Lynn Margulis (1938- ).
(Courtesy of Lynn Margulis, Boston
University Photo Services.)

Among the mitochondrial DNAs that have been se-
quenced from different organisms, we see great varia-
tion in content and organization. Yeast mitochondrial
DNA, for example, is not as economical as human mito-
chondrial DNA. Yeast mitochondrial DNA, about five
times larger than human mitochondrial DNA, has non-
coding regions as well as introns. Because mitochon-
dria are similar in structure and biochemistry to
prokaryotic cells, given the general lack of introns in

prokaryotic genes, it was surprising to find introns in
yeast mitochondrial DNA. These genes most probably
arose later as nuclear genes that were then "captured"
by the mitochondria, possibly by recombination with
nuclear DNA.

Of the many mitochondria sequenced to date (about
175 at the beginning of 2001), the sizes range from less
than 6 to more than 200 kilobases and from 3 to 97
genes. With this wide range of genes present, the only
generality we can make about mitochondrial DNA is
that the large and small segments of the mitochondrial
ribosomal RNA, as well as most of the mitochondria's
transfer RNAs, are usually coded by the mitochondria's
own genome, as are several proteins in respiratory
complexes III and IV (cytochrome c oxidase and cy-
tochrome c oxidoreductase). Once the interaction
within the mitochondrial-nuclear genetic system is
clearly understood, we might expect to see several dif-
ferent inheritance patterns — following either cytoplas-
mic or nuclear lines — for the genetic defects that lead
to interruption of cellular respiration. Among the best-
studied phenotypes with such inheritance patterns are
the petite mutations of yeast.

MET

VAL ARG

GLY

ASP LEU

THR

SER SER

vil

ILE

PRO GLU

—

LYS

GLU PRO

ILE

GLU GLN

Kinmi

PRO LEU ASN CYS ILE VAL ALA VAL SER

25 26 +

PRO TRP PRO PRO LEU ARG ASN GLU PHE

B {MlSlMMlSiMSiMm
- + 49

VAL GLU GLY LYS GLN ASN LEU VAL ILE

+ + + - +

LYS ASN ARG PRO LEU LYS ASP ARG ILE

PRO ARG GLY ALA HIS PHE LEU ALA LYS

GLN ASN MET GLY ILE GLY LYS ASN

+ +38

LYS TYR PHE GLN ARG MET THR THR

+ +

MET GLY ARG LYS THR TRP PHE SER

+ -
ASN ILE VAL LEU SER ARG GLU LEU

PRO GLU LEU ALA SER LYS VAL ASP MET

VAL TYR GLN GLU ALA MET ASN GLN PRO GLY HIS LEU
GLN GLU PHE GLU SER ASP THR PHE PHE PRO GLU ILE

SER LEU ASP ASP ALA LEU ARG LEU

VAL TRP ILE VAL GLY GLY SER SER

E OlMMi
+ +

ARG LEU PHE VAL THR ARG ILE MET

- + +

ASP LEU GLY LYS TYR LYS LEU LEU

— — — — + + +

PRO GLU TYR PRO GLY VAL LEU SER GLU VAL GLN GLU GLU LYS GLY ILE LYS TYR LYS PHE

- - + + -

GLU VAL TYR GLU LYS LYS ASP

Figure 17.5 The amino acid sequence of mouse dihydrofolate reductase. Numbers refer to sequential
amino acids. The first eighty-five amino acids serve as the signal sequence for transport into
mitochondria. Five a-helical regions exist in the protein (A-E). Positively and negatively charged amino
acids are marked with (+) and (-) Signs. (Reprinted by permission from E. C. Hurt and G. Schatz, "A cytosolic
protein contains a cryptic mitochondrial targeting signal," Nature, Volume 325, p. 499, 1987. Copyright © 1987 Macmillan
Magazines, Ltd.)

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Chapter Seventeen Non-Mendelian Inheritance

Petites

Under aerobic conditions, yeast grows with a distinctive
colony morphology Under anaerobic conditions, the
colonies are smaller, and the structures of the mitochon-
dria are reduced. Occasionally, when growing aerobi-
cally, small, anaerobiclike colonies appear; but in these
colonies, the mitochondria appear perfectly normal.
These colonies are caused by petite mutations. When
petites are crossed with the wild-type, three modes of in-
heritance emerge (fig. 17.6). The segregational petite,
caused by mutation of a chromosomal gene, exhibits
Mendelian inheritance. The neutral petite is lost immedi-
ately upon crossing to the wild-type. The suppressive pe-
tite shows variability in expression from one strain to the
next but is able to convert the wild-type mitochondria to
the petite form. All petites represent failures of mito-
chondrial function, whether the function is controlled by
the mitochondria themselves or by the cell's nucleus;
they usually lack one or another cytochrome.

Although the mechanisms that produce neutral and
suppressive petites are not known with certainty, their
DNA has supplied some interesting information. In some
petites, no change in the buoyant density of the DNA is
found. (Buoyant density, a term that describes the posi-
tion at which the DNA equilibrates during density-
gradient centrifugation, is a measure of the composition
of the molecule; see chapter 15.) In other petites, changes

Diploid

Meiosis
f

Meiosis

1 Petite : 1 Wild-type

All wild-type

Mostly petites

Segregational
petites

Neutral petites

Suppressive
petites

Figure 17.6 Petite yeasts categorized on the basis of
segregation patterns. Three types of petites are recognized
(segregational, neutral, and suppressive), depending on the
meiotic segregation pattern of petite x wild-type diploids.
Segregational petite heterozygotes segregate a 1:1 ratio of
spores; neutral petites are lost when heterozygous; and
suppressive petites act in a dominant fashion under the same
circumstances.

in buoyant density range from very small to the complete
absence of DNA.

Petites, therefore, can be the result of an approxima-
tion to a point mutation (with no measurable change in
the buoyant density of the DNA), marked changes in the
DNA, or the total absence of DNA. In most petites, pro-
tein synthesis within the mitochondrion is lacking. Any
and all of these changes produce the petite (anaerobic-
like) phenotype.

Neutral petites seem to have mitochondria that en-
tirely lack DNA. When neutral petites are crossed with
the wild-type to form diploid cells, the normal mitochon-
dria dominate. During meiosis, virtually every spore re-
ceives large numbers of normal mitochondria; the prog-
eny are, therefore, all normal.

Suppressive petites could exert their influence over
normal mitochondria in one of two ways. The suppres-
sive mitochondria might simply out-compete the normal
mitochondria and take over; they might simply repro-
duce faster within a cell. Alternatively, crossing over be-
tween the DNA of the suppressive petite and the wild-
type might affect the normal DNA if the suppressive
petite's DNA were severely damaged. Presumably, recom-
bination in mitochondrial DNAs occurs when two or
more mitochondria fuse, bringing the two different sets
of DNA in contact within the same organelle. Recombi-
nation would presumably take place by normal crossover
mechanisms.

If large portions of the DNA from the suppressive mi-
tochondria were missing or altered, recombination with
the normal mitochondria's DNA might exchange some
of this damaged DNA. Several experiments have crossed
a suppressive petite and a wild-type, each with mito-
chondrial DNA of known buoyant density. The DNAs of
the offspring colonies, which were petites, were of vari-
ous buoyant densities. For example, when a normal
strain with mitochondrial DNA with a buoyant density
of 1.684 g/cm 3 was crossed with a suppressive petite
with a buoyant density of 1.677 g/cm 3 , the offspring
colonies' mitochondrial DNA had buoyant densities of
1.671, 1.674, and 1.683 g/cm 3 . Such information sup-
ports the notion that the suppressive character takes
over a colony by way of recombination.

Human Mitochondrial Inheritance

In human beings, certain diseases trace their dysfunction
to mitochondrial pathologies. The first such disease, Luft
disease, characterized by excessive sweating and general
weakness, was reported in 1962. In 1988, Douglas Wal-
lace and his colleagues showed that Leber optic atrophy
is a cytoplasmically inherited disease. This disease causes
blindness, with a median age of onset of twenty to
twenty-four years. The onset age and phenotype are vari-
able, depending on the degree of heteroplasmy in the in-

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dividual. Apparently, defects in mitochondria are not tol-
erable in the optic nerve, which demands a great deal of
energy. The disease also does some damage to the heart.
Pedigrees showed that Leber optic atrophy is transmitted
only maternally. Sequencing of mitochondrial DNAs in af-
fected families pinned down the disease to a point muta-
tion, a change in nucleotide 1 1,778, which is in the gene
for NADH dehydrogenase subunit 4 (see fig. 17.4). A gua-
nine is changed to an adenine at codon 340, which con-
verts an arginine to a histidine. This is the first human
disease traced to a specific mitochondrial DNA mutation.
Since 1962, over one hundred diseases, including some
of the general symptoms of aging and cancer, have been
attributed to mitochondrial pathology.

Antibiotic Influences

Since the machinery of mitochondrial protein synthesis
is prokaryotic in nature, antibiotics such as chloram-
phenicol and erythromycin can inhibit it. These antibi-
otics elicit a petite-type growth response in yeast. Antibi-
otic-resistant strains can be obtained by growing yeast on
the antibiotic; only resistant mutants will grow. The re-
sistance appears to be inherited in the mitochondrial,
not the cellular, DNA. A mitochondrial inheritance pat-
tern results, with crosses between a resistant and a sensi-
tive (wild-type) yeast, as shown in figure 17.7. The result-
ing diploid colonies segregate both resistant and
sensitive cells. Although not expected on the basis of a
chromosomal gene, the random sorting of mitochondria
through cell division could result in a wild-type cell con-
taining only sensitive mitochondria. Since some yeast
have only one to ten mitochondria per cell, this random
assortment of sensitive mitochondria can be expected to
occur at a relatively high rate.

Chloramphenicol-
resistant haploid

Chloramphenicol-
sensitive haploid

Mitochondrion

Resistant diploid

Resistant cells

Sensitive cells

Figure 17.7 Inheritance of antibiotic (chloramphenicol)
resistance in yeast. Resistant and sensitive cells are produced
by a diploid cell that resulted from a cross of resistant and
sensitive haploids. The segregation is not in a simple Mendelian
ratio, but depends on the random assortment of mitochondria.
Sensitive cells have no resistant mitochondria. Resistant cells
have resistant mitochondria.

Chloroplasts

The chloroplast is the chlorophyll-containing organelle
that carries out photosynthesis and starch-grain forma-
tion in plants (fig. 17.8). Chloroplasts are referred to as
plastids before chlorophyll develops. However, when
grown in the dark (and under some other circum-
stances), plastids do not develop into chloroplasts, but
remain reduced in size and complexity. These undevel-
oped plastids, referred to as proplastids, are each about
the size and shape of a mitochondrion.

Like mitochondria, chloroplasts contain DNA and ri-
bosomes, both with prokaryotic affinities. The DNA of
chloroplasts (cpDNA) is a circle that ranges in size from
85 kilobases (kb) in the green alga Codium to as large as
2,000 kilobases in the green alga Acetabularia. Thus,
chloroplast DNA is minimally about five times the size of
an animal mitochondrial DNA. The chloroplast DNA, like
mitochondrial DNA, controls the production of transfer

RNAs, ribosomal RNAs, and some of the proteins found
within the organelle. From the more than nineteen
chloroplast DNAs that have been sequenced, there seem
to be about one hundred genes in the chloroplast
genome. About thirty code for the subunits of the five
photosynthetic protein complexes: photosystem I, photo-
system II, ribulose bisphosphate carboxylase-oxygenase,
cytochrome b6-f complex, and ATP synthase. About sixty
genes code for the protein synthesis apparatus of the
chloroplast. Scientists believe that the chloroplast evolved
from symbiotic cyanobacteria (blue-green algae), which
have many affinities with the chloroplast: The ribosomal
RNA of cyanobacteria will hybridize with the DNA of
chloroplasts.

The similarities between mitochondria and chloro-
plasts make it possible to predict the inheritance pat-
terns of chloroplast mutations on the basis of existing

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Chapter Seventeen Non-Mendelian Inheritance

Figure 17.8 Electron micrograph of lettuce
chloroplasts. The chloroplast consists of an
outer membrane, stacks of grana, lamellae,
and stroma. Magnification 3,570x. (© Dr. J.
Burgess/Science Photo Library/Photo Researchers, Inc.)

Stroma

Granum

(~ 1 (um wide)

knowledge of mitochondrial genetics: We should find
both chromosomal and plastid mutants of chloroplast
functions. Simple segregation should occur in the chro-
mosomal mutations, and cytoplasmic patterns of inheri-
tance should occur with the chloroplast DNA mutations.
Investigation of these inheritance patterns is compli-
cated by the fact that plant cells have both mitochondria
and chloroplasts. Since both have prokaryotic affinities, it
is sometimes difficult to determine whether a genetic
trait is due to a defect in the genetic system of the chloro-
plast or the mitochondrion. Like mitochondria, chloro-
plasts generally show homoplasmy and maternal inheri-
tance, although, as in mitochondria, there are exceptions.
For example, gymnosperms usually have paternal inheri-
tance of chloroplasts.

Lesions in the photosystems of the chloroplast result
in proplastid formation, with a loss of green color. When
proplastid formation occurs in a particular tissue of a
plant, variegation results. That is, there are both green and
white parts, often as stripes. Some interesting genetic
studies have focused on the inheritance of variegation,
especially in the interaction of chloroplast and chromo-
somal genes.

Zea mays

M. Rhoades worked on the variegation in corn (Zea
mays) controlled by the iojap chromosomal locus,
which, when homozygous, prevents proplastids from de-
veloping into chloroplasts and thus results in variegation.
The zq/op-affected plastids do not contain ribosomes or
ribosomal RNA; they therefore lack protein synthesis.

Marcus M. Rhoades
(1903-1991). (Courtesy of
Indiana University Office of
Communications and Marketing.)

The interaction of chromosomal and extrachromoso-
mal inheritance is shown in the reciprocal crosses de-
picted in figure 17.9. One cross produces results exactly
as would be predicted on the basis of simple Mendelian
inheritance, with the homozygous recessive genotype
(ijij) inducing variegation. When the reciprocal cross is
carried out, blotch variegation is seen in both the F : and
F 2 that carry the dominant Ij allele.

This inheritance pattern is caused by the fact that the
pollen grain in corn does not carry any chloroplasts,
whereas the ovule does. Thus, the first cross in figure
17.9 deals with the passage of normal chloroplasts only
into the F 2 generation. In the F 2 , the ijij genotype then in-
duces variegation. The chloroplasts of the pollen parent
are unimportant because they do not enter the F x . In the
reciprocal cross, however, because the stigma parent is
variegated, the F : is heterozygous but carries proplastids
from the ovule that remain proplastids even under the
dominant normal (If) allele. Therefore, regions of color-
less cells produce white spots (blotchy variegation).
Once the ij allele induces chloroplasts to become pro-

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Stigma v

III!
Green

Stigma

ijij
Variegated

i /

/y/y'X Self
Green

ijij ijij ijij

Green Green Variegated

Some

blotch

variegated

/y/y'X Self

Blotch

variegatec

Ijij
Some

blotch

variegated

ijij
Variegated

Figure 17.9 Reciprocal crosses
involving the chromosomal gene
iojap in corn. The homozygous
recessive condition (ijij) induces
variegation (representative corn
leaves are shown). (Blotch
variegation consists of irregularly
shaped white areas rather than
striping.) However, plants with the
dominant allele (Ijij, Ijij) can still be
variegated if their mothers were
variegated, since mothers pass
on their chloroplasts to their
offspring; males (pollen parents)
do not pass on their chloroplasts.
Iojap homozygotes induce
variegation. The defective
chloroplasts are then inherited in
a cytoplasmic fashion.

1:2:1

plastids, they do not revert to normal type even under
the If allele. Thus, we see the interaction of a chromo-
somal gene and the chloroplast itself, which "inherits" a
changed condition.

There is some evidence that iojap may suppress the
chloroplast rather than cause a mutation of some func-
tion. There are loci in corn and in other species that can
induce back mutation in the chloroplasts. Removal of
suppression rather than an actual reversion is more likely
to occur because the reversion rate is too high to be due
to simple back mutation.

Four-O' clocks

The first work with corn variegation was done by Carl
Correns, one of Mendel's rediscoverers. Correns also
found maternal inheritance of variegation in the four-
o'clock plant, Mirabilis jalapa. He could predict color
and variegation of offspring solely on the basis of the re-
gion of the plant on which the stigma parent was lo-
cated. A flower from a white sector, when pollinated by
any pollen, would produce white plants; a flower on a
green sector or a variegated sector produced green or
variegated plants, respectively, when pollinated by
pollen from any region of a plant. We thus see the simple
maternal nature of the inheritance of the variegation. A
chromosomal gene, like iojap, induces variegation. In-
heritance of this induced variegation follows the "mater-
nal" pattern of chloroplast inheritance.

Chlamydomonas

The single-celled green alga, Chlamydomonas rein-
hardi, has been used in the study of extrachromosomal
inheritance for several reasons. First, it has a single, large
chloroplast; it can survive by culture technique even
when the chloroplast is not functioning; and finally, it
shows some interesting non-Mendelian inheritance pat-
terns related to mating type. R. Sager has done extensive
work on the inheritance of streptomycin resistance in
Chlamydomonas.

Streptomycin resistance can be selected for in
Chlamydomonas in several ways. Normal cells, sensitive
to the antibiotic, are killed in its presence. If cells are
grown in low levels of the antibiotic (100 g/ml), some
cells show resistance to it. When these cells are crossed

Ruth Sager (1918-1997).
(Courtesy of Dr. Ruth Sager.)

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Chapter Seventeen Non-Mendelian Inheritance

with the wild-type, the resistance segregates in a 1:1 ra-
tio, indicating that streptomycin resistance is controlled
by a chromosomal locus. The same experiment can be re-
peated using high levels of the antibiotic in the medium
(500-1,600 g/ml). Again, resistant colonies grow. If they
are crossed with the wild-type, a 1 : 1 ratio does not ensue.

Chlamydomonas does not have sexes but does have
mating types mt + and mt~ . Only individuals of opposite
type can mate. Mating type is inherited as a single locus
with two alleles. When two haploid cells of opposite
mating type fuse, they form a diploid zygote, which then
undergoes meiosis to produce four haploid cells, two of
mt + and two of mt~ .The high-level resistance always seg-
regates with the mt + parent (fig. 17.10). It is as if the mt +
parent were contributing the cytoplasm to the zygote in a
manner similar to maternal plastid inheritance in plants.
The mt~ parent acts like a pollen parent by making a
chromosomal contribution but not a cytoplasmic one.

The mechanism of the extrachromosomal inheri-
tance pattern of Chlamydomonas is the preferential di-

gestion of the DNA of the chloroplast from the mt~ par-
ent. Currently, we believe that streptomycin's target is
the chloroplast.

More recent work has shown that the mt + inheri-
tance is only 99-98% effective— that is, 0.02% of the off-
spring in crosses of the type shown in figure 17.10 have
the streptomycin phenotype of the mt~ parent. Thus, we
have the possibility of studying recombination in chloro-
plast genes. Although most of the evidence is only indi-
rect and plagued by the previously mentioned problems
of separating chloroplast and mitochondrial effects,
some initial mapping studies have been done.

Infective Particles

Paramecium

Tracy Sonneborn discovered the killer trait in Parame-
cium. Before analyzing this trait, we must digress a mo-
ment to look at the life cycle of Paramecium, a ciliated

^\ _

mt~

Streptomycin

sensitive

All strepl

resis

X^ n

mt ratio

mt + : mt~
streptomycin
resistant
n

2:2

mt +

Streptomycin

sensitive

' „ f)

X Streptomycin

resistant

resistant
n

mt + /mr

Streptomycin

sensitive

Meiosis

mt : mt~

All streptomycin

sensitive

2:2

Figure 17.10 Inheritance pattern of streptomycin resistance in Chlamydomonas is dependent on the genotype
of the mt + parent. (The n and 2n refer to the ploidy of the cells.) If the mt + parent is streptomycin resistant
{red), then the diploid heterozygote, as well as the meiotic products, will be streptomycin resistant. If, however,
the mt + parent is streptomycin sensitive {green), the diploid heterozygote, as well as the meiotic products, will
be streptomycin sensitive.

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Cytoplasmic Inheritance

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Tracy M. Sonneborn
(1905-1981). (Courtesy: Indiana
University Archives.)

protozoan familiar to most biologists. Ciliates have two
types of nuclei: macronuclei and micronuclei. In Para-
mecium, there are two micronuclei, which are primarily
reproductive nuclei, and one macronucleus, which is a
polyploid nucleus concerned with the vegetative func-
tions of the cell. During cell division, termed binary fis-
sion, the micronuclei divide by mitosis and the
macronucleus constricts and is pulled in half.

Paramecium undergoes two types of nuclear re-
arrangements, during conjugation and autogamy. In
conjugation, individuals of two mating types come to-

gether and form a connecting bridge. The nuclear events
are shown in figure 17.11. Briefly, the macronucleus of
each cell disintegrates while the micronuclei undergo
meiosis. Of the resulting eight micronuclei per cell, seven
disintegrate and one remains; this one undergoes mitosis
to form two haploid nuclei per cell. A reciprocal ex-
change of nuclei across the bridge then occurs. Each cell
now has two haploid nuclei, one original and one mi-
grant. The two nuclei fuse to form a diploid nucleus. The
diploid nuclei in the two conjugating cells are genetically
identical because of the reciprocity of the process. These
nuclei then undergo two mitoses each to form four
diploid nuclei per cell. Two nuclei become macronuclei,
which separate at the next cell division; two remain as
micronuclei that divide by mitosis at the next cell divi-
sion. The two cells that separate are known as exconju-
gants. Depending primarily on the amount of time con-
jugating cells remain united, an exchange of cytoplasm
may occur along with the exchange of nuclei.

In the second type of process, autogamy, only one
Paramecium is involved (fig. 17.12). The nuclear events
are the same as in conjugation except that, at the point
where a reciprocal exchange of nuclei would take place,
the two haploid nuclei within the cell fuse. All cells after
autogamy are homozygous.

Macronucleus
breaks down;
meiosis of
micronuclei
occurs
>■

Seven
micronuclei
break down

Mitosis occurs
and is followed
by nuclear
exchange

Nuclei fuse;

conjugants

separate

Two mitoses per

exconjugant

occur

K,k

Kk Kk

Two micronuclei

become

macronuclei

Micronuclei
undergo mitosis
and macronuclei
separate as
cell divides

Kk Kk

Figure 17.11 Conjugation in Paramecium. The letters K and k represent alleles of a
gene in each micronucleus. When a KK and a kk individual conjugate, the exconjugants
have the identical Kk genotype.

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Macronucleus
breaks down;
meiosis of
micronuclei
occurs
>■

Seven
micronuclei
break down

Mitosis occurs

K,k

K,K

(ork,k)

Nuclei fuse

(orkk)

Two mitoses
occur

(orkk)

Two micronuclei

become

macronuclei

(orkk)

Micronuclei
undergo mitosis
and macronuclei
separate as
cell divides

KKandKK
(or kk and kk)

Figure 17.12 Autogamy in Paramecium. The letters K and k represent alleles of a
gene in each micronucleus. If a heterozygote undergoes autogamy, it becomes
homozygous for one of the alleles {KK or kk).

Killer Paramecium and Kappa Particles

Sonneborn and his colleagues found that when certain
stocks of Paramecium were mixed together, one stock
had the ability to cause the individuals of the other stock
to die. Those individuals causing death were called
"killers" and those dying were referred to as "sensitives."
During conjugation, the sensitives are temporarily resis-
tant to the killers. If cytoplasm is not exchanged during
the conjugation, the exconjugants retain their original
phenotypes so that killers stay killers and sensitives stay
sensitives. When an exchange of cytoplasm occurs be-
tween sensitive and killer cells, both exconjugants are
killers. The transfer of some cytoplasmic particle seems
to be implied. Indeed, Sonneborn observed such parti-
cles in the cytoplasm of killers and called them kappa
particles (fig . 17.13).

Although the occurrence of killer Paramecium does
not appear to involve chromosomal genes, Sonneborn re-
ported one case in which exconjugant killer paramecia
of hybrid origin underwent autogamy. He found that half
of the resulting cells had no kappa particles and had be-
come sensitives. He concluded that a gene is required for
the presence of kappa particles, which has subsequently

been verified by numerous crosses. Figure 17.14 illus-
trates the sequence of genetic events that would pro-
duce a heterozygous killer Paramecium that, upon au-
togamy, would have a 50% chance of becoming sensitive.

Although not yet cultured outside of a Paramecium,
kappa is presumably a bacterium because it has many
bacterial attributes including size, cell wall, presence of
DNA, and presence of certain prokaryotic reactions (fig.
17.15). J. Preer and his colleagues, who studied kappa it-
self, named it Caedobacter taeniospiralis. Kappa occurs
in at least two forms. The N form, the infective form that
passes from one Paramecium to another, does not con-
fer killer specificity on the host cell. The N form is at-
tacked by bacteriophages that induce formation of inclu-
sions, called R bodies, inside the kappa particle and thus
convert it to the B form. These R bodies are visible under
the light microscope as refractile bodies (fig. 17.15).

In the B form, kappa can no longer replicate; it is of-
ten lysed within the cell. It confers killer specificity on
the host cell, however. The sensitives are killed by the
toxin paramecin, which is released by the killer Para-
mecium into the environment. Precisely what steps are
involved in its formation are not known, although it is

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(a)

(b)

plain that the virus plays an integral role. Whether the vi-
ral DNA or the kappa DNA codes the toxin is also not
known at present.

Mate-Killer Infection and Mu Particles

Kappa is not the only infective agent known in Parame-
cium. Another agent is the mate-killer infection. Here
again, killer cells have visible, bacterialike particles,
called mu particles, in the cytoplasm. Preer and his col-
leagues have named them Caedobacter conjugatus.
Mate-killers do not release a toxin into the environment,
but instead kill their mates during conjugation. One of
two unlinked dominant genes, M 1 and M 2 , is required for
the presence of mu particles. An interesting phenome-
non occurs when a mate-killer becomes homozygous
m 1 m 1 m 2 m 2 by autogamy. Although the offspring even-
tually lose their mu particles, virtually no loss of particles
occurs until about the eighth generation, when some off-
spring lose all their mu. Up to this generation, all the cells
maintain a full complement of mu. In the fifteenth gener-
ation, only about 7% of the cells still have mu particles.

This phenomenon is explained as the diluting out
not of the mu themselves, but of a factor called meta-
gon, which is necessary for the maintenance of mu in
the cell. Once the cell becomes homozygous recessive,
no further metagon production occurs. The verifica-
tion that metagon is subsequently diluted out is evident
in fifteenth-generation cells that still have their mu. We
would expect that after fission, one daughter cell
would have a metagon and the other would not. What
we expect, in fact, happens. The rate of dilution is con-
sistent with an original number of about one thou-
sand metagons per cell. The metagon appears to be

Figure 17.13 (a) Normal (sensitive) Paramecium,
(b) Kappa-containing (killer) Paramecium. A Paramecium
is about 200 |xm long. (Source: T. M. Sonneborn, figure 29.3,
p. 373 in I. H. Herskowitz, Genetics, 2nd ed. [Boston: Little, Brown,
1965]. Reproduced by permission.)

Killer

kk
Sensitive

Kappa -\*-\

Conjugation with
cytoplasmic exchange

All killer exconjugants

Autogamy

Killer

kk
Sensitive

1 : 1

Figure 17.14 Autogamy in a heterozygous (Kk) killer
Paramecium (formed by conjugation, with cytoplasmic
exchange, of a KK killer and a kk sensitive cell). Upon
autogamy, the heterozygote has a 50% chance of becoming a
homozygous (KK) killer or a homozygous (kk) sensitive cell that
loses its kappa particles.

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Chapter Seventeen Non-Mendelian Inheritance

Figure 17.15 Electron micrograph of a
sectioned kappa particle {Caedobacter
taeniospiralis). Phage particles appear as
dark inclusions. The plane of the section
cuts through a rolled-up R body.
Magnification 61,200X. (Reproduced by
permission of J. R. Preer, Jr.)

messenger RNA because it is destroyed by RNase. Its
protein product is presently unknown.

We thus see several instances of infective particles
that interact with the Paramecium genome to produce
interesting phenotypic results. Similar interactions are
known in other organisms — for example, the killer
trait in yeast.

Drosophila

Several infective particles mimic patterns of inheritance
in insects. In Drosophila, we find forms of the sex-ratio
phenotype in which females produce mostly, if not ex-
clusively, daughters. One form is inherited as a chromo-
somal gene; another form, however, is not chromosomal.
In the nonchromosomal form, females usually produce a
few sons. These sons do not pass on the sex-ratio trait,
but the daughters of sex-ratio females do. Because the
trait persisted even after all the chromosomes had been
substituted out of the stock by appropriate crosses, it
was proven to be extrachromosomal.

In addition, about half the eggs of a sex-ratio female
fail to develop. Cytoplasm can be withdrawn from the
undeveloped eggs and used to infect other females. The
trait, then, is caused by some cytoplasmic factor that
could infect other females and is not passed on by sperm.
Detailed cytological examination of the cytoplasm of sex-
ratio females has revealed a spirochete (fig. 17.16) that
has been isolated and used to infect other female
Drosophila with the sex-ratio trait; it is, therefore, the
causal agent of this phenotype.

Prokaryotic Plasmids

In chapters 7 and 13, we discussed the role of plasmids in
the study of prokaryotic genetics and in recombinant
DNA work. They are mentioned again here because they
represent extrachromosomal genetic systems, primarily

Figure 17.16 Electron micrograph of
the spirochete associated with the
extrachromosomal sex-ratio trait in
Drosophila. Magnification 22,700x.
(K. Oishi and D. F. Poulson, "A virus associated
with SR-spirochetes of Drosophila nebulosa, "
Proceedings of The National Academy of
Sciences, USA, 67 [1970]: 1565-72.
Reproduced by permission of the authors.)

in prokaryotes. The autonomous segments of DNA
known as plasmids are, for the most part, known from
bacteria, in which they occur as circles of DNA within
the host cell (noncircular DNA is soon degraded). When
plasmids become integrated into the chromosomes, they
become indistinguishable from chromosomal material.

R and Col Plasmids

In addition to the F factor found frequently in bacteria, a
variety of other plasmids occur, including the R and Col
plasmids. The R plasmids carry genes for resistance to
various antibiotics, and the Col plasmids have genes
that are responsible for producing proteins called
colicins, which are toxic to strains of E. coli (fig. 17.17).
Plasmids containing genes for Col-like toxins specific for
other bacterial species are also known. Col and R plas-
mids can exist in two states. In one state, the plasmid has

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..•■'■»* ',. . - ■■ . . i ...,.'' 7 v. .v / .- ■

Figure 17.17 Electron micrograph of
replication of Col E1 circular plasmid. The
arrows mark the branch points of the theta
structure. Magnification 90,000x. (Source: J. I.
Tomizawa, Y. Sakakibara, and T. Kakefuda, "Replication
of Colicin E1 plasmid DNA in cell extracts: Origin and
direction of replication," Proceedings of The National
Academy of Sciences, USA, 71 [1974]:2260-64.)

a sequence of genes called the transfer operon (tra) ,
which makes the plasmids similar to F factors in that they
can transfer their genes from one bacterium to the next.
In the other state, the plasmids lack this operon and
cannot transfer their loci to another cell. Thus, Col and R
plasmids are actually made of two parts: the loci for an-
tibiotic resistance or colicin production and the part re-
sponsible for infectious transfer. In R plasmids, the infec-
tious transfer part is called the resistance transfer
factor (RTF).

The occurrence of resistance plasmids was observed
in Japan in the late 1950s, when it was discovered that
bacteria were simultaneously acquiring resistance to sev-
eral antibacterial agents. When cultures of Shigella, a
dysentery-causing bacterium, were exposed to strepto-
mycin, sulfonamide, chloramphenicol, or tetracycline,
the bacteria exhibited resistance not only to the one par-
ticular agent they were exposed to but to one or more of
the others as well. The plasmid responsible for this mul-
tiple resistance was named R222.

The Col plasmids contain loci that produce proteins
that are toxic, for various reasons, to strains of bacteria
not carrying the plasmids. Colicins attack sensitive bac-
terial cells at bacterial surface receptors. They have been
classified into twenty or more categories according to
the types of receptors they attack. Some colicins may en-
ter the cell directly, but others do not. For example, col-
icin K appears to kill sensitive cells by inhibiting DNA,
RNA, and protein synthesis, although not directly enter-
ing the cell. Colicin E3, however, acts as an intracellular
ribonuclease that cleaves off about fifty nucleotides from
the 3' end of the 16S ribosomal RNA within the ribo-
some. The cleavage inactivates the sensitive cell's ribo-
somes and is, of course, lethal.

Since many R plasmids, Col plasmids, and F factors, as
well as host chromosomes, have insertion sequences
(chapter 14), a good deal of exchange occurs among the
plasmids, and many are able to integrate into the host
chromosome. Although their mobility makes it easier to
map and study plasmids, it also poses a human health
problem. Resistance to various antibacterial agents is eas-
ily transferred among enterobacteria worldwide. This can
even occur outside of host organisms (people) where
pollution or sewage is found. In addition, resistance
found in relatively harmless enterobacteria, such as
E. coli, can easily pass to more pathogenic bacteria, such
as Shigella and Salmonella. Since we are selecting for re-
sistance every time we use antibacterial drugs, we should
not use these drugs indiscriminately. For some time,
health workers have been concerned about excessive
medical use of antibacterial drugs as well as about the
large quantities of antibiotics used in animal feed.

Uncovering Plasmids

How do we know when the phenotype is controlled by a
plasmid rather than by the chromosomal genes of a bac-
terium? Plasmids can be seen with an electron microscope
or by density-gradient centrifugation of the cell's DNA. But
several less direct lines of evidence also supply the answer.
To begin with, multiple aspects of the phenotype (e.g., re-
sistance to several antibacterial agents) change simulta-
neously, as with plasmid R222. Another clue is that the
phenotypic change is infectious: Japanese workers found
that with R222, resistant cells converted nonresistant cells.
As B. Lewin stated, "Resistance is infectious."

Several other clues point to the presence of a plas-
mid. In linkage studies, using transduction for example,

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Chapter Seventeen Non-Mendelian Inheritance

plasmid loci show no linkage to host loci; plasmids them-
selves can be mapped because their loci are linked to
each other. Since the plasmid DNA replicates at its own
speed, it can miss being incorporated into a daughter
cell. Thus, many spontaneous losses of the plasmid occur.
And finally, certain treatments — with acridine dyes, for
example — have little effect on the replication of the host
chromosome, but selectively prevent the plasmid from
replicating; thus the plasmid can be eliminated from the
cell population. The existence of plasmids in a bacterial
population can, therefore, be verified with morphologi-
cal, physiological, and analytical evidence.

IMPRINTING

Although sex linkage alters inheritance patterns, we do
not expect different inheritance patterns, dependent on
the parent of origin, from genes located on autosomal
chromosomes. That is, the genotype of an offspring
should be predicted by its alleles regardless of which par-
ent donated which allele. That understanding has now
been shown to be incorrect for a group of genes whose
phenotypic effects are determined by the parent that
donated a particular allele. This phenomenon is called
imprinting (or molecular imprinting or parental im-
printing). It falls under the general classification of an
epigenetic effect, a term that has come to mean an ef-
fect due to an environmentally induced change in the ge-
netic material but not causing a change in base pairs. It is
a phenomenon of differential expression of the alleles at
a locus depending on which parent the gene originated
with.

A striking example of imprinting in human beings in-
volves two medical syndromes, both resulting in mental
retardation. In Prader-Willi syndrome, affected persons
are extremely obese; in Angelman syndrome, those af-
fected are thin and sometimes referred to as "happy pup-
pets," because they exhibit a happy facial expression and
erratic, jerky movements. It turns out that both syn-
dromes are associated with deletions in the long arm of

chromosome 15, in bands 15qll-ql3.The effect is seen
in an individual arising from a gamete missing a
15qll-ql3 region. If the remaining region is of paternal
origin, due to a deletion of the maternal gene, the off-
spring will have Angelman syndrome; if the remaining re-
gion is of maternal origin, the offspring will have Prader-
Willi syndrome. This unusual situation indicates that the
phenotype is dependent on the parent from which the
region comes. Recently, this region of chromosome 15
has come under intense scrutiny. In males, from five to
seven genes are expressed from this area, and in females,
one gene has been identified as the cause of Angelman
syndrome, UBE3A (E3 ubiquitin protein ligase). It has
been hypothesized that there is an imprinting center
(IC), a region responsible for the control of imprinting.
The imprinting mark is almost certainly DNA methyla-
tion, which has the property of turning off gene tran-
scription. Stretches of CG repeats (called CpG islands,
in which CpG indicates sequential bases on the same
strand of DNA rather than a C-G base pair) have been
found in these imprinting centers. The imprinting center
would be the site of the erasure of past imprinting and
the initiation of new imprinting during gametogenesis.
Over twenty genes exhibit imprinting, and the epige-
netic phenomenon also appears in proteins, with differ-
ential acetylation of proteins as the imprinting mark.

The question arises as to how imprinting evolved;
that is, what evolutionary advantages come from silenc-
ing an allele from one of the parents? Although we don't
really know at this point, several hypotheses have been
suggested, including competition among maternal and
paternal alleles for expression (see chapter 21). For ex-
ample, the Igf-2 gene (insulin-like growth factor) places
demands on pregnant females to produce larger fetuses.
This is advantageous to the father (assuming that the fe-
male will have offspring from several fathers), but not the
mother. So, the mother's gene is usually methylated and
therefore inactive. It is as if the genes are in competition
with each other, with the father's gene promoting the for-
mation of a large fetus and the mother's gene promoting
the formation of a smaller fetus. Currently, the phenome-
non of imprinting is under active study.

SUMMARY

STUDY OBJECTIVE 1: To analyze the inheritance patterns
of maternal effects 509- 5 1

Patterns of non-Mendelian inheritance fall into two cate-
gories: maternal effects and cytoplasmic inheritance. Ma-
ternal effects are illustrated by snail-shell coiling. The direc-
tion of coiling is determined by the genotype of the

maternal parent, with dextral coiling dominant to sinistral
coiling.

STUDY OBJECTIVE 2: To analyze the patterns of cytoplas-
mic inheritance 511-524

Cytoplasmic inheritance is usually seen in organelles, sym-

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bionts, or parasites that have their own genetic material.
Chloroplasts and mitochondria have relatively small, circu-
lar chromosomes with prokaryotic affinities. An interaction
exists between organelles and nuclei; the organelles do not
encode all their own proteins and enzymes. Mitochondrial
defects can be inherited through nuclear genes or through
the mitochondrion itself. A similar pattern is seen in chloro-
plasts. The processes of cytoplasmic inheritance are exem-
plified by symbiotic bacteria in Paramecium.

Plasmids are autonomous segments of DNA. In prokary-
otes, R and Col plasmids, as well as the F factor, have been
well studied. Plasmids usually carry an operon for transfer
and insertion sequences for attachment to cell chromo-

somes and to each other. Hence, they represent highly mo-
bile segments of genetic material.

STUDY OBJECTIVE 3: To analyze the patterns of imprint-
ing 524

Imprinting is a phenomenon of gene activity affected by
the parent of origin. Due to a pattern of gene methylation
that differs in male and female parents, a gene may show
differential activity depending on the parent from which it
came. More than twenty genes exhibiting this epigenetic
phenomenon are known.

SOLVED PROBLEMS

PROBLEM 1: What possible phenotypes and genotypes
could the female parent of a sinistrally coiled snail have?

Answer: If a snail is sinistrally coiled, its mother must
have had the dd genotype, since sinistrality is recessive. If
the female parent is a recessive homo zygote, its mother
must have contributed a recessive d allele. Therefore its
mother (the grandmother) could have had either a Dd or
dd genotype. Its daughter could therefore be either dex-
trally or sinistrally coiled (respectively). Thus, to answer
the question, a sinistrally coiled snail could have had a
mother that was either dextrally or sinistrally coiled, but
only of the dd genotype.

PROBLEM 2: You have just noticed a petite yeast colony
growing in a petri plate under aerobic conditions. What
type of petite is it?

Answer: The simplest way to determine the nature of
the lesion resulting in the petite phenotype is to make a
cross of the petite strain with a wild-type strain. After

meiosis, isolate the four products (spores) and allow
them to grow separately under normal, aerobic condi-
tions. If the ratio of petite to wild-type is 1:1, the muta-
tion is of a nuclear gene. If progeny are wild-type, the mu-
tation is in the mitochondrial genome and is of the
neutral type. If progeny are mostly petites, the mutation
is also in the mitochondrial genome, but it is of the sup-
pressive type.

PROBLEM 3: Killer Paramecium with the genotype KK
are mated with kk cells under a situation that allows cy-
toplasmic exchange. If the exconjugants undergo autog-
amy, what types of progeny would you expect?

Answer: Both exconjugants will be Kk, and since cyto-
plasmic exchange occurred, both cytoplasms will con-
tain kappa. Autogamy will produce either KK or kk cells.
Since at least one K gene is needed for the maintenance
of kappa, the kk cells eventually lose the kappas and be-
come sensitive. Thus, we expect 1/2 sensitive: 1/2 killers.

EXERCISES AND PROBLEMS

DETERMINING NON-MENDELIAN INHERITANCE

1. J. Christian and C. Lemunyan have shown that mice
raised under crowded conditions produce two gen-
erations with reduced growth rates. What sort of
genetic control might exist, and how could this con-
trol be demonstrated?

2. Describe the types of evidence that could be gath-
ered to determine whether a trait in E. coli is con-
trolled by chromosomal or plasmid genes. (See also
CYTOPLASMIC INHERITANCE)

3. The maroon-like (ma-l) locus in Drosophila is inher-
ited in an X-linked recessive fashion. If you cross a
heterozygous female with a maroon-like male, all the
progeny are wild-type. If the female progeny from
this cross are mated again with maroon-like males,
half of the females produce all maroon-like progeny,
and the other half produce all wild-type progeny. Ex-
plain these results. (See also MATERNAL EFFECTS)

*Answers to selected exercises and problems are on page A-20.

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Chapter Seventeen Non-Mendelian Inheritance

MATERNAL EFFECTS

4. Snail coiling is called a maternal trait. Is it possible
that it is caused by an allele at a sex-linked locus?

5. How would you rule out a viral origin for snail-shell
coiling?

6. Give the genotypes involved when a sinistral female
snail produces dextral offspring. What genotypes
could the male parent of the sinistral female have?

7. A dextral snail is self-fertilized and produces only
sinistral progeny. What is the probable genotype of
this snail and its parents?

8. In corn, male sterility is controlled by a maternal
cytoplasmic element. A dominant nuclear gene,
Restorer (Rf), restores fertility to male sterile lines. If
pollen from a homozygous RfRf plant is used to pol-
linate a male sterile plant, what genotypes and phe-
notypes would you expect in the progeny?

CYTOPLASMIC INHERITANCE

9. What evidence indicates that it is not absolutely es-
sential, in an evolutionary sense, for mitochondria to
have genes for specific components of oxidative
phosphorylation?

10. How would you determine that a segregative petite
mutant in yeast is controlled by a chromosomal
gene?

11. What results would you obtain by making all possible
pairwise crosses of the three types of yeast petites?

12. An ornamental spider plant has green and white
striped leaves. How can you determine whether
cytoplasmic inheritance is responsible for the strip-
ing and whether there is interaction with an iojap-
type chromosomal gene?

13. In Chlamydomonas, 0.02% of the meiotic products
are of the mt~ parental type. How can you use this
information in mapping? (Use streptomycin sensitiv-
ity, str s , and resistance, str r , as an example.)

14. What similarities do mitochondria and plastids share?

15. What evidence is there that mitochondria and
chloroplasts originated from prokaryotes?

16. Individuals from killer and nonkiller strains of Para-
mecium are mixed together. Cytoplasmic exchange
occurs during conjugation. Approximately 25% of
the exconjugants are sensitive, and the remaining
75% are killers. What are the genotypes of the indi-
viduals of the two strains, and what ratios of sensi-
tives and killers would result if the various exconju-
gants underwent autogamy?

17. What genetic tests could you conduct to show that
the mate-killer phenotype in Paramecium requires a
dominant allele at any one of two loci?

18. Resistant and sensitive strains of Drosophila
melanogaster differ in their ability to tolerate
C0 2 — anesthetization with it kills sensitive flies.
What genetic experiments would you perform to de-
termine whether the trait is caused by a virus? How
would you rule out chromosomal genes?

19. Suppose you have identified a person who has in-
trons in his or her mitochondrial DNA. What would
you deduce about the origin of this DNA?

20. A mutation in the mitochondrial genome in people
causes blindness. If reciprocal matings between af-
fected and normal individuals occur in a family pedi-
gree, what types of children would you expect from
each cross?

21. When chloroplast DNA from Chlamydomonas is di-
gested with a particular restriction enzyme and then
hybridized with a particular probe, two bands are
detected. Some strains (type 1) yield bands of 1.5
and 3.7 kilobases; other strains (type 2) yield bands
of 2.5 and 6.0 kilobases. For the following crosses,
predict the progeny:

a. mt + , strain 1 X mt~ , strain 2

b. mt + , strain 2 X mt~ , strain 1

22. What type of asci do you expect if you cross a yeast
strain carrying an antibiotic resistance gene in its mi-
tochondria with a strain that has normal (sensitive)
mitochondria?

23. In Paramecium, the maintenance of kappa particles
requires the dominant nuclear gene K. A Kk killer
cell conjugates with a sensitive cell of the same
genotype without cytoplasmic exchange. Predict
the genotypes and phenotypes that result if each ex-
conjugant then undergoes autogamy.

24. In Neurospora, the slow-growing trait poky is inher-
ited maternally and is due to an abnormal respiratory
protein. A nuclear gene F makes poky individuals
grow faster, even though the protein is still defective.
Such strains are called fast-poky (F' is normal poky}.
Poky cytoplasm is not altered by F in a zygote, and F
has no effect on normal cytoplasm. What genotypes
and phenotypes do you expect if the maternal parent
is fast-poky and the paternal parent is normal?

25. In corn, two independent, recessive nuclear genes,
japonica (j) and iojap (ij), produce variegation
(green and white striped leaves). Matings between
individuals heterozygous fox japonica always pro-
duce 3 green: 1 striped individuals regardless of how
the cross is performed. The behavior of iojap was
described in figure 17.9. You have a variegated plant
that could be either jj or ijij. What cross can you
make to determine the genotype of this plant, and
what results do you expect in the F : generation in
each case?

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Critical Thinking Questions

527

26. If Paramecium cells heterozygous for both genes in-
volved in the maintenance of the mate-killer trait are
forced to undergo autogamy, what phenotypic ratios
do you expect?

27. A petite yeast strain is crossed with a wild-type
strain. What phenotypic ratio do you expect after
meiosis if the petite is

a. nuclear?

b. suppressive?

c. neutral?

CRITICAL THINKING QUESTIONS

1. When a eukaryotic cell divides, cell organelles such as
mitochondria and chloroplasts are distributed to the
daughter cells. What mechanisms might exist to ensure
an even distribution of these organelles?

2. Lamarckian inheritance, the inheritance of acquired
characteristics, is generally discounted as a major evo-
lutionary mechanism (chapter 21). (For example,

Lamarck suggested that the long neck of the giraffe
came about by as giraffes stretched for food, followed
by the inheritance of this longer, stretched neck.) Is the
progression of the lysogenic state of E. coli from one
generation to the next an example of Lamarckian in-
heritance? Why or why not?

Suggested Readings for chapter 1 7 are on page B-18.

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IV. Quantitative and
Evolutionary Genetics

18. Quantitative
Inheritance

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QUANTITATIVE
INHERITANCE

STUDY OBJECTIVES

1. To understand the patterns of inheritance of phenotypic traits
controlled by many loci 531

2. To investigate the way that geneticists and statisticians
describe and analyze normal distributions of phenotypes 535

3. To define and measure heritability, the unit of inheritance of
variation in traits controlled by many loci 542

STUDY OUTLINE

Traits Controlled by Many Loci 531

Two-Locus Control 531

Three-Locus Control 532

Multilocus Control 532

Location of Polygenes 533

Significance of Polygenic Inheritance 534
Population Statistics 535

Mean, Variance, and Standard Deviation 536

Covariance, Correlation, and Regression 539
Polygenic Inheritance in Beans 541
Selection Experiments 541
Heritability 542

Realized Heritability 542

Partitioning of the Variance 543

Measurement of Heritability 544
Quantitative Inheritance in Human Beings 545

Skin Color 545

IQ and Other Traits 546
Summary 547
Solved Problems 548
Exercises and Problems 549
Critical Thinking Questions 551
Box 18.1 Mapping Quantitative Trait Loci 537
Box 18.2 Human Behavioral Genetics 547

Much human variation is quantitative.

(©Jim Cummins/FPG International.)

530

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IV. Quantitative and
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18. Quantitative
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Traits Controlled by Many Loci

531

When we talked previously of genetic
traits, we were usually discussing traits
in which variation is controlled by single
genes whose inheritance patterns led to
simple ratios. However, many traits, in-
cluding some of economic importance — such as yields of
milk, corn, and beef — exhibit what is called continuous
variation.

Although some variation occurred in height in
Mendel's pea plants, all of them could be scored as either
tall or dwarf; there was no overlap. Using the same meth-
ods that Mendel used, we can look at ear length in corn
(fig. 18.1). With Mendel's peas, all of the F : were tall. In a
cross between corn plants with long and short ears, all of
the V 1 plants have ears intermediate in length between
the parents. When both pea and corn F : plants are self-
fertilized, the results are again different. In the F 2 genera-
tion, Mendel obtained exactly the same height categories
(tall and dwarf) as in the parental generation. Only the ra-
tio was different — 3:1.

Peas

Corn

plants

Dwarf X Tall

03
CD

Short X Long

CD
_Q

AA,

i_
CD

AA,

Height

Ear length

Number of plants

Tall X Self

C/)

i_
03
CD

H —

i_
CD
_Q

Intermediate X Self

Height

Ear length

-n

Number of plants

Tall 3/4
Dwarf 1/4 / 1

j\ A.

03
CD

H —

i_
CD
_Q

Whole range

Height

Ear length

In corn, however, ears of every length, from the short-
est to the longest, are found in the F 2 ; there are no dis-
crete categories. A genetically controlled trait exhibiting
this type of variation is usually controlled by many loci.
In this chapter, we study this type of variation by looking
at traits controlled by progressively more loci. We then
turn to the concept of heritability which is used as a sta-
tistical tool to evaluate the genetic control of traits deter-
mined by many loci.

TRAITS CONTROLLED
BY MANY LOCI

Let us begin by considering grain color in wheat. When a
particular strain of wheat having red grain is crossed
with another strain having white grain, all the F : plants
have kernels intermediate in color. When these plants are
self-fertilized, the ratio of kernels in the F 2 is 1 red: 2 in-
termediate:! white (fig. 18.2). This is inheritance involv-
ing one locus with two alleles. The white allele, a, pro-
duces no pigment (which results in the background
color, white); the red allele, A, produces red pigment. The
F : heterozygote, Aa, is intermediate (incomplete domi-
nance). When this monohybrid is self-fertilized, the typi-
cal 1:2:1 ratio results. (For simplicity, we use dominant-
recessive allele designations, A and a. Keep in mind,
however, that the heterozygote is intermediate in color.)

Two-Locus Control

Now let us examine the same kind of cross using two
other stocks of wheat with red and white kernels. Here,
when the resulting intermediate (medium-red) F : are
self-fertilized, five color classes of kernels emerge in a ra-
tio of 1 dark red: 4 medium dark red: 6 medium red: 4 light
red:l white (fig. 18.3). The offspring ratio, in sixteenths,
comes from the self-fertilization of a dihybrid in which
the two loci are unlinked. In this case, both loci affect the
same trait in the same way. In figure 18.3, each capital

Red

X White

Intermediate color

Red

: Intermediate : White

Aa aa

Figure 18.1 Comparison of continuous variation (ear length in
corn) with discontinuous variation (height in peas).

1:2:1

Figure 18.2 Cross involving the grain color of wheat in which
one locus is segregating.

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Chapter Eighteen Quantitative Inheritance

Red X

White

AABB

aabb

Medium red

AaBb

Dark red

: Medium dark red : Medium rec

: Light red

: White

AABB

AaBB AaBb

Aabb

aabb

AABb AAbb

aaBb

aaBB

1 : 4 : 6 : 4 : 1
Figure 18.3 Another cross involving wheat grain color in which two loci are segregating.

letter represents an allele that produces one unit of color,
and each lowercase letter represents an allele that pro-
duces no color. Thus, the genotype AaBb has two units
of color, as do the genotypes AAbb and aaBB. All pro-
duce the same intermediate grain color. Recall from
chapter 2 that a cross such as this produces nine geno-
types in a ratio of 1:2:1:2:4:2:1:2:1. If these classes are
grouped according to numbers of color-producing al-
leles, as shown in figure 18.3, the 1:4:6:4:1 ratio appears.
This ratio is a product of a binomial expansion.

Three-Locus Control

In yet another cross of this nature, H. Nilsson-Ehle in
1909 crossed two wheat strains, one with red and the
other with white grain, that yielded plants in the ¥ 1 gen-
eration with grain of intermediate color. When these
plants were self-fertilized, at least seven color classes,
from red to white, were distinguishable in a ratio of
1:6:15:20:15:6:1 (fig. 18.4). This result is explained by as-
suming that three loci are assorting independently, each
with two alleles, so that one allele produces a unit of red
color and the other allele does not. We then see seven
color classes, from red to white, in the 1:6:15:20:15:6:1
ratio. This ratio is in sixty-fourths, directly from the 8X8
(trihybrid) Punnett square, and comes from grouping
genotypes in accordance with the number of color-
producing alleles they contain. Again, the ratio is one that
is generated in a binomial distribution.

Multilocus Control

From here, we need not go on to an example with four
loci, then five, and so on. We have enough information to
draw generalities. It should not be hard to see how dis-
crete loci can generate a continuous distribution (fig.
18.5). Theoretically, it should be possible to distinguish

Red X

AA BB CC

White
aa bb cc

Intermediate color
Aa Bb Cc

Self

ABC
A B c

A B C A B c

A b C

a B C A b c

a Be

a b C

a b c

e 5

A b C 5

a B C

Abe

a B c

a b C

a b c

Phenotype

Red

White

Number of color-producing alleles 6:5:4:3:2:1 :0

Ratio

1 : 6 : 15 :20:15 :6 : 1

Figure 18.4 One of Nilsson-Ehle's crosses involving three loci
controlling wheat grain color. Within the Punnett square, only
the number of color-producing alleles is shown in each box to
emphasize color production.

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18. Quantitative
Inheritance

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Traits Controlled by Many Loci

533

Figure 18.5 The change in shape of the distribution as
increasing numbers of independent loci control grain color in
wheat. If each locus is segregating two alleles, with each allele
affecting the same trait, eventually a continuous distribution will
be generated in the F 2 generation.

different color classes down to the level of the eye's abil-
ity to perceive differences in wavelengths of light. In fact,
we rapidly lose the ability to assign unique color classes
to genotypes because the variation within each genotype
soon causes the phenotypes to overlap. For example,
with three loci, a color somewhat lighter than medium
dark red may belong to the medium-dark-red class with

three color alleles, or it may belong to the medium-red
class with only two color alleles (fig. 18.5).

The variation within each genotype is due to the en-
vironment — that is, two organisms with the same geno-
type may not necessarily be identical in color because
nutrition, physiological state, and many other variables
influence the phenotype. Figure 18.6 shows that it is pos-
sible for the environment to obscure genotypes even in a
one-locus, two-allele system. That is, a height of 17 cm
could result in the F 2 from either the aa or Aa genotype
in the figure when there is excessive variation (fig. 18.6,
column 3). In the other two cases in figure 18.6, there
would be virtually no organisms 17 cm tall. Systems such
as those we are considering, in which each allele con-
tributes a small unit to the phenotype, are easily influ-
enced by the environment, with the result that the distri-
bution of phenotypes approaches the bell-shaped curve
seen at the bottom of figure 18.5.

Thus, phenotypes determined by multiple loci with
alleles that contribute dosages to the phenotype will ap-
proach a continuous distribution. This type of trait is said
to exhibit continuous, quantitative, or metrical vari-
ation. The inheritance pattern is polygenic or quanti-
tative. The system is termed an additive model because
each allele adds a certain amount to the phenotype.

From the three wheat examples just discussed, we can
generalize to systems with more than three polygenic
loci, each segregating two alleles. From table 18.1, we
can predict the distribution of genotypes and pheno-
types expected from an additive model with any number
of unlinked loci segregating two alleles each. This table is
useful when we seek to estimate how many loci are pro-
ducing a quantitative trait, assuming it is possible to dis-
tinguish the various phenotypic classes. For example,
when a strain of heavy mice was crossed with a lighter
strain, the F : were of intermediate weight. When these F :
were interbred, a continuous distribution of adult
weights appeared in the F 2 generation. Since only about
one mouse in 250 was as heavy as the heavy parent
stock, we could guess that if an additive model holds,
then four loci are segregating. This is because we expect
l/(4) n to be as extreme as either parent; one in 250 is
roughly l/(4) 4 = 1/256.

Location of Polygenes

The fact that traits with continuous variation can be
controlled by genes dispersed over the whole genome
was shown by James Crow, who studied DDT resistance
in Drosophila. A DDT-resistant strain of flies was cre-
ated by growing them on increasing concentrations of
the insecticide. Crow then systematically tested each
chromosome for the amount of resistance it conferred.
Susceptible flies were mated with resistant flies, and the
sons from this cross were backcrossed. Offspring were

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Chapter Eighteen Quantitative Inheritance

CD
=3
D~
CD

No environmental
variation

10 15 20 25 30
Height (cm)

aa Aa AA

I I

10 15 20 25 30
Height (cm)

Some environmental
variation

10 15 20 25 30

Height (cm)

10 15 20 25 30
Height (cm)

aa Aa AA

10 15 20 25 30
Height (cm)

Much environmental
variation

10 15 20 25 30
Height (cm)

aa Aa AA

10 15 20 25 30
Height (cm)

Figure 18.6 Influence of environment on phenotypic distributions.

James F. Crow (1916- ).
(Courtesy of Dr. James F. Crow.)

then scored for the particular resistant chromosomes
they contained (each chromosome had a visible
marker) and were tested for their resistance to DDT.
Sons were used in the backcross because there is no
crossing over in males. Therefore, the sons would pass
resistant and susceptible chromosomes on intact.
Crow's results are shown in figure 18.7. As you can see,
each chromosome has the potential to increase the fly's
resistance to DDT. Thus, each chromosome contains

loci (polygenes) that contribute to the phenotype of
this additive trait (box 18.1).

Significance of Polygenic Inheritance

The concept of additive traits is of great importance to
genetic theory because it demonstrates that Mendelian
rules of inheritance can explain traits that have a contin-
uous distribution — that is, Mendel's rules for discrete
characteristics also hold for quantitative traits. Additive
traits are also of practical interest. Many agricultural
products, both plant and animal, exhibit polygenic inher-
itance, including milk production and fruit and vegetable
yield. In addition, many human traits, such as height and
IQ, appear to be polygenic, although with substantial en-
vironmental components.

Historically, the study of quantitative traits began be-
fore the rediscovery of Mendel's work at the turn of the
century. In fact, biologists in the early part of this century
debated as to whether the "Mendelians" were correct or
whether the "biometricians" were correct in regard to

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Population Statistics

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Table 18.1 Generalities from

an Additive Model of Polygenic

Inheritance

One Locus

Two Loci

Three Loci

n Loci

Number of gamete

2 n

types produced by an

(A, a)

CAB, Ab, aB, ab)

(ABC, ABc, AbC, Abe,

F x multihybrid

aBC, aBc, abC, abc)

Number of different F 2

3 n

genotypes

(AA, Aa, aa)

(AABB, AABb,

AAbb, AaBB,

AaBb, Aabb,

aaBB, aaBb,

aabb)

(AABBCC, AABBCc, AABBcc,
AABbCC, AABbCc, AABbcc,
AAbbCC, AAbbCc, AAbbcc,
AaBBCC, AaBBCc, AaBBcc,
AaBbCC, AaBbCc, AaBbcc,
AabbCC, AabbCc, Aabbcc,
aaBBCC, aaBBCc, aaBBcc,
aaBbCC, aaBbCc, aaBbcc,

aabbCC, aabbCc, aabbcc)

Number of different F 2

2n+ 1

phenotypes

Number of F 2 as extreme as

1/4

1/16

1/64

l/4 n

one parent or the other

(AA or aa)

(AABB or aabb)

(AABBCC or aabbcc)

Distribution pattern of

1:2:1

1:4:6:4:1

1:6:15:20:15:6:1

(A + a) 2n

F 2 phenotypes

Chromosomes
X 2 3

Survival (%)
1 5 10 20 40 60

i Chromosome from nonresistant nonselected strain
Chromosome from resistant strain

Figure 18.7 Survival of Drosophila in the presence of DDT.
Numbers and arrangements of DDT-resistant and susceptible
chromosomes vary. (Reproduced with permission from the Annual
Review of Entomology, Volume 2, © 1957 by Annual Reviews, Inc.)

the rules of inheritance. Biometricians used statistical
techniques to study traits characterized by continuous
variation and claimed that single discrete genes were not
responsible for the observed inheritance patterns. They
were interested in evolutionarily important facets of the
phenotype — traits that can change slowly over time.
Mendelians claimed that the phenotype was controlled
by discrete "genes." Eventually the Mendelians were
proven correct, but the biometricians' tools were the
only ones suitable for studying quantitative traits.

The biometric school was founded by F. Galton and K.
Pearson, who showed that many quantitative traits, such as
height, were inherited. They invented the statistical tools
of correlation and regression analysis in order to study the
inheritance of traits that fall into smooth distributions.

POPULATION STATISTICS

A distribution (see fig. 18.5, bottom) can be described in
several ways. One is the formula for the shape of the curve
formed by the frequencies within the distribution. A more
functional description of a distribution starts by defining its
center, or mean (fig. 18.8). As we can see from the figure,
the mean is not itself enough to describe the distribution.
Variation about this mean determines the actual shape of
the curve. (We confine our discussion to symmetrical, bell-
shaped curves called normal distributions. Many distri-
butions approach a normal distribution.)

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Chapter Eighteen Quantitative Inheritance

Normal distribution
with "more" variation

Mean

Normal distribution
with "less" variation

Mean

Figure 18.8 Two normal distributions (bell-shaped curves) with
the same mean.

Mean, Variance, and Standard Deviation

The mean of a set of numbers is the arithmetic average of
the numbers and is defined as

x = ^jx/n

(18.1)

in which

x = the mean

Xx = the summation of all values

n = the number of values summed

In table 18.2, the mean is calculated for the distribution
shown in figure 18.9. The variation about the mean is cal-
culated as the average squared deviation from the mean:

s 2 = V = —

n — 1

Table 1 8.2 Hypothetical Data Set of Ear Lengths

(x) Obtained When Corn Is Grown
from an Ear of Length 11 cm

Cx — X )

( x — x ) 2

-4.12

16.97

-3.12

9.73

-2.12

4.49

-2.12

4.49

-1.12

1.25

-1.12

1.25

-1.12

1.25

-1.12

1.25

-1.12

1.25

-1.12

1.25

-0.12

0.01

-0.12

0.01

-0.12

0.01

-0.12

0.01

-0.12

0.01

-0.12

0.01

0.88

0.77

0.88

0.77

0.88

0.77

1.88

3.53

1.88

3.53

1.88

3.53

2.88

8.29

2.88

8.29

Xx = 278

4.88

23.81

X(x - x) 2 = 9635

n = 25

x —

278
25

- 11.12

s 2 = V =

X(x -

n —

« 2 = *>.53 .

1 24

s = Vs 2

= V4

.02 = 2.0

(18.2)

This value (For s 2 ) is called the variance. Observe that the
flatter the distribution is, the greater the variance will be.

The variance is one of the simplest measures we can
calculate of variation about the mean. You might wonder
why we simply don't calculate an average deviation from
the mean rather than an average squared deviation. For
example, we could calculate a measure of variation as

X(x — x)
n — 1

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18. Quantitative
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Population Statistics

537

Mapping the location of a
standard locus is conceptu-
ally relatively easy, as we
saw in the mapping of the fruit fly
genome. We look for associations of
phenotypes that don't segregate with
simple Mendelian ratios and then
map the distance between loci by the
proportion of recombinant offspring.
However, with quantitative loci we
have a problem: We can't do simple
mapping because genes contributing
to the phenotype are often located
across the genome. Thus, a particular
continuous phenotype will be con-
trolled by loci linked to numerous
other loci, many unlinked to each
other. However, with the advent of
molecular techniques, it has become
feasible to map polygenes.

High line

RFLP1--
QTL1"

-RFLP1
--QTL1

RFLP1
QTL1

BOX 18.1

Experimental
Methods

Mapping Quantitative
Trait Loci

In chapter 13, we showed how a
locus can be discovered and mapped
in the human genome (and other
genomes) by association with molec-
ular markers. That is, as the Human
Genome Project has progressed, we
have discovered restriction fragment
length polymorphisms (RFLPs) that
mark every region of all the chromo-
somes. Conceptually, there is not

Low line

RFLP2
QTL2

-RFLP2
-QTL2

I-RFLP2
QTL2

RFLP1
QTL1 +

RFLP1 RFLP1
QTL1 QTL1+

RFLP2 RFLP2
QTL2 QTL2f

RFLP2
1-QTL2

Figure 1 Mapping a quantitative trait locus (QTL) to a particu-
lar chromosomal region using a restriction fragment length
polymorphism (RFLP) marker. A hypothetical chromosome pair
in the fruit fly is shown. The flies have been selected for a geo-
tactic score; QTL1 is the locus in the high line, and QTL2 is
the locus in the low line. RFLP1 is homozygous in the high line
and RFLP2 is homozygous in the low line.

much difference between finding the
gene for cystic fibrosis and finding
the gene that contributes to a quanti-
tative trait.

In theory, we look at a population
of organisms and note various RFLPs
or other molecular markers. We then
look for the association of a marker
and a quantitative trait. If an associa-
tion exists, we can gain confidence
that one or more of the polygenes
controlling the trait is located in
the chromosomal region near the
marker. The closer the polygenes are
to the markers, the more reliable our
estimates are, because they depend
on few crossovers taking place in
that population. With many cross-
overs, the association between a
particular marker and a particular ef-
fect diminishes. Since we don't know
immediately from this method
whether the region of interest has
one or more polygenic loci, a new
term has been coined to indicate that
ambiguity. Instead of talking about
polygenic loci directly, we talk of
quantitative trait loci.

For example, consider the search
for polygenes associated with geotac-
tic behavior in fruit flies (see fig.
18.13). As selection proceeds, flies in
the high and low lines diverge in
their geotactic scores. The lines are
also becoming homozygous for many
loci since only a few parents are cho-
sen to begin each new generation
(see chapter 19). Thus, quantitative
trait loci can become associated with
different molecular markers in each
line (fig. 1). If flies from each line are
crossed, heterozygotes will be pro-
duced of both the markers and the
quantitative trait loci. If there is very
little crossing over between the two,
three classes of F 2 offspring will be
produced. These offspring can be
grouped according to their RFLPs
and then tested for their geotactic
scores. If, as figure 1 suggests, a rela-
tionship exists between a locus influ-
encing geotactic score and an RFLP,

continued

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18. Quantitative
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Chapter Eighteen Quantitative Inheritance

then the three groups will have dif-
ferent geotactic scores. We can then
conclude that the region of the chro-
mosome that contains the RFLP also
contains a quantitative trait locus.
Finding the right RFLP is, of course, a
tedious and time-consuming task.

In a recent summary of the litera-
ture, Steven Tanksley reported that
numerous quantitative trait loci have
been mapped in tomatoes, corn, and
other organisms. For example, five
quantitative trait loci have been
mapped in tomatoes for fruit growth,
and eleven quantitative trait loci have
been mapped in corn for plant
height. Enough data seem to be pres-
ent to recognize an interesting gener-

BOX 18.1 CONTINUED

ality. That is, our definition of an addi-
tive model may need to be rethought
because it appears that in almost
every case studied so far, one or more
of the quantitative trait loci account
for a major portion of the phenotype,
whereas most of the loci had very
small effects. Thus, the additive
model that assumes that all polygenes
contribute equally to the phenotype
may be wrong. However, additive
models that allow different loci to
contribute different degrees to the
phenotype are still supported.

Also of value from locating quanti-
tative trait loci is a new ability to esti-
mate the number of loci affecting a
quantitative trait. In this chapter, we

use an estimate of extreme F 2 off-
spring to estimate the number of
polygenes. There are other methods,
including sophisticated statistical
methods, that we will not develop
here. Mapping quantitative trait loci
gives us a third method, that is, sim-
ply counting the number of quantita-
tive trait loci mapped.

As the methods of mapping quan-
titative trait loci have been devel-
oped, they have also been refined.
High-resolution techniques under de-
velopment will help us determine
whether quantitative trait loci are, in
fact, individual polygenes or clusters
of polygenes.

(We will get to why we use n — 1 rather than n in the
denominator in a moment.) Note, however, that the
above measure is zero. By the definition of the mean,
the absolute value of the sum of deviations above it is
equal to the absolute value of the sum of deviations be-
low it — one is negative and the other is positive. How-
ever, by squaring each deviation, as in equation 18.2,
we create a relatively simple index — the variance —
which is not zero and has useful properties related to
the normal distribution.

Figure 18.9 Normal distribution of ear lengths in corn. Data
are given in table 18.2.

The ear lengths measured in table 18.2 are a sample
of all ear lengths in the theoretically infinite population
of ears in that variety of corn. Statisticians call sample val-
ues statistics (and use letters from the Roman alphabet
to represent them), whereas they call population values
parameters (and use Greek letters for them). The sam-
ple value is an estimate of the true value for the popula-
tion. Thus, in the variance formula (equation 18.2), the
sample value, For s 2 , is an estimate of the population
variance, a 2 . When sample values are used to estimate pa-
rameters, one degree of freedom is lost for each parame-
ter estimated. To determine the sample variance, we di-
vide not by the sample size, but by the degrees of
freedom (n — 1 in this case, as defined in chapter 4). The
variance for the entire population (assuming we know
the population mean, |x, and all the data values) would be
calculated by dividing by n. The sample variance is cal-
culated in table 18.2.

The variance has several interesting properties, not
the least of which is the fact that it is additive. That is, if
we can determine how much a given variable con-
tributes to the total variance, we can subtract that
amount of variance from the total, and the remainder is
caused by whatever other variables (and their interac-
tions) affect the trait. This property makes the variance
extremely important in quantitative genetic theory.

The standard deviation is also a measure of varia-
tion of a distribution. It is the square root of the variance:

5 = Vv (18.3)

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Population Statistics

539

^S i i

r i i

i\
i \

i \
i \
i \.

i i >w^
i i l

i
i

i
i
i

67%

96%

99%

Figure 18.10 Area under the bell-shaped curve. The abscissa
is in units of standard deviation (s) around the mean (x).

In a normal distribution, approximately 67% of the area
of the curve lies within one standard deviation on either
side of the mean, 96% lies within two standard devia-
tions, and 99% lies within three standard deviations (fig.
18.10). Thus, for the data in table 18.2, about two-thirds
of the population would have ear lengths between 9.12
and 13.12 cm (mean ± standard deviation).

One final measure of variation about the mean is the
standard error of the mean (SE):

SE = s/Vw

The standard error (of the mean) is the standard devia-
tion about the mean of a distribution of sample means. In
other words, if we repeated the experiment many times,
each time we would generate a mean value. We could
then use these mean values as our data points. We would
expect the variation among a population of means to be
less than among individual values, and it is. Data are often
summarized as "the mean ± SE." In our example of table
18.2, SE = 2.0/ V25 = 2.0/5.0 = 0.4. We can summarize
the data set of table 18.2 as 11.1 ± 0.4 (mean ± SE).

Covariance, Correlation, and Regression

It is often desirable in genetic studies to know whether a
relationship exists between two given characteristics in a
series of individuals. For example, is there a relationship
between height of a plant and its weight, or between
scholastic aptitude and grades, or between a phenotypic
measure in parents and their offspring? If one increases,
does the other also? An example appears in table 18.3; the
same data set is graphed in figure 18.11, in what is re-
ferred to as a scatter plot. A relation does appear between

CT>

Q.
O

1.1

■

'■ "

2 3

Midparent

Figure 18.11 The relationship between two variables,
parental and offspring wing length in fruit flies, measured in
millimeters. Midparent refers to the average wing length of the
two parents. The line is the statistical regression line. (Source:
Data from D. S. Falconer, Introduction to Quantitative Genetics, 2d ed.
[London: Longman, 1981].)

the two variables. With increasing wing length in midpar-
ent (the average of the two parents: x-axis), there is an in-
crease in offspring wing length (y-axis). We can deter-
mine how closely the two variables are related by
calculating a correlation coefficient — an index that
goes from — 1.0 to +1.0, depending on the degree of re-
lationship between the variables. If there is no relation (if
the variables are independent), then the correlation coef-
ficient will be zero. If there is perfect correlation, where
an increase in one variable is associated with a propor-
tional increase in the other, the coefficient will be + 1 .0. If
an increase in one is associated with a proportional de-
crease in the other, the coefficient will be —1.0 (fig.
18. 12). The formula for the correlation coefficient (r) is

r =

covariance of x and y

(18.4)

where s x and s y are the standard deviations of x and y, re-
spectively.

To calculate the correlation coefficient, we need to
define and calculate the covariance of the two vari-
ables, cov(x, y). The covariance is analogous to the vari-
ance, but it involves the simultaneous deviations from
the means of both the x and y variables:

cov(x, y)

JXx - xXy ~ jO
n — 1

(18.5)

The analogy between variance and covariance can be
seen by comparing equations 18.5 and 18.2. The vari-
ances, standard deviations, and covariance are calculated

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Chapter Eighteen Quantitative Inheritance

in table 18.3, in which the correlation coefficient, r, is
0.78. (There are computational formulas available that
substantially cut down on the difficulty of calculating
these statistics. If a computer or calculator is used, only
the individual data points need to be entered — most
computers and many calculators can be programmed to
do all the computations.)

Many experiments deal with a situation in which we
assume that one variable is dependent on the other (in a
cause-and-effect relationship). For example, we may ask,
what is the relationship of DDT resistance in Drosophila
to an increased number of DDT-resistant alleles? With
more of these alleles (see fig. 18.7), the DDT resistance of
the flies should increase. Number of DDT-resistant alleles
is the independent variable, and resistance of the flies is
the dependent variable. That is, a fly's resistance is de-
pendent on the number of DDT-resistant alleles it has,

Table 1 8.3 The Relationship Between Two

Variables, x and y (x = the midparent —
average of the two parents — in wing
length in fruit flies in millimeters;
y = the offspring measurement)

1.5

2.2

2.3

2.4

2.7

2.9

2.7

1.7

2.3

2.2

2.4

2.7

2.9

2.7

1.9

2.2

2.3

2.6

2.7

2.9

2.4

2.6

2.7

2.8

2.2

2.4

2.3

2.6

2.8

2.2

2.4

2.6

2.9

2.1

1.9

2.4

2.6

2.8

2.7

3.1

2.1

2.2

2.4

2.6

2.8

2.7

3.2

2.4

2.1

2.5

2.4

2.6

2.9

2.5

3.2

2.8
2.9

Xx :

= 92.7

n = 37

ly = 93.2

x —

iZx

— = 2.51
n

y =

2.52

s 2 =

lZ(x -
n —

X?
1

= 0.19

2 . . K y -

n —

yf _
i

0.10

v ° X

0.44

Sy " V Sy

= 0.32

cov(x, y)

lZ(x

- x)(y
n — 1

^^ = 0.11

cov(x, y)

^x^y

0.11

— n 7£

(0.44)(0.32)

not the other way around. Going back to figure 18.11, we
could make the assumption that offspring wing length is
dependent on parental wing length. If this were so, a
technique called regression analysis could be used. This
analysis allows us to predict an offspring's wing length
(y variable) given a particular midparental wing length
(x variable). (It is important to note that regression analy-
sis assumes a cause-and-effect relationship, whereas cor-
relation analysis does not.)

The formula for the straight-line relationship (regres-
sion line) between the two variables is y = a + bx,
where b is the slope of the line (change in y divided by
change in x, or Aj/Ax) and a is the j-intercept of the line
(see fig. 18.11). To define any line, we need only to cal-
culate the slope, b, and thej intercept, a:

b = cov(x, y)/s x

y — bx

(18.6)
(18.7)

Thus equipped, if a cause-and-effect relationship does ex-
ist between the two variables, we can predict a y value
given any x value. We can either use the formula y = a +
bx or graph the regression line and directly determine
the y value for any x value. We now continue our exami-
nation of the genetics of quantitative traits.

r = 1

r = 0.5

Perfect (linear) correlation

Intermediate correlation

- •

•

- •

•

- •

•

- •

• • •
. *

•

• • •

• •

•

• • •

- •

•

- •
•

•
-•

• •

• • •

• •

r = -1

r =
No correlation

Perfect (linear)
inverse correlation

. . #

- •

• . . #

: \ ••• * y
• • •
• • • •
-• • • .

- •

_•

•
•

••

•
•

•

•
•
•
•
•

••

•
•

• •

••

Source: Data from D. S. Falconer, Introduction to Quantitative Genetics,
2d ed. (London: Longman, 1981).

Note: Data are graphed in figure 18.11.

Figure 18.12 Plots showing varying degrees of correlation
within data sets.

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Selection Experiments

541

Table 18.4

Johannsen's

Findings

of Relationship

Between Bean

Weights of Parents and Their Progeny

Weight
Parent
Beans

Weight

of Progeny

Beans (centigrams)

Mean ± SE

65-75

104

105

494

58.47 ± 0.43

55-65

103

127

102

609

54.37 ± 0.41

45-55

101

204

287

234

120

1,138

51.45 ± 0.27

35-45

139

278

498

584

372

213

2,238

48.62 ± 0.18

25-35

133

189

195

115

835

46.83 ± 0.30

15-25

180

46.53 ± 0.52

Totals

107

263

608

1,068 1,278

977

622

306

135

5,491

50.39 ±0.13

POLYGENIC INHERITANCE
IN BEANS

In 1909, W. Johannsen, who studied seed weight in the
dwarf bean plant (Phaseolus vulgaris), demonstrated
that polygenic traits are controlled by many genes. The
parent population was made up of seeds (beans) with a
continuous distribution of weights. Johannsen divided
this parental group into classes according to weight,
planted them, self-fertilized the plants that grew, and
weighed the ¥ 1 beans. He found that the parents with the
heaviest beans produced the progeny with the heaviest
beans, and the parents with the lightest beans produced
the progeny with the lightest beans (table 18.4). There
was a significant correlation coefficient between parent
and progeny bean weight (r = 0.34 ± 0.01). He contin-
ued this work by beginning nineteen lines (populations)
with beans from various points on the original distribu-
tion and selling each successive generation for the next
several years. After a few generations, the means and
variances stabilized within each line. That is, when
Johannsen chose, within each line, parent plants with
heavier-than-average or lighter-than-average seeds, the
offspring had the parental mean with the parental vari-
ance for seed size. For example, in one line, plants with
both the lightest average bean weights (24 centigrams)
and plants with the heaviest average bean weights
(47 eg) produced offspring with average bean weights of
37 eg. By selling the plants each generation, Johannsen
had made them more and more homozygous, thus low-
ering the number of segregating polygenes. Therefore,
the lines became homozygous for certain of the poly-
genes (different in each line), and any variation in bean
weight was then caused only by the environment. Jo-
hannsen thus showed that quantitative traits were under
the control of many segregating loci.

SELECTION EXPERIMENTS

Selection experiments are done for several reasons. Plant
and animal breeders select the most desirable individuals
as parents in order to improve their stock. Population ge-
neticists select specific characteristics for study in order
to understand the nature of quantitative genetic control.

For example, Drosophila were tested in a fifteen-
choice maze for geotactic response (fig. 18. 13). The maze
was on its side, so at every intersection, a fly had to make
a choice between going up or going down. The flies with
the highest scores were chosen as parents for the "high"
line (positive geotaxis; favored downward direction), and
the flies with the lowest score were chosen as parents for
the "low" line (negative geotaxis; favored upward direc-
tion). The same selection was made for each generation.
As time progressed, the two lines diverged quite signifi-
cantly. This tells us that there is a large genetic compo-
nent to the response; the experimenters are successfully
amassing more of the "downward" alleles in the high line
and more of the "upward" alleles in the low line. Several
other points emerge from this graph. First, the high and
low responses are slightly different, or asymmetrical. The
high line responded more quickly, leveled out more
quickly, and tended toward the original state more slowly
after selection was relaxed. (The relaxation of selection
occurred when the parents were a random sample of the
adults rather than the extremes for geotactic scores.) The
low line responded more slowly and erratically. In addi-
tion, the low line returned toward the original state more
quickly when selection was relaxed.

The nature of these responses (fig. 18.13) indicates that
the high line became more homozygous than the low line.
This is shown by the former's response when selection
is relaxed: It has exhausted a good deal of its variability
for the polygenes responsible for geotaxis. The low line,

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Chapter Eighteen Quantitative Inheritance

Figure 18.13 Selection for geotaxis. The dotted lines represent relaxed selection. (Source: Data from T.
Dobzhansky and B. Spassky, "Artificial and natural selection for two behavioral traits in Drosophila pseudoobscura,"
Proceedings of the National Academy of Sciences, USA, 62:75-80, 1969.)

however, seems to have much of its original genetic vari-
ability, because the relaxation of selection caused the mean
score of this line to increase rapidly. It still had enough ge-
netic variability to head back to the original population
mean. The response to a selection experiment is one way
that plant and animal breeders can predict future response.

HERITABILITY

Plant and animal breeders want to improve the yields of
their crops to the greatest degree they can. They must
choose the parents of the next generation on the basis of
this generation's yields; thus, they are continually per-
forming selection experiments. Breeders run into two
economic problems. They cannot pick only the very best
to be the next generation's parents because (1) they can-
not afford to decrease the size of a crop by using only a
very few select parents and (2) they must avoid in-
breeding depression, which occurs when plants are
self-fertilized or animals are bred with close relatives for
many generations. After frequent inbreeding, too much
homozygosity occurs, and many genes that are slightly or
partially deleterious begin to show themselves, depress-
ing vigor and yield. (Chapter 19 presents more on in-

breeding.) Thus, breeders need some index of the poten-
tial response to selection so that they can then get the
greatest amount of selection with the lowest risk of in-
breeding depression.

Realized Heritability

Breeders often calculate a heritability estimate, a value
that predicts to what extent their selection will be suc-
cessful. Heritability is defined in the following equation:

Yn-Y

gain

Y selection differential

(18.8)

in which
H

Y
Y

Y P

heritability

offspring yield

mean yield of the population

parental yield

From this equation, we can see that heritability is the
gain in yield divided by the amount of selection prac-
ticed (fig. 18.14). Y — Y is the improvement over the
population average due to Y P — Y , which is the amount
of difference between the parents and the population av-
erage. If there is no gain ( Y = Y ), then the heritability

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Heritability

543

Y Mean yield of original population

Yield (phenotype)

V p Mean yield of
selected parents

Selection
differential

V Mean yield of F 1

Y - V=Gain

Figure 18.14 Realized heritability is the gain in yield divided by
the selection differential when offspring are produced by parents
with a mean yield that differs from that of the general population.

will be zero, and breeders will know that no matter how
much selection they practice, they will not improve their
crops and might as well not waste their time. Since this
value is calculated after the breeding has been done, it is
referred to as realized heritability. Some typical values
for realized heritabilities are shown in table 18.5.

The following example may help to clarify the calcu-
lation of realized heritability. The number of bristles on
the sternopleurite, a thoracic plate in Drosophila, is un-
der polygenic control. In a population of flies, the mean
bristle number was 6.4. Three pairs of flies served as par-
ents; they had a mean of 7.2 bristles. Their offspring had
a mean of 6.6 bristles. Hence, Y = 6.6, Y = 6.4, and
Y P = 7.2. Dividing the gain by the selection differential —
that is, substituting in equation 18.8 — gives us

H =

6.6 - 6.4 0.2
7.2 -6.4 0.8

= 0.25

If both a low line and a high line were begun, and if
both were carried over several generations, the heritability
would be measured by the final difference in means of the
high and low lines (gain) divided by the cumulative selec-
tion differentials summed for both the high and low lines.

Note from figure 18.13 that the response to selection
declines with time as the selected population becomes
homozygous for various alleles controlling the trait. As
the response declines, the calculated heritability value it-

Table 18.5

Some Realized Heritabilities

Animal

Trait

Heritability

Cattle

Birth weight

0.49

Milk yield

0.30

Poultry

Body weight

0.31

Egg production

0.30

Egg weight

0.60

Swine

Birth weight

0.06

Growth rate

0.30

Litter size

0.15

Sheep

Wool length

0.55

Fleece weight

0.40

self declines. After intense or prolonged selection, heri-
tability may be zero. It does not mean that the trait is not
controlled by genes, only that there is no longer a re-
sponse to selection. Hence, heritability is specific for a
particular population at a particular time. Intense selec-
tion exhausts the genetic variability, rendering the re-
sponse to selection, and thus the heritability itself, zero.
Quantitative geneticists treat the realized heritability
as an estimate of true heritability. True heritability is ac-
tually viewed in two different ways: as heritability in the
narrow sense and heritability in the broad sense. We de-
fine these on the basis of partitioning of the variance of
the quantitative character under study.

Partitioning of the Variance

Given that the variance of a distribution has genetic and
environmental causes, and given that the variance is ad-
ditive, we can construct the following formula:

^Ph = V G + F E

(18.9)

in which

Vp h = total phenotypic variance
V G = variance due to genotype
V E = variance due to environment

Throughout the rest of this discussion, we will stay
with this model. We could construct a more complex
variance model if there are interactions between vari-
ables. For example, if one genotype responded better in
one soil condition than in another soil condition, this
environment-genotype interaction would require a sepa-
rate variance term (F GE ).

The variance due to the genotype (V G ) can be further
broken down according to the effects of additive poly-
genes (Ya), dominance (F D ), and epistasis (VJ) to give us
a final formula:

^h = V A + V D + V l + F E

(18.10)

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Chapter Eighteen Quantitative Inheritance

We can now define the two commonly used — and often
confused — measures of heritability. Heritability in the

narrow sense is

VJV,

(18.11)

This heritability is the proportion of the total pheno-
typic variance caused by additive genetic effects. It is
the heritability of most interest to plant and animal
breeders because it predicts the magnitude of the re-
sponse under selection.

Heritability in the broad sense is

VcJV,

(18.12)

This heritability is the proportion of the total phenotypic
variance caused by all genetic factors, not just additive
factors. It measures the extent to which individual differ-
ences in a population are caused by genetic differences.
This measure is the one most often used by psycholo-
gists. We are concerned primarily with i/ N , heritability in
the narrow sense.

Measurement of Heritability

Three general methods are used to estimate heritability.
First, as discussed earlier, we can measure heritability by
the response of a population to selection. Second, we can
directly estimate the components of variance by minimiz-
ing one component; the remaining variance can then be
attributed to other causes. For example, by minimizing en-
vironmental causes of variance, we can estimate the ge-
netic component directly. Or, by eliminating the genetic
causes of variance, we can estimate the environmental
component directly. Third, we can measure the similarity
between relatives. We look now at the latter two methods.
Variance components can be minimized in several
different ways. If we use genetically identical organisms,
then the additive, dominance, and epistatic variances are
zero, and all that is left is the environmental variance. For
example, F. Robertson determined the variance compo-
nents for the length of the thorax in Drosophila. The to-
tal variance (F Ph ) in a genetically heterogeneous popula-
tion was 0.366 (measured directly from the distribution
of the trait, as in tables 18.2 and 18.3). He then looked at
the variance in flies that were genetically homogeneous.
These were from isolated lines inbred in the laboratory
over many generations to become virtually homozygous.
Robertson studied the F : in several different matings of
inbred lines and found the variance in thorax length to
be 0.186 (F E ). By subtraction (0.366 - 0.186), we know
that the total genetic variance (F G ) was 0.180. From this,
we can calculate heritability in the broad sense as

#b = V G /V Ph = 0.180/0.366 = 0.49

To calculate a heritability in the narrow sense, it is neces-
sary to extract the components of the genetic variance, V G .

Genetic variance can be measured directly by mini-
mizing the influence of the environment. This is most
easily done with plants grown in a greenhouse. Under
that circumstance, environmental variables, such as soil
quality, water, and sunlight, can be controlled to a very
high degree. Hence, the variance among individuals
grown under these circumstances is almost all genetic
variance. The total phenotypic variance can be obtained
from the plants grown under natural circumstances. This
allows us to calculate heritability in the broad sense.

Several methods exist to sort out the additive from
the dominant and epistatic portions of the genetic vari-
ance. The methods rely mostly on correlations between
relatives. That is, the expected amount of genetic similar-
ity between certain relatives can be compared with the
actual similarity. The expected amount of genetic similar-
ity is the proportion of genes shared; this is a known
quantity for any form of relatedness. For example, par-
ents and offspring have half their genes in common. The
relation of observed and expected correlations between
relatives is a direct measure of heritability in the narrow
sense. We can thus define

•"N " " ^obs'^exp

(18.13)

in which r obs is the observed correlation between the
relatives, and r exp is the expected correlation. The ex-
pected correlation is simply the proportion of the genes
in common.

We must point out that the observed correlation be-
tween relatives can be artificially inflated if the environ-
ments are not random. Since we know that relatives fre-
quently share similar (or correlated) environments, they
may show a phenotypic similarity irrespective of genetic
causes. It is important to keep that in mind, especially
when we analyze human traits, where it may be almost
impossible to rule out or quantify environmental similar-
ity. Hence, r obs may be inflated, which will inflate i/ N .

In human beings, finger-ridge counts (fingerprints,
fig. 18.15) have a very high heritability; there seems to be
very little environmental interference in the embryonic
development of the ridges (table 18.6). Monozygotic
twins are from the same egg, which divides into two em-
bryos at a very early stage. They have identical geno-
types. Dizygotic twins result from the simultaneous fer-
tilization of two eggs. They have the same genetic
relationship as siblings. (However, environmental influ-
ences may be different; they may be treated differently by
relatives and friends.) The data therefore suggest that hu-
man finger ridges are almost completely controlled by ad-
ditive genes with a negligible input from environmental
and dominance variation. Few human traits are con-
trolled this simply (table 18.7).

This brief discussion should make it clear that the
components of the total variance can be estimated. For a
given quantitative trait, the total variance can be mea-

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Quantitative Inheritance in Human Beings

545

(a)

(b)

Table 1 8.7 Some Estimates of Heritabilities (// N )

for Human Traits and Disorders

(c)

Figure 18.15 The three basic fingerprint patterns. Ridges are
counted where they intersect the line connecting a triradius with
a loop or whorl center, (a) An arch; there is no triradius; the
ridge count is zero, (b) A loop; thirteen ridges, (c) A whorl; there
are two triradii and counts of seventeen and eight (the higher
one is routinely used). (From Sarah B. Holt, "Quantitative genetics of
finger-print patterns," British Medical Bulletin, 17. Copyright © 1961 Churchill
Livingstone Medical Journals, Edinburgh, Scotland. Reprinted by permission.)

Table 1 8.6 Correlations Between Relatives, and

Heritabilities, for Finger-Ridge Counts

Relationship

* obs

'exp

H N

Mother-child

0.48

0.50

0.96

Father-child

0.49

0.50

0.98

Siblings

0.50

1.00

Dizygotic twins

0.49

0.50

0.98

Monozygotic twins

0.95

1.00

0.95

Source: From Sarah B. Holt, "Quantitative genetics of finger-print patterns,"
British Medical Bulletin, 17. Copyright © 1961 Churchill Livingstone Medical
Journals, Edinburgh, Scotland. Reprinted by permission.

sured directly. If identical genotypes can be used, then
the environmental component of variance can be deter-
mined. By correlation of various relatives, it is possible to
directly measure heritability in the narrow sense. If heri-
tability is known, and if the total phenotypic variance is
known, then all that are left, assuming no interaction, are
the dominance and epistatic components. In practice,
the epistatic components are usually ignored. Thus, op-

Trait

Heritability

Schizophrenia

0.85

Diabetes Mellitus

Early onset

0.35

Late onset

0.70

Asthma

0.80

Cleft Lip

0.76

Heart Disease, Congenital

0.35

Peptic Ulcer

0.37

Depression

0.45

Stature*

1.00+

* A heritability higher than one can be obtained when the correlation among
relatives is higher than expected. This is usually the result of dominant alleles.

erationally, all that is left is the dominance variance, ob-
tained by subtracting the additive from the total genetic
variance. In addition, plant and animal breeders use so-
phisticated statistical techniques of covariance and vari-
ance analysis, techniques that are beyond our scope.

QUANTITATIVE INHERITANCE
IN HUMAN BEINGS

As with most human studies, the measurement of heri-
tability is limited by a lack of certain types of informa-
tion. We cannot develop pure human lines, nor can we
manipulate human beings into various kinds of environ-
ments or do selection experiments. However, certain
kinds of information are available that allow some esti-
mation of heritabilities.

Skin Color

Skin color is a quantitative human trait for which a sim-
ple analysis can be done on naturally occurring matings.
Certain groups of people have black skin; other groups
do not. Many of these groups breed true in the sense that
skin colors stay the same generation after generation
within a group; when groups intermarry and produce
offspring, the ¥ 1 are intermediate in skin color. In turn,
when Fi individuals intermarry and produce offspring,
the skin color of the F 2 is, on the average, about the same
as the F l3 but with more variation (fig. 18.16). The data
are consistent with a model of four loci, each segregating
two alleles. At each locus, one allele adds a measure of
color, whereas the other adds none.

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Chapter Eighteen Quantitative Inheritance

IQ and Other Traits

In human beings, twin studies have been helpful in esti-
mating the heritability of quantitative traits. One way of
looking at quantitative traits is by the concordance
among twins. Concordance means that if one twin has
the trait, the other does also. Discordance means one has
the trait and the other does not. Table 18.8 shows some
concordance values. High concordance of monozygotic
as compared with dizygotic twins is another indicator of
the heritability of a trait. Concordance values for measles
susceptibility and handedness, which are similar for both
monozygotic and dizygotic twins, demonstrate the envi-
ronmental influence on some traits.

Some monozygotic twins (M2) have been reared
apart. The same is true for dizygotic twins (DZ) and
nontwin siblings. IQ (intelligence quotient) is a measure
of intelligence highly correlated among relatives, indicat-
ing a strong genetic component. In three studies of
monozygotic twins reared apart, the average correlation
in IQ was 0.72. In thirty-four studies of monozygotic
twins reared together, the average correlation in IQ was
0.86; dizygotic twins reared together have an average
correlation of 0.60 in IQ.Thus, it is clear that there is a ge-
netic influence on IQ. However, experts disagree strongly
on the environmental role in shaping IQ and the exact
meaning of IQ as a functional measure of intelligence
(box 18.2).

At present, twin studies are emerging from the
shadow of a scandal involving a knighted British psy-
chologist, Cyril Burt (1883-1971), who did classical
twin research on the inheritance of IQ. Burt was
posthumously accused of fraud, an accusation that was
almost universally accepted and that cast doubt on all
of his data and conclusions. More recently, new infor-

Four loci (actual)

Three loci (expected)
One locus (expected)

Figure 18.16 Inheritance of skin color in human beings. Four
loci are probably involved.

mation seemed to cast doubt on the charges of fraud.
These on-again, off-again charges have been a focus of
scientific interest.

Table 18.8 Concordance of Traits Between Identical and Fraternal Twins

Identical (MZ) Twins (%)

Fraternal (DZ) Twins (%)

Hair color

Eye color

99.6

Blood pressure

Handedness (left or right)

Measles

Clubfoot

Tuberculosis

Mammary cancer

Schizophrenia

Down syndrome

Spina bifida

Manic-depression

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IV. Quantitative and
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Inheritance

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Summary

547

BOX 18.2

The study of human behavioral
genetics was at first associ-
ated with the eugenics move-
ment, founded in the late nineteenth
century by Francis Galton, one of the
founders of quantitative genetics. Eu-
genics was a movement designed to
improve humanity by better breed-
ing. This movement was tainted by
bad science done by people with
strong prejudices. However, although
still controversial, the study of human
behavioral genetics is back in vogue.
The political climate has changed,
and scientific methods to study hu-
man behaviors have improved. Even
some of the strongest critics against
these studies have changed their
minds when confronted with the dis-
covery of particular behavioral genes
through the mapping of quantitative
trait loci. In addition, advocates for
people with many human conditions,
such as homosexuality and mental ill-
ness, feel that if these traits are
shown to be genetic in origin, then
those who have them will be treated
as people with medical conditions
rather than as social outcasts.

Ethics and Genetics

Human Behavioral
Genetics

Even with better methods and
more objective practitioners, the
study of human behavioral genetics is
still difficult to achieve. Some traits
are very poorly denned and may be
complex mixtures of phenotypes,
such as schizophrenia. Other traits
are just difficult to define, such as al-
coholism, criminal tendency, and ag-
gressiveness. To make things more
complicated, several recent studies
that seemed to isolate genes for spe-
cific behavioral traits were not veri-
fied or were later retracted. In one
case, a gene for manic-depressive be-
havior was isolated in an Amish pop-
ulation. However, when the study
was expanded, new cases were dis-
covered that were not linked to the
particular marker locus. The result
was that a "found" genetic locus was

lost. To their credit, the workers were
quick to retract their conclusions.

Currently, there are intriguing re-
sults suggesting that divorce, aggres-
sion, and dyslexia are under genetic in-
fluence. For example, in a recent study,
investigators measured a heritability of
0.52 for divorce. This doesn't mean
that "divorce genes" exist, but rather
that genes for certain personality
traits might predispose a person to
divorce. We should make it clear that
genetic control does not mean that
the environment does not play a role
in these traits, just that there are
genes that are influential also, some-
times very significantly.

In retrospect, it should not be sur-
prising that genes influence much of
our behavior. There are numerous an-
imal studies confirming genetic con-
trol of behaviors, indicating that the
same would be found in people. As
long as the research is done in a com-
petent fashion and the results are not
"politicized," human behavior genet-
ics should not only be a reasonable
area of study, but an exciting one as
we learn more about ourselves.

SUMMARY

STUDY OBJECTIVE 1: To understand the patterns of in-
heritance of phenotypic traits controlled by many loci
531-535

Some genetically controlled phenotypes do not fall into dis-
crete categories. This type of variation is referred to as
quantitative, continuous, or metrical variation. The genetic
control of this variation is referred to as polygenic control.
If the number of controlling loci is small, and offspring fall
into recognizable classes, it is possible to analyze the ge-
netic control of the phenotypes with standard methods.
Polygenes controlling DDT resistance are located on all
chromosomes in Drosophila.

STUDY OBJECTIVE 2: To investigate the way that geneti-
cists and statisticians describe and analyze normal distri-
butions of phenotypes 535-542

When phenotypes fall into a continuous distribution, the
methods of genetic analysis change. We must describe a dis-
tribution using means, variances, and standard deviations.
Then we must describe the relationship between two vari-
ables using variances and correlation coefficients.

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548

Chapter Eighteen Quantitative Inheritance

STUDY OBJECTIVE 3: To define and measure heritability,
the unit of inheritance of variation in traits controlled by
many loci 542-547

Equipped with statistical tools, we analyzed the genetic
control of continuous traits. The heritability estimates tell
us how much of the variation in the distribution of a trait
can be attributed to genetic causes. Heritability in the nar-
row sense is the relative amount of variance due to additive

loci. Heritability in the broad sense is the relative amount of
variance due to all genetic components, including domi-
nance and epistasis. In practice, heritability can be calcu-
lated as realized heritability — gain divided by selection dif-
ferential. Estimates of human heritabilities can be
constructed from correlations among relatives, concor-
dance and discordance between twins, and studies of
monozygotic twins reared apart.

SOLVED PROBLEMS

PROBLEM 1: In a certain stock of wheat, grain color is
controlled by four loci acting according to an additive
model. How many different gametes can a tetrahybrid
produce? How many different genotypes will result if
tetrahybrids are self-fertilized? What will be the pheno-
typic distribution of these genotypes?

Answer: Assume the A, B, C, and D loci with A and a, B
and b, C and c, and D and d alleles, respectively A tetrahy-
brid will have the genotype Aa Bb Cc Dd. A gamete can
get either allele at each of four independently assorting
loci, so there are 2 4 = 16 different gametes. Three geno-
types are possible for each locus, two homozygotes and a
heterozygote. Therefore, for four independent loci, there
are 3 4 = 81 different genotypes. Phenotypes are distrib-
uted according to the binomial distribution. Thus, there
will be a pattern of (A + d) 2n = (A + df\ a ratio of
1:8:28:56:70:56:28:8:1 of phenotypes with decreasing
red color from left to right, eight red colors plus white.

PROBLEM 2: In horses, white facial markings are inher-
ited in an additive fashion. These markings are scored on
a scale that begins at zero. In a particular population, the
average score is 2.2. A group of horses with an average
score of 3.4 is selected to be parents of the next genera-
tion. The offspring of this group of selected parents have
a mean score of 3.1. What is the realized heritability of
white facial markings in this herd of horses?

Answer: This is a simple selection experiment; the data
fit our equation for realized heritability (equation 18.8).
In this case:

Y = offspring yield = 3.1

Y = mean yield of the population =2.2

Y P = parental yield = 3.4

Substituting into equation 18.8:

Y -Y 3.1 - 2.2

H =

Y P -Y 3.4 - 2.2

09
1.2

= 0.75

PROBLEM 3: Corn growing in a field in Indiana had a ly-
sine (amino acid) content of 2.0%, with a variance of
0.16. When grown in the greenhouse under controlled
and uniform conditions, the mean lysine content was
again 2.0%, but the variance was 0.09. What measure of
heritability can you calculate?

Answer: We use equation 18.9 for the calculation of her-
itability by partitioning of the variance (V Fh = V G + F E ).
In this case:

Vph
V G

total phenotypic variance = 0.16
variance due to genotype = 0.09
variance due to environment = ?

In the greenhouse, we have minimized environmental
variance, meaning the total genotypic variance = 0.09. If
we subtract this from the total variance, we get the origi-
nal environmental variance: 0.16 — 0.09 = 0.7. Heritabil-
ity in the broad sense is the genetic variance divided by
the total phenotypic variance, or 0.09/0.16 = 0.56.

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Evolutionary Genetics

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Inheritance

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Exercises and Problems

549

EXERCISES AND PROBLEMS

TRAITS CONTROLLED BY MANY LOCI

1. A variety of squash has fruits that weigh about
5 pounds each. In a second variety, the average weight
is 2 pounds. When the two varieties are crossed, the F :
produce fruit with an average weight of 35 pounds.
When two of these are crossed, their offspring pro-
duce a range of fruit weights, from 2 to 5 pounds. Of
two hundred offspring, three produce fruits weighing
about 5 pounds and three produce fruits about 2
pounds in weight. Approximately how many allelic
pairs are involved in the weight difference between
the varieties, and approximately how much does each
effective gene contribute to the weight?

2. In rabbit variety 1, ear length averages 4 inches. In a
second variety, it is 2 inches. Hybrids between the va-
rieties average 3 inches in ear length. When these hy-
brids are crossed among themselves, the offspring ex-
hibit a much greater variation in ear length, ranging
from 2 to 4 inches. Of five hundred F 2 animals, two
have ears about 4 inches long, and two have ears
about 2 inches long. Approximately how many allelic
pairs are involved in determining ear length, and how
much does each effective gene seem to contribute to
the length of the ear? What do the distributions of P l5
F l5 and F 2 probably look like?

3. Assume that height in people depends on four pairs
of alleles. How can two persons of moderate height
produce children who are much taller than they are?
Assume that the environment is exerting a negligible
effect.

4. How do polygenes differ from traditional Mendelian
genes?

5. If skin color is caused by additive genes, can matings
between individuals with intermediate-colored skin
produce light-skinned offspring? Can such matings
produce dark-skinned offspring? Can matings be-

tween individuals with light skin produce dark-
skinned offspring? (See also QUANTITATIVE INHERI-
TANCE IN HUMAN BEINGS)

6. The tabulated data from Emerson and East ("The In-
heritance of Quantitative Characters in Maize," 1913,
Univ. Nebraska Agric. Exp. Sta. Bull, no. 2) show the
results of crosses between two varieties of corn and
their F 2 offspring (see the table on ear length in corn).
Provide an explanation for these data in terms of num-
ber of allelic pairs controlling ear length. Do all the
genes involved affect length additively? Explain.

7. In Drosophila, a marker strain exists containing domi-
nant alleles that are lethal in the homozygous condi-
tion on both chromosome 2, 3, and 4 homologues.
These six lethal alleles are within inversions, so there
is virtually no crossing over. The strain thus remains
perpetually heterozygous for all six loci and therefore
all three chromosome pairs. (Geneticists use a short-
hand notation in these "balanced-lethal" systems in
which only the dominant alleles on a chromosome are
shown, with a slash separating the two homologous
chromosomes.) The markers are: chromosome 2, Curly
and Plum (Cy/Pm, shorthand for CyPm + /Cy + Pm);
chromosome 3, Hairless and stubble (H/S); and chro-
mosome 4, Cell and Minute(4) (Ce/M[4]). With this
strain, which allows you to follow particular chromo-
somes by the presence or absence of phenotypic
markers, construct crosses to give the strains Crow
used (see fig. 18.7) to determine the location of poly-
genes for DDT resistance.

8. A red-flowered plant is crossed with a yellow-
flowered plant to produce V 1 plants with orange flow-
ers. The F : offspring are selfed, and they produce
plants with flowers in a range of seven different col-
ors. How many genes are probably involved in color
production?

Ear Length in Corn (cm)

10 11

13 14 15 16 17 18 19 20 21

Variety P 60
Variety P 54

F 2 (Fj X FO

15 10
9 1

Answers to selected exercises and problems are on page A-20.

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IV. Quantitative and
Evolutionary Genetics

18. Quantitative
Inheritance

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550

Chapter Eighteen Quantitative Inheritance

9. A plant with a genotype of aabb and a height of
40 cm is crossed with a plant with a genotype of
AABB and a height of 60 cm. If each dominant allele
contributes to height additively what is the ex-
pected height of the ¥ 1 progeny?

10. If the Fj generation in the cross in problem 9 is
selfed, what proportion of the F 2 offspring would
you expect to be 50 cm tall?

11. Two strains of wheat were compared for the time re-
quired to mature. Strain X required fourteen days,
and strain Y required twenty-eight days. The strains
were crossed, and the F : generation was selfed. One
hundred F 2 progeny out of 6,200,000 matured in
fourteen days or less. How many genes may be in-
volved in maturation?

POPULATION STATISTICS

12. A geneticist wished to know if variation in the num-
ber of egg follicles produced by chickens was inher-
ited. As a first step in his experiments, he wished to
determine if the number of eggs laid could be used
to predict the number of follicles. If this were true,
he could then avoid killing the chickens to obtain
the data he needed. He obtained the following data
from fourteen chickens.

Chicken

Eggs

Ovulated

Number

Laid

Follicles

Calculate a correlation coefficient. Graph the data,
and then calculate the slope and ^-intercept of the
regression line. Draw the regression line on the same
graph.

13. The following table (data from Ehrman and Parsons,
1976, The Genetics of Behavior, 121, Sunderland,
Mass.: Sinauer Associates) gives heights in centime-
ters of eleven pairs of brothers and sisters. Calculate
a correlation coefficient and a heritability. Is this re-
alized heritability, heritability in the broad sense, or
heritability in the narrow sense?

Pair

Brother

Sister

Pair

Brother

Sister

180

175

178

165

173

162

186

163

168

165

183

168

170

160

165

160

178

165

168

157

180

157

How can environmental factors influence this heri-
tability value? (See also HERITABILITY)

14. You determine the following variance components
for leaf width in a particular species of plant:

Additive genetic variance (F A ) 4.0

Dominance genetic variance (F D ) 1.8

Epistatic variance (V{) 0.5

Environmental variance ( F E ) 2 . 5

Calculate the broad sense and narrow sense heri-
tabilities. (See also HERITABILITY)

SELECTION EXPERIMENTS

15. Psychologists refer to defecation rate in rats as
"emotionality." The data shown in the accompanying
figure (data modified from Broadhurst, I960, Experi-
ments in Personality, vol. 1, London: Eysenck) show
mean emotionality scores during five generations in
high and low selection lines. In the final generation,
the parental mean was 4 for the high line and 0.9 for
the low line. The cumulative selection differential is
5 for each line. Calculate realized heritability overall,
and separate heritabilities for each line. Do these
differ? Why? Why was the response to selection
asymmetrical? (See also HERITABILITY)

i —

-1— '

"05

2
1

I I I

1 2 3

Generation

16. Data were gathered during a selection experiment
for six-week body weight in mice. Graph these data
and calculate a realized heritability. (See also HERI-
TABILITY)

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IV. Quantitative and
Evolutionary Genetics

18. Quantitative
Inheritance

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Critical Thinking Questions

551

High Line

Low Line

Generation

1
2

4
5

HERITABILITY

17. Outstanding athletic ability is often found in several
members of a family. Devise a study to determine to
what extent athletic ability is inherited. (What is
"outstanding athletic ability"?)

18. Variations in stature are almost entirely due to hered-
ity. Yet average height has increased substantially
since the Middle Ages, and the increase in the height
of children of immigrants to the United States, as
compared with the height of the immigrants them-
selves, is especially noteworthy. How can these ob-
servations be reconciled?

19. Would you expect good nutrition to increase or de-
crease the heritability of height?

20. Two adult plants of a particular species have ex-
treme phenotypes for height (1 foot tall and 5 feet
tall), a quantitative trait. If you had only one uni-
formly lighted greenhouse, how would you deter-
mine whether the variation in plant height is envi-
ronmentally or genetically determined? How would
you attempt to estimate the number of allelic pairs
that may be involved in controlling this trait?

21. The components of variance for two characters of
D. melanogaster are shown in the following table
(data from A. Robertson, "Optimum Group Size in
Progeny Testing," Biometrics, 13:442-50, 1957).

Estimate the dominance and epistatic components,
and calculate heritabilities in the narrow and broad
sense.

Variance
Components

Thorax
Length

Eggs Laid
in Four Days

V A

v E

Vu + V,

100

43
51

100
18
38

22. In a mouse population, the average tail length is 10
cm. Six mice with an average tail length of 15 cm are
interbred. The mean tail length in their progeny is
135 cm. What is the realized heritability?

23. The narrow sense heritability of egg weight in chick-
ens in one coop is 0.5. A farmer selects for heavier
eggs by breeding a few chickens with heavier eggs.
He finds a difference of 9 g in the mean egg weights
of selected and unselected chickens. By how much
can he expect egg weight to increase in the selected
chickens?

24. If, in a population of swine, the narrow sense heri-
tability of maturation weight is 0.15, the phenotypic
variance is 100 lb 2 , the total genetic variance is
50 lb 2 , and the epistatic variance is 0, calculate the
dominance genetic variance and the environmental
variance.

25. A group of four-month-old hogs has an average
weight of 170 pounds. The average weight of se-
lected breeders is 185 pounds. If the heritability of
weight is 40%, what is the expected average weight
of the first generation progeny?

QUANTITATIVE INHERITANCE IN HUMAN BEINGS

26. Does schizophrenia seem to have a strong genetic
component (see table 18.8)? Explain.

CRITICAL THINKING QUESTIONS

1. Several cases mentioned in the text reported and then
retracted the discovery of human genes controlling
specific traits. Barring fraud, what might cause a scien-
tist to retract a study of this type?

2. Monozygotic twins share identical genes. Under what
conditions could they show discordance of traits?

Suggested Readings for chapter 18 are on page B-19.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

19. Population Genetics:
The Hardy-Weinberg
Equilibrium and Mating
Systems

©TheMcGraw-Hil
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POPULATION
GENETICS

The Hardy-Weinberg
Equilibrium and Mating

Systems

The cheetah (Acinonyx jubatus) is in peril of extinction;
it has very low genetic variability.

(Gregory G. Dimijian, MD/Photo Researchers, Inc.)

STUDY OBJECTIVES

1 . To understand the concept of population-level genetic
processes 553

2. To learn the assumptions and nature of the Hardy-Weinberg
equilibrium and its extensions 554

3. To test whether a population is in Hardy-Weinberg
equilibrium 557

4. To analyze the process and consequences of nonrandom
mating in diploid populations 560

STUDY OUTLINE

Hardy-Weinberg Equilibrium 553

Calculating Allelic Frequencies 553

Assumptions of Hardy-Weinberg Equilibrium 554

Proof of Hardy-Weinberg Equilibrium 555

Generation Time 556

Testing for Fit to Hardy-Weinberg Equilibrium 557
Extensions of Hardy-Weinberg Equilibrium 558

Multiple Alleles 558

Multiple Loci 559
Nonrandom Mating 560

Inbreeding 560

Pedigree Analysis 562

Population Analysis 564
Summary 566
Solved Problems 566
Exercises and Problems 567
Critical Thinking Questions 569
Box 19.1 The Determination of Lethal Equivalents 562

552

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IV. Quantitative and
Evolutionary Genetics

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Systems

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Hardy-Weinberg Equilibrium

553

Evolution is a process that takes place in popu-
lations of organisms. To study evolution, we
need to shift our focus to population genetics,
the algebraic description of the genetic
makeup of a population and the changes in al-
lelic frequencies in populations over time. This chapter is
the first of three that looks at what population genetics
can tell us about the way evolution proceeds.

Almost all of the mathematical foundations of
genetic changes in populations were developed in a
short period of time during the 1920s and 1930s by
three men: R. A. Fisher, J. B. S. Haldane, and S. Wright.
Some measure of disagreement emerged among these
men, but they disagreed on which evolutionary pro-
cesses were more important, not on how the processes
worked. Since the 1960s, excitement has arisen in the
field of population genetics, primarily on three fronts.
First, the high-speed computer has made it possible to
do a large amount of arithmetic in a very short period of
time; thus, complex simulations of real populations can
be added to the repertoire of the experimental geneti-
cist. Second, electrophoresis has provided a means of
gathering the large amount of empirical data necessary
to check some of the assumptions used in mathematical
models. The information and interpretation of the elec-
trophoretic data have generated some controversy
about the role of "neutral" evolutionary changes in nat-
ural populations. Last, newer techniques of molecular
genetics are being used to analyze the relationships
among species and the rate of evolutionary processes.
We consider these studies later.

Sir Ronald A. Fisher

(1 890-1 962). (Courtesy of The

National Portrait Gallery, England.)

Sewall Wright (1889-1988).
(Courtesy of Dr. Sewall Wright.)

HARDY-WEINBERG
EQUILIBRIUM

Let us begin with a few definitions. For the most part, we
define a species as a group of organisms potentially capa-
ble of interbreeding. Most species are made up of popu-
lations, interbreeding groups of organisms that are usu-
ally subdivided into partially isolated breeding groups
called denies. As we will see, it is these demes, or local
populations, that can evolve.

In 1908, G. H. Hardy, a British mathematician, and
W. Weinberg, a German physician, independently discov-
ered a rule that relates allelic and genotypic frequencies
in a population of diploid, sexually reproducing individu-
als if that population has random mating, large size, no
mutation or migration, and no selection. The rule has
three aspects:

1 . The allelic frequencies at an autosomal locus in a pop-
ulation will not change from one generation to the
next (allelic-frequency equilibrium).

2. The genotypic frequencies of the population are de-
termined in a predictable way by the allelic frequen-
cies (genotypic-frequency equilibrium).

3. The equilibrium is neutral. That is, if it is perturbed, it
will be reestablished within one generation of random
mating at the new allelic frequencies (if all the other
requirements are maintained).

Calculating Allelic Frequencies

If we consider an autosomal locus in a diploid, sexually re-
producing species, allelic frequencies can be measured in
either of two ways. The first is simply by counting genes:

frequency of the a allele, q, =

number of a alleles
total number of alleles

The expression "frequency of" can be shortened to/( ).
For example, the frequency of the a allele is written as
f(d). Since the homozygotes have two of a given allele and
heterozygotes have only one, and since the total number
of alleles is twice the number of individuals (each individ-
ual carries two alleles), we can calculate allelic frequen-
cies in the following manner. Consider, for example, the
phenotypic distribution of MN blood types (controlled by
the codominant M and N alleles) among two hundred per-
sons chosen randomly in Columbus, Ohio:

type M (MM genotype) =114

type MN (MN genotype) = 76

type N (NN genotype) = 10

200

Tamarin: Principles of
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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

Then,

P=f(M} =

2(114) + 76 304

2(200)

400

= 0.76

Similarly,

q=JVO

2(10) + 76 96

2(200)

400

= 0.24

Alternatively, because the frequencies of the two alleles,
M and N, must add up to unity (p + q = 1, q = 1 — p,
andp = 1 — q), if we know thatp = 0.76, then q = 1 —
0.76 = 0.24.

Another way of calculating allelic frequencies is
based on knowledge of the genotypic frequencies. In this
example, the frequencies are

114

f(MM) = = 0.57

7 200

/(MV)

/(AW) =

76
200

10
200

0.38

= 0.05

We derive an expression for calculating^? and q based on
genotypic frequencies as follows:

p=f(M) =

2 X number of MM + number of MN
2 X total number

2 X number of MM number of MN

2 X total number 2 X total number
= f(MM) + (l/2)/(MW)

and,

q=/cm =

2 X number of NN + number of MN
2 X total number

2 X number of AW number of MN

2 X total number 2 X total number

= /(WW) + (l/2)/(MW)

Thus, allelic frequencies can be calculated as the fre-
quency of homozygotes, plus half the frequency of het-
ero zygotes, as follows:

p = /(M) = /(MM) + (l/2)/(MW)

= 0.57 + (1/2)0.38 = 0.76
q = /(W) = /(WW) + (l/2)/(MW)

= 0.05 + (1/2)0.38 = 0.24

q=l-p=l- 0.76 = 0.24

Note that these two methods (counting alleles and using
genotypic frequencies) are algebraically identical and
thus give identical results.

Assumptions of Hardy-Weinberg
Equilibrium

We will consider a population of diploid, sexually repro-
ducing organisms with a single autosomal locus segre-
gating two alleles (i.e., every individual is one of three
genotypes — MM, MN, or WW). Later on, we generalize
the discussion to include multiple alleles and multiple
loci. For the moment, the focus is on a genetic system
such as the MW locus in human beings. The following ma-
jor assumptions are necessary for the Hardy-Weinberg
equilibrium to hold.

Random Mating

The first assumption is random mating, which means
that the probability that two genotypes will mate is the
product of the frequencies (or probabilities) of the geno-
types in the population. If the MM genotype makes up
90% of a population, then any individual has a 90%
chance (probability = 0.9) of mating with a person with
an MM genotype. The probability of an MM by MM mat-
ing is (0.9X0.9), or 0.81.

Deviations from random mating come about for two
reasons: choice or circumstance. If members of a popu-
lation choose individuals of a particular phenotype as
mates more or less often than at random, the population
is engaged in assortative mating. If individuals with
similar phenotypes are mating more often than at ran-
dom, positive assortative mating is in force; if matings
occur between individuals with dissimilar phenotypes
more often than at random, negative assortative mating,
or disassortative mating, is at work.

Deviations from random mating also arise when mat-
ing individuals are either more closely related genetically
or more distantly related than individuals chosen at ran-
dom from the population. Inbreeding is the mating of
related individuals, and outbreeding is the mating of ge-
netically unrelated individuals. Inbreeding is a conse-
quence of pedigree relatedness (e.g., cousins) and small
population size.

One of the first counterintuitive observations of pop-
ulation genetics is that deviations from random mating al-
ter genotypic frequencies but not allelic frequencies.
Envision a population in which every individual is the
parent of two children. On the average, each individual
will pass on one copy of each of his or her alleles. Assor-
tative mating and inbreeding will change the zygotic
(genotypic) combinations from one generation to the
next, but will not change which alleles are passed into

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the next generation. Thus genotypic, but not allelic, fre-
quencies change under nonrandom mating.

Large Population Size

Even when an extremely large number of gametes is pro-
duced in each generation, each successive generation is
the result of a sampling of a relatively small portion of the
gametes of the previous generation. A sample may not be
an accurate representation of a population, especially if
the sample is small. Thus, the second assumption of the
Hardy-Weinberg equilibrium is that the population is infi-
nitely large. A large population produces a large sample
of successful gametes. The larger the sample, the greater
the probability that the allelic frequencies of the off-
spring will accurately represent the allelic frequencies in
the parental population. When populations are small or
when alleles are rare, changes in allelic frequencies take
place due to chance alone. These changes are referred to
as random genetic drift, or just genetic drift.

No Mutation or Migration

Allelic and genotypic frequencies may change through the
loss or addition of alleles through mutation or migration
(immigration or emigration) of individuals from or into a
population. The third and fourth assumptions of the Hardy-
Weinberg equilibrium are that neither mutation nor migra-
tion causes such allelic loss or addition in the population.

No Natural Selection

The final assumption necessary to the Hardy-Weinberg
equilibrium is that no individual will have a reproductive
advantage over another individual because of its geno-
type. In other words, no natural selection is occurring.
(Artificial selection, as practiced by animal and plant
breeders, will also perturb the Hardy-Weinberg equilib-
rium of captive populations.)

In summary, the Hardy-Weinberg equilibrium holds
(is exactly true) for an infinitely large, randomly mating
population in which mutation, migration, and natural se-
lection do not occur. In view of these assumptions, it
seems that such an equilibrium would never be charac-
teristic of natural populations. However, this is not the
case. Hardy-Weinberg equilibrium is approximated in
natural populations for two major reasons. First, the con-
sequences of violating some of the assumptions, such as
no mutation or infinitely large population size, are small.
Mutation rates, for example, are on the order of one
change per locus per generation per 10 6 gametes. Thus,
there is virtually no measurable effect of mutation in a
single generation. In addition, populations do not have to
be infinitely large to act as if they were. As we will see, a

relatively small population can still closely approximate
Hardy-Weinberg equilibrium. In other words, minor devi-
ations from the other assumptions can still result in a
good fit to the equilibrium; only major deviations can be
detected statistically. Second, the Hardy-Weinberg equi-
librium is extremely resilient to change because, regard-
less of the perturbation, the equilibrium is usually
reestablished after only one generation of random
mating. The new equilibrium will be, however, at the
new allelic frequencies — the Hardy-Weinberg equilib-
rium does not "return" to previous allelic values.

Proof of Hardy-Weinberg Equilibrium

The three properties of the Hardy-Weinberg equilibrium
are that (1) allelic frequencies do not change from genera-
tion to generation, (2) allelic frequencies determine geno-
typic frequencies, and (3) the equilibrium is achieved in
one generation of random mating. We will concentrate for
a moment on the second properly. In a population of indi-
viduals segregating the A and a alleles at the A locus, each
individual will be one of three genotypes: AA, Aa, or aa. If
p = f(A) and q = f(a), then we can predict the genotypic
frequencies in the next generation. If all the assumptions
of the Hardy-Weinberg equilibrium are met, the three
genotypes should occur in the population in the same fre-
quencies at which gametes would be randomly drawn in
pairs from a gene pool. A gene pool is defined as all of the
alleles available among the reproductive members of a
population from which gametes can be drawn. Thus,

f(AA) = {pxp)=p 2

f(Aa) = (pXq) + (qXp) = 2pq

f(ad) = (q X q) = q 2

demonstrates the second property of the Hardy-
Weinberg equilibrium (fig. 19. 1).

A
f(A) = p

a
f(a) = q

A
f(A) = p

AA
f(AA) = p 2

f(Aa) = pq

f(a) = q

Aa
f(Aa) = pq

aa
f(aa) = q 2

Figure 19.1 Gene pool concept of zygote formation. Males
and females have the same frequencies of the two alleles:
f (A) = p and f (a) = q. After one generation of random
mating, the three genotypes, AA, Aa, and aa, have the
frequencies of p 2 , 2pq, and q 2 , respectively.

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

Another way of demonstrating the properties of the
Hardy-Weinberg equilibrium for the one-locus, two-allele
case in sexually reproducing diploids is by simply ob-
serving the offspring of a randomly mating, infinitely
large population. Let the initial frequencies of the three
genotypes be any values that sum to one; for example, let
X, Y, and Z be the proportions of the AA, Aa, and aa
genotypes, respectively The proportions of offspring
after one generation of random mating are as shown in
table 191. For example, the probability that an AA indi-
vidual will mate with an AA individual is X X X, or X 2 .
Since all the offspring of this mating are AA, they are
counted only under the AA column of offspring in table
191. When all possible matings are counted, the off-
spring with each genotype are summed. The proportion
of AA offspring is X 2 + XY + (1/4) Y 2 , which factors to
(X + [1/2] F) 2 . Recall that the frequency of an allele is
the frequency of its homozygote plus half the frequency
of the heterozygote. Hence, X + (l/2)Fis the frequency
of A, since X = f(AA) and Y = f(Aa). If p = f(A), then
(X + [1/2] F) 2 is^? 2 .Thus, after one generation of random
mating, the proportion of AA homozygotes is p 2 . Simi-
larly, the frequency of aa homozygotes after one genera-
tion of random mating is Z + YZ + (1/4)F , which
factors to (Z + [1/2] F) 2 , or q 2 . The frequency of het-
erozygotes when summed and factored (table 19.1) is
2(X + [1/2] F) (Z + [1/2] F), or 2pq. Therefore, after one
generation of random mating, the three genotypes (AA,
Aa, and aa) occur as^? 2 , 2pq, and q 2 .

Looking at the first property of the Hardy-Weinberg
equilibrium, that allelic frequencies do not change gen-
eration after generation, we can ask, Have the allelic fre-
quencies changed from one generation to the next (from

the parents to the offspring)? Before random mating, the
frequency of the A allele is, by definition, p:

f(A) =p= f(AA) + (V2)f(Aa) = X + (1/2)F

After random mating, the frequency of the A homozygote
is^? 2 , and the frequency of the heterozygote is 2pq. Thus,
the frequency of the A allele, the frequency of its ho-
mozygote plus half the frequency of the heterozygotes, is

f(A) = f(AA) + a/2)f(Aa)

= p 2 + (l/2)(2^)

= p 2 +pq= p(p + q)

= p (remember, p + q = 1)

Thus, in a randomly mating population of sexually repro-
ducing diploid individuals, the allelic frequency, p, does
not change from generation to generation. Here, by ob-
serving the offspring of a randomly mating population,
we have proven all three properties of the Hardy-
Weinberg equilibrium.

Generation Time

Although generation interval is commonly thought of as
the average age of the parents when their offspring are
born, the statistical concept of a generation is more com-
plex. Demographers use formulas that relate generation
time to the age of reproducing females, the reproductive
level of each age group, and the probability of survival in
each age group. Here, to avoid these complexities, we will
use discrete generations, unless otherwise noted. That
is, we will assume that all the individuals drawn in a sam-

TablG 19.1 Proportions of Offspring in a Randomly Mating Population Segregating the A and a Alleles at

the A locus: X = f(AA), Y = f(Aa), and Z = f(aa)

ting

Proportion

Offspring

X AA

X 2

X Aa

(1/2)XF

X aa

X AA

(1/2)XF

(\/2)XY

X Aa

Y 2

(1/4)F 2

(1/2)F 2

(1/4)F 2

X aa

(1/2)FZ

X AA

X Aa

(1/2)FZ

X aa

Z 2

Sum

+ Y +

Z) 2

+ [1/2]F) 2

2(X +

[1/2] F)(Z + [1/2] F)

(Z + [1/2] F) 2

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pie, for purposes of determining allelic and genotypic fre-
quencies, are drawn from the same generation, and that, in
resampling the population, the second sample represents
the offspring of the first generation. The discrete-generation
model holds for organisms such as annual plants and fruit
flies maintained under laboratory conditions, with no
breeding among individuals of different generations. Gen-
erations that overlap, as in populations of human beings
and many other organisms, usually are better described by
somewhat more complex mathematical models.

Testing for Fit to Hardy-Weinberg Equilibrium

There are several ways to determine whether a given
population conforms to the Hardy-Weinberg equilibrium
at a particular locus. However, the question usually arises
when there is just a single sample from a population, rep-
resenting only one generation. Can the existence of the
Hardy-Weinberg equilibrium be determined with just
one sample? The answer is that we can determine
whether the three genotypes (AA, Aa, and ad) occur
with the frequencies^? 2 , 2pq, and q 2 . If they do, then the
population is considered to be in Hardy-Weinberg pro-
portions; if not, then the population is not considered to
be in Hardy-Weinberg proportions.

MN Blood Types

To determine whether observed and expected allelic fre-
quencies are the same, we can use the chi-square statisti-
cal test. In a chi-square test, we compare an observed
number with an expected number. In this case, the ob-
served values are the actual numbers of the three geno-
types in the sample, and the expected values come from
the prediction that the genotypes will occur in the p 2 ,
2pq, and q 2 proportions. An analysis for the Ohio MN
blood-type data is presented in table 19. 2. The agreement
between observed and expected numbers is very good,
obvious even before the calculation of the chi-square
value. Since the critical chi-square for one degree of free-
dom at the 0.05 level is 3.841 (see table 4A), we find that

the Ohio population does not deviate from Hardy-
Weinberg proportions at the MN locus.

Earlier (chapter 4), we used the chi-square statistic to
test how well real data fit an expected data set based on
a ratio predicted before the test. For example, we tested
the data against a 3:1 ratio in table 4.2. In that case, the
number of degrees of freedom was simply the number of
independent categories: the total number of categories
minus one. Here, however, our expected ratio is derived
from the data set itself. The values p 2 , 2pq, and q 2 came
from p and q, which were estimated from the data. In this
case, we lose one additional degree of freedom for every
independent value we estimate from the data. If we cal-
culate p from a sample, we lose one degree of freedom.
However, we do not lose a degree of freedom for esti-
mating q, since q is no longer an independent variable:
q = 1 — p. So in the previous case, we lose two degrees
of freedom — one for estimating p and one for indepen-
dent categories. The general rule of thumb in using chi-
square analysis to test for data fit to Hardy-Weinberg
proportions is that the number of degrees of freedom
must equal the number of phenotypes minus the number
of alleles (in this case, 3 — 2 = 1).

The chi-square analysis in table 19.2 may seem par-
adoxical. Because the observed allelic frequencies cal-
culated from the original genotypic data are used to
calculate the expected genotypic frequencies, it may
appear to some individuals that the analysis must, by
its very nature, show that the population is in Hardy-
Weinberg proportions. To demonstrate that this is not
necessarily the case, a counterexample appears in table
19.3. We use data similar to the Ohio sample, except
that the original number of heterozygotes has been dis-
tributed equally among the two homozygote classes.
The same allelic frequencies are maintained, yet the
genotypic distribution differs. The chi-square value of
200.00 for these data demonstrates that the population
represented in table 19.3 is not in Hardy-Weinberg pro-
portions. Thus, a chi-square analysis of fit to the Hardy-
Weinberg proportions by no means represents circular
reasoning.

Table 19.2 Chi-Square Test of Goodness-of-Fit to the Hardy-Weinberg Proportions of a Sample of 200

Persons for MN Blood Types for Which/? = 0.76 and q = 0.24

Total

Observed Numbers

114

200

Expected Proportions

P 2

2pq

q 2

1.0

(0.5776)

(0.3648)

(0.0576)

1.0

Expected Numbers

115.52

72.96

11.52

200.0

X 2 = (O - Ef/E

0.020

0.127

0.201

0.348

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

Table 1 9.3 Chi-Square Test of Goodness-of-Fit to the Hardy-Weinberg Proportions of a Second Sample of 200

Persons for MN Blood Types for Which p = 0.76 and q = 0.24 and Heterozygotes Are Absent

Total

Observed Numbers

152

200

Expected Proportions

P 2

2pq

q 2

1.0

(0.5776)

(0.3648)

(0.0576)

1.0

Expected Numbers

115.52

72.96

11.52

200.0

X 2 = (O - Ef/E

11.52

72.96

115.52

200.00

PKU

Circumstances sometimes do not allow us to test for Hardy-
Weinberg proportions. In the case of a dominant trait, for
example, allelic frequencies cannot be calculated from the
genotypic classes because the homozygous dominant indi-
viduals cannot be distinguished from the heterozygotes.
However, we can estimate allelic frequencies by assuming
that the Hardy-Weinberg equilibrium exists and, thereby, as-
suming that the frequency of the recessive homozygote is
q 2 , from which q and then^? can be estimated.

If, for example, Hardy-Weinberg equilibrium is as-
sumed for a disease such as phenylketonuria (PKU),
which is expressed only in the homozygous recessive
state, it is possible to calculate the proportion of the pop-
ulation that is heterozygous (carriers of the PKU allele).
But is it fair to assume Hardy-Weinberg equilibrium here?
Until recent medical advances allowed intervention,
there was a good deal of selection against individuals
with PKU, who were usually mentally retarded. Thus the
assumption of no selection, required for equilibrium, is
violated. However, only one child in ten thousand live
births has PKU. When a genotype is as rare as one in ten
thousand, selection has a negligible effect on allelic fre-
quencies. Therefore, because of the rarity of the trait, we
can assume Hardy-Weinberg equilibrium and calculate

frequency of recessive homozygote = q 2 =

so,

and

1/10,000 = 0.0001

q = Vo.0001 = 0.01

p = 1 - q = 0.99
Therefore,

frequency of normal homozygote = p 2 = (0.99) 2

= 0.98 or 98 in 100

frequency of heterozygote = 2pq

= 2(0.01) (0.99) = 0.02 or 2 in 100.

By assuming the Hardy-Weinberg equilibrium, we have
discovered something not intuitively obvious: A reces-
sive gene causing a trait as rare as one in ten thousand
is carried in the heterozygous state by one individual
in fifty. Obviously, the chi-square test cannot be used
to verify the Hardy-Weinberg proportions since we
derived the allelic frequencies by assuming Hardy-
Weinberg proportions to begin with. In statistical terms,
the number of phenotypes minus the number of alleles
= 2 — 2 = degrees of freedom, which precludes
doing a chi-square test.

EXTENSIONS OF HARDY-
WEINBERG EQUILIBRIUM

The Hardy-Weinberg equilibrium can be extended to in-
clude, among other cases, multiple alleles and multiple loci.

Multiple Alleles

Multinomial Expansion

The expected genotypic array under Hardy-Weinberg
equilibrium is p 2 , 2pq, and q 2 , which form the terms of
the binomial expansion (/> + q) 2 . If males and females
each have the same two alleles in the proportions of p
and q, then genotypes will be distributed as a binomial
expansion in the frequencies p 2 , 2pq, and q 2 (see fig.
191). To generalize to more than two alleles, one need
only add terms to the binomial expansion and thus create
a multinomial expansion. For example, with alleles a, b,
and c with frequencies p, q, and r, the genotypic distri-
bution should be (p + q + r) 2 , or

p 2 + q 2 + r 2 + 2pq + 2pr + 2qr

Homozygotes will occur with frequencies^? 2 , q 2 , and r 2 ,
and heterozygotes will occur with frequencies 2pq, 2pr,
and 2qr. The ABO blood-type locus in human beings is an
interesting example because it has multiple alleles and
dominance.

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ABO Blood Groups

The ABO locus has three alleles: I A , I B , and *', with the
I A and I B alleles codominant, and both dominant to
the i allele. These alleles control the production of a
surface antigen on red blood cells (see fig. 2.13). Table
19.4 contains blood-type data from a sample of five
hundred persons from Massachusetts. Is the popula-
tion in Hardy-Weinberg proportions? The answer is not
apparent from the data in table 19.4 alone, since there
are two possible genotypes for both the A and the B
phenotypes. No estimate of the allelic frequencies is
possible without making assumptions about the num-
ber of each genotype within these two phenotypic
classes. Is it possible to estimate the allelic frequencies?
The answer is yes, if we assume that Hardy-Weinberg
equilibrium exists.

One procedure follows. Let us assume that^? =f(I A ),
q = f(I B ), and r = f(i). Blood type O has the ii genotype;
if the population is in Hardy-Weinberg proportions, this
genotype should occur at a frequency of r 2 . Thus

fiii) = 231/500 = 0.462 = r'

and

r = fit) = Vo.462 = 0.680

From table 19.4, we see that blood type A plus blood
type O include only the genotypes I A I A , I A i, and //. If the
population is in Hardy-Weinberg proportions, these to-
gether should be (p + r) 2 , in which p 2 = f(I A I A ), 2pr =
f(I A 0,<indr 2 =f(ii):

Cp + rf = (199 + 231)/500 = 0.860

Then, taking the square root of each side

and

p + r = V0.860 = 0.927

p = 0.927 - r = 0.927 - 0.680 = 0.247

Table 1 9.4 ABO Blood -Type Distribution in 500

Persons from Massachusetts

Blood

Type

Genotype

Number

I A I A or I A i

199

I B I B or I B i

I A I B

231

Total

500

The frequency of allele / , q, can be obtained by
similar logic with blood types B and O, or simply by sub-
traction:

q=\-(p + r)=\- 0.927 = 0.073

Thus, the Hardy-Weinberg equilibrium can be
extended to include multiple alleles and can be used to
make estimates of the allelic frequencies in the ABO
blood groups. With ABO, it is statistically feasible to do
a chi-square test because there is one degree of
freedom (number of phenotypes — number of alleles =
4 — 3 = 1). We are really testing only the AB and B cat-
egories; if we did our calculations as shown, the ob-
served and expected values of phenotypes A and O
must be equal.

Multiple Loci

The Hardy-Weinberg equilibrium can also be extended
to consider several loci at the same time in the same pop-
ulation. This situation deserves mention because
the whole genome is likely involved in evolutionary
processes and we must, eventually, consider simul-
taneous allelic changes in all loci segregating alleles
in an organism. (Even with a high-speed computer,
simultaneous consideration of many loci is a bit far off in
the future.) When two loci, A and B, on the same
chromosome are in equilibrium with each other, the
combinations of alleles on a chromosome in a gamete
follow the product rule of probability. Consider the A
locus with alleles A and a and the B locus with alleles B
and b, respectively, with allelic frequencies p A and q A
for A and a, respectively, and p B and q B for B and b,
respectively. Given completely random circumstances,
the chromosome with the A and B alleles should occur at
the frequency p A p B . This is referred to as linkage
equilibrium. When alleles of different loci are not in
equilibrium (i.e., not randomly distributed in gametes),
the condition is referred to as linkage disequilibrium.
The approach to linkage equilibrium is gradual and is
a function of the recombination distance between the
two loci.

For example, let's start with a population out of
equilibrium so that all chromosomes are AB (70%) or ab
(30%). Then p A = 0.7, q A = 0.3, p B = 0.7, and q B = 0.3.
We expect the Ab chromosome to occur 0.7 X 0.3 =
0.21, or 21% of the time. The frequency of the Ab chro-
mosome is zero. Assume the map distance between the
two loci is 0.1; in other words, 10% of chromatids in
gametes are recombinant. Initially, we consider that
each locus is in Hardy-Weinberg proportions, or the fre-
quency of AB/AB individuals = 0.49 (0.7 X 0.7); the fre-
quency of ab/ab individuals is 0.09 (0.3 X 0.3); and the
frequency of AB/ab individuals is 0.42 (2 X 0.7 X 0.3).

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

After one generation of random mating, gametes will be
as follows:

from AB/AB individuals (49%): only AB gametes, 49%
of total

from ab/ab individuals (9%): only ab gametes, 9% of
total

from AB/ab individuals (42%):

AB gametes, 18.9% of total (0.45 X 0.42)

ab gametes, 18.9% of total (0.45 X 0.42)

Ab gametes, 2.1% of total (0.05 X 0.42)

aB gametes, 2.1% of total (0.05 X 0.42)

(The values of 18.9% and 2.1% for the dihybrids result
from the fact that since map distance is 0.1, 10% of ga-
metes will be recombinant, split equally between the
two recombinant classes — 5% and 5%. Ninety percent
will be parental, split equally between the two parental
classes — 45% and 45%. Each of these numbers must be
multiplied by 0.42 because the dihybrid makes up 42% of
the total number of individuals.)

Although we expect 21% of the chromosomes to be
of the Ab type, only 2.1%, 10% of the expected, appear in
the gene pool after one generation of random mating.
You can see that linkage equilibrium is achieved at a rate
dependent on the map distance between loci. Unlinked
genes, appearing 50 map units apart, also gradually ap-
proach linkage equilibrium.

Although we will not derive these extensions here, we
note two others. If the frequencies of alleles at an autoso-
mal locus differ in the two sexes, it takes two generations
of random mating to achieve equilibrium. In the first
generation, the allelic frequencies in the two sexes are
averaged so that each sex now has the same allelic
frequencies. Genotypic frequencies then come into
Hardy-Weinberg proportions in the second generation.
However, if the allelic frequencies differ in the two sexes
for a sex-linked locus, Hardy-Weinberg proportions are
established only gradually. The reasoning is straightfor-
ward. Females, with an X chromosome from each parent,
average the allelic frequencies from the previous genera-
tion. However, males, who get their X chromosomes from
their mothers, have the allelic frequencies of the females
in the previous generation. Hence, the allelic frequencies
are not the same in the two sexes after one generation of
random mating, and equilibrium is achieved slowly

NONRANDOM MATING

The Hardy-Weinberg equilibrium is based on the as-
sumption of random mating. Deviations from random
mating come about when phenotypic resemblance or re-

latedness influences mate choice. When phenotypic re-
semblance influences mate choice, either assortative or
disassortative mating occurs, depending on whether in-
dividuals choose mates on the basis of similarity or dis-
similarity, respectively. For example, in human beings, as-
sortative mating occurs for height — short men tend to
marry short women, and tall men tend to marry tall
women.When relatedness influences mate choice, either
inbreeding or outbreeding occurs, depending on whether
mates are more or less related than two randomly chosen
individuals from the population. An example of inbreed-
ing in human beings is marriage between first cousins.
Both types of nonrandom mating (assortative-disassortative
mating and inbreeding-outbreeding) have the same
qualitative effects on the Hardy-Weinberg equilibrium:
assortative mating and inbreeding increase homozygosity
without changing allelic frequencies, whereas disassortative
mating and outbreeding increase heterozygosity without
changing allelic frequencies.

Two differences are apparent, however, between the
effects of phenotypic resemblance and relatedness on
mate choice. First, assortative or disassortative mating
disturbs the Hardy-Weinberg equilibrium only when the
phenotype and genotype are closely related. That is, if as-
sortative mating occurs for a nongenetic trait, then the
Hardy-Weinberg equilibrium will not be distorted. In-
breeding and outbreeding affect the genome directly. A
second difference between the two types of mating is
that the effects of inbreeding or outbreeding are felt
across the whole genome, whereas the disturbances to
the equilibrium caused by assortative and disassortative
mating occur only for the particular trait being consid-
ered (and for closely linked loci). Given the similarities in
the consequences of the two types of matings, we will
concentrate our discussion on inbreeding.

Inbreeding

Inbreeding comes about in two ways: (1) the systematic
choice of relatives as mates and (2) the subdivision of a
population into small subunits, leaving individuals little
choice but to mate with relatives. We will concentrate on
inbreeding as the systematic choice of relatives as mates.
The consequences of both are similar.

Common Ancestry

An inbred individual is one whose parents are related —
that is, there is common ancestry in the family tree.
The extent of inbreeding thus depends on the degree of
common ancestry that the parents of an inbred individ-
ual share. When mates share ancestral genes, each may
pass on copies of the same ancestral allele to their off-
spring. An inbred individual can then carry identical
copies of a single ancestral allele. In other words, an in-

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561

dividual of aa genotype is homozygous and, if it is possi-
ble that the a allele from each parent is a length of DNA
originally copied from a common ancestor, the aa indi-
vidual is said to be inbred.

The first observable effect of inbreeding is the
expression of hidden recessives. In human beings, each
individual carries, on the average, about four lethal-
equivalent alleles, alleles that kill when paired to form
a homozygous genotype (box 19.1). In many, and proba-
bly most, human societies, zygotes are generally het-
erozygous for these lethal alleles because of a cultural
pattern of outbreeding, mating with nonrelatives. Rarely
does an outbred zygote receive the same recessive lethal
from each parent. Dominance acts to mask the expres-
sion of deleterious recessive alleles. But, in the process of
inbreeding, when the zygote may receive copies of the
same ancestral allele from each parent, there is a sub-
stantial increase in the probability that a deleterious al-
lele will pair to form a homozygous genotype (fig. 19.2).
Inbreeding can result in spontaneous abortions (miscar-
riages), fetal deaths, and congenital deformities. In many
species, however, inbreeding — even self-fertilization —
occurs normally. These species usually do not have the
problem with lethal equivalents that species that nor-
mally outbreed do. Through time, species that normally
inbreed have had these deleterious alleles mostly elimi-
nated, presumably by natural selection. Inbreeding has
even been used successfully for artificial selection in live-
stock and crop plants.

From our previous discussion, you can see that there
are two types of homozygosity — allozygosity, in which
two alleles are alike but unrelated (not copies of the
same ancestral allele) and autozygosity, in which two
alleles have identity by descent (i.e., are copies of the
same ancestral allele). An inbreeding coefficient, i 7 ,
can be defined as the probability of autozygosity, or the
probability that the two alleles in an individual at a given
locus are identical by descent. This coefficient can range
from zero, at which point there is no inbreeding, to one,
at which point it is certain an individual is autozygous.

Increased Homozygosity from Inbreeding

What are the effects of inbreeding on the Hardy-
Weinberg equilibrium? Let us for a moment return to
the gene pool concept to produce zygotes. Assume that
an allele drawn from this gene pool is of the A type,
drawn with a probability of p. On the second draw, the
probability of autozygosity, that is, of drawing a copy of
the same allele^, is F, the inbreeding coefficient. Thus the
probability of an autozygous AA individual is pF. On
the second draw, however, with probability (1 — i 7 ), either
the A or a allele can be drawn, with probabilities of
p 2 (\ — i 7 ) and^(l — i 7 ), respectively. Note that a second
A allele produces a homozygote that is not inbred

Ch\

O OrO QhU

O OtO

Figure 19.2 Homozygosity by descent of copies of the same
ancestral allele, a. The individual at the bottom of the pedigree
is inbred with the aa genotype.

(allozygous). If the first allele drawn was an a allele, with
probability q, then the probability of drawing the same
allele (copy of the same ancestral allele) is F, and thus the
probability of autozygosity is qF. However, the probabil-
ity of drawing an a or A allele that does not contribute to
inbreeding is (1 — i 7 ) and, therefore, the probability of an
aa or Aa genotype is q 2 (\ — i 7 ) and pq(\ — i 7 ), respec-
tively. These calculations are summarized in table 19. 5, a
summary of the genotypic proportions in a population
with inbreeding.

Several points emerge from table 19.5. First, when the
inbreeding coefficient is zero (completely random mat-
ing), the table reduces to Hardy-Weinberg proportions.
Second, compared with Hardy-Weinberg proportions, in-
breeding increases the proportion of homozygotes in the
population (identity by descent implies homozygosity).
With complete inbreeding (i 7 = 1), only homozygotes
will occur in the population.

How does inbreeding affect allelic frequencies? Recall
that an allelic frequency is calculated as the frequency of
homozygotes for one allele plus half the frequency of the
heterozygotes. Here we let^ +1 be the frequency of the
A allele after one generation of inbreeding:

Pn+i = P 2 ^ - F^+pF+ (\/2)(2pq)(l ~ i 7 )

= p\\ - i 7 ) + pF + pq(l - i 7 )
= p 2 +pq + F(p ~ p 2 ~ pq)
= pip + q) +pF(l ~ p ~ q)
= pO) + pFQS)
= P

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

BOX 19.1

The average person carries
about four lethal-equivalent
alleles that are hidden be-
cause they are recessive. Four lethal
equivalents means four alleles that
are lethal when homozygous, or
eight alleles conferring a 50% chance
of mortality when homozygous, or
any similar combination of lethal
and semilethal alleles. The exact
arrangement cannot be determined
with current analytical methods. We
arrive at the estimate of hidden de-
fective and lethal alleles by using in-
breeding data.

J. Crow and M. Kimura, in 1970,
analyzed data showing that in
Swedish families in which marriages
occurred between first cousins, be-
tween 16 and 28% of the offspring
had genetic diseases. For unrelated
parents, the comparable figure is be-
tween 4 and 6%. Therefore, it is esti-
mated that the offspring of first
cousins have an added risk of 12
to 22% of having a genetic defect.
The children of first cousins have
an inbreeding coefficient of one-

Experimental
Methods

The Determination of Lethal
Equivalents

sixteenth. Hence, a theoretical indi-
vidual who is completely inbred has
the risk of genetic defect increased
sixteenfold over an individual whose
parents are first cousins. If 100% risk
is considered 1 lethal equivalent,
then a completely inbred individual
would carry 2 to 3.5 lethal equiva-
lents (16 X 12%-16 X 22%). How-
ever, a completely inbred individual
is, in essence, a doubled gamete.
Since our interest is in the number of
deleterious alleles a normal person
carries, it is necessary to further
multiply the risk by a factor of two
to determine the number of lethal-
equivalent alleles carried by a normal
individual. The conclusion is that the
average person carries the equivalent

of four to seven alleles that would, in
the homozygous state, cause a ge-
netic defect.

A similar calculation can be
made using viability data rather than
genetic defects to determine the oc-
currence of lethal equivalents. A
study from rural France, also ana-
lyzed by Crow and Kimura, showed
that the mortality rate of offspring
of first cousins was 25%, whereas
the analogous figure for the off-
spring of unrelated parents was
about 12%, an increased risk of 13%
for the offspring of cousins. Multi-
plying this risk figure of 0.13 by 32
(16 X 2) presents a figure of four
lethal equivalents per average per-
son in the population. In 1971, L.
Cavalli-Sforza and W. Bodmer, using
data primarily from Japanese popu-
lations, reported an estimate of
about two lethal equivalents per av-
erage person. Despite some inter-
population differences in these esti-
mates, they are about the same
order of magnitude — two to seven
lethal equivalents per person.

Thus, inbreeding does not change allelic frequencies.
We can also see intuitively that inbreeding affects
zygotic combinations (genotypes), but not allelic
frequencies: Although inbreeding may determine the
genotypes of offspring, inbreeding does not change
the numbers of each allele that an individual transmits
into the next generation.

In summary, inbreeding causes an increase in ho-
mozygosity, affects all loci in a population equally, and, in
itself, has no effect on allelic frequencies, although it can
expose deleterious alleles to selection. The results of in-
breeding are evident in the appearance of recessive
traits that are often deleterious. Inbreeding increases
the rate of fetal deaths and congenital malformations in
human beings and in other species that normally out-
breed. In outbred agricultural crops and farm animals,
decreases in size, fertility, vigor, and yield often result
from inbreeding. Once deleterious traits appear due to
inbreeding, natural selection can cause their removal
from the population. However, in species adapted to in-
breeding, including many crop plants and farm animals,

inbreeding does not expose deleterious alleles because
those alleles have generally been eliminated already.

Pedigree Analysis

Path Diagram Construction

The inbreeding coefficient, F, of an individual (the prob-
ability of autozygosity) can be determined by pedigree
analysis. This is done by converting a pedigree to a path
diagram by eliminating all extraneous individuals, those
who cannot contribute to the inbreeding coefficient of
the individual in question. A path diagram shows the di-
rect line of descent from common ancestors. An example
of the conversion of a pedigree to a path diagram is
shown in figure 19. 3, in which individuals C and F are
omitted from the path of descent because they are not re-
lated to anyone on the other side of the family tree and,
therefore, do not contribute to the "common ancestry" of
individual I. The pedigree in figure 19. 3 shows an off-
spring who is the daughter of first cousins. Since first

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Table 19.5

Genotypic Proportions

in a

Populat

ton

with

Inbreeding

Genotype

Due to Random Mating

(1-/0

Due to Inbreeding
(F)

Observed
Proportions

AA
Aa
aa
Total

p 2 a-F)

2pq(\ - F)
q\l ~ F}

+
+

P 2 + Fpq

2pq(\ - F)

q 2 + Fpq

(P 2

+ 2pq + q 2 } (1

-iO

+
+

(P + q)F
F

cousins are the offspring of siblings, they share a set of
common grandparents. Thus, individual I can be autozy-
gous for alleles from either ancestor A or B, her great-
grandparents. The path diagram shows the only routes by
which autozygosity can occur.

The inbreeding coefficient of the offspring of first
cousins can be calculated as follows. The path diagram of
figure 193 is shown again in figure 19.4, with lowercase
letters designating gametes. Two paths of autozygosity
appear in this diagram, one path for each grandparent as
a common ancestor: A to D and E, then to G and H, and
finally to I; or B to D and E, then to G and H, and finally
to I.

In the path with A as the common ancestor, A con-
tributes a gamete to D and a gamete to E. The probability
is one-half that D and E each carry a copy of the same al-
lele. That is, there are four possible allelic combinations
for the two gametes, a 1 and a 2 : A-A; A-a; a-A; and a-a. Of
these combinations, the first and last (A-A and a-a) give a
copy of the same allele to the two offspring, D and E, and
can thus contribute to autozygosity. The probability that
gametes a x and d carry copies of the same allele is one-
half, and the probability that d and g carry copies of the
same allele is also one-half. Similarly, on the other side of
the pedigree, the probability is one-half that a 2 and e
carry copies of the same allele and one-half that e and h
carry copies of the same allele. Thus, the overall proba-
bility that the alleles that g and h carry are identical by
descent (autozygous) is (1/2) 5 . In general, it would be
(\/2) n for each path, where n is the number of ancestors
in the path.

You may have spotted an additional factor here. Of
the possible combinations of allelic copies passed on to
D and E, one-half (A-A and a-d) are autozygous combina-
tions. However, the other half of the combinations, A-a
and a-A, can lead to autozygosity if A is itself inbred. If
we let F A be the inbreeding coefficient of A (the prob-
ability that any two alleles at a locus in A are identical
by descent), then F A is the probability that the A-a and
a-A combinations are also autozygous. Thus, the prob-
ability that a common ancestor, A, passes on copies of an

Pedigree

Path diagram

Figure 19.3 Conversion of a pedigree to a path diagram.
This pedigree depicts the mating of first cousins. In the path
diagram, all extraneous individuals are removed, leaving only
those who could contribute to the inbreeding of individual I.
Individuals in the line of descent are connected directly with
straight lines, indicating the paths along which gametes are
passed.

Figure 19.4 The path diagram of the mating of first cousins
with gametes labeled in lowercase letters.

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

identical ancestral allele is 1/2 + (1/2)/% or (1/2) (1 + F A ).
In other words, there is a one-half probability that the al-
leles transmitted from A to D and E are copies of the
same allele. In the other half of the cases, these alleles
can be identical if A is inbred. The probability of identity
of As two alleles is F A . The expression for the inbreeding
coefficient of I, F h can now be changed from (l/2) n by
substituting (1/2)(1 + ,F A ) for one of the (l/2)s to

F l = (l/2) n (l + i? A )

This equation accounts only for the inbreeding of I by
the path involving the common ancestor, A, and does not
account for the symmetrical path with B as the common
ancestor. To obtain the total probability of inbreeding,
the values from each path must be added (because these
are mutually exclusive events; see chapter 4). Thus the
complete formula for the inbreeding coefficient of the
offspring of first cousins is

Fj = 2 [(1/2/(1 + Fj)]

(19.1)

in which F 1 is the probability that the two alleles in I are
identical by descent, n is the number of ancestors in a
given path, Fj is the inbreeding coefficient of the com-
mon ancestor of that path, and all paths are summed.

In the example of the mating of first cousins
(fig. 19.4)

F, = (1/2) 5 (1 + F A ) + (1/2) 5 (1 + F B )

If we assume that F A and F B are zero (which we must as-
sume when the pedigrees of A and B are unknown), then

F l = 2(l/2) 5 = (1/2) 4 = 0.0625

This can be interpreted to mean that about 6.25% of in-
dividual I's loci are autozygous, or that there is a 6.25%
chance of autozygosity at any one of I's loci.

The inbreeding coefficient of the offspring of siblings
(fig. 195) can also be calculated, assuming that A and B
are not themselves inbred (F A and F B are zero), as

Fi = 2(l/2) 3 = 0.25

Pedigree

Path diagram

Figure 19.5 Conversion of a sib-mating pedigree to a path
diagram. Individual I is inbred.

Thus, about 25% of the loci in an offspring of siblings are
autozygous.

Path Diagram Rules

The following points should be kept in mind when cal-
culating an inbreeding coefficient:

1 . All possible paths must be counted. A path is possible
if gametes can actually pass in that direction. Paths
that violate the rules of inheritance cannot be used.
For example, in figure 19.4, the following path is un-
acceptable: I G E A D H I.

2. In any path, an individual can be counted only once.

3. Every path must have one and only one common an-
cestor. The inbreeding coefficient of any other indi-
vidual in the path is immaterial.

In figure 19.6, we present a complex pedigree pro-
duced from repeated sib mating, a pattern found in
livestock and laboratory animals. This pedigree has two
interesting points. First, common ancestors occur in sev-
eral different generations. Second, some of the paths are
complex. Thus, we must be sure to count all paths
(paths 5 and 6 might not be immediately obvious). Al-
though not shown in figure 19.6, one of the common an-
cestors, A, is also inbred (F A = 0.05) — a fact that we
must take into consideration in paths 3 and 5. Thus, F I is
as follows:

From path 1: (1/2) 3

From path 2: (1/2) 3

From path 3: (1/2) 5 (1 + 0.05)

From path 4: (1/2) 5

From path 5: (1/2) 5 (1 + 0.05)

From path 6: (1/2) 5

= 0.1250
= 0.1250
= 0.0328

= 0.0313
= 0.0328
= 0.0313

F 1 = 0.3782

Population Analysis

It is also possible to define the inbreeding coefficient, F,
of a population as the relative reduction in heterozygos-
ity in the population due to inbreeding. In an individual,
F is the probability of autozygosity; it represents an in-
crease in homozygosity, which is therefore a decrease in
heterozygosity. In a population, it also represents the re-
duction in heterozygosity. From the definition, we can
calculate the population F as follows:

F =

Qpq - H}
2pq

where H is the actual proportion of heterozygotes in a
population, and 2pq is the expected proportion of het-

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erozygotes based on Hardy-Weinberg proportions. This
equation reduces to

F = 1 -

2pq

(19.2)

This equation shows that when H = 2pq, F is zero,
meaning that there is no decrease in heterozygotes
and therefore, apparently, no inbreeding. When there
are no heterozygotes, F = 1. This could be the case in
a completely inbred population — for example, a self-
fertilizing plant species.

As an example of an intermediate case, take the sam-
ple of one hundred individuals segregating the^ and^4 2
alleles at the A locus: A 1 A 1 , fifty-four; A^ 2 , thirty-two; and
A 2 A 2 , fourteen. In this example, p = 0.7, q = 0.3, and
H = 0.32. Since 2pq = 0.42, H/2pq = 0.32/0.42 = 0.76,
and F = 1 — 0.76, or 0.24. Thus, the inbreeding coeffi-
cient of this population is 0.24; there is a 24% reduction
in heterozygotes, due presumably to inbreeding.

K)b

kO°

I
Pedigree

A | J
C| [
E

Path diagram

Paths

D C
F E

/
/

A I I

(1)

(2)

(4)

f Vi

—^->

f f

(6)

Figure 19.6 Pedigree and path diagram of two generations of sib matings. The six paths involving the
potential for autozygosity are shown. F A = 0.05. The paths involve common ancestors in two generations.

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Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

SUMMARY

STUDY OBJECTIVE 1: To understand the concept of pop-
ulation-level genetic processes 553-554

In a large, randomly mating population of sexually repro-
ducing diploid organisms, not subject to the influences of
mutation, migration, or selection, an equilibrium will be
achieved for an autosomal locus with two alleles.

STUDY OBJECTIVE 2: To learn the assumptions and na-
ture of the Hardy-Weinberg equilibrium and its exten-
sions 554-557

The Hardy-Weinberg equilibrium predicts that (1) allelic
frequencies (p, q) will not change from generation to
generation; (2) genotypes will occur according to the bino-
mial distribution p 2 = f(AA), 2pq =f(Ad), and q 2 = fiaay,
and (3) if perturbed, equilibrium will reestablish itself in
just one generation of random mating.

STUDY OBJECTIVE 3: To test whether a population is in
Hardy-Weinberg equilibrium 557-560

To determine whether a population is in Hardy-Weinberg
proportions, the observed and expected distribution of
genotypes can be compared by the chi-square statistical
test. In some circumstances, when it is reasonable to as-
sume equilibrium, we can estimate allelic and genotypic
frequencies even when dominance occurs. The Hardy-
Weinberg equilibrium is easily extended to the prediction
of the frequencies of multiple alleles, multiple loci, and dif-
ferent frequencies of alleles in the two sexes, for both sex-
linked and autosomal loci.

STUDY OBJECTIVE 4: To analyze the process and conse-
quences of nonrandom mating in diploid populations
560-565

Random mating is required for the Hardy-Weinberg equilib-
rium to hold. Deviations from random mating fall into two
categories, depending on whether phenotypic resemblance
or relatedness is involved in mate choice. Phenotypic re-
semblance is the basis for assortative and disassortative mat-
ing, in which individuals choose similar or dissimilar mates,
respectively. Assortative mating causes increased homozy-
gosity only among loci controlling the traits that influence
mate choice. There are no changes in allelic frequencies.
Similarly, disassortative mating causes increased heterozy-
gosity without changing allelic frequencies.

Mating among relatives, or inbreeding, is represented
by F, the inbreeding coefficient, which measures the prob-
ability of autozygosity (homozygosity by descent). It can be
calculated from pedigrees by using the formula

F=£[(l/2)f(l + Fj)]

where n is the number of ancestors in a given path and Fj is
the inbreeding coefficient of the common ancestor of that
path. Inbreeding exposes recessive deleterious traits al-
ready present in the population and causes homozygosity
throughout the genome. It does not, by itself, change allelic
frequencies. F can also be calculated from the reduction in
heterozygosity in a population.

SOLVED PROBLEMS

PROBLEM 1: One hundred fruit flies (Drosophila
melanogaster) from California were tested for their
genotype at the alcohol dehydrogenase locus using
starch-gel electrophoresis. Two alleles were present, S
and F, for slow and fast migration, respectively. The fol-
lowing results were noted: SS, sixty-six; SF, twenty; FF,
fourteen. What are the allelic and genotypic frequen-
cies in this population?

Answer: Since the sample size is one hundred, the pro-
portions of the three genotypes, SS, SF, and FF, are 0.66,
0.20, and 0.14, respectively. We can calculate allelic fre-
quencies directly from these genotypes, remembering
that the frequency of an allele is the frequency of its ho-
mo zygote plus half the frequency of the heterozygote, or

P = /W = f(SS) + (l/2)/CSiO
= 0.66 + (l/2)(0.20) = 0.76

q = f^ = f(FF) + (l/2)/CSiO
= 0.14 + (l/2)(0.20) = 0.24

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Exercises and Problems

567

Alternatively, we could get allelic frequencies by count-
ing alleles. Thus,

2 X number of SS + number of SF

2 X total number
2(66) + 20 152

2(100)

200

= 0.76

2 X number of FF + number of SF

2 X total number
2(14) + 20 48

2(100)

200

= 0.24

PROBLEM 2: Is the population described in problem 1 in
Hardy-Weinberg equilibrium?

Answer: We can determine whether the numbers of the
three genotypes (SS, SF, and FF} are in Hardy-Weinberg
proportions through the chi-square statistical test. The ob-
served numbers of the three genotypes are sixty-six,
twenty, and fourteen, respectively. Using allelic frequen-
ciesofp = f(S) = 0.76 and q = f(F) = 0.24, we expect^? ,
2pq, and q 1 , respectively, of the three genotypes. That is,

p 2 = (0.76) 2 = 0.5776, or 57.76 in 100

2pq = 2(0.76)(0.24) = 0.3648, or 36.48 in 100

q 2 = (0.24) 2 = 0.0576, or 5.76 in 100

We can now set up a chi-square table as follows:

Total

Observed Numbers

100

Expected Proportions

P 2

2pq

q 2

1.0

(0.5776)

(0.3648)

(0.0576)

1.0

Expected Numbers

57.76

36.48

5.76

100

X 2 = (O - Ef/E

1.176

7.445

11.788

20.408

The critical chi-square value (0.05 at one degree of free-
dom) is 3841, so we reject the hypothesis that this pop-
ulation is in Hardy-Weinberg proportions. From inspec-
tion of the table, it appears that there are too few
heterozygotes and too many homozygotes, indicating
that inbreeding could be the cause of the discrepancy.

PROBLEM 3: Convert the pedigree in figure 19.2 into a
path diagram, and determine the inbreeding coefficient
of the inbred individual, assuming that the common an-
cestors are not themselves inbred.

Answer: There are two paths (see the figure), each with
seven ancestors. Thus, the inbreeding coefficient is

F = 2 [(l/2) w (l + Fj)] = 2(l/2) 7 = 0.016

Hence, the inbreeding coefficient is 0.016; about 1.6% of
the loci of the inbred individual are autozygous.

EXERCISES AND PROBLEMS

HARDY-WEINBERG EQUILIBRIUM

1. One hundred persons from a small town in Pennsyl-
vania were tested for their MN blood types. Is the
population they represent in Hardy-Weinberg pro-
portions? The genotypic data are: MM, forty-one;
MN, thirty-eight; and NN, twenty-one.

2. From the following two sets of data, calculate allelic
and genotypic frequencies, and determine whether

Answers to selected exercises and problems are on page A-21.

the populations are in Hardy-Weinberg proportions.
Do a statistical test if one is appropriate.

a. Allele A is dominant to a; A-, 91; aa, 9.

b. Electrophoretic alleles F and S are codominant at
the malate dehydrogenase locus in Drosophila;
FF, 137; FS, 196; SS, 87.

3. The dominant ability to taste PTC comes from the al-
lele T. Among a sample of 215 individuals from a
population in Vancouver, 150 could detect the taste
of PTC, and 65 could not. Calculate the allelic fre-
quencies of T and t. Is the population in Hardy-
Weinberg proportions?

Tamarin: Principles of
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IV. Quantitative and
Evolutionary Genetics

19. Population Genetics:
The Hardy-Weinberg
Equilibrium and Mating
Systems

568

Chapter Nineteen Population Genetics: The Hardy-Weinberg Equilibrium and Mating Systems

4. The frequency of children homozygous for the re-
cessive allele for cystic fibrosis is about one in
twenty-five hundred. What is the percentage of het-
erozygotes in the population?

5. PTC tasting is dominant in human beings.

a. Should most human populations be heading to-
ward a 3:1 ratio of tasters to nontasters? Explain.

b. Confronted with a population sample of human
beings of unknown origin, would you expect
more or less than half the sample to be tasters?

6. Graph the relationship of the proportions of geno-
type (AA, Aa, ad) as allelic frequencies change.

7. A particular recessive disorder is present in one in
ten thousand individuals. If the population is in
Hardy-Weinberg equilibrium, what are the frequen-
cies of the two alleles?

8. What allelic frequency will generate twice as many
recessive homozygotes as heterozygotes?

9. Assume brown eye color is the result of a dominant
allele at one locus. Attack or defend mathematically
the following statement: With time, the frequency of
brown-eyed individuals will increase, until about
three out of four individuals are brown-eyed.

10. A particular human population has five hundred MM
individuals, three hundred MN, and seven hundred
NN. Calculate the allelic frequencies, and determine
whether the population is in Hardy-Weinberg equi-
librium.

11. Assume random mating occurs among the individu-
als of the population described in problem 10. What
will be the frequency of each type of individual in
the next generation?

12. On a small island, 235 mating individuals are all true-
breeding for brown eyes. An epidemic eliminates all
the population except ten young women, two young
men, and four older (postmenopausal) women. A
boatload of foreigners arrives; the foreign popula-
tion consists of six heterozygous brown-eyed fe-
males, four homozygous brown-eyed males, and ten
blue-eyed males. Assuming that one locus controls
eye color, that mating is random with respect to eye
color, and that each male and female capable of
breeding does so, calculate the genotypic frequen-
cies of their offspring.

13. In a given population, only the I A and I B alleles are
present in the ABO system; there are no individuals
with type O blood or with i alleles. If two hundred
people have type A blood, seventy-five have type AB
blood, and twenty-five have type B blood, what are
the allelic frequencies in this population?

EXTENSIONS OF HARDY-WEINBERG EQUILIBRIUM

14. The following data are ABO phenotypes from a pop-
ulation sample of one hundred persons. Determine
the frequencies of the three alleles: type A, seven;
type B, seventy-two; type AB, twelve; type O, nine.
What do you have to assume? Is the population in
Hardy-Weinberg proportions?

15. How quickly and in what manner is Hardy-Weinberg
equilibrium achieved under the following initial con-
ditions (assuming a diploid, sexually reproducing
population)?

a. One locus, five alleles

b. Two unlinked loci, two alleles each

16. A sample of fruit flies was testcrossed to determine
the allelic arrangements of two linked loci in the ga-
metes of that generation. With the following data,
can you determine whether linkage equilibrium
holds? Gametic arrangements are AB, fifty-eight; ab,
eight; Ab, twelve; and aB, twenty-two.

17. In a large, randomly mating human population, the
frequencies of the I A , I B , and i alleles are 0.7, 0.2,
and 0.1, respectively. Calculate the expected fre-
quencies for each blood type.

18. In a human population of one hundred people, sev-
enteen have type A blood, seventeen have type B,
two have type AB, and sixty-four have type O. If this
population is in equilibrium, what are the allelic fre-
quencies?

NONRANDOM MATING

19. Under what circumstances is inbreeding deleterious?

20. What is the inbreeding coefficient of I in the follow-
ing pedigree? Assume that the inbreeding coeffi-
cients of other members of the pedigree are zero un-
less other information tells you differently.

21. What is the inbreeding coefficient of individual I in
this pedigree? F A = 0.01; F B = 0.02; F c = 0.02.

) ^N.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

19. Population Genetics:
The Hardy-Weinberg
Equilibrium and Mating
Systems

Critical Thinking Questions

569

22. The following is the pedigree of an offspring pro-
duced by the mating of half siblings. Individuals A
and C have inbreeding coefficients of 0.2; all others
are zero. Convert the pedigree to a path diagram and
determine the inbreeding coefficient of individual G.

23. Given the population in Exercises and Problems
problem 1 , what is its inbreeding coefficient?

24. In a sample of one hundred people, are fourteen
MM, thirty-two MN, and fifty-four NN individuals.
Calculate the inbreeding coefficient.

25. If, in a population with two alleles at an autosomal
locus, p = 0.8, q = 0.2, and the frequency of het-
erozygotes is 0.20, what is the inbreeding coeffi-
cient?

CRITICAL THINKING QUESTIONS

1. Prove that two generations are needed for the establish-
ment of Hardy-Weinberg proportions when an autoso-
mal locus with two alleles in a sexually reproducing
species has frequencies of the two alleles that differ in
the two sexes.

2. What might the ramifications to conservation efforts be
of zoos maintaining captive breeding programs for rare
and endangered species?

Suggested Readings for chapter 19 are on page B-19.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

POPULATION
GENETICS

Processes That

Change Allelic

Frequencies

Natural selection works on the variation found in

nature, shown here by different banding patterns in

tree snails (Liguus fasciatus), found mainly in southern

STUDY OBJECTIVES

1. To develop ways to analyze population genetics
problems 571

2. To analyze the effects of mutation, migration, and population
size on the Hardy-Weinberg equilibrium 571

3. To study the ways in which natural selection results in
organisms adapted to their environments 577

STUDY OUTLINE

Models for Population Genetics 571
Mutation 571

Mutational Equilibrium 571

Stability of Mutational Equilibrium 571
Migration 573
Small Population Size 574

Sampling Error 574

Simulation of Random Genetic Drift 575

Founder Effects and Bottlenecks 576
Natural Selection 577

How Natural Selection Acts 577

Selection Against the Recessive Homozygote 578

Selection-Mutation Equilibrium 581

Types of Selection Models 581
Summary 585
Solved Problems 585
Exercises and Problems 586
Critical Thinking Questions 587

Box 20.1 A General Computer Program to Simulate the
Approach to Allelic Equilibrium Under
Heterozygous Advantage 583

570

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Mutation

571

We continue our discussion of the genetics
of the evolutionary process. This chapter
is devoted to a discussion of some of the
effects of violating, or relaxing, the as-
sumptions of the Hardy-Weinberg equi-
librium other than random mating, which we discussed in
chapter 19. Here we consider the effects of mutation, mi-
gration, small population size, and natural selection on the
Hardy-Weinberg equilibrium. These processes usually
change allelic frequencies.

MODELS FOR POPULATION
GENETICS

The steps we need to take to solve for equilibrium in
population genetics models follow the same general pat-
tern regardless of what model we are analyzing. We em-
phasize that these models were developed to help us
understand the genetic changes taking place in a popula-
tion. The models shed light on nonintuitive processes
and help quantify intuitive processes. The steps in the
models can be outlined as follows:

1. Set up an algebraic model.

2. Calculate allelic frequency in the next generation,

3. Calculate change in allelic frequency between genera-
tions, Aq.

4. Calculate the equilibrium condition, q (^-hat), at
Aq = 0.

5. Determine, when feasible, if the equilibrium is stable.

MUTATION

Mutational Equilibrium

Mutation affects the Hardy-Weinberg equilibrium by
changing one allele to another and thus changing allelic
and genotypic frequencies. Consider a simple model in
which two alleles,^ and a, exist. A mutates to a at a rate
of fx (mu), and a mutates back to A at a rate of v (nu):

A^a
v

lfp n is the frequency of A in generation n and q n is the
frequency of a in generation n, then the new frequency
of a, q n+1 , is the old frequency of a plus the addition of
a alleles from forward mutation and the loss of a alleles
by back mutation. That is,

tfn + 1 ~ <ln "r \^P

(20.1)

in which [xp n is the increment of a alleles added by for-
ward mutation, and vq n is the loss of a alleles due to
back mutation. Equation 20.1 takes into account not
only the rate of forward mutation, |x, but also^, the fre-
quency of A alleles available to mutate. Similarly, the loss
of a to A alleles is the product of both the rate of back
mutation, v, and the frequency of the a allele, q n . Equa-
tion 20.1 completes the second modeling step, deriva-
tion of an expression for q n+ i, allelic frequency after
one generation of mutation pressure. The third step is to
derive an expression for the change in allelic frequency
between two generations. This change (A#) is simply the
difference between the allelic frequency at generation
n + 1 and the allelic frequency at generation ^.Thus, for
the a allele

Aq = q n+1 - q n = (q n + \Lp n - vqj - q n

which simplifies to

Aq = [Lp n

(20.2)

(20.3)

The next step in the model is to calculate the equilib-
rium condition q, or the allelic frequency when there is
no change in allelic frequency from one generation to
the next — that is, when Aq (equation 20.3) is equal to
zero:

Aq = \Lp n - vq n =

(20.4)

Thus,

VPn = V <ln (20.5)

Then, substituting (1 — q„) forp n (since p = 1 — q), gives

|x(l-#„) = vq,
or, by rearranging:

M-

[X + V

(20.6)

And, since p + q = 1

|JL + V

(20.7)

We can see from equations 20.6 and 20.7 that an equilib-
rium of allelic frequencies does exist. Also, the equilib-
rium value of allele a iq) is directly proportional to the
relative size of \x, the rate of forward mutation toward a.
If fi = v, the equilibrium frequency of the a allele (q) will
be 0.5. As fi gets larger, the equilibrium value shifts to-
ward higher frequencies of the a allele.

Stability of Mutational Equilibrium

Having demonstrated that allelic frequencies can reach
an equilibrium due to mutation, we can ask whether the
mutational equilibrium is stable. A stable equilibrium is

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

572

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

one that returns to the original equilibrium point after
being perturbed. An unstable equilibrium is one that will
not return after being perturbed but, rather, continues to
move away from the equilibrium point. As we mentioned
in the last chapter, the Hardy-Weinberg equilibrium is a
neutral equilibrium: It remains at the allelic frequency it
moved to when perturbed.

Stable, unstable, and neutral equilibrium points can
be visualized as marbles in the bottom of a concave sur-
face (stable), on the top of a convex surface (unstable), or
on a level plane (neutral; fig. 20.1). Although more so-
phisticated mathematical formulas exist for determining
whether an equilibrium is stable, unstable, or neutral, we
will use graphical analysis for this purpose.

Figure 20.2 introduces the process of graphical analy-
sis, which provides an understanding of the dynamics of
an event or process by representing the event in graphi-
cal form. In figure 20.2, we have graphed equation 20.3,
the Aq equation of mutational dynamics. The ordinate, or
y-axis, is Aq, the change in allelic frequency. The abscissa,
or x-axis, is q, or allelic frequency. The diagonal line is the
Aq equation, the relationship between Aq and q. Note
that Aq can be positive (q is increasing) or negative (q is
decreasing), whereas q is always positive (0-1.0). Graph-
ical analysis can provide insights into the dynamics of
many processes in population genetics.

The diagonal line in figure 20.2 crosses the Aq =
line at the equilibrium value (#) of 0.167. This line also
shows us the changes in allelic frequency that occur in
a population not at the equilibrium point. We will look

Stable

at two examples of populations under the influence of
mutation pressure, but not at equilibrium: one at q =
0.1 (below equilibrium) and one at q = 0.9 (above
equilibrium).

If we substitute q = 0.1 into equation 20.3, we get a
Aq value of 4 X 10~ 6 . If we substitute q = 0.9 into the
equation, we get a Aq value of —4A X 10~ 5 . In other
words, when the population is below equilibrium, q in-
creases (Aq = + 4 X 10~ 6 ); if the population is above
equilibrium, q decreases (Aq = — 4A X 10~ 5 ). We can
read these same conclusions directly from the graph in
figure 20.2.

We can see that the mutational equilibrium is a stable
one. Any population whose allelic frequency is not at the
equilibrium value tends to return to that equilibrium
value. A shortcoming of this model is that it provides no
obvious information revealing the time frame for reach-
ing equilibrium. To derive the equations needed to deter-
mine this parameter is beyond our scope. (We could use
computer simulation or integrate equation 20.3 with re-
spect to time.) In a large population, any great change in
allelic frequency caused by mutation pressure alone
takes an extremely long time. Most mutation rates are on

/JL = 10" 5

v =5X10" 5

q =jjL/{fi + v) = 1/6 = 0.167

Neutral

Figure 20.1 Types of equilibria: stable, unstable, and neutral.

Figure 20.2 Graphical analysis of mutational equilibrium. The
graph of the mutational Ag equation shows that when the
population is perturbed from the equilibrium point (g = 0.167),
it returns to that equilibrium point. At q values above
equilibrium, change is negative, tending to return the population
to equilibrium. At q values below equilibrium, change is
positive, also tending to return the population to equilibrium.

Tamarin: Principles of
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IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Migration

573

the order of 10~ 5 , and equation 20.3 shows that change
will be very slow with values of this magnitude. For ex-
ample, if fx = 10~ 5 , v = 10~ 6 , and p = q = 0.5, Aq =
(0.5 X 10" 5 ) - (0.5 X 10" 6 ) = 4.5 X 10" 6 , or 0.0000045.
It usually takes thousands of generations to get near equi-
librium, which is approached asymptotically.

As you can see from the low values of mutation rates,
it would usually be nearly impossible to detect perturba-
tions to the Hardy-Weinberg equilibrium by mutation in
any one generation. The mutation rate can, however, de-
termine the eventual allelic frequencies at equilibrium if
no other factors act to perturb the gradual changes that
mutation rates cause. Mutation can also affect final allelic
frequencies when it restores alleles that natural selection
is removing, a situation we will discuss at the end of the
chapter. More important, mutation provides the alterna-
tive alleles that natural selection acts upon.

Natives

Migrants

MIGRATION

Migration is similar to mutation in the sense that it adds
or removes alleles and thereby changes allelic frequen-
cies. Human populations are frequently affected by mi-
gration.

Assume two populations, natives and migrants, both
containing alleles A and a at the A locus, but at different
frequencies (p N and q N versus p M and # M ), as shown in
figure 20.3. Assume that a group of migrants joins the na-
tive population and that this group of migrants makes up
a fraction m (e.g., 0.2) of the new conglomerate popula-
tion. Thus, the old residents, or natives, will make up a
proportionate fraction (1 — m; e.g., 0.8) of the combined
population. The conglomerate a-allele frequency, q c , will
be the weighted average of the allelic frequencies of the
natives and migrants (the allelic frequencies weighted —
multiplied — by their proportions):

q c = mq M + (1 - m)q n
q c = #n + w(#m - #n)

(20.8)
(20.9)

The change in allelic frequency, a, from before to after
the migration event is

kq = q c ~ q N = [q N + m(q M - # N )]

Aq = m(q m - # N )

<2n

(20.10)
(20.11)

We then find the equilibrium value, q (at Aq = 0). Re-
membering that, in a product series, any multiplier with
the value of zero makes the whole expression zero, Aq
will be zero when either

or q M - q N = 0; q M = q N

The conclusions we can draw from this model are in-
tuitive. Migration can upset the Hardy-Weinberg equilib-

Conglomerate

Figure 20.3 Diagrammatic view of migration. A group of
migrants enters a native population, making up a proportion,
m, of the final conglomerate population.

rium. Allelic frequencies in a population under the influ-
ence of migration will not change if either the size of the
migrant group drops to zero (m, the proportion of the
conglomerate made up of migrants, drops to zero) or
the allelic frequencies in the migrant and resident groups
become identical.

This migration model can be used to determine the
degree to which alleles from one population have entered
another population. It can analyze the allele interactions
in any two populations. We can, for example, analyze the
amount of admixture of alleles from Mongol populations
with eastern European populations to explain the rela-
tively high levels of blood type B in eastern European
populations (if we make the relatively unrealistic as-
sumption that each of these groups is homogeneous).
The calculations are also based on a change happening
all in one generation, which did not happen. Blood type
and other loci can be used to determine allelic frequen-
cies in western European, eastern European, and Mongol
populations. We can rearrange equation 20.9 to solve for
m, the proportion of migrants:

ffc — #N

(20.12)

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

574

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

From one sample, we find that the B allele is 0.10 in west-
ern Europe, taken as the resident or native population (# N );
0.12 in eastern Europe, the conglomerate population (q^;
and 0.21 in Mongols, the migrants (# M )- Substituting these
values into equation 20.12 gives a value for m of 0.18. That
is, given the stated assumptions, 18% of the alleles in the
eastern European population were brought in by genetic
mixture with Mongols.

When a migrant group first joins a native group, be-
fore genetic mixing (mating) takes place, the Hardy-
Weinberg equilibrium of the conglomerate population is
perturbed, even though both subgroups are themselves
in Hardy-Weinberg proportions. A decrease will occur in
heterozygotes in the conglomerate population as com-
pared to what we would predict from the allelic fre-
quencies of that population (the average allelic frequen-
cies of the two groups). This is a phenomenon of
subdivision referred to as the Wahlund effect. The rea-
son this happens is because the relative proportions of
heterozygotes increase at intermediate allelic frequen-
cies. As allelic frequencies rise above or fall below 0.5,
the relative proportion of heterozygotes decreases.

In a conglomerate population, the allelic frequencies
will be intermediate between the values of the two
subgroups because of averaging. This generally means
the predicted proportion of heterozygotes will be higher
than the actual average proportion of heterozygotes in
the two subgroups. An example is worked out in table
20.1. Assume that the two subgroups each make up 50%
of the conglomerate population. In subgroup \,p = 0.1
and q = 0.9; in subgroup 2,p = 0.9 and q = 0.1. Each
subgroup will have 18% heterozygotes. The average,
(0.18 + 0.18)/2 = 0.18, is the proportion of heterozy-
gotes actually in the population. However, the conglom-
erate allelic frequencies are^? = 0.5 and q = 0.5, leading
to the expectation that 50% of the population will be
heterozygotes. Hence, the observed frequency of net-

Table 20.1 The Wahlund Effect: Heterozygote

Frequencies Are Below Expected
in a Conglomerate Population

Subgroup I

Subgroup 2

Conglomerate

0.1

0.9

0.5

0.9

0.1

0.5

P 2

0.01

0.81

Expected

Observed

0.25

0.41

2pq

0.18

0.50

0.18

q 2

0.81

0.01

0.25

0.41

erozygotes is lower than the expected frequency (i.e.,
the Wahlund effect).

We should note that the same logic holds even if both
populations have allelic frequencies above or below 0.5.
Also, this effect happens when an observer samples what
he or she thinks is a single population but is actually a
population subdivided into several demes. When most
population geneticists sample a population and find a de-
ficiency of heterozygotes, they first think of inbreeding
and then of subdivision, the Wahlund effect. (A further
complication is that inbreeding leads to subdivision, and
subdivision leads to inbreeding. Statistics have been
developed to try to separate the effects of these two phe-
nomena.) As soon as random mating occurs in a subdi-
vided population, Hardy-Weinberg equilibrium is estab-
lished in one generation. We refer to a population in
which the individuals are mating at random as unstruc-
tured or panmictic.

Note: In this example, the subgroups are of equal sizes.

SMALL POPULATION SIZE

Another variable that can upset the Hardy-Weinberg equi-
librium is small population size. The Hardy-Weinberg equi-
librium assumes an infinitely large population because, as
defined, it is deterministic, not stochastic. That is, the
Hardy-Weinberg equilibrium predicts exactly what the al-
lelic and genotypic frequencies should be after one gener-
ation; it ignores variation due to sampling error. Obviously,
every population of organisms on earth violates the Hardy-
Weinberg assumption of infinite population size.

Sampling Error

The zygotes of every generation are a sample of gametes
from the parent generation. Sampling errors are the
changes in allelic frequencies from one generation to
the next that are due to inexact sampling of the alleles of
the parent generation. Toss a coin one hundred times, and
chances are, it will not land heads exactly fifty times. How-
ever, as the number of coin tosses increases, the percent-
age of heads will approach 50%, a percentage reached
with certainly only after an infinite number of tosses. The
same applies to any sampling problem, from drawing
cards from a deck to drawing gametes from a gene pool.

If small population size is the only factor causing devi-
ation from Hardy-Weinberg equilibrium, it will cause the
allelic frequencies of a population to fluctuate from gener-
ation to generation in the process known as random ge-
netic drift. In other words, an Aa heterozygote will some-
times produce several offspring that have only the A allele,
or sometimes random mortality will kill a disproportionate
number of aa homozygotes. In either case, the next gen-
eration may not have the same allelic frequencies as the

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Small Population Size

575

Figure 20.4 Random genetic drift. Ten populations, each
consisting of two individuals with initial q = 0.5, all go to fixation
or loss of the a allele (four or zero copies) within ten generations
due to the sampling error of gametes. Once the a allele has been
fixed or lost, no further change in allelic frequency will occur
(barring mutation or migration). We show a population of only two
individuals to exaggerate the effects of random genetic drift.

present generation. The end result will be either fixation
or loss of any given allele (q = 1 or q = 0; fig. 20.4), al-
though which will be fixed or lost depends on the original
allelic frequencies. The rate of approach to reach the fixa-
tion-loss endpoint depends on the size of the population.

Simulation of Random Genetic Drift

We can investigate the process of random genetic drift
mathematically by starting with a large number of popula-
tions of the same finite size and observing how the distri-
bution of allelic frequencies among the populations
changes in time due only to random genetic drift. For ex-
ample, we can start with one thousand hypothetical pop-
ulations, each containing one hundred individuals, with
the frequency of the a allele, q, 0.5 in each (fig. 20.5). We
measure time in generations, t, as a function of the popu-
lation size, N (one hundred in this example). For instance,
t = N is generation one hundred, t = N/5 is generation
twenty, and t = 3N is generation three hundred. Then, by
using computer simulation (or the Fokker-Planck equa-
tion, which physicists use to describe diffusion processes
such as Brownian motion), we generate the series of
curves shown in figure 20.6. These curves show that as the
number of generations increases, the populations begin to
diverge from q = 0.5. Approximately the same number of
populations go to q values above 0.5 as go to q values be-
low 0.5. Therefore, the distribution spreads symmetrically.
When the distribution of allelic frequencies reaches the
sides of the graph, some populations become fixed for the
a allele and some lose it. In a sense, the sides act as sinks:

Figure 20.5 Initial conditions of random drift model. One
thousand populations, each of size one hundred, and each
with an allelic frequency (q) of 0.5.

Figure 20.6 Genetic drift in small populations: q = 0.5. After
time passes, the populations of figure 20.5 begin to diverge in
their allelic frequencies. Time is measured in population size
(A/), showing that the effects of random genetic drift are
qualitatively similar in populations of all sizes; the only difference
is the timescale. (From M. Kimura, "Solution of a process of random
genetic drift with a continuous model," Proceedings of the National Academy
of Sciences, USA, 41:144-50, 1955. Reprinted by permission.)

Any population that has the a allele lost or fixed will be
permanently removed from the process of random genetic
drift. Without mutation to bring one or the other allele

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

576

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

2/V

a
o

Q.
O

►

) 0.25

0.5 0.75 1

Figure 20.7 Continued genetic drift in the one thousand
populations, each numbering one hundred in size, shown in
figures 20.5 and 20.6. After approximately 2/V generations, the
distribution is flat, and populations are going to loss or fixation of
the a allele at a rate of 1/2/V populations per generation. (From
S. Wright, "Evolution in Mendelian Populations," Genetics, 97:114. Copyright ©
1931 Genetics Society of America.)

back into the gene pool, these populations maintain a con-
stant allelic frequency of zero or 1.0.

At a point between N (one hundred) and 27V (two
hundred) generations, the distribution of allelic frequen-
cies flattens out and begins to lose populations to the
edges (fixation or loss) at a constant rate, as shown in fig-
ure 20.7. The rate of loss is about 1/2/V (1/200), or 0.5% of
the populations per generation. If the initial allelic fre-
quency was not 0.5, everything is shifted in the distribu-
tion (fig. 20.8), but the basic process is the same — in all
populations, sampling error causes allelic frequencies to
drift toward fixation or elimination. If no other factor
counteracts this drift, every population is destined to even-
tually be either fixed for or deficient in any given allele.

The amount of time the process takes depends on
population size. The example used here was based on
small populations of one hundred. If we substitute one
million for one hundred in figure 20.6, a flat distribution
of populations would not be reached for two million gen-
erations, rather than two hundred generations. Thus, a
population experiences the effect of random genetic
drift in inverse proportion to its size: Small populations
rapidly fix or lose a given allele, whereas large popula-
tions take longer to show the same effects. Genetic drift
also shows itself in several other ways.

Founder Effects and Bottlenecks

Several well-known genetic phenomena are caused by
populations starting at or proceeding through small num-

Figure 20.8 Random genetic drift in small populations with
q = 0.1. Compare this figure with figure 20.6. In this case, the
probability of fixation of the a allele is 0.1, and the probability

of its loss is 0.9. (From M. Kimura, "Solution of a process of random
genetic drift with a continuous model," Proceedings of the National Academy
of Sciences, USA, 41:144-50, 1955. Reprinted by permission.)

bers. When a population is initiated by a small, and there-
fore genetically unrepresentative, sample of the parent
population, the genetic drift observed in the subpopula-
tion is referred to as a founder effect. A classic human
example is the population founded on Pitcairn Island by
several of the Bounty mutineers and some Polynesians.
The unique combination of Caucasian and Polynesian
traits that characterizes today's Pitcairn Island popula-
tion resulted from the small number of founders for the
population.

Sometimes populations go through bottlenecks, pe-
riods of very small population size, with predictable ge-

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Natural Selection

511

netic results. After the bottleneck, the parents of the next
generation have been reduced to a small number and
may not be genetically representative of the original pop-
ulation. The field mice on Muskeget Island, Massachu-
setts, have a white forehead blaze of hair not commonly
found in nearby mainland populations. Presumably, the
island population went through a bottleneck at the turn
of the century, when cats on the island reduced the num-
ber of mice to near zero. The population was reestab-
lished by a small group of mice that happened by chance
to contain several animals with this forehead blaze.

NATURAL SELECTION

Although mutation, migration, and random genetic drift
all influence allelic frequencies, they do not necessarily
produce populations of individuals that are better
adapted to their environments. Natural selection, how-
ever, tends to that end. The consequence of natural se-
lection, Darwinian evolution, is considered in detail in
the next chapter. We discuss here the algebra behind
the process of natural selection. Artificial selection, as
practiced by animal and plant breeders, follows the
same rules.

How Natural Selection Acts ■

Selection, or natural selection, is a process whereby
one phenotype and, therefore, one genotype leaves rela-
tively more offspring than another genotype, measured
by both reproduction and survival. Selection is thus a
matter of reproductive success, the relative contribu-
tion of that genotype to the next generation. It is impor-
tant to remember that selection acts on whole organisms
and thus on phenotypes. However, we analyze the
process by looking directly at the genotype, usually only
at one locus.

Fitness

A measure of reproductive success is the fitness, or
adaptive value, of a genotype. A genotype that, com-
pared with other genotypes, leaves relatively more off-
spring that survive to reproduce has the higher fitness.
(Note that this use of the word fitness differs from our
common notion of physical fitness.)

Fitness is usually computed to vary from zero to one
(0-1) and is always related to a given population at a
given time. For example, in a normal environment, fruit
flies with long wings may be more fit than fruit flies with
short wings. But in a very windy environment, a fruit fly
with limited flying ability may survive better than one
with the long-winged genotype, which will be blown

around by the wind. Thus, fitness (usually assigned the
letter W) is relative to a given circumstance. In a given
environment, the genotype that leaves the most offspring
is usually assigned a fitness of W = 1 , and a lethal geno-
type has a fitness of W = 0. Any other genotype has a fit-
ness value between zero and one. A number of factors
can decrease this fitness value, W, below one. A selec-
tion coefficient measures the sum of forces acting to
prevent reproductive success. It is usually represented
by the letter s ov t and is defined by the fitness equation

W = 1 - s

and

s = 1 - W

(20.13)

(20.14)

Thus, as the selection coefficient increases, fitness de-
creases, and vice versa.

Components of Fitness

Natural selection can act at any stage of the life cycle of
an organism. It usually acts in one of four ways. (1) The
reproductive success of a genotype can be affected by
prenatal, juvenile, or adult survival. Differential survival
of genotypes is referred to as viability selection or zy-
gotic selection. (2) A heterozygote can produce ga-
metes with differential success when one of its alleles
fertilizes more often than the other. This is termed ga-
metic selection. A well-studied case is the ^-allele (tail-
less) locus in house mice; the gametes of as many as 95%
of the heterozygous males of the Tt genotype carry the t
allele. (This phenomenon is also referred to as segrega-
tion distortion or meiotic drive.) Selection can also
take place in two areas of the reproductive segment of an
organism's life cycle. (3) Some genotypes may mate more
often than others (have greater mating success), resulting
in sexual selection. Sexual selection usually occurs
when members of the same sex compete for mates or
when females have some form of choice. Adaptations for
fighting, such as antlers in male elk, or displaying, such as
the peacock's tail, are the results of sexual selection.
(4) Finally, some genotypes may be more fertile than
other genotypes, resulting in fecundity selection. The
particular variable of the life cycle that selection acts
upon is termed a component of fitness.

Effects of Selection

Figure 20.9 shows the three main ways that the sum total
of selection can act. Directional selection works by con-
tinuously removing individuals from one end of the phe-
notypic (and therefore, presumably, genotypic) distribution
(e.g., short-necked giraffes are removed). Removal means
disappearance through death or failure to reproduce (ge-
netic death). Thus, the mean is constantly shifted toward

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

578

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

Mean

Original
distribution

Phenotype
(e.g., height)

Directional
selection

Stabilizing
selection

Disruptive
selection

Before
selection

After
selection

Figure 20.9 Directional, stabilizing, and disruptive selection.
Colored areas show the groups being selected against. At the
top is the original distribution of individuals. The final
distributions after selection appear in the bottom row.

the other end of the phenotypic distribution; in our exam-
ple, the mean shifts toward long-necked giraffes. The evo-
lution of neck length in giraffes, presumably by directional
selection, has been documented from the geologic record.
Stabilizing selection (fig. 20.9) works by constantly
removing individuals from both ends of a phenotypic dis-
tribution, thus maintaining the same mean over time. Sta-
bilizing selection now works on giraffe neck length — it is
neither increasing nor decreasing. Disruptive selection
works by favoring individuals at both ends of a pheno-
typic distribution at the expense of individuals in the mid-
dle. It, like stabilizing selection, should maintain the same
mean value for the phenotypic distribution. Disruptive se-
lection has been carried out successfully in the laboratory
for bristle number in Drosophila. Starting with a popula-
tion with a mean number of sternopleural chaeta (bristles
on one of the body plates) of about eighteen, investiga-
tors succeeded after twelve generations of getting a fly
population with one peak of bristle numbers at about six-
teen and another at about twenty-three (fig. 20.10).

Selection Against the Recessive
Homozygote

We can analyze selection by using our standard model-
building protocol of population genetics — namely, de-

fine the initial conditions; allow selection to act; calculate
the allelic frequency after selection (q n+1 ); calculate A#
(change in allelic frequency from one generation to the
next); then calculate equilibrium frequency, q, when Aq
becomes zero; and examine the stability of the equilib-
rium. In the analysis that follows, we consider a single au-
tosomal locus in a diploid, sexually reproducing species
with two alleles and assume that selection acts directly
on the phenotypes in a simple fashion (i.e., it occurs at a
single stage in the life of the organism, such as larval mor-
tality in Drosophila). After selection, the individuals re-
maining within the population mate at random to form a
new generation in Hardy-Weinberg proportions.

Selection Model

In table 20.2, we outline the model for selection against the
homozygous recessive genotype. The initial population is
in Hardy-Weinberg equilibrium. Even with selection acting
during the life cycle of the organism, Hardy-Weinberg pro-
portions will be reestablished anew after each round of
random mating, although presumably at new allelic fre-
quencies. All selection models start out the same way. They
diverge at the point of assigning fitness, which depends on
the way natural selection is acting. In the model in table
20.2, the dominant homozygote and the heterozygote have

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Natural Selection

579

c
o

-1— '

_CD
CD

c
o

-1—'

i_
CD

CD
O

2
3

6
7
8
9

11
12

i i i i_

15 20 25

Bristle number

Figure 20.10 Disruptive selection in Drosophila melanogaster.
After twelve generations of selection for flies with either many
or few bristles (chaetae) on the sternopleural plate, the
population was bimodal. In other words, many flies in the
population had either few or many bristles, but few flies had an
intermediate bristle number. (Reprinted with permission from Nature,
Vol. 193, J. M. Thoday and J. B. Gibson, "Isolation by Disruptive Selection."
Copyright © 1962 Macmillan Magazines Limited.)

the same fitness (W = 1). Natural selection cannot differ-
entiate between the two genotypes because they both
have the same phenotype.The recessive homozygote, how-
ever, is being selected against, which means that it has a
lower fitness than the two other genotypes (W = 1 — s).

After selection, the ratio of the different genotypes is
determined by multiplying their frequencies (Hardy-
Weinberg proportions) by their fitnesses. The procedure
follows from the definition of fitness, which in this case
is a relative survival value. Thus, only 1 — 5 of the aa
genotype survives for every one of the other two geno-
types. For example, if 5 were 0.4, then the fitness of the
aa type would be 1 — s, or 0.6. For every ten AA and Aa

Table 20.2 Selection Against the Recessive

Homozygote: One Locus with
Two Alleles, A and a

Genotype

Total

Initial genotypic

P 2

2pq

q 2

frequencies

Fitness (W^>

\-s

Ratio after

P 2

2pq

q 2 a~s)

1-sq 2 = W

selection

Genotypic fre-

2pq

q 2 (l-s)

quencies after

selection

individuals that survive to reproduce, only six aa indi-
viduals would survive to reproduce. The total of the
three genotypes after selection is 1 — sg 2 . That is,

p 2 + 2pq + q\l - s) = p 2 + 2pq + q 2
= 1 - sq 2

sq'

Mean Fitness of a Population

The value (1 — sq 2 } is referred to as the mean fitness of
the population, W, because it is the sum of the

fitnesses of the genotypes multiplied (weighted) by the
frequencies at which they occur. Thus, it is a weighted
mean of the fitnesses, weighted by their frequencies. The
new ratios of the three genotypes can be returned to
genotypic frequencies by simply dividing by the mean fit-
ness of the population, IF, as in the last line of table 20.2.
(Remember that a set of numbers can be converted to
proportions of unity by dividing them by their sum.) The
new genotypic frequencies are thus the products of their
original frequencies times their fitnesses, divided by the
mean fitness of the population.

After selection, the new allelic frequency (q n+1 ) is
the proportion of aa homozygotes plus half the propor-
tion of heterozygotes, or

a n+\

q\l

gig - sq + p)
1 - sg 2

g(\ - sg)
1 - sg 2

(20.15)

This model can be simplified somewhat if we assume
that the aa genotype is lethal. Its fitness would be zero,

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

580

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

and s, the selection coefficient, would be one. Equation
20.15 would then change to

tfn+l

#(i - q)

l-q 2

(20.16)

Since (1 — q ) is factorable into (1 — q)(\ + q), equation
20.16 becomes

<2n+l

q(\ - q)

(1 - q)(\ + q)

(20.17)

(1 +q)
The change in allelic frequency is then calculated as

Atf = q n+1 ~ q =

1 + q

To solve this equation, q is multiplied by (1 + q)/(l + q)
so that both parts of the expression are over a common
denominator:

q - q(l + q)

1 + q

(20.18)

This is the expression for the change in allelic frequency
caused by selection. Since selection will not act again un-
til the same stage in the life cycle during the next gener-
ation, equation 20.18 is also an expression for the change
in allelic frequency between generations.

Two facts should be apparent from equation 20.18.
First, the frequency of the recessive allele (q) is declin-
ing, as indicated by the negative sign of the fraction. This
fact should be intuitive because of the way selection was
defined in the model (eliminating aa homozygotes). Sec-
ond, the change in allelic frequency is proportional to
q 2 , which appears in the numerator of the expression. In
other words, allelic frequency is declining as a relative
function of the number of homozygous recessive indi-
viduals in the population. This fact is consistent with the
premise of the selection model (with selection against
the homozygous recessive genotype). This final formula
supports the methodology of the model.

Equilibrium Conditions

Next we calculate the equilibrium q by setting the Aq
equation equal to zero, since a population in equilibrium
will show no change in allelic frequencies from one gen-
eration to the next:

1 + q

(20.19)

For a fraction to be zero, the numerator must equal zero.
Thus, q 2 = 0, and q = 0. At equilibrium, the a allele
should be entirely removed from the population. If the
aa homozygotes are being removed, and if there is no
mutation to return a alleles to the population, then even-
tually the a allele disappears from the population.

Time Frame for Equilibrium

One shortcoming of this selection model is that it is not im-
mediately apparent how many generations will be
required to remove the a allele. The deficiency can be com-
pensated for by using a computer simulation or by intro-
ducing a calculus differential into the model. Either method
would produce the frequency-time graph of figure 20.11.
This figure clearly shows that the a allele is removed more
quickly when selection is stronger (when 5 is larger) and
that the curves appear to be asymptotic — the a allele is not
immediately eliminated and would not be entirely removed
until an infinitely large number of generations had passed.
There is a reason for the asymptotic behavior of the graph:
As the a allele becomes rarer and rarer, it tends to be found
in heterozygotes (table 20.3). Since selection can remove
only aa homozygotes, an a allele hidden in an Aa het-
erozygote will not be selected against. When q = 0.5, there
are two heterozygotes for every aa homozygote. When

1.0

0.9

0.8

0.7

0.6

-1 \

q 0.5

- 1 \

0.4

- \ \»

0.3

- V >s = 0.5

0.2

"•C ^***>

0.1
(

) 1 2 3 4 5 6 7 8 910111213141516171819 20

Generations

Figure 20.11 Decline in q (the frequency of the a allele) under
different intensities of selection against the aa homozygote.
Note that the loss of the a allele is asymptotic in both cases,
but the drop in allelic frequency is more rapid with the larger
selection coefficient.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Natural Selection

581

Table 20.3 Relative Occurrence of Heterozygotes

and Homozygotes as Allelic Frequency
Declines: q = /*(«); p = f(A)

f(Aa)

f(aa)

(2pq)

(tf 2 )

f(Aa)/f(aa)

0.5

0.50

0.25

0.2

0.32

0.04

0.1

0.18

0.01

0.0198

0.0001

198

0.001

0.001998

0.000001

1,998

q = 0.001, there are almost two thousand heterozygotes
per aa homozygote. Remember, only the recessive ho-
mozygote is selected against. Natural selection cannot dis-
tinguish the dominant homozygote from the heterozygote.

Selection-Mutation Equilibrium

Although a deleterious allele is eliminated slowly from a
population, the time frame is so great that there is op-
portunity for mutation to bring the allele back. Given a
population in which alleles are removed by selection and
added by mutation, the point at which no change in al-
lelic frequency occurs, the selection-mutation equilib-
rium, may be determined as follows. The new frequency
(q n +i) of the recessive a allele after nonlethal selection
(s < 1) against the recessive homozygote is obtained by
equation 20.15:

<ln+l

q(\ - sq)
1 - sq 2

Change in allelic frequency under this circumstance will
thus be

Atf = q n+1 ~ q =

q(\ - sq) q(\

sq 2 }

(1 - sq 2 ) (1 - sq 2 )

q — sq — q + sq
(1 " sq 2 )

sq\\

(20.20)

1 - sq'

Equation 20.20 is the general form of equation 20.18 for
any value of 5. The change in allelic frequency due to mu-
tation can be found by using equation 20.4:

A# = fxp — vq

where |x and v are the rate of forward and back muta-
tion, respectively. When equilibrium exists, the change

from selection will just balance the change from muta-
tion. Thus,

jjl/7 — vq +

sq\\

and

|xp — vq =

sq\\

1 - sq'

(20.21)

Now, some judicious simplifying is justified, because
in a real situation, q will be very small because the a al-
lele is being selected against. Thus, vq will be close to
zero, and 1 — sq 2 will be close to unity. Equation 20.21,
therefore, becomes:

\Lp = sq (1
^(1 - q) = sq\\
q 2 = \Xj/s

(20.22)

q = VuA

In the case of a recessive lethal, 5 would be unity, so

q 2 = fx and q = V|jl

If a recessive homozygote has a fitness of 0.5 (s = 0.5)
and a mutation rate, jx, of 1 X 10~ 5 , the allelic frequency
at selection-mutation equilibrium will be

q = y/yJs = Vl X 10 5 /0.5 = V2 X 10 5
= 0.004
If the recessive phenotype were lethal, then

q = VyJs = Vl X 10 5 /l
= 0.003
These are very low equilibrium values for the a allele.

Types of Selection Models

In view of the limited ways that fitnesses can be assigned,
only a limited number of selection models are possible.
Table 20.4 lists all possible selection models if we assume
that fitnesses are constants and the highest fitness is one.
(You might now go through the list of models and deter-
mine the equilibrium conditions for each.) Note that two
possible fitness distributions are missing. There is no model
in which fitnesses are 1 — s, 1 , and 1 for the A l A l ,A 1 A 2 , and
A 2 A 2 genotypes, respectively (remembering that^? =f[A 1 ]
and q = /[^4 2 ])That model is for selection against the A 1 A 1
homozygote. Some reflection should show that this is the
same model as model 1 of table 20.4, except that the^ al-
lele is acting like a recessive allele. In other words, natural
selection acts against A 1 A 1 homozygotes, but not against
the AiA 2 and A 2 A 2 genotypes. Thus, the model reduces to
model 1 if we treat A 1 as the recessive allele and^4 2 as the

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

582

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

Table 20.4 All Possible One-Locus, Two-Allele

Selection Models (Assuming All
Selection Coefficients Are Constants)

Genotypic Fitness

Type of Selection

A X A-

A ± A

AyA

2 ^2

1. Against recessive 1
homozygotes

2. Against heterozygotes 1

3. Against one allele 1

4. Against homozygotes 1 — s 1

1 - 5

1 - Sl
1

1 - 5

1
1 - s 2
1 - s 2

dominant allele. Similarly, the (1 — s l3 1 — s 2 , 1) model is
eliminated for the same reason (allele A 2 is acting like the
dominant allele and A 1 like the recessive allele). We now
describe the outcome of each of the models in the table.

In both models 1 and 3 (table 20.4), selection is
against genotypes containing the A 2 allele. Model 1,
which we just derived in detail, is the model for a delete-
rious recessive allele. Almost any enzyme defect in a meta-
bolic pathway fits this model, such as PKU, alkaptonuria,
Tay-Sachs disease, and so on. In model 3, however, natural
selection can detect the heterozygote, as is the case with
deleterious alleles that are not completely recessive. An
example would be the hemoglobin anomaly called thal-
assemia, a disorder common in some European and Asian
populations, that produces a severe anemia in homozy-
gotes and a milder anemia in heterozygotes. It should be
clear that selection can more quickly eliminate a partially
recessive allele than a completely recessive allele because
the allele can no longer "hide" in the heterozygote.

Dominant or semidominant alleles (model 3) are usu-
ally more quickly removed from a population because
they are completely open to selection. It takes an infinite
number of generations to remove a recessive lethal allele,
but only one generation for natural selection to remove a
completely dominant lethal allele (see model 3, where
s 1 = s 2 = 1). Examples of dominant deleterious traits in

people are Huntington disease, facioscapular muscular
dystrophy, and chondrodystrophy.

Model 2 is interesting because selection against the
heterozygote leads to an unstable equilibrium at q = 0.5.
If one heterozygote is removed by selection, one each of
the two alleles is eliminated. However, if p and q are not
equal (and thus not equal to 0.5), then one^ allele is not
the same proportion of the A 1 alleles as one^4 2 allele is of
all the A 2 alleles. In other words, in a population of fifty
individuals with q = 0.1 andp = 0.9, one^4 2 allele is 10%
(1/10) of the A 2 alleles, whereas one^ allele is only 1.1%
(1/90) of the A 1 alleles. Removing one each of the two al-
leles causes a decrease in q. Therefore, a population fol-
lowing model 2 is at equilibrium at^? = q = 0.5. How-
ever, this is an unstable equilibrium. Any perturbation
that changes the allelic frequencies causes the rarer allele
to be selected against and eventually removed from the
population. An example is the maternal-fetal incom-
patibility at the Rh locus in human beings. The disease
erythroblastosis occurs only in heterozygous fetuses
(Rh + Rh~) in Rh-negative (Rh~Rh~) mothers. Heterozy-
gotes are, therefore, selected against.

In model 4, selection is against homozygotes. This
model is called the heterozygote advantage, and we will
derive its equilibrium condition because the results are im-
portant to evolutionary theory (table 20.5). At equilibrium

pq(s x p- s 2 q)
W

(20.23)

For this expression to be zero, either

p = 0, q = 0, or (S!p ~ s 2 q)

If p = or q = 0, the result is trivial; the equilibrium ex-
ists only because of the absence of one of the alleles. The
more meaningful equilibrium occurs when s 1 p — s 2 q
= 0. In that case

S\P = s 2 q or s^l - q) = s 2 q

and

Sl + s 2

(20.24)

Table 20.5 Selection Model of Heterozygote Advantage: The A Locus with A x andA 2 Alleles

Genotype

Total

A 1 A 1

A X A 2

A 2 A 2

Initial genotypic frequencies

Fitness (W)

Ratio after selection

Genotypic frequencies after selection

P 2
1 - Sl

p\l ~ sO
P<1 ~ O

2pq

1
2pq

2pq
W

q 2
1 - s 2

q\\ - 5 2 ) l
W

1
1 - srf 2 - s 2 q 2

= W

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Natural Selection

583

BOX 20.1

It is surprising how much insight
we can gain into the processes of
population genetics by modeling
them on a computer. The simple
computer program presented here
calculates changing allelic frequencies
due to random mating when alleles
at a locus are under a heterozygote-
advantage selection regime. The pro-
gram is written in the Microsoft®
Visual Basic language. You can simu-
late any of the selection models de-
scribed in chapter 20 by simply
changing the variables. Also, this pro-
gram can model many of the other
processes discussed in this and the

Experimental
Methods

A General Computer

Program to Simulate the

Approach to Allelic

Equilibrium Under

Heterozygote Advantage

last chapter; usually, only a few lines
need to be changed to look at an en-
tirely different process. Other com-

puter programs can substitute. Out-
put should be graphed. The program
should be rerun several times with
various sets of values for the allelic
frequencies and fitnesses. If the out-
come isn't clear by twenty-five gener-
ations, the number of generations can
be increased with a few small
changes in the program.

In the computer program (fig. 1),
p is set to 0.9, q is 1 — p (0. 1), and the
three fitnesses are named wll, wl2,
and w22 for the AA,Aa, and aa geno-
types, respectively. In this case, wll
is set to 0.4, wl 2 to l,andw22 to 0.6,
a model of heterozygote advantage;

continued

Sub Command1_Click ()
Static q(25)
Static p(25)
Picturel.CIs

'Set variables

P(1)
w11

w12

w22

q(1)

1-p(1)

'Calculate p and q values

For i = 2 To 25

wbar = p(i - 1) A 2 * w11 + 2 * p(i - 1) * q(i - 1)*w12 + q(i- 1) A 2*w22

q(i) = (q(i- 1) A 2*w22 + p(i - 1)*q(i- 1)*w12)/wbar

p(i) = 1 - q(i)
Next i

'Draw axes and grid

Picture"!. Scale (-1, 1.1)-(26, -.1)
Picturel.Line (0, 0)-(0, 1)
For i = 0To 10

Picturel.Line (0, .1 * i)-(25, .1 * i)
Next i
For i = 5 To 25 Step 5

Picturel.Line (i, 0)-(i, 1)
Next i

'Draw q values

Picturel.DrawWidth = 5

For i = 1 To 25

Picture1.PSet(i,q(i))

Next i
End Sub

Figure 1 A Microsoft® Visual Basic computer program for the simulation of heterozygote advantage. The first statement
indicates that the program is run by clicking a command button. Twenty-five values of q and p are calculated and stored
for printing. The program also prints a grid of lines at increments of q = 0.1 and generations = 5.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

584

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

BOX 20.1 CONTINUED

the number of generations is twenty- lection (the new proportion of aa ure 2 results. As you can see, q is ap-
five. The program calculates the homozygotes plus half the propor- proaching 0.6. If you would like to
mean fitness of the population, wbar, tion of heterozygotes). The program see the values generated by the pro-
as p 2 (w\Y) + 2pq(wl2) + # 2 (w22); then repeats this process twenty-five gram, appropriate print statements
it then calculates the new allelic fre- times, storing each new q in the array can be added,
quencies after one generation of se- #(i).The graphic output shown infig-

1 n

I .u

0.9
0.8
0.7
0.6

q 0.5
0.4
0.3
0.2
0.1

Figure 2

drawn in

■
■

■

5 10 15 20 21

Generations

Graphical output of the computer program from figure 1 , with axis labels
. The frequency of the a allele, q, begins at 0.1 and asymptotes toward 0.6

Since p + q = 1,

P =

s 2

s 1 + s 2

(20.25)

Several interesting conclusions follow. First, unlike
the other models of selection, this model allows a popu-
lation to maintain both alleles. We can demonstrate that
this equilibrium is stable by graphing the Aq value
against q. Such a graph appears in figure 20.12, in which
q is the frequency of allele A 2 and the fitnesses of geno-
types A 1 A 1 , A X A 2 , and A 2 A 2 are assumed to be 0.8, 1, and
0.7, respectively. Note that if the equilibrium is perturbed
by an increase or decrease in q, the population returns to
the point of equilibrium. Second, the equilibrium is inde-
pendent of the original allelic frequencies since it in-
volves only the selection coefficients, s 1 and s 2 . Last, the
equilibrium for each allele (equations 20.24 and 20.25) is
directly proportional to the selection coefficient against
the other allele. As the selection against^ increases (s t
increases), the equilibrium shifts toward a higher value
of q (more A 2 alleles; box 20.1).

Aq -0.01 h

-0.02
-0.03
-0.04 L

Figure 20.12 Plot of allelic frequency (q) versus change in
allelic frequency (Aq) for a polymorphism maintained by
heterozygote advantage. In this case, s 1 = 0.2 and s 2 = 0.3;
the equilibrium value, q, is 0.4. When perturbed, the population
tends to return to this value unless the perturbation brings q to
either 1 .0 or 0.0, in which case the population is either fixed for
the a allele or has lost it. In both cases, no further change in
allelic frequency will take place, barring mutation or migration.

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IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Solved Problems

585

SUMMARY

STUDY OBJECTIVE 1: To develop ways to analyze popula-
tion genetics problems 571

A five-step protocol is presented to determine equilibrium
allelic frequencies.

STUDY OBJECTIVE 2: To analyze the effects of mutation,
migration, and population size on the Hardy-Weinberg
equilibrium 571-577

The effects of relaxing some of the assumptions of the
Hardy-Weinberg equilibrium are analyzed. Both mutation
and migration transport alleles in and out of a population.
Mutation provides the variability on which natural selec-
tion acts, but it usually does not directly affect the equilib-
rium because mutation rates are usually very low. If two
randomly mating populations merge, or if two randomly
mating demes are mistakenly treated as a single deme, the
conglomerate will be deficient in heterozygotes.This devia-
tion is called the Wahlund effect.

Finite population size is a source of sampling error. It
results in changes in allelic frequencies known as random
genetic drift. The smaller the population, the more rapidly
allelic frequencies change. The dynamics of random genetic
drift were studied graphically.

STUDY OBJECTIVE 3: To study the ways in which natural
selection results in organisms adapted to their environ-
ments 577-584

Natural selection is defined by differential reproductive
success. Depending upon which phenotypes are most fit,
natural selection can act in several ways to change allelic
and genotypic frequencies. Selection against the recessive
homozygote acts to remove the allele from the population.
Mutation brings the allele back into the population. Thus, a
selection-mutation equilibrium maintains the unfavorable
allele at a relatively low frequency. Heterozygote advantage
maintains both alleles in a population.

SOLVED PROBLEMS

PROBLEM 1: At a particular locus, there are two alleles, B
and b. The mutation rate of B to b is 3.5 X 10~ , whereas
the mutation rate of b to B is 6 X 10~ 8 . What is the equi-
librium frequency of the b allele, assuming no other fac-
tor is operating in this population to disturb the Hardy-
Weinberg equilibrium?

Answer: We let q = /(&), |x = 3.5 X 10" 4 , and v = 6 X
10~ 8 . We then simply substitute jjl and v into equation
20.6:

q = |x/(|i + v) = 3.5 X 10" 4 /
(3.5 X 10" 4 + 6 X 10" 8 )
= 0.9998

PROBLEM 2: Given a population of about one million ci-
cadas with a frequency of the a allele at the A locus of
0.75, what is the probability that the a allele will be lost
due to random genetic drift? How much longer will the
possible loss of the allele take than the loss of the allele
would take in a population of one thousand?

Answer: Regardless of the size of a finite population,
random genetic drift takes place. The probability of the
loss of an allele with a frequency of 0.75 is 0.25; the prob-

ability of its fixation is 0.75 (see fig. 20.8). Since it is con-
venient to measure time (number of generations) within
populations of finite size in units of population size, we
can see that an event that takes N generations will be one
thousand generations in the small population, but one
million generations in the large population. Thus, random
genetic drift occurs in the larger population at about one-
thousandth the rate of the small population.

PROBLEM 3: In a laboratory colony of fruit flies, the fit-
nesses of the genotypes of an electrophoretic locus
(malate dehydrogenase) are determined. Three geno-
types, FF, FS, and SS, have fitnesses of 0.85, 1.0, and 0.6,
respectively. What is the equilibrium frequency of the
slow allele (5 1 )?

Answer: If the fitnesses of the three genotypes FF, FS,
and SS are as given, then the locus is exhibiting heterozy-
gote advantage with selection coefficients of the two ho-
mozygotes of s t = 0.15 (1 - 0.85) ands 2 = 0.4 (1 - 0.6).
If q is the frequency of the slow allele, then, using equa-
tion 20.24,

4 = V(si + s 2 ) = 0.15/(0.15 + 0.4) = 0.27

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IV. Quantitative and
Evolutionary Genetics

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Process that Change
Allelic Frequencies

586

Chapter Twenty Population Genetics: Processes That Change Allelic Frequencies

EXERCISES AND PROBLEMS

MUTATION

1. Consider a locus with alleles A and a in a large, ran-
domly mating population under the influence of mu-
tation.

a. If the mutation rate of A to a is 6 X 10~ 5 , and the
back-mutation rate to A is 7 X 10~ 7 , what is the
equilibrium frequency of a?

b. If q = 0.9 in generation n, what would it be one
generation later, under only the influence of
mutation?

2. Derive an expression for mutation equilibrium when
no back mutation is occurring.

3. Consider a population in which ^? = 0.9 and # = 0.1.
If the forward mutation rate, A — » «, is 5 X 10~ 5 and
the reverse mutation rate, a —> A, is 2 X 10~ 5 , calcu-
late the equilibrium frequency, q , of the a allele.

4. If the forward mutation rate, A — > a, is five times the
reverse mutation rate, what is the equilibrium fre-
quency of the a allele?

MIGRATION

5. The following data refer to the R° allele in the Rh
blood system:

frequency in western Europeans = 0.62

frequency in eastern Europeans = 0.45

frequency in Mongols = 0.03

What is the total proportion of alleles that have en-
tered the eastern European population?

6. Given the data from problem 1 of chapter 19, what
factors could have caused the population to leave
Hardy-Weinberg equilibrium? (See also SMALL POP-
ULATION SIZE and NATURAL SELECTION)

7. In a population of nine hundred butterflies, the fre-
quency (p) of the fast allele of the enzyme phospho-
enol pyruvate is 0.6, and the frequency of the slow
form (q) is 0.4. Ninety butterflies migrate to this pop-
ulation, and the migrants have a slow-allele fre-
quency of 0.8. Calculate the allelic frequencies of
the new population.

8. If the frequency of the N allele is 0.25 in a native
population, 0.32 in a conglomerate population, and
0.4 in a migrant population, what percentage of the
N alleles in the conglomerate population were de-
rived from the migrant population?

9. In a particular population, the frequency of allele t
was 0.25 in a migrant population and 0.45 in the

Answers to selected Exercises and Problems are on page A-22.

conglomerate population. If the migration rate was
0.1, calculate the frequency of t in the original, na-
tive population.

SMALL POPULATION SIZE

10. In a population of five hundred individuals with a
frequency of allele A of 0.7, what is the ultimate fate
of the A allele? What is the probability that the pop-
ulation will eventually lose the A allele? How many
are N/5 generations? AN generations?

NATURAL SELECTION

11. Differentiate among stabilizing, directional, and dis-
ruptive selection.

12. Derive a model of selection in which the fitness of
the heterozygote is half the fitness of one of the ho-
mozygotes and twice the fitness of the other. Give
expressions for the following:

a. Mean population fitness

b. Equilibrium allelic frequency (stable?)

13. Derive an expression for the equilibrium allelic fre-
quencies under a model in which selection acts
against heterozygotes. Is the equilibrium stable?

14. Table 20.6 describes selection at the A locus in a
given diploid species in which p = f(A) and q =
f(a).

a. Describe the type of selection occurring here.
Why does the total equal one before selection but
W, after?

b. Derive an equation for q after one generation of
selection (q n+1 ).

c. This system will reach equilibrium, with p =
s 2 /(s 1 + s 2 ). If selection is twice as strong
against aa as against AA, what are the equilib-
rium allelic frequencies? If s 1 = 0.1 ands 2 = 0.3,
what percentage of heterozygotes is at equilib-
rium?

15. Given a locus with alleles A and a in a sexually re-
producing, diploid population in Hardy-Weinberg
equilibrium, set up a model and the initial formula
for the frequency of the dominant allele after one
generation (p n + 1) if selection acts against the dom-
inant phenotype. What are the equilibrium condi-
tions?

16. There is a locus with alleles A and a in a large, ran-
domly mating, diploid, sexually reproducing popula-
tion. Allele A mutates to a at a rate of jx, and no back
mutation takes place. However, the aa homo zygote
is selected against with a fitness of 1 — s. Give a for-
mula for the equilibrium condition. If jx = 5 X 10~ 5

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

20. Population Genetics:
Process that Change
Allelic Frequencies

Critical Thinking Questions

587

Table 20.6

Genotypes

Total

Before selection
Fitness (IF)
After selection

1 - Sl

2pq

1
2pq

q 2
1 - s 2

q\l ~ S2)

1
W=l- Sl p 2 - s 2 q 2

and 5 = 0.15, what are the equilibrium allelic fre-
quencies?

17. If a locus has alleles A 1 and A 2 , what is the equilib-
rium frequency ofA 1 if both homozygotes are lethal?

18. The following data were collected from a population
of Drosophila segregating sepia (s) and wild-type
(s + ) eye colors. One sample was taken when the
eggs were deposited, and another was taken later
among adults. Reconstruct the mode of selection.

+ +

Egg

Adult

19. The data in table 20.7 come from T. Dobzhansky's
work with chromosomal inversions in Drosophila
pseudoobscura. They represent four samples from
various altitudes in the Sierra Nevada Mountains in
California. What would you say about, and what
would you do in the lab to determine, the fitnesses
of the inversions? What factors could cause the
changes in fitness?

20. In a particular population with two alleles at a locus,
the frequency of AA individuals = 0.25, Aa = 0.5,
and aa = 0.25. If the AA genotype fitness = l,Aa =
0.8, and aa = 0.6, what will the frequencies of A and
a be in the next generation? Assume mutations do
not occur.

Table 20.7

Data from Dobzhansky's

Work

Inversion

Elevation

of Sample

Others

6,800 ft

4,600 ft

3,000 ft

800 ft

Note: ST — standard; AR — Arrowhead; CH — Chiricahua

21. Calculate the frequency of the recessive b allele in a
population one generation after selection if in the
original population q = f(b) = 0.7 and the relative
fitness of bb homozygotes is 0.4.

22. A type of dwarfism in dogs is caused by a recessive
allele. The mutation rate from the normal to the mu-
tant allele has been estimated at 5 X 10~ 5 , and the
fitness of the dwarf is 0.2 when compared with nor-
mal individuals. Calculate the equilibrium frequency
of the dwarf allele.

23. A recessive allele (q = 0.5) was initially neutral, but
suddenly the environment changed and the recessive
homozygote became lethal. What is q one generation
after selection begins? What is the expected frequency
of the recessive allele two generations after selection?

CRITICAL THINKING QUESTIONS

1. If the selection model of heterozygous disadvantage
leads to the elimination of the rarer allele, why would
such systems (e.g., the Rh blood system) still exist.

2. A scientist studied the distribution of electrophoretic
genotypes in a sample of an insect species and found a
deficiency of heterozygotes. How could this come
about?

Suggested Readings for chapter 20 are on page B-20.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

EVOLUTION AND

SPECIATION

STUDY OBJECTIVES

1. To analyze the mechanisms of evolution and speciation 589

2. To investigate the mechanisms of the maintenance of genetic
variation in natural populations, both selective and
neutral 596

3. To discuss sociobiology, the evolution of social behavior 603

STUDY OUTLINE

Darwinian Evolution 589
Evolution and Speciation 589

Mechanisms of Cladogenesis 592

Phyletic Gradualism Versus Punctuated Equilibrium 594
Genetic Variation 596

Maintaining Polymorphisms 596

Maintaining Many Polymorphisms 598

Which Hypothesis Is Correct? 599

Grand Patterns of Variation 600
Sociobiology 603

Altruism 603

Kin Selection and Inclusive Fitness 605
Summary 607
Solved Problems 607
Exercises and Problems 608
Critical Thinking Questions 609
Box 21.1 Attacks on Darwinism 590
Box 21.2 Mimicry 604

The cactus ground-finch (Geospiza scandens) from

Photo Researchers, Inc.)

588

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

Evolution and Speciation

589

Populations change, or evolve, through natural
selection and the other forces that perturb the
Hardy-Weinberg equilibrium. The merger of
population genetics theory with the classical
Darwinian view of evolution is known as neo-
Darwinism, or the "new synthesis." In the two previous
chapters, we laid the theoretical groundwork for an un-
derstanding of the process of evolution in natural popu-
lations. In this chapter, we concern ourselves with long-
term evolution and speciation.

DARWINIAN EVOLUTION

Charles Darwin (fig. 21.1) was a British naturalist who
published his theory of evolution in 1859 in a book enti-
tled The Origin of Species by Means of Natural Selection,
or the Preservation of Favored Races in the Struggle for
Life. This book provided overwhelming support for evo-
lution as well as a mechanism for the process. Darwin had
been greatly influenced by the writings of the Reverend
Thomas Malthus, who is best known for his theory that
populations increase exponentially, whereas their food
supplies increase arithmetically. Malthus, who proposed
his theory in An Essay on the Principle of Population in
1798, was referring specifically to human populations and
was trying to encourage people to reduce their birthrate

rather than let their offspring starve to death. Malthus s
writings impressed upon Darwin the realization that un-
der limited resources — the usual circumstance in
nature — not all organisms survive. In nature, organisms
compete for the resources needed to survive.

Darwin sailed aboard the HMS Beagle, a ship that cir-
cled the world from 1831 to 1836 with the primary pur-
pose of charting the coast of South America. During his
travels on the Beagle, Darwin amassed great quantities
of observations (especially on South America and the
Galapagos Islands) that led him to suggest a theory. Dar-
win proposed that organisms become adapted to their en-
vironment by the process of natural selection. In outline,
the process works according to the following principles:

1 . Variation is a characteristic of virtually every group
of animals and plants. Darwin saw variation as an in-
herent property among individuals of all populations.

2. Every group of organisms overproduces offspring.
Most populations maintain a relatively constant den-
sity over time. Thus, every parent, on average, just re-
places itself. Therefore, most of the offspring the indi-
viduals of a population produce will die before they
reproduce. Hence, in every group of organisms, there
is an overabundance of young.

3. Those that do survive and reproduce will pass on
their genes in greater proportion. This step is the
cornerstone and the best-known part of Darwin's the-
ory. Among all the organisms competing for a limited
array of resources, only the organisms best able to ob-
tain and utilize these resources survive (survival of
the fittest). If the favorable characteristics of these
individuals are inherited, these traits pass on to the
next generation. These organisms then have the great-
est reproductive success (box 21.1).

Thus, over time, if advantageous mutations arise, or if the
environment changes, the characteristics of a population
should change through the process of natural selection
(directional or disruptive selection). A particularly well-
adapted population in a stable environment may maintain
its numbers through the forces of stabilizing selection
(see fig. 20.9). Nonrandom mating, genetic drift, and mi-
gration may also play a role in population differentiation.

Figure 21.1 Charles Darwin (1809-1882). Darwin was an
English naturalist who first established the theory of organic
evolution by natural selection. (Painting by George Richmond, 1840.
Downe House, Downe, Kent. © Archiv/Photo Researchers, Inc.)

EVOLUTION AND
SPECIATION

The term evolution describes a change in genotypic
frequencies, which usually results in a population of in-
dividuals better adapted to the environment than their
ancestors were. Speciation comes in two different
forms. (1) It may be the evolution of a population over
time until the current population cannot be classified as

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IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

590

Twenty-One Evolution and Speciation

BOX 2 1.1

From time to time, attacks on
neo-Darwinism are mounted,
usually by persons who either
view evolutionary theory as antireli-
gious or who misunderstand Dar-
win's theory. One attack, entitled
"Darwin's Mistake," by Tom Bethell,
was published in Harper's magazine
in 1976.

Bethell began by pointing out
that Darwinian theory is a tautology
rather than a predictive theory. (The
term tautology means a statement
that is true by definition.) That is,
evolution is the survival of the
fittest. But who are the fittest? Obvi-
ously, the individuals who survive.
Thus, without an independent crite-
rion for fitness other than survival,
we are left with the statement that
evolution is the survival of the sur-
vivors. This, indeed, is a tautology.
But it is possible to assign indepen-
dent criteria for fitness. Darwin
wrote extensively about artificial se-
lection in pigeons, in which the
breeders' choice was the criterion
for fitness. (Many novel breeds of pi-
geon have been created this way.)

Ethics and Genetics

Attacks on Darwinism

Plant and animal breeders have prac-
ticed artificial selection extensively.
Here, survival is not the criterion for
fitness; productivity is.

It is more difficult to establish a
priori independent criteria of fit-
ness in nature. Often, uncontrolled
or unseen vagaries have major im-
pacts on the course of events. Surely
the temperature became colder be-
fore the mammoths became woolly.
Is it then reasonable to predict that
elephants would get woolly if the
climate became colder in Africa to-
day? The answer is no, for several
reasons. First, the elephants might
adapt to colder weather in any of a
large number of different ways —
they could get fatter, they could mi-
grate, and so on. To some extent,

adaptation depends not only on the
changing environment, but also on
the reserve variation within the
gene pool of the species. Second,
the elephants could become extinct;
they might not be able to adapt at
all. And third, if the climatic changes
were not severe, the elephants
might not change at all.

Predicting the exact course of
evolution is nearly impossible. To pro-
vide independent criteria for fitness
in nature is, therefore, very difficult.
Some modern evolutionary biolo-
gists, although not doubting neo-
Darwinism, do worry to some extent
about the difficulties in testing mod-
ern evolutionary theory. However,
lower-level experiments are done to
test various aspects of evolution in
specific systems. For example, in
1993, B. Grant and P. Grant hypothe-
sized that changes in bill size would
occur in the finch Geospiza fortis
(see fig. 21.6) because of changing
food size due to changing weather on
the Galapagos Islands. Their proposal
seems to be correct. In addition, the
support for Darwinism (the fossil

belonging to the same species as the original popula-
tion. This process is known as anagenesis, or phyletic
evolution (an is Latin for without, genesis is Latin for
birth or creation). (2) Speciation may also be the diver-
gence of a population into two distinct forms (species)
that exist simultaneously. This branching process is
known as cladogenesis (clado is Greek for branch; fig.
21.2). What do we mean by the term species?

Before Darwin's time, typological thinking pre-
vailed, and a species was defined as a group of organisms
that were morphologically similar. All variants were con-
sidered imperfections of the model or type. One of Dar-
win's greatest contributions to modern biological theory
was to treat variation as a normal phenomenon in a
group of organisms. The modern biological species
concept groups together as members of the same
species organisms that can potentially interbreed. A
species, therefore, is a group of organisms that can mate
among themselves to produce fertile offspring.

Species 2

Time

Species 2

Species 3

Species 1

(a)

(b)

Figure 21.2 Forms of speciation. In anagenesis (a), a species
changes over time until it is so different from its progenitor that
it is classified as a new species. In cladogenesis (£>), speciation
takes place as a branching process wherein one species
becomes two or more.

Tamarin: Principles of
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IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

Evolution and Speciation

591

record, embryology, comparative
anatomy, geographic distributions,
etc.) is so overwhelming that the
general nature of evolution is not in
doubt. We can clearly trace its path,
although we cannot make exact pre-
dictions for its future.

From a philosophical point of
view, neo-Darwinism is the general
paradigm (broad concept) denning
"normal" biology. Every scientific en-
deavor works under the umbrella of a
paradigm. When enough inconsisten-
cies appear, a new paradigm is sought
to replace the old in what Thomas
Kuhn called a "scientific revolution."
In physics, relativity overthrew New-
tonian principles. In biology, Darwin-
ism overthrew the concept of a re-
cent, biblically described origin of
animals and plants. Darwinism be-
came the paradigm because it ex-
plained many things in a consistent
fashion that a recent origin of all
forms of life could not. Neo-
Darwinism will remain the current
paradigm unless it is overthrown by a
better theory that explains previous
inconsistencies. To date, no major

inconsistencies suggest that neo-
Darwinism is not correct.

In his article, Bethell went on to
try to refute neo-Darwinism using
the following argument: Survival of
the fittest can be redefined to mean
that some organisms have more off-
spring than others. Thus, natural se-
lection cannot be a creative force be-
cause the only thing it works on is
organisms alive now, some having
more offspring than others. How,
asks Bethell, can this possibly give us
tigers and horses from ancestors
that did not look like tigers and
horses? The answer is that mutation
produces variants in the population.
The organism best able to compete
will leave the most offspring. With
an array of different genotypes in a
population, natural selection deter-
mines which genotypes will in-
crease in future generations. Traits
that give the bearer an advantage in-
crease in the population, and evolu-
tion takes place. Natural selection
was the force behind the evolution
from the small Eocene horse to the
modern Equus.

Misinterpretation of mutation is
the basis for other attacks on Darwin-
ism. For example, Darwinian evolu-
tion has been attacked as not feasible,
since most mutations are deleterious.
How, the argument goes, can evolu-
tion proceed by a combination of
deleterious events? The answer is
that although most mutations are
deleterious, some are not. This is es-
pecially true in changing environ-
ments; yesterday's deleterious mutant
may be today's favored mutant.

The most recent attacks on Dar-
winism have been launched by cre-
ationists, who have attempted to
pass laws in many states requiring
schools to teach the biblical version
of creation as an alternative to Dar-
winism. The courts have rejected
this position because creationism is
not a scientific theory. It does not
follow the rules of the scientific
method wherein empirical evidence
can refute it.

Unfortunately, the definition of species on the basis of
interbreeding cannot be used in many places, mostly due
to the technical problems of applying it. Taxonomists and
paleontologists, who often use nonliving specimens (pre-
served or fossilized), use the morphological species
concept as a working definition. Under this concept, two
organisms are classified as belonging to the same species
if they are morphologically similar. They are classified as
belonging to two different species if they are as different
as two organisms belonging to two recognized species.
Other problems arise for taxonomists since speciation is a
dynamic process. For example, isolated subgroups of a
population may be in various stages of becoming new
species; the rate of successful interbreeding among indi-
viduals from these subgroups may range from to 100%.
How should the in-betweens be classified? There is no
correct answer. It depends on the circumstances.

Still other problems make it necessary to turn to the
morphological species concept. Haploid and asexual

species are hard to classify. Also, two organisms that will
not interbreed in nature may do so in a laboratory setting.
Thus, the interbreeding test carried out in the laboratory
(as is done frequently) is not necessarily an adequate crite-
rion for speciation. Other problems arise in classifying
groups that are geographically isolated from each other,
such as populations on islands. These individuals are physi-
cally isolated, but in many cases they can interbreed freely
when brought together with their mainland counterparts.
So, although there is a good theoretical definition of a
species (potentially interbreeding individuals), more often
than not it is necessary for biologists to apply the morpho-
logical species concept to determine whether two popula-
tions belong to the same species. In some cases, no deci-
sion can be made about the species status of a population.
It is clear that a population has evolved, but it is not clear
whether it has evolved enough to be called a new species.
However, this is more of a problem for taxonomists and
evolutionary biologists than for the organisms themselves.

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IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

592

Twenty-One Evolution and Speciation

Mechanisms of Cladogenesis

Reproductive Isolation

How does one species become two? Basically, repro-
ductive isolating mechanisms must evolve to prevent
two subpopulations from interbreeding when they come
into contact. Reproductive isolating mechanisms are en-
vironmental, behavioral, mechanical, and physiological
barriers that prevent individuals of two species from pro-
ducing viable offspring. Following is a modification of
the classification system of isolating mechanisms sug-
gested by evolutionary biologist G. L. Stebbins:

1. Prezygotic mechanisms prevent fertilization and zy-
gote formation.

a. Residential — The populations live in the same re-
gion, but occupy different habitats.

b. Seasonal or temporal — The populations exist in
the same region, but are sexually mature at differ-
ent times.

c. Ethological (in animals only) — The populations
are isolated by incompatible premating behavior.

d. Mechanical — Cross-fertilization is prevented or re-
stricted by incompatible differences in reproduc-
tive structures.

2. Postzygotic mechanisms affect the hybrid zygotes af-
ter fertilization has taken place.

a. F : hybrid breakdown — ¥ 1 hybrids are inviable or
weak.

b. Developmental hybrid sterility — Hybrids are ster-
ile because gonads develop abnormally or because
meiosis breaks down before it is completed.

c. Segregational hybrid sterility — Hybrids are sterile
because of abnormal distribution to the gametes of
whole chromosomes, chromosome segments, or
combinations of genes.

d. F 2 breakdown — F : hybrids are normal, vigorous,
and fertile, but the F 2 generation contains many
weak or sterile individuals.

Allopatric, Parapatric, and Sympatric Speciation

Reproductive isolating mechanisms are barriers to gene
flow, the spread of genes between populations. These
isolating mechanisms can evolve in three different ways,
each of which defines a different mechanism of specia-
tion. Usually, the mode of speciation is dictated by both
the properties of the genetic systems of the organisms
and stochastic (random) or accidental events. For exam-
ple, vertebrates tend to have different speciation modes
than phytophagous (plant-feeding) insects.

The appearance of a geographic barrier, such as a
river or mountain, through the range of a species physi-
cally isolates populations of the species. Physical isola-
tion can also occur if migrants cross a particular barrier

and begin a new population (founder effect). The physi-
cally isolated populations can then evolve independently.
If reproductive isolating mechanisms evolve, then two
distinct species are formed, and if they come together in
the future, they remain distinct species. Speciation that
occurs because reproductive isolating mechanisms
evolve during physical separation of the populations is
called allopatric speciation (fig. 21.3). As evolutionary
biologist Guy Bush pointed out, "Although examples in
nature are difficult to substantiate ... it [allopatric speci-
ation] has been convincingly demonstrated in frogs . . .
and lizards."

Reproductive isolating mechanisms usually originate
incidentally to the speciation process. That is, they arise in-
cidentally during the process of evolution in isolated pop-
ulations rather than being selected for. When isolated
populations come together again, incomplete isolating
mechanisms may allow hybrids to form. If the hybrids are
normal and viable and can freely interbreed with individ-
uals of each parent population, then no speciation has
taken place. However, if the hybrids are at a disadvan-
tage, natural selection may favor stronger isolating mech-
anisms. In this case, organisms that mate with individuals
from the other population leave fewer offspring. The re-
sult is a more effective barrier to hybridization. Regions
in which previously isolated populations come into con-
tact and produce hybrids are called hybrid zones.

Until recently, evolutionary biologists believed that al-
lopatric speciation was the general rule. Many now be-
lieve that two other modes of speciation may occur fre-
quently in certain groups of organisms. Parapatric
speciation occurs when a population of a species that
occupies a large range enters a new niche or habitat (fig.
21.3). Although no physical barrier arises, the new niche
acts as a barrier to gene flow between the population in
the new niche and the rest of the species. Here again, re-
productive isolating mechanisms evolve to produce two
species where there was only one before. Parapatric spe-
ciation is believed to have occurred often in relatively
nonvagile animals such as snails, flightless grasshoppers,
and annual plants. Sympatric speciation occurs when
a polymorphism, which is the occurrence of alternative
phenotypes in the same population, arises within an in-
terbreeding population before a shift to a new niche. This
mode of speciation may be common in parasites and phy-
tophagous insects. For example, if a polymorphism arises
within a parasitic species that allows an individual with a
certain genotype to adapt to a new host, this genotype
may be the forerunner of a new species. If the parasite
not only feeds on the new host but also mates on the
new host, a barrier to gene flow arises, although the par-
asite may be surrounded by other members of its species
with the original genotype. Sympatric speciation can
thus occur in the middle of a species range rather than at
the edges (fig. 21.3).

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Allopatric
speciation

Parapatric
speciation

Original
population

Initial step of
speciation process

Barrier forms

New niche
entered

Sympatric
speciation

Figure 21.3 The three general
mechanisms of speciation. In
allopatric speciation, reproductive
isolation evolves after the population
has been geographically divided. In
parapatric speciation, reproductive
isolation evolves when a segment of
the population enters a new niche.
In sympatric speciation, reproductive
isolation evolves while the incipient
group is still in the vicinity of the
parent population.

Polymorphism
occurs

Evolution of
reproductive
isolating mechanisms

In isolation

In new niche

New, genetically
distinct species
after equilibration
of new ranges

Within the
population

An example of incipient sympatric speciation has
been seen recently in host races of the apple maggot fly
{Rhagoletis pomonelld) in North America (fig. 21.4).
This fly was found originally only on hawthorn plants.
However, in the nineteenth century it spread as a pest to
newly introduced apple trees. In fact, races are now
known on pear and cherry trees and on rose bushes.
These races have developed genetic, behavioral, and eco-
logical differences from the original hawthorn-dwelling
parent. Evolutionary biologists view this as an opportu-
nity to observe sympatric speciation as it occurs.

Another form of sympatric speciation occurs when
cytogenetic changes take place that result in "instanta-
neous speciation." These cytogenetic changes include
polyploidy and translocations. For example, if polyploid
offspring cannot produce fertile hybrids with individuals
from a parent population, then the polyploid is repro-
ductively isolated. This mechanism is much more com-
mon in plants because they can exist vegetatively despite
odd ploidy and they usually do not have chromosomal
sex-determining mechanisms, which are especially vul-
nerable to ploidy problems (see chapter 8).

The end result of cladogenesis is the divergence of
a homogeneous population into two or more species.
One of the classic examples of cladogenesis appears in

Figure 21.4 The apple maggot fly, Rhagoletis pomonella. This
species has exhibited host range expansion since the
nineteenth century from hawthorn to apple, cherry, and roses.
Host races are presumably the initial step in sympatric
speciation. Magnification 10X. (Source: Jeffrey L Feder and Guy L
Bush, Zoology Department, Michigan State University.)

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Twenty-One Evolution and Speciation

the ground finches of the Galapagos Islands. These
birds are very well studied not only because they
present a striking case of speciation, but also because
Darwin studied them and was strongly influenced by
them in his views. Figure 21.5 is a map of the Galapa-
gos Islands, and figure 21.6 is a diagram of the species
of Darwin's finches.

An original flock of finches somehow reached the
Galapagos Archipelago from South America, 700 miles
away, and with time spread to the various islands of the
Galapagos Archipelago. Given the limited ability of the
birds to get from island to island, allopatric speciation
took place. On each island, the finch population evolved
reproductive isolating mechanisms while evolving to fill
certain niches not already filled on the islands. For ex-
ample, in South America, no finches have evolved to be
like woodpeckers because many woodpecker species al-
ready live there. But the Galapagos Islands, being iso-
lated from South America, have what is called a depau-
perate fauna, a fauna lacking many species found on
the mainland. The islands lacked woodpeckers, and a
very useful food resource for birds — insects beneath the
bark of trees — was going unused. Finches that could

92°W

91 °W

90°W

<p Culpepper
o Wenman

1°N

North
Albemarle

Narborough

£j Abingdon
BindloeP

1°S

South
Albemarle

James

1°N

q Tower

Dunci e n r ^l ndefati 9 able

Barrington

£*>

Chathany^

Charles

1°S

H °°d

92°W

91 °W

90°W

10 20 30 40 50

Scale in land miles

Figure 21.5 The Galapagos Archipelago is located about 700
miles west of Ecuador. This isolated chain of islands is a
natural laboratory for the study of evolutionary processes.
(From David Lack, Darwin's Finches. Copyright © 1947 by Cambridge
University Press, New York, NY. Reprinted by permission.)

make use of this resource would be at an advantage and
would thus be favored by natural selection. On one is-
land, a finch did evolve to use this food resource. The
woodpecker finch acts like a woodpecker by inserting
cactus needles into holes in dead trees to extract insects.
Darwin wrote: "Seeing this gradation and diversity of
structure in one small, intimately related group of birds,
one might really fancy that from an original paucity of
birds in this archipelago, one species had been taken
and modified for different ends."

Phyletic Gradualism Versus Punctuated
Equilibrium

Darwin visualized cladogenesis as a gradual process,
which we refer to as phyletic gradualism. However, an
alternative view arose in 1972, when N. Eldredge and
S. J. Gould suggested that speciation itself, and the mor-
phological changes accompanying speciation, occur rap-
idly, separated by long periods of time when little change
occurs (stasis). They called their model punctuated
equilibrium (periods of stasis punctuated by rapid evo-
lutionary change). Although figure 21.7 presents what
appear to be two clear alternatives, in practice the mod-
els are very hard to tell apart. They both start with the
same ancestral species and predict the same number of
modern species. Allopatric, parapatric, and sympatric
speciation mechanisms apply to both punctuated equi-
librium and phyletic gradualism. The only major differ-
ence between the models is the rate of change, and this
can only be discovered from an almost complete fossil
record. The punctuated equilibrium model has brought
much excitement to modern evolutionary biology. We
await a time in the near future when we can decide
which model has predominated in evolutionary history.

Stephen J. Gould (1943- ).
(Courtesy of Dr. Stephen J. Gould
and the Harvard University News
Office.)

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Large ground-finch

Geospiza magnirostris

Medium ground-finch

Geospiza fortis

Cactus ground-finch

Geospiza scandens

Large cactus
ground-finch

Geospiza conirostris

Small ground-finch

Geospiza fuliginosa

Woodpecker-finch

Camarhynchus pallidus

Sharp-beaked ground-finch

Geospiza difficilis

Mangrove-finch

Camarhynchus heliobates

Warbler-finch

Certhidea olivacea

Large insectivorous tree-finch

on Charles

Camarhynchus pauper

Large insectivorous tree-finch

Camarhynchus psittacula

Cocos-finch

Pinaroloxias inornata

Small insectivorous tree-finch

Camarhynchus parvulus

Vegetarian tree-finch

Camarhynchus crassirostris

Migrants from
South American mainland

Figure 21.6 Species of Darwin's finches. These birds apparently evolved from a single group of migrants from the South
American mainland. Isolated on the different islands, the birds evolved to fill many vacant niches.

Time

(a)

(b)

Figure 21.7 Diagrammatic interpretation of cladogenesis.
(a) Phyletic gradualism is depicted as a gradual divergence
over time, (b) Punctuated equilibrium is depicted as a rapid
divergence of two groups after long periods of no change.
The horizontal axis is some arbitrary measure of species
differences.

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Twenty-One Evolution and Speciation

GENETIC VARIATION

Darwinian evolution depends on the variation within a
population. E. B. Ford, a British evolutionary biologist,
applied the term genetic polymorphism to the oc-
currence of more than one allele at a given locus. Usu-
ally, we consider a locus polymorphic if a second allele
occurs in the population at a frequency of 5% or more.
Before the mid-1960s, the general belief was that only a
few loci were polymorphic in any individual or any
population.

In 1966, two researchers found a way to sample the
genome in what they perceived was a random manner.
R. C. Lewontin and J. L. Hubby used acrylamide-gel elec-
trophoresis (see chapter 5) to investigate variability in a
fruit fly species, Drosophila pseudoobscura. (H. Harris
reported independent, similar work with human DNA.)
Lewontin and Hubby reasoned that choosing enzymes
and general proteins that are amenable to separation by
electrophoresis, is, in fact, choosing a random sample of
the genome of the fruit fly. If this is the case, then the de-
gree of polymorphism found by electrophoretic sam-
pling would provide an estimate of the amount of vari-
ability occurring in the individual organism and in the
population. Their results were startling.

Lewontin and Hubby found that the species was
polymorphic at 39% of eighteen loci examined, the aver-
age population was polymorphic at 30% of its loci, and
the average individual was heterozygous at 12% of its
loci. The high rate of polymorphism sparked two inter-
related controversies. The first was whether elec-
trophoresis does, in fact, randomly sample the genome.
The second was whether most electrophoretic alleles
are maintained in the population by natural selection.
Let us return to the arguments after looking at ways in
which genetic polymorphisms could be maintained in
natural populations.

Edmund Brisco Ford
(1901-1988). (Courtesy of
Professor Edmund Brisco Ford.)

Richard C. Lewontin (1929- ).
(Courtesy of Dr. Richard C. Lewontin.)

Maintaining Polymorphisms

Heterozygote Advantage

When selection acts against both homozygotes, an equi-
librium is achieved, dependent solely on the selection co-
efficients, that maintains both alleles (see chapter 20).
The classic example of heterozygote advantage in human
beings is sickle-cell anemia. Sickle-cell hemoglobin (Hb s )
differs from normal hemoglobin (Hb A ) because it has a
valine in place of a glutamic acid in position number 6 of
the beta chain of the globin molecule. When the avail-
ability of oxygen is reduced, the erythrocytes containing
sickle-cell hemoglobin change from round to sickle-
shaped cells (see fig. 2.28). There are two unfortunate
consequences: (1) sickle-shaped cells are rapidly broken
down, which causes anemia as well as hypertrophy of
the bone marrow, and (2) the sickle cells clump, which
blocks capillaries and produces local losses of blood flow
that result in tissue damage.

This condition of reduced fitness would lead one to
predict that the sickle-cell allele would be selected against
in all populations and, therefore, would be rare. But this is
not the case. The sickle-cell allele is common in many
parts of Africa, India, and southern Asia. What could pos-
sibly maintain this detrimental allele? In the search for an
answer to this question, biologists discovered that the dis-
tribution of the sickle-cell allele coincided well with the
distribution of malaria. The following facts have now
been uncovered. The sickle-cell homozygote (Hb s Hb s ^) al-
most always dies of anemia. The sickle-cell heterozygote
(Hb A Hb s ) is only slightly anemic and has resistance to
malaria. The normal homozygote (Hb A Hb A ) is not anemic
and has no resistance to malaria. Thus, in areas where
malaria is common, the most fit genotype of the three
appears to be the sickle-cell heterozygote, which has
resistance to malaria and only a minor anemia.

This conclusion is supported by the changes in allelic
frequencies that occur when a population from a malarial
area moves to a nonmalarial area. Since the normal ho-
mozygote is no longer at risk for malaria, selection acts
mainly on the sickle-cell homozygote and, to a slight ex-
tent, on the heterozygote. Table 21.1 shows data for
African blacks versus African Americans. The African pop-
ulation is, of course, under malarial risk, whereas the
American population is not. The sickle-cell hemoglobin al-
lele (Hb s ) is reduced in frequency in African Americans.

Heterozygote advantage is an expensive mechanism
for maintaining a polymorphism. Losses must occur in
both homozygous groups in order for the polymorphism
to exist. Thus, part of the reproductive output of a popu-
lation is lost each generation to maintain each polymor-
phism under heterozygote advantage. In the case of
sickle-cell anemia, this means a tragic loss of human life
due to either anemia or malaria. (The loss of individuals

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Table 21.1 Sickle-Cell Anemia Frequencies in African Blacks and African Americans

Percentage of
Homozygotes (Hb A Hb A ^)

Percentage of
Heterozygotes (Hb A Hb s ^)

Frequency of Hb s (q)

African Blacks (Midcentral Africa)
African Americans

82
92

18
8

0.09
0.04

to maintain genetic variation at a particular locus is called
genetic load. In the sickle-cell case, it is due to the seg-
regation of individuals with lowered fitness and is there-
fore called segregational load.) Very few other exam-
ples of heterozygote advantage have been documented.

Frequency-Dependent Selection

All the selection models discussed so far (chapter 20) have
had selection coefficients that were constants. This is not
always the case. For example, L. Ehrman has shown that
when a female fruit fly has a choice between mates with
different genotypes, the female fly chooses to mate with a
male with a rare genotype. Frequency-dependent selec-
tion is selection in which the fitnesses of genotypes
change according to their frequencies in the population.

The population geneticist Bruce Wallace has coined
the terms hard selection and soft selection to deal with
cases of frequency and density dependence. (Density-
dependent selection exists when the fitness of a genotype
changes as population density changes. We will not deal
with that here.) Wallace defined soft selection as selection
in which the selection coefficients depend on the fre-
quency and density of genotypes. Hard selection is selec-
tion that is independent of both frequency and density.
For example, the low fitness of sickle-cell anemia ho-
mozygotes involves hard selection because of the objec-
tively deleterious effects of the anemia. Soft selection

Lee Ehrman (1935- ). (Courtesy of Dr. Lee Ehrman. Photo by Jan
Robert Factor.)

could be envisioned as selection that might act on aggres-
sive behavioral genotypes in some lemming and field
mouse species. When population density and frequency
of the genotypes are low, these animals survive and re-
produce. As population density increases, there can be a
selection for more aggressive genotypes because they
may be more successful in obtaining resources. As density
increases further and the frequencies of the aggressive
genotypes increase, they may be selected against because
of the preoccupation of these aggressive individuals with
territory defense under crowded conditions. This has
been suggested as a mechanism of wildlife's" lemming cy-
cle," rapid declines in the density of lemming and field
mouse populations every three to five years.

A model for frequency-dependent selection can be
constructed by assigning fitnesses that are not constants.
One way to do this is to assign fitnesses that are a function
of allelic frequencies. Thus, the assigned fitnesses for one
locus with two alleles could be (1.5 — p),l, and (1.5 — q)
for the AA, Aa, and aa genotypes, respectively (table
21.2). An interesting outcome of this model is that ?Xp =
q = 0.5, the system is in equilibrium, and no selection
takes place because all the fitnesses are equal to 1 .

Another way of looking at frequency-dependent selec-
tion is to look at the situation in which each genotype ex-
ploits a slightly different resource. As a genotype becomes
rare, competition for the resource that genotype uses will
likely decrease, and the genotype will thus have an advan-
tage over the common genotypes, which are competing for
resources. This type of selection is probably very common.

Transient Polymorphism

A genetic polymorphism can result when an allele is be-
ing eliminated either by random or selective mecha-
nisms. If a population starts out homozygous for the a al-
lele, for example, and a mutation brings in a more favored
A allele, the population gradually becomes all A through
directional selection. However, during the process of re-
placement, both alleles are present.

Other Systems

Selection at one stage in the life cycle of an organism can
balance a different form of selection at another stage in
the life cycle. For example, an allele can be favored in a

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Twenty-One Evolution and Speciation

Table 21 .2 Selection Model of Frequency-Dependent Selection: The A Locus with the A and a Alleles

Genotype

Total

Initial genotypic frequencies

2pq

q 2

Fitness (IF)

1.5 -p

1.5 - q

Ratio after selection

p\\5 ~ P)

2pq

tf 2 (1.5 - q)

W -

- 0.5 + 2pq

Genotypic frequencies after selection

p\l3 ~P)

2pq

tf 2 (1.5 - q)

larva but selected against in an adult. There can also be a
balance of selection in different parts of the habitat in a
heterogeneous environment. For instance, an allele can
be favored in a wet part of the habitat but selected
against in a dry part.

Maintaining Many Polymorphisms

In summary, allelic polymorphisms in a population were
classically accounted for by heterozygote advantage,
frequency-dependent selection, or, infrequently, some
other mechanism. Until Lewontin and Hubby did their
work, heterozygote advantage was considered the most
common method of maintaining a polymorphism at a
given locus. The maintenance of an allele by heterozy-
gote advantage costs the population a certain number of
its offspring due to the mortality (or sterility) of the ho-
mo zygotes. Most populations can afford the loss if poly-
morphisms are maintained at only a few loci. After
Lewontin and Hubby reported that polymorphisms
seemed to exist at a large proportion of loci, new expla-
nations were needed to account for them. Three expla-
nations were considered:

1 . Electrophoresis (the technique used in Lewontin and
Hubby's research) does not randomly sample the
genome, and thus the large amount of variability they
found does not really exist.

2. New population genetic models can be derived that
explain how natural selection maintains this large
amount of variability.

3. Electrophoretic alleles are not under selective pres-
sure. That is, allozymic forms of an enzyme all perform
the function of the enzyme equally well. This idea is
called the neutral gene hypothesis.

Sampling the Genome

Does electrophoresis randomly sample the genome?
Since, on the basis of DNA content, the genome of higher
organisms has the potential to contain half a million
genes, this question may be difficult to resolve. Since the

original reports of Lewontin and Hubby and Harris, nu-
merous studies on many different organisms agree, for
the most part, on the high amount of polymorphism in
natural populations (table 21.3). However, several lines
of evidence suggest that the results from electrophoresis
are actually underestimates of the true amount of genetic
variability present in a population.

The majority of amino acid substitutions, for exam-
ple, do not change the charge of the protein. Thus, what
appear to be single bands on an electrophoretic gel
could actually be heterogeneous mixtures of the prod-
ucts of several alleles. Also, we now know that glycolytic
enzymes are less polymorphic than other enzymes. Since
glycolysis is a limited process in which most enzymes are
not involved, it follows that the average heterozygosity
over all loci should be slightly higher than the original es-
timates that included glycolytic enzymes. Recent techni-
cal advances of multidimensional electrophoresis and
DNA sequencing support the hypothesis that elec-
trophoresis does randomly sample the genome. How-
ever, DNA sequencing studies have shown that abundant
variation exists, especially in the third (wobble) position
of codons, and in parts of introns. Heterozygosity at the
DNA sequence level seems to approach 100%.

Multilocus Selection Models

Can standard genetic models account for the high degree
of variability in natural populations? If each locus is con-
sidered independently, then for each polymorphic locus,
offspring in a population lost to maintain that polymor-
phism by heterozygote advantage are independent of off-
spring lost due to selection at other loci. The losses would
soon outstrip the reproductive capacity of any species.
Models proposed since Lewontin and Hubby's report
have suggested that natural selection favors the individu-
als that are the most heterozygous overall. Individuals se-
lected against because of their homozygosity would be in-
dividuals with many homozygous loci. In other words,
natural selection acts on the entire genome, not on each
locus separately. We can show algebraically that the large

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21. Evolution and
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599

Table 21.3 Survey of Genie Heterozygosity

Proportion

of Loci

Number of

Number

Polymorphic

Heterozygosity

Standard Error

Species

Populations

of Loci

per Population

per

Locus

of Heterozygosity

Homo sapiens

0.28

0.067

0.018

Mus musculus musculus

0.29

0.091

0.023

M. m. brevirostris

0.30

0.110

—

M. m. domesticus

0.20

0.056

0.022

Peromyscus polionotus

7 (regions)

0.23

0.057

0.014

Drosophila pseudoobscura

0.43

0.128

0.041

D. persimilis

0.25

0.106

0.040

D. obscura

3 (regions)

0.53

0.108

0.030

D. subobscura

0.47

0.076

0.024

D. willistoni

2-21

0.86

0.184

0.032

0.81

0.175

0.039

D. melanogaster

0.42

0.119

0.037

D. simulans

0.61

0.160

0.052

Limulus polyphemus

0.25

0.061

0.024

Source: The Genetic Basis of Evolutionary Change by R. C. Lewontin, (New York: Columbia University Press, 1974). Reprinted with permission of the publisher.
Note: See source (Lewontin, 1974) for individual references.

number of polymorphisms that exist in natural popula-
tions could be maintained according to these models.

Neutral Alleles

The high incidence of polymorphism that electrophoresis
reveals may not be important from an evolutionary point
of view. If all or most electrophoretic alleles are neutral
(i.e., if no allele is more fit than its alternative) or only very
slightly deleterious, there is virtually no selection at these
loci, and the variation observed in the population is
merely a chance accumulation of a combination of muta-
tion and genetic drift. This model, proposed by M. Kimura
of Japan, is an alternative to the natural selection model.

Motoo Kimura
(1924-1994). (Courtesy of
Dr. Motoo Kimura.)

Which Hypothesis Is Correct?

Researchers who favor the concept that most elec-
trophoretic alleles are neutral do not deny that selection
exists. They do not hold that evolution is non-adaptive,
but say merely that most of the molecular variation
(electrophoretic) found in nature is not related to fit-
ness — it is neutral. Thus, the demonstration that selec-
tion actually exists, in electrophoretic systems or other-
wise, is not proof against the neutralist view. No one
denies the explanation for the maintenance of sickle-cell
anemia. Selection at several other electrophoretic sys-
tems is also known.

For example, R. Koehn showed that different alleles
of an esterase locus in a freshwater fish in Colorado pro-
duced proteins with different enzyme activities at differ-
ent water temperatures. Koehn then showed that the al-
leles were distributed as one would predict on the basis
of the water temperature. In other words, the distribu-
tion of alleles correlated with the distribution of water
temperature. The enzyme produced by the ES-l a allele
functioned best at warm temperatures, whereas the en-
zyme produced by the ES-1 allele functioned best at
cold temperatures. The cold-adapted enzyme was preva-
lent in the fish in colder waters (higher latitudes), and
the warm-adapted enzyme was prevalent in the fish in
warmer waters (lower latitudes; fig. 21.8).

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Twenty-One Evolution and Speciation

38
37

CD
■D

B 36

S 35

5? 34
33

•
•

•

31
C

i i i

) 0.5

1.0

Frequency of Es-I a

Figure 21.8 Relation of latitude and frequency of the warm-
adapted esterase allele Es-I a in populations of the fish
Catostomus clarki. Note how the frequency of the allele
increases as latitude decreases (warmer water). (From Richard

Koehn, "Functional and evolutionary dynamics of polymorphic esterases in
catostomid fishes," Transactions of the American Fisheries Society, 99:223.
Copyright © 1970 American Fisheries Society, Bethesda, MD.)

Isolated instances of selection, however, do not ade-
quately prove the case for maintaining variation by means
of natural selection or disprove the case for maintaining
variation of neutral alleles. Both theories recognize natu-
ral selection as the guiding force in producing adapted or-
ganisms. What is needed is proof that the majority of poly-

morphic loci are either being selected or are neutral. For
this proof, many loci must be examined independently —
a very difficult undertaking — or some grand pattern must
emerge supporting one hypothesis or the other.

Grand Patterns of Variation

Clinal Selection

Data on the geographic distribution of alleles fail to ade-
quately support either theory. Often, a single allele pre-
dominates over the range of a species (fig. 21.9). Changes
in allelic frequency from one geographic area to another
can often be attributed to clinal selection, selection
along a geographic gradient, in which allelic frequencies
change as altitude, latitude, or some other geographic at-
tribute changes. Note in figure 21.9 the general increase
in the Es-5 b frequency from west to east in the southern
United States. But, in line with the neutralist view, geo-
graphic patterns similar to those in figure 21.9 can also
be produced by neutral alleles with a very low level of
migration, as little as one individual per one thousand per
generation.

Molecular Evolutionary Clock

The advancing technology that made it possible to detect
the sequence of amino acids in a protein also made it
possible to discover how much the proteins and DNA of
various species differ. In chapter 17, we discussed the
use of mitochondrial DNA (mtDNA) to determine evolu-
tionary relationships. Currently, protein, nuclear DNA,
and mtDNA clocks are being studied.

Niihau

Kauai

Molokai

Lanai o

Kahoolawe
Es-5 a ■ 1 Es-5 b

Hawaii

N = 24
Maui

Denmark

50 mi.
I 1

Qj M. m.jriusculus
N = 79 L

M. m. domesticus

Figure 21.9 Frequency distribution of the
Es-5 alleles of an esterase locus in house
mice. Each circle represents allelic
frequencies at that geographic location.
Note the general tendency for the Es-5 b
allele to increase from west to east in the
continental United States. (From L Wheeler

and R. Selander, "Genetic Variation in Populations of
the House Mouse, Mus musculus, in the Hawaiian
Islands," Studies in Genetics, VII, 1972. University of
Texas Publication 7213. Reprinted with permission of
M. R. Wheeler.)

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601

Knowledge of the changes in amino acid sequences
can be used to estimate the rate of evolutionary change.
That is, the data show how many amino acid substitu-
tions have occurred between two known groups of or-
ganisms. The genetic code dictionary allows us to esti-
mate the minimum number of nucleotide substitutions
required for this change. For example, if one protein con-
tains a phenylalanine in position 7 (codons UUU, UUC),
and the same protein in a different species has an
isoleucine in the same position (AUU, AUC, AUA), we can
see that the minimum number of substitutions to convert
a phenylalanine codon to an isoleucine codon is one
(UUU — > AUU). When we know the minimum number of
substitutions, we can calculate molecular evolutionary
rates, nucleotide substitutions per million years. In a
sense, these rates provide us with a molecular evolu-
tionary clock that measures evolutionary time in nu-
cleotide substitutions.

Many studies of the rate of amino acid and nucleotide
substitutions have been done on hemoglobin, on cy-
tochrome c, on a class of proteins involved in blood clot-
ting called fibrinopep tides, and on many others. Fig-
ure 21.10 shows the way in which an amino acid
sequence differs among species. From comparisons of
this type, we can calculate the actual number of amino
acid differences, as well as percentage differences. Table
21.4 is a compilation of percentage differences between
various species based on the cytochrome c protein. This
type of information can be used two ways.

First, we can construct a phylogenetic tree that tells
us the evolutionary history of the species under consid-
eration (fig. 21.11). This tree can be compared with phy-
logenetic trees constructed by more classical means us-
ing fossil evidence and evidence from morphology,
physiology, and development. From the comparisons, we
can look at areas of disagreement in an attempt to find
out the best way to create phylogenetic trees. In addi-
tion, molecular phytogenies can give us information un-
attainable in any other way, as, for example, when the fos-
sil record is incomplete or ambiguous.

A second use of DNA or amino acid difference data is
to determine average rates of substitution. Once we

Human being

(Tyr

fl_euj

iLys

Lysj

(Ala

.ThrYAsn

(gIu)

Dog

(Tyr

(Leu)

^ Lys

. LysJ

[Ala)

> ThrlLys,

(Glu)

Chicken

(Tyr

iLeul

J-ys-

, Asp)

,Ala)

.ThrYser

^Lys)

know the current amino acid differences in the proteins
of two species, it is possible to estimate the actual num-
ber of nucleotide substitutions that have taken place
over evolutionary time using the statistical Poisson distri-
bution, which deals with rare events. The index, K, is the
average number of amino acid substitutions, per site, be-
tween two proteins:

K = -ln(l - p)

in which In is the natural logarithm (to the base e), and
p = d/n in which d is the number of amino acid differ-
ences and n is the total number of amino acid sites being
compared. For example, in figure 21.10, n = 8 and d = 3
between the dog and chicken. Thus

K = -ln(l - 0.375) = 0.47

Therefore, the average number of amino acid substitu-
tions, per site, between dog and chicken is 0.47.

We can take this calculation one step further by de-
termining the per-year rate:

k = K/2T

in which k is the amino acid substitution rate per site per
year, and T is the number of years since the two species
diverged from a common ancestor. We divide by 2T
because each side of the tree has evolved independently
for T years. When ks are calculated for many proteins
over many species, they cluster around 10~ 9 (table 21.5).
In fact, Kimura has suggested the unit of a pauling to be
equal to 10~ 9 amino acid substitutions per year per site in

Human

Gorilla

Gibbon

Monkey

Rodent

Rabbit

Dog

Horse

Donkey

Pig

Llama

Sheep

Goat

Cow

Kangaroo

Chicken

Frog

Fish

Figure 21.10 The amino acids making up the terminal portion
of cytochrome c in three species. Note the similarities and
differences.

Figure 21.11 Composite evolution of hemoglobin, cytochrome c,
and fibrinopeptide A. The total number of nucleotide substitutions
appears on the horizontal axis. Note how the tree groups similar
organisms and generally agrees with classical systematics. (From
C. H. Langley and W. M. Fitch, "An examination of the constancy of the rate of
molecular evolution," Journal of Molecular Evolution, 3:168. Copyright © 1974
Springer-Verlag, Heidelberg. Reprinted by permission.)

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Twenty-One Evolution and Speciation

Table 21 .4 Amino Acid Differences (By Percentage)

in Cytochrome c

Between Different

Organisms

••H

5/3

J)
u

2
u

S3
H

u
O

5/3

•d
5/3

5fi
U
5/3

d
5/3

5*5
5*5

Human being

Pig, bovine, sheep

Horse

Chicken, turkey

Snapping turtle

Bullfrog

Tuna

Carp

Lamprey

Fruit fly

Screwworm fly

Silkworm moth

Sesame

Sunflower

Wheat

Candida krusei

Baker's yeast

Neurospora crassa

Rhodospirillum rubrum

Source: From M. O. Dayhoff, ed., Atlas of Protein Sequence and Structure," National Biomedical Research Foundation, Washington, D.C., 1972. Reprinted with
permission.

honor of Linus Pauling, who, along with E. Zuckerkandl,
first proposed the concept of a molecular clock in 1963.
If the values of k (such as those in table 21.5) form a nor-
mal distribution around 10~ 9 , then 10~ 9 would be the
rate of "the" molecular evolutionary clock. So far, the data
have been too limited to determine the distribution.

Although controversy still exists, the neutralists have
interpreted the relative constancy of the molecular evolu-
tionary clock as strong evidence in support of the neutral
gene hypothesis. A constant rate of molecular evolution
over many groups of organisms over many different time
intervals implies that the substitution rate is a stochastic
or random process rather than a directed or selectional
process. This is not to say that no adapted changes occur
in proteins or that there are no constraints. In fact, the ev-
idence suggests that three classes of amino acids can be
grouped in terms of substitution rate: invariant, moder-
ately variant, and hypervariant. It seems possible that vir-
tually no substitutions of amino acids will occur in and
around the active site of the enzyme since any amino acid
change in that area might be deleterious or lethal. For ex-

ample, a segment of cytochrome c that runs from amino
acids 70 to 80 is invariant in all organisms tested. This area
includes a binding site of the protein.

DNA Variation

If the neutralist view of molecular evolution is correct,
we should be able to make some predictions about rates
of change in DNA. For example, we predict that DNA un-
der greater constraint should amass fewer base changes
than DNA under lesser constraint. We could test this by
looking at the accumulation of mutations in the three po-
sitions of the codon, or we could look at DNA that is not
directly translated, such as pseudogenes (see chapter 15)
or introns, which are probably under lesser constraint.
Let us first look at the three positions of the codon.

A reexamination of the codon dictionary (see table
11.4) shows that the third, or wobble, position of the
codon should be under less constraint. Eight amino acids
belong to unmixed families; their amino acids are defined
by the first and second positions coupled with any of the

Tamarin: Principles of
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21. Evolution and
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603

Table 21.5 Evolutionary Rates 00 as 10" 9

Substitutions Per Amino Acid Site Per
Year for Various Proteins

Protein

Fibrinopeptide

8.3

Pancreatic ribonuclease

2.1

Lysozyme

2.0

Hemoglobin alpha

1.2

Myoglobin

0.89

Insulin

0.44

Cytochrome c

0.3

Histone H4

0.01

Source: From M. Kimura, The Neutral Theory of Molecular Evolution, Cam-
bridge University Press, 1983- Reprinted with the permission of Cambridge
University Press.

four bases in the third position of the codon.The remaining
amino acids belong to mixed families; the first two posi-
tions and the purine or pyrimidine nature of the third posi-
tion in their codons is important. Hence, the wobble (third)
position of the codon is under the least constraint and
should build up the most neutral or near-neutral mutations.

In addition, analysis of changes in the first and second
positions indicates that more drastic change takes place
by mutation of the second rather than the first position
of the codon. Thus, we predict that evolutionary dis-
tance, as measured by base substitutions, should be great-
est for the third codon position and least for the second
position. This turns out to be generally true (table 21.6).

It should be clear that a major problem facing those
who study evolutionary clocks is how to calibrate them.
Are average changes uniform throughout lineages? Do
clocks speed up, slow down, or show other unpre-
dictable changes through time? There is evidence, for ex-
ample, that both the nuclear and mitochondrial DNA
clocks have slowed down in the hominid lineage as corn-
Table 21.6 Evolutionary Distance of Codons,

Measured in Base Substitutions Per
Nucleotide Site

Codon Site

1 3

Beta globin, human being vs. mouse
Beta globin, chicken vs. rabbit
Rabbit, alpha vs. beta globin

0.13
0.19
0.44

0.17 0.34
0.30 0.64
0.54 0.90

pared with old world monkeys. If the clocks change
speed in different lineages, at different times, and for dif-
ferent parts of the genome, there will be errors in inter-
preting lineages and errors in using averages to under-
stand the general patterns of change.

At this point, it is probably safe to say that while nat-
ural selection acts to create organisms that are adapted to
their environments (see box 21.2 on mimicry), many
nucleotide and amino acid changes may not have measur-
able effects on the fitness of the organism, and hence
their frequencies may be determined by the stochastic
processes of mutation and genetic drift. Adaptation is by
natural selection, but neutral variation most certainly also
occurs in organisms.

SOCIOBIOLOGY

We close this chapter by looking at a level of evolution
only recently addressed. In 1975, E. O. Wilson published
a mammoth tome entitled Sociobiology: The New Syn-
thesis. This book has been at the center of major contro-
versies that have spread to the fields of sociology, psy-
chology, anthropology, ethology, and political science.
The basic premise of the book is that social behavior is
under genetic control. Although Wilson's book contains
twenty-six chapters concerned with the animal king-
dom, controversies have arisen because of the one chap-
ter that applies the theory to human beings.

Altruism

V C.Wynne-Edwards published a book in 1962 entitled A n-
imal Dispersion in Relation to Social Behavior. In it, he
suggested that animals regulate their own population den-
sity through altruistic behavior. For example, under
crowded conditions, many birds cease reproducing. The in-
terpretation of this phenomenon was that these birds were
being altruistic: Their failure to breed was for the ultimate
good of the species. (Altruism means risking loss of fitness
in an act that could improve the fitness of another individ-
ual.) Wynne-Edwards suggested a mechanism called group

Edward O.Wilson (1929- ).
(Courtesy of Dr. Edward 0. Wilson.
Photo by Pat Hill/OMNI Publications
International, Ltd.)

Source: From M. Kimura, The Neutral Theory of Molecular Evolution, Cam-
bridge University Press, 1983- Reprinted with the permission of Cambridge
University Press.

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Twenty-One Evolution and Speciation

BOX 2 1.2

Mimicry is a phenomenon
whereby an individual of
one species gains an advan-
tage by resembling an individual of a
different species. There are at least
two types of mimicry.

In Miilerian mimicry, named af-
ter F. Miiller, several groups of organ-
isms gain an advantage by looking
like one another. This mimicry occurs
among organisms in which all the
mimetic species are offensive and ob-
noxious. The classic example is the
general similarity among bees, wasps,
and hornets.

In Batesian mimicry, named af-
ter H. W. Bates, a vulnerable organism
(mimic) gains a selective advantage
by looking like a dangerous or dis-
tasteful organism (model). The classic
example of Batesian mimicry was, un-
til 1991, the monarch (Danaus plex-
ippus) and viceroy (Limenitis archip-
pus) butterflies (fig. 1). Although the
viceroy is smaller and, on close exam-
ination, looks different from the
monarch, the resemblance is striking
at first glance. Monarch butterflies
feed on milkweed plants, obtaining
noxious chemicals called cardiac gly-
cosides, which the monarchs store in
their bodies. When a bird tries to eat a
monarch, it becomes sick and regur-
gitates what it has eaten. Thereafter,
the bird will not only avoid eating
monarchs, but it will also avoid eating
any butterflies that look anything like
monarchs. Previously it was believed
that the mimetic viceroy butterfly
gained a selective advantage by look-
ing like the monarch and fooling bird
predators into thinking that the
viceroy was bad to eat. However, D.

Experimental
Methods

Mimicry

Ritland and L. Brower demonstrated
a previously unrealized fact: The
viceroys taste as bad as the monarchs
to birds. This fact changes the mim-
icry of these two species from Bates-
ian to Miillerian mimicry.

Examples of Batesian mimicry do
occur in numerous butterfly species.
For example, in West Africa,
Pseudacraea species mimic species
of the genus Bematistes (fig. 2).
These species are primarily black and
white or black and orange, and in
some the sexes differ, each having a
different mimic. Upwards of twenty
species can be involved in these
mimicry complexes in one area. Both
forms of mimicry depend on the se-
lective pressure generated by preda-
tion. Certain requirements must be
met for each system to work prop-
erly. Batesian mimicry has the follow-
ing requirements:

1. The model species must be con-
spicuous and inedible or danger-
ous.

2. Both model and mimic species
must occur in the same area, with
the model being very abundant. If
the model is rare, predators do not
have sufficient opportunity to
learn that its pattern is associated
with a bad taste. In fact, the re-
verse can happen; the model can

be at a selective disadvantage if it
is rare because the predators will
learn from the mimic that the pat-
tern is associated with something
good to eat.
3. The mimic should be very similar
to the model in the morphological
characteristics predators perceive
but not necessarily similar in
other traits. The mimic is not
evolving to be the model, only to
look like it.

Miillerian mimicry requires that all
the species be similar in appearance
and distinctly colored. They can,
however, be equally numerous. And,
as the British geneticist P. M. Shep-
pard pointed out, the resemblance
among Miillerian mimics need not be
as good as between the mimic and
model of a Batesian pair because Miil-
lerian mimics are not trying to de-
ceive a predator, only to remind the
predator of the relationship.

Although there have been some
critics of mimicry theory, especially
critics of the way in which the sys-
tem could evolve, the general model
put forth by population geneticist
and mathematician R. A. Fisher is gen-
erally accepted. According to Fisher,
any new mutation that gave a mimic
any slight advantage would be se-
lected for. As time proceeded, other
loci that might favorably modify the
expression of mimetic genes would
also be selected for in order to in-
crease the similarity of mimic and
model. This mechanism surmounts
the criticism that a single mutation
could not produce a mimic that so
closely resembled its model.

selection: groups that had altruistic behavior would have
a survival advantage over groups that did not.

In 1966, G.Williams, in his book Adaptation and Nat-
ural Selection: A Critique of Some Current Evolution-
ary Thought, refuted the altruistic view with the charge
that individuals that performed altruistic acts would be
selected against. In other words, organisms not perform-
ing altruistic acts would have a higher degree of fitness.

Williams held that apparent altruism had to be inter-
preted on the basis of benefits accruing to the individual
performing the altruistic act. After his book, the idea of
doing something for the good of the species became
passe. How, then, can apparent altruism be accounted
for? How can we explain why ground squirrels appear to
put themselves at risk to predators by giving alarm calls,
and why female workers in ant, wasp, and bee colonies

Tamarin: Principles of
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(a)

(b)

Figure 1 Mullerian mimicry, (a) Monarch butterfly and
(b) viceroy butterfly. Both have similar colors {orange and
black) and a generally similar color pattern, ([a] © Robert
Finke/Photo Researchers, Inc. [b] © Richard Parker/Photo
Researchers, Inc.)

Figure 2 Batesian mimicry seen in West African butter-
flies that live in the same places. Those on the left are
species belonging to the genus Bematistes. Those on
the right that mimic them are different species belonging
to the genus Pseudacraea. (© J. A. L Cooke/Oxford Scien-
tific Films/Animals, Animals.)

forsake reproduction in order to work for the colony?
Sociobiology, the study of the evolution of social be-
havior, attempts to answer these questions.

Kin Selection and Inclusive Fitness

In 1964, W. D. Hamilton developed concepts that
explained altruistic acts without resorting to group

selection. Starting with the known fact that relatives have
alleles in common, Hamilton suggested that natural se-
lection would favor an allele that promoted altruistic be-
havior toward relatives because the result might be an in-
crease in copies of that allele in the next generation. The
proportion of alleles shared by two individuals can be
defined as a coefficient of relationship, r. If an indi-
vidual has a certain allele, the probability that a particular

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Twenty-One Evolution and Speciation

relative also has that allele is r. Siblings have an r = 1/2. A
squirrel is likely to have virtually all its alleles still viable
if it sacrifices itself for two or more siblings. In fact, natu-
ral selection should definitely favor altruism of an indi-
vidual toward three siblings because, in a sense, natural
selection is weighing 1 copy of an individual's alleles (the
individual itself) versus 1.5 copies (three siblings).

This sort of reasoning has been termed the calculus of
the genes. It does not imply that individuals actually think
these things out; rather, natural selection has favored the in-
dividuals that behave this way. Hamilton referred to the sum
of an individual's fitness plus the fitness effects of alleles that
relatives share as inclusive fitness. He referred to the way
natural selection acts on inclusive fitness as kin selection.

Hamilton applied his ideas of inclusive fitness and kin
selection to explain sterile castes in the eusocial (truly so-
cial) hymenoptera (bees, ants, and wasps). The workers in
these colonies are sterile females. Why do they forsake
their ability to reproduce in order to help maintain the
hive or colony? The answer seems to come from hap-
lodiploidy, the unusual sex-determining mechanism of
these species. In the eusocial hymenoptera with sterile
castes, fertilized eggs produce diploid females, whereas
unfertilized eggs produce haploid males (drones). The dif-
ference between a reproductive queen and a sterile
worker in bees is larval nutrition: larvae fed "royal jelly"
can become queens. Hamilton showed that since a worker
is more closely related to her sisters than to her own po-
tential offspring, kin selection could favor a worker who
helps her sisters at the expense of her own reproduction.

Figure 21.12 shows a queen (female) with alleles A 1
and A 2 at the A locus and a haploid drone (male) with the
A 3 allele. A daughter will have either the A 1 A 3 or A 2 A 5
genotype. If we compare one of these daughters with her
sisters, we see that the average r = 0.75 — half of the time,
r = 1.0, and the other half of the time, r = 0.5 . A queen and
her daughters have an r = 0.5. Thus, we see that workers
(females) are more closely related to their sisters, and
hence are at a reproductive advantage by raising them
rather than their own young. Wilson has pointed out that
sterile caste systems have evolved among insects in only
one other group beside the eusocial hymenoptera, the ter-
mites. Although eusocial hymenoptera make up only 6% of
insects, sterile castes have independently evolved at least
eleven times. This is compelling evidence for the validity of
Hamilton's analysis. Only one noninsect example of a caste
has been discovered: the naked mole rat, a small subter-
ranean rodent living in Africa, has this type of social system.

Many studies concerned with apparently altruistic
acts have provided a large body of support for Hamilton's
theory of kin selection and inclusive fitness. P. Sherman,
working with ground squirrels, for example, has ob-
served that the individuals that make the alarm calls have
the most to gain from the standpoint of inclusive fitness;
these individuals are resident females surrounded by kin.

One other explanation for altruism is also consistent
with benefits to individual fitness. It is that many appar-
ently altruistic acts are in reality selfish — they just look
altruistic. To be altruistic, an individual must risk reduc-
ing its fitness to potentially benefit the fitness of others.
We may, in fact, misinterpret some acts as altruistic that
simply are not.

This turnaround in thought, from group selection to
individual selection, has been an intellectual revolution
in modern evolutionary biology. Before this revolution,
many of the behaviors in nature that involved apparent
altruism were difficult to explain. Now sociobiological
reasoning provides an explanation.

The reason so much controversy has sprung up over
the theory of genetic control of social behavior is because
of the implications the theory has for human social, politi-
cal, and legal issues. Human husband-wife, parent-child,
and child-child conflicts, for example, may be built into the
genes. Altruism, our highest form of nobility, may be mere
selfishness. Many critics fear that sociobiological concepts
can be used to support sexism and racism. For human be-
ings, the alternative to the theory of sociobiology is the
theory that most human behavior, including cultural learn-
ing, is determined by the environment. At present, al-
though much evidence remains to be gathered, the socio-
biology concept is compelling to many evolutionists.

Queen
A^A 2

Drone
A n

Daughters

A 1 A 3

A 2 A 3
Sisters of daughter 1

A^A 3

A 2 A 3

Daughter 1
A^A 3

^=0.75

Sisters of daughter 2

A 1 A 3

A 2 A 3

Daughter 2

A 2 A 3

r=0.5

r= 1.0

T=0.75

Figure 21.12 Haplodiploidy in eusocial hymenoptera produces
sisters with an average r of 0.75. Because drones (males) are
haploid, queens produce daughters of only two genotypes at
any locus. A given daughter has an r of 1 .0 with sisters of
identical genotype and an r of 0.5 with sisters of the other
genotype, for an average r of 0.75. In other words, females
have a 75% genetic similarity with their sisters but only a 50%
similarity with their own offspring.

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IV. Quantitative and
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Solved Problems

607

SUMMARY

STUDY OBJECTIVE 1: To analyze the mechanisms of evo-
lution and speciation 589-595

The theory of evolution by natural selection was put for-
ward by Charles Darwin, who recognized the natural varia-
tion among individuals within a population of similar or-
ganisms. He noted also that offspring are overproduced in
nature, and this overproduction inevitably leads to compe-
tition for scarce resources. Darwin assumed that, when
competition occurs, the most fit will survive; through time,
then, a population will become better adapted to its envi-
ronment through the process of natural selection. Applying
the algebra of population genetics to this theory leads to
the modern concept of evolution, neo-Darwinism.

Cladogenic speciation occurs when reproductive iso-
lating mechanisms arise, usually after gene flow in a popu-
lation is blocked. Different populations of a species can
then evolve independently. When individuals from the iso-
lates can no longer interbreed, speciation has taken place.
If the isolates then come in contact again, they will remain
as separate species. Speciation may occur gradually or in a
punctuated manner; it can be by allopatric, parapatric, or
sympatric mechanisms.

STUDY OBJECTIVE 2: To investigate the mechanisms of
the maintenance of genetic variation in natural popula-
tions, both selective and neutral 596-603

Evolution depends on variation. In 1966, Lewontin and
Hubby, using electrophoresis, showed that a tremendous

amount of heterozygosity occurred in natural populations.
Attempts to explain this variation have led to two major
competing theories: (1) variation is maintained selectively
and (2) variation is not under selective pressure, but is in-
stead neutral. Two areas of evidence support the neutralist
view.

First, the molecular evolutionary clock (the per-year,
per-amino acid, substitution rate) appears to be fairly con-
stant at 10 -9 . This constancy implies that the majority of
amino acid changes are the result of stochastic processes.
Second, there have been greater numbers of nucleotide
substitutions in DNA under lesser constraint than in DNA
under greater constraint. For example, the third, or wobble,
position of the codon has accumulated more mutations
than the other two positions. We conclude that natural se-
lection creates adapted organisms, but the majority of base
and amino acid changes may be neutral.

STUDY OBJECTIVE 3: To discuss sociobiology the evolu-
tion of social behavior 603-606

Sociobiology is another term for evolutionary behavioral
ecology. It attempts to provide evolutionary explanations
for social behaviors. Apparent altruistic behavior can be ex-
plained either as kin selection or as selfishness. Sterile in-
sect castes have come about because of the unusual hap-
lodiploid sex-determining mechanism in the eusocial
hymenoptera. There is much controversy about and little
information for applying sociobiological principles to hu-
man behavior.

SOLVED PROBLEMS

PROBLEM 1: What are the roles of reproductive isolating
mechanisms in the process of evolution?

Answer: Reproductive isolating mechanisms prevent in-
dividuals in two populations from mating with each
other or producing viable offspring. These mechanisms
can be prezygotic or postzygotic. They usually evolve
while populations are isolated from each other, either
physically or during parapatric or sympatric speciation.
For example, if a species is split by a new river, the pop-
ulations on either side of the river can evolve in isolation
from each other. Reproductive isolating mechanisms usu-
ally evolve irrespective of the other facets of evolution
taking place. Thus, if, after time, the two populations
come into contact (the river dries up), reproductive iso-
lating mechanisms may have evolved to prevent mating.

If weak reproductive isolating mechanisms have evolved,
natural selection usually favors strengthening them by se-
lecting against hybrids and against any mating behavior
that leads to the formation of hybrids.

PROBLEM 2: What is our modern evolutionary concept
of altruism?

Answer: An altruistic act is one in which an individual
risks the loss of fitness in order to benefit another indi-
vidual. Human beings value these "selfless" acts; however,
they are not favored in natural animal populations, ex-
cept under very specific circumstances, because altruis-
tic acts should be selected against. In other words, all
other things being equal, an individual that did not do al-
truistic acts would have a higher fitness than one that did

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

608

Twenty-One Evolution and Speciation

do these acts. Therefore, fitness is higher for "selfish" in-
dividuals. Altruistic acts, however, are expected if the
beneficiary of the acts shares genes in common with the

benefactor performing the acts. Generally, altruism can
be expected among relatives, following the rules of kin
selection.

EXERCISES AND PROBLEMS

DARWINIAN EVOLUTION

1. Outline the Darwinian mechanism of the process of
evolution. What is meant by neo-Darwinism?

EVOLUTION AND SPECIATION

2. Population geneticist Hampton Carson has defined
a "population flush" as a period of reduced selection
during population increase. Why should there be re-
duced selection during a flush?

3. Describe how the processes of allopatric, para-
patric, and sympatric speciation could take place.

4. Can information on evolutionary rates gained from
molecular techniques shed light on the punctuated
equilibrium-phyletic gradualism controversy? What
additional data are needed to decide this controversy?

5. What is meant by "constraint" in the molecular evo-
lution of DNA and proteins?

6. Recently, a vial of bull semen was stolen from an ar-
tificial insemination facility. Your friend is about to
undergo artificial insemination and is concerned
she may give birth to a Minotaur, or a cow-human
hybrid. Provide two explanations for why she
should not worry about this possibility.

7. In Drosophila, females in populations A and B pro-
duce an average of 250 offspring each. When the
two populations are crossed, AB females produce
only about 100 offspring each. Are populations A
and B in the process of becoming different species?

8. A few plants of species Q (2n = 14) suddenly dou-
ble their chromosomes (2n = 28) and immediately
become a new species, R. Why are QR hybrids
sterile?

9. One of the arguments creationists use to refute evo-
lution is the presence of gaps in the fossil record.
How can you explain the gaps from an evolutionary
standpoint?

GENETIC VARIATION

10. The following electrophoretic data are from a sam-
ple of one hundred field mice for their salivary
amylase- 1 genotypes. The two alleles are F and S,

* Answers to selected Exercises and Problems are on page A-24.

for fast and slow migration in an electric field: FF,
forty-three, FS, fifty-four, and SS, three. Is selection
acting? What would you look for in data to deter-
mine whether frequency-dependent selection, het-
erozygote advantage, or transient polymorphism is
at work?

11. What mechanisms permit the maintenance of ge-
netic variability in natural populations? Give exam-
ples where possible.

12. Discuss the "neutral gene hypothesis." What are its
alternatives? What data are needed to distinguish
among these views?

13. Koehn showed that differently functioning alleles of
an esterase system in fish correlated with water tem-
perature. What sorts of selection can you imagine
that could affect the same type of alleles in mam-
mals, which are homeothermic (warm-blooded) and
hence maintain a relatively constant internal tem-
perature?

14. P. Niemala and J. Tuomi have suggested that the ir-
regular leaf outlines in some plant species are a form
of mimicry. What would the leaves be mimicking?
What form of mimicry might this be?

15. From figure 21.10, what is K (the average number of
amino acid substitutions per site) between human
beings and chickens? between dogs and human be-
ings? Do all three possible comparisons support
known evolutionary relationships?

16. How does the acceptance of the neutral mutation
theory change our basic view of neo-Darwinism?

17. In a given population, the frequencies of AA, Aa,
and aa genotypes are 0.36, 0.48, and 0.16, respec-
tively. If the assigned fitnesses are 1.5 — p, 1.0, and
1.5 — q, what will the genotypic frequencies be af-
ter one generation of selection?

18. If the rate of amino acid substitution per site per
year is 2 X 10~ 9 , and the average number of amino
acid substitutions per site is 0.2, how long has it
been since the two species diverged?

19. Scientists have examined one thousand amino acids
in the proteins of human beings and chimps and have
found a difference in twenty-three. Calculate the av-
erage number of amino acid substitutions per site.

Tamarin: Principles of
Genetics, Seventh Edition

IV. Quantitative and
Evolutionary Genetics

21. Evolution and
Speciation

Critical Thinking Questions

609

20. Scientists are now using DNA sequences to show
phylogenetic relationships between or among
species. In many cases, cDNA is made from isolated
mRNA and then sequenced. Is the method a reason-
able approach to show evolutionary relationships?

21. In which codon position should the greatest abun-
dance of variation occur? Why?

23. If the "calculus of the genes" suggests sacrificing
oneself for two siblings, for how many first cousins
should one sacrifice oneself?

24. In certain animal populations, infanticide is prac-
ticed by one or more males. Do you think this infan-
ticide is random, or would you expect specific indi-
viduals to be eliminated?

SOCIOBIOLOGY

22. What are the differences among individual selec-
tion, group selection, and kin selection? How could
each type of selection explain altruistic acts?

CRITICAL THINKING QUESTIONS

1. The peppered moth (Biston betularid) has two pheno-
typic forms, melanic (dominant) and peppered (reces-
sive). The moths face predation by birds, and the preda-
tion is selective against different-colored tree trunks. In
an industrialized area, one in which the tree trunks are
dark like the melanic form (and thus "hide" the melanic
forms from the birds), a sample of moths indicated that
the frequency of the allele for peppering was 0.6; the
next year, it was 0.5. What is the fitness of the peppered
genotype under this circumstance?

2. V. C. Wynne-Edwards suggested that birds form flocks
so that they can assess their population numbers. When
they assess that their numbers are high, they decide not
to breed for the good of the species, so that they do not
exhaust their resources. Edwards called this process
group selection. Why can't this mechanism work, given
that it involves behavior that is for the good of the
species?

Suggested Readings for chapter 21 are on page B-20.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

APPENDIX A

Brief Answers to Selected Exercises, Problems,
and Critical Thinking Questions

Chapter 2 Mendel's Principles

1. Dwarf F 2 (1/4 of total F 2 ), when selfed, produce all dwarf progeny
(ft). Tall F 2 (3/4 of total F 2 ), when selfed, fall into two categories:
1/3 (TT, 1/4 of total F 2 ) produce all tall, and 2/3 (Tt, 1/2 of total F 2 )
produce tall and dwarf progeny in a 3:1 ratio. (The 3:1 ratio is from
1/2 the F 2 , so the tall component is 3/8 of the total F 3 [3/4 X 1/2],
and the dwarf is 1/8 of the total F 3 [1/4 X 1/2].) Overall, the F 3 are
3/8 TT (tall), 2/8 Tt (tall), and 3/8 tt dwarf (see fig. 2.7).

3. Rule of segregation: adult diploid organisms possess two copies of
each gene. Gametes get one copy. Fertilization restores the diploid
number to the zygote. Rule of independent assortment: alleles of
different genes segregate independently of each other.

5. Black is dominant, white is recessive, and both parents were het-
erozygous. The progeny are in an approximate 3:1 ratio. Since both
parents had the same phenotype, the simplest cross is Bb X Bb.

7. The disease is recessive at the individual level but incompletely
dominant at the enzymatic level. Check the glossary for definitions.

9- LI X LI. Let L = long ears and / = no ears. We see three phenotypes
in an approximate 1:2:1 ratio. One of the phenotypes (short) is in-
termediate between long ears and no ears. Therefore, we have in-
complete dominance.

11. Washed eye mutant, We; wild-type, We + . (W is already the allelic
designation for the wrinkled phenotype.)

13. All. Since the child was type A, it must have gotten the T A allele
from its mother. The other allele in the child is either I A or /. A type
A (T A I A or A), type B (A B or A), type O, or type AB man could
have supplied either an I A or i allele.

15. Universal donor, type O (no red-cell antigens); universal recipient,
type AB (no serum antibodies).

17. All AB; or 1/2 AB, 1/2 A; or 1/2 AB, 1/2 B; or 1/4 A: 1/4 AB: 1/4 B: 1/4
O. Crosses can be I A I A X f / B , I A I A X A, A X A B , or A X A.

A A x A B
i

All A 15

A x A B

1/2 A 6 : 1/2 A

(AB)

A A x A

1/2 A 8 : 1/2 A

(AB) (B)

A X A

1/4 A 8 : 1/4 A: 1/4 A: 1/4 ii

(AB) (A)

(AB) (A) (B) (O)

19. Steve and his fiance could be related. Both the dean and Steve's fa-
ther must be A to produce O children, and each could have con-
tributed M to produce M offspring. If the dean and Steve's father
each contributed an S allele, the daughter would be SS. Note that if
the daughter had B blood, she and Steve could not be related.

21. RrTtX self yields:

1/16 RRTT red, tall

2/16 RRTt red, medium

1/16 RRtt red, dwarf

2/16 RrTT pink, tall

4/16 RrTt pink, medium

2/16 Rrtt pink, dwarf

1/16 rrTT white, tall

2/16 rrTt white, medium

1/16 rrtt white, dwarf

23. Choice (b) is preferred because although each will give the correct
genotype, generally, testcrossing has the greatest probability of ex-
posing the recessive allele in a heterozygote. For example, an Aa
genotype, when selfed, produces aa offspring one-fourth of the
time; when testcrossed, aa offspring appear one-half of the time;
and when crossed with the Aa type (backcross), aa offspring oc-
cur one-fourth of the time. Thus, with a limited number of off-
spring examined per cross, testcrossing most reliably exposes the
recessive allele.

25. The F : are tetrahybrids (Aa Bb Cc Dd). If selfed, an F : would form
2 4 = 16 different types of gametes; 2 4 different phenotypes would
appear in the F 2 , which would be made up of 3 4 = 81 different
genotypes; 1/(16) 2 = 1/256 of the F 2 would be of the aa bb cc dd
genotype.

27. a. In the first cross, look at the yellow-to-green ratio, 120:43 —
almost exactly 3:1. Therefore, yellow is dominant and both par-
ents must be heterozygous. Now look at the tall-to-short ratio,
122:41. Again, we see a 3:1 ratio, which indicates that tall is
dominant and each parent is heterozygous. Thus, the cross is
probably YyTt X YyTt.

b. In the second cross, there are no tall progeny. Therefore, either
the short phenotypes are homozygous, or short is dominant
and at least one parent is homozygous. In the absence of the
first cross, we can't determine the mode of inheritance of
height. We can, however, conclude that yellow is dominant (we
got a 3:1 ratio) and that each parent is heterozygous. Based on
the first cross, we can conclude that this cross is Yytt X Yytt.

c. In the third cross, we see 41 yellow:46 green, and 45 tall:42
short. Both of these ratios are 1:1, and all we can conclude is
that these ratios result from matings between a heterozygote
and a recessive homozygote. With only this cross, we can't de-
termine dominance. However, we can if we use all three
crosses. The cross is yyTt X Yytt.

A-l

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-2

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

29. From the ¥ 1 progeny, we can see that long and tan must be domi-
nant, and the F 2 result confirms this assumption. We see a 3:1 ratio
for both tan:dark and long:short. The total number of flies is 80. An
ideal 9:3:3:1 ratio would be 45:15:15:5. Our results are very close
to this. Therefore, we conclude that tan and long are dominant, and
that the F x flies were heterozygous.

31. First list all possible genotypes for colored plants:

AACCRR

AACCRr

AACcRR

AACcRr

AaCCRR

AaCCRr

AaCcRR

AaCcRr

The first genotype can be eliminated because all progeny should
be colored, regardless of the tester strain. AACCRr can be elimi-
nated because the progeny of the first cross would have all been
colored. AACcRR can be eliminated because the progeny of the
second cross would have all been colored. AaCCRR can also be
eliminated because the progeny of the third cross would have all
been colored.

We are now left with AACcRr, AaCCRr, AaCcRR, and AaCcRr. Try
AACcRr X aaccRR (cross 1), which will give 1/2 colored (A-C-R-):
1/2 colorless (A-ccR-) progeny. This could be the genotype, so try it
in the second cross: AACcRr X aaCCrr. This too will give 1/2 col-
ored and 1/2 colorless offspring (A-C-rr). Since this does not fit the
observed result, the unknown genotype is not AACcRr. Now try
AaCCRr X aaccRR (cross 1). This fits the results. Try AaCcRR X
aaccRR. This will give 3/4 colorless (aaccRR, A-ccRR, or aaC-RR),
which does not fit the results; therefore, the genotype is not AaC-
cRR. Now try AaCcRr X aaccRR. This, too, will give 3/4 colorless
(aaccR-, A-ccR-, or aaC-R-), which is not seen. Therefore, the geno-
type must be AaCCRr. Confirm this with the other two crosses:

AaCCRr X aaCCrr^ 1/4 colored : 3/4 colorless, which fits.

AaCCRr X AAccrr^ 1/2 colored : 1/2 colorless, which fits.

33. a. all normal

b. 9 normal: 7 dark

c. 1/2 ebony: 1/2 normal; 1/2 black: 1/2 normal

If we let e = ebony, e + = wild-type, b = black, and b + = wild-
type, the first cross is:

ee b b

X e e bb
I

all e e b b
i selfing

9/l6e + -b + - wild-type

3/16 eeb + — ebony

3/l6e + -bb black

1/16 ee bb ebony, black

Since it is difficult to distinguish black and ebony, 7/16 will be dark-
bodied.

In c, the crosses are:

l.e e b b X ee b b

2.e e b b X e e bb
In each case we have a testcross situation for only one gene.

35. 9 wild-type:3 orange-l:3 orange-2: 1 pink.

The first two crosses indicate that wild-type is dominant to both or-
anges, and the fourth indicates that orange-2 is dominant to pink.
The fifth cross produces four phenotypes, indicating we are deal-
ing with at least two genes. The presence of two genes is also sug-
gested by orange-1 X orange-2. If these two traits were allelic, all
the progeny should have been orange. The F x X pink produces
progeny that resemble those from a double testcross. If A-B- =
wild-type, A-bb = orange-1, aaB- = orange-2, aabb is probably
pink. The crosses in question are then:

AAbb

aaBB

(orange-1)

AaBb

(orange-2)

(wild-type)
i self

9/16,4-5-
3/16 A-bb
3/16 aaB-
1/16 aabb

wild-type
orange-1
orange-2
pink

37. Two loci with epistasis. AaBb X self yields 9/16 A-B-.6/16 A-bb +
aaB-: 1/1 6 aabb. Verify by testcrossing the various classes.

39. See for example figure 2.25. Other ratios are 10:3:3; 10:6; 12:3:1;
12:4.

41. 9 nonworkers: 7 hard workers. The F x indicates nonworker is dom-
inant; therefore, a worker must be a recessive homozygote. If one
gene is involved, the cross of the F x female X worker male (Ww X
wiv) should produce 1 worker: 1 nonworker, for this is a testcross.
This result is not seen, so we must have more than one gene in-
volved. Perhaps a worker can result from more than one gene. Let
A-B- = nonworker, and A-bb, aaB-, or aabb = workers. The origi-
nal worker is aabb, and the cross is:

aabb

AaBb X
(nonworker)

1/4 A-B-: 1/4 A-bb:

AABB

aabb
(worker)

1/4 aaB-: 1/4 aabb

nonworker worker
If AaBb

worker
AaBb

worker

x
i
9/16 A-B- nonworkers

3/16 aaB-

3/16 A-bb
1/16 aabb .

workers

43. 2 1 3 4

? —> indole — > tryptophan — > 3-hydroxyanthranilic — > niacin
or kynurenine acid

Accumulation: 1, indole; 2, ?; 3, tryptophan or kynurenine; 4, 3-
hydroxyanthranilic acid. Without serine, the pathway would be
blocked before the point of tryptophan production, after indole.

45. a. Maple sugar urine disease is recessive. If two individuals with
the same phenotype produce offspring, some of whom have a
different phenotype, both parents must be heterozygous. Since
they are heterozygous, their phenotype must be dominant.

b. 3/4. Let M = normal and m = maple sugar urine alleles. The
cross is Mm X Mm. At each conception, the probability is three-
fourths that each child is of the M- genotype.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-3

Critical Thinking Question:

1. The 15:1 ratio indicates that all genotypes except the recessive ho-
mozygote (aabb) produce a triangular capsule. The rounded cap-
sule results from the recessive homozygote. One way to look at this
is that the rounded form is a "default" form when neither locus has
functional — dominant — alleles. However, a dominant allele at ei-
ther of two loci is adequate to form the triangular seed capsule.
The loci can be considered redundant in the pathway of seed cap-
sule shape since a functional allele at one, the other, or both will
provide a dominant (triangular) phenotype. At this point in time, it
is impossible to know precisely what the enzymatic function of
each dominant allele is.

Chapter 3 Mitosis and Meiosis

1. See table 3.1 for a summary answer.

3. When a eukaryotic chromosome replicates during the S phase of
the cell cycle, one chromosome becomes two chromatids, at-
tached near the centromere. These are sister chromatids. Chro-
matids of different chromosomes are nonsister chromatids. Homol-
ogous chromosomes are members of a pair of essentially identical
chromosomes. In diploid organisms, one member from each pair
comes from each parent. Nonhomologous chromosomes do not
share this relationship.

Metaphase,
mitosis

Anaphase,
mitosis

7. S phase

.3 _

9. 2=8 different gametes can arise. A crossover between the A lo-
cus and its centromere does not alter gametic combinations.

Metaphase I,
meiosis

/ One possible
* anaphase I,

meiosis

11. The intent of this problem is to make you think about the essential
steps of meiosis, primarily the necessity to separate members of
homologous pairs of chromosomes. Presumably, any method you
devise will force you through that process.

13. Meiosis apportions homologous chromosomes the same way
Mendel's rules apportion alleles. Each gamete gets one member of
a homologous pair of chromosomes. Segregation predicts the same
about alleles. The separation of homologues of one chromosome
pair at meiosis is independent of the separation of other homolo-
gous pairs. Independent assortment makes the same prediction
about alleles.

15. A gamete from wheat will have twenty-one chromosomes, and a
gamete from rye will have seven chromosomes. Even if the seven
rye chromosomes could pair with seven wheat chromosomes, a
highly unlikely possibility, the remaining fourteen wheat chromo-
somes could not pair and would segregate randomly during meio-
sis. Almost every gamete would get an incomplete set; if fertiliza-
tion did occur, the zygotes would have extra chromosomes
(trisomic) or would be missing some chromosomes (monosomic
or nullosomic).

17. 64. The number of combinations is 2 n where n — the number of
different chromosomes in the set.

19.

DNA
(Number of
Chromatids) Ploidy

Spermatogonium or Oogonium

Primary Spermatocyte or Primary

Oocyte

Secondary Spermatocyte or

Secondary Oocyte

Spermatid or Ovum

Sperm

21. a. 50 b. 50. The primary oocyte is diploid and will undergo
meiosis, but only one functional ovum results from each primary
oocyte. The secondary oocyte will divide to produce an ovum and
a polar body.

23. Any possible genotype, from AAABBB through aaabbb can occur
in the endosperm. If at a given locus the endosperm is homozy-
gous, so is the embryo. If the locus is heterozygous (e.g., AAa or
Aaa), so is the embryo. Thus, an AAabbb endosperm is associated
with an Aabb embryo.

25. A greater maternal influence in Drosophila and corn than in Neu-
rospora means the sexes (mating types) do not show the disparity
in size between male and female cells as in Drosophila and corn.

27. b. Homologous chromosomes will pair during meiosis. Each ga-
mete gets one of each chromosome, A, B, C, D, and E. Fertilization
fuses two cells with the chromosome complement given. Since
root cells are somatic tissue, these cells will be diploid.

a. 2 50 or about 1.1 X 10 15 b. 2 2 . The number of gametes pro-
duced is 2 n ,where n = number of independently behaving entities.
If the genes are completely independent, we expect 2 50 , and if they
are completely linked, we expect 2 2 . In reality, the number falls be-
tween these extremes.

29.

Telophase II,
meiosis

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-4

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

Critical Thinking Question:

1. Both meiosis and mitosis are processes that initiate under certain
circumstances of cell cycle and place. Neither is actually depen-
dent on the chromosomal content of the cell. Thus, meiosis could
begin in a haploid cell, but it would not be a successful process be-
cause there is no homologue for any chromosome to pair with.
Mitosis would, however, be successful because there is no pairing
(synapsis) required for successful completion of the process.

Chapter 4 Probability and Statistics

1. a. (5!/3!2!)(l/2) 3 (l/2) 2 = 0.3125

b. (1/2) 5 = 0.03125 (SDSDS, in which S = son, D = daughter)

c. 2(l/2) 5 = 0.0625 (SDSDS + DSDSD)

d. (1/2) 5 = 0.03125

e. 2(l/2) 5 = 0.0625 (all sons + all daughters)

f. 4 daughters, 1 son + 5 daughters: (5!/4!l!)(l/2) 4 (l/2) + (1/2) 5
= 0.1875

g. (1/2) 2 = 0.25 (DXXXS, in which X is either a daughter or a son,
withp= 1;P= [l/2][l][l][l][l/2])

3. Remember that albinos have blue eyes. Therefore, 7/16 of the off-
spring will have blue eyes. If we let B = brown, b = blue, C = nor-
mal color expression, and c = albinism, the following genotypes
are blue-eyed: C-bb and cc-.

a. (1/4) 5 = 0.0009765

b. (1/8) 5 = 0.0000305

c. (5!/4!l!)(7/32) 4 (9/32) = 0.00322

d. (4!/2!2!)(l/8) 2 (l/8) 2 = 0.0014648

5. a. 2(l/2) 4 = 0.125

b. (l/2) 4 (l/2) 4 = 0.0039063 (Probability that sperm and egg cre-
ating the zygote each had only paternal centromeres.)

7. One-half.

9. a. 81/256 b. 108/256 c. 9/256. In (a), since all children have
the same phenotype, each child will have the same probability of
having no molars. Therefore, (3/4) 4 = 81/256. In (b),

P =

4!
3!1!

(3/4) 3 (l/4) = 108/256.

When order is given, we multiply the chance of each event, 1/4 X

1/4 X 3/4 X 3/4 = 9/256.

11. One-eighth. B must be heterozygous (Gg), as must A's father. We as-
sume A's mother is GG, since there is no mention of the disease in
her family. Therefore, A has a one-half chance of getting g from his
father. If two heterozygotes mate, the chance of a recessive child is
one-fourth, so P = 1/4 X 1/2.

13. 1/512. The ¥ 1 progeny are Aa Bb Cc Dd Ee. The chance of getting
any individual with a particular homozygous genotype is (1/4) 5 .
Since we are looking for two different possibilities, we have
2(l/4) 5 = 2/1024 = 1/512.

15. 0.049. Since the order is not specified, we use the multinomial for-
mula. We have six mice, so n = 6. If p = chance of agouti, p =
3/4 X 1/2 = 3/8; q (black coat color) = 1/4 X 1/2 = 1/8, and r (al-
bino) = 1/2. The equation becomes:

6!
2!2!2!

(3/8) 2 (l/8) 2 (l/2) 2

6X5X4X3X2X1X9X1X1
2X2X2X 64 X 64 X4

17. Hypothesis: RrYy X RrYy produces R-Y-.R-yy.rrY-.rryy in a 9:3:3:1
ratio. The critical chi-square, three degrees of freedom at probabil-
ity of 0.05, = 7.815.

R-Y-

R-yy

rrY-

rryy

Sum

Observed

315

108

101

556

Expected

9/16

3/16

1/16

312.75

104.25

34.75

556

O- E

2.25

3.75

-3.25

-2.75

(O - Ef

5.06

14.06

10.56

7.56

(O - Ef/E

0.016

0.135

0.101

0.218

0.470 = x 2

Since this chi-square, 0.470, is less than the critical chi-square, we
fail to reject our hypothesis of two-locus genetic control with dom-
inant alleles at each locus.

19. We reject the 3:1 ratio as an appropriate null hypothesis. If we cal-
culate the chi-square using 3: 1 as the expected ratio, we expect 72
and 24 (3/4 and 1/4 of 96, respectively):

O-E (0-E) :

(0-E) 2 /E

Curly-winged

flies
Straight-winged 35 24

flies

61 72 -11 121
11 121

121/72 = 1.681
121/24 = 5.042

Chi-square = 6.723 (1.681 + 5.042). With one degree of freedom,
p > 0.05 (critical chi-square = 3. 841).

Critical Thinking Question:

1. You should change your choice because the box you chose origi-
nally has a 1/3 chance of containing the prize, whereas the re-
maining box has a 2/3 chance of containing the prize. The 1/3
chance of your choice is set by the fact that there were three
equally likely choices at the beginning. When your friend elimi-
nated an empty box, she left two choices: your original box and
the third box. Since the probability of your original choice has not
changed, the probability of the remaining box must be 2/3 to give
a combined probability of 1.0 that a box contains a prize.

Chapter 5 Sex Determination, Sex Linkage,
and Pedigree Analysis

1. The differences are in terminology only, not in shape or size of the
chromosomes. In species in which females have a homomorphic
pair of sex chromosomes, the members of the pair are called X
chromosomes. In species in which males have a homomorphic sex
chromosome pair, the members of the pair are called Z chromo-
somes.

3. 3/8 males, 3/8 females, 2/8 intersexes (dsx dsx homozygotes).The
1/4 dsx dsx flies will be intersexes, whereas half of the 3/4 will be
normal males and half will be normal females.

5. The protein is probably a dimer, which, in the heterozygote, can be
of fast-fast, fast-slow, or slow-slow subunit combinations. A female
heterozygous for a sex-linked gene controlling a dimeric enzyme
should show the pattern of lane 3 in whole blood (mixture of slow-
slow and fast-fast dimers) and lanes 1 or 2 in individual cells.

0.049

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-5

7. a. 0; human female, male fly

b. 1; human female, female fly

c. 0; human male, male fly

d. 1; human male, female fly

e. 2; human female, female fly

f. 4; human female, female fly

g. 1/0 mosaic; human male-female mosaic, male-female fly mosaic

Cross Reciprocal

Now diagram the second cross:

Pj female

X + X +

-^rlZ-srrlZ

male

X lz Y

X + Y

Fj female

X + X lz

x + x lz

male

X + Y

X lz Y

F 2 females

x + x + ,

x + x lz

X + X lz , X lz X lz

males

X + Y,X

fey

X + Y, X fe Y

11. Exemptions should be made minimally for hemophilia in brother,
sister's son, mother's brother, mother's sister's son, mother's father,
and others more distantly related.

13. Pi My x+x+ (female) X fy + fy + X ct Y (male)

F i fy + fy X + X ct (female) X fy + fy X + Y (male)

Male

fy + x +

fyx +

fy + Y

fyY

Female

fy + * +

fy + fy + X + X +

fy + fy X + X +

fy + fy + X + Y

fy + fyX + Y

fy + x rt

fy + fy + x + x rt

fy + fy X + X ct

fy + fy + x c? y

fy + fy X ct Y

fy* +

fy + fy X + X +

fyfy X + X +

fy + fy x + y

fyfy x + y

fyX ct

fy + fy X + X ct

fyfy X + X ct

fy + fy X ct Y

fyfy x c 'y

F 2 : females, 3/4 wild-type, 1/4 fuzzy; males, 3/8 wild-type, 3/8 cut,
1/8 fuzzy, and 1/8 cut and fuzzy.

15. a. X linked

b. gray

c. 1/2 gray: 1/2 yellow in both sexes

In both crosses, we see a difference in the phenotypes of the sexes,
suggesting sex linkage. The F x offspring from the first cross indi-
cate that gray is dominant to yellow. The F x females from this cross
must be heterozygous, and the two phenotypes in the F 2 males re-
sult from each of the X chromosomes in the F x female being hemi-
zygous in the F 2 males. The first cross is therefore (calling gray the
wild-type):

x^r

X^Y

x+x'

X + Y

gray

X + X +

X + X y

X + Y

X y Y

gray

yellow

x y x y

X"Y

x + x y

X V Y

gray

yellow

x+x^

x y x y

X + Y

X^Y

gray

yellow

gray

yellow

17. Yes. Begin by determining genotypes of the two individuals. The

woman must be heterozygous XX. A man with normal vision

must be X Y, and all his daughters must receive his X chromosome

CVC

rCx

and should be normal, either XX or X X . Since color blindness
is recessive, the daughter must have two X c chromosomes. A very
rare possibility is that the man is the father and nondisjunction oc-
curred in both parents: at meiosis I in the male and at meiosis II in
the female (see chapter 8).

19- a. F x : wild-type females, white-eyed males.

b. F 2 : 3 wild-type:3 white-eyed: 1 ebony: 1 ebony, white-eyed in
both sexes.

c. Reciprocal F 2 : 3 wild-type females: 1 ebony female: 3 wild-type
males: 3 white-eyed males: 1 ebony male:l ebony, white-eyed
male. Let X + = red, X w = white, e + = wild-type, e = ebony.

X W X W e + e +

X
I

X w Yee

X + X w e + e

X W Y e + e

(wild-type)

(white-eyed)

Use probability for the F 2 generation rather than the Punnett square:

(1/2)X + X (l/2)X u; X (3/4)e + -

X (l/4>£?

X (1/2)Y X (3/4)e + -

X (l/4><?

(1/2)X W X (1/2)X W X (3/4> + -

X (l/4>e

X (1/2)Y X (3/4)e + -

X (l/4>e

For the reciprocal cross,

3/16 wild-type females

1/16 ebony females

3/16 wild-type males

1/16 ebony males

3/16 white-eyed females

1/16 white-eyed, ebony females

3/16 white-eyed males

1/16 white-eyed, ebony males

X + X + ee

X + X w e + e
(wild-type)

X
i
X

X"Y

e e

X^Y e e
(wild-type)

All F 2 females will get X
the males shown.

e :e will be 3: 1 . The males will be as in

21. The female is heterozygous for an X-linked color gene (one of the
X chromosomes in the cells of female cats is inactivated, leading to
the black and yellow spots). We immediately deduce X-linkage be-
cause of the different phenotypes in the sexes. Finding two types
of males indicates that the female was heterozygous. The patches
of yellow and black come from X-inactivation. The cross is:

X & X°

X^Y

x b x b

black

X & X°

black
and yellow

X^Y
black

X°Y
yellow

23. Penetrance is the proportion of individuals of a particular geno-
type that shows the appropriate phenotype; expressivity is the de-
gree to which a trait is expressed.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-6

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

25. a. The phenotype is the propensity to have twin offspring. It
could be caused by a recessive or dominant, sex-linked or auto-
somal allele.

b. Autosomal dominant or possibly autosomal recessive inheritance.

c. Autosomal, or sex-linked, recessive inheritance.

d. Autosomal recessive inheritance.
27. Assuming 100% penetrance:

(a)

(b) W (c) W (d)

Critical Thinking Question:

1. The immediate effect of a null allele is to make heterozygous geno-
types appear to be homozygotes.That is, in the simplest system, we
expect one band in a homozygote and two bands in a heterozygote.
If we see only one band, we assume that the individual is homozy-
gous for that allele, when in fact that individual might be heterozy-
gous for the allele that produces the particular band and the allele
that produces no band (null allele). The null allele can be verified by
both the absence of any bands in the null homozygote and the re-
sults of breeding experiments when the null allele is suspected.

Chapter 6 Linkage and Mapping in Eukaryotes
1. a.

P x groucho

grogro ro ro

¥ l female gro gro ro ro

F 2 grogro ro ro

gro + gro roro

grogro roro

gro gro ro ro

(6+ 5)/ 1,000 = 0.011 = 1.1% recombination

= 1.1 map units apart

rough

gro gro roro

male grogro roro

518

471

b. Given the map units, F x gametes are produced on the average
by females as follows: gro ro + , 49.45% = 0.4945 (98.9%/2);
gro + ro, 49.45% = 0.4945; gro ro, 0.55% (l.l%/2) = 0.0055;
and gro + ro + , 0.55% = 0.0055. Males, lacking crossing over,
produce only two gamete types: gro ro + and gro + ro, each
50% = 0.50. Summing from the Punnett square following, the
phenotypes of the offspring would be as follows: wild-type,
50%; groucho, rough, 0%; groucho, 25%; and rough, 25%.

Male

Female

gro ro + (0.5)

gro + ro (0.5)

gro ro
(0.4945)
gro + ro
(0.4945)
gro ro
(0.0055)
gro ro
(0.0055)

groucho

0.24725

wild-type

0.24725

groucho

0.00275

wild-type

0.00275

wild-type

0.24725

rough

0.24725

rough

0.00275

wild-type

0.00275

A dihybrid female is testcrossed (with a hemizygous male having
both recessive alleles). Each recombinant class will make up about
5% of the offspring. Each parental class will make up about 45% of

7. a.

the offspring. Phenotypic classes will be equally distributed be-
tween the two sexes. The same results will be found for an autoso-
mal locus if the dihybrids are females (no crossing over in males). A
reciprocal cross cannot be done for X-linked genes because males
cannot be dihybrid. Males dihybrid for an autosomal gene produce
only two classes of offspring when testcrossed — parentals.

a. The hotfoot locus is in the middle (compare, for example, hot-
foot, a double crossover, with the wild-type, a parental); there
are 16.0 map units from hotfoot to either end locus: 74 + 66 +
11 + 9 recombinants between hotfoot and waved and 79 +
61 + 11 + 9 recombinants between hotfoot and obese.

b. The trihybrid parent was o h wa/o + h + wa + .

The coefficient of coincidence is 20/25.6 (20/[0.l6 X 0.16 X
1,000]); interference is 1 - (20/25.6) = 0.22, or 22%.

Work backward from the 0.61% double recombinants (0.100 X
0.061 X 100). Thus, there would be 6 of 1,000 double recombi-
nants. In the an-sple region, we need the total of single + dou-
ble recombinants = 100 of 1,000 (10 map units). Thus, 100 -
6 = 94; divided by 2 (two phenotypes) is 47 each. For the sple-
at region, the total of single and double recombinants = 61 (6. 1
map units). Thus, 61 — 6 = 55; divided by 2 is 27 and 28. The
parentals make up the remainder for a total of 1,000.

b. With a coefficient of coincidence of 0.60, only 0.366% (0.61 X
0.60) of the expected double recombinants will occur, that is,
4 instead of 6. Thus:

Coefficient of Coincidence

1.0

0.6

ancon, spiny, arctus oculus

wild-type

ancon, spiny

arctus oculus

ancon

spiny, arctus oculus

ancon, arctus oculus

spiny

Total

422

423

1,000

421

422

1,000

9. a. linked; b. trans; c. 28.7%. The cross is a testcross. If the
genes were not linked, we would expect a 1:1:1:1 ratio of off-
spring; we don't see that. The alleles that are linked will appear
as the majority classes, which are Trembling, long-haired and
normal, Rex. Therefore, Trembling and Rex are in the trans po-
sition. If we let T = Trembling, and R = Rex, the cross is

Tr tr
— X —
tR tr

Recombinants are Trembling, Rex and normal, long-haired;

42 + U
300

X 100 = 28.7%

11. a.

k e cd
k e cd

b. k 6.9 e 5.1 cd

k e cd k e cd

The initial cross is z X -z,

k e cd k e cd

k e cd
Producing a trihybrid F x female: -zjt - + , in any order.

k e cd

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-7

The last two classes (3 + 4 offspring in the F 2 ) are double
crossovers and allow us to determine order by comparison with
the parentals (880 + 887). If the order is k cd e, a double crossover
in the F x females yields k cd + e + and k + cd e. If the order is cd k e,
a double crossover yields cd k + e + and cd + k e. Therefore, the or-
der must be k e cd, which gives the correct double recombinants
in the F 2 generation. After reconstructing the trihybrid

k e cd

and scoring each of the offspring for crossovers in the k-e and e-cd
regions: map units, k-e = ([64 + 67 + 4 + 3]/2000) X 100 = 6.9
and map units, e-cd = ([49 + 46 + 4 + 3]/2000) X 100 = 5.1.

13. 0.0125. This problem requires the manipulation of equations. We
know that interference = 1 — coefficient of coincidence, so coef-
ficient of coincidence = 1 — interference = 1 — (—1.5) = 2.5.
Since coefficient of coincidence =

observed double crossovers

expected double crossovers

observed double crossovers = (coefficient of coincidence) X (ex-
pected double crossovers). The expected double crossover fre-
quency is (2.5X0.005) = 0.0125.

15. PD, 1, 2, 4, 6, 8-10; NPD, 3; TT, 5, 7. Map units = ([NPD +
U/2}TT]/Total) X 100 = ([1 + {l/2}2]/10) X 100 = 20. The loci
are 20 map units apart.

17. FDS, 3-5, 8, 10; SDS, 1,2,6, 7, 9. The distance between the arg locus
and its centromere is (1/2)%SDS = (1/2)50% = 25 map units.

19. For example, the only variant of the type 2 pattern of table 6.7 is
a b, a b, a b, a b, ab , ab , ab , ab . Other patterns that are
variants of the remaining five categories are derived by inverting
the eight spores of a pattern (bottom to top) or by switching spores
1 and 2 with spores 3 and 4 or spores 5 and 6 with spores 7 and 8.

21. a X a + — > a/a + , which undergoes meiosis. Twelve map units
means that the SDS pattern makes up 24% of the asci.

23. For example: 1, 408; 2, 42; 3, 250; 4, 250; 5, 30; 6, 5; 7, 15. To make
the numbers work out, classes 4-7 must equal 300, as must classes

3 + 5-7, in order to make (1/2)%SDS = 150. Thus, if classes 3 and

4 are 250 each, 50 must be spread out among classes 5-7. NPD +
(1/2)TT should equal 300 to confirm the arrangement (a-b dis-
tances). These numbers will give an a-b distance of 30.

25. a. yes; PD > > NPD (actually, no NPD). b. a: 2.5 map units; b:
7.5 map units. Classify each ascus— I: PDT, FDS for both; II: TT, FDS
for a, SDS for b; III: TT, FDS for a, SDS for b; IV: TT, FDS for a, SDS
for b; V TT, FDS for a, SDS for b; VI: PDT, SDS for a and b; VII, PDT,
SDS for a and b; VIII: PDT, SDS for a and b. We see no NPDs, so
genes are linked. For gene to centromere distances, use the formula

1/2 (number of SDS asci)
100

X 100:

a to centromere = (l/2)(3 + 1 + 1)% = 2.5 map units; b to cen-
tromere = (l/2)(2 + 3 + 2 + 3 + 3 + 1 + 1)% = 7.5 map units.

27. 12.5 map units. The only genotype that grows on minimal medium is
arg + ade + . If the two genes were unlinked, 1/4 of the progeny should
have this genotype; this is not seen. The genes must be linked; wild-
type results from recombination between these two genes. The re-
ciprocal class, arg~ade~ ', which has not been selected for, should be
equally frequent, so: map units = (2 X 25)/400 X 100 = 12.5.

pab

29. Construct a pedigree of the Duffy alleles (Fy a , Fy)> Arbitrarily as-
sign one allele to a normal chromosome 1 and the other allele to
the coiled chromosome. Then, accompanying the pedigree, the al-
leles and their morphologically proper chromosomes would be as-
sociated.

AB
UN

BB
NN

\^y nn

BB
NN

AB
UN

Fy a = A
Fy b = B

Normal
chromosome = N

Uncoiled
chromosome = U

31. Twenty map units apart (two often recombinant sons according to
the "grandfather method").

33. Thirty-three map units. The woman is heterozygous in trans con-
figuration for color blindness and hemophilia: hc + /h + c. Recombi-
nation between these two markers yields b + c + and hc.We can only
detect recombinants in sons, so

# normal sons + # double mutant sons

total sons

X 100

map distance = 2/6 =0.33

35. Enzyme A on 1 1; B on 15; C on 18; D on 3; E on 7. Enzyme A is pres-
ent in clones X and Y, and chromosome 1 1 is common to these two
clones. Enzyme B is present only in X, and 15 is the only chromo-
some unique to X. Similar logic allows the assignment of the other
genes.

Critical Thinking Question:

1. Three-point crosses capture (allow us to see) double crossovers
that have taken place in the two regions defined by the three loci.
However, any double crossovers that occur within one region or
any crossovers involving more than two events will not be indi-
cated correctly by random-strand analysis.

Chapter 7 Linkage and Mapping in Prokaryotes and
Bacterial Viruses

1. The prokaryotic chromosome is a double-stranded DNA circle that
is small compared with most eukaryotic chromosomes. Viral chro-
mosomes can be DNA or RNA. Viruses are obligate intracellular
parasites. Whether they are alive depends on the definition of the
term alive.

3. A colony is a visible mass of cells derived usually from a single pro-
genitor. A plaque is the equivalent growth of phages on a bacterial
lawn, producing a cleared area lacking intact bacteria.

5. The bacterium could have survived and produced a colony if it was
on a X-free area, it became lysogenic (and thus resistant to further
phage attack), or it was genetically resistant to phage X.

7. 1, his~ arg~; 2, leu~; 3, lys ~; 4, his~ met~ or his met~\ 5, arg~ .

9- Where phages cannot grow: E. coli ton r , phage h + . Where phage
can grow: E. coli ton s , phage h + or h, or E. coli ton r , phage h.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-8

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

11. Far. If the selective locus is near, it passes into the F~ cell very early
during conjugation. Consequently, there is a great reduction in the
recovery of loci distal to the selective marker because both the Hfr
and F~ members of a conjugation event can be killed by the selec-
tive agent (e.g., an antibiotic such as streptomycin).

13. See figures 7.8, 7.9, 7.15, 7.17, 7.18, and 7.26.

15. 9 min 1 min 8 min 7 min

Origin az ton lac galB

17. For example, use an Hfr that is wild-type but str s with the F factor
integrated at minute 20. Use an F~ strain that is pyrD~ , purB~ ,
man~ , uvrC~ ', his~ , and str r . Interrupt mating at one-minute inter-
vals; plate cells on complete medium with streptomycin to kill Hfr
cells and grow recombinant and nonrecombinant F~ cells. The
next day, after colonies have grown up, replica-plate onto selective
media. The following data would be generated:

Colony Growth on Media Selective for
pyrD + purB + man + uvrC + his +

Minute

—

19. The order is a c b, and c is close to a. Genes c and a are cotrans-
formed 76% of the time, suggesting that these two genes are very
close and b is far away (a-b cotransduction is 10%, or 0.1). Two or-
ders are possible: a-c b and c-a b. If the first order

is correct, a + b + c~ results from a double crossover; this class
should be the least frequent. If the second order is correct, a single
exchange between a and c would yield a + b + c~ , but this fre-
quency should be similar to a + b~ c~, and it is not.

21. thr leu pro his. We see that cells that are tbr + are the most fre-
quent. The chance of interruption in the conjugation increases
with the length of time for the mating. Therefore, genes farther
from the origin of transfer appear less frequently. We can order the
genes based merely upon the frequency of genotypes seen. The or-
der must be thr leu pro his. Since we see no his + , and since we
stopped the mating at 25 minutes, his must be after minute 25 on
the map of this Hfr strain.

23. a and c are close; b is farther away; c is probably in the middle. The
numbers of the first three transformant classes indicate that each
gene, by itself, is readily transformed. We notice that classes with b
and any other gene are quite rare, a situation that indicates b is far
from a and c. We notice that a and c are cotransformed about 13%
of the time, b and c about 3% of the time, and a and b about 2% of
the time.

25. a. lys his val

lys his val~

lys + his~ val +

lys his~ val~

Since there is no lysine in the medium, lys + must be present to al-
low growth.

b. lys val his

lys + val + his~

Both lys + and val + must be present to allow growth.

c. lys val his
lys val~ his

Both lys + and his + must be present to allow growth.

d. lys val his~
lys val~ his

We see no lys + val + his + cells.

e. lys + and val + are close together; they are cotransformed 75% of
the time. Order could be lys val his or val lys his.

f. val lys his. If the order is val lys his, val + lys~ his + should be
rare, since this genotype results from a double exchange; and in-
deed, this class is the least frequent.

27. Mix the phages together with bacteria with increasing quantities of
the two phages. Knowing the numbers of each in a particular case,
it is possible to predict the proportion of cells doubly infected
(product of probabilities). The recovery of recombinants should in-
crease with that probability. In other words, recombination should
occur only in doubly infected cells.

I 0.6 1 0.2 I 1.2 I

Small recombination frequencies should be approximately addi-
tive. Note that recombination distances are twice the value of wild-
type plaques since the double mutant recombinants were not
counted. Thus, the data table should be:

Percent

Cross

Wild-Type Plaques

Recombinants

a X b

0.3

0.6

a X c

1.0

2.0

a X d

0.4

0.8

bX c

0.7

1.4

bX d

0.1

0.2

cXd

0.6

1.2

The largest distance is between a and c; therefore a and c must be
at opposite ends. Since a-b = 0.6, b must be 0.6 units to the right
of a. This position gives b-c as 1.4, the observed distance. We now
have the following map

0.6

1.4

If d is to the left of b, then d-c should be greater than 1.4, a result
not seen. Therefore, d is 0.2 units to the right of b.

31. I 0.04 I 0.02 I

To calculate map distance, you must have the number of recombi-
nations and the total number of progeny. Since all phages grow on
strain B, this number must equal the total number of progeny; this
is 250 X 10 7 . Since only wild-type phages grow on K12, and since
wild-types result from recombination between two genes,

a b
X

h> a b and a b ,

the number that grow on K12 must be recombinants. But this num-
ber represents only half of the recombinants, for the double mu-
tant will not grow on K 12. Total recombinants are:

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-9

1X2 (2 X 50)(10 4 ) = 10°
1X3 (2 X 25X10 4 ) = 5 X 10 5
2X3 (2 X 75)(10 4 ) = 1.5 X 10 6

(100 X 10 4 )
map distance 1-2 = ^- X 100

(250 X 10 7 )

(0.4 X 10 3 )(100) = 4 X 10

0.04

map distance 1-3

(50 X 10 4 )
(250 X 10 7 )

X 100

= (0.2 X 10 3 )(100) = 2 X 10 2

= 0.02

(150 X 10 4 )
map distance 2-3 = -r~ X 100

^ (250 X 10 7 )

(0.6 X 10 3 )(100) = 6 X 10

0.06

33. Use replica-plating on selective media with arabinose as the sole
carbon source, thus selecting for ara^ cells. Although all three loci
can be cotransduced, the rarity of ara + leu~ ilvH + indicates leu is
the middle locus (ara leu ilvH). Cotransductance frequencies:

ara to leu = (9 + 340)/(9 + 340 + 32) = 0.92

ara to ilvH = 340/(340 + 32 + 9) = 0.89

35. azf leu^ tbr + . The cotransduction frequency of leu and azi indi-
cated leu is closer to azi than to thr. Thus, two orders are possible:
leu azi thr or azi leu thr. If the first order is correct, and leu + is se-
lected, leu + azf > leu + thr + ; this prediction fits. However, with
the same gene order, and selecting for thr + , thr + azf > thr + leu + ;
this result is not seen, and the order must be azi leu thr. The sec-
ond order predicts thr + leu + > thr + azf.

First
Second

^g Third
=g Fourth

Alternate segregation Adjacent-1 segregation Adjacent-2 segregation

First with Second with First with

fourth

third

Second with
fourth

First with Third with
second fourth

I a

b I

) c

j> (

> c

j> (

' C

j) (

> (

)

2 |

9
h

2
1

o o

Normal

and

Reciprocal Duplication Duplication Duplication Duplication
translocation deficiency deficiency deficiency deficiency

and and and and and

a
b
c

d
e
f

9
h

6
5
4
3
2
1

a
b
c

d
e
3
2
1

6
5
4
f

9
h

a
b
c

9
h

a
b
c

d
e
3
2
1

O O

Duplication Duplication
deficiency deficiency

Normal

6
5

Reciprocal Duplication Duplication
translocation deficiency deficiency

Critical Thinking Question:

1. At first, with much smaller data sets, scientists came up with branch-
ing models of the bacterial chromosome. However, with the very
large data set of table 7.4, it is almost impossible to come up with a
reasonable mechanism other than a circular bacterial chromosome.

Chapter 8 Cytogenetics

1. All chromosomes form linear bivalents. The cross-shaped figure is
seen only in heterozygotes.

3. No, there are no inversion loops formed in homozygotes.

5. A diagram will show that a crossover between a centromere and
the center of the cross can change the consequences of the pattern
of centromere separation. For example, in the figure at left, we dia-
gram a crossover between loci 4 and 5, as in figure 8. 1 1.

7. Reciprocal translocation (some effects occur only in the heterozy-
gous condition). Look for the cross-shaped figure at meiosis or in
salivary gland chromosomes.

9. Assume crossovers as shown (following page, top left).

11. a. All females should get a wild-type X chromosome from their fa-
ther. Irradiation produces chromosomal breaks, so a deletion of
part of the X is possible, producing a situation of pseudodomi-
nance. (Alternatively, the offspring could have gotten a mutant
X w from the father.)

b. Diagram the crosses (a "/" represents a deleted part of the chro-
mosome and = X chromosome)

llll

w
Y

inn

wild-typ<

i wild-typ(

white

dead

We expect all wild-type females and all white-eyed males, but in
a ratio of 2 female: 1 male.

13. We see that the F 2 offspring from cross A X B yields fewer progeny
than the other crosses. Something unusual must be involved. One
explanation is that one of the strains is homozygous for a recipro-
cal translocation. The translocation, when heterozygous, results in
some inviable gametes or progeny, and thus reduces the number of
progeny.

15. We expect to see about 32% recombination between these two
genes, but we see only 2%. The most likely explanation is that an in-
version occurred so that these two genes came to lie close to each
other; the stocks are homozygous. Since semisterility is not re-
ported, we are probably not dealing with crossover suppression in
inversion heterozygotes.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-10

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

a b c

First !

Second 1

a b c

2
1

First with
fourth

a
b
c

Second with
third

a
b
c

Second with
third

d
e
f

9
h

6
5
4
f

9
h

d
e
3
2
1

6
5
4
f

9
h

Normal

Duplication
deficiency

Reciprocal
translocation

Duplication
deficiency

17. A translocation from the tip of the normal X in the male to the Y.
We expect all males to receive an X chromosome with the white-
eye allele from the female. For the male to be wild-type, we still
must have part of the wild-type X chromosome. To test, cross this
wild-type male with white-eyed females. All the female progeny
should be white-eyed and all the male progeny red-eyed. Cytologi-
cal examination of the chromosomes should reveal the transloca-
tion.

19. 4n = 92;2n - 1 = 45.

21. n = 8 + n = 6 equals 14 X 2 = 28; 20 + 20 = 40

23. An XO/XYY mosaic can occur by nondisjunction of the Y chromo-
some in a cell during early cleavage in an XY individual. An
XX/XXY mosaic can come about if one of the cells during early
cleavage in an XX zygote is fertilized again by a Ybearing sperm.
Trisomy 21 usually comes about from an egg with two copies of
the chromosome; the egg had two copies because of meiotic
nondisjunction.

25. The father. The allele for color blindness can only come from the
mother. If meiosis in her is normal, an egg could get the X chro-
mosome carrying the mutant allele. The daughter has only one X
chromosome, so the sex chromosomes failed to separate in the
man, and a sperm with neither X nor Y fertilized the egg.

27. The first meiotic division in the father is normal, producing cells
with either two X or two Y chromatids. During the second meiotic
division in the cell with the two Y chromatids, both Y chromatids
move to the same pole and end up in the same sperm cell.

Third
Fourth

Critical Thinking Question:

1. The numbers are all multiples of 14 (1, 2, 3, 4, 5, 6, 7, and 8 copies).
We can thus hypothesize that the original diploid chromosome
number (2n) is 14. The other species would be polyploids, multi-
ples of the original 14 (tetraploid, hexaploid, etc.). These are all the
even ploids up to 112 chromosomes. As we saw, even ploids have
the potential to succeed in meiosis, whereas odd ploids rarely do.

Chapter 9 Chemistry of the Gene

1. The genetic code would somehow be read in number of tetranu-
cleotide units, in which each unit consists of one each of the four
bases (G, C, T, A). For example, one unit might be the amino acid
alanine, two units might be the amino acid arginine, and so on.

3. Sugars: DNA has deoxyribose, RNA has ribose; and bases: DNA has
thymine in place of uracil, RNA has uracil in place of thymine.

5. See figures 9.18 and 9.19.

7. a. 28% G; 28% C; 22% A; 22% T b. Same percentages except 22%
U, 0% T. Chargaff 's rule states that the quantity of A = T and the
quantity of G = C. If G = 28%, then C = 28% and G + C = 56%.
The sum of all bases must equal 100%. Therefore, (A + T) = 100 —
56 = 44. Since A = T, 1/2(44) = 22%. This is the amount of both A
and T For an RNA molecule, proceed the same way, except re-
member that U replaces T, so we have 22% U.

9. There must be regions of complementarity within the single-
stranded regions. A melting temperature indicates some regions
that are double-stranded. We can envision at least two different
possible configurations:

1 . Whole molecule complementary

2 . Fragment complementary

In fact, most single-stranded molecules have some regions that are
complementary.

11. 191 billion base pairs. One base takes up 3.4 A. Since 1A is
1/10,000,000 millimeter, there are 10,000,000/3.4 = 2.94 million
bases per millimeter. Multiply by 1,000 to get to meters, and finally

by 6.5 for 6.5 meters: = 1.91 X 10 iU bases.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

All

13.

Radioactivity

Conservative

Dispersive

15. A primosome is a helicase plus a primase; it opens the DNA and
creates RNA primers on lagging strands and is part of the repli-
some. A replisome includes a primosome plus two copies of DNA
polymerase III; it coordinates replication on both the leading and
lagging strands at the Yj unction.

17. See figure 9.28.

19. At one time molecular swivels, presumably protein in nature, lo-
cated periodically along the DNA, were suggested.

21.

RNA

RNA
DNA

host

Insert into host

RNA

This question really asks how double-stranded DNA can be formed
from single-stranded RNA. First, we could synthesize a comple-
mentary DNA strand, then begin to make a second DNA strand
complementary to the first. While synthesizing the second DNA
strand, we begin to degrade the RNA. The double-stranded DNA
molecule now inserts into the host DNA. In order to get more
viruses, a single-stranded RNA molecule must be made from the
DNA. We would need, at a minimum, an enzyme to make a single-
stranded DNA to form a hybrid with the RNA; an enzyme to de-
grade the RNA and to make the second, complementary DNA
strand; an enzyme to cut the host DNA to allow the viral DNA to
insert itself; an enzyme to ligate the two molecules; and an enzyme
to make more viral RNA. The earlier functions are performed by
one viral enzyme, reverse transcriptase.

23. Finding small pieces or fragments of DNA suggests the Okazaki
pieces are only slowly, if at all, joined, a function of DNA ligase. The
fact that not many long DNA molecules are seen also suggests that
the DNA is being broken, implicating a nuclease as well.

25. It is unlikely that bases are added faster in developing embryos. So
we must look for another mechanism. If there are more replicons,
and hence more origins of DNA replication, each replicon will be
shorter and be able to duplicate faster. Alternatively, and more
likely, the process is regulated to slow down adult division.

Critical Thinking Question:

1. One way to study mutations that are generally lethal is by isolating
temperature-sensitive mutations. These mutations involve amino
acids that disrupt the functioning of the enzymes at some critical
temperature but are phenotypically normal at other temperatures.
Thus, the mutant organisms can be kept alive by growing them at
one temperature (the permissive temperature) but their mutant ef-
fect can be studied at the temperature in which the protein func-
tion is disabled (the restrictive temperature). (For additional dis-
cussion of these mutations, see chapter 12.)

Chapter 10 Gene Expression: Transcription

1. See figure 10.3. Complementarity is achieved between messenger
RNA and ribosomal RNA and between messenger RNA and transfer
RNA.

3. Transcription has higher error rates. Errors of DNA polymerase
tend to become permanent, whereas errors of RNA polymerase do
not.

5. A consensus sequence is made up of the nucleotides that appear in
a significant proportion of cases when similar sequences are
aligned. A conserved sequence consists of nucleotides found in all
cases when similar sequences are aligned. For example, the Prib-
now box (fig. 10.6) is the consensus sequence TATAAT.

7. See figure 10.8 for a promoter and figure 10.10 for a terminator.
The transcript starting from the promoter would be
5'-CUUAUACGGU....The transcript from the terminator is shown in
figure 10.10.

9. A stem-loop structure can form when a single strand of DNA or
RNA has a double helical section (see fig. 10.10). An inverted re-
peat is a sequence read outward on both strands of a double helix
from a central point (see fig. 10.10). A tandem repeat is a segment
of nucleic acid repeated consecutively; that is, the same sequence
repeats in the same direction on the same strand:

5 '-TCCGGTCCGGTCCGG-3 '

3 '-AGGCCAGGCCAGGCC-5 '

A DNA sequence with a seven-base inverted repeat is

5 '-ATTACCGCGGTAAT-3 '

3 '-TAATGGCGCCATTA-5 '

11. Footprinting is a technique in which DNA in contact with a protein
is exposed to nucleases; only DNA protected by the protein is undi-
gested. Promoters could be isolated by protection with RNA poly-
merase in the absence of ribonucleotides — the polymerase will
not move — and then sequenced.

13. The superscripts of the sigma factors refer to their molecular
weights (e.g., a 70 is 70,000 daltons). Different sigma factors usually
recognize different prokaryotic promoters.

15. 3'-GGTAGTACTGTCTGGGAACGATTGCG-5'<-

5'-CCATCATGACAGACCCTTGCTAACGC-3'

Begin by writing the strand that is complementary to the RNA. This
will be the transcribed strand. Remember, U in RNA pairs with A in
DNA. Since transcription proceeds 5' — > 3\ the 5' end of the RNA
is opposite the 3' end of the DNA.

17. The bottom strand is transcribed and the molecule is arranged as

5'
3'.

.5'

Begin by writing the RNA that could be transcribed from each
strand. Since the DNA represents the beginning portion of the gene,
the RNA must have an AUG to start protein synthesis. Unfortunately,

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-12

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

both strands yield RNAs with one or more AUGs. The RNA from the
top strand has AUGs in both directions, but in each case the AUG is
followed by a termination signal, UAA, or UAG. This RNA could not
make a protein. The bottom strand produces an RNA with only one
AUG. Since transcription and protein synthesis both proceed 5' — >
3', the left end of the bottom strand must be 3'.

19. If transcription of the genes is rho-dependent, the RNAs made at
40° C will be longer than those made at 30° C. Since the rho cannot
function at the high temperature, RNA polymerase will read past
the termination region. If transcription is rho-independent, a rho
mutant will have no effect on transcription, and hence, the size of
the RNAs.

21. The double helix must unwind in order for transcription to occur.
A-T pairs, because they have only two H-bonds, are more easily dis-
rupted than G-C pairs.

23. Removing one base too many or too few would result in a shift in
the reading frame during translation (see chapter 1 1), thus radically
altering the protein product.

25. Five introns

mRNA
DNA

No introns

mRNA
DNA

27. Group I introns are self-splicing introns that require a guanine-
containing nucleotide for splicing. Group II introns are similar but
do not require an external nucleotide for splicing. Group I and II
introns are released as linear and lariat-shaped molecules, respec-
tively.

29. Spliceosomes are composed of small nuclear ribonucleoproteins
(see fig. 10.37).

31.

There are three introns, so we expect three single-stranded DNA
loops. The coding regions (exons) form RNA-DNA hybrids and ap-
pear thicker.

33. Enhancers [E] bind activator proteins that also bind proteins of
the polymerase at the promoter [P]. In some eukaryotic genes,
there are many enhancers, allowing numerous levels of control
and forcing enhancers further and further upstream. For activa-
tors bound to enhancers to bind polymerase proteins, the DNA
must loop around.

Critical Thinking Question:

1. Given that a gene controls the production of a protein, there are
both realistic and theoretical limits to the size of a gene. From what
we know about the functioning of the centromere, a gene would
have to occupy no more than the length of a chromosomal arm.
However, given that human chromosomes must contain about fifty
thousand genes, it is unlikely that any gene is that large. We also
know that many functioning proteins are made up of subunits,
each controlled by its own gene. Thus, large proteins tend to be
conglomerates of smaller ones rather than large functional units. Fi-
nally, the larger the protein the more time it takes to transcribe and
translate it, making very large size inefficient. As we mentioned be-
fore, the average protein is about 300 to 500 amino acids; with in-
trons, and control elements, the gene for an average protein could
be quite large. The largest known gene is the human dystrophin
gene that codes for a cytoskeletal protein. It is 2,300,000 bases
long, has 79 exons, and takes 16 hours to be transcribed.

Chapter 11 Gene Expression: Translation

1. The messenger RNA is 5'-AUGUUACCGGGAAAAUAG-3'; the anti-
codons are 3'-UAC-5', 3'-AAU-5', 3'-GGC-5\ 3'-CCU-5', 3'-UUU-5';
the amino acids are methionine, leucine, proline, glycine, lysine
(see the figure below).

3. See figure 11.16. Use the messenger RNA of problem 1 and be sure
to include EF-Tu and EF-Ts.

5. There are approximately twenty aminoacyl-tRNA synthetases in an
E. colt cell, one for each amino acid. Recognition signals can occur
at any point on a given transfer RNA, although the anticodon fig-
ures prominently in most.

7. See figure 11.7.

9. Three; see figure 11.29.

11. 5'-UAA-3' —> 5'-UUA-3' (leucine). The consequence is the failure to
terminate the particular protein leading to continued chain elon-
gation to the next nonsense codon or to the end of the messenger
RNA. The result is probably a nonfunctioning enzyme or protein.

H 3 N +

CH 3
S

CH 2

— CH — C —

ChL Ch

CH 2

CH 2
NH— CH-

-c

CH 9 CH 9

\ /

— N— CH — C-

-NH-

-CH 9 -

-c

N + H,

CH,

— C — NH — CH — C

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-13

13. EF-Tu brings a charged transfer RNA to the A site at the ribosome.
EF-Ts is involved in recharging EF-Tu (see fig. 11.14). The eukary-
otic equivalents are eEFla and eEFl|3y.

15. A signal peptide is a sequence of amino acids at the amino-terminal
end of a protein that signals that the protein should enter a mem-
brane (see fig. 11.25). Although the concept is the same, the situa-
tion in eukaryotes is somewhat more complex because there are
many different membrane-bound organelles, each having their
own membrane-specific requirements. Signal peptides are usually
cleaved off the protein after the protein enters or passes through
the membrane.

17. NH 2 -FGKICABHLNOEDJM-COOH

19. a. 5 'AUG AUU GAA UGC GAG CGG AGU-3'

b. N-met-ile-glu-cys-glu-arg-ser

First determine the sequence of the RNA complementary to the
given DNA strand. Don't forget about polarity; as the strand is writ-
ten, the 5 ' end of the RNA will be on the left. Blocking off succes-
sive groups of three bases allows the determination of the codons.
Use the code to determine the amino acid sequence.

21. 12/27 phenylalanine (UUU, [2/3] 3 ; UUC, [2/3] 2 [ 1/3]); 6/27 serine
(UCU, [2/3] 2 [l/3]; UCC, [2/3] [1/3] 2 ); 6/27 leucine (CUU,
[2/3] 2 [l/3]; CUC, [2/3] [1/3] 2 ); 3/27 proline (CCU, [2/3] [1/3] 2 ;
CCC, [1/3] 3 ).

23. The table could look the same (see table 11.4) except that the po-
sition would be left side = first position (5' end); top = third posi-
tion (3' end); right side = second position. For example, the codons
for valine (currently 5'-GUU-3', 5'-GUC-3', 5'-GUA-3', and 5'-GUG-
3') would be 5'-GUU-3', 5'-GCU-3', 5'-GAU-3\ and 5'-GGU-3'.

25. We are mixing two RNA strands that are complementary; these
strands will form a double-stranded RNA molecule. Since we ob-
served the incorporation of no amino acids, the ribosome must not
be able to read a double-stranded molecule.

27. If we write out (GUA) n as GUA GUA GUA GUA . . ., we see that we
could use any of three different reading frames: GUA, UAG, or AGU.
Since we see only two amino acids incorporated, either two of the
possible codons code for the same amino acid, or one of the
codons is a stop codon. If you look at the code, you will see that
UAG is a stop codon.

29. The stop codon has probably mutated to give a codon for the
amino acid leucine. The longer-than-normal protein suggests that
the original stop was not read. Numerous possibilities exist. If the
second letter of a stop codon were changed to a U, we would have
UUA or UUG leucine codons. Alternatively, the insertion of a C be-
fore the U would yield CUA, CUG, or CUA as leucine codons. Simi-
larly, an insertion of a U next to the first G yields UUA or GUG as
leucine codons. Since the next amino acid is phenylalanine, the
next codon must be UUC or UUU. If a base were added, as above,
the next codon would have to begin with A or G, and phenylala-
nine does not begin with A or G. Therefore, the most likely expla-
nation is a change of the second letter from an A to a U.

31. Either CAU or CAC. Write down all possible codons for each amino
acid.

his

tyr

gin

pro

leu

\U/C

CAA/G

ccx

CUX

For leu, note that UUA/G cannot result from a single change in the
his codon. Therefore, leucine must be CUX. All of the other amino
acids could result from single changes in either the first or second
base, and we are left with either codon being the one for his.

Critical Thinking Question:

1. In general, transcriptional and translational signals are indepen-
dent. We could look at this by asking the question, how does
changing one of the signals affect the other process? In other
words, if we changed a translational signal such as a start or stop
codon, would that affect the transcription of that gene? In general,
the answer is no.

Chapter 12 DNA: Its Mutation, Repair, and Recombination

1. For example, if the twenty individual cultures of table 12.1 had val-
ues of 15, 13, 15, 20, 17, 14, 21, 19, 16, 13, 27, 14, 15, 26, 12, 21, 14,
17, 12, 14, then the mutation theory would not have been sup-
ported because the variation between the individual and bulk cul-
tures would not have been different.

3. Reading across each row, we gather more and more information.

Row 1: 1,6, and 7 are part of one complementation group.

Row 2: 2 and 5 are part of one complementation group.

Row 3: 3 and 4 are part of one complementation group.

Row 4: no new information.

Row 5: no new information.

Row 6: reinforces that 6 and 7 are part of the same complementa-
tion group.

Row 7: no new information.

Thus, we conclude that there are three complementation groups
present: 1,6, and 7 are mutually noncomplementing, as are 2 and 5,
and 3 and 4. The half-table missing is a mirror image because a
cross of 1 and 3 is the same as a cross of 3 and 1 (reciprocity). The
diagonal always contains negative elements because every mutant
is a functional allele of itself.

5. All mutants should be crossed in pairwise combinations yielding
heterozygote daughters. (Presumably, earlier crosses indicated that
these are X-linked loci.) All F x daughters will be wild-type: The mu-
tations complement and therefore are not alleles. When eosin flies
are crossed in a similar fashion, daughters will be wild-type except
when the parents were eosin and white. In that case, daughters
will be mutant, showing the lack of complementation and hence
that the mutations are alleles.

7. Prokaryotic and phage genes generally do not have intervening se-
quences. Benzer and Yanofsky worked at a time when introns were
unknown and it was assumed that the length of a gene was tran-
scribed and then translated. If the genes had introns, Benzer and
Yanofsky would have been unaware of them; introns would not af-
fect colinearity or mapping. Introns would affect physical mea-
sures of the lengths of DNA, with which Benzer and Yanofsky were
not involved.

9. The rex gene of phage A. represses growth of phage T4 rll mutants.

11. In replicating DNA, a transition mutation can occur by tautomer-
ization of a base in the template strand (template transition) or en-
tering the progeny strand (substrate transition).

13. 5-bromouracil (pyrimidine analogue) and 2-aminopurine (purine
analogue) are incorporated into DNA as thymine and adenine, re-
spectively. However, each undergoes tautomeric shifts more fre-
quently than the normal base. Both cause transitions. Nitrous acid
also promotes transitions by converting cytosine into uracil, which
acts like thymine, and adenine into hypoxanthine, which acts like
guanine. Proflavin induces insertions and deletions by intercalating
and buckling DNA. Ethyl ethane sulfonate removes purine rings
and thus promotes transitions and transversions.

15. See figure 12.25.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-14

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

17. Three genes. Gene A: mutants 1, 4, 8; gene B: mutants 2,5; gene C:
mutants 6, 7. Mutant 3 probably contains a deletion that spans
genes A and C. Begin by finding mutants that do not complement.
These should have mutations in the same gene. Mutants 2 and 5 are
in the same gene. Initially we suspect that mutants 1,3,4, and 8 are
in the same gene, which is different from the gene that contains 6
and 7. If mutant 3 is in gene A, it should complement 6 and 7, and
it does not. One explanation is that 3 is a deletion spanning genes
A and C Alternatively, mutant 3 could be in gene A but be a polar
mutation. Either possibility implies that the order is B A C

19. I 2

Begin with deletions that yield mostly "-"s. These must be large dele-
tions that cover most of the other deletions. Mutations 1 and 5 are
such mutations. Since they give no wild-type, they must overlap:

Now look at mutant 2. It gives wild-type recombinants with 1 but
not 5. Therefore it must overlap the region deleted in 5. Mutant 3,
by similar logic, must cover part of deletion 1. We can draw these
results as follows. Broken lines indicate we do not know yet how
long the deletion is:

Now look at 4. It gives "-" (no wild-type recombinants) with both
1 and 5 and therefore must be in the region that 1 and 5 overlap. If
4 extended to and overlapped either 2 or 3, we expect to see "-"
with them. Since this prediction is not met, 4 must be a small dele-
tion spanning at least part of the overlap of 1 and 5. Since mutant
6 gives no wild-type with 1, 4, or 5, it must be within the common
region deleted in all three strains.

21. We know that the anticodon pairs with the codon, and we expect
nonsense suppressors to contain an altered anticodon. The fact
that the nonsense codon can be read by a transfer RNA with a nor-
mal anticodon but altered dihydrouridine loop suggests that the
way in which this loop interacts with the ribosome causes the an-
ticodon sequence to be misread.

23. x + — > x: AT — > GC. y + — > y: GC — > AT. This problem requires logic
and a knowledge of how mutagens work. For x, the key is the re-
sponse to HA, which only causes GC — > AT transitions. Mutant x is
reverted by HA, therefore, x must be GC, and the normal x + was
AT. AP-induced mutations can also be reverted by AP. Since y is not
reverted by HA, y must be AT. Therefore, y + must be GC.

25. A deletion that spans regions of more than one gene. Consider the
following two genes:

By definition, mutations in gene 1 will complement mutations in
gene 2 but not other mutations in gene 1 . If we have a deletion, x,
that covers part of both 1 and 2 (slashes),

-///////////////-

presumably gene 1 will be nonfunctional because it is missing the
last part of the protein. Gene 2 will be nonfunctional because the

beginning portion of the gene is missing. The following genotypes
will give no complementation:

-/////////-

-////////-

No functional product of gene 1 No functional product of gene 2

We could also get a lack of complementation when mutants are in
two genes if we have a bacterial operon in which one of the genes
contains a polar mutation creating a transcription stop signal. Such
a mutation eliminates all distal functions. Thus, if the operon is

and we construct the partial diploid

A~ (polar) B + C + D

B C + D

we will get no complementation because the top DNA is effec-
tively^" B~ C~ D~.

27. The auxotroph probably contains a deletion. If a few bases are
missing, nothing is available to cause transitions or transversions. It
is highly unlikely that the correct number of missing bases could
be spontaneously and correctly inserted.

29. Excision repair endonucleases can recognize dimerizations, mis-
matched bases, and apurinic-apyrimidinic sites.

31. See figure 12.38 for a diagram of recombination. Branch migration
is shown in figure 12.39.

Critical Thinking Question:

1. The gene is a linear entity that specifies the linear order of amino
acids in a protein in a colinear fashion. Although scientists in the
1960s were convinced of colinearity, there were other alternatives
possible. For example, DNA could be a branching structure. Or,
transcription of DNA could take place such that the beginning,
middle, and end of the gene were not in order. Thus colinearity
supported our understanding of the shape and functioning of
DNA.

Chapter 13 Genomics, Biotechnology, and Recombinant DNA

1. Type II endonucleases are valuable because they cut DNA at spe-
cific points and many leave overlapping or "sticky" ends.

3. In DNA with a random sequence, a four-cutter will find sites ap-
proximately once in 4 4 bases (= 1/256 = 0.0039). A six-cutter will
find sites approximately once in 4 6 bases (= 1/4096 = 0.0002).
An eight-cutter will find sites approximately once in 4 8 bases
(= 1/65,536 = 0.000015).

5. DNA can be joined by having compatible ends to begin with or by
blunt-end ligation (linkers combine these methods). The appropri-
ateness of a method depends on what DNA is to be cloned and
how that DNA can be obtained. Having DNA with "sticky" ends cre-
ated by the same restriction enzyme would be easiest but some-
times is not available. Adding linkers by blunt-end ligation with a
particular restriction site is usually the best compromise.

7. A plasmid is a self-replicating circle of DNA found in many cells.
Foreign DNA inserted into a vector forms an expression vector if
that foreign DNA produces a protein product. Cosmids are plas-
mids that contain cos sites and are useful for cloning large seg-
ments of DNA (up to 50 kb). YACs, yeast artificial chromosomes,
have the loci to replicate in yeast (centromere, replication origin,
and telomeres). They can be used to clone very large pieces of
DNA, upwards of one million bases.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-15

9. Chromosome walking is a technique for cloning overlapping chro-
mosomal regions starting from an arbitrary point (see fig. 13. 38). It
is useful for determining relative locations of genes in uncharted
regions as well as cloning regions too big to fit in a single vector.

11. Southern and northern blotting are gel transfer techniques used to
probe for DNA and RNA sequences, respectively. Western blotting
is a technique used for locating a protein by antibody recognition.
Dot blotting is a probing technique for cloned DNA that eliminates
the electrophoretic separation step.

13. Plasmids of E. coli origin survive in yeast when a yeast centromere
(CEN region) is added, allowing them to replicate within the yeast
cell. Inactivated SV40 viruses can function in the presence of intact
"helper" viruses that allow them to complete their life cycles.
Phage A. has parts of its chromosome that can be removed while
still allowing it to complete its life cycle.

15. Partial digestion of molecule 2 leads to the following molecule:

AAAAAAAA

^ ■ *'■ *'i *'i *'i *'i *'i *'i *

Some of these molecules will form a circle with the single-stranded
Ts paired with single-stranded As. The circle eliminates the free 5'
phosphate, and the enzyme can no longer work.

17. We must insert DNA that has no introns into bacterial plasmids.
This DNA can be obtained by isolating mature, cytoplasmic mes-
senger RNA and then using reverse transcriptase to make double-
stranded cDNA. Plasmids with cDNA inserted can then be used to
produce human proteins (expression vectors).

19. Electrophoretic bands of the total digest are (* indicates end label) 50,
100*, 150*, 250, 300 bp. Bands of the partial digest are 50, 100*, 150*,
250, 300(X2), 350, 400*(X2), 450*(X2), 600, 700*, 750*, 850* bp.

Total

Partial

900
850
800
750
700
650
600
550
500
450
400
350
300
250
200
150
100
50

*
*

21. Mutant A: elimination of site between the 300- and 50-bp segments.
Mutant B: elimination of site between 100- and 300-bp segments.
Mutant C: creation of a new site within the 300-bp segment, divid-
ing it into 75- and 225-bp segments.

23. This problem begins as a trial and error attempt to overlay two re-
striction maps, made more difficult by the fact that one enzyme,
BamHI, has made three cuts that are unordered, leaving many pos-
sibilities. However, a bit of thought beforehand makes this problem
much easier. If you compare the double digest with the BamHI di-
gest, they share 200-, 250-, and 400-bp segments. The double digest
has 50- and 100-bp segments replacing the 1 50-bp segment in the
BamHI digest. The inference is that there is an EcoRI cut in the 1 50-
bp segment leading to the 50- and 100-bp segments, with all other
segments of the BamHI digestion left uncut. That leaves only two
possibilities, as shown below; the data are insufficient to distin-
guish between the two choices.

EcoW

300 |

700

^coRI

300 |

700

200 t 150 t
BamHI BamHI

[any order]

250 t 150 1
BamHI BamHI

[any order]

25. Ecom
BamHI

6.2 | 2.8 | 4.6 | 7.4 | 8.0

10.0

13.0

| 6.0

We know that the 6.2 and 8.0 kb EcoKI fragments are at opposite
ends, and that the 10.0 and 6.0 kb BamHI fragments are at the
ends. Therefore, the BamHI 130 kb fragment must be in the mid-
dle. If the 6.2 and 6.0 kb fragments are at the same end, a double di-
gest should produce a fragment of 0.2 kb. This is not seen, so they
are at opposite ends:

6.2 | (2.8,4.6,7.4) | 8.0

10.0

13.0

| 6.0

If 7.4 is next to 6.2, we should see a 38 fragment. Similarly, if 4.6 is
next to 6.2, we should see a 3.8 fragment. We see neither of these
fragments, so 2.8 is next to 6.2. The 4.6 fragment must be next.

27. Since EcoHI does not eliminate either resistance, its site must be be-
tween the tef and amp r genes. The PstI site must be within the
amp r gene, since insertion of DNA into this site eliminates ampi-
cillin resistance. By similar logic, the other sites must be in the tef
gene. In the double digests, the smaller fragment must represent
the distance from the Eco RI site to the other site. We can draw part
of the plasmid as:

EcoRI

Sail

r amp

(a)

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-16

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

EcoR\

tef

amp'

0.7

0.08

Pstl

0.3

H/ndIII

Bglll
0.85

Sail

(b)

29.

31.

If this arrangement is correct, digestion with Bglll, EcoBI, and Pstl
should yield 0.3 kb (Bglll + .EcoRI), 0.7 kb (EcoRl + Pstl) + 2.0 kb
(Pst I + Bglll), and this is, in fact, observed.

The second possibility predicts that a 1.5-kb fragment should be
seen in the double digestion. This fragment would result from di-
gestion of the 35-kb EcoHl fragment with Bglll. Since we don't see
a 1.5-kb fragment in the double digest, the second possibility does
not agree with the observed results.

Two of the three number 21 chromosomes present in the child
came from the father, not the mother. Since the probe produces dif-
ferent bands in the mother and the father, all of these bands must
also be present in the child. The intensity of a band is proportional
to the amount of DNA present. The bands that are of paternal
origin are more intense than the maternal bands. The father
contributed two number 21 chromosomes.

33. For the steps in the dideoxy sequencing method, see figures 1332
and 13- 33. Use of fluorescent dyes has allowed for the automation
of the process and the elimination of radioactive tags.

35. The DNA can be inserted into the M13 general sequencing vector.

37. What will appear in the gel are fragments of the newly synthesized
strand. Since DNA synthesis proceeds 5' — > 3', the 5' base will beT
in the new strand. Proceed up the gel by indicating the base com-
plementary to the sequence given.

ddCTP

ddTTP

ddATP

ddGTP

14
13
12
11
10
9
8

7
6

5
4
3
2

39. Hypervariable DNA is DNA showing a great deal of interindividual
variation. A RFLP (restriction fragment length polymorphism) is a
polymorphism (variation) that shows up after Southern blotting
and probing of restriction digests. A VNTR (variable number of tan-
dem repeats) locus is one that is hypervariable due to unequal
crossing over among the tandem repeats. VNTR loci are a hyper-
variable subset of RFLPs. Sequence-tagged sites are unique sites in
the genome that can be amplified with polymerase chain reaction.
Microsatellite DNA, repeats of very short segments such as CA,
forms VNTR loci that are usually examined by polymerase chain re-
action (PCR) if primer sequences are known.

Critical Thinking Question:

1. Sticky ends would exist if the plasmid DNA had a 3' overhang of
one nucleotide residue, while the foreign DNA had a 3' overhang
of its complement. The exact number of bases would not need to
be the same because repair enzymes plus ligase could close the
gap. For example, the plasmid could have a 3' tail of thymines
added, whereas the foreign DNA could have a 3 ' tail of adenines
added. This method (see figure next page) was called the poly-
dA/poly-dT technique.

Chapter 14 Gene Expression: Control in Prokaryotes
and Phages

1. a. inducible (wild-type); b-d. constitutive; e. neither, super-
repressed; f. inducible

3. One mutant could fail to bind to operator DNA but could still bind
the inducer (op~ ', m + ).This mutant would have constitutive tran-
scription of the operon. The reverse situation could also be true;
the repressor could bind the operator but not the inducer (qp + ,
in~). This mutant would be off all the time. A third mutant could
fail to bind both (pp~, in~), being constitutive.

5. a. Operator and repressor, b. Make a partial diploid with wild-type;
the operator mutant will make /3-galactosidase constitutively, and
the repressor mutant will make it only in the presence of lactose.
Four mutations are possible in the lac operon: mutations in the z
gene or the promoter never make the enzyme. Mutations in the re-
pressor always make the enzyme because the repressor cannot bind
DNA. In operator mutations, a good repressor can never bind DNA.
In a partial diploid, i~o + /i + o + , the wild-type repressor is trans act-
ing and can bind to both operators, creating an inducible situation.
In i + o~/i + o + , repressor cannot bind to o~, and this DNA is always
on, even though the wild-type DNA is off in the absence of lactose.

7. Cyclic AMP, combined with CAP protein, attaches to CAP sites en-
hancing transcription of nonglucose, sugar-metabolizing operons in
E. coll Glucose inhibits its formation by inhibiting adenylcyclase.

9. We must think about how these operons are controlled. Not only do
they need inducer, but they also require the catabolite repression-
activation system. These mutants could be unable to make cAMP be-
cause the adenylcylase gene is defective. Alternatively, they could
be making a defective catabolite activating protein (CAP).

11. b = tryptophan synthetase gene, a = operator, c = repressor. Look
first for the single mutation that never gives enzyme; this genotype
will give the letter of the structural gene. The genotype a + b~c + fits
this requirement, so b is the structural gene, and a and c represent
control regions. Genotypes 4 and 5 tell us nothing. Look at geno-
type 6. The right DNA will never make the enzyme. If c is the op-
erator, the left DNA should always make the enzyme, and this is not
seen. Therefore, a must be the operator. Check these assignments
with genotype 7. The right DNA will never make the enzyme. If a
is the operator, the left DNA is always on. If a is the repressor, the
right DNA makes a good repressor that will bind to the top DNA
and regulate it.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-17

Foreign DNA

Plasmid

Cleave with

restriction

endonuclease

Cleave by shearing

Treat with

deoxynucleoside terminal
transferase plus dTTP

I I I I I

AAAAA

Treat with

deoxynucleoside terminal
transferase plus dATP

AAAAA

Treat with repair enzymes
(for gap filling) and ligase

Hybrid
plasmid

13. The E. colt trp operon functions as a normal repressible operon. In
addition, attenuation, based on secondary structure and stalling of
the ribosome on the leader transcript, can further prevent tran-
scription (see fig. 14.16). Attenuator control can be exerted based
on other amino acids if their codons appear in the leader tran-
script, causing ribosome stalling.

15. Assuming that the mutants produce inactive proteins: cl, ell, and
cIII, lytic response; A^ neither lytic nor lysogenic responses possi-
ble; cro, no lytic response possible; att, no lysogenic response pos-
sible; Q, no lytic response possible.

17. The A chromosome has one circular and two linear forms. The
circular form is the infective cellular form. A break at one point
(cos site) takes place during packaging into the phage head, and

a break at another point forms the linear integrative prophage
(see fig. 14.18).

19. The prophage region of the Hfr chromosome enters the F~ cell
with no repressor present. The situation is thus similar to regular
phage infection, which can go either way (lysogenic or lytic cycles).

21. The cells that form colonies do not contain a prophage. The initial
heat shock inactivated the repressor and allowed the prophage to
excise. The lower temperature activated the repressor, and the re-
pressor bound to the excised phage DNA, preventing gene expres-
sion and reintegration of some of the phages. These "cytoplasmic"
phages failed to replicate and were lost during cell division. The
cells that retained the virus were lysed when the temperature was
raised because the phage could now make RNA.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-18

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

23. An IS element is a simple transposon, which is a segment of DNA
that can make a copy of itself to be inserted at another place in the
genome. An intron is an intervening sequence, a region excised
from messenger RNA before expression. A plasmid is an au-
tonomous, self-replicating genetic particle. A cointegrate is an in-
termediate structure in transposition.

25. See figure 14.29.

27. See figure 14.35.

29. Transcription is the level at which most control mechanisms work.
These include sigma factors, efficiency of promoter recognition,
catabolite repression, operon-repressor systems, attenuation, and
transposition. Translational control mechanisms include polarity
placement, antisense RNA, differences in the efficiency of
processes due to nucleotide sequence differences, codon prefer-
ence, and the stringent response. Posttranslational mechanisms
include feedback inhibition and differential rates of protein degra-
dation.

31. Heat shock proteins are normally induced by the presence of a
specific sigma factor, which itself is induced by heat shock.

33. The stringent response is the response of a prokaryotic cell to
amino acid starvation. The idling reaction of the ribosome results
in production of 3'-ppGpp-5', whose appearance is associated
with the cessation of transcription, especially of transfer RNAs and
ribosomal RNAs, through an unknown mechanism.

35. Feedback inhibition: allosteric enzymes in some synthetic path-
ways can be inhibited by the end product of that pathway. In addi-
tion, many repressor proteins are allosteric.

Critical Thinking Question:

1. One might think that a bacterium should metabolize any sugar in
its environment. However, glucose is the most efficient sugar to
metabolize and if it is present, it should be metabolized first. Even
under maximal growth, it takes an E. coli cell about 20 minutes to
grow and divide. If the cell were to begin to take up and metabo-
lize other sugars, there would be a cost for the manufacture of new
enzymes and the inefficiency of the initial steps of the sugar me-
tabolism. In fact, the growth of the cell would slow down under
these circumstances and be at an evolutionary disadvantage.

Chapter 15 The Eukaryotic Chromosome

1. In general, prokaryotes are small, have a relatively small circular
chromosome, and have little internal cellular structure compared
to eukaryotes. Most prokaryotic messenger RNAs are polycistronic,
under operon control; eukaryotic messenger RNAs are highly
processed, monocistronic, and usually not under operon control.
Prokaryotes are mostly single-celled organisms, whereas eukary-
otes are mostly multicellular. Eukaryotes have repetitive DNA, ab-
sent for the most part in prokaryotes. Prokaryotic chromosomes
are not complexed with protein to anywhere near the same extent
that eukaryotic chromosomes are.

3. Assume that each chromosome contained two complete copies of
the same DNA. Following the protocol of figure 15.1, the final re-
sults would be chromosomes, before separation, that consisted of
either two labeled chromatids or only one labeled chromatid, in a
1:1 ratio (barring sister chromatid exchanges). The labeled chro-
matid in chromosomes with just one chromatid labeled will have
twice the label of each chromatid in the chromosomes in which
both chromatids are labeled (see the following figure).

DNA (in the
chromosome)

Chromosome
appearance

Replication in
3 H-thymidine

Color represents
3 H-thymidine label

Separation

Replication in
unlabeled medium

50%

5. The length of DNA associated with nucleosomes was determined
by footprinting, in which free DNA was digested, leaving only
those segments protected by nucleosomes. Nucleosome hypersen-
sitive sites are sites not in a nucleosomal state; they seem to be sites
involved in the initiation of replication, transcription, and other
DNA activities.

7. See figures 15.10 and 15.11 for the relationship of the 110, 300, and
2,400 A chromosome fibers.

9. See figure 15.19.

11. Polytene chromosomes are chromosomes that underwent endomi-
tosis: They consist of numerous copies of the same chromatid
(e.g., in the Drosophila salivary glands). Regions of active tran-
scription in polytene chromosomes form diffuse areas called puffs
or Balbiani rings (see fig. 15.14). Lampbrush chromosomes occur
in amphibian oocytes (see fig. 15.18).

13. Satellite DNA differs in its base sequence from the main quantity of
DNA and thus forms a satellite band during buoyant density analy-
sis. It is usually centromeric heterochromatin, composed of a
highly repetitive DNA.

15. Telomeres are repetitive DNA sequences at chromosomal ends.
They are repetitions of a five- to eight-base sequence. Most telo-
meres are G-rich. Telomeres protect the ends of chromosomes and
probably provide signals on senescence of cells.

17. Highly repetitive DNA usually makes up the centromeric and
telomeric regions of the chromosome. Unique DNA, making up the
bulk of structural genes, has a large component that is transcribed.
Repetitive DNA is composed of dispersed DNA (e.g., short and
long interspersed elements — SINES and LINES), multiple copies of

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-19

transcribed DNA (e.g., ribosomal RNA, histories), and diverged
copies of ancestral genes (e.g., globin family genes).

19. The most direct method of determining the direction of transcrip-
tion of the histone genes would be to clone and sequence the re-
gion, from which transcriptional information can be ascertained.

21. Cloning and then sequencing the region would provide the an-
swer. Analysis would show genes of similar sequence to the active
genes but lacking the sequences for transcription.

23. It is very rich in A-T sequences. Density is proportional to G-C con-
tent. Since the molecule has a low density, it has low G-C, and
therefore high A-T.

25. Spaces between the nucleosomes must contain many promoter se-
quences. For the DNA to be digested, it must be unprotected. Since
we see little transcription, the promoters must be missing and
must have been destroyed by the nucleases.

27. Highly repetitive DNA must be located in these regions. Since most
highly repetitive DNA is not transcribed, the results suggest that
centromeric and telomeric regions are not transcribed.

29. The C-value paradox involves the issues of the excessive amounts
of DNA in eukaryotic cells and the difference between eukaryotic
species that seem to have similar complexity. It is explained by the
large amount of structural DNA in chromosomes as well as the
large amounts of short and long interspersed elements (SINEs and
LINEs).

Critical Thinking Question:

1. Comparative DNA studies can be helpful in understanding the
roles of the various types of DNA in the eukaryotic chromosomes
if there are cases in which there are remarkably large differences in
the amount of DNA in similar species. It can then be inferred that
the basic developmental plan of an organism is contained in the
one with the lower amount of DNA, and the extra DNA in the
species with more DNA may be superfluous. We do have cases in
which amphibians differ by as much as one hundred times the
amount of DNA found in similar species. The puffer fish has only
one-sixth the amount of DNA as other higher eukaryotes.

Chapter 16 Gene Expression: Control in Eukaryotes

1. See figure 16.5.

3. Since 5-azacytidine prevents methylation, the observed increase in
transcription suggests that the presence of methyl groups inhibits
transcription.

5. Genomic equivalence means that all of the cells of a multicellular
eukaryotic organism are genetically identical. Yet, cells in different
tissues and different regions of the organism are phenotypically dif-
ferent, expressing different suites of genes. Explaining differences
in gene expression among cells that are genetically identical is a
major question of eukaryotic genetics.

7. The three classes of segmentation genes in Drosophila are gap,
pair-rule, and segment polarity. Mutations in gap genes leave gaps
of missing segments. Mutations in pair-rule genes leave gaps of
even or odd sets of segments. Mutations of segment polarity genes
cause changes in all segments, generally the change in anterior or
posterior portions of each.

9. In the development of the early Drosophila embryo, a syncitial
blastoderm stage is achieved after thirteen cell divisions. The nu-
clei are near the surface of the embryo but not surrounded by cell
membranes. Thereafter, membranes form, creating a cellular blas-
toderm.

11. Maternal-effect genes determine four regions of the developing
embryo (major gene in parentheses): anterior (bicoid), posterior
(nanos), dorso-ventral (7b//), and terminal (torso).

13. The helix-turn-helix motif (see box 16.1, fig. 1) consists of two al-
pha helices separated by a short turn within the protein, providing
the structure to interact with DNA. Two other motifs are the zinc
finger and the leucine zipper. Another motif, a combination of
helix-turn-helix and leucine zipper, is shown in box 16.1, figure 4.

15. Amphibians have very large eggs, development is external to the
female, a ready supply of zygotes is available, and they are easily
manipulated experimentally.

17. Assuming that each cancer might be controlled by a single locus
and assuming that breast cancer appears only in women and
prostate cancer appears only in men (sex limited), pancreatic and
prostate cancer are probably controlled by autosomal recessive
genes; colon cancer is probably controlled by a dominant gene (au-
tosomal or sex linked); and breast cancer by a recessive gene, ei-
ther autosomal or sex linked.

19. The protein product of the retinoblastoma gene, pi 05, binds with
oncogene proteins. Thus, the protein may somehow suppress
transformation; when bound by oncogene proteins, pl05 may be
rendered ineffective. Hence, pl05 seems to act to suppress trans-
formation and thus the gene is called an anti-oncogene.

21. Animal viruses can have DNA or RNA, either single- or double-
stranded. They can be enveloped or nonenveloped.They can have
simple or complex protein coats.

23. The following are translation mechanisms: normal translation; read-
through translation; and splice and then translation.

25. The v and c refer to viral and cellular, respectively. A proto-
oncogene is a cellular oncogene within a nontransformed cell.

27. The v-src gene has no introns and the virus can function without
the gene.

29. We see one band that is common to both cell lines; this band must
represent the normal oncogene. The fact that this band is present
in both lines indicates that the insertion of a virus has occurred in
only one of the two copies of the gene present in the clone 1 cell.
If it had inserted within both genes, we should not have seen the
normal band. We see a larger fragment in clone 1, indicating that
the DNA of the virus does not contain a site for the restriction en-
zyme used and that the virus has inserted within the restriction
sites that define the band, lengthening the region probed. Alterna-
tively, the virus could contain a restriction site and has still inserted
in such a way as to lengthen the band probed by inserting between
the sequence probed and the original restriction site.

31. The components of an immunoglobulin light chain are V, J, and C
regions; the components of an immunoglobulin heavy chain are V,
D, J, and C regions.

33. The V-J joining recognition signal is a heptamer and nonamer sepa-
rated by twenty-three and twelve base pairs; see figure 16.35.

35. AT-cell receptor is an immunoglobulinlike molecule located on the
surfaces of T cells, enabling them to identify infected host cells.

37. One explanation is a defect in the maturation process of B cells. An-
other explanation is a defect in the process of V(D)J joining, which
could involve five or more genes.

39. The simplest interpretation of these results is that something is dif-
ferent in the organization of antibody genes in embryonic cells and
in B lymphocytes. If the genes were in the same place in both
cases, we should have seen identical patterns for the two types of
cells. The probe recognizes both variable and constant regions of
the antibody gene, since it is made from the mature mRNA. In the
B lymphocyte, the variable and constant regions are adjacent, but
in the embryonic cells, there is some extra DNA between these
two genes. This result led to the notion that variable genes are re-
arranged during the development of the immune system.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-20

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

Critical Thinking Question:

1. Adenovirus attacks normal cells by binding to p53; it also attacks
cancerous cells that lack p53. If we remove the gene for the pro-
tein that binds p53, the E1B gene, then the modified adenovirus
will not be able to attack normal cells but will be able to attack can-
cerous ones lacking p53. This modification of adenovirus as a po-
tential tool in treating cancer was published in 1996.

Chapter 17 Non-Mendelian Inheritance

1. Persistence of an environmentally induced trait into later genera-
tions is known as dauermodification and does not imply genetic
control. After a suitable number of generations, the phenotype re-
turns to normal, indicating an environmental rather than genetic
response.

3. Maroon-like must affect the cytoplasm of the egg. The first cross is
unusual and alerts us to maternal inheritance. A true Mendelian
factor should produce one-half maroon-like males and females. The
genotypes of the F x females are ma-l + /ma-1 and ma-l/ma-1; half
of the females should be of each genotype. If the wild-type allele
produces wild-type cytoplasm for the progeny, regardless of the
genotype of the progeny, any female that is ma-1 ^ /ma-1 will pro-
duce all wild-type progeny.

5. Although the genetic scheme predicts shell coiling perfectly, one
could do experiments involving the injection of cytoplasm into
eggs to test the viral hypothesis.

7. Female parent: Dd. Male parent: d-. (A): dd. Since selfing produces
only sinistral snails, individual (A) must be homozygous for sinistral
coiling, dd. The individual must get one d allele from each parent,
so each parent must be at least heterozygous. Since (A) has dextral
coiling, the mother must be heterozygous. The father could be ei-
ther D/d or d/d.

9- By looking at different species, it is clear that few genes for oxida-
tive phosphorylation are found in all mitochondrial genomes.

11. The rule of thumb is that suppressive petite mitochondria will
dominate a cell, whereas neutral petite mitochondria will be lost in
a competitive situation. Therefore

segregational petite X segregational petite — > segregational petites

segregational petite X neutral petite — > segregational petites

segregational petite X suppressive petite — > suppressive petites

neutral petite X neutral petite — > neutral petites

neutral petite X suppressive petite — > suppressive petites

suppressive petite X suppressive petite —> suppressive petites

13. In 0.02% of the offspring cells, the mt~ allele of the streptomycin
locus is inherited. In essence, these cells seem to have inherited
chloroplast genes from the mt~ parent. These cells thus provide us
with a window on the possibility of having chloroplast genotypes
from both parents viable within the same cell. Thus, interaction
among other chloroplast genes can be looked for in this class of off-
spring (0.02% of total). If recombination occurs, map distances can
be calculated by the usual methods, keeping in mind that we are
taking data only from within this 0.02% of offspring.

15. Mitochondria and chloroplasts have prokaryotic affinities. They
both have circular chromosomes, and their metabolism is affected
by prokaryotic inhibitors (e.g., antibiotics). Certain prokaryotic
messenger RNAs will hybridize with the organelle's DNA. There are
numerous other aspects of biochemistry, morphology, and physiol-
ogy that help to demonstrate affinities.

17. One way to determine that two loci are involved is to look at the
proportion of offspring that lose mu particles after autogamy. In
some strains, one-half the offspring will lose mu particles, indicat-
ing one locus was initially segregating (autogamy of M 1 m l m 2 m 2

yields M 1 M 1 m 2 m 2 or m 1 m l m 2 m 2 in a 1:1 ratio). In other strains,
one-fourth of the offspring will lose mu after autogamy, indicating
that two unlinked loci were segregating (autogamy in M l m l M 2 m 2
yields M 1 M 1 M 2 M 2 , M 1 M l m 2 m 2 , m l m 1 M 2 M 2 , or m 1 m l m 2 m 2 in a
1:1:1:1 ratio).

19. Human mitochondrial DNA does not have introns. Finding an in-
tron would suggest that the mitochondria had acquired a nuclear
gene.

21. a. All type 1, 1.5 and 3.7 kilobases. b. All type 2, 2.5 and 6.0 kilo-
bases. Recall that the chloroplast DNA from mt~ cells does not ap-
pear in progeny.

23. Conjugation produces exconjugants with the same genotype, but
which haploid micronucleus survives in each cell is random. There-
fore, one-fourth of the time, the genotypes of the exconjugants are
expected to be KK, one-half of the time Kk, and one-fourth of the
time kk. The sensitive exconjugant remains sensitive, regardless of
which of the genotypes is present. Autogamy does not affect the
genotype of homozygous exconjugants, but it does affect the het-
erozygote. Among the killer exconjugants we expect:

Killer

Autogamous

Phenotypic

Exconjugants

Products

Ratio

1/4 KK

a&KK

1/4 killer

1/2 Kk

1/2 KK

1/4 killer

1/2 kk

1/4 sensitive
(Kappa are lost)

1/4 kk

allfcfc

1/4 sensitive
(Kappa are lost)

25. Use the striped plant as the egg parent, and get pollen from a plant
with the following genotype: Ijljjj. If the striped plant is iojap (ijij
JJ), all progeny will be heterozygous for both genes and will also
contain iojap cytoplasm. The ¥ 1 plants will segregate green,
striped, and white sections within the plant. If the original plant is
Ijljjj (and thus japonica), all F x plants will have striped leaves.

27. a. 2 normal:2 petite b. normal:4 petite c. 4 normal: petite.
A nuclear gene should segregate 2:2 for each allele. Cytoplasmic
factors will produce four spores with identical cytoplasm.

Critical Thinking Question:

1. We believe there are two mechanisms to ensure the distribution of
cellular organelles during cytokinesis: stochastic and ordered in-
heritance. Stochastic inheritance simply means that no real mecha-
nism exists; rather, the cell depends on the large number of the or-
ganelles to ensure an even distribution during the dividing of the
cell. Ordered inheritance requires the even distribution of or-
ganelles in small numbers. This can be accomplished by special
structures that divide a large organelle (e.g., a single, large chloro-
plast), or by other mechanisms that insert part of the mitochon-
drial system into new buds in budding yeast.

Chapter 18 Quantitative Inheritance

1. Three in two hundred is approximately 1 in 64 = l/(4) 3 ; therefore,
three loci (see table 18.1). Each effective allele contributes about
1/2 pound over the 2-pound base (3-pound difference divided by
six effective alleles: AA BB CC = 5 pounds, aa bb cc = 2 pounds).

3. Independent assortment: for example, Aa Bb Cc Dd parents can
have AA BB CC DD offspring.

5. Individuals of intermediate color can produce both lighter and
darker offspring by independent assortment. That is, Aa Bb Cc Dd

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-21

parents can produce AA BB CC DD and aa bb cc dd offspring.
However, if each effective allele adds color, then individuals with
the base color (white) who mate with each other (aabbccdd) pre-
sumably cannot have children with darker skin.

7. The marker stock is exposed to DDT over many generations, se-
lecting for DDT resistance. It is then crossed with the wild-type
and F : offspring are backcrossed to generate flies with various
combinations of selected chromosomes. These flies are then tested
for their DDT resistance. For example:

P x Cy/Pm H/S Ce/M(4) X + (wild-type)

¥ 1 (male) Cy/+ H/+ Ce/+ X (backcross) + (wild-type female)

F 2 Cy/+ +/+ +/+; +/+ H/+ +/+,etc.

or,

F x (male) Cy/+ H/+ Ce/+ X (backcross) Cy/Pm H/S Ce/M(4)

(female)
F 2 Cy/Pm H/+ Ce/+; +/Pm +/S +/M(4)\ etc.

9- 50 cm. The difference between the two heights is 20 cm. This dif-
ference must result from the presence of effective (uppercase) al-
leles. The tall plant has four effective alleles, so each effective allele
contributes an average of 5 cm to the height of the plant. The het-
erozygote has two effective alleles; 2 X 5 = 10 cm above the base
of 40 cm, or a total of 50 cm.

11. Eight. The frequency of individuals in the F 2 that resemble one par-
ent is l/4 n , where n = the number of genes involved. In this case,
100 of 6,200,000 were like one parent. 100/6,200,000 is approxi-
mately 1/64,000, which approximates 1/4 8 . So we probably have
eight genes involved.

13. r = 0.43; H N = 0.43/0.50 = 0.86 (narrow-sense heritability); envi-
ronmental variance, a component of the total phenotypic variance
appears in the denominator of the heritability equations (18.11
and 18. 12). Thus, environmental factors that lower environmental
variance (more uniform environments or environmental effects)
increase heritability. And, factors that raise the environmental vari-
ance (less uniform environments or environmental effects) de-
crease heritability.

15. H=(4- 0.9)/(5 + 5) = 0.31; H (high line) = (4 - 3)/5 = 0.2; H
(low line) = (3 — 0.9)/5 = 0.42. Part of the difference may be due
to the number of alleles available for selection in each direction
and nonadditive factors.

17. First, "outstanding athletic ability" must be denned. It can be de-
fined subjectively by accomplishment or more objectively with a
physiological measure. Then genetic effects must be assessed
through heritability analyses such as twin studies and correlations
among relatives.

19. Uniformly good nutrition should increase the heritability of height
by eliminating some of the environmental variance; it affects only
the denominator in a heritability equation.

21. Thorax length: V D + V l = (100 - [43 + 51]) = 6;H N = 43/100 =
0.43; H B = 49/100 = 0.49. Eggs laid: V D + V I = 44; H N = 0.18;
H B = 0.62.

23. 4.5 g.

gain

selection differential

Thus (// N ) (selection differential) = gain. Since H N = 0.5 and
selective differential = 9, (0.5X9.0) = 4.5 g.

25. 1 76 pounds. To solve this problem, use the formula for realized her-
itability:

H =

0.4 =

gain

(Y - F)

selection differential (F P — F)

(F - F)

F - F

(F P - F) 185 - 170

F - F = (0.4)(15) = 6.0

6.0 = F - 170
F = 176 lbs

Critical Thinking Question:

1. The simplest cause for retraction of a study is the failure of that
study to be replicated by others. That would come about when the
conclusion isn't generally supportable, a phenomenon that could
have two major causes. First, the study might have been done well
but the phenomenon was specific to that study, due possibly to
small sample sizes or a "private mutation," an effect found in one
family or a small group of individuals but not a general phenome-
non. Second, there could have been an inadvertent error in the
study. For example, a size difference in a small region of the brain of
homosexual and heterosexual males was reported but has not been
verified. One investigator, who is following up the study, suggested
that the difference could be an effect of differences in the way the
brains were preserved and thus represent no real (biological) effect.

Chapter 19 Population Genetics: The Hardy- Weinberg
Equilibrium and Mating Systems

1. The frequencies of the three genotypes are f(MM) = 41/100 =
0Al;f(MN) = 38/100 = 038;f(NN) = 21/100 = 0.21. The fre-
quency of M, p, is the frequency of MM homozygotes plus half the
frequency of heterozygotes:

p =f(M) + (l/2)f(MN) = 0.41 + (l/2)(0.38) = 0.41 + 0.19 = 0.60
q= 1 -p= 1 - 0.60 = 0.40

Alternatively

p=f(M)

2(#MM) + #MN 2(41) + 38

2 X total

200

120

= 0.60

200

1 - p = 0.40

We do the following chi-square test:

Total

Observed

100

Expected

p 2 X 100

2pq X 100

q 2 X 100

100

Chi-square

0.694

2.083

1.563

4.340

The critical chi-square (0.05, one degree of freedom) = 3. 841. We
thus reject the null hypothesis that this population is in Hardy-
Weinberg proportions.

3. Here we must assume Hardy-Weinberg equilibrium because of

dominance. The f(tf) = 65/215 = 0.302. Thus q = f(f> =

Vfttt) = Vo.302 = 0.55; andp =fCO = 1 - 0.55 = 0.45. Since

there are zero degrees of freedom (number of phenotypes

— number of alleles = 2 — 2 = 0), we cannot do a chi-square test

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-22

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

to determine if the population is mating at random (in Hardy-
Weinberg proportions).

5. a. Most human populations should be in Hardy-Weinberg propor-
tions at the taster locus regardless of what the allelic frequen-
cies are. 3: 1 is a family ratio when the parents are heterozygotes
or a population ratio when/? = q = 0.5, given dominance.

b. With no other information, it is probably safest to assume p =
q = 0.5. At equal frequencies of alleles, the dominant pheno-
type (tasting) occurs in 75% of people (p 2 + 2pq).

7. p = 0.99, q = 0.01. If the population is in equilibrium, there should
be p 2 of AA + 2pq of Aa + q 2 of aa individuals. Since 1/10,000
shows the recessive trait, this is q 2 . Therefore,

q = Vl/10,000 = VO.0001 = 0.01

Since/? + q = l,p = 1 - 0.01 = 0.99.

9. The expected frequency of brown-eyed individuals will depend on
the allelic frequencies of the original population. If we assume that
mating is random with respect to eye color, and there is no selec-
tion, allelic frequencies will not change with time. We can, for ex-
ample, calculate the frequencies of brown-eyed individuals for two
populations at equilibrium. Let/? = frequency of the brown-eye al-
lele and q = frequency of the blue-eye allele.

Population p q Frequency of brown Cp 2 + 2pq)

0.7
0.5

0.3
0.5

0.91
0.75

We see that the original premise will be met only if the alleles are
equally frequent.

11. 0.187 MM, 0.491 MN, 0.321 NN. The next generation will achieve
equilibrium and there will be (0.43) 2 MM + 2(0.43)(0.57)M7V +
(0. 57 f MM

13. f(A) = 0.792; /(5) = 0.208. The easiest way to calculate frequen-
cies is to do it empirically. We have three hundred people, so we
have six hundred alleles.

2 X 200(A4) + 75045) 475

f(A) = = -^= 0.792

J 600 600

2 X 25(55) + 75045) 125
/(5) = = 0.208

600

15. a. Hardy-Weinberg proportions are achieved in one generation of
random mating (multiple-allelic extension) if the locus is auto-
somal. If the locus is autosomal but there are different frequen-
cies in the two sexes, then Hardy-Weinberg proportions are
achieved in two generations. If the locus is sex linked, with dif-
ferent initial frequencies in the two sexes, then approach to
equilibrium is gradual.

b. If the loci are not in equilibrium to begin with (linkage disequi-
librium), then equilibrium is achieved asymptotically.

17. 0.63 type A, 0.08 type B, 0.28 type AB, 0.01 type O. Since the popu-
lation is in equilibrium, the genotypic frequencies can be calculated
as (/? + q + rf = p 2 + 2pq + q 2 + 2pr + 2qr + r 2 . Let/? = 0.7,
q = 0.2, and r = 0. 1 . Blood types will be represented by the following:

A:/? 2 + 2pr
= 0.49 + 0.14

B: q 2 + 2qr
= 0.04 + 0.04

AB:2pq
= 0.28

0:r z
= 0.01

19. Inbreeding is disadvantageous when it causes recessive deleterious
alleles to become homozygous. This occurs in normally outbred
populations of diploids that have built up these harmful alleles. In
species that normally inbreed, these deleterious alleles are proba-
bly no longer present; they were either removed by selection long
ago or cannot build up in the population because of the regular
pattern of inbreeding.

21. There are four paths passing through A, the only common ancestor
(JF A = 0.01). All have ABCI as one side of the path. The second legs
of the four other paths are ADEHI, ADGHI, AFGHI, and AFEHI. (A
path such as IHEDAFGHI is invalid, passing through H twice.)
Since each path has six ancestors, the inbreeding coefficient is
F x = 4(1/2) 6 (1.01) = 0.063.

23. Using the formula F = (2pq - H)/2pq, we calculate that 2pq = 0.48,
and#= 38/100 = 0.38. Therefore, 5= (0.48 - 0.38)/0.48 = 0.208.

(0.32 - 0.20) 0.12

25. 0.375. F = (2pq - H)/2pq

0.32

0.375

Critical Thinking Question:

1. Let us use the superscripts m and/? for male and female, respec-
tively. Then, we can construct the following Punnett square creat-
ing the next generation:

Males, A;f(A) = p m Males, a;f(a) = q m

Females, A;f(A) = p f f(AA) = p m p f
Females, a;f(a) = q f f(Aa) = p m q f

f(Aa) = p f q m
f(aa) = q m q f

Since the distribution of offspring in the table is independent of
sex, it is the same in both sexes. The frequency of the A allele,/* (in
both sexes), will be the sum of the frequencies of the homozygotes
and half the heterozygotes, or:

p =p m p f + il/2Xp m q f + />V)
We then substitute (1 — p) for all q's:

p =p m p f + (l/2)p m Cl -p^ + (1/2)^(1 -p m )

which simplifies to:

p = (l/2Xp m + P f )

In other words, after one generation of random mating, the allelic
frequencies in each sex are the averages for both sexes. Now the
frequency is the same in both sexes, and a second generation of
random mating will achieve Hardy-Weinberg proportions. The pop-
ulation is not in equilibrium after one generation because the pro-
portion of genotypes is notp 2 , 2pq, and q 2 ifp m does not equalp f .
In other words, p m p f does not equal/? 2 .

Chapter 20 Population Genetics: Processes That Change
Allelic Frequencies

1. a. The equilibrium frequency of a is q = — —~. Therefore

q = (6 X 10" 5 )/(6 X 10" 5 + 7 X 10" 7 ) =
0.00006/0.0000607 = 0.988

b. If q = 0.90 at generation n, then

q n + 1 = In + PPn - va n =

0.90 + (6 X 10" 5 )(0.10) - (7 X 10" 7 )(0.90) = 0.9000054

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

A-23

3. 0.714.

5 X 10

0u + v) 5 X 10 5 + 2 X 10 5 7

0.714

5. We use equation 20.12 to calculate the migration rate, m:

m = (q c - # n )/(#m - #n)

In this case, q c = 0.45, q N = 0.62, and q M = 0.03. Thus

m = (0.45 - 0.62)/(0.03 - 0.62) = (-0.17)/(-0.59) = 0.288

7. Fast: 0.56; slow: 0.44. With 900 butterflies we have 1,800 alleles. 0.6
(1,800) = 1,080 fast alleles, and 0.4 (1,800) = 720 slow alleles. In
the migrant population, (0.8)(180) = 144 slow and (0.2)(180) =
36 fast alleles. Therefore, the frequency of the fast allele is (1,080
+ 36)/l,980 = 0.56. The frequency of the slow allele is 1 - f(fast
allele) = 1 - 0.56 = 0.44.

9. 0.47. We again use equation 20.12:

m =

qc ~ #n

qM ~ #n

where m = 0.1, q c = 0.45, and q M = 0.25.

0.1 =

0.45 - ff N
0.25 - q N

0.1(0.25 - tf N ) = 0.45 - q N

0.025 - 0.lq N = 0.45 - q N

0.9q N = 0.425

0.425

q N = = 0.47

* 0.9

11. In stabilizing selection, extremes of a distribution are selected
against. In directional selection, one extreme is favored over the
other. In disruptive selection, both extremes are favored over the
middle of the distribution.

13. Heterozygote disadvantage:

Total

Before selection

p 2 Pq

Fitnesses (W)

1 I - s

Frequencies after

selection

p 2 /W 2pq(\

- s)/W q

Then

q n + l = (pq[l - s] + q l yW

Aq = q„ + 1 - q = (pq[l - s] + q 2 yW - qW/W

= ipq[l - s] + q 2 - q[l - 2pqs])/W

which simplifies to

Aq = spq(2q - Y)/W

At Aq = 0, q = 0, 1, or 0.5 (2q - 1 = 0, therefore q =0.5). The equi-
librium points of zero and one are stable — if perturbed slightly
(less than 0.5), the population will return to these values. The value
q =0.5 is, however, unstable — if perturbed, it will continue away
from the equilibrium point. This can be seen by either substituting
into or graphing the Aq equation.

15.

Total

Before selection

Fitnesses (W}

Frequencies
after selection

p 2 a

2pq
1 -

1 — p 2 s — 2pqs

s)/ W 2pq{\ - s)/ W q A l W

p n+1 = ip 2 [l - s] +pq[l - sWW
Ap=p n + l ~p = ip 2 [l ~ s] +pq[l - sWW-pW/W

= spcp 2 + 2pq - \yw = sp(- p 2 + 2p - \yw

And/> = or 1 (from the root of the quadratic). Only zero is stable.

-0.005

^ -0.01

-0.015

-0.02

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

P
17. Since only heterozygotes survive, q = 0.5. This can also be derived
from equation 20.24:

q = VOi + S2)

If s 1 = s 2 = 1, then q = 1/2

19. The ST inversion seems to do best at lower elevations and the AR
at higher elevations. CH (and others) do not appear to be affected
by altitude. To test this hypothesis, we would grow caged popu-
lations of flies with different initial frequencies of the various in-
versions at different simulated elevations, simulated by tempera-
ture, pressure, oxygen content, or other. We predict that,
regardless of initial conditions, they would eventually equilibrate
at the values in the table for the given parameter of the altitude
(temperature, pressure, oxygen content, or other) that is acting as
a selective agent. We would thus identify the selective agent.
Since the inversions isolate various allelic combinations, our next
step (a potentially long-term step) would be to determine which
loci the selective agent is acting on.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

App. A: Brief Ans. to
Selected Exercises,
Problems, and Critical
Thinking Ques.

A-24

Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions

21. 0.575. We can use the formula in equation 20.15, which is a sim-
plified form of the sum of homozygotes + one-half the proportion
of heterozygotes of the b allele.

q — sq 2
1 — sq

Since the relative fitness, W = 0.4, s = 1 - W= 0.6.

a n+l

0.7 - (0.6X0.49) 0.406
1 - (0.6)(0.49) 0.706

= 0.575

23. 0.33, 0.25. Since the fitness is zero, 5=1, and we can use equation
20.17: q x = qo/Ctfo + 1) = 0.5/1.5 = 0.33. For the second genera-
tion, we substitute the first generation numbers: q 2 = 0.33/1.33 =
0.248.

Critical Thinking Question:

1. There could be several reasons why these systems are in existence.
First, they could be in selection-mutation equilibrium. However,
that would not account for the high frequencies of both alleles in
human populations in the Rh system. Second, the polymorphism
could be relatively new, somehow maintaining both alleles as the
human population increased in recent times with natural selection
not having enough time to eliminate one of the alleles. Third,
although the Rh blood system could follow the heterozygous
disadvantage model, selection could also be acting in other ways
that might maintain the polymorphism. That is, aside from Rh in-
compatibility eliminating heterozygotes, other genotypic combina-
tions could be favored under other circumstances. Finally, although
one or the other allele is being eliminated in any one population
due to heterozygous disadvantage, the constant mixing of human
populations could be reintroducing the rarer allele.

Chapter 21 Evolution and Speciation

1. Neo-Darwinism is the application of population genetics to Dar-
winian evolution. Darwinian evolution works as natural selection
favors the most fit organisms in competition among the overpro-
duced young of any species.

3. Each process lets reproductive isolating mechanisms evolve while
some barrier to breeding arises (see fig. 21.3).

5. Constraint refers to the limitations on changes that can take place.
Some changes result in nonfunctional proteins and enzymes and
thus cannot be in a successful lineage. For example, many base
changes that lead to new amino acids in enzyme active sites dis-
rupt enzymatic activity. If these mutations take place, they are elim-
inated by natural selection.

7. Yes. Two distinct species should not yield fertile progeny. We see a
great reduction in the numbers of offspring from hybrids, indicat-
ing that hybrid inviability is one isolating mechanism operating.

9. Punctuated equilibrium proposes that species remain unchanged
for long periods of time and that major changes occur only period-
ically. If species A existed for ten million years and suddenly (geo-
logically speaking) changed dramatically to species X and Y, there
would be few fossils because of the relatively short time in which
intermediate forms were present. Another argument is, barring an
incredibly detailed and complete sequence (of which there are al-
most none), there will always be gaps in the fossil record.

11. Genetic variability can be maintained by heterozygote advantage
(e.g., sickle-cell anemia in people); frequency-dependent selection

(rare-male mating advantage in Drosophila); transient polymor-
phism (industrial melanism in moths during an increase or de-
crease in industrialization); life-stage selection, which often hap-
pens when comparing egg, larval, pupal, and adult mortalities in
Drosophila; differential selection in heterogeneous environments,
common in some land snails; and neutrality.

13. Presumably, in mammals, selection could involve types of sub-
strates acted upon by electrophoretic variants; functioning at dif-
ferent pHs and ionic strengths in various cellular compartments;
resistance to enzyme inhibitors; interaction with other proteins
and membrane components; and others.

15. Use the formula K = — ln(l — d/ri), in which d is the number of
amino acid differences and n is the total number of amino acid
sites. Thus:

human being-dog: K = -ln(l - 1/8) = 0.133

human being-chicken: K = — ln(l — 3/8) = 0.470

and dog-chicken: K = -ln(l - 3/8) = 0.470

These values place human beings and dogs very close and both
equally far from chickens, which is consistent with the known evo-
lutionary relationships.

17. 0.33 AA, 0.49 Aa, 0.18 aa. The mean fitness of the population after
selection is 0.5 + 2pq = 0.98 (table 2 1.2). The new frequency of a
genotype is its original frequency times its fitness, all divided by
the mean fitness of the population. Or:

f(AA)

(0.36X1.5 - 0.6)
0.98

0.33

19. 0.023.

f(Aa) = 0.48/0.98 = 0.49
fiad) = (0.16)(1.5 - 0.4)/0.98 = 0.18
K= -ln(l - p)
K= -ln(l - [23/1,000]) = -ln(0.0977)

0.023

21. The third. For many amino acids, the third position can be any of
the four bases. In addition, wobble allows for some variation here
as well. Changing the first or second base almost always produces
a new amino acid.

23. Eight first cousins, on average, carry the complete genome of an in-
dividual. Therefore, from an evolutionary point of view, an individ-
ual and his or her eight cousins include the same alleles.

Critical Thinking Question:

1. Given that the gene for peppering is recessive and that one year
equals one generation, the moths should be following a selection
model in which natural selection acts against the recessive ho-
mozygote. In that case, we have already developed a selection
model for this in chapter 20. Equation 20.15 relates the new allelic
frequency to the old, given a particular selection coefficient:

+ i

qi\ - sq)/(l - sq )

We can solve this equation for the selection coefficient:

s = iq - q n+ JKq 2 - q 2 q n + x )

Here, q = 0.6 and q n + x = 0.5. When we solve the equation, we
get: ^ = 0.56, or W = 1 — s = 0.44, which is the fitness of the pep-
pered moth.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Appendix B: Suggestions
for Further Reading

APPENDIX B

Suggestions for Further Reading

Chapter 1 Introduction

Crow, J. 1993. N.I. Vavilov, martyr to genetic

truth. Genetics 134:1-4.
Hull, D. 1996. A revolutionary philosopher of

science. Nature 382:203-4.
Kuhn,T. 1962. The Structure of Scientific

Revolutions. Chicago: University of

Chicago Press.
Maddox, J. 1999. The Unexpected Science to

Come. Scientific American, December,

62-67.
Moore, J. 1993. Science as a Way of Knowing.

Cambridge, Mass.: Harvard University

Press.
Popper, K. 1962. Conjectures and

Refutations: The Growth of Scientific

Knowledge. New York: Basic Books.
Ruse, M., ed. 1989 . Philosophy of Biology.

New York: Macmillan.
Soyfer,V 1994. Lysenko and the Tragedy of

Soviet Science. New Brunswick, N.J. :

Rutgers University Press.
Sterelny K., and P. Griffiths. 1999. Sex and

Death: An Introduction to Philosophy of

Biology. Chicago: University of Chicago

Press.
Stubbe, H. 1972. History of Genetics.

Cambridge, Mass.: MIT Press.
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Back Matter

Glossary

GLOSSARY

acentric fragment A chromosomal piece

without a centromere.
acrocentric chromosome A chromosome

whose centromere lies very near one end.
acron The anterior end of the arthropod

embryo from which eyes and antennae

develop.
activation energy See free energy of

activation (AG*).
activator A eukaryotic specific transcription

factor that binds to an enhancer, often far

upstream of a promoter.
active site The part of an en2yme where

the actual en2ymatic function is performed.
adaptive mutation See directed mutation.
adaptive value See fitness.
additive model A mechanism of quantitative

inheritance in which alleles at different loci

either add an amount to the phenotype or

add nothing.
adenine See purines.
adjacent- 1 segregation A separation of

centromeres during meiosis in a reciprocal

translocation heterozygote so that

unbalanced zygotes are produced.
adjacent- 2 segregation Separation of

centromeres during meiosis in a

translocation heterozygote so that

homologous centromeres are pulled to

the same pole.
A DNA The form of DNA with high water

content; it has tilted base pairs and more

base pairs per turn than does B DNA.
affected Individuals in a pedigree that exhibit

the specific phenotype under study.
allele Alternative form of a gene.
allelic exclusion A process whereby only

one immunoglobulin light chain and one

heavy chain gene are transcribed in any one

cell; the other genes are repressed.
allopatric speciation Speciation in which

the evolution of reproductive isolating

mechanisms occurs during the physical

separation of the populations.
allopolyploidy Polyploidy produced by the

hybridization of two species.
allosteric protein A protein whose shape is

changed when it binds a particular molecule.

In the new shape, the protein's ability to

react to a second molecule is altered.
allotype Mutant of a nonvariant part of an

immunoglobulin gene that follows the rules

of simple Mendelian inheritance.
allozygosity Homozygosity in which the

two alleles are alike but unrelated. See

autozygosity.

allozymes Forms of an enzyme, controlled
by alleles of the same locus, that differ in
electrophoretic mobility. See isozymes.

alternate segregation A separation of
centromeres during meiosis in a reciprocal
translocation heterozygote so that balanced
gametes are produced.

alternative splicing Various ways of splicing
out introns in eukaryotic premessenger RNAs
so that one gene produces several different
messenger RNA and protein products.

altruism A form of behavior in which an
individual risks lowering its fitness for the
benefit of another.

Alu family A dispersed, intermediately
repetitive DNA sequence found in the human
genome about 300,000 times. The sequence
is about 300 bp long. The name Alu comes
from the restriction endonuclease that
cleaves it.

aminoacyl-tRNA synthetases Enzymes
that attach amino acids to their proper
transfer RNAs.

amphidiploid An organism produced by
hybridization of two species, followed by
somatic doubling. It is an allotetraploid that
appears to be a normal diploid.

anagenesis The evolutionary process
whereby one species evolves into another
without any splitting of the phylogenetic
tree. See cladogenesis.

anaphase The stage of mitosis and meiosis
in which sister chromatids or homologous
chromosomes are separated by spindle
fibers.

anaphase A The stage of anaphase in
which chromatids are separated by the
shortening of kinetochore microtubules.

anaphase B The stage of anaphase in
which chromatids are separated by the
general elongation of the spindle.

anaphase-promoting complex (APC)
Protein complex that breaks down cyclin B
and promotes anaphase among its various
roles in controlling the cell cycle. (Also
called the cyclosome.)

aneuploids Individuals or cells exhibiting
aneuploidy

aneuploidy The condition of a cell or of an
organism that has additions or deletions of
whole chromosomes.

angiosperms Plants whose seeds are
enclosed within an ovary. Flowering plants.

antibody A protein produced by a

B lymphocyte that protects the organism
against antigens.

anticoding strand The DNA strand that
forms the template for both the transcribed
messenger RNA and the coding strand.

anticodon The three-base sequence on
transfer RNA complementary to a codon on
messenger RNA.

antigen A foreign substance capable of
triggering an immune response in an
organism.

antimutator mutations Mutations of DNA
polymerase that decrease the overall
mutation rate of a cell or of an organism.

anti-oncogene A gene that represses
malignant growth and whose absence
results in malignancy (e.g., retinoblastoma).

antiparallel strands Strands, as in DNA,
that run in opposite directions with respect
to their 3' and 5 'ends.

antisense RNA RNA product of mic
(wRNA-mterfering complementary RNA)
genes that regulates another gene by base
pairing with, and thus blocking, its
messenger RNA.

antisense strand See anticoding strand.

anti-sigma factor A protein that interferes
with the action of a sigma factor.

antiterminator protein A protein that,
when bound at its normal attachment sites,
lets RNA polymerase read through normal
terminator sequences (e.g., the N- and
Q-gene products of phage A).

AP endonucleases Endonucleases that
initiate excision repair at apurinic and
apyrimidinic sites on DNA.

apoptosis Programmed cell death.

archaea Highly specialized, bacterialike
organisms that make up the third kingdom
of life on earth along with the bacteria and
the eukaryotes. Identified by Carl Woese in
1977 based on ribosomal RNA sequences.
Most are thermophilic, halophilic, or
methanogenic.

ascospores Haploid spores found in the
asci of Ascomycete fungi.

ascus The sac in Ascomycete fungi that
holds the ascospores.

A (aminoacyl) site The site on the

ribosome occupied by an aminoacyl-tRNA
just prior to peptide bond formation.

assignment test A test that determines
whether a locus is on a specific chromosome
by observing the concordance of the locus
and the specific chromosome in hybrid
cell lines.

assortative mating The mating of
individuals with similar phenotypes.

G-l

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-2

Glossary

aster Configuration at the centrosome with
microtubules radiating out in all directions.

ataxia-telangiectasia A disease in human
beings caused by a defect in X-ray-induced
repair mechanisms.

attenuator region A control region at the
promoter end of repressible amino acid
operons that exerts transcriptional control
based on the translation of a small leader
peptide gene.

attenuator stem See terminator stem.

autogamy Nuclear reorganization in a single
Paramecium cell similar to the changes
that occur during conjugation.

autonomously replicating sequence
(ARS) Eukaryotic site of the initiation of
DNA replication consisting of an 11 base-pair
consensus sequence and several other
sequences covering 100-200 base pairs.

autopolyploidy Polyploidy in which all the
chromosomes come from the same species.

autoradiography A technique in which
radioactive molecules make their locations
known by exposing photographic plates.

autosomal set A combination of nonsex
chromosomes consisting of one from each
homologous pair in a diploid species.

autosomes The nonsex chromosomes.

autotrophs Organisms that can utilize
carbon dioxide as a carbon source.

autozygosity Homozygosity in which the
two alleles are identical by descent (i.e.,
they are copies of an ancestral gene).

auxotrophs Organisms that have specific
nutritional requirements.

bacillus A rod-shaped bacterium.

backcross The cross of an individual with
one of its parents or with an organism with
the same genotype as a parent.

back mutation The process that causes
reversion. A change in a nucleotide pair in a
mutant gene that restores the original
sequence and hence the original phenotype.

bacterial artificial chromosomes (BACs)
Artificial chromosomes used for sequencing
that are derived from bacterial fertility
factors (F plasmids).

bacterial lawn A continuous cover of
bacteria on the surface of a growth medium.

bacteriophages Bacterial viruses.

Balbiani rings The larger polytene

chromosomal puffs. Generally synonymous
with puffs. See chromosome puffs.

Barr body Heterochromatic body
(X chromosome) found in the nuclei of
normal female mammals but absent in the
nuclei of normal males.

basal bodies Microtubule organizing
center for cilia and flagella, composed of
centrioles.

base excision repair The DNA excision
repair mechanism that replaces a nucleotide
lacking its base with a complete nucleotide.
Glycosylases or the environment create the
AP (apurinic and apyrimidinic) nucleotides.

base flipping A process whereby enzymes
gain access to bases within the DNA double
helix by first flipping the bases out of the
interior to the outside.

basic/helix-loop-helix/leucine zipper A
motif of proteins that bind DNA that consists
of a series of basic amino acids followed by
a helix-loop-helix domain and then a
leucine zipper. See helix-turn-helix motif;
leucine zipper.

Batesian mimicry Form of mimicry in
which an innocuous model gains protection
by resembling a noxious or dangerous host.

B DNA The right-handed, double-helical form
of DNA described by Watson and Crick.

fj-galactosidase The enzyme that splits
lactose into glucose and galactose (coded
by a gene in the lac operon).

fj-galactoside acetyltransferase An enzyme
that is involved in lactose metabolism and
encoded by a gene in the lac operon.

pj-galactoside permease An enzyme
involved in concentrating lactose in the cell
(coded by a gene in the lac operon).

binary fission Simple cell division in
single-celled organisms.

binomial expansion The terms generated
when a binomial expression is raised to a
particular power.

binomial theorem The theorem that gives
the terms of the expansion of a binomial
expression raised to a particular power.

biochemical genetics The study of the
relationships between genes and enzymes,
specifically the role of genes in controlling
the steps in biochemical pathways.

bioinformatics The science of storing,
retrieving, and analyzing genomic data.

biolistic A method (biological ballistic) of
transfecting cells by bombarding them with
microprojectiles coated with DNA.

biological species concept The idea that
organisms are classified in the same species
if they are potentially capable of interbreed-
ing and producing fertile offspring.

bivalents Structures, formed during
prophase of meiosis I, consisting of the
synapsed homologous chromosomes.
Equivalent to a tetrad of chromatids.

blastoderm The outer layer of cells in an
insect embryo after cleavage but before
gastrulation. A syncitial blastoderm gives
way to a cellular blastoderm when cell
walls form.

blunt-end ligation The ligating or attaching
of blunt-ended pieces of DNA by, for
example, T4 DNA ligase. Used in creating
hybrid vectors.

bottleneck A brief reduction in the size of a
population, which usually leads to random
genetic drift.

bouquet stage A stage during zygonema in
which chromosome ends, attached to the
nuclear membrane, come to lie near each
other.

branch migration The process in which a
crossover point between two duplexes
slides along the duplexes.

breakage and reunion The general mode
by which recombination occurs. DNA
duplexes are broken and reunited in a
crosswise fashion according to the
Holliday model.

breakage-fusion-bridge cycle Damage
that a dicentric chromosome goes through
during each cell cycle.

buoyant density of DNA A measure of the
density or size of DNA determined by the
equilibrium point reached by DNA after
density gradient centrifugation.

calculus of the genes Apparent calculation
by the genes to determine when a particular
altruistic behavior is beneficial to inclusive
fitness and hence worth doing.

cancer An informal term for a diverse class
of diseases marked by abnormal cell
proliferation.

cancer-family syndromes Pedigree
patterns in which unusually large numbers
of blood relatives develop certain kinds of
cancers.

cap A methylated guanosine added to the 5'
end of eukaryotic messenger RNA.

capsid The protein shell of a virus.

capsomere Protein clusters making up
discrete subunits of a viral protein shell.

carcinoma Tumor arising from epithelial
tissue (e.g., glands, breasts, skin, linings of
the urogenital and respiratory systems).

cassette mechanism The mechanism by
which homothallic yeast cells alternate
mating types. The mechanism involves two
silent transposons (cassettes) and a region
where these cassettes can be expressed
(cassette player).

catabolite activator protein (CAP)

A protein that, when bound with cyclic AMP,
can attach to sites on sugar-metabolizing
operons to enhance transcription of these
operons.

catabolite repression Repression of certain
sugar-metabolizing operons in favor of
glucose utilization when glucose is present
in the environment of the cell.

cDNA See complementary DNA.

cell cycle The cycle of cell growth,
replication of the genetic material, and
nuclear and cytoplasmic division.

cell-free system A mixture of cytoplasmic
components from cells, lacking nucleic
acids and membranes. Used for in vitro
protein synthesis and other purposes.

cellular immunity Immunity controlled by
killer and helper T cells, which recognize
infected body cells and either cause the
infected cells to destroy internal invaders
or destroy the infected cells directly.

centimorgan A chromosome-mapping unit.
One centimorgan equals 1% recombinant
offspring.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-3

central dogma The original postulate of
the way that information can be transferred
between DNA, RNA, and protein, barring
any transfer originating in protein.

centric fragment A piece of chromosome
containing a centromere.

centrioles Cylindrical organelles, found in
eukaryotes (except in higher plants), that
reside in the centrosome. Also called basal
bodies when they organize flagella or cilia.

centromere Constrictions in eukaryotic
chromosomes on which the kinetochore
lies. Also, the DNA sequence within the
constriction that is responsible for
appropriate function.

centromeric fission Creation of two
chromosomes from one by splitting at the
centromere.

centrosome The spindle-microtubule
organizing center in eukaryotes except for
those, such as fungi, that use spindle pole
bodies to organize the spindles.

chaperone See molecular chaperone.

Chargaff s rule Chargaff 's observation that
in the base composition of DNA, the quantity
of adenine equals the quantity of thymine,
and the quantity of guanine equals the
quantity of cytosine (equal purine and
pyrimidine content).

Charon phages Phage lambda derivatives
used as vehicles in DNA cloning.

checkpoint Used to describe points in the
cell cycle that can be stopped if certain
conditions are not met.

chemiluminescent techniques Techniques
in which various molecules are made visible
when exposed to ultraviolet or laser light if
the molecules have a fluorescent tag.

chiasmata X-shaped configurations seen in
tetrads during the later stages of prophase I
of meiosis. They represent physical
crossovers (singular: chiasma).

chimeras Individuals made up of two or
more cell lines in which the cells originated
in different zygotes. See mosaics.

chimeric plasmid Hybrid, or genetically
mixed, plasmid used in DNA cloning.

chi site Sequence of DNA at which the
RecBCD protein cleaves one of the DNA
strands during recombination.

chi-square distribution The sampling
distribution of the chi-square statistic.
A family of curves whose shapes depend
on degrees of freedom.

chloroplast The organelle that carries out
photosynthesis and starch grain formation.

chromatids The subunits of a chromosome
prior to anaphase of meiosis or mitosis. At
anaphase of meiosis II or mitosis, when the
sister chromatids separate, each chromatid
becomes a chromosome.

chromatin The nucleoprotein material of
the eukaryotic chromosome.

chromatin assembly factors Proteins
involved in the construction of nucleosomes.

chromatin remodeling The change in the
structure or positioning of nucleosomes,
usually to allow transcription.

chromatosome The core nucleosome
plus the HI protein, a unit that includes
approximately 168 base pairs of DNA.

chromomeres Dark regions of chromatin
condensation in eukaryotic chromosomes
at meiosis, mitosis, or endomitosis.

chromosomal painting A variant of the
technique known as fluorescent in situ
hybridization. Fluorescent dyes, attached
to numerous nucleotide probes, give each
human chromosome a different fluorescent
signature.

chromosomal theory of inheritance
The theory that chromosomes are linear
sequences of genes.

chromosome The form of the genetic
material in viruses and cells. A circle of
DNA in most prokaryotes; a DNA or an RNA
molecule in viruses; a linear nucleoprotein
complex in eukaryotes.

chromosome jumping A technique for
isolating clones from a genomic library that
are not contiguous but skip a region between
known points on the chromosome. This is
usually done to bypass regions that are
difficult or impossible to "walk" through or
regions known not to be of interest.

chromosome puffs Diffuse, uncoiled
regions in polytene chromosomes where
transcription is actively taking place.

chromosome walking A technique for
studying segments of DNA, larger than can
be individually cloned, by using overlapping
cloned DNA.

cis Meaning "on the near side of"; refers to
geometric configurations of atoms or
mutants on the same chromosome.

cis- dominant Mutants (e . g . , of an operator)
that control the functioning of genes on the
same piece of DNA.

cis-trans complementation test A mating
test to determine whether two mutants on
opposite chromosomes complement each
other; a test for allelism.

cistron Term Benzer coined for the smallest
genetic unit that exhibits the cis-trans
position effect; synonymous with gene.

cladogenesis The evolutionary process
whereby one species splits into two or
more species. See anagenesis.

classical linkage map A chromosomal
map, measured in centimorgans, based on
genetic crosses to locate the relative
distances between genes and their relative
locations on chromosomes.

clinal selection Selection that changes
gradually along a geographic gradient.

clonal evolution theory The theory that
cancer develops from sequential changes
(mutations) in the genome of a single cell.

clone A group of cells arising from a single
ancestor.

coccus A spherical bacterium.

coding strand The DNA strand with the
same sequence as the transcribed messenger
RNA (given U in RNA and T in DNA).
Compare with "anticoding strand."

codominance The relationship of alleles in
a heterozygote that shows the individual
expression of each allele in the phenotype.

codon preference The idea that for amino
acids with several codons, one or a few are
preferred and are used disproportionately.
They would correspond with abundant
transfer RNAs.

codons The sequences of three RNA or
DNA nucleotides that specify either an
amino acid or termination of translation.

coefficient of coincidence The number of
observed double crossovers, divided by the
number expected based on the independent
occurrence of crossovers.

coefficient of relationship, r The
proportion of alleles held in common by
two related individuals.

cohesin Proteinaceous complex that holds
sister chromatids together until anaphase of
mitosis.

cointegrate A fusion of two elements. An
intermediate structure in transposition.

colicinogenic factors See col plasmids.

col plasmids Plasmids that produce

antibiotics (colicinogens) that the host uses
to kill other strains of bacteria.

combinatorial control Transcriptional
control in eukaryotes, which involves a
large number of polypeptides, many of
which recognize specific DNA sequences.

common ancestry The shared genetic in-
heritance of two individuals who are blood
relatives. When two parents have a common
ancestor, their offspring will be inbred.

compensasome The multisubunit dosage
compensation complex in Drosophila with
attendant RNAs.

competence factor A surface protein that
binds extracellular DNA and enables the
bacterial cell to be transformed.

complementarity The correspondence of
DNA bases in the double helix so that
adenine in one strand is opposite thymine
in the other strand and cytosine in one
strand is opposite guanine in the other.
This relationship explains Chargaff 's rule.

complementary DNA (cDNA) DNA
synthesized by reverse transcriptase using
RNA as a template.

complementation The production of the
wild-type phenotype by a cell or an organism
that contains two mutant genes. If comple-
mentation occurs, the mutants are almost
certainly nonallelic.

complementation group Cistron
(determined by the cis-trans
complementation test).

complete medium A culture medium that
is enriched to contain all of the growth
requirements of a strain of organisms.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-4

Glossary

component of fitness A particular aspect

in the life cycle of an organism upon which

natural selection acts.
composite transposon A transposon

constructed of two IS elements flanking a

control region that frequently contains host

genes.
concordance The amount of phenotypic

similarity between individuals.
condensin A protein complex including

SMC (structural maintenance of

chromosomes) proteins needed for the

condensation of interphase chromosomes

to mitotic chromosomes.
conditional-lethal mutant A mutant that

is lethal under one condition but not lethal

under another.
confidence limits A statistical term for a

pair of numbers that predicts the range of

values within which a particular parameter

lies.
conjugation A process whereby two cells

come in contact and exchange genetic

material. In prokaryotes, the transfer is a

one-way process.
consanguineous Meaning "between blood

relatives"; usually refers to inbreeding or

incestuous matings.
consensus sequence A sequence of the

common nucleotides found in many different

DNA or RNA samples of homologous

regions (e.g., promoters).
conservative replication A postulated

mode of DNA replication in which an intact

double helix acts as a template for a new

double helix; known to be incorrect.
conserved sequence A sequence found in

many different DNA or RNA samples (e.g.,

promoters) that is invariant in the sample.
constitutive heterochromatin

Heterochromatin that surrounds the

centromere. See satellite DNA.
constitutive mutant A mutant whose

transcription is no longer under regulatory

control.
contigs Genomic libraries of overlapping,

contiguous clones that cover complete

regions of a chromosome.
continuous replication In DNA,

uninterrupted replication in the 5' to 3'

direction using a 3' to 5' template.
continuous variation Variation measured

on a continuum rather than in discrete units

or categories (e.g., height in human beings).
corepressor The metabolite that when

bound to the repressor (of a repressible

operon) forms a functional unit that can

bind to its operator and block transcription.
correlation coefficient A statistic that

gives a measure of how closely two variables

are related.
cosmid A hybrid plasmid that contains cos

sites at each end. Cos sites are recognized

during head filling of lambda phages.

Cosmids are useful for cloning segments of

foreign DNA up to 50 kb.

cotransduction The simultaneous
transduction of two or more genes.

coupling Allelic arrangement in which
mutants are on the same chromosome and
wild-type alleles are on the homologue.

covariance A statistical value measuring
the simultaneous deviations of x and y
variables from their means.

CpG islands Stretches of CG repeats (in
which CpG indicates sequential bases on
the same strand of DNA, rather than a C-G
base pair). These repeats, found in imprinting
centers, are important in regulation.

crisscross pattern of inheritance The
phenotypic pattern of inheritance controlled
by X-linked recessive alleles.

critical chi-square A chi-square value for a
given degree of freedom and probability
level, to which we can compare an
experimental chi-square.

crossbreed To facilitate fertilization between
separate individuals.

cross-fertilization See crossbreed.

crossing over A process in which

homologous chromosomes exchange parts
by a breakage-and-reunion process.

crossovers See chiasmata.

crossover suppression The apparent lack
of crossing over within an inversion loop in
heterozygotes. Usually due to mortality of
zygotes carrying defective crossover
chromosomes rather than to actual
suppression.

C-value paradox Structural and junk DNA
create large eukaryotic genomes and large
differences in DNA content between
eukaryotic species.

cyclic AMP A form of AMP (adenosine
monophosphate) used frequently as a
second messenger in eukaryotic hormone
nets and in catabolite repression in
prokaryotes.

cyclin Family of proteins involved in cell
cycle control.

cyclin-dependent kinase (CDK) Family
of kinases (phosphorylating enzymes) that,
when combined with cyclin, are active in
controlling checkpoints in the cell cycle.

cyclosome See anaphase-promoting
complex (APC).

cytogenetics The study of cells from the
perspective of genetics. In practice, the
study of changes in the gross structure and
number of chromosomes in cells.

cytokinesis The division of the cytoplasm
of a cell into two daughter cells. See
karyokinesis.

cytoplasmic inheritance

Extrachromosomal inheritance controlled
by nonnuclear genomes.

cytosine See pyrimidines.

cytotoxic T lymphocytes T cells
responsible for attacking host cells that
have been infected with an invading
bacterium or virus.

dauermodification The persistence for

several generations of an environmentally

induced trait.
degenerate code A code in which several

code words have the same meaning. The

genetic code is degenerate because many

different codons may specify the same

amino acid.
degrees of freedom An estimate of the

number of independent categories in a

particular statistical test or experiment.
deletion chromosome A chromosome

with part deleted.
deme A locally interbreeding population.
denatured Loss of natural configuration (of

a molecule) through heat or other treatment.

Denatured DNA is single-stranded.
denominator elements Genes on the

autosomes of Drosophila that regulate the

sex switch (sxV) to the off condition

(maleness). Refers to the denominator of

the X/A genie balance equation.
density-gradient centrifugation

A method of separating molecular entities

by their differential sedimentation in a

centrifugal gradient.
depauperate fauna A fauna, especially

common on islands, lacking many species

found in similar habitats.
derepressed The condition of an operon

that is transcribing because repressor

control has been lifted.
deterministic Referring to events that have

no random or probabilistic aspects but

proceed in a fixed, predictable fashion.
development The process of orderly

change an individual goes through in the

formation of structure.
diakinesis The final stage of prophase I of

meiosis.
dicentric chromosome A chromosome

with two centromeres.
dictyotene A prolonged diplonema of

primary oocytes that can last many years.
dideoxy method A method of DNA

sequencing that uses chain-terminating

(dideoxy) nucleotides.
dihybrid An organism heterozygous at two

loci.
dimerization The chemical union of two

similar molecules.
diploid Having each chromosome in two

copies per nucleus or cell.
diplonema (diplotene stage) The stage of

prophase of meiosis I in which chromatids

appear to repel each other.
directed mutation A form of mutation that

appears to respond to the needs of the cell

but may, in fact, be due to the cell's

hypermutable state under duress.
directional selection A type of selection

that removes individuals from one end of a

phenotypic distribution and thus causes a

shift in the distribution.
disassortative mating The mating of two

individuals with dissimilar phenotypes.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-5

discontinuous replication In DNA, the

replication in short 5' to 3' segments, using
the 5' to 3' strand as a template while going
backward, away from the replication fork.

discontinuous variation Variation that
falls into discrete categories (e.g., the green
and yellow color of garden peas).

discrete generations Generations that
have no overlapping reproduction. All
reproduction takes place between
individuals of the same cohort.

dispersive replication A postulated mode
of DNA replication combining aspects of
conservative and semiconservative
replication; known to be incorrect.

disruptive selection A type of selection
that removes individuals from the center of
a phenotypic distribution and thus causes
the distribution to become bimodal.

D-loop Configuration found during DNA
replication of chloroplast and mitochondrial
chromosomes wherein the origin of
replication is different on the two strands.
The first structure formed is a displacement
loop, or D-loop.

DNA cloning See gene cloning.

DNA-DNA hybridization The process of
taking DNA from the same or different
sources and heating and then cooling it,
causing double helices to re-form at
homologous regions. This technique is useful
for determining sequence similarities and
degrees of repetitiveness among DNAs.

DNA fingerprint A pattern of bands created
on an electrophoretic gel of a DNA digest
probed for a variable locus.

DNA glycosylases Endonucleases that
initiate excision repair at the sites of various
damaged or improper bases in DNA.

DNA gyrase A topoisomerase that relieves
supercoiling in DNA by creating a transient
break in the double helix.

DNA ligase An enzyme that closes nicks or
discontinuities in one strand of double-
stranded DNA by creating an ester bond
between adjacent 3'-OH and 5'-P0 4 ends
on the same strand.

DNA polymerase One of several classes of
enzymes that polymerize DNA nucleotides
using single-stranded DNA as a template.

DNA-RNA hybridization The process of
heating and then cooling a mixture of DNA
and RNA so that the RNA can hybridize
(form a double helix) with DNA with a
complementary nucleotide sequence.

docking protein Responsible for attaching
(docking) a ribosome to a membrane by
interacting with a signal particle attached to
a ribosome destined to be membrane bound.

dominant An allele that expresses itself
even when heterozygous. Also, the trait
controlled by that allele.

dosage compensation A mechanism by
which species with sex chromosomes ensure
that one sex does not have differential
activity of alleles on the sex chromosomes.

dot blotting A blotting technique, used on
DNA already cloned, that eliminates the
electrophoretic separation step.
Autoradiographs reveal dots rather than
bands on a gel, indicating a probed sequence.

double digest The product formed when
two different restriction endonucleases act
on the same segment of DNA.

double helix The normal structural
configuration of DNA consisting of two
helices rotating about the same axis.

downstream A convention on DNA related
to the position and direction of transcription
by RNA polymerase (5 '— >3'). Downstream
(in the 3 ' direction) is in the direction of
transcription, whereas upstream (in the 5'
direction) is in the direction from which
the polymerase has come.

downstream promoter element (DPE)
A consensus sequence at about +28 to +34
of RNA polymerase II promoters that have
initiator elements but not TATA boxes.

dyad Two sister chromatids attached to the
same centromere.

dynein Microtubule motor protein.

dysplasia Excessive cell growth that involves
pathological changes to the cells and their
nuclei.

electrophoresis The separation of molecular
entities by electric current.

electroporation A technique for transfecting
cells by applying a high-voltage electric
pulse.

elongation complex The form of RNA
polymerase II that actively carries out basal
transcription.

elongation factors (EF-Ts, EF-Tu, EF-G)
Proteins necessary for the proper elongation
and translocation processes during trans-
lation at the ribosome in prokaryotes.
Replaced by eEFIa and eEFip^ in eukaryotes.

endogenote Bacterial host chromosome.

endomitosis Chromosomal replication
without nuclear or cellular division that
results in cells with many copies of the
same chromosome, such as in the salivary
glands of Drosophila.

endonucleases Enzymes that make
nicks internally in the backbone of a
polynucleotide. They hydrolyze internal
phosphodiester bonds.

enhancer A eukaryotic DNA sequence that
increases transcription of a gene by binding
specific transcription factors.

enriched medium See complete medium.

enzyme Protein catalyst.

epigenetic effect An environmentally
induced change in the genetic material that
does not cause a change in base pairs.
Generally, a phenomenon of differential
expression of alleles of a locus depending
on the parent of origin. Also applied to an
effect in proteins.

epistasis The masking of the action of
alleles of one gene by allelic combinations
of another gene.

equational division A division, such as the
second meiotic division, that does not reduce
chromosomal numbers.

E (exit) site Site on the ribosome that
depleted transfer RNAs pass through during
ejection.

euchromatin Regions of eukaryotic
chromosomes that are diffuse during
interphase. Presumably the actively
transcribing DNA of the chromosomes.

eugenics A social movement designed to
improve humanity by encouraging those
with beneficial traits to breed and
discouraging those with undesirable traits
from breeding.

eukaryotes Organisms with true nuclei.

euploidy The condition of a cell or organism
that has one or more complete sets of
chromosomes.

evolution A change in phenotypic
frequencies in a population.

evolutionary rates The rate of divergence
between taxonomic groups, measurable as
number of amino acid substitutions per
million years.

excision repair A process whereby cells
remove part of a damaged DNA strand and
replace it through DNA synthesis, using the
undamaged strand as a template.

exconjugant Each of the two cells that
separate after conjugation has taken place.

exogenote DNA that a bacterial cell has
taken up through one of its sexual processes.

exon In a gene that has intervening

sequences (introns), a region that is actually
exported from the nucleus to be expressed
or become part of a transfer or ribosomal
RNA.

exon shuffling The hypothesis put forward
by Walter Gilbert that exons code for the
functional units of a protein, and that the
evolution of new genes proceeds by
recombination or the exclusion of exons.

exonucleases Enzymes that digest

nucleotides from the ends of polynucleotide
molecules. They hydrolyze phosphodiester
bonds of terminal nucleotides.

experimental design A branch of statistics
that attempts to outline the way in which
experiments should be carried out so the
data gathered has statistical value.

expression vector A hybrid vector
(plasmid) that expresses its cloned genes.

expressivity The degree of expression of a
genetically controlled trait.

F 2 See filial generation.

factorial The product of all integers from
the specified number down to one (unity).

Fanconi's anemia A disease in human
beings with a syndrome of congenital
malformations; associated with various
cancers.

fate map A map of the developmental fate
of a zygote or early embryo showing the
adult organs that will develop from a given
position on the zygote or early embryo.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-6

Glossary

F-duction See sexduction.

fecundity selection The forces causing
one genotype to be more fertile than another
genotype.

feedback inhibition A posttranslational
control mechanism in which the end
product of a biochemical pathway inhibits
the activity of the first enzyme of the same
pathway.

fertility factor The plasmid that allows a
prokaryote to engage in conjugation with,
and pass DNA into, an F~ cell.

F factor See fertility factor.

filial generation Offspring generation. ¥ 1 is
the first offspring, or filial, generation; F 2 is
the second; and so on.

fimbriae See pili.

first-division segregation (FDS) The allelic
arrangement of ordered spores that indicates
the lack of recombination between a locus
and its centromere.

fitness, W The relative reproductive success
of a genotype as measured by survival,
fecundity, or other life-history parameters.

5' untranslated region (5' UTR) See
leader.

floral meristem The shoot apical meristem
sets aside this tissue that gives rise to flowers.

floral-meristem identity genes At least
five genes known to establish the identity
of the floral meristem.

fluctuation test An experiment by Luria
and Delbruck that compared the variance
in number of mutations among small cultures
with that among subsamples of a large
culture to determine the mechanism of
inherited change in bacteria.

fluorescent in situ hybridization (FISH)
A technique in which a fluorescent dye is
attached to a nucleotide probe that then
binds to a specific site on a chromosome
and makes itself visible by its fluorescence.

Fokker-Planck equation An equation that
describes diffusion processes. It is used by
population geneticists to describe random
genetic drift.

footprinting A technique to determine the
length of nucleic acid in contact with a
protein. While in contact, the free DNA is
digested. The remaining DNA is then isolated
and characterized.

founder effect Genetic drift observed
in a population founded by a small,
nonrepresentative sample of a larger
population.

F-pili Sex pili. Hairlike projections on an F +
or Hfr bacterium involved in anchorage
during conjugation.

fragile site A chromosomal region that has
a tendency to break.

fragile-X syndrome The most common
form of inherited mental retardation.
Named for its association with an X
chromosome with a tip that breaks or
appears uncondensed. Inheritance involves
imprinting.

frameshift A mutation in which there is an

addition or deletion of nucleotides that

causes the codon reading frame to shift.
free energy of activation (AG*) Energy

needed to initiate a chemical reaction.
frequency-dependent selection

A selection whereby a genotype is at an

advantage when rare and at a disadvantage

when common.
functional alleles Mutations that fail to

complement each other in a cis-trans

complementation test.
fundamental number (NF) The number

of chromosome arms in a somatic cell of a

particular species.
gamete A germ cell having a haploid

chromosomal complement. Gametes from

parents of opposite sexes fuse to form

zygotes.
gametic selection The forces acting to

cause differential reproductive success of

one allele over another in a heterozygote.
gametophyte The haploid stage of a plant

life cycle that produces gametes (by mitosis).

It alternates with a diploid, sporophyte

generation.
G-bands Eukaryotic chromosomal bands

produced by treatment with Giemsa stain.
gene Inherited determinant of the

phenotype. See cistron; locus.
gene amplification A process or processes

by which the cell increases the number of

repeats of a particular gene within the

genome.
gene cloning Production of large numbers

of a piece of DNA after that piece of DNA

is inserted into a vector and taken up by a

cell. Cloning occurs as the vector

replicates.
gene conversion In Ascomycete fungi, a

2:2 ratio of alleles is expected after meiosis,

yet a 3:1 ratio is sometimes observed. The

gene conversion mechanism is explained

by repair of heteroduplex DNA produced

by recombination.
gene family A group of genes that has

arisen by duplication of an ancestral gene.

The genes in the family may or may not

have diverged.
gene flow The movement of genes from

one population to another by interbreeding

between individuals in the two populations.
gene pool All of the alleles available among

the reproductive members of a population

from which gametes can be drawn.
generalized transduction Form of

transduction in which any region of the

host genome can be transduced. See

specialized transduction.
general transcription factors Eukaryotic

proteins that form part of the RNA

polymerase holoenzymes.
genetic code The linear sequences of

nucleotides that specify the amino acids

during the process of translation at the

ribosome.

genetic engineering Popular term for
recombinant DNA technology. See
recombinant DNA technology.

genetic load The relative decrease in the
mean fitness of a population due to the
presence of genotypes that have less than
the highest fitness.

genetic polymorphism The occurrence
together in the same population of more
than one allele at the same locus, with the
least frequent allele occurring more
frequently than can be accounted for by
mutation.

genie balance theory Bridges 's theory that
the sex of a fruit fly is determined by the
relative number of X chromosomes and
autosomal sets.

genome The entire genetic complement of
a prokaryote or virus or the haploid genetic
complement of a eukaryote.

genomic equivalence The concept that
differentiated cells in a eukaryotic organism
have identical genetic contents.

genomic library A set of cloned fragments
making up the entire genome of an organism
or species.

genomics The study of the mapping and
sequencing of genomes. Bioinformatics is
the science of mining the data from these
DNA sequences obtained from sequencing.

genophore The chromosome (genetic
material) of prokaryotes and viruses.

genotype The genes that an organism
possesses.

Giemsa stain A complex of stains specific
for the phosphate groups of DNA.

Goldstein-Hogness box See TATA box.

green fluorescent protein A reporter
system that uses the gene from a jellyfish
that specifies a protein that fluoresces
green when ultraviolet light is shined on it,
indicating the success of a transfection
experiment.

group I introns Self-splicing introns that
require a guanine-containing nucleotide for
splicing; the intron is released in a linear
form.

group II introns Self-splicing introns that
do not require an external nucleotide for
splicing; the intron is released in a lariat form.

group selection Selection for traits that
would be beneficial to a population at the
expense of the individual possessing the trait.

G-tetraplex A structure of four guanines
that can base pair to form a planar structure
that may be involved in novel structures at
the end of eukaryotic chromosomes.

guanine See purines.

guide RNA (gRNA) RNA that guides the
insertion of uridines (RNA editing) into
messenger RNAs in trypanosomes. Found in
transcripts from minicircles and maxicircles
of DNA in kinetoplasts.

gynandromorphs Mosaic individuals
having simultaneous aspects of both the
male and the female phenotype.

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Back Matter

Glossary

G-7

hammerhead ribozyme A catalytic RNA,
shaped like a hammerhead, capable of
splitting other RNA molecules with
appropriate complementary sequences.

haplodiploidy The sex-determining
mechanism found in some insect groups
among which males are haploid and females
are diploid.

haploid The state of having one copy of
each chromosome per nucleus or cell.

HAT medium A selection medium for hybrid
cell lines; contains hypoxanthine,
aminopterin, and thymidine. HPRT + TK +
cell lines can survive in this medium.

heat shock proteins Proteins that appear
in a cell after the cell has been subjected to
elevated temperatures.

helicase A protein that unwinds DNA, usually
at replicating Yj unctions.

helix-turn-helix motif Configuration
found in DNA-binding proteins consisting
of a recognition helix and a stabilizing helix,
separated by a short turn.

hemizygous The condition of loci present
in only one copy in a diploid organism,
such as loci on the X chromosome of the
heterogametic sex of a diploid species.

heritability A measure of the degree to
which the variance in the distribution of a
phenotype is due to genetic causes. In the
broad sense, it is measured by the total
genetic variance divided by the total
phenotypic variance. In the narrow sense,
it is measured by the genetic variance due
to additive genes divided by the total
phenotypic variance.

hermaphrodite An individual with both
male and female genitalia.

heterochromatin Chromatin that remains
tightly coiled (and darkly staining)
throughout the cell cycle.

heteroduplex DNA See hybrid DNA.

heterogametic The sex with heteromorphic
sex chromosomes; during meiosis, it
produces different kinds of gametes in
accordance with these sex chromosomes.

heterogeneous nuclear mRNA (hnRNA)
The original RNA transcripts found in
eukaryotic nuclei before posttranscriptional
modifications.

heterokaryon A cell that contains two or
more nuclei from different origins.

heteromorphic chromosome pair
Members of a homologous pair of
chromosomes that are not morphologically
identical (e.g., the sex chromosomes).

heteroplasmy The existence within an
organism of genetic heterogeneity within
the populations of mitochondria or
chloroplasts.

heterothallic A botanical term used for
organisms in which the two sexes reside in
different individuals.

heterotrophs Organisms that require an
organic form of carbon as a carbon source.

heterozygote A diploid or polyploid with

different alleles at a particular locus.
heterozygote advantage A selection

model in which heterozygotes have the

highest fitness.
heterozygous DNA See hybrid DNA.
Hfr High frequency of recombination. A

strain of bacteria that has incorporated an F

factor into its chromosome and can then

transfer the chromosome during conjugation.
histone acetyl transferases (HATs)

Proteins that remodel chromatin by

acetylating histones.
histones Arginine- and lysine-rich basic

proteins making up a substantial portion of

eukaryotic nucleoprotein.
hnRNA See heterogeneous nuclear mRNA.
Hogness box See TATA box.
holandric trait Trait controlled by a locus

found only on the Y chromosome. Involves

father-to-son transmission.
Holliday junction A junction point between

two cross-linked DNA double helices. It is

an intermediate stage in DNA recombination.
holoenzyme The complete enzyme,

including all subunits. Often used in

reference to RNA and DNA polymerases.
homeo box A consensus sequence of

about 180 base pairs discovered in

homeotic genes in Drosophila. Also found

in other developmentally important genes

from yeast to human beings.
homeo domain The sixty amino acid

polypeptide translated from the homeo box.
homeotic gene Gene that controls the

developmental fate of a cell type; mutations

of the homeotic gene cause one cell type to

follow the developmental pathway of

another cell type.
homogametic The sex with homomorphic

sex chromosomes; it produces only one kind

of gamete in regard to the sex chromosomes.
homologous chromosomes Members of a

pair of essentially identical chromosomes

that synapse during meiosis.
homologous recombination Breakage

and reunion between homologous lengths

of DNA mediated by RecA and RecBCD.
homomorphic chromosome pairs

Morphologically identical members of a

homologous pair of chromosomes.
homoplasmy The existence within an

organism of only one type of plastid;

usually referring to the genetic identity

of mitochondria or chloroplasts.
homothallic A botanical term used for

groups whose individuals are not of

different sexes.
homozygote A diploid or a polyploid with

identical alleles at a locus.
humoral immunity Immunity due to

antibodies in the serum and lymph.
H-Y antigen The histocompatibility

Yantigen, a protein found on the cell

surfaces of male mammals.

hybrid Offspring of unlike parents.
hybrid DNA DNA whose two strands have

different origins.
hybridoma A cell resulting from the fusion

of a spleen cell and a multiple myeloma cell.

These cells can be maintained indefinitely

in cell culture, in which they produce

monoclonal antibodies.
hybrid plasmid A plasmid that contains an

inserted piece of foreign DNA.
hybrid vector See hybrid vehicle.
hybrid vehicle A plasmid or phage

containing an inserted piece of foreign

DNA.
hybrid zone Geographical region in which

previously isolated populations that have

evolved differences come into contact and

form hybrids.
hyperplasia Excessive cell growth that

does not involve pathological changes to

the cells.
hypervariable loci Loci with many alleles;

especially those whose variation is due to

variable numbers of tandem repeats.
hypostatic gene A gene whose expression

is masked by an epistatic gene.
identity by descent The state of two alleles

when they are identical copies of the same

ancestral allele (autozygous).
idiogram A photograph or diagram of the

chromosomes of a cell arranged in an

orderly fashion. See karyotype.
idiotypic variation Variation in the variable

parts of immunoglobulin genes.
idling reaction The production of guanosine

tetraphosphate (3'-ppGpp-5') by the

stringent factor when a ribosome encounters

an uncharged transfer RNA in the A site.
immunity The ability of an organism to

resist infection.
immunoglobulins (Igs) Specific proteins

produced by derivatives of B lymphocytes

that protect an organism from antigens.
imprinting See molecular imprinting.
imprinting center (IC) A region

responsible for the control of imprinting.

The imprinting mark is almost certainly

DNA methylation, which is able to turn

off gene transcription.
inbreeding The mating of genetically related

individuals.
inbreeding coefficient, F The probability

of autozygosity
inbreeding depression A depression of

vigor or yield due to inbreeding.
incestuous A mating between blood relatives

who are more closely related than the law

of the land allows.
inclusive fitness The expansion of the

concept of the fitness of a genotype to

include benefits accrued to relatives of an

individual since relatives share parts of

their genomes. An apparently altruistic act

toward a relative may thus enhance the

fitness of the individual performing the act.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-8

Glossary

incomplete dominance The situation in
which both alleles of the heterozygote
influence the phenotype.The phenotype is
usually intermediate between the two
homozygous forms.

independent assortment, rule of
Mendel's second rule, describing the
independent segregation of alleles of
different loci.

inducible system A system (a coordinated
group of enzymes involved in a catabolic
pathway) is inducible if the metabolite it
works upon causes transcription of the
genes controlling these enzymes. These
systems are primarily prokaryotic operons.

induction In regard to temperate phages,
the process of causing a prophage to
become virulent.

informatics See bioinformatics.

initiation codon The messenger RNA
sequence AUG, which specifies methionine,
the first amino acid used in the translation
process. (Occasionally, GUG is recognized
as an initiation codon.)

initiation complex The complex formed
for initiation of translation. It consists of the
30S ribosomal subunit, messenger RNA,
N-formyl-methionine transfer RNA, and
three initiation factors.

initiation factors (IFI, IF2, IF3) Proteins
(prokaryotic with eukaryotic analogues)
required for the proper initiation of
translation.

initiator element (inr) A CT-rich area
found in RNA polymerase II promoters
without TATA boxes.

initiator proteins Proteins that recognize
the origin of replication on a replicon and
take part in primosome construction.

insertion mutagenesis Change in gene
action due to an insertion event that either
changes a gene directly or disrupts control
mechanisms.

insertion sequences (IS) Small, simple
transposons. See transposable genetic
element.

inside marker The middle locus of three
linked loci.

intercalary heterochromatin

Heterochromatin, other than centromeric
heterochromatin, dispersed throughout
eukaryotic chromosomes.

intergenic suppression A mutation at a
second locus that apparently restores the
wild-type phenotype to a mutant at a first
locus.

interkinesis The abbreviated interphase
that occurs between meiosis I and II. No
DNA replication occurs here.

internal ribosome entry site Sequence in
eukaryotic messenger RNAs that allows
ribosomes to initiate translation at a point
other than the 5' cap.

interphase The metabolically active,
nondividing stage of the cell cycle.

interpolar microtubules Microtubules

extending from one pole of the spindle

and overlapping spindle fibers from the

other pole, but not in contact with

kinetochores.
interrupted mating A mapping technique

in which bacterial conjugation is disrupted

after specified time intervals.
intersex An organism with external sexual

characteristics of both sexes.
intervening sequences (introns) DNA

sequences within a gene that are transcribed

but removed prior to translation.
intra-allelic complementation The

restoration of activity in an enzyme made of

subunits in a heterozygote of two mutants

that, when homozygous, are not active.

Caused by the interaction of the subunits in

the protein.
intragenic suppression A second change

within a mutant gene that results in an

apparent restoration of the original

phenotype.
intron See intervening sequences.
inversion The replacement of a section of

a chromosome in the reverse orientation.
inverted repeat sequence A nucleotide

sequence read in opposite orientations on

the same double helix.
in vitro Biological or chemical work done

in the test tube (literally, "in glass") rather

than in living systems.
IS elements See insertion sequences.
isochromosome A chromosome with two

genetically and morphologically identical

arms.
isozymes Different electrophoretic forms

of the same enzyme. Unlike allozymes,

isozymes are due to differing subunit

configurations rather than allelic

differences.
junctional diversity Variability in

immunoglobulins caused by variation in the

exact crossover point during V-J,V-D, and

D-J joining.
kappa particles The bacterialike particles

that give a Paramecium the killer

phenotype.
karyokinesis The process of nuclear

division. See cytokinesis.
karyotype The chromosome complement

of a cell. See idiogram.
kinesin Microtubule motor protein.
kinetochore The chromosomal attachment

point for the spindle fibers, located on the

centromere.
kinetochore microtubules Microtubules

radiating from the centrosome and attached

to kinetochores of chromosomes during

mitosis and meiosis.
kin selection The mode of natural

selection that acts on an individual's

inclusive fitness.

Klenow fragment Proteolytic fragment —
obtained by treatment with a protease, or
protein-cleaving enzyme — of E. colt DNA
polymerase I with both 5 '— »3' polymerase
activity and 3 '— >5' exonuclease activity. (It
has been studied extensively because it has
been easier for X-ray crystallographers and
biochemists to work with this fragment
than with the whole enzyme.)

knockout mice Transgenic mice that have
been made homozygous for a nonfunctioning
allele at a particular locus.

lac operon The inducible operon, including
three loci involved in the uptake and
breakdown of lactose.

ladder gel See stepladder gel.

lagging strand Strand of DNA being
replicated discontinuously

lampbrush chromosomes Chromosomes
of amphibian oocytes having loops that are
suggestive of a lampbrush.

leader The length of messenger RNA from
the 5 ' end to the initiation codon, AUG.

leader peptide gene A small gene within
the attenuator control region of a repressible
amino acid operon. Translation of the gene
tests the concentration of amino acids in
the cell.

leader transcript The messenger RNA
transcribed by the attenuator region of a
repressible amino acid operon. The
transcript is capable of several alternative
stem-loop structures, depending on the
translation of a short leader peptide gene.

leading strand Strand of DNA being
replicated continuously.

leptonema (leptotene stage) The first
stage of prophase I of meiosis, in which
chromosomes become distinct.

lethal-equivalent alleles Alleles whose
summed effect is that of lethality — for
example, four alleles, each of which would
be lethal 25% of the time (or to 25% of their
bearers), are equivalent to one lethal allele.

leucine zipper Configuration of a DNA-
binding protein in which leucine residues on
two helices interdigitate, in zipper fashion,
to stabilize the protein.

leukemia Cancer of the bone marrow
resulting in excess production of leukocytes.

level of significance The probability
value in statistics used to reject the null
hypothesis.

linkage The association of loci on the same
chromosome.

linkage disequilibrium The condition
among alleles at different loci such that
allelic combinations in a gamete do not
follow the product rule of probability.

linkage equilibrium The condition among
alleles at different loci such that any allelic
combination in a gamete occurs as the
product of the frequencies of each allele at
its own locus.

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Back Matter

Glossary

G-9

linkage groups Associations of loci on the
same chromosome. In a species, there are
as many linkage groups as there are
homologous pairs of chromosomes.

linkage number The number of times one
strand of a helix coils about the other.

linker A small segment of DNA that contains
a restriction site. It can be added to blunt-
ended DNA to give that DNA a particular
restriction site for cloning.

liposomes Transfecting DNA is delivered to
target cells by way of these membrane-bound
vesicles.

locus The position of a gene on a
chromosome (plural: loci}.

lod score method A technique (/ogarithmic
odds) for determining the most likely
recombination frequency between two loci
from pedigree data.

long interspersed elements (LINEs)
Sequences of DNA, up to seven thousand
base pairs in length, interspersed in
eukaryotic chromosomes in many copies.

lymphoma Cancer of the lymph nodes and
spleen that causes excessive production of
lymphocytes.

Lyon hypothesis The hypothesis that
suggests that the Barr body is an inactivated
X chromosome.

lysate The contents released from a lysed
cell.

lysis The breaking open of a cell by the
destruction of its wall or membrane.

lysogenic The state of a bacterial cell that
has an integrated phage (prophage) in its
chromosome.

major histocompatibility complex A
group of highly polymorphic genes whose
products appear on the surfaces of cells,
imparting to them the property of "self"
(belonging to that organism). Some other
functions are also involved.

mapping The process of locating the
positions of genes on chromosomes.

mapping function The mathematical
relationship between measured map
distance in a given experiment and the
actual recombination frequency.

map unit The distance equal to 1%
recombination between two loci.

mate-killer A Paramecium phenotype
induced by intracellular bacterialike mu
particles.

maternal effect The effect of the maternal
parent's genotype on the phenotype of her
offspring.

maternal-effect gene A gene expressed in
maternal tissue that influences a developing
embryo.

mating type In many species of

microorganisms, individuals can be divided
into two mating types. Mating can take
place only between individuals of opposite
mating types due to the interaction of cell
surface components.

maturation-promoting factor (MPF)

A protein complex of cyclin B and p3 <icdc2
that initiates mitosis during the cell cycle.
Also called the mitosis-promoting factor.

mean The arithmetic average; the sum of
the data divided by the sample size.

mean fitness of the population, W The
sum of the fitnesses of the genotypes of a
population weighted by their proportions;
hence, a weighted mean fitness.

meiosis The nuclear process in diploid
eukaryotes that results in gametes or spores
with only one member of each original
homologous pair of chromosomes per
nucleus.

meiotic drive See gametic selection.

merozygote A bacterial cell having a second
copy of a particular chromosomal region in
the form of an exogenote.

messenger RNA (mRNA) A complementary
copy of a gene that is translated into a
polypeptide at the ribosome.

metacentric chromosome A chromosome
whose centromere is located in the middle.

metafemale A fruit fly with an X/A ratio
greater than unity.

metagon An RNA necessary for the

maintenance of mu particles in Paramecium.

metamale A fruit fly with X/A ratio
below 0.5.

metaphase The stage of mitosis or meiosis
in which spindle fibers are attached to kinet-
ochores, and the chromosomes are posi-
tioned in the equatorial plane of the cell.

metaphase plate The plane of the equator
of the spindle into which chromosomes are
positioned during metaphase.

metastasis The migration of cancerous cells
to other parts of the body.

metrical variation See continuous variation.

microsatellite DNA Repeats of very short
sequences of DNA, such as CACACACA,
dispersed throughout the eukaryotic
genome. The loci can be studied by
polymerase chain reaction amplification.

microtubule organizing center Active
center from which microtubules are
organized. The spindle is organized by the
centrosome, which may or may not contain
a centriole.

microtubules Hollow cylinders made of the
protein tubulin (a and (3 subunits) that make
up, among other things, the spindle fibers.

mimicry A phenomenon in which an
individual gains an advantage by looking
like the individuals of a different species.

minimal medium A culture medium for
microorganisms that contains the minimal
necessities for growth of the wild-type.

mismatch repair A form of excision repair
initiated at the sites of mismatched bases
in DNA.

missense mutation Mutations that change
a codon for one amino acid into a codon for
a different amino acid.

mitochondrion The eukaryotic cellular

organelle in which the Krebs cycle and

electron transport reactions take place.
mitosis The nuclear division producing two

daughter nuclei identical to the original

nucleus.
mitosis-promoting factor See maturation-
promoting factor (MPF).
mitotic apparatus See spindle.
mixed families Groups of four codons

sharing their first two bases and coding for

more than one amino acid.
modern linkage map A chromosomal map

based on the positions of RFLP markers

along its length.
molecular chaperone A protein that aids

in the folding of a second protein. The

chaperone prevents proteins from forming

structures that would be inactive.
molecular evolutionary clock

A measurement of evolutionary time in

nucleotide substitutions per year.
molecular imprinting The phenomenon

in which there is differential expression

of a gene depending on whether it was

maternally or paternally inherited.
molecular mimicry The situation in which

one type of molecule resembles another

type in order to function. For example, the

prokaryotic ribosomal release factors, RF1

and RF2, mimic the structure of a transfer

RNA.
monocistronic Usually referring to a

messenger RNA that carries the information

for only one gene (cistron).
monoclonal antibody The antibody from

a clone of cells producing the same antibody.

An individual with multiple myeloma usually

produces monoclonal antibodies.
monohybrids Offspring of parents that

differ in only one genetic characteristic.

Usually implies heterozygosity at a single

locus under study.
monosomic A diploid cell missing a single

chromosome.
monovalent A single chromosome

composed of two sister chromatids.

Equivalent to a dyad.
morphogen A substance transported into

or produced in a developing embryo that

diffuses to form a gradient that helps

determine cell differentiation.
morphological species concept The idea

that organisms are classified in the same

species if they appear similar.
mosaicism The condition of being a mosaic.

See mosaics.
mosaics Individuals made up of two or

more cell lines in which the cells originated

in the same zygote.
mRNA See messenger RNA (mRNA).
Miillerian mimicry A form of mimicry in

which noxious species evolve to resemble

each other.
multihybrid An organism heterozygous at

numerous loci.

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Back Matter

Glossary

G-10

Glossary

multinomial expansion The terms

generated when a multinomial is raised to a

power.
mu particles Bacterialike particles found in

the cytoplasm of Paramecium that impart

the mate-killer phenotype.
mutability The ability to change.
mutants Alternative phenotypes to the

wild-type; the phenotypes produced by

alternative alleles.
mutation The process by which a gene or

chromosome changes structurally; the end

result of this process.
mutation rate The proportion of mutations

per cell division in bacteria or single-celled

organisms or the proportion of mutations

per gamete in higher organisms.
mutator mutations Mutations of DNA

polymerase that increase the overall

mutation rate of a cell or of an organism.
muton A term Benzer coined for the smallest

mutable site within a cistron.
natural selection The process in nature

whereby one genotype leaves more offspring

than another genotype because of superior

life history attributes such as survival or

fecundity.
negative interference The phenomenon

whereby a crossover in a particular region

facilitates the occurrence of other apparent

crossovers in the same region of the

chromosome.
N-end rule The life span of a protein is

determined by its amino-terminal

(N-terminal) amino acid.
neo-Darwinism The merger of classical

Darwinian evolution with population

genetics.
neoplasm New growth of abnormal tissue.
neutral gene hypothesis The hypothesis

that most genetic variation in natural

populations is not maintained by natural

selection.
NF See fundamental number.
nickase See DNA gyrase.
noncoding strand See anticoding strand.
nondisjunction The failure of a pair of

homologous chromosomes to separate

properly during meiosis.
nonhistone proteins The proteins

remaining in chromatin after the histones

are removed.
nonparental ditype (NPD) A spore

arrangement in Ascomycete fungi that

contains only the two recombinant-type

ascospores (assuming two segregating loci).
nonparentals See recombinants.
nonrecombinants In mapping studies,

offspring that have alleles arranged as in the

original parents.
nonsense codon One of the messenger

RNA sequences (UAA, UAG, UGA) that

signals the termination of translation.
nonsense mutations Mutations that

change a codon for an amino acid into a

nonsense codon.

normal distribution Any of a family of
bell-shaped frequency curves whose relative
positions and shapes are defined on the basis
of the mean and standard deviation.

northern blotting A gel transfer technique
used for RNA. See Southern blotting.

N segments Sequences of nucleotides added
in a template-free fashion at the joining
junctions of heavy-chain antibody genes.

nuclease-hypersensitive site A region of a
eukaryotic chromosome that is specifically
vulnerable to nuclease attack because it is
not wrapped as nucleosomes.

nucleolar organizer The chromosomal
region around which the nucleolus forms;
site of tandem repeats of the major ribosomal
RNA gene.

nucleolus The globular, nuclear organelle
formed at the nucleolar organizer. Site of
ribosome construction.

nucleoprotein The substance of eukaryotic
chromosomes consisting of proteins and
nucleic acids.

nucleoside A sugar-base compound that
is a nucleotide precursor. Nucleotides are
nucleoside phosphates.

nucleosomes Arrangements of DNA and
histones forming regular spherical structures
in eukaryotic chromatin.

nucleotide Subunits that polymerize into
nucleic acids (DNA or RNA). Each nucleotide
consists of a nitrogenous base, a sugar, and
one or more phosphate groups.

nucleotide excision repair The DNA
excision repair mechanism responsible for
repairing thymine dimers and other lesions.
Enzymes excise a short segment of one of
the DNA strands and then repair and ligate
the DNA.

null hypothesis The statistical hypothesis
that there are no differences between
observed data and those data expected based
on the assumption of no experimental effect.

nullisomic A diploid cell missing both
copies of the same chromosome.

numerator elements Genes on the X
chromosome in Drosophila that regulate
the sex switch gene (sxO to the on condition
(femaleness). Refers to the numerator of the
X/A genie balance equation.

nutritional-requirement mutants
See auxotrophs.

Okazaki fragments Segments of newly
replicated DNA produced during
discontinuous DNA replication.

oncogene Genes capable of transforming a
cell. They are found in the active state in
retroviruses and transformed cells and in
the inactive state in nontransformed cells,
in which they are called proto-oncogenes.

one-gene-one-enzyme hypothesis

Hypothesis of Beadle and Tatum that states
that one gene controls the production of
one enzyme. Later modified to the concept
that one cistron controls the production of
one polypeptide.

oogenesis The process of ovum formation
in female animals.

oogonia Cells in females that produce
primary oocytes by mitosis.

open reading frames (ORFs) Sequence
of codons between the initiation and
termination codons in a gene.

operator A DNA sequence recognized by a
repressor protein or repressor-corepressor
complex. When the operator is complexed
with the repressor, transcription is
prevented.

operon A sequence of adjacent genes all
under the transcriptional control of the
same operator.

origin recognition complex (ORC)
A complex of six proteins that bind to the
eukaryotic autonomously replicating
sequences (ARS). Needed for the initiation
of DNA replication in concert with other
proteins.

outbreeding The mating of genetically
unrelated individuals.

ovum Egg. The one functional product of
each meiosis in female animals.

pachynema (pachytene stage) The stage
of prophase I of meiosis in which chromatids
are first distinctly visible.

palindrome A sequence of words, phrases,
or nucleotides that reads the same regardless
of the direction from which one starts; the
sites of recognition of type II restriction
endonucleases.

panmictic Referring to unstructured
(random mating) populations.

paracentric inversion A chromosomal
inversion that does not include the
centromere.

paramecin A toxin liberated by a "killer"
Paramecium.

parameters Measurements of attributes of
a population; denoted by Greek letters.

parapatric speciation Speciation in which
reproductive isolating mechanisms evolve
when a population enters a new niche or
habitat within the range of the parent
species.

parasegment The first series of segments
that form in a developing insect embryo;
they form after about 5.5 hours in the
developing Drosophila embryo.

parental ditype (PD) A spore arrangement
in Ascomycete fungi that contains only the
two nonrecombinant-type ascospores.

parental imprinting See molecular
imprinting.

parentals See nonrecombinants.

parthenogenesis The development of an
individual from an unfertilized egg that did
not arise by meiotic chromosomal reduction.

partial digest A restriction digest that has
not been allowed to go to completion, and
thus contains pieces of DNA that have re-
striction endonuclease sites that have not
been cleaved.

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Back Matter

Glossary

G-ll

partial dominance See incomplete

dominance.
Pascal's triangle A triangular array of

numbers made up of the coefficients of the

binomial expansion.
path diagram A modified pedigree showing

only the direct line of descent from common

ancestors.
pedigree A representation of the ancestry

of an individual or family; a family tree.
penetrance The normal appearance of

genetically controlled traits in the phenotype.
peptidyl transferase The enzymatic center

responsible for peptide bond formation

during translation at the ribosome.
pericentric inversion A chromosomal

inversion that includes the centromere.
permissive temperature A temperature at

which temperature-sensitive mutants are

normal.
PEST hypothesis Degradation of a protein

in less than two hours is signaled by a region

within the protein rich in proline (P),

glutamic acid (E), serine (S), or threonine (T).
petite mutations Mutations of yeast that

produce small colonies, like those grown

under anaerobic conditions.
phages See bacteriophages.
phenocopy A phenotype that is not

genetically controlled but looks like a

genetically controlled phenotype.
phenotype The observable attributes of an

organism.
pheromone A chemical signal, analogous

to a hormone, that passes information

between individuals.
phosphodiester bond A diester bond

linking two nucleotides together (between

phosphoric acid and sugars) to form the

nucleotide polymers DNA and RNA.
photocrosslinking A technique used to

determine which moieties (proteins, DNA)

are in close proximity during a particular

process.
photoreactivation The process whereby

dimerized pyrimidines (usually thymines)

in DNA are restored by an enzyme

(deoxyribodipyrimidine photolyase) that

requires light energy.
phyletic evolution See anagenesis.
phyletic gradualism The process of gradual

evolutionary change over time.
phylogenetic tree A diagram showing

evolutionary lineages of organisms.
physical map Chromosomal map in which

distances are in physical units of base pairs.

These maps can be of microsatellite markers

or of sequence-tagged sites.
pili (fimbriae) Hairlike projections on the

surface of bacteria; Latin for "hair."
plaques Clear areas on a bacterial lawn

caused by cell lysis due to viral attack.
plasmid An autonomous, self-replicating

genetic particle, usually of double-stranded

DNA.

plastid A chloroplast prior to the develop-
ment of chlorophyll.

pleiotropy The phenomenon whereby a
single mutant affects several apparently
unrelated aspects of the phenotype.

point centromere The type of centromere,
such as that found in Saccharomyces
cerevisiae, that has defined sequences large
enough to accommodate one spindle
microtubule.

point mutations Small mutations that
consist of a replacement, addition, or
deletion of one or a few bases.

polar bodies The small cells that are the
by-products of meiosis in female animals.
One functional ovum and as many as three
polar bodies result from meiosis of each
primary oocyte.

polarity Meaning "directionality" and
referring either to an effect seen in only
one direction from a point of origin or to
the fact that linear entities (such as a single
strand of DNA) have ends that differ from
each other.

polar mutant An organism with a mutation,
usually within an operon, that prevents the
expression of genes distal to itself.

pollen grain The male gametophyte in
higher plants.

poly-A tail A sequence of adenosine

nucleotides added to the 3' end of eukaryotic
messenger RNAs.

polycistronic Referring to prokaryotic
messenger RNAs that contain several genes
within the same messenger RNA transcript.

polygenic inheritance See quantitative
inheritance.

polymerase chain reaction (PCR) A
method to amplify DNA segments rapidly
in temperature-controlled cycles of
denaturation, primer binding, and replication.

polymerase cycling The process by which
a DNA polymerase III enzyme completes an
Okazaki fragment, releases it, and begins
synthesis of the next Okazaki fragment.

polymerized Formed into a complex
compound by linking together smaller
elements.

polynucleotide phosphorylase An enzyme
that can polymerize diphosphate nucleotides
without the need for a primer. The function
of this enzyme in vivo is probably in its
reverse role as an RNA exonuclease.

polyploids Organisms with greater than
two chromosome sets.

polyribosome See polysome.

polysome The configuration of several
ribosomes simultaneously translating the
same messenger RNA. Shortened form of
the term polyribosome.

polytene chromosome Large chromosome,
seen, for example, in Drosophila salivary
glands, consisting of many chromatids
formed by rounds of endomitosis. Synapsis
of homologous chromosomes occurs
during the process.

population A group of organisms of the
same species relatively isolated from other
groups of the same species. See deme.

position effect An alteration of phenotype
caused by a change in the relative
arrangement of the genetic material.

positive interference When the occurrence
of one crossover reduces the probability
that a second will occur in the same region.

postreplicative repair A DNA repair
process initiated when DNA polymerase
bypasses a damaged area.

posttranscriptional modifications
Changes in eukaryotic messenger RNA
made after transcription has been completed.
These changes include additions of caps
and tails and removal of introns.

preemptor stem A configuration of leader
transcript messenger RNA that does not
terminate transcription in the attenuator-
controlled amino acid operons.

pre-initiation complex (PIC) The form of
the RNA polymerase II enzyme with general
transcription factors bound equivalent to the
E. coli holoenzyme. Phosphorylation of the
enzyme then allows transcription to begin.

Pribnow box Consensus sequence of
TATAAT in prokaryotic promoters centered
at the position — 10.

primary oocytes The cells that undergo
meiosis in female animals.

primary spermatocytes The cells that
undergo meiosis in male animals.

primary structure The sequence of
polymerized amino acids in a protein.

primary transcript The product of
eukaryotic transcription before
posttranscriptional modification takes place.

primase An enzyme that creates a messenger
RNA primer for Okazaki fragment initiation.

primer In DNA replication, a length of
double-stranded DNA that continues as a
single-stranded template in the 3' to 5'
direction.

primosome A complex of two proteins, a
primase and helicase, that initiates RNA
primers on the lagging DNA strand during
DNA replication.

prion Infectious agent responsible for
several neurological diseases (scrapie, kuru,
Creutzf eld-Jakob syndrome, mad-cow
disease). It is a protein that lacks DNA
or RNA.

probability The expectation of the
occurrence of a particular event.

probability theory The conceptual
framework concerned with quantification
of probabilities. See probability.

proband See propositus.

probe In recombinant DNA work, a

radioactive nucleic acid complementary to
a region being searched for in a restriction
digest or genomic library.

processivity The ability of an enzyme to
repetitively continue its catalytic function
without dissociating from its substrate.

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Back Matter

Glossary

G-12

Glossary

product rule The rule that states that the

probability that two independent events

will both occur is the product of their

separate probabilities.
progeny testing Breeding of offspring to

determine their genotypes and those of

their parents.
prokaryotes Organisms that lack true

nuclei.
promoter A DNA region that RNA

polymerase binds to in order to initiate

transcription.
proofread Technically to read for the

purpose of detecting errors for later

correction. DNA polymerase has 3' to 5'

exonuclease activity which it uses during

polymerization to remove incorrect

nucleotides it has recently added.
prophage A temperate phage replicating

with the host that can later initiate the lytic

cycle.
prophase The initial stage of mitosis or

meiosis in which chromosomes become

visible and the spindle apparatus forms.
proplastid Mutant plastids that do not

grow and develop into chloroplasts.
propositus (proposita) The person

through whom a pedigree was discovered.
proteasome A barrel-shaped cellular

organelle for protein breakdown involving

the ubuiquitin pathway.
proteome From proteins of the genome;

the complete set of proteins from a

particular genome. It is the protein

analogue to "genome."
proteomics The study of the complete set

of proteins from a particular genome. It is

the protein analogue to "genomics."
proto-oncogene A cellular oncogene in an

untransformed cell.
prototrophs Strains of organisms that can

survive on minimal medium.
pseudoalleles Genes that are functionally

but not structurally allelic. Within gene

families, pseudoalleles are alleles that are

not expressed.
pseudoautosomal gene A gene that occurs

on both sex-determining heteromorphic

chromosomes.
pseudodominance The phenomenon in

which a recessive allele shows itself in the

phenotype when only one copy of the

allele is present, as in hemizygous alleles or

in deletion heterozygotes.
P (peptidyl) site The site on the ribosome

occupied by the peptidyl-tRNA just before

peptide bond formation.
punctuated equilibrium The evolutionary

process involving long periods without

change (stasis) punctuated by short periods

of rapid speciation.
Punnett square A diagrammatic

representation of a particular cross used to

predict the progeny of the cross.
purines Nitrogenous bases of which guanine

and adenine are found in DNA and RNA.

pyrimidines Nitrogenous bases of which

thymine is found in DNA, uracil in RNA, and

cytosine in both.
quantitative inheritance The mechanism of

genetic control of traits showing continuous

variation.
quantitative trait loci Chromosomal

regions contributing to the inheritance of a

quantitative trait. These regions may contain

one or more polygenes that contribute to

the phenotype.
quantitative variation See continuous

variation.
quaternary structure The association of

polypeptide subunits to form the final

structure of a protein.
random genetic drift Changes in allelic

frequency due to sampling error.
random mating The mating of individuals

in a population such that the union of

individuals with the trait under study occurs

according to the product rule of probability.
random strand analysis Mapping studies

in organisms that do not keep all the

products of meiosis together.
read-through Transcription or translation

beyond the normal termination signals in

DNA or RNA, respectively.
realized heritability Heritability measured

by a response to selection.
recessive An allele (or phenotype) that

does not express itself in the heterozygous

condition.
reciprocal cross A cross with the

phenotype of each sex reversed as compared

with the original cross. Made to test the

role of parental sex on inheritance pattern.
reciprocal translocation A chromosomal

configuration in which the ends of two

nonhomologous chromosomes are broken

off and become attached to the

nonhomologues .
recombinant DNA technology Techniques

of gene cloning. Recombinant DNA refers

to the hybrid of foreign and vector DNA.

See gene cloning.
recombinant plasmid A plasmid that

contains an inserted piece of foreign DNA.
recombinants In mapping studies,

offspring with allelic arrangements made

up of a combination of the original

parental alleles.
recombination The nonparental

arrangement of alleles in progeny that can

result from either independent assortment

or crossing over.
recombination nodule Proteinaceous

nodules found on bivalents during

zygonema and pachynema associated with

crossing over.
recon A term Benzer coined for the smallest

recombinable unit within a cistron.
reductional division The first meiotic

division. It reduces the number of

chromosomes and centromeres to half that

in the original cell.

regional centromere The type of
centromere found in higher eukaryotes
that can accommodate several spindle
microtubules.

regulator gene A gene primarily involved
in control of the production of another
gene's product.

reinitiation The initiation of translation by a
ribosome that has just completed translation
of a region of the messenger RNA upstream
of the current point of initiation.

relative Darwinian fitness See fitness.

relaxed mutant A mutant that does not
exhibit the stringent response under amino
acid starvation.

release factors (RFl and RF2) Proteins in
prokaryotes responsible for termination
of translation and release of the newly
synthesized polypeptide when a nonsense
codon appears in the A site of the ribosome.
Replaced by eRF in eukaryotes.

repetitive DNA DNA made up of copies of
the same nucleotide sequence.

replica-plating A technique to rapidly
transfer microorganism colonies to numerous
petri plates.

replication-coupling assembly factor
A protein complex in fruit flies that
assembles new nucleosomes.

replicons A replicating genetic unit
including a length of DNA and its site for
the initiation of replication.

replisome The DNA-replicating structure at
the Yjunction, consisting of two DNA
polymerase III enzymes and a primosome
(primase and DNA helicase).

reporter systems Genetic constructs that
allow an investigator to determine that a
specific locus is active by measuring the
phenotypic expression of an associated
locus, such as the luciferase reporter, which
glows if watered with luciferin.

repressible system A coordinated group
of enzymes involved in a synthetic pathway
(anabolic) is repressible if excess quantities
of the end product of the pathway lead to
the termination of transcription of the
genes for the enzymes. These systems are
primarily prokaryotic operons.

repressor The protein product of a regulator
gene that acts to control transcription of
inducible and repressible operons.

reproductive isolating mechanisms
Environmental, behavioral, mechanical, and
physiological barriers that prevent two
individuals of different populations from
producing viable progeny.

reproductive success The relative
production of offspring by a particular
genotype.

repulsion Allelic arrangement in which
each homologous chromosome has mutant
and wild-type alleles.

resistance transfer factor Infectious
transfer part of R plasmids.

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Back Matter

Glossary

G-13

restricted transduction See specialized
transduction.

restriction digest The results of the action of
a restriction endonuclease on a DNA sample.

restriction endonucleases Endonucleases
that recognize certain DNA sequences, then
cleave them. They protect cells from viral
infection; they are useful in recombinant
DNA work.

restriction fragment length

polymorphism (RFLP) Variations
(among individuals) in banding patterns
of electrophoresed restriction digests.

restriction map A physical map of a piece
of DNA showing recognition sites of specific
restriction endonucleases separated by
lengths marked in numbers of bases.

restriction site The sequence of DNA
recognized by a restriction endonuclease.

restrictive temperature A temperature at
which temperature-sensitive mutants display
the mutant phenotype.

retinoblastoma A childhood cancer of
retinoblast cells caused by the inactivation
of an anti-oncogene.

retrotransposons Transposable genetic
elements found in eukaryotic DNA that
move through the reverse transcription of
an RNA intermediate.

reverse transcriptase An enzyme that can
use RNA as a template to synthesize DNA.

reversion The return of a mutant to the
wild-type phenotype by way of a second
mutational event.

R factors See R plasmids.

rho-dependent terminator A DNA
sequence signaling the termination of
transcription; termination requires the
presence of the rho protein.

rho-independent terminator A DNA
sequence signaling the termination of
transcription; the rho protein is not
required for termination.

rho protein A protein involved in the
termination of transcription.

ribosomal RNA (rRNA) RNA components
of the subunits of the ribosomes.

ribosome recycling factor (RRF) A protein
needed to prepare ribosomal subunits that
have just finished translating a messenger
RNA for another cycle of translation.

ribosomes Organelles at which translation
takes place. They are made up of two
subunits consisting of RNA and proteins.

ribozyme Catalytic or autocatalytic RNA.

RNA editing The insertion of uridines into
messenger RNAs after transcription is
completed; controlled by guide RNA. May
also involve insertion of cytidines in some
organisms or possible deletions of bases.

RNA phages Phages whose genetic material
is RNA. They are the simplest phages known.

RNA polymerase The enzyme that
polymerizes RNA by using DNA as a
template. (Also known as transcriptase
or RNA transcriptase.^

RNA replicase A polymerase enzyme that

catalyzes the self-replication of

single-stranded RNA.
Robertsonian fusion Fusion of two

acrocentric chromosomes at the centromere.
rolling-circle replication A model of DNA

replication that accounts for a circular DNA

molecule producing linear daughter double

helices.
R plasmids Plasmids that carry genes that

control resistance to various drugs.
rRNA See ribosomal RNA (rRNA).
rule of independent assortment

See independent assortment, rule of.
rule of segregation See segregation, rule of.
sampling distribution The distribution

of frequencies with which various possible

events could occur, or a probability

distribution defined by a particular

mathematical expression.
sarcoma Tumor of tissue of mesodermal

origin (e.g., muscle, bone, cartilage).
satellite DNA Highly repetitive eukaryotic

DNA primarily located around centromeres.

Satellite DNA usually has a different buoyant

density than the rest of the cell's DNA.
scaffold The eukaryotic chromosomal

structure that remains when DNA and

histones have been removed.
scanning hypothesis Proposed mechanism

by which the eukaryotic ribosome

recognizes the initiation region of a

messenger RNA after binding the 5 ' capped

end of it. The ribosome scans the messenger

RNA for the initiation codon.
scientific method A procedure scientists

use to test hypotheses, making predictions

about the outcome of an observation or

experiment before the experiment is

performed. The results provide support or

refutation of the hypothesis.
screening technique A technique to isolate

a specific genotype or phenotype of an

organism.
secondary oocytes The cells formed by

meiosis I in female animals.
secondary spermatoctyes The products

of the first meiotic division in male animals.
secondary structure The flat or helical

configuration of the polypeptide backbone

of a protein.
second-division segregation (SDS) The

allelic arrangement in the spores of

Ascomycete fungi with ordered spores that

indicates a crossover between a locus and

its centromere
securin An inhibitory protein that prevents

separin from acting on cohesin to separate

sister chromatids.
segmentation genes Genes of developing

embryos that determine the number and

fate of segments.
segregation, rule of Mendel's first principle,

which describes how genes pass from one

generation to the next.

segregational load Genetic load caused
when a population is segregating less fit
homozygotes because of heterozygote
advantage.

segregation distortion See gametic
selection.

selection See natural selection.

selection coefficients, s, t The sum of
forces acting to lower the relative
reproductive success of a genotype.

selection-mutation equilibrium

An equilibrium allelic frequency resulting
from the balance between selection against
an allele and mutation re-creating this allele.

selective medium A culture medium
enriched with a particular substance to
allow the growth of particular strains of
organisms.

selfed See self-fertilization.

self-fertilization Fertilization in which
the two gametes come from the same
individual.

selfish DNA A segment of the genome with
no apparent function, although it can control
its own copy number.

semiconservative replication The mode
by which DNA replicates. Each strand acts
as a template for a new double helix. See
template.

semisterility Nonliability of a proportion
of gametes or zygotes.

sense strand See coding strand.

separin An enzyme that breaks down
cohesin and allows sister chromatids to
separate at the start of anaphase of mitosis.

sequence-tagged sites (STSs) DNA lengths
of 100-500 base pairs that are unique
in the genome. They are created by
polymerase chain reaction amplification
of primers that are then tested to be sure
the sequence is unique.

sex chromosomes Heteromorphic
chromosomes whose distribution in a
zygote determines the sex of the organism.

sex-conditioned traits Traits that appear
more often in one sex than in another.

sex-determining region Y (SRY) The sex
switch, or testis-determining factor, in human
beings, located on the Y chromosome (Sry
in mice).

sexduction A process whereby a bacterium
gains access to and incorporates foreign
DNA brought in by a modified F factor
during conjugation.

sex-influenced traits See sex-conditioned
traits.

Sex-lethal A gene in Drosopbila, located
on the X chromosome, that is a sex switch,
directing development toward femaleness
when in the "on" state. It is regulated by
numerator and denominator elements
that act to influence the genie balance
ratio (X/A).

sex-limited traits Traits expressed in only
one sex. They may be controlled by sex-
linked or autosomal loci.

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Genetics, Seventh Edition

Back Matter

Glossary

G-14

Glossary

sex-linked The inheritance pattern of loci
located on the sex chromosomes (usually
the X chromosome in XY species); also
refers to the loci themselves.

sex-ratio phenotype A trait in Drosophila
whereby females produce mostly, if not
only daughters.

sex switch A gene in mammals, normally
found on the Y chromosome, that directs
the indeterminate gonads towards
development as testes.

sexual selection The forces, determined
by mate choice, that act to cause one
genotype to mate more frequently than
another genotype.

Shine-Dalgarno hypothesis A proposal
that prokaryotic messenger RNA is aligned
at the ribosome by complementarity
between the messenger RNA upstream
from the Initiation codon and the 3 ' end
of the 16S ribosomal RNA.

shoot apical meristem The major

meristematic tissue of the plant; surrounds
the shoot.

short interspersed elements (SINEs)
Sequences of DNA interspersed in eukaryotic
chromosomes in many copies. Alu, a three-
hundred base-pair sequence, is found about
half a million times in human DNA.

shotgun cloning The random cloning of
pieces of the DNA of an organism without
regard to the genes or sequences present in
the cloned DNA.

shunting Process in which the first initiation
codon on a messenger RNA is bypassed for
an initiation codon further down the
messenger. The process is probably guided
by secondary structure in the messenger.

siblings (sibs) Brothers and sisters.

sigma factor The protein that gives

promoter-recognition specificity to the RNA
polymerase core enzyme of bacteria.

signal hypothesis The major mechanism
whereby proteins that must insert into or
across a membrane are synthesized by a
membrane-bound ribosome. The first
thirteen to thirty-six amino acids synthesized,
termed a signal peptide, are recognized by
a signal recognition particle that draws
the ribosome to the membrane surface.
The signal peptide may be removed later
from the protein.

signal peptide See signal hypothesis.

signal recognition particle See signal
hypothesis.

signal transduction pathway A pathway
in which the action of kinase enzymes that
free transcription factors, or some other
action, translates an environmental signal
into some form of gene action.

single-nucleotide polymorphisms (SNPs)
Differences between individuals involving
single base pairs that are located about
every 1,000 bases along the human genome.
SNPs are useful for mapping disease genes.

single-strand binding proteins Proteins
that attach to single-stranded DNA, usually
near the replicating Yjunction, to stabilize
the single strands.

sister chromatids See chromatids.

site-specific recombination A crossover
event, such as the integration of phage X,
that requires homology of only a very short
region and uses an enzyme specific for that
recombination.

small nuclear ribonucleoproteins
(snRNPs) Components of the
spliceosome, the intron-removing apparatus
in eukaryotic nuclei. See snRNPs.

small nucleolar ribonucleoprotein

particles (snoRNPs) Particles composed
of RNA and protein found in the nucleolus
that modify ribosomal RNAs, particularly by
converting some uridines to pseudouridines
and methylating some ribose sugars.

small nucleolar RNAs (snoRNAs) RNAs
found in small nucleolar ribonucleoprotein
particles (snoRNPs) that take part in
modifying ribosomal RNA in the nucleolus.

SMC proteins For structural maintenance
of chromosomes; proteins that aid mitotic
segregation, sister-chromatid adhesion,
dosage compensation, recombination, and
other chromosomal activities.

snRNPs See small nuclear
ribonucleoproteins.

sociobiology The study of the evolution of
social behavior in animals.

somatic doubling A disruption of the
mitotic process that produces a cell with
twice the normal chromosome number.

somatic hypermutation The occurrence
of a high level of mutation in the variable
regions of immunoglobulin genes.

SOS box The region of the promoters of
various genes that the LexA repressor
recognizes. Release of repression results in
the induction of the SOS response.

SOS response Repair systems (RecA, Uvr)
induced by the presence of single-stranded
DNA that usually occurs from postreplicative
gaps caused by various types of DNA
damage. The RecA protein, stimulated by
single-stranded DNA, is involved in the
inactivation of the LexA repressor, thereby
inducing the response.

Southern blotting A method, first devised
by E. M. Southern, used to transfer DNA
fragments from an agarose gel to a
nitrocellulose gel for the purpose of DNA-
DNA or DNA-RNA hybridization during
recombinant DNA work.

specialized transduction Form of

transduction based on faulty looping out by
a temperate phage. Only neighboring loci
to the attachment site can be transduced.
See generalized transduction.

speciation A process whereby, over time,
one species evolves into a different species
(anagenesis) or one species diverges to
become two or more species (cladogenesis).

species A group of organisms capable of

interbreeding to produce fertile offspring.
specific transcription factors Proteins

needed for activation of transcription at

specific eukaryotic promoters. Also, negative

factors that can inhibit transcription at a

specific eukaryotic promoter.
spermatids The four products of meiosis in

males that develop into sperm.
spermatogenesis The process of sperm

production.
spermatogonium A cell type in the testes

of male vertebrates that gives rise to primary

spermatocytes by mitosis.
sperm cells The gametes of males.
spermiogenesis The process by which

spermatids mature into sperm cells.
spindle The microtubule apparatus that

controls chromosomal movement during

mitosis and meiosis.
spindle pole body Spindle microtubule

organizing center found in fungi.
spiral cleavage The cleavage process in

mollusks and some other invertebrates

whereby the spindle is tipped at mitosis in

relation to the original egg axis.
spirillum A spiral bacterium.
spliceosome Protein-RNA complex that

removes introns in eukaryotic nuclear RNAs.
sporophyte The stage of a plant life cycle

that produces spores by meiosis and

alternates with the gametophyte stage.
stabilizing selection A type of selection

that removes individuals from both ends of

a phenotypic distribution, thus maintaining

the same distributional mean.
standard deviation The square root of the

variance.
standard error of the mean The standard

deviation divided by the square root of the

sample size. It is the standard deviation of a

sample of means.
statistics Measurements of attributes of a

sample from a population; denoted by

Roman letters. See parameters.
stem-loop structure A lollipop-shaped

structure formed when a single-stranded

nucleic acid molecule loops back on itself

to form a complementary double helix

(stem), topped by a loop.
stepladder gel A DNA-sequencing gel.

The numerous bands in each lane give the

appearance of a stepladder.
stochastic A process with an indeterminate

or random element as opposed to a

deterministic process with no random

element.
stringent factor A protein that catalyzes

the formation of an unusual nucleotide

(guanosine tetraphosphate) during the

stringent response under amino acid

starvation.
stringent response A translational control

mechanism in prokaryotes that represses

transfer RNA and ribosomal RNA synthesis

during amino acid starvation.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-15

structural alleles Alleles whose alterations
include some of the same base pairs.

submetacentric chromosome A
chromosome whose centromere lies
between its middle and its end, but closer
to the middle.

subtelocentric chromosome

A chromosome whose centromere lies
between its middle and its end, but closer
to the end.

sum rule The rule that states that the
probability that two mutually exclusive
events will occur is the sum of the
probabilities of the individual events.

supercoiling Negative or positive coiling of
double-stranded DNA that differs from the
relaxed state.

supergenes Several loci, which usually
control related aspects of the phenotype,
in close physical association.

suppressor gene A gene that, when
mutated, apparently restores the wild-type
phenotype to a mutant of another locus.

surveillance mechanism Used to describe
mechanisms that oversee checkpoints in the
cell cycle, points in which the process can
be halted if certain conditions are not met.

survival of the fittest In evolutionary
theory, survival of only those organisms best
able to obtain and utilize resources (the
fittest). This phenomenon is the cornerstone
of Darwin's theory of evolution.

Svedberg unit A unit of sedimentation
during centrifugation. Abbreviation is S,
as in 50S.

swivelase See DNA gyrase.

sympatric speciation Speciation in which
reproductive isolating mechanisms evolve
within the range and habitat of the parent
species. This type of speciation may be
common in parasites.

synapsis The point-by-point pairing of
homologous chromosomes during 2ygonema
or in certain dipteran tissues that undergo
endomitosis.

synaptonemal complex A proteinaceous
complex that apparently mediates synapsis
during the zygotene stage and then
disintegrates.

syncitium A cell that has many nuclei not
separated by cell membranes.

synexpression group A group of
eukaryotic genes that are involved in the
same function or pathway and are induced
together.

synteny test A test that determines

whether two loci belong to the same link-
age group by observing concordance in
hybrid cell lines.

synthetic medium A chemically defined
substrate that microorganisms are grown
upon.

TACTAAC box A consensus sequence
surrounding the lariat branch point of
eukaryotic messenger RNA introns.

TATA-binding protein A protein, part of

TFIID, that binds the TATA consensus

sequence in eukaryotic promoters.
TATA box An invariant DNA sequence at

about —25 in the promoter region of

eukaryotic genes; analogous to the Pribnow

box in prokaryotes.
tautomeric shift Reversible shifts of

proton positions in a molecule. The bases

in nucleic acids shift between the keto and

enol forms or between the amino and

imino forms.
TBP-associated factors (TAFs) Proteins

that bind with the TATA-binding protein

to form TFIID. They aid in the selectivity

of TFIID.
T-cell receptors Surface proteins of T cells

that allow the T cells to recognize host cells

that have been infected.
telocentric chromosome A chromosome

whose centromere lies at one of its ends.
telomerase An enzyme that adds telomeric

sequences to the ends of eukaryotic

chromosomes.
telomeres The ends of linear chromosomes.
telophase The terminal stage of mitosis or

meiosis in which chromosomes uncoil, the

spindle breaks down, and cytokinesis

usually occurs.
telson The posterior end of the arthropod

embryo, where the end of the alimentary

canal is located.
temperate phage A phage that can enter

into lysogeny with its host.
temperature-sensitive mutant An organism

with an allele that is normal at a permissive

temperature but mutant at a restrictive

temperature.
template A pattern serving as a mechanical

guide. In DNA replication, each strand of

the duplex acts as a template for the

synthesis of a new double helix.
template strand See anticoding strand.
terminator sequence A sequence in DNA

that signals the termination of transcription

to RNA polymerase.
terminator stem A configuration of the

leader transcript that signals transcriptional

termination in attenuator- controlled amino

acid operons.
tertiary structure The further folding of a

protein, bringing a helices and (3 sheets

into three-dimensional arrangements.
testcross The cross of an organism with a

homozygous recessive organism.
testing of hypotheses The determination

of whether to reject or fail to reject a

proposed hypothesis based on the likelihood

of the experimental results.
testis-determining factor (TDF) General

term for the gene determining maleness in

human beings (Tdf in mice).
tetrads The meiotic configuration of four

chromatids first seen in pachynema. There

is one tetrad (bivalent) per homologous

pair of chromosomes.

tetranucleotide hypothesis

The hypothesis, based on incorrect
information, that DNA could not be the
genetic material because its structure was
too simple — that is, that repeating subunits
contain one copy each of the four DNA
nucleotides.

tetraploids Organisms with four whole sets
of chromosomes.

tetratype (TT) A spore arrangement in
Ascomycete fungi that consists of the two
parental and two recombinant spores.

theta structure An intermediate structure
formed during the replication of a circular
DNA molecule.

three-point cross A cross involving three
loci.

thymine See pyrimidines.

t-loop A loop that forms at the end of
mammalian telomeres by the interdigitation
of the 3' free end into the DNA double
helix.

topoisomerase An enzyme that can relieve
(or create) supercoiling in DNA by creating
transitory breaks in one (type I) or both
(type II) strands of the helical backbone.

topoisomers Forms of DNA with the same
sequence but differing in their linkage
number (coiling).

totipotent The state of a cell that can give
rise to any and all adult cell types, as
compared with a differentiated cell whose
fate is determined.

trailer The length of messenger RNA from
the nonsense codon to the 3' end or, in
polycistronic messenger RNAs, from a
nonsense codon to the next gene's leader.

trans Meaning "across" and referring
usually to the geometric configuration of
mutant alleles across from each other on a
homologous pair of chromosomes.

t rans -acting Referring to mutations of, for
example, a repressor gene, that act through
a diffusible protein product; the normal
mode of action of most recessive mutations.

transcription The process whereby RNA is
synthesized from a DNA template.

transcription factors Eukaryotic proteins
that aid RNA polymerase in recognizing
promoters. See general transcription factors
and specific transcription factors.

transducing particle A defective phage,
carrying part of the host's genome.

transduction A process whereby a cell can
gain access to and incorporate foreign DNA
brought in by a viral particle.

transfection The introduction of foreign
DNA into eukaryotic cells.

transfer operon (fra) Sequence of loci
that impart the male (F-pili-producing)
phenotype on a bacterium. The male cell
can transfer the F plasmid to an F~cell.

transfer RNA (tRNA) Small RNA molecules
that carry amino acids to the ribosome for
polymerization.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Glossary

G-16

Glossary

transformation A process whereby
prokaryotes take up DNA from the
environment and incorporate it into their
genomes, or the conversion of a eukaryotic
cell into a cancerous one.

transgenic Eukaryotic organisms that have
taken up foreign DNA.

transition mutation A mutation in which a
purine-pyrimidine base pair replaces a
base pair in the same purine-pyrimidine
relationship.

translation The process of protein synthesis
wherein the nucleotide sequence in
messenger RNA determines the primary
structure of the protein.

translocase (EF-G) Elongation factor in
prokaryotes necessary for proper
translocation at the ribosome during the
translation process. Replaced by eEF2 in
eukaryotes.

translocation A chromosomal configuration
in which part of a chromosome becomes
attached to a different chromosome. Also a
part of the translation process in which the
messenger RNA is shifted one codon in
relation to the ribosome.

translocation channel (translocon)
A protein-lined pore or channel in a
membrane through which nascent proteins
are transported during translation.

transposable genetic element A region of
the genome, flanked by inverted repeats, a
copy of which can be inserted at another
place; also called a transposon or a jumping
gene.

transposon See transposable genetic
element.

transversion mutation A mutation in which
a purine replaces a pyrimidine, or vice versa.

trihybrid An organism heterozygous at
three loci.

triploids Organisms with three whole sets
of chromosomes.

trisomic A diploid cell with an extra
chromosome.

tRNA See transfer RNA (tRNA).

trp RNA-binding attenuation protein
(TRAP) Protein that can bind to the
attenuation region of the messenger RNA of
the tryptophan operon in Bacillus subtilis,
causing a terminator stem to form and
halting further transcription.

true heritability See heritability.

tumor Abnormal growth of tissue.

tumor-suppressor genes Genes that
normally prevent unlimited cellular growth.
When both copies of the gene are mutated,
cellular transformation follows. Examples
are the p53 gene and the genes for
retinoblastoma and Wilm's tumor.

two-point cross A cross involving two loci.

type I error In statistics, the rejection of a
true hypothesis.

type II error In statistics, the accepting of a
false hypothesis.

typological thinking The concept that
organisms of a species conform to a specific
norm. In this view, variation is considered
abnormal.

ubiquitin A peptide of twenty-six amino
acid residues that enzymes attach to proteins
that the proteasome will degrade.

unequal crossing over Nonreciprocal
crossing over caused by the mismatching of
homologous chromosomes. Usually occurs
in regions of tandem repeats.

uninemic chromosome A chromosome
consisting of one DNA double helix.

unique DNA A length of DNA with no
repetitive nucleotide sequences.

unmixed families Groups of four codons
sharing their first two bases and coding for
the same amino acid.

unusual bases Other bases, in addition to
adenine, cytosine, guanine, and uracil,
found primarily in transfer RNAs.

UP element See upstream element.

upstream A convention on DNA related to
the position and direction of transcription
by RNA polymerase (5 '— »3'). Downstream
(or 3 ' to) is in the direction of transcription
whereas upstream (5' to) is in the direction
from which the polymerase has come.

upstream element A sequence of about
twenty AT-rich bases centered at — 50 in
promoters of prokaryotic genes that are
expressed strongly.

uracil See pyrimidines.

variable-number-of-tandem-repeats
(VNTR) loci Loci that are hypervariable
because of tandem repeats. Presumably,
variability is generated by unequal crossing
over.

variance The average squared deviation
about the mean of a set of data.

variegation Patchiness; a type of position
effect that results when particular loci are
contiguous with heterochromatin.

virion A virus particle.

viroids Bare RNA particles that are plant
pathogens.

V(D)J joining The process of joining

variable, diversity, and joining gene segments
(V-J,V-D, or D-J joining) in the formation of
a functioning immunoglobulin gene.

Wahlund effect The effect of subdivision
on a population, causing it to contain fewer
heterozygotes than predicted despite the
fact that all subdivisions are in
Hardy-Weinberg proportions.

western blotting A technique for probing
for a particular protein using antibodies. See
Southern blotting.

whole-genome shotgun method

A method of sequencing entire genomes by
breaking up the genomes into small pieces,
sequencing the pieces, and then using
computers to establish order by overlapping
the sequences.

wild-type The phenotype of a particular
organism when it is first seen in nature.

Wilm's tumor A childhood kidney
cancer caused by the inactivation of
an anti-oncogene.

wobble Referring to the reduced constraint
over the third base of an anticodon as
compared with the other bases, thus
allowing additional complementary base
pairings.

xeroderma pigmentosum A disease in
human beings caused by a defect in the UV
mutation repair system.

X-inactivation center (XIC) Locus at
which inactivation is initiated on the X
chromosome in mammals.

X linked See sex-linked.

X-ray crystallography A technique, using
X rays, to determine the atomic structure of
molecules that have been crystallized.

yeast artificial chromosome (YAC)

Originating from a bacterial plasmid, a YAC
contains additionally a yeast centromeric
region (CEN) and a yeast origin of DNA
replication (ARS).YACs are capable of
including large pieces of foreign DNA
during cloning.

Y-junction The point of active DNA

replication where the double helix opens up
so that each strand can serve as a template.

Y linked Inheritance pattern of loci located
on the Y chromosome only. Also refers to
the loci themselves.

2 DNA A left-handed form of DNA found
under physiological conditions in short GC
segments that are methylated. It may be
involved in regulating gene expression in
eukaryotes.

ZFY gene Originally believed to be the
human male sex-switch gene, located on
the short arm of the Y chromosome. ZFY
stands for 2inc/inger on the Y chromosome.

zinc finger Configuration of a DNA-binding
protein that resembles a finger with a base,
usually cysteines and histidines, binding a
zinc ion. Discovered in a transcription
factor in Xenopus.

zygonema (zygotene stage) The stage of
prophase I of meiosis in which synapsis
occurs.

zygotic induction The beginning of
vegetative growth when a prophage is
passed into an F~ cell during conjugation.

zygotic selection The forces acting to cause
differential mortality of an organism at any
stage (other than gametes) in its life cycle.

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

INDEX

ABC excinuclease, 342
AbelsonJ., 268

ABO blood groups, 25-26, 32, 559
Ac-Ds system, 468
Acentric fragment, 178, 1 79
Acetabularia (green alga), 515
Acinonyx jubatus. See Cheetah
Acquired immunodeficiency syndrome.

See AIDS
Acridine dyes, 33 7
Acrocentric chromosome, 48, 49
Acron, 470-71

Activators, for transcription, 263
Active site, of enzyme, 206, 208
Acute lymphocytic leukemia, 485
Adaptation and Natural Selection: A Cri-
tique of Some Current Evolutionary
Thought (Williams, 1966), 604-5
Adaptive mutations, 339
Adaptive value, 577

Additive models, of variation, 533, 538, 548
Adenines

chemistry of nucleic acids, 213,2/5

dideoxy method of DNA sequencing, 385

tautomeric shifts, 329, 330
Adenosine deaminase, 397
Adjacent-1 and adjacent-2 types, of segrega-
tion, 184-85
A DNA, 219

Affected individuals, and pedigree analysis, 98
Affymetrix, Inc., 396
Africa, and AIDS, 502
African sleeping sickness, 275
AGAMOUS gene, 483
Agriculture

genetically modified crops, 371, 398

life cycle of plants, 66

polyploidy in plants and, 199

recombinant DNA techniques and, 13
Agrobacterium tumefaciens, 371-72, 376-77
AIDS, 13, 276, 502-4
Alberts, B., 234, 252
Albinism, 33, 34, 37, 97-98
Alcoholism, 103-4
Alkaptonuria, 37, 38, 582
Allele. See also Allelic frequency; Allelism

classical genetics and, 9

dominant and recessive traits, 18

Hardy-Weinberg equilibrium, 558-59

lethal-equivalent, 561

multiple, 25-26, 558-59
Allelic exclusion, 495
Allelic frequency. See also Allele

Hardy-Weinberg equilibrium, 553-54, 555,
561-62, 566-67

heterozygote advantage, 584

mutational equilibrium, 572-73

selection-mutation equilibrium, 581
Allelism, 319, 354. See also Allele
Allium cepa. See Onion
Allolactose, 407, 408, 409
Allopatric speciation, 592,5.92, 594
Allopolyploidy, 198
Allosteric protein, 409
Allotypes, 493
Alio zygosity, 561
Alternate segregation, 184
Alternative energy, and biotechnology, 398
Alternative splicing, 271-72
Altman, Sidney, 265, 268
Altruism, and sociobiology, 14, 603-5, 606,

607-8
Alu family, 459
American Red Cross, 26
Ames, Bruce, 332
Ames test, for carcinogens, 332
Amino acids

experimental methods and sequencing of,
284-86

mutations, 313

proteins and structure of, 282

rate of evolutionary change and sequences

of, 601-602, <56>3

RNA and, 12, 281

sequence of dihydrofolate reductase, 513
Aminoacyl site (A site), 292
Aminoacyl-tRNA synthetases, 257, 286, 287
ampC gene, 389
Amphidiploids, 198
Ampicillin, 427
Anagenesis, 590

Anaphase, of mitosis, 52, 54, 56,57
Anaphase A and anaphase B, 54
Anaphase I, and meiosis, 59, 60, 62
Anaphase-promoting complex (APC), 51
Anemia. See Thalassemia
Aneuploids and aneuploidy, 83, 190, 192-97
Anfinsen, Christian, 303
Angelman syndrome, 524
Angiogenesis, 492
Angiosperms, 64

Animal Dispersion in Relation to Social
Behav ior (Wynne-Edwards, 1962),
603-4
Animals. See also Cattle; Mouse; Sheep

life cycle, 47

meiosis, 63-64

polyploidy, 198-99

vectors and cloning, 371
Antennapedia complex (ANT-C), 478

Antibiotics and antibiotic resistance. See also
Drug resistance

antibacterial mechanisms, 153

Chlamydomonas, 517-18

composite transposons, 427

mitochondrial protein synthesis, 515

resistance plasmids, 523

translation and, 294-95
Antibodies

ABO blood group system, 25

definition of, 492

diversity of, 494-97

molecular genetics and, 13
Anticoding strand, of DNA, 249
Anticodons, 256, 258, 309. See also Codons
Antigens, 25, 492
Antimutator mutations, 338
Anti-oncogenes, 13, 485
Antiparallel strands, of DNA, 218
Antirepressor, 42 1
Antirrhinum majus (Snapdragon),

35-36,484
AntisenseRNA, 431
Anti-sigma factor (AsiA), 430
Antiterminator protein, 420
AP endonuclease, 341, 354
Apolipoprotein-B (apoE) gene, 275
Apoptosis, 484

Apple maggot fly (Rhagoletis pomonelld), 593
Arabidopsis thaliana (Meadow weed)

dwarf form of, 3 73

flower formation, 482-83

methylation of DNA, 466-67

study of plant genetics and, 372
Arabinose, 435
Arberg,W,4,359
Archaea, 149

Armillaria bulbosa (fungus), 441
Aromatic amines, 492
Arsenic, 492
Asbestos, 492
Ascaris spp., 50, 87
Ascopores, 123
Ascus, 123

AsiA (anti-sigma factor), 430
Aspartate transcarbamylase, 433, 434
Aspergillus nidulans (fungus), 132
Assignment test, 136-37
Assortative mating, 554, 560
Aster, 53
Asthma, 545
Ataxia-telangiectasia, 484
Attenuator-controlled system, of gene

expression, 415-18
Attenuator region, 415

1-1

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-2

Index

Attenuator stem, 4 16
Autogamy, 519, 520,521
Autonomously replicating sequences

(ARS), 239
Allopolyploidy, 198
Autoradiography, and DNA replication,

222-24
Autosomal dominant inheritance, 102, 104
Autosomal linkage, 134
Autosomal recessive inheritance, 102
Autosomal set, 84
Autosomes, 83
Autotrophs, 150
Autozygosity, 561, 564
Auxiliary factors, and introns, 271
Auxotrophs, 150
Avery, Oswald, 4, 154, 209
AZT, 504

Bacillus, 149

Bacillus subtilis (soil bacterium)

DNA-RNA complementarity, 247

gene transcription, 253

transformation and transformation
mapping, 155, 156-57, 172

trp oepron, 41 7, 418
Bacillus thuringiensis (Bt), 398
Bacitracin, 153
Backcross, 22
Back mutations, 327
Bacteria. See also Escherichia coli;
Prokaryotes

antibiotic resistance, 294-95

cultivation of, 150-51

genetic research and, 149-50

life cycle, 64

phenotypes of, 151-54

recombinant DNA technology, 349-51

sexual processes in, 154-62

size of cells, 441
Bacterial artificial chromosomes (BACs),

393, 395
Bacterial lawn, 151

Bacterial viruses, 66, 149-50. See also Viruses
Bacteriophages

genetic research and, 8, 149-50

life cycle of, 163-65

sexual processes in, 154-62

transduction, 165-68
Balaenoptera musculus (Blue whale), 441
Balbiani, E. G., 449
Balbiani rings, 449
Baltimore, David, 276
BamHI restriction endonuclease, 360
Barr, M.,90
Barr body, 90
Bar system, 185-86, 188
Basal bodies, 52
Base excision repair, 340-41
Base flipping, 341
Basic/helix-loop-helix/leucine zipper,

480, 481
Bates, H.W., 604
Batesian mimicry, 604, 605

Bateson, William, 2 1

B-cell chronic lymphocytic leukemia, 485

BDNA, 219,220

Beadle, George, 10, 38-39

Beagle (ship), 589

Bean (Phaseolus vulgaris}, 541. See also

Broad bean
Behavioral genetics, human, 547. See also

Sociobiology
Bennett, J. C.,494

Benzer, Seymour, 318, 320, 321, 322-23, 324
Benzine, 492
Berg, Paul, 4, 370, 383
Bertram, E., 90
BgH restriction enzyme, 360
Bicoid gene, 472, 474, 475
Binary fission, 519
Binomial expansion, 73
Binomial theorem, 72
Biochemical genetics, 37-39
Bioinformatics, 396
Biolistic transfer, 374, 375
Biological containment, 370-71
Biological species concept, 590
Biomedical applications

AIDS and retroviruses, 502-4

Ames test for carcinogens, 332

antibiotics and antibiotic resistance,
294-95

of multiple-stranded DNA, 221, 222

of prions, 213
Biometrics, 534-35
Bithorax gene, 478, 479
Bivalents, and meiosis, 58
Blackburn, Elizabeth H., 454, 455
Blastoderm, 470
Blobel, Gunter, 301
Blood coagulating factor III, 136
Blood pressure, and concordance, 546
Blood transfusions, 26
Blood types

Hardy-Weinberg equilibrium, 557,558, 559

migration, 573-74

phenotypes, 25-26, 32
Blunt-end ligation, 363-64
Boa constrictor {Constrictor constrictor), 50
Bobbed gene, 87, 96
Bodmer,W.,562
Body plan, 471-75
Bolivar, E, 363
Bos taurus. See Cattle
Bottlenecks, genetic, 576-77
Bounty (ship), 576
Bouquet stage, 56
Boveri, Theodor, 3
Boyer,H.,360
Branch migration, 348, 350
Brassica oleracea (cabbage), 198
Breakage-fusion-bridge cycle, 178
Breakage-and-reunion process, 347
Breast cancer, 390, 391, 392, 546
Brenner, Sydney, 325, 483
Brewer, B., 252

Bridges, Calvin B., 84, 121, 199
Britten, Roy J., 457-58

Broad bean (Viciafaba), 50

Brody, E., 268

Broker, Thomas, 265, 266

Brower, L.,604

Brown, Robert, 3

Buoyant density, of DNA, 451, 452, 514

Burkitt's lymphoma, 485

Burt, Cyril, 546

Bush, Guy, 592

Butterflies, and mimicry, 604, 605

Cabbage. See Brassica oleracea

Cactus ground-finch (Geospiza scandens),

588,595
Cactus protein, 467

Caedobacter conjugatus (kappa particle), 521
Caedobacter taeniospiralis (kappa particle),

520, 522
Caenorhabditis elegans (Roundworm)

genetic control of development, 483-84

genome sequencing, 396
Cairns, John, 222-24, 339
Calculus of genes, 606
Cameleo pardalis. See Chameleon
Cancer. See also Breast cancer

Ames test for carcinogens, 332

environmental causes of, 492

gene expression, 484

gene mapping, 390, 591,392

mutational nature of, 484-87, 506

recombinant DNA technology, 13

viral nature of, 487-92, 506
Cancer-family syndromes, 484-85
Cannabalism, and kuru, 213
Cap, and eukaryotic transcription, 265
Capsids, 149
Capsomeres, 149
Carcinogens

Ames test, 332

environmental causes of cancer, 492
Carcinomas, 484
Carpel whorl, 479, 482
Cassette mechanism, 469
Cat, tortoiseshell, 91, 95
Catabolite activator protein (CAP), 412-13
Catabolite repression, 412-13
Catostomus clarki (fish), 599, 600
Cat's eye syndrome, 195
Cattle (Bos taurus), 66, 543
Cavalli-Sforza, L., 562
C-bands, 451, 452
CCR5 protein, 502
Cdc2 gene, 50-51
CD4 protein, 502
Cech, Thomas, 265-66, 268
Celera Genomics, 396
Cell. See also Meiosis; Mitosis

chromosomes and complements of, 48-49

daughter cells, 49,57

diploid cells, 9, 48

disomic cells, 197

eukaryotic compared to prokaryotic, 440

expression of foreign DNA and eukaryotic,
372-75

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-3

F cells, 160, 161

follicle cells, 470, 474

haploid cells, 9, 48

helper T cells, 502

mitosis and lymphocytic cells, 15

monosomic cells, 190

nullisomic cells, 190

nurse cells, 470

pole cells, 470

size of, 441

sperm cells, 63

totipotent cells, 469

trisomic cells, 190

upregulation of growth, 487
Cell cycle, 50-51
Cell-free system, 306
Cellular immunity, 492
Centimorgan, 111
Central dogma, of genetic information, 244,

275-76
Centric fragment, 178
Centrioles, 52-53
Centromeres, 48, 453, 454
Centromeric breaks, 185
Centromeric fission, 185
Centrosome, 52, 53

Chain-terminating dideoxy nucleotides, 383-84
Chameleon (Cameleo pardalis), 2
Chaperone proteins, 303
Chaperonins, 303
Chapeville, E, 287
Chargaff, Erwin, 216
Chargaff's ratios, 216
Charon phages, 360, 363
Chase, Martha, 209-10
Checkpoints, and cell cycle, 51
Cheetah (Acinonyx jubatus), 552
Chemical method, of DNA sequencing,

383, 390
Chemical mutagenesis, 330-31, 336-37
Chemiluminescent techniques, 367
Chiasmata, 59
Chicken. See also Poultry

epistatic interactions and phenotype of, 3 7

genotypic interactions and combs of,
30-32,34

sex-linked traits, 96, 98
Chimeras, 190
Chimeric plasmid, 360
Chirality, on amino acid, 281
Chironomus pallidivittatus (midge), 449
Chironomus tentans (midge)

chromosome puffs, 449, 450

messenger RNA, 300
Chi site, 350

Chi square analysis, 76-77, 79, 557, 558, 567
Chlamydomonas reinhardi (green alga),

517-18
Chloramphenicol, 294, 515
Chloroplasts, and cytoplasmic inheritance,

515-18
Chondrodystrophy, 582
Chorthippus parallelus. See Grasshopper
Chow, Louise T., 265, 266
Chromatids, 52, 67

Chromatin

definition of, 48

higher-order structure of, 448

nucleosome structure, 443-44

types of in eukaryotic chromosomes, 453
Chromatin assembly factors, 444
Chromatin remodeling, 446, 465
Chromatosome, 443
Chromium, 492
Chromomeres, 48

Chromosomal banding, 134, 451-53
Chromosomal breaks, 178-85, 200
Chromosomal maps. See also Mapping

of Drosophila melanogaster, 121, 122,
169-71

of humans, 132-40
Chromosomal painting, 486
Chromosomal rearrangements, in humans,

186-90
Chromosomal theory, 3-4, 66
Chromosome

attachment point, 48

cancer and defects in, 484, 485

chromatids compared to, 67

combinations of maternal and paternal in
gametes, 67

complements of cell, 48-49

cytogenetics and variations in structure of,
178-90

cytological crossing over, 120-22

definition of, 154

Down syndrome, 177

eukaryotic, 440-61, 462

sex determination, 83

species and numbers of, 49-50

three-point cross, 119

two-point cross, 111, 114
Chromosome jumping, 390-91
Chromosome number, 190-99
Chromosome puffs, 449-50, 462
Chromosome walking, 390-91, 392
Chronic myelogenous leukemia, 485
Chronic wasting disease, 213
Chymotrypsin, 284
Cigarettes, and cancer, 492
C/s-acting mutants, 411
Cis configuration, 111
C/s-dominant mutation, 411, 412
6Ys-trans arrangement, and X linkage, 133
Cis-trans complementation test, 318
Cistron, 318

Cladogenesis, 590, 592-94, 595
Clamp-loader complex, 232
Classical genetics, 4-5, 9-10, 66
Classical linkage maps, 393
Cleft lip, 545
Clinal selection, 600
Clonal cancers, 484
Clonal evolution theory, of cancer, 484
Clones and cloning. See also Recombinant
DNA technology

assignment test, 137

benefits from, 397-98

chromosome location of gene, 141

ethical debate on, 374-75

particular genes, 564-66

probing for cloned gene, 368-77

restriction enzymes, 360-61

of sheep, 1
Clubfoot, and concordance, 546
C-myc gene, 490

Coal products, as carcinogens, 492
Coccus, 149

Coding strand, of DNA, 249
Codium (green alga), 515
Codominance, 23
Codon library, 602-3
Codon preference, 431
Codons. See also Anticodons

common and alternative meanings of, 311

synthetic, 306-7

transcription, 11-12

transfer RNA, 256

wobble hypothesis, 307-8, 309
Coefficient of coincidence, 118
Coefficient of relationship, 605-6
Cohen, S., 360
Cohesin, 53

Coiling, direction of in snails, 509-10, 525
Cointegrate state, of transposition, 428
Colicins, 522-23

Colinearity, and mutations, 324-25
Collins, Francis S., 358
Colony morphology

bacteria, 151

fungi, 124
Color. See also Pigment and pigmentation;
Skin color

flowers and inheritance of, 23, 37

metabolic pathways, 36

mouse and genotypes, 33-34

multilocus control of in wheat, 531-33

proplastids and variegation in plants, 516
Color blindness, 133
Col plasmids, 522-23
Columba livia. See Pigeon
Combinatorial control, 264-65
Comings, David E., 452
Common ancestry, 560-61
Comparative studies, 8
Compensasome, 94
Competence factor, 155
Complement, 492
Complementarity

DNA-RNA hybridization, 246-47

double helix of DNA, 1 1
Complementary DNA (cDNA), 365-66
Complementation, and mutations, 318-19
Complementation groups, 321
Complete linkage, 115
Complete medium, 150
Complementarity, and structure of DNA,

218,219
Complementation matrix, 320, 323
Component of fitness, 577
Component numbers, and transfer RNA,

281,286
Composite transposons, 427
Computer programs, and population genetics,
583-84

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-4

Index

Concordance, among twins, 546

Condensin, 444

Conditional-lethal mutant, 150, 327-28

Confidence limits, 74

Congenital malformations, and inbreed-
ing, 562

Congenital heart disease, 545

Conjugation, 154, 157, 158, 161

Consanguineous degrees, of relationship, 99

Consensus sequence, 248

Conservative replication, of DNA, 220

Conserved sequence, 248, 277-78

Constitutive heterochromatin, 452-53

Constitutive mutants, 410-12

Constitutive puffs, 450

Constrictor constrictor. See Boa constrictor

Containment, and recombinant DNA technol-
ogy, 370-71

Contigs, 393, 394,395

Continuous replication, of DNA, 225

Continuous variation, 17, 531, 533

Control, and experimental design, 6

Controlling elements, and transposons, 468

Corepressor, 414

Corey, R. B., 206

Corn. See Zea mays

Correlation and correlation coefficient,
539-40

Correns, Carl, 4, 17

Cotransduction, 166, 167

Coulson, Alan, 383

Coupling, of alleles, 111

Covariance, 539-40

CpG islands, 524

Creationism, 591

Creighton, Harriet, 120-22, 123

Crick, Francis, 4, 10, 11,205,215,216,244,
272, 276, 304, 328, 330

Cricket (Gryllus domesticus), 50

Cri du Chat syndrome, 188-90

Crisscross pattern of inheritance, 96

Critical chi-square, 76

Crogene product, 423

Crossbreeding, 17

Cross-fertilization, 17-18

Crossing over, and meiosis, 58-59

Crossover suppression, 179-80

Crow, James, 88, 533, 534, 562

Crown gall tumor, 371, 372, 377

C-sis oncogene, 491

Cucurbita pepo . See Summer squash

Cultivation

of bacteria, 150-51
of viruses, 151

CURLY LEAF gene, 483

CUX family, of codons, 309

C-value paradox, 457

Cyanobacteria, and introns-early hypothesis of
exon shuffling, 274

Cyclic AMP (cAMP), 412-13

Cyclin, 50-51

Cyclin-dependent kinase (CDK), 50

Cyclosome, 51

Cysteines, 152, 283

Cystine, 283

Cystic fibrosis, 13

Cytogenetics. See also Molecular genetics

chromosomal structure, 178-90

chromosome number, 190-99

definition of, 178
Cytokinesis, 47, 49-50
Cytological crossing over, 120-22, 123
Cytoplasmic inheritance. See also Inheritance

chloroplasts, 515-18

definition of, 509

infective particles, 518-22

mitochondria, 511-15

prokaryotic plasmids, 522-24
Cytosine, 213,2/5
Cytotoxic T lymphocytes, 492

Damage reversal, and DNA repair, 340

Danaus plexippus. See Monarch butterfly

Danna,K., 359

Darnell, J., 273

Darwin, Charles, 3, 4, 5, 13, 17, 589, 594

Dauermodification, 509

Daughter cells, 49, 57

Davies, David, 221

DDT, resistance to in Drosophila, 533-34,
535, 540

Dean, D, 252

Deer mouse (Peromyscus maniculatus), 92

Degenerate code, 307

Degrees of freedom, 76, 79

Delbruck, Max, 4, 163, 164, 316, 317

Deleted in colorectal cancer (DCC) gene, 491

Deletion chromosome, 178, 179

Deletion mapping, 321-23

Deletions, and mutations, 336-37

Demes, 553

Denaturation studies, of DNA, 218

Denominator elements, 85

Density-gradient centrifugation, 221

Depauperate fauna, 594

Depression, and heritability, 545

Derepressed operon, 414

Dervan,R,221

Deterministic population size, 574

Development
definition of, 469

gene expression and patterns, 469-84
hemoglobins and human, 459-61

Diabetes mellitus, 545

Diakinesis, 56

Dicentric chromosome, 178, 179

Dictyotene stage, of meiosis, 64

Dideoxyinosine, 504

Dideoxy method, of DNA sequencing,
383-86, 390

Dideoxynucleotides (dd), 384-85

Diener, Theodor, 272

Dihybrids, 26, 30, 33, 36

Dihydrofolate reductase, 513

Dimerization, 340

Dioecious flowers, 87

Diploid cells, 9, 48

Diploid mapping, 110-11, 114-22

Diplonema, 56

Directed mutation, 339
Directional selection, 577-78
Disassortative mating, 560
Discontinuous DNA replication, 225,227
Discontinuous variation, 17, 531
Discrete generations, 556-57
Disomic cells, 197
Dispersive replication, of DNA, 220
Disruptive selection, 578, 579
Dissortative mating, 554
Disulfide bridge, and cysteines, 283
Dizygotic twins, 546

D-loop model, of DNA replication, 238, 239
DNA. See also Complementary DNA; Mito-
chondrial DNA; Mutation; Repetitive
DNA; Satellite DNA; Transcription;
Translation; Triplex DNA; Z DNA

alternative forms of, 219

biologically active structure, 214-15

chemistry of , 2 1 1 - 1 9

cloning and creation of, 564-66

computer-generated images, 204, 315

control of proteins, 205-7

denaturation studies, 218

double helix structure, 10-11, 206-7,
216-18

enzymes and, 218-19

eukaryotic cells and expression of foreign,
372-75

eukaryotic chromosome and arrangement
of, 440-42

genomic library, 366

location of, 208-9, 219

methylation of, 466-67

misalignment mutagenesis, 33 7

normal and tautomeric forms of bases,
328,329

partitioning of, 238

phage labeling, 209-10

plasmids, 357

protein motifs of recognition, 480-81

RecA protein, 345

recombinant DNA technology, 12-13,
347-51,352

repair, 339-46, 348

replication of, 207, 219, 220-39, 382

translation, 276

transcription, 246-55, 260-75

transformation, 155, 156, 209

variation, 602-3

X-ray crystallography, 215-16
DNA-DNA hybridization, 457-58
DNA fingerprints, 381, 382. See also Finger-
prints
DNA glycosylases, 340-41
DNA gyrase, 236
DNA helicase, 230
DNA ligase, 227-29
DNA polymerase, 225, 227, 229, 230
DNA-RNA hybridization, 246-47
DNA sequencing, 383-90
DNA tumor viruses, 371
Docking protein (DP), 301
Dominance and dominant traits

heterozygotes and homozygotes, 22-23

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-5

Mendel's experiments, IS, 19

pedigree analysis, 99, 102
Doolittle,W. E.,273
Dorsal gene, 466, 467
Dosage compensation, 90-95
Dot blotting, 368
Doublebar gene, 185-86, 188
Double breaks, in chromosome, 179
Double crossovers, 117, 118
Double digests, and restriction mapping,

379-80
Double helix, structure of DNA, 10-11,

206-7,216-18
Double-strand break model, of recombination,

347-49, 365
Double-strand break repair, 344
Double-stranded DNA, 221, 225
Double transformation, 157
Douglas, L., 30
Down, John Langdon, 192
Downstream direction, of DNA transcrip-
tion, 249
Downstream promoter element (DPE), 262
Down syndrome, 177, 192-94, 546
Dreyer,W.J.,494
Drosphila ananassae, 182-84
Drosophila melanogaster (fruit fly)

allelism, 354

chromosomal breaks, 200

chromosomal map, 121, 122, 169-71

chromosome number, 50

chromosome puffs, 450

complementational analysis, 319, 320

DDT resistance, 533-34, 535, 540

development, 469-79

diagram of chromosome 2, 9

disruptive selection, 578, 579

dosage compensation, 94-95

duplications of chromosomal segments,
185-86

epistasis, 33

gene mapping, 396

generation interval, 66

genie balance, 84-85

geotactic response, 541, 542

giant banded chromosomes, 449

giant salivary gland chromosome, 119-20,
179, 439, 455

gynandromorph, 192

heterozygosity, 599

homeobox genes, 482

infective particles, 522

inversion, 180, 181

linkage groups of, 110

as model organism, 8, 65

mutants and mutations, 24

nomenclature, 23-24

nondisjunction, 190, 191

nuclease-hypersensitive sites of DNA, 446

nucleolus, 260

population genetics, 566-67

realized heritability, 543

satellite DNA, 452

scanning electron micrograph of, 109

sex determination, 87

sex-linked inheritance, 95-96, 103, 104

size of chromosome, 441-42

testcrossing of, 110-11, 114-16, 118-20,
140-41

twin spots, 132

unequal crossing over, 188
Drosophila pseudoobscura, 596, 599
Drug resistance. See also Antibiotics and
antibiotic resistance

bacteria and, 153

fungi and, 124
Drug sensitivity, 154
Duffy blood group, 134
Duplications, of chromosomal segments,

185-86
Dusts, as carcinogens, 492
Dyad chromatid pair, 59
Dynein, 52
Dysplasia, 491

Ectoderm, 470
Edman method, 284-85
Edward syndrome, 194, 195
Ehrman, Lee, 597
Eicher, Eva, 86
Eichwald, Ernst, 85
Either-or rule, 72
Electrophoresis

evolution and empirical data, 14, 553

experimental methods, 92-94

glucose-6-phosphate dehydrogenase
system, 91

polymorphisms, 14, 598
Electroporation, 374
Ellis, J., 303

Elongation, and translation, 288, 292-93, 296
Elongation complex, 262-63, 264
Elongation factors (EF-Ts and EF-Tu), 292, 293
Eldredge,N,594

Encephalopathies, and animal diseases, 213
Endoderm, 470
Endogenote, 159
Endomitosis, 119-20
Endonucleases, 226, 334
Endosomic vesicles, 500
Endotoxins, 398
Energy cost

of protein biosynthesis, 313

of translation, 297
Energy industry, and biotechnology, 398
Enhancers, and transcription, 263
Enhancer of zeste gene, 483
Enriched medium, 150
Env (envelope) gene, 503
Environment

carcinogens, 492

evolutionary role of sexual reproduction, 88

induced versus inherited traits, 509

phenotypic distributions, 533, 534
Environmentally induced puffs, 450
Enzymes

active site, 206, 208

classical genetics, 10

DNA and control of, 218-19

DNA repair in E. colt, 348

DNA replication in E. coli, 231

Tay-Sachs disease, 23

three-dimensional structure, 206-7
Enzymology, of DNA replication, 225-38
Eosinea gene, 35-36
Ephestia kuhniella (flour moth), 510,511
Epidemiology, of AIDS, 502-3
Epistasis, 32-37, 40-41
Epstein, R. H., 296

Epuliscium fishelsoni (bacteria), 441
Equational division, 60
Erythromycin, 153, 515
Escherich, Theodor, 149
Escherichia coli. See also Bacteria

autoradiography of DNA, 223

conditional-lethal mutants, 328

conjugation, 157, 158

DNA repair, 348

DNA replication in, 231, 233, 238

DNA-RNA hybridization, 246

F factor, 16 1

fine-structure mapping, 321

gene expression, 406-12, 435-36

generation interval, 66, 149

genetic variation, 316

heat shock proteins, 430

host-range mutations, 163

lacZ gene and directed mutations, 339

medium for cultivation, 150

as model organism, 8

nucleotide excision repair, 341

N-formyl methionine and protein syn-
thesis, 288

nucleolus, 278

phage X, 420

pili, 159

plasmids and cloning, 357, 361

promoter of ribosomal RNA gene,
rrnB, 249

recombinant DNA techniques, 13

ribosome, 257

RNA polymerase, 250

scanning electron micrograph, 148

T4 bacteriophage, 405

transduction experiments, 168, 172

transformation, 155

tRNAs for methionine, 289
Essay on the Principle of Population, An

(Malthus, 1798), 589
Ethics, and genetics

cloning, 374-75

evolution, 590-91

human behavioral genetics, 547

human genome sequencing, 397

Lysenko affair, 6-7

recombinant DNA technology, 370-71

sociobiology, 606
Euchromatin, 48, 452
Eugenics, 547
Eukaryotes

chromosome, 440-61, 462

clones and cloning, 369-70

control of transcription, 465-69

daughter cells, 49

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-6

Index

Eukaryotes — Cont.

DNA replication, 231-32, 238-39

DNA transcription, 260-75, 278

foreign DNA, 372-75

gene expression, 12

initiation factor, 290

life cycle, 64

nucleotide excision repair, 343

prokaryotes compared to, 47, 261-62,
298, 440

size of cells, 441
Eukaryotic vectors, 369-70
Euploidy, 197-98
Eusocial hymenoptera, 606
Even-skipped gene, 476, 477
Evolution

Darwinian, 5, 589, 590-91

DNA synthesis, 228-29

genetic code, 311-12

hemoglobin genes, 460

homeotic control, 479

imprinting, 524

intron function, 272-74

inversions, 182-84

mutations, 339

population genetics, 553

proto-oncogenes, 491

reproductive isolating mechanisms, 607

sexual reproduction, 88-89

sociobiology, 603-6

speciation, 589-94, 595

supergenes, 181

transposons, 429-30
Evolutionary genetics, 4, 5, 13-14
Evolutionary rates, 601, 603
Ewing's sarcoma, 485
Excisionase, 418
Excision repair, 340-46
Exogenote, 159
Exons, 265

Exon shuffling, 272-74
Exonucleases, 226-27
Experimental design, 74
Experimental methods

adaptive mutations, 339

amino acid sequencing, 284-86

chromosomal painting, 486

computer program for allelic equilibrium
under heterozygote advantage, 583-84

gene sequencing, 388-89

high-speed chromosomal sorting, 444-45

inversions and evolutionary sequences,
182-84

in vitro site-directed mutagenesis, 333-35

lethal equivalents, 562

mapping of quantitative trait loci, 537-38

mimicry, 604

polymerase collisions, 252

protein motifs of DNA recognition, 480-81

size of cells, 441

transcription in real time, 250-51

viroids and introns, 272-73
Expression vectors, 369
Expressivity, of phenotype, 97, 98
Extrachromosomal inheritance, 509
Eye color, and concordance, 546

Facioscapular muscular dystrophy, 582

Factorials, 72

Familial Down syndrome, 193-94

Familial insomnia, 213

Family tree, 98-99

Fate map, 471

F cells, 160, 161

F-duction, 16 1

Fecundity selection, 577

Feedback inhibition, and posttranslational

control, 433
Felsenfeld, Gary, 221
Female-lethal gene, 85
Fertility factor (F factor), 158-59
Fetus, human

hemoglobin, 460

inbreeding and death rates, 562

Rh locus and maternal incompatibility, 582
F factor, 1 61, 162

Field mouse. See Peromyscus polionotus
Filial generation, 18
Fine-structure mapping, 320-23
Fingerprints, heritability of human, 544, 545.

See also DNA fingerprints
First-division segregation (FDS), 128-29, 130
Fisher, Ronald A., 4, 13, 553, 604
Fitness, and natural selection, 577, 585,

605-6. See also Survival of the fittest
5-bromouracil (5BU), 331
5' untranslated region (5' UTR), 291
Flemming,W., 3
Floral induction, 479, 482
Floral meristem, 479
Floral-meristem identity genes, 483
Flow cytometry, 445
Flowers. See also Plants

genetic control of development, 479,
482-83

inheritance of color, 23, 37

sex determination, 87, 90
Flow karyotype, 444
Fluctuation test, 316-17, 318
Fluorescent in situ hybridization (FISH), 486
FMR-1 gene, 187-88
Fokker-Planck equation, 575
Follicle cells, 470, 474
Follicular lymphoma, 485
Food processing, and biotechnology, 398
Footprinting, 248
Ford, Edmund B., 596
Forensics, and DNA fingerprint, 381
Four o'clock plants (Mirabilis jalapa), 23, 517
Founder effect, 576
Four-stranded DNA, 221
F-pili, 159

Fraenkel-Conrat, H. , 2 1 0- 1 1
Fragile site, of chromosome, 490
Fragile-X syndrome, 186-88
Fragment length polymorphisms (RFLPs), 537
Frameshift, and genetic code, 304-5
Frameshift mutation, 326-27, 338
Franklin, Rosalind, 205, 215
FrdC gene, 389
Free energy of activation, 205

Frequency-dependent selection, 597
Fruit fly See Drosophila melanogaster
Functional alleles, 318, 320
Fundamental number (NF), 185
Fungi

haploid mapping, 123

heterokaryon test, 509

phenotypes, 124

size of cells, 441
Furberg, S., 206
Fushi tarazu gene, 477
Futch, David, 182-84

Gag (group antigen gene), 503
p-Galatosidase, 406, 407
(3-Galatoside acetyltransf erase, 406
(3-Galatoside permease, 406
Galapagos Islands, 594
Gallo, Robert C., 502
Galton, Francis, 535, 547
Gametes

combinations of chromosomes in, 67

meiosis, 55

rule of independent assortment, 27

rule of segregation, 18
Gametic selection, 577
Gametophyte, 64
Gancylovir, 376
Gap genes, 475
Garden pea (Pisum sativuni)

chromosome number, 50

classical genetics, 9

independent assortment, 27

Mendel's experiments, 17-18, 65

phenotypic and genotypic ratios of
dihybrid, 32
Garrod, A. E., 37
Gastrulation, 470

G-banded chromosomes, 134, 138-39, 451
Gehring, Walter J. , 478
Gene. See also Gene expression; Gene
mapping; Gene order

classical genetics, 9

HIV and, 503-4

segregation, 18
Gene amplification, 459

Gene cloning. See Clones and cloning; Recom-
binant DNA technology
Gene conversion, 351
Gene expression. See also Transcription;
Translation

cancer, 484-92

catabolite repression, 412-13

immunogenetics, 492-501

inducible system, 406-12, 435

lytic and lysogenic cycles in phage X,
418-24

operon model, 12, 406, 408, 418

patterns in development, 469-84

posttranslational control, 433

transcription factors, 430

translational control, 430-33

transposable genetic elements, 425-30

trp operon, 413-18

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-7

Gene family, 458-59

Gene flow, 592

Gene mapping. See also Mapping

human mitochondrial chromosome, 512

phage \,419

sequencing of human genome, 390-97
Gene order

haploid mapping, 129-32

Hfr strains, 162

three-point cross, 118

two-factor cotransduction, 167
Gene overlap, 388-89
Gene pool, 555
General interval, 66
Generalized transduction, 166
General transcription factors, 262, 263
Generation time, 556-57
Gene therapy, 397

Genetically modified crops (GM crops), 371
Genetic code

breaking of, 305

DNA control of enzymes, 218-19

evolution and, 311-12

overlap and punctuation of, 305, 3 06

synthetic codons, 306-7

synthetic messenger RNAs, 306

triplet nature of, 304-5

universality of, 308-11

wobble hypothesis, 307-8
Genetic diseases

gene therapy, 397

hemoglobins, 460-61

imprinting, 524

inbreeding, 562

mitochondrial inheritance, 514-15
Genetic drift, 555, 575-76, 585
Genetic engineering. See Recombinant DNA

technology
Genetic fine structure, 317-24
Genetic load, 597
Genetic polymorphism, 596
Genetics. See also Biomedical applications;
Cytogenetics; Ethics; Experimental
methods; Heredity; Inheritance;
Molecular genetics

bacteria and bacterial viruses and research
in, 149-50

classical, evolutionary, and molecular, 4-5,
9-14

definition of, 3

fruit flies and colon bacteria as subjects of
experiments, 7-8

history of, 3-4

Nobel laureates, 113-14

scientific method, 5-7

techniques of study, 8
Genetic variation, 316, 596-603
Genie balance, in Drosophila, 84-85
Genome

electrophoresis and sampling of, 598

HW-1,504

mapping, 110

mitochondrial, 512

sequencing of human, 390-97
Genomic equivalence, 469

Genomic library, 366, 367
Genomics. See also Recombinant DNA
technology

benefits of cloning, 397-98

DNA sequencing, 383-90

polymerase chain reaction, 381-83

probing for cloned gene, 368-77

prokaryotic vectors, 360-66

restriction endonucleases, 359-60

restriction mapping, 377-81

southern blotting, 366-68
Genophore, 154
Genotypes

evolutionary genetics, 14

genotypic interactions, 30-37

independent assortment, 27

pedigree analysis, 97-98

segregation, 18, 21

transposition, 429-30
Genotypic interactions, 30-37
Geochelone elephantopus Vanderburgh. See

Giant land tortoise
Geospiza fortis (Ground finch), 590-91,595
Geotactic response, 541,542
Gerald, Park, 197

Gerstmann-Straussler-Scheinker syn-
drome, 213
Giant gene, 476
Giant land tortoise (Geochelone elephantopus

vanderburgh), 529
Giant salivary gland chromosomes, 119-20,

179, 439, 455
Giemsa stain, 451

Gilbert, Walter, 272-73, 274, 383, 390, 412
(3-Globin gene, 367-68
Globin gene family, 459-61
Glucose-6-phosphate dehydrogenase (G-6-PD),

90-91
Glycolytic pathway, 10
Goodfellow, Peter, 86
Gooseberry gene, 477
Gould, Stephen J., 594
Grandfather method, and X linkage, 132-33
Grant, B. & P., 590

Grasshopper (Chorthippus parallelus), 59
Green fluorescent protein, 377
Griffith, E, 154,209
Ground finches, of Galapagos Islands, 588,

590-91,594,595
Ground squirrels, 606
Group I and Group II introns, 266
Group selection, 603-4, 606
Grunberg-Manago, M., 306
Gryllus domesticus. See Cricket
G-tetraplex, 456, 457
Guanine, 213,275
Guide RNA (gRNA), 275
Gurken gene, 474-75
Gynandromorphs, 191, 192

Haemanthus katherinae, 53, 54
Haemophilus influenzae, 396
Hair color, and concordance, 546
Hairless mutation, of Drosophila, 24

Haldane,J.B.S.,4, 13, 553
Hall, B., 246
Hamilton, W. D., 605-6
Hammerhead ribozyme, 266, 270
Hamster, chromosome of, 448
Handedness (left or right), and concor-
dance, 546
Haploid cells, 9, 48
Haploidiploidy, 606
Haploid mapping, 122-32
Hard selection, 597
Hardy, G.H., 13, 553
Hardy-Weinberg equilbrium

alleleic frequencies, 553-54

definition of, 553

extensions of, 558-59

generation time, 556-57

migration, 573

multiple loci, 559-60

mutation, 572, 573

natural selection, 555

proof of, 555-56

random mating, 554-55, 560

selection models, 578

small population size, 574

testing for fit of, 557-58
HAT medium, 135
Heart disease, congenital, 545
Heat, and trans versions, 336
Heat shock proteins, 249-50, 430
Helicase II, 342
Helix-turn-helix motif, 480
Helper T cells, 502
Hemizygous genes, 96
Hemoglobin

human development, 459-61

sickle-cell anemia, 39
Hemophilia, 95, 100-102
Heredity, chromosomal theory of, 66. See also

Genetics; Inheritance
Heritability. See also Inheritance

measurement of, 544-45, 548

partitioning of variance, 543-44

realized, 542-43
Hermaphrodites, 191
Hershey, A. D., 209-10
Hertwig, O., 3
Heterochromatin, 48
Heteroduplex DNA, 351, 352
Heterogametic chromosomes, 83
Heterogeneous nuclear mRNAs (hnRNAs), 265
Heterokaryon test, 134, 509
Heteromorphic chromosome pair, 49, 83
Heteroplasmy, 511
Heterothallic mating type, 469
Heterotrophs, 150

Heterozygote advantage, 582, 583-84, 596-97
Heterozygotes, 21, 22-23
Heterozygosity, 564-65, 599
Heterozygous DNA, 351
Hexoseaminidase-A, 23
Hfr strains, 158-59, 160, 161, 162
High-speed chromosomal sorting, 444-45
Hindll restriction enzyme, 360
Histocompatibility Y antigen (H-Y antigen), 85

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-8

Index

Histone acetyl transferases (HATs), 446

Histone genes, 459

Histone proteins, 443

Histones, composition of, 447

History, of genetics, 3-4, 17, 28-29, 30-31, 47

HIV, and genes, 503-4. See also AIDS

Ho, David, 504

Hogness box, 262

Holandric traits, 95

Holland, J. J., 276

HollidayR.,347

Holliday junction, 347, 350

Holoenzyme, 231

Homeo box, 478-79, 482, 505

Homeo domain, 478, 479, 505

Homeotic genes, 477-78

Homogametic chromosomes, 83

Homogentisic acid, 38

Homologous chromosomes, 48

Homologous recombination, 347, 375

Homology-directed recombination, 344

Homomorphic chromosome pairs, 48, 83

Homoplasmy, 511

Homo sapiens. See Humans

Homothallic mating type, 469

Homozygosity, and inbreeding, 561-62,

562, 564
Homozygotes, 21, 22-23
Horseshoe crab. See Limulus polyphemus
Host-range mutations, 163
Hot spots, and mutations, 323
Hox clusters, 479
HubbyJ.L.,596, 598
Human Genome Project, 6, 390-97, 537
Humans (Homo sapiens). See also Fetus

aneuploidy 192-97

behavioral genetics, 547

chromosomal maps, 132-40

chromosomal rearrangements, 186-90

chromosome number, 50

composition of DNA, 216

fingerprints and heritability, 544, 545

flow karyotype of chromosomes, 444

gene mapping and sequencing of genome,
390-97

generation interval, 66

hemoglobins, 459-61

heterozygosity, 599

karyotype, 49

linkage groups of, 110

melanin synthesis, 36

metacentric, submetacentric, and acrocen-
tric chromosomes, 49

mitochondria and mitochondrial inheri-
tance, 512, 514-15

monosomy, 200

nondisjunction of sex chromosomes, 19 1

nuclease-hypersensitive sites of DNA, ^6

quantitative inheritance, 545-46

Rh locus and maternal-fetal incompati-
bility, 582

sex determination, 83, 84, 85-86, 87

telomeric sequence, 455

xeroderma pigmentosum, 343
Humoral immunity, 492

Hunchback gene, 475, 476

Hungry codons, 432

Huntington disease, 188, 582

Hybrid DNA, and recombinant DNA technol-
ogy 351

Hybridomas, 494

Hybrid plasmid, 360

Hybrids and hybridization, 18, 199

Hybrid vectors, 361, 363

Hybrid vehicle, 360

Hybrid zones, 592

Hydrogen bonding, in DNA, 218

Hyperplasia, 491

Hypervariable loci, 380-81

Hypothesis testing, 74-76

Hypotrichosis, 100

Hypoxanthine phosphoribosyl transferase
(HPRT), 134-35

Identity by descent, 561
Idiogram, 48, 49
Idiotypic variation, 493
Idling reaction, 432
Ijij genotype, 517-18
Imino acid, 281

Immune system, and adenosine deami-
nase, 397
Immunity, definition of, 492
Immunity reactions, of ABO blood

types, 25
Immunogenetics, 492-501
Immunoglobulins

definition of, 492

formation of, 498

genes and antibodies, 497

hybridomas, 494

variability of, 493, 506
Imprinting, 509, 524
Imprinting center (IC), 524
Inborn Errors of Metabolism (Garrod

1909), 37
Inbreeding

definition of, 554

genetic diseases, 562

genotypic proportions in population
with, 563

homozygosity from, 561-62

nonrandom mating, 560
Inbreeding coefficient, 561, 567
Inbreeding depression, 542
Incestuous unions, 99
Inclusive fitness, 605-6
Incomplete dominance, 23
Independent assortment

classical genetics, 10

example of, 34

meiosis and mitosis, 63, 67

rule of, 26-30
Independent events, 71-72
Indian fern (Ophioglossum reticulatum), 50
Induced mutation, 325-26
Inducer, and gene expression, 408
Inducible system, of gene expression,
406-12,435

Induction, and gene expression, 165, 409,

489-91
Industry, and genetic engineering, 398
Infective particles, and cytoplasmic inheri-
tance, 518-22
Information transfer, and translation, 281-303
Inheritance. See also Cytoplasmic inheritance;
Genetics; Heredity; Heritability; Quanti-
tative inheritance

determination of non-mendelian, 509

heritability, 542-45

polygenic inheritance, 533-35, 541

population statistics, 535-40

selection experiments, 541-42

traits controlled by multiple loci, 531-35
Initiation codon, 288

Initiation complex, and translation, 288-91
Initiation factors (IF1, IF2, IF3), 289
Initiation signals, for transcription, 248
Initiator Element (Inr), 262
Initiator proteins, 230
Insecticides, 398
Insertion mutagenesis, 488
Insertion sequences (IS elements), 425, 426
Insertions, and mutations, 336-37
Insomnia, familial, 213
Instantaneous speciation, 593
Interbreeding, 591
Intercalary heterochromatin, 453
Intergenic suppression, and mutations,

337-38
Integrase, 418

Intelligence quotient (IQ), and herita-
bility, 546
Interkinesis, 59

Internal ribosome entry site, 291
Interphase, of mitosis, 48, 53, 56, 62
Interpolar microtubules, 53
Interrupted mating, 159-61, 162
Intersex, 85

Intervening sequences. See Introns
Intestinal bacterium. See Escherichia coli
Intra-allelic complementation, 323-24
Intragenic suppression, 327
Introns

eukaryotic transcription, 265, 266

evolution, 272-74

self-splicing of, 271

single-stranded DNA loops, 268

Tetrahymena ribozyme, 269

viroids, 272-73
Introns-early and introns-late views, of exon

shuffling, 273, 274
Inversion, and chromosome breaks,

179-81,200
Inverted-repeat sequence, 251
In vitro research, 209
In vitro site-directed mutagenesis, 333-35
Iojap chromosomal locus, 516
Ionizing radiation, 492
Iron oxide, 492
IS elements, 425, 426, 427
Isochromosome, 194, 195
Isopropyl oil, 492
Isozymes, 94

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-9

Jacob, Francois, 159-60, 161, 409
Jeffreys, A., 381
Johannsen, Wilhelm, 21, 541
Junctional diversity, 495
Junk DNA, 458-59

Kaposi's sarcoma, 502

Kappa particles, 520-21, 522

Karotype, 48, 49

Karpechenko, G. D., 198

Karyokinesis, 47

Kavenoff, Ruth, 440-41

Khruschev, Nikita, 7

Killer Paramecium, 520-21, 525

Kimura, Motoo, 562, 599

Kinesin, 52

Kinetochore, 48, 55, 453

Kinetochore microtubules, 53

Kinetoplasts, 275

King, Mary-Claire, 391

Kin selection, 605-6

Kleckner, Nancy, 429, 430

Klenow fragment, 232

Klinefelter syndrome, 85, 197

Klotz,L., 440-41

Klug,A.,480

Knirps gene, 476

Knockout mice, 375-76

Koehn,R.,599

Kornberg, Arthur, 225

Kruppel gene, 476

Kuhn, Thomas, 591

Kuru, 213

Kynurenin, 39, 510, 511

Lac operon, and gene expression, 406-12, 436

Lactate dehydrogenase (LDH), 95,94

Lactose metabolism, 406, 407, 436

LacZ gene, 339

Ladder gels, 386, 399

Lagging strand, 225

Lahn, Bruce, 89

Lamarck, Jean-Baptiste, 5

Lampbrush chromosomes, 450-51

Landsteiner, Karl, 25

Lathyrus odorata. See Sweet pea

Leader, and RNA transcription, 255

Leader peptide gene, 416-19

Leader transcript, 415-16

Leading strand, 225

Leber optic atrophy, 514-15

Leder, Phillip, 307

Lederberg, Joshua, 157, 158, 161, 165, 166

Leishmania tarentolae , 27 5

Lejeune, Jerome, 192-93

Lemmings, and population density, 597

Leptonema, 56

Lethal-equivalent alleles, 561, 562

Leucine zipper, 480, 481

Leukemia, 484, 485

Level of significance, 11 -IS

Lewin, B., 523

Lewis, Edward B., 478

Lewontin, Richard C, 596, 598
LexA protein, 346, 347
Life cycle

of bacteriophages, 163-65

of eukaryotes, 64

generalized for animals and plants, 47

of Paramecium, 518-19

of plants, 64-66

of yeast, 126
Ligation, and DNA replication, 227-29
Liguus fasciatus (Tree snail), 570
Lily (Lilium longiflorum), 50
Limenitis archippus. See Viceroy butterfly
Limnea peregra (Pond snail), 51
Limulus polyphemus (Horseshoe crab), 599
Linear measurement, metric units of, 48
Linkage

classical studies of, 66

equilibrium and disequilibrium, 559-60
Linkage groups, 110
Linkage number, 236

Linkers, and blunt-end ligation, 363-64, 365
Liposarcoma, 485

Liposome-mediated transfection, 374
Liposomes, 374
Literature, scientific, 6-7
LittlefieldJ.W., 134
Liver, and sickle-cell anemia, 39
Location, of DNA, 208-9, 219
Lock-and-key model, of enzyme functioning,

206-7
Lod score method, 135
Long interspersed elements (LINES), 458
Lovell-Badge, Robin, 86
Luciferase gene, 377
Luciferin, 377
Luft disease, 514
Luria, Salvador E., 4, 316, 317
Lwoff, Andre, 423
Lymphocytic cell, and mitosis, 15
Lymphomas, 484
Lyon, Mary, 90
Lyon hypothesis, 90-91
Lysate, 163, 164
Lysenko,T D.,6-7

Lysogenic cycles, in phage X, 418-24
Lysogeny, 165
Lytic cycles, in phage X, 418-24

MacLeod, C, 209

Major histocompatibility complex (MHC), 492
Malaria, 596

Male-specific lethal (msl) gene, 94
Malthus, Thomas, 589
Maniatis, Tom, 461

Manic-depression, and concordance, 546
Map distances, 116-18, 131
Mapping. See also Chromosomal maps; Gene
mapping; Restriction mapping

classical studies of, 66

conjugation, 161

diploid, 110-11, 114-22

haploid, 122-32

human chromosomal maps, 132-40

quantitative trait loci, 537-38

sequencing of human genome, 390-97

transduction, 166-68

transformation, 155-57
Mapping function, 117-18
Map units, 10, 111
Margulis, Lynn, 513
Mariculture, and biotechnology, 398
Mass production methods, and Human

Genome Project, 396
Master-switch genes, 478, 505
Mate-killer infection, 521-22
Maternal effects

definition of, 509

maternal-effect genes, 472, 473, 475

moth pigmentation, 510, 511

snail coiling, 509-10
Maternal-effect genes, 472, 473, 475
Mating types, 124, 469
Matthei,J.H.,306

Maturation-promoting factor (MPF), 50
Maxam, Allan, 383
McCarthy, B. J., 276
McCarty, M., 209

McClintock, Barbara, 120-22, 123, 425, 468
McKusick, Victor A., 99, 137
Meadow weed. See Arabidopsis thaliana
Mean, statistical, 74, 535, 536, 538-39
Mean fitness, of population, 579-80
Measles, and concordance, 546
Measurement

heritability, 544-45, 548

metric units of linear, 48
Medicine, and genetic engineering, 397. See

also Biomedical applications
Meiosis, 62

animals and, 63-64

definition of, 47

origins of term, 49

process of, 55-61

rules of segregation and independent
assortment, 67

significance of, 61, 63

triploids, 197
Meiosis II, 60-61
Meiotic drive, 577
Melanin, 33,34,3^
Melanoma, 485
Mendel, Gregor

classical genetics, 9-10

dominance, 22-23

experiments in genetics, 17-18, 97

genotypic interactions, 30-37

history of genetics, 3, 4, 65

independent assortment, 26-30

multiple alleles, 25-26

nomenclature, 23-24

quantitative traits, 534

rediscovery of, 47

rule of segregation, 18-22, 63

statistics, 71, 74

techniques of study, 8
Merozygote, 159, 409-10
Meselson, Matthew, 220-21
Mesoderm, 470

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

Messenger RNA (mRNA)

cloning, 364-65, 366

function of, 245

polycistronic and monocistronic, 298,
299,300

prokaryotes, 277

synthetic, 306, 307
Messing, J., 386
Metabolic pathways, of color production in

dihybrids, 36. See also Metabolism
Metabolism, and biochemical genetics, 37-38
Metacentric chromosome, 48, 49
Metafemales, 85
Metagon, 521
Metamales, 85

Metaphase, of mitosis, 52, 54, 57
Metaphase I, and meiosis, 59, 60, 62
Metaphase plate, 54
Metastasis, 484
Methionine, 152, 288, 289
Methylation, of DNA, 466-67
Metrical variation, 533
Metric measurements, linear, 48
MHC proteins, 498-500, 501
Michigan Technological University, 441
Microsatellite DNA, 382
Microscope, history of, 3
Microtubule organizing centers, 52
Microtubules, 48, 52
Midparent, and wing length, 539
Migration

Hardy-Weinberg equilibrium, 555

population genetics, 573-74
Mimicry, 604, 605
Minimal medium, 124, 150
Mirabilis jalapa. See Four o'clock plants
Misalignment mutagenesis, 337
Mismatch repair, 343-44, 354
Missense mutations, 338
Mitochondrial DNA (mtDNA), 600
Mitochondrial transfer RNAs, 309
Mitochondrion

cytoplasmic inheritance, 511-15

genetic code of yeast, 311

scanning electron microscope image of, 508

signal hypothesis, 303
Mitosis

definition of, 47

late anaphase, 56

lymphocytic cell, 15

origins of term, 49

process of, 52-55

rules of segregation and independent
assortment, 67

significance of, 55

stages of, 46,57
Mitosis-promoting factor, 50
Mitotic crossing over, 132
Mitotic spindle, 52-53, 55
Mixed families, of codons, 307
Model organisms, 8
Modern linkage map, 393, 395
Molecular chaperones, 303
Molecular evolutionary clock, 600-602
Molecular genetics. See also Cytogenetics

Down syndrome, 193

evolutionary processes, 553

as general area of study in genetics, 4, 5,
10-13

growth in field of, 358

search for genetic material, 205-11
Molecular imprinting, 524
Molecular mimicry, 297
Moloney murine leukemia virus, 374
Monarch butterfly (Danaus plexippus),

604, 605
Monocistronic messenger RNAs, 299
Monoclonal antibody, 493
Monod, Jacques, 409
Monoecious flowers, 87
Monohybrids, 18
Monosomic cells, 190
Monosomy, 195, 200
Monovalent chromosome, 59
Monozygotic twins, 546
Montagnier, Luc, 502
Morgan, Thomas Hunt, 83, 95, 110, 111
Morphogen, 472

Morphological species concept, 591
Morton, Newton E., 135
Mosaicism, 90, 190-92, 194
Moths, and pigmentation, 510, 511
Mouse (Mus musculus). See also Deer mouse

allelic frequency, 600

c-banding of chromosomes, 452

chromosome number, 50

epistasis and color of, 33-34, 35,37

generation interval, 66

genetic bottlenecks, 577

heterozygosity, 599

knockout, 375-76

nuclease-hypersensitive sites of DNA, 446

population density, 597

transgenic, 373

Turner syndrome, 196
Muller,F.,604

Muller, H. J., 88, 121, 325-26, 330
Muller-Hill, B., 412
Mullerian mimicry, 604, 605
Muller's ratchet, 88

Multihybrids, 30,3^

Multilocus control, of inheritance, 532-33

Multilocus selection models, 598-99

Multinomial expansion, 73, 558

Multiple alleles, 25-26

Multiple loci, 532-33, 559-60

Mu particles, 521-22

Muscular dystrophy, 13, 188

Mustard gas, 492

Mutability, of DNA, 207

Mutational equilibrium, 571-73

Mutation rates, 326, 572-73

Mutations and mutants

adaptive, 339

AIDS virus, 504

amino acids, 313

bicoid gene, 472

cancer, 484-87, 506

chemical mutagenesis, 330-31, 336-37

classical genetics, 10

colinearity, 324-25

definition of, 316

evolutionary theory, 5, 88, 312

fluctuation test, 316-17

genetic fine structure, 317-24

Hardy-WEinberg equilibrium, 555

homeotic genes, 477-78

intergenic suppression, 337-38

in vitro site-directed mutagenesis, 333-35

lac operon, 409-12

misalignment mutagenesis, 337

mutator and antimutator, 338

nomenclature, 24

point mutations, 326-28

population genetics, 571-73

selection-mutation equilibrium, 581

sexual reproduction, 88

spontaneous mutagenesis, 329-30

spontaneous versus induced, 325-26
Mutator mutations, 338
Muton, 320

Mutually exclusive events, 71
Mycobacterium tuberculosis, 216, 546
Mycoplasmas, 441

Nail-patella syndrome, 134, 135
Nanos gene, 472-73, 474
Nathans, D., 4, 359
National Academy of Sciences, 370
National Institutes of Health

gene therapy, 397

guidelines on recombinant DNA technol-
ogy, 370

Human Genome Research Institute, 358
Natural selection

effects of, 577-78

evolution, 5, 14

fitness, 577

Hardy-Weinberg equilibrium, 555

models of, 581-82, 584

recessive homozygote, 578-81

selection-mutation equilibrium, 581
Ate/ gene, 504

Negative assortative mating, 554
Negative interference, 118
N-end rule, 433
Neo-Darwinism, 13, 589
Neomycin, 375, 376
Neoplasms, 484
Neurospora crassa (Bread mold)

biosynthesis of niacin, 38-39

chromosome number, 49

generation interval, 66

haploid mapping and, 123, 141

mutations of, 124

ordered spores, 125-27

phenylalanine synthesis, 41
Neutral alleles, 599
Neutral gene hypothesis, 598
Neutral petite, 514

Newt. See Notophthalmus viridescens; Triturus
N-formyl methionine, 288
Niacin, 38, 39
Nickel, as carcinogen, 492
Nicotindamide adenine dinucleotide

(NADH), 93
Nilsson-Ehle,H.,532

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

Nirenberg, Marshall W., 305, 306

Nitrous acid, 331

Nivea gene, 35-36

Nobel, Alfred, 112

Nobel Prize, in genetics, 112-14

Nomenclature, 23-24

Noncoding strand, of DNA, 249

Nondisjunction

mosaicism, 190-92

sex chromosomes, 83
Nonhistone proteins, 448
Non-Hodgkin's lymphoma, 485
Nonhomologous chromosomes, 182-85
Nonhomologous end joining, 344
Non-Mendelian inheritance, 509
Nonparental ditype (NPD), 125, 12 7
Nonparental phenotypes, 111
Nonrandom mating, 560-65
Nonreciprocity, and sex linkage, 96
Nonrecombinant phenotypes, 111, 118
Nonsense codons, 296-97
Nonsense mutations, 337-38
Normal distributions, 535, 536, 538
Northern blotting, 368
Notophthalmus viridescens (newt), 451
Novitski, E.,30
N segments, 496-97
Nuclease-hypersensitive sites, 445-46
Nucleic acids, chemistry of, 211-19
Nucleolar organizer, 53
Nucleolus, 53, 260-61, 278
Nucleoprotein, 47, 440, 442-61
Nucleoside, 212,274
Nucleosome structure, 442-47
Nucleotides, 8

biomedical applications of triple-stranded
chains, 221

conserved sequence, 277-78

DNA replication, 226

excision repair, 341-44

nomenclature, 215

nucleic acids, 211-12, 213-14

techniques of genetic research, 8
Null hypothesis, 77-78
Nullisomic cells, 190
Numerator elements, 85
Nurse cells, 470

Nusslein-Volhard, Christiane, 471
Nutritional-requirement mutants, 327
Nutritional requirements

of bacteria, 151, 153

of fungi, 124

Ochoa, Severo, 305, 306

Okazaki fragments, 225, 228, 229, 230

Oligonucleotide primer, 387

Oncogenes, 13, 484, 488-92

One-gene-one-enzyme rule, 10, 38-39

Onion (Allium cepd), 46

Oocyte, nucleus and follicle cells, 474

Oogenesis, 64

Oogonia, 64

Open reading frames (ORFs), 291

Operator, and gene expression, 408-9, 414

Operon model, 12, 406, 408, 418

Ophioglossum reticulatum. See Indian fern

Ordered spores, 125-27

Original literature, 6-7

Origin recognition complex (ORC), 239

Origin of Species, The (Darwin), 3, 589

Ostreococcus tauri (green alga), 441

Outbreeding, 554, 560

Ovary determining (Od) gene, 86

Overlap, of genetic code, 305, 3 06

Ovum, 64

Oxidative phosphorylation, 512

Oxytricha nova, 456, 457

Pachynema, 56

Page, David, 85, 89

Pair-rule genes, 475

Palindrome, 3 59,3 61

Palmer, J., 274

Panmictic population, 574

Paracentric inversion, 180

Paramecin, 520

Paramecium, 518-19, 525

Parameters, statistical, 538

Parapatric speciation, 592, 593

Parasegments, 470, 471

Parental ditype (PD), 125, 127

Parental imprinting, 524

Parental phenotypes, 111

Parthenogenesis, 64

Partial digest, 379

Partial dominance, 23

Partitioning, of variance, 543-44

Pascal's triangle, 73

Patau syndrome, 194-95

Path diagram, 562-64, 565

Pattern formation, and flower development,
479, 482

Pauling, Linus, 206, 214-15, 602

pBR322 cloning plasmid, 363, 364,372

Pea. See Garden pea; Sweet pea

Pearson, K., 535

Pedigree analysis, 97-102, 562-64, 565, 567

Peer review, and scientific journals, 7

Penetrance, of genetic trait, 97-98, 103, 104

Penicillin, 153, 154, 172, 294

Pepsin, 284

Peptic ulcer, 545

Peptide bond formation, 292

Peptide fingerprint, 284

Peptide map, 284

Peptidyl site (P site), 292

Peptidyl transferase, 292

Pericentric inversion, 180

Permissive temperature, and mutation, 327

Peromyscus maniculatus. See Deer mouse

Peromyscus polionotus (Field mouse), 599

PEST hypothesis, 433

Petal whorl, 479, 482

Petite mutations, 514, 525

Petroleum, as carcinogen, 492

P53 gene, 487, 488

Phage, and cloning with restriction en-
zymes, 561. See also Bacteriophages

Phage labeling, 209-10

PhageM13,3iW

Phage \, lytic and lysogenic cycles in, 418-24,

435-36
Phage resistance, 163
Phage T4, 430

Phaseolus vulgaris. See Bean
Phenocopy, 98
Phenotypes

of bacteria, 151-54

blood groups, 25-26

Drosophila melanogaster, 8

of fungi, 124

independent assortment, 29

lac operon, 436

segregation, 21

transposition, 429-30

of viruses, 154
Phenylalanine, 41
Phenylisothiocyanate (PITC), 285
Phenylketonuria (PKU), 558, 582
Pheromones, 469
Phosphodiester bonding, 214
Phosphoglycerate kinase, 283
Photocrosslinking, 250
Photoreactivation, 340
Phyletic evolution, 590
Phyletic gradualism, 594, 595
Phylogenetic tree, 601
Physical crossover, 120-22
Physical map, and Human Genome Project,

393, 395
Pigeon (Columba livid), 50, 59
Pigment and pigmentation. See also Color;
Skin color

epistasis in mice, 33-34

maternal effects in moths, 510, 511
Pili (fimbriae), 159

Pink bread mold. See Neurospora crassa
Pistil, 87

Pisum sativum. See Garden pea
Plants. See also Arabidopsis thaliana;
Flowers; Garden pea; Zea mays

genetically altered crops, 398

genetic control of development, 479, 482-83

life cycle, 47, 64-66

polyploidy, 198-99

proplastid formation and variegation, 516

sex determination, 87, 90

vectors and cloning of, 371-72
Plaques, bacterial, 151
Plasmids

bacteria in genetic research, 149

E. coli and, 357

F factor, 158

prokaryotic and cytoplasmic inheritance,
522-24

recombinant DNA technology, 13
Plastids, 515
Pleiotropy, 39
Plus-and-minus method, of DNA sequencing,

383, 388
Pneumocystis carinii, 502
Pneumonia, and AIDS, 502
Point centromere, 453
Point mutations, 326-28
Polar body, 64
Polarity, of DNA structure, 218

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-12

Index

Polar mutants, 425

Pole cells, 470

Pol (polymerase) gene, 503

Politics, and ethics in genetics, 6-7

Pollen grain, 65

Poly-A tail, 265, 267

Polycistronic messenger RNAs, 297,300

Polydactyly, 99, 100

Polygenes and polygenic inheritance,

533-35,541
Polymerase chain reaction (PCR), 381-83
Polymerase collisions, 252
Polymerase cycling, 231
Polymerization, of nucleotides, 214, 215
Polymorphisms, 14, 596-99. See also Variation
Polynucleotide phosphorylase, 306
Polyploids, 83, 197-98
Polyribosome, 297
Polysome, 297, 300
Polytene chromosomes, 439, 449, 455
Pond snail. See Limnea peregra
Population, definition of, 553
Population analysis, 564-65
Population density, 597
Population genetics

computer programs to simulate, 583-84

evolution, 553, 589

Hardy-Weinberg equilibrium, 553-60

migration, 573-74

models for, 571

mutation, 571-73

natural selection, 577-84

nonrandom mating, 560-65

small population size, 574-77
Population statistics, 535-40
Populus tremuloides (Quaking aspen), 441
Position effect, 179
Positive assortative mating, 554
Positive interference, 118
Postreplicative repair, of DNA, 344-46
Posttranscriptional modifications, 261-62
Posttranslational control, 433
Posttranslocational state, of ribosome,

293, 298
Postzygotic mechanisms, 592
Potato spindle tuber viroid (PSTV), 272-73
Poultry, and realized heritability, 543- See also

Chicken
PPL Therapeutics, 374
Prader-Willi syndrome, 524
Preemptor stem, 417
PreerJ.,520, 521
Preinitiation complex (PIC), 262
Pretranslocational state, of ribosome, 293, 298
Prezygotic mechanisms, 592
Pribnow box, 248
Primary oocytes, 64
Primary spermatocytes, 63
Primary structure, of protein, 281
Primary transcript, 265
Primase, 225-26
Primers

creation of general-purpose, 386-87, 390

DNA replication, 225-27, 228, 232
Prions, 213

Probability, 71-74, 78
Probability theory, 71
Proband, 98
Product rule, 72
Progeny testing, 21-22
Prokaryotes. See also Bacteria;
Escherichia coli

DNA transcription, 246-55, 277, 465

eukaryotes compared to, 47, 261-62,
298, 440

genomics, 360-66

mitochondrial ribosomal RNA, 512-13

operon model and, 1 2

origins of term, 149

plasmids, 522-24

ribosomes, 301
Prokaryotic vectors, 360-66
Prolactin, and signal peptide, 302
Promoters

efficiency of, 430

lac operon, 408

repressor operon of phage X, 435-36

transcription in eukaryotes, 262-65

transcription in prokaryotes, 248-51
Proofreading, and DNA polymerase, 227
Prophage, 165

Prophase, of mitosis, 52, 53, 57
Prophase I and II, and meiosis, 56-60, 62
Proplastids, 515, 516
Proposita and propositus, 98
Proteasome, 499
Protein degradation, and posttranslational

control, 433
Protein electrophresis, 92-94
Protein-folding problem, 303
Protein-mediated splicing, 268-72
Proteins

DNA and control of, 205-7

DNA repair in E. coli, 348

energy requirements of biosynthesis, 313

methods of depiction, 283

motifs of DNA recognition, 480-81

primary, secondary, tertiary, and quaternary
structures, 281

synthesis of, 245, 283, 287
Proteomics, 396-97
Proto-oncogenes, 13, 489, 491
Prototrophs, 150
Prusiner, Stanley, 213
Pseudoalleles, 319
Pseudoautosomal regions, 87, 95
Pseudodominance, 96, 178
Ptashne, Mark, 412
Punctated equilibrium, 14, 594,595
Punctuation, of genetic code, 305, 306
Punnett, Reginald C, 21, 26
Punnett square diagram, 21, 26, 28, 30, 532
Purine nucleotides, 213,214
Puromycin, 294, 295
Pyrimidine nucleotides, 213,214

Quantitative inheritance, 533, 534, 537-38,

545-46. See also Inheritance
Quaternary structure, of proteins, 281

Rabl, K., 3

Racism, and sociobiology, 606

RAG recombinase, 495, 496

Random genetic drift, 555, 575-76, 585

Random mating, 554-55, 556

Random strand analysis, 122

Raphanobrassica (cabbage-radish cross), 198

Raphanus sativus (radish), 198

Ras oncogene, 491

R-bands, 452

Read-through process, and terminators, 251

Realized heritability, 542-43

Real time, and observation of transcription,

250-51
RecA protein, 345-46, 435
RecBCD protein, 349-51, 354
Recessive homozygote, and natural selection,

578-81
Recessive inheritance and recessive traits, 18,

99-100, 102
Reciprocal cross, 18, 115
Reciprocal translocation, 182, 186, 187
Recombinant DNA technology. See also
Genomics

bacteria, 349-51

bacteriophages, 163-64

double-strand break model of, 347-49

ethical debate on, 370-71, 374-75

genome sequencing, 110

heteroduplex DNA, 351, 352

hybrid DNA, 351

meiosis, 61,63

overview of techniques, 358

restriction endonucleases, 12-13
Recombinant phenotypes, 111, 116
Recombinant plasmids, 360, 362
Recombination nodules, 59
Recombination signal sequences, 494
Recon, 320
Red clover, 26

Red eye, mutation of Drosophila, 24
Reductional division, 59
Redundant controls, and amino acid

operons, 418
Regional centromeres, 453
Regression and regression analysis, 539-40
Regulator gene, 406
Reintiation, of translation, 291
Relaxes mutant, 432
Release factors (RF), 296
Repair, of DNA

categories of systems, 339-40

damage reversal, 340

in E. coli, 348

excision repair, 340-46
Repetitive DNA, 458
Replication, of DNA

control of enzymes, 219

enzymology of, 225-38

in eukaryotes, 231-32, 238-39

mutability, 207

process of, 220-24

structures and, 238
Replication-coupling assembly factor, 444

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

Replicons, of DNA, 224

Replisome model, of DNA replication,

234, 235
Reporter systems, 376-77, 399
Repressible system, 406
Repression, gene expression and maintenance

of, 421-23
Repressor, and gene expression, 406, 408, 409
Repressor transcription, 42 1
Reproductive isolating mechanisms, 592, 607
Reproductive success, 577
Repulsion, of alleles, 111
Rescue experiment, 472
Residues, of amino acids, 281
Resistance transfer factor (RTF), 523
Resolution, of transposition, 428
Resolvase, 428
Restricted transduction, 165
Restriction digest, 378
Restriction endonucleases, 12-13, 359-60
Restriction enzymes, 360-61, 400
Restriction fragment length polymorphisms

(RFLPs), 378, 380-81
Restriction mapping, 377-81
Restriction sites, 359

Restrictive temperature, and mutation, 327
Retinoblastoma, 486-87
Retro transposons, 458
Retroviruses

AIDS, 502-4

cancer, 487-88
Reverse bands, 452
Reverse transcription, 276
Reverse transcriptionase, 276
Reversion, and mutation rates, 326
Rev (regulation of expression of virion

proteins) gene, 504
Rhesus system, and blood types, 32
Rh locus, and maternal-fetal incompati-
bility, 582
Rhoades, M., 516

Rho-dependent terminators, 251, 254, 255
Rho-independent terminators, 251, 255
Rhomboid gene, 466
Rho protein, 251
Ribosomal RNA (rRNA), 245, 246
Ribosome recylcing factor (RRF), 297
Ribosomes

messenger RNA, 245

transcription, 256

translation, 290-91, 292, 299, 301
Ribozyme, 266

Rice, and genetic modified varieties, 398
Rice, William, 89
Rich, Alexander, 219, 221
Rickets, vitamin-D-resistant, 97-98, 102, 509
Ritland,D.,604

RNA. See also Messenger RNA; Ribosomal
RNA; Transfer RNA

antisense, 431

computer model of serine transfer, 243

editing, 275

genetic code dictionary of, 12

genetic material and, 210-11

guide, 275

priming of DNA synthesis, 228, 230, 232

self-replication, 276

types of, 245-46

viruses, 213
RNA phages, 276
RNA polymerase

observation of, 250

prokaryotic and eukaryotic, 247, 260

transcription, 11, 12
RNA replicase, 276
RNA tumor viruses, 276
Roberts, Richard J., 265, 266
Robertson, E, 544
Robertson, W., 185
Robertsonian fusion, 185
Rodriguez, R., 363
Rolling-circle model, of DNA replication,

238, 239
Roslin Institute, 374
Roundworm. See Ascaris spp.
Rous, Peyton, 487
R plasmids, 522-23
RRE (rev response element) gene, 504
R II screening techniques, 321, 322, 323
R222 plasmid, 523

Rule of independent assortment, 27-29, 67
Rule of segregation, 18, 21-22, 63, 67
Rumex acetosa. See Sorrel

Saccharomyces cerevisiae (Baker's yeast)

centromeres, 453

genetically modified, 398

haploid mapping, 123

telomeres, 456

unordered spores, 124-25
Sager, Ruth, 517
Salmonella typhimurium

Ames test for carcinogens, 332

generalized transduction, 166

transposon orientation, 429, 468
Sampling distribution, 74
Sampling error, 574-75
Sanger, Frederick, 284, 383
Sarcomas, 484
Satellite DNA, 451, 452, 458
Scaffold structure, of chromatin, 448, 449
Scanning hypothesis, 291
Schizophrenia, 545, 546
Schizosaccharomyces pombe (fission

yeast), 111
Scientific journals, 7
Scientific method, 5-7, 71
Scrapie, 213

Screening, for drug resistance, 154
Sea urchin, 216
Secondary oocyte, 64

Secondary sources, and scientific literature, 7
Secondary spermatocytes, 64
Secondary structure, of proteins, 281
Second-division segregation (SDS), 128-29,

130, 131
Second messenger, 412
Securin, 54
Segmentation genes, 475-77

Segment-polarity genes, 475
Segregation

classical genetics, 10

meiosis and mitosis, 67

rule of, 18-22,63,67

translocation, 183-85
Segregational load, 597
Segregational petite, 514
Segregation distortion, 577
Selection coefficient, 577
Selection experiments, 541-42
Selection model, 578-79
Selection-mutation equilibrium, 581
Selective medium, 150
Selenocysteine, 310-11
Selfed. See Self-fertilization
Self-fertilization, 17, 21-22,3^
Selfish DNA, 430
Self-splicing, 265-66, 268
Semiconservative replication, of DNA, 220
Semisterility, 181
Sendai virus, 134
Sepal whorl, 479, 482
Separin, 54

Sequence-tagged sites (STSs), 393
Sequoiadendron giganteum (giant

redwood), 441
Seryl tRNA synthetase, 280
Serratia marcescens (bacteria), 151
Sex chromosomes, 83, 191, 197
Sex-conditioned traits, 96
Sex determination, 83-90
Sex-determining region Y (SRY), 86
Sexduction, 16 1
Sex-influenced traits, 96
Sexism, and sociobiology, 606
Sex-lethal (Sxt) gene, 85, 95
Sex-limited traits, 96
Sex linkage, 95-96, 104
Sex-linked inheritance, 100-102, 103, 104
Sex-ratio phenotype, 522
Sex-reversed individuals, 85-86
Sex-switch gene, 85

Sexual reproduction, and evolution, 88-89
Sexual selection, 577
Shapiro, J. A., 427-28
Sharp, Philip A., 265, 266
Sheep (Ovus aries)

cloning of, 1, 374-75

generation interval, 66

realized heritability, 543
Shepherd's purse (Capsella bursa-

pastoris), 3 7
Sheppard,E M.,604
Sherman, P., 606
Sherman Paradox, 187
Shigella, and resistance plasmids, 523
Shine-Dalgarno hypothesis, 290-91, 300
Shoot apical meristem, 479
Short interspersed elements (SINES), 458-59
Shotgun cloning, 364, 367
Shub,D.,274

Shunting, and translation initiation, 291
Sickle-cell anemia, 39, 460-61, 596, 597
Sigma factor, 247

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-14

Index

Signal hypothesis, 301-3

Signal peptidase, 301

Signal peptide, 501,302

Signal recognition particle (SRP), 301

Signal-sequence receptors, 302

Signal sequences, and mitochondrial

genomes, 512
Signal transduction, 467-69
Signal transduction pathway, 467
Silene latifolia. See White campion
Simian immune deficient viruses (SIVs), 502
Singer, B., 211
Single breaks, 178

Single-nucleotide polymorphisms (SNPs), 393
Single-strand binding proteins, 232, 234
Sister chromatids, 52
Site-specific recombination, 418
Site-specific variation, in codon reading,

309-10
Size

of cells and chromosomes, 441-42
of population, 555, 574-77
Skin color, and quantitative inheritance in

humans, 545, 546
Skolnick,M.,391
Sleeping sickness, 275
Small nuclear ribonucleoprotein particles

(snoRNPs), 261, 268, 270
Small nucleolar RNAs (snoRNAs), 261
Small population size, 574-77
SMC proteins, 444
Smith, H., 4, 359
Smoking, and cancer, 492
Snails, and direction of coiling, 509-10, 525.

See also Pond snail; Tree snails
Snapdragon. See Antirrhinum majus
Sobrero, Ascanio, 112
Sociobiology, 14, 603-6
Sociobiology: The New Synthesis (Wilson,

1975), 603
Soft selection, 597

Solenoid model, for chromatin fiber, 448
Somatic-cell hybridization, 134-40
Somatic crossing over, 132
Somatic doubling, 198
Somatic hypermutation, 497
Sonneborn, Tracy M., 519, 520
Sorrel (Rumex acetosa), 90
SOS box, 346
SOS response, 346
Southern, Edwin, 368
Southern blotting, 366-68
Soviet Union, and Lysenko affair, 6-7
Spatzle protein, 467
Specialized transduction, 165
Speciation, and evolution, 589-94, 595
Species

allopolyploidy and cross-fertilization

between, 198
definition of, 553
inversion process and evolutionary history

of groups, 181
speciation, 589-94, 595
Specific transcription factors, 263
Speckles, and nuclear messenger RNAs, 268

Spermatids, 64

Spermatogenesis, 63-64

Spermatogonium, 63

Sperm cells, 63

Spiegelman, S., 246

Spina bifida, and concordance, 546

Spindle pole body, 52

Spiral cleavage, 510

Spirillum, 149

Spliceosome, 268-72

Splicing, of exons, 265

Splicing factors, and introns, 271

Spontaneous mutagenesis, 328-30

Spontaneous mutation, 325-26

Sporophyte, 64

Stability, of mutational equilibrium, 571-72

Stabilizing selection, 578

Stadler,L. J., 325-26

Stage-specific puffs, 449-50

Stahl, Franklin W., 220-21

Stalin, Joseph, 6

Stamen, 87

Stamen whorl, 479, 482

Stanbridge, E., 487

Standard deviation, 74, 536, 538-39

Standard error of the mean (SE), 539

Standard method, and Human Genome

Project, 392
Stanford University, 371
Starvation, and attenuator regulation, 417
Statistics and statistical methods, 74-78
Stature, and heritability of human traits, 545
Steizjoan A., 268, 269
Stem-loop structure, 252
Stepladder gels, 386, 387
Stern, Curt, 132
Stochastic events, 71
Streptococcus pneumoniae , 209
Streptomyces antibioticus, 151
Streptomyces griseus, 151
Streptomycin, 153, 294, 517-18
Stringent factor, 432
Stringent response, 432
Strobel, S.,221

Strong inference, and scientific method, 5
Structural alleles, 318-19, 320
Sturtevant, Alfred H., 4, 121, 186
Submetacentric chromosome, 48, 49
Substrate transition, and DNA replication, 329
Subtelocentric chromosome, 48
Sucrose density-gradient centrifugation, 256
Sulfur bacteria, 441

Summer squash (Cucurbita pepo), 37, 40-41
Sum rule, 72

Supercoiling, 234, 236, 237
Supergenes, 181

Superinfection, and repression, 422
Superrepressed mutation, 412
Suppressive petite, 514
Suppressor gene, 337
Suppressor of Stellae (Su or Ste) gene, 87
Surveillance mechanisms, and cell cycle, 51
Survival of the fittest, 589. See also Fitness
Sutton, Walter, 4, 66, 83, 1 10
SV40 virus, 371,372, 378

Svedberg,T.,256

Svedberg unit (S), 256

Sweet pea (Lathyrus odoratd), 3 7

Swine, and realized heritability, 543

SWI/SNF complex, 446-47

Sympatric speciation, 592-94

Synapsis, and meiosis, 56

Synaptonemal complex, 58

Syncitium, 470

Syn-expression group, 465

Synteny test, 136

Synthetic codons, 306-7

Synthetic medium, 150

Synthetic messenger RNAs, 306, 307

SzostakJ.,347

TACTAAC box, 268

Tanksley, Steven, 538

Tat (trans-activating transcription factor),

503-4
TATA-binding protein (TBP), 262
TATA box, 262, 263
Talmud, 95

TAR (trans-activating response element), 504
Tatum, Edward L., 10, 38-39, 157, 161
Tautomeric shifts, 328, 329
Taylor, J., 440
Tay-Sachs disease, 23, 582
T/B-cell lymphoma, 485
TBP-associated factors (TAFs), 262
T-cell receptors, 492, 498, 499
Telocentric chromosome, 48
Telomerase, 454-55, 456-57
Telomeres, 178, 454-57
Telophase, of mitosis, 52, 54-55, 56,57
Telophase I, of meiosis, 59-60, 62
Telson, 471
Temin, Howard, 276
Temperate phages, 165
Temperature-sensitive mutants, 327
Template, and DNA replication, 220
Template strand, of DNA, 249
Template transition, and DNA replication, 329
Termination

of DNA replication, 236, 238

of translation, 296-301
Termination signals, for transcription, 248
Terminators, and transcription, 248, 251-55
Terminator sequence, 248
Terminator stem, 416
Terramycin, 153

Tertiary structure, of proteins, 281
Testcross and testcrossing. See also Three-
point cross; Two-point cross

dihybrids, 33

examples of, 140-41

multihybrids, 30

segregation rule, 22
Testing, for AIDS, 504
Testis-determining factor (TDF), 85
Tetracycline, 153, 294

Tetrad analysis, and haploid mapping, 122-32
Tetrads, 59
Tetrahymena thermophila, 269, 272-73

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-15

Tetranucleotide hypothesis, 216
Tetraploids, 83, 197
Tetratype (TT), 125, 12 7
Thalassemia, 460, 461, 582
Thermus aquaticus (hot-spring bacteria), 382
Theta structure, of DNA, 223-24
Thiomargarita namibiensis (sulfur bacte-
ria), 441
3-hydroxyanthranilic acid, 38, 39
Three-locus control, 532
Three-point cross, 114-16, 118-20
Thymidine kinase (TK), 134-35
Thymine, 213,275

Time frame, for population equilibrium, 580-81
Tissue-specific puffs, 450
T-loop, and telomeres, 456, 458
T-lymphocytes, 464
Tobacco mosaic virus, 210-11, 212
Tobacco plants, and recombinant DNA

technology, 377
Toll gene, 473

Topoisomers and topoisomerases, 236, 237
Torpedo gene, 474-75
Torso gene, 473
Tortoise. See Geochelone elephantopus

vanderburgh
Totipotent cells, 469
Touchette, Nancy, 213
Trailer, and RNA transcription, 255
Trans-acting mutations, 411
Trans configuration, 111, 133
Transcription. See also Gene expression

classical genetics, 11-12

control of in eukaryotes, 465-69

definition of, 244

early and late in phage infection of E. coli,
420-21

eukaryotic DNA, 260-75

flow of genetic information and, 244,
275-76

observation of in real time, 250-51

prokaryotic DNA, 246-55

prokaryotic initiation and termination sig-
nals for, 248

transfer RNA, 256-60

translation, 297, 299

types of RNA, 245-46
Transcription factors, 430, 466-69
Transducing particle, 166
Transduction, 154, 165-68, 172
Transfection, 372
Transfer operon (tra), 523
Transfer RNA (tRNA)

function of, 245

transcription, 256-60

translation, 281-83, 286-88, 292
Transfer RNA loops, 258, 260
Transformation

bacteria and bacteriophages, 154-57

DNA and, 155,156, 209

in eukaryotes, 371
Transformation mapping, 155-57, 172
Transformed cancers, 484
Transgenic organisms, 372, 397
Transient polymorphism, 597

Transition mutation, 329, 331, 336
Translation. See also Gene expression

classical genetics, 12

control of, 430-33

definition of, 244

DNA involvement, 276

energy cost of, 297

genetic code, 304-12

information transfer, 281-303

initiation of, 288-91

start and stop signals of, 313

transcription, 297, 299
Translational control, 430-33
Translocation

segregation, 183-85

translation and elongation, 292-93, 296
Translocation channel (translocon), 301
Transposable genetic elements, 425-30
Transposase, 428

Transposition, mechanism of, 427-28
Transposons, 425-30, 468, 469
Transversion mutation, 329-30, 336
Tree snail. See Liguus fasciatus
Trihybrids, 30,33,40
Triplet nature, of genetic code, 304-5
Triplex DNA, 221, 222
Triple-X female, 197
Triploids, 83, 197
Trisomic cells, 190
Trisomy 8, 195
Trisomy 13, 194-95
Trisomy 18, 194, 195
Trisomy 21, 192-94
Triticum wheat, 198
Triturus (newt), 260
Trp operon, 413-18
Trp RNA-binding attenuation protein (TRAP),

417-18
True heritability, 543
Trypanosomes, and RNA editing, 275
Trypsin, 284

Tryptophan, 39, 413-14, 416-17
Tuberculosis, 216, 546
Tumor-inducing (Ti) plasmid, 371
Tumors, and cancer, 484
Tumor-suppressor genes, 485-87
Turner syndrome, 85, 195-96
Twin spots, and mitotic recombination, 132
Twin studies, of concordance, 546
Twist gene, 466
2-aminopurine (2AP), 331
Two-dimensional chromatography, 284-86
Two-locus control, 531-32
Two-point cross, 110-11, 114
Type I and type II errors, 75
Typological thinking, 590
Tyrosine, 36
Tyrosine kinases, 491

UAA, UAG, and UGA stop codons, 309
Ubiquitin, 5 1
Ultraviolet (UV) light

as carcinogen, 492

dimerization, 340

Unequal crossing over, 186
Uninemic chromosome, 440, 442
Unique DNA, 458

Universality, of genetic code, 308-11
University of California, 371
Universisty of Toronto, 441
Unmixed families, of codons, 307
Unordered events, 72
Unordered spores, and yeast, 124-25
Unusual bases, of transfer RNA, 257-58
Upregulation, of cell growth, 487
Upstream direction, of DNA transcrip-
tion, 249
Upstream element (UP element), 249
Uracil, 213, 215, 344
Uracil-DNA glycosylase, 342
U-tube experiment, 158
UvrC and uvrD genes, 341, 342

Van Breden, E., 3

Vancomycin, 294

Van Leeuwenhoek, Anton, 3

Variable-number-of-tandem-repeats

(VNTR), 381
Variance

partitioning of, 543-44

statistical methods, 536, 538-39
Variation. See also Polymorphisms

additive models, 533, 538, 548

causes of, 316

chromosome number, 190-99

codon-reading and site-specific, 309-10

continuous, 17, 531, 533

Darwinian evolution, 313, 589, 596

discontinuous, 17, 531

frequency-dependent selection, 597

hetero2ygote advantage, 596-97

multilocus selection models, 598-600

patterns of, 600-603
Variegation, of color, 179
Vavilov, Nikolai, 6
V(D)J joining, 496
VenterJ. Craig, 358, 396
Viceroy butterfly (Limenitis archippus),

604, 605
Viciafaba. See Broad bean
Victoria (Queen of England), 100-102
Vif gene, 504
Vinyl chloride, 492
Virchow, Rudolf, 3
Virion, 154

Viroids, and introns, 272-73
Viruses. See also Bacterial viruses; Retro-
viruses

cancer, 13,487-92

cultivation of, 151

life cycle, 64

operon model, 12

phenotypes of, 154

recombinant DNA technology, 371

RNA, 213
Vitamin-D-resistant rickets, 97-98, 102
V-J joining, 494, 495, 497
Y-myc gene product, 491

Tamarin: Principles of
Genetics, Seventh Edition

Back Matter

Index

1-16

Index

Von Mohl, Hugo, 3

Von Nageli, C, 3, 48

Vonnegut, Kurt, 213

Von Tschermak, Erich, 4, 17

Vpr gene, 504

Vpu gene, 504

Vries, Hugo de, 4, 17

Wahlund effect, 574

Waldeyer,W.,3,48

Wallace, Douglas, 514, 597

Walzer, Stanley, 197

Washburn, Linda, 86

Watson, James, 4, 10, 11, 205, 215, 216,
328, 330

Weiling,F.,31

Weinberg, W, 13, 553

Weismann, August, 3

Western blotting, 368, 369

Wheat, and grain color, 531-33. See also
Triticum wheat

White campion (Silene latifolia), 90

Whole-genome shotgun method, 392

Wieschaus, Eric E, 471

Wild-type

metabolic pathways in Neurospora, 41
phenotypes of Drosophila, 23-24

Wilkins, Maurice, 205

Williams, G., 604-5

Williams, R., 210-11

Wilm's tumor, 487

Wilson, Edward O., 603

Wobble hypothesis, 307-8, 309, 310, 312

Wollman,E., 159-60, 161

Wood and leather dust, 492

Wright, Sewall, 4, 13, 553

Wynne-Edwards, V C, 603-4

X chromosome, 90-95, 100

Xeroderma pigmentosum, 343, 484

X-gal, 386-87

X inactivation center (XIC), 91

X linkage, 95-96, 132-33

XO-XX system, of sex determination, 86

X-ray crystallography, of DNA, 215-16

XY system, of sex chromosomes, 83, 87, 90

XYY karyotype, 196-97

Yanofsky, Charles, 324, 415
Y chromosome, 87, 88-89
Yeast. See also Saccharomyces cerevisiae;
Schizosaccharomyces pombe

antibiotic resistance, 515

centromeres, 454

computer model of transcriptional fac-
tor, 244

genetic code and mitochondria, 311

mating type, 469

mitochondrial DNA, 513, 515

nucleotide excision repair, 343

petite mutations, 514, 525

testcrossing, 111, 141

transfer RNA, 259

unordered spores, 124-25

vectors and cloning, 370
Yeast artificial chromosomes (YACs), 370
Y-junctions, of DNA, 223, 232, 234, 236
Ylinked chromosomes, 95
Yunis, J., 486-87

ZDNA, 219,220,467
Zea mays (corn)

Ac-Ds system, 468

cytoplasmic inheritance, 516-17

epistasis, 34-35,37

generation interval, 66

genotypic interactions and phenotypes, 32

life cycle of, 65, 66

meiosis, 60, 62

testcross of trihybrid, 40
Zeigler,D.,252
ZFY gene, 86
Zimm, B., 440
Zinc finger, 480, 481
Zinder,N D., 166
Zuckerkandl, E., 602
ZW system, 87
Zygonema, 56

Zygotes, and gene pool concept of forma-
tion, 555
Zygotic induction, 165
Zygotic selection, 577

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