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Full text of "Fundamentals of Genetics"

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R.)K K I M . IMlAdN 







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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 



XV 



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: ^ 

o 

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. 

(© Art Wolfe/Tony Stone Images.) 



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? 



7 




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 
Companies, 2001 



8 



Chapter One Introduction 








.^- 



f 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



I. Genetics and the 
Scientific Method 



1. Introduction 



©TheMcGraw-Hil 
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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 



o 



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 



A 



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 



V 



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 
Genetics, Seventh Edition 



I. Genetics and the 
Scientific Method 



1. Introduction 



©TheMcGraw-Hil 
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10 



Chapter One Introduction 




ATP 



ADP 



a/ 



Hexokinase 



Glucose-6-phosphate 



Phosphoglucose 
isomerase 



Fructose-6-phosphate 



ATP 



ADP 



j/ 



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 
Genetics, Seventh Edition 



I. Genetics and the 
Scientific Method 



1. Introduction 



©TheMcGraw-Hil 
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Classical, Molecular, and Evolutionary Genetics 



11 





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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I. Genetics and the 
Scientific Method 



1. Introduction 



©TheMcGraw-Hil 
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12 



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 


He 


ACU 


Thr 


AAU 


Asn 


AGU 


Ser 


AUC 


He 


ACC 


Thr 


AAC 


Asn 


AGC 


Ser 


AUA 


He 


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 




Ribosomes 



RNA 



Nascent protein 



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- 



Tamarin: Principles of 
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I. Genetics and the 
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©TheMcGraw-Hil 
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Classical, Molecular, and Evolutionary Genetics 



13 



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 




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 



Tamarin: Principles of 
Genetics, Seventh Edition 



I. Genetics and the 
Scientific Method 



1. Introduction 



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14 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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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 



28 



16 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
Companies, 2001 



Mendel's Experiments 



17 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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18 



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, 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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Segregation 



19 



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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II. Mendelism and the 


2. Mendel's Principles 




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Genetics, Seventh Edition Chromosomal Theory 



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20 



Chapter Two Mendel's Principles 




X 




Tall 



Dwarf 




X Self 



Tall 





Tall 



Dwarf 



3 : 1 



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



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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Segregation 



21 



Gametes 




X Self 



Tt 



Gametes 



T or t 



or 



1 


r 


> 


r 




> 


r 


i 77 


Tt 


", 


tt 




i 
3/4 




1/4 






Tc 


ill 




Dw 


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 


t 






Ovule 
Tt 


f + + 

T t T 

T T 
TT Tt 

1:2: 

Diagrammatic 

(Punnett square) 

Pollen 

T t 


+ 
t 

i 

tt 
1 


TT 
1 




W T 

3 


TT 


Tt 


Tt tt 


> 

o t 


Tt 


tt 


: 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 



1 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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22 



Chapter Two Mendel's Principles 



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 



x 



Gamete of aa 



Offspring 



„ . Gamete 

AA >- ( a ) X (a 





Aa 

(dominant phenotype) 



Aa 



Gamete 





X a 




Aa 

(dominant phenotype) 

aa 

(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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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Nomenclature 



23 





Red 



x 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
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24 



Chapter Two Mendel's Principles 




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. 








dp 




D 




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. 



Tamarin: Principles of 
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Chromosomal Theory 



2. Mendel's Principles 



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Multiple Alleles 



25 



Table 2.2 ABO Blood Types with Immunity Reactions 



Blood Type Corresponding 






Reaction of Red 


Reaction of Red 


to Antigens on Red 






Cells to Anti-A 


Cells to Anti-B 


Blood Cells 


Antibodies in Serum 


Genotype 


Antibodies 


Antibodies 


O 


Anti-A and anti-B 


ii 


— 


— 


A 


Anti-B 


I A I A or I A i 


+ 


— 


B 


Anti-A 


ff or A 


— 


+ 


AB 


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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II. Mendelism and the 
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2. Mendel's Principles 



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Chapter Two Mendel's Principles 



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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Chromosomal Theory 



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Independent Assortment 



27 





Round, yellow 
{RRYY) 



X 



Wrinkled, green 
(rryy) 




XSelf 



Round, yellow 
{RrYy) 




Q 



Round, yellow 
(315) 

(RRYY; RRYy; 
RrYY;RrYy) 



Round, green 
(108) 
(RRyy; Rryy) 



Round, yellow 
RRYY 



X 



Wrinkled, green 
rryy 



Gametes 







.0 



RrYy 



Gametes 



Qty) (V) 



1:1:1:1 



Pollen 



RY 



Ry 



rY 



F 2 RY 



Ry 



CO 
_CD 

> 

O 



rY 



ry 



RRYY 




RRYy 




RrYY 




RrYy 




RRYy 




RRyy 

c 



RrYy 




Rryy 

C 



RrYY 




RrYy 




rrYY 



wfi/ 



rrYy 



ry 



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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II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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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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II. Mendelism and the 
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2. Mendel's Principles 



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Independent Assortment 



29 




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 


1 


2,001 


3.01 


1 


224 


3.15 


1 


299 


2.95 


1 


152 


2.82 


1 


207 


3.14 


1 


277 


2.84 


1 


5,010 


2.98 


1 



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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Chromosomal Theory 



2. Mendel's Principles 



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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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2. Mendel's Principles 



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Genotypic Interactions 



31 




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 


4 


8 


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 


4 


8 


2 n 


Number of different genotypes (or phenotypes, if no 


3 


9 


27 


r 


dominance exists) 











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Chapter Two Mendel's Principles 



Genotype 



RRYY 



RRYy 



RRyy 



RrYY 



RrYy 



Rryy 



rrYY 



rrYy 



rryy 




Q 







Ratio of 
genotype 
in F„ 



16 



Ratio of 
phenotype 
in F n 





O 






16 



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 



o 

E 



l A d 



l a D 



fB 



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 



33 



Gametes 



RrYy 


X 


rryy 


® 




© 


© 


© 




© 




RY Ry 


rY ry 


RrYy 


Rryy 


rrYy 


rryy 


(55) 


(51) 


(49) 


(52) 



ry 



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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Chapter Two Mendel's Principles 





Rose 



Pea 





Walnut Single 

Figure 2.20 Four types of combs in fowl. 



First white variety X Second white variety 
AAbb aaBB 



AB 



AB 



Ab 



aB 



ab 



Purple 
AaBb X Self 



Ab 



aB 



ab 



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. 



RP 



Ftp 



rP 



rp 



Summary 



RP 



Walnut comb 
RrPp 

F 1 X F 1 



Rp 



rP 



rp 



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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35 




Albino 
AAc a c a 



Black hair 




White hair 



Agout 
AaCc a 




Black hair with 
yellow stripes 



AC 



AC 



A& 



aC 



ac 



AACC 




Agouti 



AACcf 




Agouti 



AaCC 




Agouti 



AaCcf 




Agouti 



A& 



aC 



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" 



+ 



OH 




NH 



<t>. 



<t> 



Thyroxine 



P-Hydroxyphenyl 
pyruvate 



Homogentisic 
acid 







CH 2 — CH — COCT 



NH. 



OH 




C0 2 + H 2 



■> Melanins 



CH 2 — CH — COO" 
NH+ 

3 



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 



37 



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 



X 



White 
nnee 




Red X 
NnEe 



Self 




Red 


Pink 


White 


NNEE 


NNee 


nnEE 


NNEe 


or 


nnEe 


NnEE 


Nnee 


or 


or 




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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Chapter Two Mendel's Principles 



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 




r 


acid 














Kvni irpnin^ 


I *w 


9 






^ 







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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Biochemical Genetics 



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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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2. Mendel's Principles 



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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- 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



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Companies, 2001 



Exercises and Problems 



41 



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 






& 

& 



A 



<F 






,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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
Companies, 2001 



42 



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 



y 

Hw 

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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
Companies, 2001 



Exercises and Problems 



43 



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 



a. 


Yellow, tall X 
yellow, tall 


89 


b. 


Yellow, short X 
yellow, short 





c. 


Green, tall X 
yellow, short 


21 



31 



42 



20 



33 







24 



10 



15 



22 



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 


P 






i 


a. Orange, 


83 


26 












crossveins 














X orange, 














crossveins 














b. Red, 


20 


18 




65 




63 


crossveins 














X red, 














crossveinless 














c. Red, 










74 




81 


crossveinless 














X red, 














crossveins 














d. Red, 


28 


11 




93 




34 


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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
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44 



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 



Fi 



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 wild-type 

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 


32 


White-bulbed 


9 



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 



X 

I 



nonworker 



Fi: 



all nonworkers of both sexes 
Fj female X worker male 

4 



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? 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



2. Mendel's Principles 



©TheMcGraw-Hil 
Companies, 2001 



Critical Thinking Questions 



45 



Mutants 



Additives 


1 


2 


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 

A 

B 

C 

D 

E 



+ 



+ 
+ 
+ 

+ 



+ 



+ 



+ 


+ 


+ 


— 


+ 


+ 


+ 


— 


+ 


+ 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



©TheMcGraw-Hil 
Companies, 2001 




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. 

(© Andrew Syred/Tony Stone Images.) 



46 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



©TheMcGraw-Hil 
Companies, 2001 



Chromosomes 



47 



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 



4^ 



Growth 



1 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



©TheMcGraw-Hil 
Companies, 2001 



48 



Chapter Three Mitosis and Meiosis 



Table 3.2 


Metric Units of Linear Measurement 


Unit 


Abbreviation 


Size 


meter 


m 


39.37 U.S. inches 


centimeter 


cm 


10~ 2 meter 


millimeter 


mm 


10~ 3 meter 


micrometer 


|jim 


10~ 6 meter 


nanometer 


nm 


10~ 9 meter 


Angstrom 


A 


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 



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♦ 



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 

11 



-?, 



"4 



u 



U 11 It M U 



U il if 



*! 

»* 




M 



io 


11 

ft 


12 

II 


16 


17 


Ifi 




e 






> • 


' t 




19 


20 
F 




« » 


a* 




21 


22 



<s 



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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II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



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Chapter Three Mitosis and Meiosis 



Table 3.3 Chromosome Number for Selected 
Species 



Species 


In 


Human being (Homo sapiens) 


46 


Garden pea (Pisum sativum) 


14 


Fruit fly (Drosophila melanogaster) 


8 


House mouse (Mus musculus) 


40 


Roundworm (Ascaris sp.) 


2 


Pigeon (Columba livia) 


80 


Boa constrictor (Constrictor constrictor) 


36 


Cricket (Gryllus domesticus) 


22 


Lily (Lilium longiflorum) 


24 


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 



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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 

of 

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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3. Mitosis and Meiosis 



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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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3. Mitosis and Meiosis 



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Mitosis 



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Centrosome 

\\ i// 



Aster 






c 



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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3. Mitosis and Meiosis 



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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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3. Mitosis and Meiosis 



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Meiosis 



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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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Chromosomal Theory 



3. Mitosis and Meiosis 



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56 



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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Meiosis 



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- 






I 



^^^p."™ 




t 







(a) Interphase 



(b) Early prophase 



(c) Late 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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3. Mitosis and Meiosis 



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Meiosis 



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E 



_s\_ 



3 



3 



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. 

















V 




to 


k 









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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3. Mitosis and Meiosis 



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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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3. Mitosis and Meiosis 



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Meiosis 



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Metaphase II 






\ 



/ 



K 



1 > 




Anaphase II 





Telophase II 




#•♦♦♦♦♦♦ .•'••^ &i 




* 



f 




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 



4 



,/ 



Pachytene 



r 



*% 



Diakinesis 





Metaphase I 



•./* 




i»v 



Telophase I 



% 



Anaphase I 



Interphase 




Prophase II 
(early) 


r 

V 

■ 

1 

Prophase II 
(late) 


• 
*** 

Metaphase II 


1 

* 

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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3. Mitosis and Meiosis 



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Meiosis in Animals 



63 



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. 





M 



Meiosis 
II 



or 




8! 



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 





A 



* 



Second 
meiotic 
division 



Differentiation 



A 









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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3. Mitosis and Meiosis 



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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 



65 



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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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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3. Mitosis and Meiosis 



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Exercises and Problems 



67 



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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Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



©TheMcGraw-Hil 
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68 



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) 



^ 



N 



k 






L 



s I/, 

i \ 



(f) 



k, 



\ i 

IN 



/A 



i \ 








Figure 1 Stages in nuclear division. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



3. Mitosis and Meiosis 



©TheMcGraw-Hil 
Companies, 2001 



Critical Thinking Questions 



69 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
Companies, 2001 




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 



70 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
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Probability 



71 



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 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
Companies, 2001 



72 



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, 



2! 



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 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
Companies, 2001 



Use of Rules 



73 



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



6! 



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 






1 


2 


3 


4 


5 


6 


B 


G 


G 


G 


G 


G 


G 


B 


G 


G 


G 


G 


G 


G 


B 


G 


G 


G 


G 


G 


G 


B 


G 


G 


G 


G 


G 


G 


B 


G 


G 


G 


G 


G 


G 


B 



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 = 



4! 



(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 

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 = 



n\ 



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 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



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74 



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 = 



4! 



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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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
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Statistics 



75 



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



n 


= 4 








w = 8 






n 


= 40 




No. Tall Plants 




Probability* 


No. 


Tall Plants 


Probability* 


No. 


Tall Plants 




Probability* 


4 




81 

256 


- 0.32 




8 


0.10 




40 




0.00001 


3 




108 
256 


= 0.42 




7 
6 


0.27 
0.31 




39 
38 




0.0001 
0.0009 


2 




54 
256 


= 0.21 




5 

4 


0.21 
0.09 




30 




0.14 


1 




12 

256 

1 


= 0.05 




3 

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 





J 



\ i i r 

25 50 75 100 
Percent tall 



1r- 

0.9 - 
S 0.8 - 



CO 



0.7 - 



2 0.6 - 



0.5 - 
0.4 - 



o 

c 

CD 

o- 0.3 h 

(D 

£ 0.2 h 




A7 = 8 



T 








J 



i i i r 

25 50 75 100 
Percent tall 



1 




0.9 




& 0.8 




1 0.7 


A7 = 40 


.Q 




2 0.6 




Q. 




- 0.5 




o 




S 0- 4 




o- 0.3 




o 




lL 0.2 




0.1 



i 


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 



X 



X 



(Q - By 

E 



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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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



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Statistics 



11 



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 


Plants 


Total 


Observed numbers (O) 


787 


277 


1,064 


Expected ratio 


3/4 


1/4 




Expected numbers (£) 


798 


266 


1,064 


O- E 


-11 


11 




(O - Ef 


121 


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 


Plants 


Total 


Observed numbers (O) 


787 


277 


1,064 


Expected ratio 


1/2 


1/2 




Expected numbers (i?) 


532 


532 


1,064 


O-E 


255 


-255 




(O - Ef 


65,025 


65,025 




(O - Ef/E 


122.23 


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 


1 


0.000 


0.004 


0.064 


0.455 


1.642 


3.841 


6.635 


2 


0.020 


0.103 


0.446 


1.386 


3.219 


5.991 


9.210 


3 


0.115 


0.352 


1.005 


2.366 


4.642 


7.815 


11.345 


4 


0.297 


0.711 


1.649 


3.357 


5.989 


9.488 


13-277 


5 


0.554 


1.145 


2.343 


4.351 


7.289 


11.070 


15.086 


6 


0.872 


1.635 


3.070 


5.348 


8.558 


12.592 


16.812 


7 


1.239 


2.167 


3.822 


6.346 


9.803 


14.067 


18.475 


8 


1.646 


2.733 


4.594 


7.344 


11.030 


15.507 


20.090 


9 


2.088 


3.325 


5.380 


8.343 


12.242 


16.919 


21.666 


10 


2.558 


3.940 


6.179 


9.342 


13-442 


18.307 


23.209 


15 


5.229 


7.261 


10.307 


14.339 


19.311 


24.996 


30.578 


20 


8.260 


10.851 


14.578 


19.337 


25.038 


31.410 


37.566 


25 


11.524 


14.611 


18.940 


24.337 


30.675 


37.652 


44.314 


30 


14.953 


18.493 


23.364 


29.336 


36.250 


43.773 


50.892 



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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



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78 



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: 



X 



= 2 



(Q - Ey 

E 



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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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



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Exercises and Problems 



79 



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 


87 


31 


30 


12 


160 


Numbers (O) 












Expected Ratio 


9/16 


3/16 


3/16 


1/16 




Expected 


90 


30 


30 


10 


160 


Numbers (E) 












O- E 


-3 


1 





2 




(O - Ef 


9 


1 





4 




(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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II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



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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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



4. Probability and Statistics 



©TheMcGraw-Hil 
Companies, 2001 



Critical Thinking Questions 



81 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
Companies, 2001 




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 




87 



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 



97 



99 
99 
100 



104 



Three generations of a family. 

(© Frank Siteman/Tony Stone Images.) 



82 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
Companies, 2001 



Sex Determination 



83 



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 \\ 



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II II It 



It 



13 



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19 


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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 

X 



Two autosomal 


Two autosomal 


sets plus 


sets plus 


XX 


XY 


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.) 



9 





Figure 5.3 Chromosomes of Drosophila melanogaster. 



Tamarin: Principles of 
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Chromosomal Theory 



5. Sex Determination, Sex 
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Sex Determination 



85 



Table 5.1 


Data 


Supporting Bridges's 


Theory 


of Sex Determination 


by 


Genie Balance in 


Drosophila 


Number of 

X Chromosomes 


Number of 
Autosomal Sets (A) 


Total Number 
of Chromosomes 


— Ratio 
A 


Sex 


3 




2 




9 




1.50 


Metafemale 


4 




3 




13 




1.33 


Female 


4 




4 




16 




1.00 


Female 


3 




3 




12 




1.00 


Female 


2 




2 




8 




1.00 


Female 


1 




1 




4 




1.00 


Female 


2 




3 




11 




0.67 


Intersex 


1 




2 




7 




0.50 


Male 


1 




3 




10 




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 



Tamarin: Principles of 
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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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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 



KJ 



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 



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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) 



X 



>< Degeneration 
of the Y 

? 

chromosome 



X 



X 



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. 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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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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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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Dosage Compensation 



91 



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. 



Tamarin: Principles of 
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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 



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.) 







1 


2 
M 


3 
M 


4 
J 


5 
J 


6 
H 


7 
H 


8 


9 


10 


J 


G 


G 


J 


Q 


Q 


M 


M 


L 


J 


M 


M 


J 


J 



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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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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Dosage Compensation 



93 




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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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



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94 



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, 



LDH, 



LDH, 




LDH, 



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



Normal Myocardial Infectious Acute 
serum infarction hepatitis leukemia 



LDH. 



LDH, 



LDH, 



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 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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Sex Linkage 



95 




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.) 




Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
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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 



x 



6 
White eye 



6* and 9 
Wild-type 



Wild-type 



6 

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. 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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Pedigree Analysis 



97 



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 



x+x w 

Wild-type 9 



6 



Y 



X + Y 

Wild-type S 






F, » 



X 



w 



x + 



O* 



x+x w 

Wild-type 9 



X^Y 

White eye 6* 






X + 



F 2 9 



X 



w 



9 6 

Wild-type X Wild-type 



6 



x 



+ 



x + x + 

Wild-type 9 



x+x w 

Wild-type 9 



X + Y 

Wild-type 6 






White eye 6 



X + 



F 2 9 



X 



w 



9 6 

Wild-type X White eye 



Y 

6 



x 



w 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
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Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis 




Barred 
z b z b 



X 




Nonbarred 
Z b W 



9 

Wild-type 

X + X + 



X 



6 
Bobbed 

vbbybb 



X H 



X bb 


ybb 


X+X bb 

Wild-type 9 


X+Y bb 

Wild-type 6 



x + 



6 



<bb 



B 



B 



W 



W 



Z B z b 

Barred $ 


Z B W 
Barred 9 




6 
Barred 

Z B d 


9 

X 

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 



bb 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
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Analysis 



©TheMcGraw-Hil 
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Pedigree Analysis 



99 



9 

Bobbed 

ybbybb 



X 



6 

Wild-type 
X + Y + 



X 



bb 



X bb x+ 

Wild-type 9 


x bb Y+ 

Wild-type 6 



6 



x 



bb 



Y + 



X 



+ 



9 



x 



bb 



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 



O 



o 



OK) 



Male 
Female 
Affected male 

Affected female 



Mating (marriage 
line) 




Identical twins 





U) 



Parents 



Siblings 




OO 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
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100 



Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis 



oa 



Ofl 



OA 



kO 





I 
II 



ChH 

5 



5 



O 

2 3 4 



6 



3tO 

7 




^ 



III 



» o 



MO 



8 9 



10 



11 




C12 



5 



IV 



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° 






K 

1 




K 

r 


> 


3 

3 


o 


J 


2 


3 




c 


X 


JOO 


H 


)^]( 1 )6 J 



II 



III 



IV 



1 23456789 10 



Figure 5.21 Part of a pedigree of hypotrichosis (early hair loss). 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
Companies, 2001 



Pedigree Analysis 



101 



Leopold Duke 
of Albany 
(1853-84) 



K) 



GhO 



6 6 



Edward Duke 
of Kent 
(1767-1820) 



U) 



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 



t 



J 



King Edward VII of England 
(1841-1910)+ 



O 

Victoria 
(1840-1901) 



©■ 



O O 
I 

Alix (Alexandra) 
(1872-1918) 



6 
t 

Olga ' 

(1895-1918) 



t 



6 (0iH 



Tsar Nicholas II of Russia 
(1868-1918) 



o o o 

/ 1 \ 

Marie 
(1899-1918) 



I 

Alexis 
(1904-18) 



Tatiana Anastasia 

(1897-1918) (1901-18) 



Beatrice 
(1857-1944) 



Irene 
(1866-1953) 



t 



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.) 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
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102 



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^ 



1 



o 

2 3 



1 



O 

2 



OrO 6 

5 6 



O 

4 



II 



O 

7 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
Companies, 2001 



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 



9 



xy 



x+ 



9 



x^ 



9 




8 


fellow 


X 


Wild-type 


X*X* 




X + Y 



s 



x 



+ 



xyx + 

Wild-type 9 


X^Y 

Yellowed 



3 



xy 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



©TheMcGraw-Hil 
Companies, 2001 



104 



Chapter Five Sex Determination, Sex Linkage, and Pedigree Analysis 



(V 



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 



II 



8 



9 



10 



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° 




X 


r Y (male) 




orange 


1 




red 




X + X° 






X°Y 




red 


1 




orange 




x + x° x°x° 






X + Y 


X°Y 


red orange 






red 


orange 



Fi 



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 



©TheMcGraw-Hil 
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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- 



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II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



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106 



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 


70 



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 


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? 



ao 



kO 



6 



ChO 



(a) 



6 



ChU 



O O DK30 



O D O 



OO 



(b) 



ChO 



o 



6 ChD 



ChO 



O 



(c) 



6 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 



©TheMcGraw-Hil 
Companies, 2001 



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\^ 



o 




71 / A 



1 2 




1 2 



7 8 





3 4 5 



3 4 



(b) 



2 



aK3 



-o 



o 

1 2 



o 



■o 



6ft 



o 



(c) 



5 



oo 



hOODW 

12 3 4 



CM 

7 



6 



J O 



8 9 10 11 



ChU 



1 



IK) O O 

12 3 4 5 6 



D O 

7 



CM3 



O 

2 



o □ o 



8 9 10 11 12 



oo 



ChU 



13 



o 

1 



OK 

2 



CM 

3 



a o o 



CM 

4 



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 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



5. Sex Determination, Sex 
Linkage, and Pedigree 
Analysis 



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Companies, 2001 



108 



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 



6 



(a) 



Dkj> 6 D 



OA 



(b) 



© 



o 



O-o (y-B 



6 6 6 



c^ 



6 




CH 



OH 



6 



6 6 6 6 



\-o o 



6 




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 



©TheMcGraw-Hil 
Companies, 2001 




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. 

(© Dr. Jeremy Burgess/SPL/Photo Researchers.) 



109 



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Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



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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 



6. Linkage and Mapping in 
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Diploid Mapping 



111 



Banded 

bn bn det + det + 

(homozygous for the bn 
allele and the det + allele) 



x 



Detached 

bn + bn + det det 
(homozygous for the bn + 
allele and the det allele) 



Teste ross 



Wild-type 




Banded, detached 


bn + bn det + det 


X 


bn bn det det 


9 




8 



bn 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 



I 



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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Chromosomal Theory 



6. Linkage and Mapping in 
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112 



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 


77 


46 




88 


20 


9 


30 


270 


Britain 


20 


24 




25 


9 


8 


5 


91 


Germany 


18 


27 




15 


4 


6 


1 


71 


France 


12 


7 




7 


8 


12 


1 


47 


Sweden 


4 


4 




7 


5 


7 


2 


29 



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Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



Diploid Mapping 



113 

















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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Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



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114 



Chapter Six Linkage and Mapping in Eukaryotes 







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 + 


X 


bn + det 


bn det + 

— i 1 — 


i 


r 


bn + det 

— i 1 — 




Wild 


-type 






bn 


det* 


X 




bn + 

— i — 


det 

— i — 





-I 1- 



bn det + 

— i 1 — 

bn det 



9 



9 



Banded, detached 



bn det 



bn det 



bn det* bn + det 



bn det 



bn + det 



-i 1- 



i + de* + 

-i 1- 



bn + det 

— i 1 — 

bn det 



bn det 

— i 1 — 

bn det 



bn + det 



i + de*- 

-i 1- 



bn det 

— i 1 — 



S 



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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115 



bn 
bn 
bn + 



bn- 



bn 





det + 
det + 
det 



det 



det + 
det + 
det 



det 



det + 
det + 
det 



det 



Gametes 



bn 


det + 




bn 


det 


bn + 


det + 


^i 


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 



or 



or 



b pr c 

b pr c 

b pr c 

b pr c 



trans 
(repulsion) 



bn 



bn 



+ 



det' 



det 



bn 



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 




388 


Black 


bb pr + pr c + c 


367 


b pr + c + 


367 




367 


Curved 


b + b pr + pr cc 


1,412 


b + pr + c 




1,412 


1,412 


Black, purple 


bb prpr c + c 


1,383 


b pr c + 




1,383 


1,383 


Purple 


b + b prpr c + c 


60 


b pr c 


60 


60 




Black, curved 


bb pr + pr cc 


72 


b pr + c 


72 


72 




Total 




15,000 




887 


2,927 


3,550 


Percent 








5.9 


19.5 


23.7 



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111 



b 


Meiotic tetrad 
pr 


c 


b 


Gametes 
pr 


c 
















Non recombinant 


b 


pr 


c 


Gametes 


pr + 


c + 




u + 


pr + 


c + 


"fc L + 


pr 


c 


- Recombinant 


b 


> 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 + 



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) 



b 



5.9 



pr 

-\— 



19.5 



c 

-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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Chapter Six Linkage and Mapping in Eukaryotes 



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 


47 


Hairy, rosy, thread 


h ry th/h ry th 


4 


Hairy, thread 


h ry + th/h ry th 


98 


Rosy 


h + ry th + /h ry th 


92 


Hairy, rosy 


h ry th /h ry th 


351 


Wild-type 


h ry th /h ry th 


6 


Hairy 


h ry + th + /h ry th 


43 



Trihybrids 










Gamete 


Number 


h-th 


th—ry 


h—ry 


h th ry 


359 








h th ry 


351 








h th ry 


98 


98 




98 


h + th + ry 


92 


92 




92 


h + th ry 


47 




47 


47 


h th + ry + 


43 




43 


43 


h th ry 


4 


4 


4 




h th ry 


6 


6 


6 




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 



th 



10.0 



ry 

— 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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Chapter Six Linkage and Mapping in Eukaryotes 




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 



e 



W 

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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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 



v 



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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123 



c 
C 



Knob 




Wx 

wx 

Meiosis 
Wx 

Wx 

wx 
wx 



Normal chromosome 9 

Knobbed interchange chromosome 9 



Extra piece 





C 


> 


( Wx 






C 




wx 




c 




Wx 


^^ ^^H 




C 




wx 



X 



Wx 



wx 



Gametes 



Gametes 



Wx 



wx 



Nonrecombinant 



Recombinant 



Recombinant 



Nonrecombinant 



Wx 



wx 



Wx 



wx 



c 


Wx 


1 


c 


Wx 


2 










c 


Wx 


c 


wx 








^^ 


Colorless 


Nonwaxy 


Colorless 


Nonwaxy 


c 


wx 


3 


c 


wx 


4 










c 


Wx 


c 


wx 










Colorless 


Nonwaxy 


Colorless 


Waxy 


C 


Wx 


5 


C 


Wx 


6 


C^^^^^^^^H 




<~~^^^^^^^m 




c 


Wx 


C 


wx 










Colored 


Nonwaxy 


Colored 


Nonwaxy 


C 


wx 


7 


C 


wx 


8 


c 


Wx 


c 


wx 








^^ 


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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125 



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 



H 



/ 



OH 



:— C— C— C 



HN— C— C- 

H H H H 
Ornithine 



^ 



O 



Step 2 

Gene 2 



■^ Enzyme 2 



H H 

H I I 

HN HN— C— C 

\ / II 

C H H 



Y H 
H NH 



/ 



OH 



— C— C— C 



^ 



H H 



O 



O 



Citrulline 



Step 3 

Gene 3 



■>- Enzyme 3 



H H 



V H 
H NH 



H 



/ 



OH 



HN HN— C— C 

\ / II 

C H H 



— C— C— C 



^ 



H H 



O 



NH 



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 + 



1 


2 


3 


(PD) 


(NPD) 


(TT) 


ab 


ab + 


ab 


ab 


ab + 


ab + 


a b 


a + b 


a + b 


a + b + 


a + b 


a + b + 


75 


5 


20 



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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127 



If linked 



a 


b 




a 


b 




a + 


b + 


a + 


b + 





If not linked 


a 


b 


a 


b 



-,+ 



o 
o 



-,+ 



o 
o 



PD 

ab 
ab 

a + b + 
a + b + 



Nonrecombinant 



Independent assortment 



NPD 



ab + 
ab + 

a + b 
a + b 



O 
O 



a 
a 



a 
a 



b 

b 




b + 



-i+ 



6 



Independent assortment 



TT 



ab 

ab + 

a + b 

a + b + 



O 



a 
a 



a 
a 



a + 



b 
b 




or 



a 
a 



b 
b 




-i+ 



a + 



6 + 




a 
a 



5 
6 





a 
a 




ft 

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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Haploid Mapping (Tetrad Analysis) 



129 



a 





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 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


1 


a 


a 


+ 
a 


a 


a 


+ 
a 


a 


+ 
a 


+ 
a 


a + 


2 


a 


a 


a + 


a 


a 


+ 
a 


a 


a + 


+ 
a 


a + 


3 


a 


a 


a + 


a + 


a + 


a + 


a 


a 


a 


a + 


4 


a 


a 


a + 


a + 


a + 


a + 


a 


a 


a 


a + 


5 


a + 


a + 


a 


a + 


a 


a 


a + 


a 


a + 


a 


6 


+ 
a 


a + 


a 


+ 
a 


a 


a 


a + 


a 


+ 
a 


a 


7 


+ 
a 


+ 
a 


a 


a 


a + 


a 


a + 


+ 
a 


a 


a 


8 


+ 
a 


+ 
a 


a 


a 


a + 


a 


a + 


+ 
a 


a 


a 




FDS 


FDS 


FDS 


SDS 


SDS 


FDS 


FDS 


SDS 


SDS 


FDS 



Note: Map distance (a locus to centromere) = (1/2)% SDS 

= (1/2) 40% 
= 20 map units 



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Chapter Six Linkage and Mapping in Eukaryotes 



a + 



O O 



a + a 



First-division segregation with 
no crossover between a locus 
and centromere 



/ \ 



,+ 



a 



6 6 6 6 



a + 



a + a 



a^+Ya+Ya+YaXaYaYal 4:4 



or 



alalalala + la + la + la + )) 4:4 



a + \a + a\ a 

/ \ 



Second-division segregation with 
crossover between a locus 
and centromere 



a 



i i <!> i 



a + 



a a 



+ 



ia + Xa + X a la Ia + la + \ a\a 




or 



aiaia + Ya + YaYaYaira* 



or 



ana^ia la la laia^a 



or 



2:2:2:2 



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 


1 


2 


3 


4 


5 


6 


7 


8 


1 


ab 


ab + 


ab 


ab + 


a b 


a b 


a + b 


a + b 


2 


ab 


ab + 


ab 


ab + 


a b 


a b 


a + b 


a + b 


3 


ab + 


ab 


ab + 


ab 


a + b 


a + b 


a + b + 


a + b + 


4 


ab + 


ab 


ab + 


ab 


a + b 


a + b 


a + b + 


a + b + 


5 


a + b 


a + b 


a + b + 


a + b + 


ab + 


ab 


ab + 


ab 


6 


a + b 


a + b 


a + b + 


a + b + 


ab + 


ab 


ab + 


ab 


7 


a b 


a b 


a + b 


a + b 


ab 


ab + 


ab 


ab + 


8 


a b 


a b 


a + b 


a + b 


ab 


ab + 


ab 


ab + 



Table 6.7 The Seven Unique Classes of Asci 

Resulting from Meiosis in a Dihybrid 
Neurospora, ab/a + b + 









Ascus Number 






Spore 
















Number 


1 


2 


3 


4 


5 


6 


7 


1 


ab 


ab + 


ab 


ab 


ab 


ab + 


ab 


2 


ab 


ab + 


ab 


ab 


ab 


ab + 


ab 


3 


ab 


ab + 


ab + 


a + b 


a b 


a + b 


a b 


4 


ab 


ab + 


ab + 


a + b 


a b 


a + b 


a b 


5 


a + b + 


a + b 


a + b + 


a + b + 


a + b + 


a + b 


a + b 


6 


a + b + 


a + b 


a + b + 


a + b + 


a + b + 


a + b 


a + b 


7 


a + b + 


a + b 


a + b 


ab + 


ab 


ab + 


ab + 


8 


a + b + 


a + b 


a + b 


ab + 


ab 


ab + 


ab + 




729 


2 


101 


9 


150 


1 


8 


SDS for a locus 








9 


150 


1 


8 


SDS for b locus 






101 




150 


1 


8 


Unordered: 


PD 


NPD 


TT 


TT 


PD 


NPD 


TT 



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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Haploid Mapping (Tetrad Analysis) 



131 



a a + 



(1) 



b + b 



>+ 



b + 



or 



a 



(2) 



b* 



a a + 
b b 



■s+ 



b 



+ 



or 



a 



(3) 



C3 

a 



a a + 



b + b + 



,+ 



or 

C3 C3 



(4) 



^ 



a 



a + 



6 tf 



,+ 



b 



a'\aay a 



6 6 



+ 



+ 



5 



i+ 




(5) 



,+ 



6 + 



a + a 



6 b + 



or 



to 



a + 



C3 

a a 

a + a 



(6) 



b 



+ 



a 



or 



C3 

o 



(7) 



to 



a + a 



b + b + 



a 



to 



or 

C3 C3 



(8) 



a + 



to 



O 



a + a 



to + to 



a 



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 



a 

-i- 



Centromere 
I 

O — 



(1) 



8.4 



r - 

21.4 

Centromere 
I 

O 



13.0 



a 

-h 



8.4 



4.6 



(2) 



13.0 



b 

-h 



b 

-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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132 



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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Chromosomal Theory 



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Human Chromosomal Maps 



133 



O 
O 



sn+ 
sn+ 

sn 



y 
y 



sn 



y+ 



Dihybrid at prophase 
(mitosis) 



O 



O 



Yellow-spot 
cell 



Singed-spot 
cell 



O 



Crossover 




sn+ 
sn+ 
sn 



sn 



▼ sn+ 
sn 




sn + 
sn 



▼ sn+ 



sn 



+ 



sn 



sn 



y 
y 
y+ 



y+ 



y 
y+ 



y 
y±_ 



y 
y 



y* 



y+ 



Figure 6.26 Formation of twin spots by somatic crossing over. 



No crossover 



A 



o 
o 



o 



o 



sn+ 
sn+ 
sn 



sn 



t 
sn+ 

sn 



sn+ 
sn 



y 
y 
y+ 



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 



D 



O 



a 



Mother 



6 




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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Chromosomal Theory 



6. Linkage and Mapping in 
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134 



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- 



O 

B 



II 



AB 



III 



O 

o 



A 

O 

3 4 



B \k 



B 





6 



B 



A 

O O 

7 8 



,o 



Nail-patella syndrome 



A, AB, B, O Blood types 



b 



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 



135 



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 



w8 



(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. 



Tamarin: Principles of 
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Chromosomal Theory 



6. Linkage and Mapping in 
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136 



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 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


13 


14 


15 


16 


17 


18 


19 


20 


21 


22 


X 


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 







32 


17 


24 


31 


21 


21 


31 


21 


24 


30 


21 


21 


28 


14 


24 


21 


28 


17 


34 


41 


21 


27 



Source: Reprinted with permission from S.D. Carson, et al., "Tissue Factor Gene Localized to Human Chromosome 1 (after lp21)," Science, 229:229-291. 

Copyright © 1985 American Association for the Advancement of Science. 

* 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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



Human Chromosomal Maps 



137 



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 


9 


AG 


Alpha globin gene family 


16 


ALB 


Albumin 


4 


AMY1 


Amylase, salivary 


1 


AMY2 


Amylase, pancreatic 


1 


BCS 


Breast cancer susceptibility 


16 


C2 


Complement component-2 


6 


CAT 


Catalase 


11 


CBD 


Color blindness, deutan 


X 


CBP 


Color blindness, protan 


X 


CML 


Chronic myeloid leukemia 


22 


DMD 


Duchenne muscular dystrophy 


X 


FES 


Feline sarcoma virus oncogene 


15 


FY 


Duffy blood group 


1 


GLB1 


Beta-galactosidase- 1 


3 


HI 


Histone-1 


7 


HBB 


Hemoglobin beta chain 


11 


HEMA 


Classic hemophilia 


X 


HEXA 


Hexosaminidase A 


15 


HLA 


Human leukocyte antigens 


6 


HP 


Haptoglobin 


16 


HYA 


Y histocompatibility antigen, locus A 


Y 


IDDM 


Insulin-dependent diabetes mellitus 


6 


IFF 


Interferon, fibroblast 


9 



Locus Protein Product 



Chromosome 



IGH 


Immunoglobulin heavy-chain gene 
family 


14 


IGK 


Immunoglobulin kappa-chain gene 
family 


2 


INS 


Insulin 


11 


LDHA 


Lactate dehydrogenase A 


11 


MDI 


Manic depressive illness 


6 


MHC 


Major histocompatibility complex 


6 


MN 


MN blood group 


4 


MYB 


Avian myeloblastosis virus oncogene 


6 


NHCP1 


Nonhistone chromosomal protein- 1 


7 


NPS1 


Nail-patella syndrome 


9 


PEPA 


Peptidase A 


18 


PVS 


Polio virus sensitivity 


19 


Rh 


Rhesus blood group 


1 


RN5S 


5S RNA gene(s) 


1 


RNTMI 


Initiator methionine tRNA 


6 


RWS 


Ragweed sensitivity 


6 


SI 


Surface antigen 1 


11 


SIS 


Simian sarcoma virus oncogene 


22 


STA 


Stature 


Y 


TF 


Transferrin 


3 


XG 


Xg blood group 


X 


XRS 


X-ray sensitivity 


13 



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). 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



138 



Chapter Six Linkage and Mapping in Eukaryotes 



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 




21 




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 

C2 

BF 

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 



11 




1 12 








21 




2 22 




23 




24 





— CRH 

CYP11B1 CYP11B2 

CA1 
-CACHCA2 



CA3 



MYC 

>TG 



8 






1 



12 



14 



^ATP1AL1 
-ESD 



q 2 



31 
3 32 








34 


^, / 



13 



COL4A1 
-ICOL4A2 
LAMP1 



TCRA 

MYH6 
MYH7 




14 



centromere 
3' a CA2 

CE 

CG4 

CG2 

CA1 

CEP1 

CG1 

CG3 

CD 

CM 

J H (<4) 

D (many) 
5' t V H (-250) 
qter 



1 -1 



11 



1 



21 
22 



J 



r GABRA5 MANA1 
-EPB42 
-FBN1 




26 



15 



LI PC 

^PKM2 
^B2M 
-MPI 
-IDH2 
-FES 
ANPEP 
IGF1R 



P 1 



I 



q 1 13 



Hi 



EF2 

LW 

FUT1 FUT3 

MANB 

LU FUT2 



APOE 
APOLP2 ^APOC1 
FTL APOC2 

CGB 

LHB 



PRNP 
CHGB 

CST3 










p 


1 12 = 
11 " 








q 


21 
O 


_ r 




"22 





SOD1 



BCEI 

COL6A1 

COL6A2 

CRYA1 

ETS2 



19 



20 



21 



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.) 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



Human Chromosomal Maps 



139 



THE HUMAN GENE MAP 
(selected 'anchor' loci) 

A confirmed assignment EN01 

A provisional assignment DHPR 

Gene cluster. MHC 



50 



100 



150 



SCALE 
(in megabases) 



Mb 




— ADD1 PDEB 

— QDPR 

~\- GABRA2 GABRB1 

PEPS 

GC 

-AFP ALB r- 

ADH1 

-ADHC1-^3 
^EGF ADH4 



-IL2 
FGC 



FGA 
FGB 
FGG 
GYPA GYPB 



GRIA2 



P 1 



11 








1 13 




14 








21 




2 




23 




31 




3 33~ 




35 





DAT1 NHE3 
SRD5A1 

HMGCS1 
\—PRLR 
-TARS 

-LARS 
-ARSB DHFR 



y 



h 



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 



w 



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 



10 



11 



12 



p 


13 






1 12 








11 






1 11 




q 


12 










2 231 

24 1 


■ 




"L 



PGP 

~PRKCB1 
^PN 
CETP 
MT1 
MT2 
-DIA4 HP 
— CTRB 




MYH1 
TP53 



MYH1 
MYH2 
MYH3 
MYH4 
MYH8 

HOX2 - 

NGFR 

TK1 

CACNLG SSTR2 

P4HB 



p 


1 11 


7\ 




11 P 




12 | 


q 


^ 


2 22 | 




\J 



-YES1 

\—JK 

-GRP 
KMBP 
-PEPA 



16 



17 



18 





1 






p 


12 


H 




1 


11 


/ \ 


q 




^s 


13 


F 



centromere 
PVALB 5' a IGLV (many) 

^vY IGLC-' -9 J " C du P lexes 



>PDGFB 
— MB 



qter 



22 



B 



P 1 11 nT M|C2Y 

n TDF 



11 



12 



-XG 

h— POLA 



— ARAF1 
- — MAOA MAOB 



Kar 

-PGK1 




^PLP 
PRPS1 



HPRT 

F9 

FRAXA 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



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 


26 


fluted, claret 


24 


fluted, curled 


167 


claret, curled 


6 


fluted, claret, curled 


298 


wild-type 


302 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



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 



62 



30 



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 


1 


2 


3 


4 


5 


6 


A 


+ 


+ 


+ 


+ 


— 


— 


B 


+ 


+ 


— 


— 


+ 


+ 


C 


+ 


— 


+ 


— 


+ 


— 



7 



+ 



8 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



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 


6 


wild-type 


5 



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 


43 


45 


brown 


37 


35 


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 

11 

14 

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 


74 


waved 


66 


obese 


79 


wild-type 


343 


hotfoot, waved 


61 


obese, waved 


11 


hotfoot 


9 



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 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



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 

X 

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? 



Tamarin: Principles of 
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Chromosomal Theory 



6. Linkage and Mapping in 
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144 



Chapter Six Linkage and Mapping in Eukaryotes 



1 

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 


4 


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 + 


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: 



1 


try 


met 


■? 




■? 




try met 


2 


? 




try~ 


met~ 


try + 


met + 


try + met 


3 


try" 


met + 


try~ 


met~ 


try + 


met~ 


try + met 


4 


try" 


met~ 


try + 


met + 


■? 




try + met 


5 


try~ 


met + 


? 




? 




try + met 


6 


try 


met + 


try 


met + 


try~ 


met~ 


try~ met 


7 


try + 


met + 


try + 


met~ 


■? 




try~ met 


8 


try 


met + 


try~ 


met~ 


■? 




try met 


9 


try~ 


met + 


try + 


met~ 


try" 


met + 


try + met 


10 


try~ 


met~ 


try + 


met + 


try~ 


met~ 


try + met 


11 


try 


met + 


try 


met + 


■? 




? 


12 


? 




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. 

l 

2 

3 

4 



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 
arg 
arg 



arg 
arg 
arg 
arg 



+ 



arg 
arg 
arg 
arg 



arg 
arg 
arg 
arg 



+ 



arg 
arg 
arg 
arg 
arg 
arg 



+ 



arg 
arg 
arg 
arg 
arg 
arg 



arg 
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: 



1 


? 


/ 


/ 


? 


? 


r 


r 


/ 


2 


/ 


/ 


r 


/ + 


/ + 


/ + 


i 


/ 


3 


/ 


? 


? 


? 


t 


? 


■? 


? 


4 


r 


? 


? 


? 


r 


/ 


f 


/ 


5 


/ 


/ 


? 


? 


? 


f 


? 


/ 


6 


? 


/ 


/ 


? 


? 


? 


? 


? 


7 


r 


r 


/ 


/ 


/ 


/ 


/ + 


/ 


8 


/ 


/ 


/ 


? 


? 


r 


/ + 


/ 


9 


/ + 


? 


? 


? 


? 


/ 


/ 


? 


10 


/ 


/ 


/ + 


r 


/ 


/ 


/ + 


/ 


11 


/ 


/ 


/ 


/ 


/ + 


r 


/ + 


/ 


12 


/ 


/ 


? 


? 


? 


? 


/ + 


/ 



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 


5 


Class 2 





Class 6 


5 


Class 3 


190 


Class 7 


10 


Class 4 


90 







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). 



a 

+- 



cm 

-e- 



b 

-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 



II 



III 



IV 



VI 



VII VIII 



1,2 


a + b 


a + b 


a + b 


a + b + 


a + b + 


a + b 


a + b 


ab + 


3,4 


a + b 


a b 


a b 


a + b 


a + b 


ab + 


ab + 


a + b 


5,6 


ab + 


ab 


ab + 


ab 


ab + 


a + b 


ab + 


a + b 


7,8 


ab + 


ab + 


ab 


ab + 


ab 


ab + 


a + b 


ab + 




85 


2 


3 


2 


3 


3 


1 


1 



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 



©TheMcGraw-Hil 
Companies, 2001 



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 


I 


II 


III 


Spores 


I 


II 


III 


1,2 


ab + 


ab 


ab + 


1,2 


bc + 


b + c + 


b + c 


3,4 


ab + 


ab 


a b 


3,4 


bc + 


b c 


b c 


5,6 


a + b 


a + b + 


a + b 


5,6 


b + c 


be 


be 


7,8 


a + b 


a b 


ab 


7,8 


b + c 


be 


bc + 




45 


43 


12 




70 


4 


26 








cd 


+ Xc + d 








Spores 


I 


II 


III 


IV 


V 


VI 


VII 


1,2 


cd + 


cd 


cd 


cd 


cd + 


cd 


cd + 


3,4 


cd + 


cd 


cd + 


c + d 


c + d 


c + d + 


c + d 


5,6 


c + d 


c + d + c + d + 


c + d + 


c + d 


c + d + 


c + d + 


7,8 


c + d 


c + d + c + d 


cd + 


cd + 


cd 


cd 




42 


2 


30 


15 


5 


1 


5 



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 



II 



III 



IV 



VI 



VII 



VIII 



1,2 


pab pk + ad + 


pab pk + ad + 


pab pk + ad + 


pab pk + ad + 


pab pk + ad + 


pab pk + ad + 


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 


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 + 


pab + pk ad + 


pabpk + ad + 


pab+pk ad 




34 


35 


9 


7 


2 


2 


1 


3 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



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Companies, 2001 



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 


A 


B 


C 


D 


E 


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 


— 


— 


— 


+ 



+ + + 



+ 



+ 



+ 


— 


+ 


+ 


— 


— 


— 


— 


+ 


— 


— 


+ 


— 


+ 


+ 



+ 

+ 



+ 



+ 



+ 



— 


+ 


+ 


— 


— 


— 


— 


+ 


+ 


+ 


— 


— 


+ 


— 


— 


+ 


— 


— 


+ 


+ 


— 


+ 


— 


+ 


— 


— 


+ 


+ 


+ 


— 


— 


+ 


+ 


+ 


+ 


— 


+ 


— 


+ 


— 


+ 


— 


+ 


+ 


+ 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



6. Linkage and Mapping in 
Eukaryotes 



©TheMcGraw-Hil 
Companies, 2001 



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 


3 


7 




9 


11 


15 




18 


20 


X 


— 


+ 




— 


+ 


+ 




— 


+ 


Y 


+ 


+ 




— 


+ 


— 




+ 


— 


Z 


— 


+ 




+ 


— 


— 




+ 


+ 












Enzyme 










Clone 


A 




B 




C 




D 




E 


X 


+ 




+ 




— 




— 




+ 


Y 


+ 




— 




+ 




+ 




+ 


Z 


— 




— 




+ 




— 




+ 



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 



14 



15 



18 



A 


+ 


+ 


B 


+ 


+ 


C 


+ 


— 



+ 



+ 



+ 



+ 
+ 



+ 



+ 



If a particular gene is located on chromosome 3, 
which clones should be positive for the enzyme 
from that gene? 



CRITICAL THINKING QUESTIONS 



i. 



2. 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



©TheMcGraw-Hil 
Companies, 2001 




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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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
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©TheMcGraw-Hil 
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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.) 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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150 



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 


lg 


Glucose 


5g 


NaCl 


5g 


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.) 



Tamarin: Principles of 
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Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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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.) 



Tamarin: Principles of II. Mendelism and the 

Genetics, Seventh Edition Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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152 



Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses 



Adenine 



O 



H 
O 
O 
C- 



Methionine 
adenosyltransferase 



CH CrL CrL S CH, 



NH, 



ATP + H 2 



rr 



H 
O 
O 
■>- C— CH — CHU- 



CK 



H 



H 



H 




H 



P.+ PP. 



NH, 



CH2 S CHo 



OH 



OH 



Methionine 



S-Adenosylmethionine 



H 
Methyltransferase Q 

rv — * c - 

Methyl-group Methylated 
acceptor acceptor 



Adenine 



CH, 



H 



H 



O 




H 



H 



CH — CH 2 — CH 2 — S 



NH, 



S-Adenosylhomocysteine 



H 

Adenosylhomocysteinase O 

O 



rr 



-► 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 



C 

H 



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). 



Tamarin: Principles of 
Genetics, Seventh Edition 



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Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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Bacterial Phenotypes 



153 




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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Genetics, Seventh Edition 



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Chromosomal Theory 



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Prokaryotes and Bacterial 
Viruses 



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Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses 



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 



Tamarin: Principles of 
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Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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Sexual Processes in Bacteria and Bacteriophages 



155 



Bacterial sexual processes 

Exogenous DNA 



^ 



o 



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. 



Tamarin: Principles of 
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Chromosomal Theory 



7. Linkage and Mapping in 
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Viruses 



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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 



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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<^> 



Strain A 

met~ bio~ thr + leu + 



Complete medium 



^=^> 



^^ Centrifuge and wash cells 



Strain B 
met + bio + thr' 



leu' 



\J 



2X 10 8 cells 



1 X 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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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 


F 


str s 


str r 


azf 


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- 



c 

CD 
O 

CD 

O 

CD 

CT 
CD 



100 i- 



80 - 



60 - 



40 - 



20 - 




► . 



J I I I lb I I L 



galB 



J I L 






10 



20 30 40 

Time (minutes) 



50 



60 



'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 



I 



I 



I 



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 





T 


L 


Az 


Ti 


Pro 


Lac 


Ad 


Gal 


Try 


H 


S-G 


Sm 


Mai 


Xyl 


Mtl 


Isol 


M 


Bi 


1 





L 


T 


B x 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


Ad 


Lac 


Pro 


Tj 


Az 


2 





Pro 


T x 


Az 


L 


T 


Bi 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


Ad 


Lac 


3 





Ad 


Lac 


Pro 


Ti 


Az 


L 


T 


Bi 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


4 





Bi 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


Ad 


Lac 


Pro 


Ti 


Az 


L 


T 


5 





M 


Bi 


T 


L 


Az 


Ti 


Pro 


Lac 


Ad 


Gal 


Try 


H 


S-G 


Sm 


Mai 


Xyl 


Mtl 


Isol 


6 





Isol 


M 


Bi 


T 


L 


Az 


Ti 


Pro 


Lac 


Ad 


Gal 


Try 


H 


S-G 


Sm 


Mai 


Xyl 


Mtl 


7 





Ti 


Az 


L 


T 


B x 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


Ad 


Lac 


Pro 


AB311 





H 


Try 


Gal 


Ad 


Lac 


Pro 


T x 


Az 


L 


T 


Bi 


M 


Isol 


Mtl 


Xyl 


Mai 


Sm 


S-G 


AB312 





Sm 


Mai 


Xyl 


Mtl 


Isol 


M 


Bi 


T 


L 


Az 


T x 


Pro 


Lac 


Ad 


Gal 


Try 


H 


S-G 


AB313 





Mtl 


Xyl 


Mai 


Sm 


S-G 


H 


Try 


Gal 


Ad 


Lac 


Pro 


Ti 


Az 


L 


T 


Bi 


M 


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. 



F 



E. co//' chromosome 




to 



oA 



/* 






Hfr chromosome 

F 








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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hr 



.+„+ 



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 



46 

34 



h + r + 



hr 



+ 



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 



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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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
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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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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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166 



Chapter Seven Linkage and Mapping in Prokaryotes and Bacterial Viruses 




Bacterial chromosome 



X prophage 



Exogenote 



Endogenote 



4 



'/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 



B 






C + C 






Two crossovers 




Four crossovers 





B 






i+ 



B 



+ 



A + 



B 






Figure 7.26 The rarest transductant requires four crossovers. 



Tamarin: Principles of 
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Chromosomal Theory 



7. Linkage and Mapping in 
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Transduction 



167 



Table 7.5 Gene Order Established by Two-Factor 
Cotransduction* 



Transductants 


Number 


A + B + 


30 


A + C + 





B + C + 


25 


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 


1 




+ 


+ 


— 


A + B + C~ 


2 




+ 


— 


— 


A + B~ C~ 


3 




+ 


— 


— 


A + B~ C~ 


4 




+ 


+ 


+ 


A + B + C + 


5 




+ 




+ 


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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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
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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 + 






50 


A + B + C~ 






75 


A + B~ C + 






1 


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.) 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



©TheMcGraw-Hil 
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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 


r 


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 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
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170 



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 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



©TheMcGraw-Hil 
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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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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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172 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



©TheMcGraw-Hil 
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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 



1 


2 


3 


4 


5 


6 





— 


+ 


— 








+ 


— 


+ 
+ 
+ 


+ 


— 


— 


— 


+ 


+ 


— 


— 


— 


— 


+ 


— 


— 


+ 



- - + 



+ 






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 



Tamarin: Principles of 
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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 


11 


67 


thr + 


94 


50 


xyl + 


73 


29 


lac + 


2 


58 


his + 


38 


94 


ilv + 


77 


33 


argG + 


62 


18 



Hfr Ra-2 



70 
87 

8 
79 

43 

4 

19 



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 


21 


a + b~ c + 


69 


a + b + c~ 


3 


a b c 


7 



+ 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 




12 

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 



a? 


b~ 


c 


a~ 


b + 


c 


a~ 


b~ 


c 


+ 
a 


b + 


c 


a + 


b + 


c 


a + 


b~ 


c 


a~ 


b + 


c 



214 

231 

206 

11 

6 

93 

14 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



©TheMcGraw-Hil 
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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 



a 



d 



5 


— 


10 


+ 


15 


+ 


20 


+ 


25 


+ 


30 


+ 


35 


+ 


40 


+ 


45 


+ 


50 


+ 


55 


+ 


60 


+ 


65 


+ 


70 


+ 


75 


+ 



+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 



+ 
+ 
+ 



+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 
+ 



+ 



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. 



.+ 



.+ 



+ 



m 


m 


m 


m 


m 


m 


m 


m 


r 


r 


+ 
r 


r 


+ 
r 


r 


+ 
r 


+ 
r 


tu 


tu 


tu 


tu + 


tu 


tu + 


tu + 


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 



Tamarin: Principles of 
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Chromosomal Theory 



7. Linkage and Mapping in 
Prokaryotes and Bacterial 
Viruses 



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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 

75 



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 


ara 


ara 


ara 


leu~ 


leu + 


leu~ 


leu + 


ilvH~ 


ilvH~ 


ilvH + 


ilvH 


32 


9 





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~ 


1 


ara-\~ 


ara-2~ 






48.5 


2 


ara-2~ 


ara-\~ 






2.4 


3 


ara-\~ 


ara-3 






4.0 


4 


ara-^~ 


ara-\~ 






19.1 


5 


ara-2~ 


ara-^~ 






1.5 


6 


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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
Companies, 2001 




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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
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178 



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. 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
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Variation in Chromosomal Structure 



179 




a b c 



d e f g h 



i J 



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 





^ 



x) 



O 

gh duplication 



f 



9 



V 



f e d 



O 

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 



or 




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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8. Cytogenetics 



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Chapter Eight Cytogenetics 



c d 

O 
O 



O 



9 



Normal chromatids 



Deletion chromatids 



J 



Figure 8.4 A bulge can occur in a meiotic tetrad if a large 
deletion has occurred. 



Heterochromatin 



w 



o 



t 



/ — 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 



w 



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 



9 



a b c d a 

o 



Synapsis 
occurs 



a b c 



o 



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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Chromosomal Theory 



8. Cytogenetics 



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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 



a b c 



o 
o 




h i i 




a b c d e 

o 



a b c d g 

o 



J 



9 



h 



g d c b a 

o 



h 



9 



i J 



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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Chromosomal Theory 



8. Cytogenetics 



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182 



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 



Tamarin: Principles of 
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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
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Variation in Chromosomal Structure 



183 




(a) 



39 

— 38 



37 



^ 36 



_^ 35 



34 
33 



32 
31 
30 



= 29 



28 
27 



26 



g 25 

=■ 24 

^ 23 

= 22 

~ 21 



20 



— 19 



— 18 



17 



= 16 



™ 15 



14 



— 13 
^ 12 



11 



- 10 



_9_ 
8_ 

7 



_6_ 
_5_ 

A 

_3_ 

_2_ 

1 







— — - 


39 
38 
37 




^ 


36 


r 


30 


— - 


35 
34 


CO 
I 


31 


==si 




CM 


3? 







I - 


33 


^^ 






29 



28 
27 



O 



CM 



9 



10 



11 — 



I- 12 
(b) 



26 



g 25 

=■ 24 

^ 23 

= 22 

~ 21 



20 



— 19 



— 18 







17 




— 


16 
15 
14 




sSJSi 


13 


3 

4 






5 






6 


= 




7 


^^ 




8 







= 2 



1 



39 

— 38 



37 



30 = 



31 



32 



33 = 



Q 

_i 

CM 



15 __ 



16 
10 



11 = 



(c) 



36 



™ 35 



34 



= 29 



28 
27 



26 



H 25 
24 





= 


7 
6 
5 
4 




^^ 


3 


■* 


^ 




13 ■ 




14 - 







12 = 2 



1 



(M 



LU 



— 


23 ° 





22 


^^^ 


21 





20 
19 
18 


17 


= 


— 9 


— 8 



r 

o 

_i 

CM 

L 



39 
38 



.r**~~ 


37 


^— 


36 


30 


35 
34 


31 ::::: 




15 — 




16 





17 



18 — 



19 = 



20 



21 



22 



23 
24 



25 = 



26 
27 



28 
29 



= 33 



32 



= 14 



= 13 



3 — 



4 


= 


5 
6 


= 


7 


^E 


8 





9 



10 



11 = 



12 



(d) 



1 



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 <^ 



! * 



r 



(b) 



40 




f 



•%^^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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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



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184 



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 



28 



m 27 

— 26 



25 



= 24 



23 



22 



— 21 

— 20 

19 



18 



17 



— 16 

= 15 

_ ^ 



~ 13 
= 12 



11 



= 10 

— 9 



8 






= 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. 


31 


^ 30 


= 29 

=: 28 

= 27 


= 26 
= 25 


= 24 

m. 23 

22 

= 21 

20 

19 

; 18 

_ 1? 

16 


= 15 
14 


3 — 

4 

5 = 

6 = 




7 = 

8 = 




9 — 
10 = 




11 




| I III: 
CO CO 




= 


2 
■ 1 





-_^- 


31 




^ 30 




= 29 




:=: 28 




= 27 
—.. 26 




= 25 




= 24 




= 23 




" 22 




::::::::: 8 




= 7 




6 




— 5 






CQ 
DC 
CO 


15 = 

16 

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 

oo 




3 27 




= 26 




= ?5 


20 — 




21 = 


Q 
DC 


22^ 


CO 


23 ^ 




24 ^ 




= 19 




18 




;;= 1 7 




— 16 




= 15 




= 14 




3 ■= 




4 =l 
h — : 
6 - 






lull 1 
!|i>i| 

CO 




_ 13 


O 


— 12 


DC 


11 


CO 


= 10 




9 




= 8 




= 7 


(d) 


1 



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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Chromosomal Theory 



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Variation in Chromosomal Structure 



185 



c d 




9 



abed 



abed 



a b c 
j i h 



e f g h i i 

o 



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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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



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186 



Chapter Eight Cytogenetics 



c d 

o 

c° d 



c d 

o 

c° d 



9 



9 



9 



1 2 3 4 5 6 




J 



j 



1 2 3 4 5 6 



o 



h Q f 4 5 6 



1 2 3 4 5 6 

Synapses occur 



o 



9 



9 



a b c d e 

O 
O 

o 
o 

a b c d e 



~N\ 



f 



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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Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
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Variation in Chromosomal Structure 



187 



9 



9 



First 
Second 



O 
O 

o 
o 



Alternate segregation 



^ 



tf 



Adjacent -1 segregation 



Third 



X Fourth 



Adjacent -2 segregation 



First with 
fourth 

a 
b 



o 










d 6 
e 15 

4 

9 |3 

h 12 

1 



1 

Normal 



J 



Second with 
third 



c 

° 



d 

e 
3 
2 



L 



o 



6 
5 






9 



J 



Reciprocal 
translocation 



First with 
third 

a 

b 

c 

o o 






9 



L 






6 
5 
4 
f 

9 



Duplication 
deficiency 



Second with 
fourth 

a 

b 

c 

o o 



d 
e 
3 
2 



6 
5 
4 
3 



J L 



J 



Duplication 
deficiency 



First with 
second 

a 

b 



L 



a 
b 



c c 

o o 

d d 



f 3 

I 

9 2 



Duplication 
deficiency 



Third with 
fourth 



o o 



6 
5 
4 
f 

9 






6 
5 
4 
3 



J L 



J 



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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Chromosomal Theory 



8. Cytogenetics 



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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. 



r 



70 facets 



45 facets 



0.0 



Bar 



Bar 



Doublebar 



56.7 
f 

—\— 



: + 



57.0 
B B 



59.5 
fu 

— I— 



66.0 



o 



Crossover point 



B B 



Mismatch 



B B B 



fu + 



fu + 

-r- 



o 



o 



Wild-type 



f + 



B 



fu 



o 



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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Genetics, Seventh Edition Chromosomal Theory 



8. Cytogenetics 



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Variation in Chromosomal Structure 



189 



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.) 





•*\ 




u 



* »« 



ift U 



II IS II 



B 



* 1 



U 



U 



8 



10 



1 1 



12 



14 »* 



13 



U 



D 



M 



15 



* * 



16 



* & 



17 



• * 



I I 



18 



♦ * 



19 




20 




F 




k ft 




• # 


21 




22 



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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Chromosomal Theory 



8. Cytogenetics 



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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 


2n 


8 


X, X, 2, 2, 3, 3, 4, 4 


Monosomic 


2n - 1 


7 


X, X, 2, 2, 3, 4, 4 


Nullisomic 


2n — 2 


6 


X, X, 2, 2, 4, 4 


Double monosomic 


2n - 1 - 1 


6 


X, X, 2, 3, 4, 4 


Trisomic 


2n + 1 


9 


X, X, 2, 2, 3, 3, 4, 4, 4 


Tetrasomic 


2n + 2 


10 


X, X, 2, 2, 3, 3, 3, 3, 4, 4 


Double trisomic 


2n + 1 + 1 


10 


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 




or 




): 



>: 



\\ 



At meiosis II 



XY 



XY 











)) xx 







I YY 







XX 



)) 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 





XY 


XX 


YY 





XX 







9x 


XXY 


XXX 


XYY 


XO 








CO 

£ Sx 

o 

z 




XXX 


XO 


0*Y 




XXY 


YO 



Figure 8.18 Results of fusion of a nondisjunction gamete (top) 
with a normal gamete {side). 



6 






X 



XYY 



X 




x ; x 



/ 




x r 
\ 

XXX 



X 



/ 



Figure 8.19 Mitotic nondisjunction of the sex chromosomes. 



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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



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Chapter Eight Cytogenetics 



Metaphase 



Anaphase 



Telophase 






XX 

Figure 8.20 Chromosomal lagging at mitosis in the X chromosomes of a female Drosophila. 




X 



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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Chromosomal Theory 



8. Cytogenetics 



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Variation in Chromosome Number 



193 






ti 



it 



8 



10 



B 



ft ft It IS II \* II 



11 



12 



I 



X 



ft ft* II 



13 



14 



15 



D 

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 



16 



17 



• • 



19 



18 



u 



20 





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 . 



* 

21 



Ik « 



22 



• 

Y 



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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II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



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Chapter Eight Cytogenetics 




Figure 8.23 Individual with trisomy 21 
Young/SPL/ Photo Researchers.) 



(© Hattie 




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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Genetics, Seventh Edition Chromosomal Theory 



8. Cytogenetics 



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Variation in Chromosome Number 



195 



Isochromosome 




Split 




Fragments 



CO 



Isochromosome 



V 




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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Chromosomal Theory 



8. Cytogenetics 



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Chapter Eight Cytogenetics 










8 



10 



B 



11 *) II u w 



11 



I 



12 



M 4| II 



13 



14 



D 



15 



It * * %l 



16 



17 



t I 



19 



18 



St 



20 



* 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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8. Cytogenetics 



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Variation in Chromosome Number 



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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 



i 



\r 









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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Chromosomal Theory 



8. Cytogenetics 



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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 



AA 



BB 



Autopolyploidy 



V 
AAAA 



AB 



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 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
Companies, 2001 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
Companies, 2001 



200 



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? 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
Companies, 2001 



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 



10 

//////- 



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 




A 




100 


B 




45 


C 




54 


D 




98 


E 




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: 



v 



m 



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 


30 


clot 


20 



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? 



Tamarin: Principles of 
Genetics, Seventh Edition 



II. Mendelism and the 
Chromosomal Theory 



8. Cytogenetics 



©TheMcGraw-Hil 
Companies, 2001 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
Companies, 2001 




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 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
Companies, 2001 



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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
Companies, 2001 



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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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, 



NH 



V 



Threonine dehydratase 



CH 3 — CH 2 — C- -COOH 



O 

Pyruvate ->v 



CO, 



V 



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, 



H 



CH 3 — CH 2 — CH — C- - COOH 



NH, 



Isoleucine 



ATP 



Enzyme 



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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Molecular Genetics 



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In Search of the Genetic Material 



209 



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) 




(No change) 



Heat-killed 
S-type cells 



OoV 

o 

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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Chemistry of Nucleic Acids 



211 



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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Molecular Genetics 



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Chapter Nine Chemistry of the Gene 




(a) 



Common type 



Type M 





Degradation 



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 

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 



O 



o 



O ...also 

represented ^^ 

® 



Sugar 



CD 
CO 
O 
-Q 

ir 



5' 
HOCH 2 OH 

4'l/°<l , 

C C ■■■ a ' so 

iViWtyi represented 



OH OH 



5' 
HOCK 



OH 

ii 




cd 4' y \i 

£ ? X H H/V 
§ H c — C H 

CD 

Q I I 

OH H 



Base 



NH, 



N 



^6\ ( 



.N 



O 
,C. 



CO 
CD 

_c 
Q_ 



HC 



5 
4 



HINT 



-N 



8CH 



L 



c^ 

Adenine 






9/ c^ 

H H 2 N 



:ch 



NT 
Guanine 



N 
H 



CD Ng 






NH, 
5 



O 
,C S 



CH 



CH HN' 



Q_ 



// 



Cytosine 



3 

HN' 



O 



CH 



^N 
H 
Thymine 



// 



*CH 



.CH 



O 



^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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Molecular Genetics 



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Chemistry of Nucleic Acids 



213 



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 



O 



O 



O P — o — 

-o 



o 



o 



O CH 




OH H 



1 

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 



O 



O 



O CH 




2' 
OH H 
Guanosine monophosphate 



O 



O 



O 



Cytosine 

^N 
5 3 



O CH 




Pyrimidine nucleotides 



2' 

OH H 
Cytidine monophosphate 



O 



O 



O 



O CH 




2' 

OH H 
Thymidine monophosphate 



Figure 9.10 Structure of the four deoxyribose nucleotides. 



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Chemistry of Nucleic Acids 



215 



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 



O 



O 



0" 



O 



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.) 



$ 






i 






V 






^^JH* 


' 









m 



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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Molecular Genetics 



9. Chemistry of the Genel 



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Chemistry of Nucleic Acids 



217 




(a) 




/ 



Sugar 



Phosphate 



Sugar 



Phosphate 



Phosphate 



/ 



Sugar 



Sugar 



Phosphate 



Sugar 



Phosphate 



\ 



Sugar 



Phosphate 
(b) 



Base 



Base 



Base 



Base 



Base 



Base 



Base 








Base 










Base 










Base 










Base 










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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Molecular Genetics 



9. Chemistry of the Genel 



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218 



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 



\ 

CH 

\ / 

,C N 

/ \ 

O To deoxyribose 



Adenine 



Thymine 



: ISL 



/ 
To deoxyribose 



\ / 



\ 



o 



H 
-H-N 



N-H N 



\ 
C- 

/ 



CH 

Y 



N : 



\ / 



CH 



N-H- 
H 



O 



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- 




3' 



3' 




Base- - -Base- 






Base- - -Base 




Base- - -Base 




Y 
3' 




y 

5' 



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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Molecular Genetics 



9. Chemistry of the Genel 



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Chemistry of Nucleic Acids 



219 









100 






80 


•Jf 




+ 60 


— jik 




O 






■i— > 
c 

CD 
O 

S. 40 






20 








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 


c- 




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. 



Tamarin: Principles af 
Genetics, Seventh Editinn 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
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220 



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.) 



3' 



5' 



3' 



5' 



B 



^^:-:-& 



/=&■-- -<=& 



^^:-:-& 



&tY:^3- 



fi=^yym 






4&-:-rf 



^:-:-a 






&tY:^3- 



5' 



v 

3' 



^^:-:-ffl 



3: 



l 



S 



^t::^3- 



-^^::Sr 



3: 



Z 



-~&T 



m 



=E 






^:Y^^_ 



- B 



- Z 



- B 



5' 



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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Molecular Genetics 



9. Chemistry of the Genel 



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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 



14 



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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Genetics, Seventh Edition 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
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222 



Chapter Nine Chemistry of the Gene 



BOX 9-3 CONTINUED 



3' 




~ZL 



5' 



3' 



Fe 



Fe 



C 



(a) 



(b) 



T 
T 
T 



T 
T 



T 
T 
T 
T 



3' 

I 

T 

G 

C 

C 

T 

A 

G 

G 

A 

A 

A 

A 

A 



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 - 

A - 

G - 

G - 

C - 

C - 

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 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



9. Chemistry of the Genel 



©TheMcGraw-Hil 
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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 



S 



DNA 

I 






;' f ■■': 



/ \ 



I 



I 




•*-*' 



i 



. « _ : *; 



/ \ / \ / \ 



S §§ «S«S %% S 

i ii 11 II 1 



i\ ■!- . 



■ ; s.- 






-T: 



, - V 






y^>.+ 



■\ * 



« ■;■'■*¥.:;. "-■ 



». * 









•vr 






"V^-wi."^ 



:-'■ - : '-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). 



*7- .-■■ - .• .: ., i > <■ 

■ ■ ■ f 1 ■' - ■ - i r ■*■ J 

■,' \ f . * . ■ .. ■ ,', - f 

• -■ A ' .- ili ■''..* 



." •" 



(a) 






wr: 



V ■■' ^ 



^-:-m 



* ■-,.; ■'.■••- 



■■*. _ -.■ 



. - I." 

;■ 
. - j* 

i. 

■ i 



i 



'&•*<*£. 



t 

» I 
% 

■ 4 

■■ v. ■ 



.■:. v 



. V m-i - r « 





.■ 


^.:;:v::-, 


:- 


f 






. ■ '- : - *, 






.1 ' ' " ■ 


•■'f*' 


- " "*, . ■ ■; 

. ■•■ \ r.' ■ - . 




4. 




- ** 








■ s --■ ■ 






\. 


m-* 


\V». 


**-.-- 


' : ' ■; 


" V: " -■ "' 


\. '- 






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 



Origin 



Origin 






V 







(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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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 



Base- - -Base 




Synthesis 



U V 7 




Polymerase cannot act 



Base- - -Base 



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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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 


replication 


(leading 


(lagging 


strand) 


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 




5' 



3' 



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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3' 



y 



3' 



Leading strand 




5' 



Previous 

Okazaki 

fragment 

3'* 



Primer 



Okazaki fragment 



Primer 



^1 




3' 



5' 
3' 



3' 



Removed primer 
fragments 





DNA polymerase I 




3' 



5' 
3' 




Nick 



3' 




3' 



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 



Previous 

Okazaki 

fragment 



New Okazaki 
fragment 




Base 



Base 



Base 



Base 



Base 



Base 




DNA ligase 



Previous 

Okazaki 

fragment 



New Okazaki 
fragment 




Base 



Base 



Base 



Base 



Base 



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 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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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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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 



5' 




JT 71 

3' 



1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 



Okazaki primer 



3' 







■ mil iii 



Okazaki primer 



5' 








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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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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DNA Replication — The Enzymology 



237 



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 


D 


E ) 


' E 




-*v 




When the tail is complete 
it can be removed by 
a nuclease 



C D E 




3' 



The linear DNA can then be 
rolled into a closed 
double-stranded 
circle 




5' A B C D E 3' 



5' 



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 



241 



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 


%A 


°/oT 


%G 


%C 


%U 


a. 


28 


28 


22 


22 





b. 


31 





31 


17 


21 


c. 


15 


15 


35 


35 






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 


%A 


%G 


%T 


%C 


%U 


a. 


33 


17 


33 


17 





b. 


33 


33 


17 


17 





c. 


26 


24 





24 


26 


d. 


21 


40 


21 


18 





e. 


15 


40 





30 


15 


f. 


30 


20 


15 


20 


15 















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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10. Gene Expression: 
Transcription 



©TheMcGraw-Hil 
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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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Molecular Genetics 



10. Gene Expression: 
Transcription 



©TheMcGraw-Hil 
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244 



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 



\ 



y 



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=^ 



K) 



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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Genetics, Seventh Edition 



Molecular Genetics 



10. Gene Expression: 
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246 



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 



51 

35 

43 



5' 



— A 

— T 

— A 

— A 
— G 

— C 

— C 
— G 

— T 



3' 



5' 



— A 

— T 

— A 

— A 
— G 

— C 

— C 
— G 

— T 



3' 



DNA 
5' 3' 



3' 



— A 


T — 


— T 


A — 


— A 


T — 


— A 


T — 


— G 


c— 


— C 


G — 


— C 


G — 


— G 


c— 


— T 


A — 



5' 

Heat (denature) and 
add RNA 



3' 



5' 



+ 



T — 
A — 

T — 
T — 

c — 

G — 
G — 

c — 

A — 



+ 



RNA 

— A 

— U 

— A 

— A 
— G 

— C 

— C 
— G 

— U 



5' 



3' 



Cool (renature) 

5' 3' 

—A T — 

— U A — 

—A T — 

—A T — 

— G C — 

-C G — 



-C G — 
— G C — 
— U A — 



3' 



5' 



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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Molecular Genetics 



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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 


24 


22 


32 


22 


Euglena 


26 


19 


31 


24 


Poliovirus 


30 


25 


25 


20 



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 



5' 



• 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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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 



3' 



DNA 



5' 



Coding strand 
TGGCGGTGATAATGGTTGCATGT • • • 3' 



3' 



ACCGCCACTATTACCAACGTACA • • • 
Pribnow box 



5' 



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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Genetics, Seventh Edition 



Molecular Genetics 



10. Gene Expression: 
Transcription 



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250 



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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Molecular Genetics 



10. Gene Expression: 
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©TheMcGraw-Hil 
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Prokaryotic DNA Transcription 



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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Molecular Genetics 



10. Gene Expression: 
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252 



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: ^ ^ 

■ ' ^ ^ § § £ 



>* 



>*. 



X 







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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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) 



V 



o 




/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 

i 



DNA 



5' ' ' 3' 

TT AAAGGCTCCTTT T GGAGCCTTTTTTTT 



AATTTCCGAGGAAA A CCTCGG A A AA AAA A 



3' 



L 



Template 
strand 



Inverted repeat 

Transcription 



5' 



Last base 
transcribed 




c 


G 




c 


G 




u 


A 


Stem-loop 


c 


G 


structure in 


G 


C 


RNA 


G 


C 




A 


U 




A 


U 






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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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 



5' 




AUG 



Leader 



UAA 



3' 



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 



5' 



16S 



23S 



5S 



tRNAs 

/ \ 



3' 



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. 



S 



O 



c 

CH, 

ch! 



o 



Glutamic acid 
(amino acid) 



^ 



NH 3 + 



O 






c 












































L 


V J 
























t 


r -s 




















Gl 




























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 



3' 

O — A 

i 

C 

c 

i 

A 

i 



5' 



T-loop 



G 
\ 



u 



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 



/ 



G 



D-loop 



G 



D 



■ y 



¥" 



G-G-C-C-U 

/ 
D. 



Variable loop ^g 



, G — C-MG 
A A — U 



C-G-C-G 



■D-G 



.G 



G — 

i 

G — 

i 

G — 

/ 



i 

C 

i 

c 

I 

c 



\ 



¥ 



u 



/ 

Ml 

\ 



\ 
U 

/ 



Anticodon loop 
Anticodon 



G 



5'- 



G 



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 



H' 



H 



O 






^ 



N- 



C 
C 



Inosine (I) 



•N 



/ 



"N 



\ 



H 



O 








ISl' 




,^ C \ N / C 



Ribose 



Ribose 



,CH, 



^H 



CH 



3 

N' 



O 



H 



^ 



N' 



C 
C 



•N 



/ 



"N 



\ 



H 



Ribose 



1-Methylinosine (Ml) 



H- 



S N' 



O 



C H 

N 



O 



,^ c \ n ^ c \ 



H 



Ribose 



H. 



O 



^3 ^°- 



N 



"N 
H 



^ 



N- 






•N 



> 



H 



"N 



\ 



Ribose 



1-Methylguanosine (MG) 



H~ 



O 

a 



*N' 



H 

C OH 



O 



^ 



■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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10. Gene Expression: 
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Eukaryotic DNA Transcription 



261 



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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263 




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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Eukaryotic DNA Transcription 



265 



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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Molecular Genetics 



10. Gene Expression: 
Transcription 



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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 



o 



B 



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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Eukaryotic DNA Transcription 



267 



10 



20 



30 



40 



50 



60 



70 



cAp 



90 



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 

ZC 



79 



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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Molecular Genetics 



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268 



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.) 



5' 



UCUA 



Exon II 
G |U 3' 



GTP 



"> 




ucu 



u 



GA 




UCU 



U 



Exon I 



Exon II 



Intron + G 



UCUU 



+ 



GA 



G 



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 



A 

A 

A 

A 

A 

U • G 

C-G 

U-A 

C-G 

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 

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 

A A 



G 
G 
G 
C 
G 



C 
C 
C 
G 
C 



PA 

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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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 


Ul 


3'-UCCAUUCAUA 


5 ' end of intron 


Recognizes and binds 5' site of intron 


U2 


3'-AUGAUGU 


Branch point of intron 


Binds branch point of intron 


U4 


3'-UUGGUCGU . . . 
AAGGGCACGUAUUCCUU 


U6 


Binds to (inactivates) U6 


U5 


3'-CAUUUUCCG 


Exon 1 and exon 2 


Binds to both exons 


U6 


3 -CGACUAGU . . . ACA 


U2, 5' site 


Displaces Ul and binds 5' site and 
U2 at branch point 



Source: With permission from the Annual Review of Genetics, Volume 28 © 1994 by Annual Reviews www.AnnualReviews.org. 



Stem III 



L2.4 

A 

L2.3A 



3' 



A13 
11.4 11.1 G12 

G G C C 



L2.2A 



L2.1 



.CCGG 

b 10 - 4 10.1^9 



5' 



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 



Cs 

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 



3' 



5' 



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 



5' 



GG 



Intron Exon II 
A — GET 



3' 



Lariat 




Intron 



Exon I 



Exon II 



GC 



+ 




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 



5' 



AGPUAUGU • 
5' site 



Intron 
••UACUAAO««§AG 



Exon II 



3' site 



3' 



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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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 



u 



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 



U 



GGUUUAAAGGC 

I I I I I I I I I I I 

CCAAAUUUCUG 



AU A AG AU A 



A 

C 
U 



U A GU C 



GGAc R u xV 

\_A G 



U C 



u 



u 



u 



5' 



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 



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" 



U 



qGUUlZ/Cu. 



A G 



U 



D-stem 



A G— C 

G— U 

G— C 

A — U 

G— C 

G— U 

U — A 

U — A 

U— A 

c c u 



U 



A 
C 
U 
C 



Au u[/> U , 

A G 



U 



//An 

u u A '/ u 



3' 



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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273 




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. 



u 



CUy 



u 



c 
u 
c 



u 



u 
u 

G 

c 
u 



u 



u 



Group 1 consensus 



U— G 

U— G 

U— G 

C— G 

G — C 



A C C— G 



u 



c 
c 



u P s puc 

U GGG U 
I I I I I 

° u cc c 





Box 2 



&\ 



U 



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 



u 



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, 



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 




3' 



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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Transcription 



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278 



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? 



70 



.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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Molecular Genetics 



10. Gene Expression: 
Transcription 



©TheMcGraw-Hil 
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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 

t 

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 
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 
Companies, 2001 



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 299 

The Signal Hypothesis 301 

The Protein-Folding Problem 303 
The Genetic Code 304 

Triplet Nature of the Code 304 

Breaking the Code 305 

Wobble Hypothesis 307 

Universality of the Genetic Code 308 

Evolution of the Genetic Code 311 
Summary 312 
Solved Problems 313 
Exercises and Problems 313 
Critical Thinking Questions 314 
Box 11.1 Amino Acid Sequencing 284 
Box 11.2 Antibiotics 294 



280 



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Genetics, Seventh Edition 



Molecular Genetics 



11. Gene Expression: 
Translation 



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Information Transfer 



281 



In this chapter, we continue our discussion of gene 
expression, concentrating on protein biosynthesis. 
This process, which translates the nucleotide infor- 
mation in messenger RNA into amino acid se- 
quences in proteins, is the final step of the central 
dogma. Nucleotide sequences in DNA are transcribed 
into nucleotide sequences in RNA, which are then trans- 
lated into amino acid sequences in proteins. 

All proteins are synthesized from only twenty naturally 
occurring amino acids (fig. 11.1). (There is one exception, 
selenocysteine, which we discuss at the end of the chap- 
ter.) These are called a-amino acids because one carbon, 
the a carbon, has four specific groups attached to it: an 
amino group, a carboxyl (acidic) group, a hydrogen, and 
one of the twenty different R groups (side chains), impart- 
ing the specific properties of that amino acid. (Technically, 
proline is termed an imino acid because of its structure.) 
Having these four groups attached imparts a property 
known as chirality on the amino acid: like left- and right- 
handed gloves, the mirror images cannot be superim- 
posed. Because of optical properties, the two forms of 
each amino acid are referred to as D and L, in which D 
comes from dextrorotatory (right turning) and L comes 
from levorotatory (left turning). All biologically active 
amino acids are of the L form, and hence we need not re- 
fer to this designation. Proteins (polypeptides) are synthe- 
sized when peptide bonds form between any two amino 
acids (fig. 11.2). In this manner, long chains of amino 
acids — called residues when incorporated into a 
protein — can join, and all chains will have an amino 
(N-terminal) end and a carboxyl (C-terminal) end. 

The sequence of polymerized amino acids determines 
the primary structure of a protein. Included in the pri- 
mary structure is the formation of disulfide bridges be- 
tween cysteine residues (fig. 11.3). Polypeptides can fold 
into several structures, the most common of which are a 
helices and p sheets. These folding configurations consti- 
tute the secondary structure of the protein. In some 
proteins, the folding is spontaneous; in some, it is guided 
by other proteins. Further folding, bringing a helices and 
p sheets into three-dimensional configurations, creates 
the tertiary structure of the protein (fig. 11.4). Many 
proteins in the active state are composed of several sub- 
units that together make up the quaternary structure of 
the protein. Translation is the process in which the pri- 
mary structure of a protein is determined from the nu- 
cleotide sequence in a messenger RNA (box 11.1). 




INFORMATION 
TRANSFER 



Before proceeding to the details of translation, a sketch of 
the beginning of the process may be helpful (fig. 11.5). 



The ribosome with its ribosomal RNA and proteins is the 
site of protein synthesis .The information from the gene is 
in the form of messenger RNA, in which each group 
of three nucleotides — a codon — specifies an amino acid. 
The amino acids are carried to the ribosome attached to 
transfer RNAs, and these transfer RNAs have anticodons, 
three nucleotides complementary to a codon, located 
at the end opposite the amino acid attachment site. A 
peptide bond will form between the two amino acids 
present at the ribosome, freeing one transfer RNA (at 
codon 1 in fig. 11.5) and lengthening the amino acid 
chain attached to the second transfer RNA (at codon 2 in 
fig. 1 1.5). The messenger RNA will then move one codon 
with respect to the ribosome, and a new transfer RNA 
will attach at codon 3. This cycle is then repeated, 
with the polypeptide lengthening by one amino acid 
each time. We can begin looking at the details of transla- 
tion by looking at the transfer RNAs. As before, we con- 
centrate on the prokaryotic system, noting details about 
eukaryotes as appropriate. 

Transfer RNA ^l* 

Attachment of Amino Acid to Transfer RNA 

The function of transfer RNA is to ensure that each amino 
acid incorporated into a protein corresponds to a partic- 
ular codon (a group of three consecutive nucleotides) in 
the messenger RNA. The transfer RNA serves this func- 
tion through its structure: It has an anticodon at one end 
and an amino acid attachment site at the other end. The 
"correct" amino acid, the amino acid corresponding to 
the anticodon, is attached to the transfer RNA by en- 
zymes known as aminoacyl-tRNA synthetases (e.g., 
arginyl-tRNA synthetase, leucyl-tRNA synthetase). A trans- 
fer RNA with an amino acid attached is said to be 
"charged." 

An aminoacyl-tRNA synthetase joins a specific amino 
acid to its transfer RNA in a two-stage reaction that takes 
place on the surface of the enzyme. In the first stage, the 
amino acid is activated with ATP. In the second stage of the 
reaction, the amino acid is attached with a high-energy 
bond to the 2' or 3' carbon of the ribose sugar at the 3' 
end of the transfer RNA (fig. 1 1 .6). In the figure, we denote 
high-energy bonds, bonds that liberate a lot of free energy 
when hydrolyzed, as "-."Thus, during the process of pro- 
tein synthesis, the energy for the formation of the peptide 
bond will be present where it is needed, at the point of 
peptide bond formation. 



Component Numbers 

In bacteria, there are twenty aminoacyl-tRNA synthetases, 
one for each amino acid. A particular enzyme recognizes a 
particular amino acid, as well as all the transfer RNAs that 



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Molecular Genetics 



11. Gene Expression: 
Translation 



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Chapter Eleven Gene Expression: Translation 






General amino acid 

H H 

// 



O 



H — N + — C — C 



\ 



H R 



O 






Acidic 

L-Aspartic acid (Asp,D) 

H H 



L-Glutamic acid (Glu,E) 



i+ 



H— N— C — C 



// 



.0 



\ 



H CH 2 

C 

/ % 

-o o 



o 



H H 
H— N + — C — C 
H CH 2 

CH 2 

C 

/ % 

-o o 









.0 



o 



Basic 

L-Lysine (Lys.K) 

H H 



H— N + — C — C 






.0 



L-Arginine (Arg,R) 
H H 



i+ 



H— N— C — C 



// 



.0 



H CH, 



CH, 



O 



\ 






CH, 



H 3 N + — CH 2 



L-Histidine (His.H) 
H H 



H CH, 



CH, 



NH CH, 



H 3 N + — C— NH 



O 



H— N + — C — C 



S 



O 



\ 



H CH, 



O 



C — N 



HC— N 



t 
\ 



CH 



H 



Figure 11.1 The twenty amino acids found in 
proteins and their three- and one-letter abbreviations. 
At physiological pH, the amino acids usually exist as 
ions. Note the classification of the various R groups. 



Nonpolar 

L-Alanine (Ala,A) 

H H 












i+ 



H — N— C — C 



// 



.0 



L-Valine (Val.V) 
H H 



i+ 



\ 



H CH, 



O 



\ 



H— N — C— C 

H CH 

/\ 

H 3 C CH 3 



// 



.0 



o 



L-Leucine (Leu,L) L-Phenylalanine (Phe,F) 
H H _ H H 



i+ 



H — N— C — C 



// 



.0 



\ 



H — N + — C — C 



// 



.0 



H CH 2 

CH 

/\ 

H 3 C CH 3 



O 



\ 



H CH, 



O 




L-Methionine (Met,M) L-lsoleucine (Me, I) 
H H ^ H H 



i+ 



H — N— C — C 



// 



O 



i+ 



\ 



H CH, 



CH, 



O 



S— CH, 



H— W— C— C 

H CH 

/\ 

CH 9 CH, 

H 3 C 



// 



O 



\ 



o 



Polar (uncharged) 

L-Cysteine (Cys,C) 
H H 



L-Proline (Pro,P) 
H H 



H — N + — C — C 

/ \ \ 

CH CH 
^X 2 
CH, 



.0 



O 



L-Tryptophan (Trp,W) 



H H 






H — N + — C — C 



H 



// 



.0 







L-Glycine (Gly.G) 



H H 






H — N + — C — C 



// 



O 



\ 



H H 



O 






H — N + — C — C 



\ 



.0 



L-Serine (Ser.S) 
H H 



L-Threonine (Thr,T) 
H H 



H CH, 



SH 



O 



H — N + — C — C H — N + — C— C 

I I V I I V 

H CH 2 H HC— CH 3 



OH 



L-Asparagine (Asn,N) L-Glutamine (Gln,Q) 
H H _ H H 



H — N + — C — C 



// 



O 



\ 



H — N + — C — C 



// 



.0 



OH 

L-Tyrosine (Tyr,Y) 
H H 









i+ 



H CH, 



O 



\ 



H — N— C — C 



// 



O 



C— NH, 



O 



H CH, 



CH, 



O 



\ 



H CH, 



O 




C— NH, 



O 




OH 



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Molecular Genetics 



11. Gene Expression: 
Translation 



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283 



H R 



i+ 



H — N— C — C 



H H 




ATP 



AMP + PP 



H 




H 2 




H — N + — C — C — N — C— C 



H H 



// 
\ 



O 



o 



N 
Terminal 



C 

Terminal 



Figure 11.2 Protein synthesis: formation of a peptide bond 
between two amino acids. The bond is between the carboxyl 
group of one amino acid and the amino group of the other. 



Cysteine 



H H 



H — N + — C — C 



S 



O 



H 



\ 



O 



H-C-H 



S — H 



Cysteine 



H— S 



H-C-H 



O 



\ 



H 



O 



/ 



C — C— N— H 



H H 



H H 



H — N + — C — C 



// 



.0 



H 



\ 



O 



H-C-H 

S 

s 

H-C-H 



O 



% 



H 



O 



/ 



C — C— N— H 

H H 
Cystine 



Figure 11.3 A disulfide bridge can form when two cysteines 
are brought into apposition. If the two amino acids are in the 
free form, the new structure is called cystine. When the two 
cysteines are in the same or different polypeptides, the disulfide 
bridge creates stability. 





(a) 



(b) 




(c) 



Figure 11.4 Three different ways of depicting a protein, the enzyme phosphoglycerate kinase. At left is a bond diagram; all the lines 
shown represent bonds between the various atoms of the molecule. In the middle is a ribbon diagram that emphasizes the 
secondary structure of the protein. Shown are alpha helices (spiral ribbons) and beta pleated sheets (flat arrows). Finally, on the right 
is a space-filling diagram that emphasizes the volume the molecule fills. The space-filling diagram is what the molecule would 
generally look like if it were magnified eight million times. (Images by David S. Goodsell, the Scripps Research Institute.) 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



11. Gene Expression: 
Translation 



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284 



Chapter Eleven Gene Expression: Translation 



BOX 11.1 



Protein-sequencing techniques 
have been known since 1953, 
when F. Sanger worked out the 
complete sequence of the protein 
hormone insulin. The basic strategy is 
to purify the protein and then se- 
quence it, beginning at one end. 
However, since most proteins contain 
too many amino acids to do this suc- 
cessfully, proteins are first broken 
into small peptides in several differ- 
ent ways. These peptides are se- 
quenced, and the whole protein se- 
quence can be determined by the 
overlap pattern of the sequenced 
subunits. 

A protein can be broken into pep- 
tide fragments by many different 
methods, including acid and alkaline 
hydrolysis. For the most part, pro- 
teolytic enzymes (proteases) that hy- 
drolyze the peptides at specific 



Experimental 
Methods 



Amino Acid Sequencing 



points are used. Pepsin, for example, 
preferentially hydrolyzes peptide 
bonds involving aromatic amino 
acids, methionine, and leucine; chy- 
motrypsin hydrolyzes peptide bonds 
involving carboxyl groups of aro- 
matic amino acids; and trypsin hy- 
drolyzes bonds involving the car- 
boxyl groups of arginine and lysine. 

The proteolytic digest is usually 
separated into a peptide map, or pep- 
tide fingerprint, by using a two- 
dimensional combination of paper 



chromatography, electrophoresis, or 
column chromatography. In two- 
dimensional chromatography, a sam- 
ple is put onto a piece of paper that is 
then placed in a solvent system. After 
an allotted time, the paper is dried, 
turned 90 degrees, and placed in a 
second solvent system for another 
allotted time (fig. 1). In each solvent, 
different peptides travel through the 
paper at different rates. The spots are 
then developed using ninhydrin, 
which reacts with the N-terminal 
amino acid and produces a colored 
product when heated. 

The spots, which represent small 
peptides, can be cut out of a second, 
identical chromatogram that has not 
been sprayed with ninhydrin. These 
spots can then be sequenced by, for 
example, the Edman method, whereby 
the peptide is sequentially degraded 




Chromatography 
plate 



Solvent 




Solvent system I 



Dry; then rotate 





Fingerprint 



Solvent system II 

Figure 1 Two-dimensional paper chromatography of a protease digest. Chromatography is done first in one 
solvent system. The paper is then dried, rotated, and placed into a second solvent system. The pattern on the 
resulting plate is called a peptide fingerprint. 



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Translation 



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from the N-terminal end. Phenyliso- 
thiocyanate (PITC) reacts with the 
amino end of the peptide. When acid 
is added, the N-terminal amino acid is 
removed as a PITC derivative and can 
be identified. The process is then re- 



peated until the whole peptide has 
been sequenced (fig. 2). 

If the fingerprint pattern is 
worked out for two different digests 
of the same polypeptide, the unique 
sequence of the original polypeptide 



HNH 



R H — CH 



C=0 



NH 



R n-? H 



C=0 

OH 
Peptide 




Base 

— >■ 




— CH Treat with acid 



R — 



CH 
C=0 



PITC 



OH 




PITC derivative 



HNH 



R 2 — CH 



C=0 

NH 

R n-CH 
C=0 

OH 
Shortened peptide 



Identification 



Repeat process 



Figure 2 Isolation of amino acids from a peptide for sequencing purposes. 
First, the peptide reacts with PITC (phenylisothiocyanate) at the amino end. Acid 
treatment produces a PITC derivative of the amino-terminal amino acid and a 
peptide one amino acid shorter than the original. The PITC derivative can be 
identified. These steps are then repeated, isolating one amino acid at a time. 



can be determined by overlap. In 
figure 3, the letters A-J represent the 
ten amino acids in a polypeptide. A is 
known to be the first (N-terminal) 
amino acid since the Edman method 
sequences peptides from this end.We 
can thus summarize the methodol- 
ogy as follows: 

1 . A protein is purified. If it is made 
up of several subunits, these sub- 
units are separated and purified. 
(If disulfide bridges exist within a 
peptide, they must be reduced. 
The bridges are later determined 
by digestion, keeping the bridges 
intact, and then resequencing.) 

2. Different proteolytic enzymes are 
used on separate subsamples so 
that the protein is broken into dif- 
ferent sets of peptide fragments. 

3. Two-dimensional chromatography, 
electrophoresis, or column chro- 
matography can be used to isolate 
the peptides. 

4. The Edman method of sequen- 
tially removing amino acids from 
the N-terminal end is used to se- 
quence each peptide. 

5. The amino acid sequence from 
the N- to C-terminal ends of the 
protein is deduced from the over- 
lap of sequences in peptide di- 
gests generated with different pro- 
teolytic enzymes. 

Today, a machine known as an amino 
acid sequencer (sequenatof) can au- 
tomatically sequence protein. Taking 
about two hours per amino acid 
residue, sequenators can carry out 
Edman degradation on polypeptides 
up to about fifty amino acids long. 



continued 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



11. Gene Expression: 
Translation 



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286 



Chapter Eleven Gene Expression: Translation 



Protein 



(AXS©(DtiXFYgHXD® 



BOX 11.1 CONTINUED 



Peptides 
Method 1 

®®©®dXE)©®0® 

Method 2 



Figure 3 The overlap of peptides 
digested two different ways provides 
the sequence of the original peptide. 



Anticodon 




Ribosome movement 

>■ 



mRNA movement 



J •••mRNA 



Figure 11.5 The initiation of the translation process at the 
ribosome. Note the two charged transfer RNAs and the 
messenger RNA. They are in position to form the first 
peptide bond between the two amino acids attached to the 
transfer RNAs. 



code for that amino acid. In eukaryotes, there are separate 
sets of twenty cytoplasmic and twenty mitochondrial syn- 
thetases, all coded in the nucleus. 

Aminoacyl-tRNA synthetases are a heterogeneous 
group of enzymes. In E. coli, they vary from monomeric 
proteins (one subunit) to tetrameric proteins, made up of 
two copies each of two subunits.The enzymes fall into 
two categories based on sequence similarity, structural 
features, and whether the amino acid is attached at the 
2'-OH (in class I enzymes) or 3 -OH (in class II enzymes) 
of the 3 '-terminal adenosine of the transfer RNA. 

To add its appropriate amino acid to the appropriate 
transfer RNA, a synthetase recognizes many parts of the 
transfer RNA. This can be shown by experiments that in- 
troduce specific changes in transfer RNAs by site- 
directed mutagenesis (see chapter 13). In seventeen of 



the twenty E. coli synthetases, recognition involves part 
of the anticodon itself. This makes sense since the anti- 
codon is the defining element of a transfer RNA in pro- 
tein synthesis. 

A synthetase can initially make errors and attach the 
"wrong" amino acid to a tRNA. For example, isoleucyl- 
tRNA synthetase will attach valine about once in 225 
times. This type of error occurs because a similar, but 
smaller, amino acid can sometimes occupy the active site 
of the enzyme (compare isoleucine and valine in fig. 
11.1). However, because of a proofreading step, only 1 in 
270 to 1 in 800 of the errors are released intact from the 
enzyme. The amino acids on the rest of the incorrectly 
charged transfer RNAs are hydrolyzed before the transfer 
RNAs are released.The overall error rate is the product of 
the two steps; this means only about one incorrectly 
charged transfer RNA occurs per 60,000 to 80,000 
formed. 

In several cases, the number of amino acyl-tRNA syn- 
thetases in a particular organism is below twenty. For ex- 
ample, in some archaea, there is no cysteinyl-tRNA syn- 
thetase. However, the prolyl-tRNA synthetase activates 
the tRNAs for both cysteine and proline with their ap- 
propriate amino acids. Similarly, in some eubacteria, there 
is no glutaminyl-tRNA synthetase; the glutaminyl tRNA is 
charged with glutamic acid, rather than glutamine. An 
amido transferase enzyme then converts the glutamic 
acid to glutamine (see fig. 11.1). 

There are sixty-four possible codons in the genetic 
code (four nucleotide bases in groups of three = 4x4x4 
= 64). Three of these codons are used to terminate trans- 
lation. Thus, sixty-one transfer RNAs are needed because 
there are sixty-one different nonterminator codons. 
About fifty transfer RNAs are known in E. coli. The num- 
ber fifty can be explained by the wobble phenomenon, 
which occurs in the third position of the codon. We ex- 
amine this phenomenon in the section on the genetic 
code. The transfer RNAs for each amino acid are desig- 
nated by the convention tRNA Leu (for leucine), tRNA His 
(for histidine), and so on. 



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(1 ) Amino acid (aa) + ATP (adenosine — ®®(p) ^ 



aa~(F) — adenosine + (p)(p) 



H R 



H — N + — C — CC + adenosine 

I I x o- 

H H 




— (PXPXP 



(2) aa~(p) — adenosine + tRNA 



H R O 

H — N + —C—C — 0~(p) — adenosine +@(? 

H H 
■>• aa~tRNA + adenosine + (p) 

OH OH 




C— s. /—Adenine 
O 



H R O 

I I II ^ 

H — N + — C — C — 0~(P) — adenosine 

H H 




V_y tRNA 



H R O 



H R 



O 



H — N— C— C~ 



N + — C 

I I 
H H 




C - <\ />— Adenine 
O 



H — N + - 
H 



C~o 




OH, 

H / 2" 

C—^-yS— Adenine 



or 




+ adenosine 



-® 



Charged tRNA 

Figure 11.6 It takes a two-step process to attach a specific amino acid to its transfer RNA by an aminoacyl 
synthetase. High-energy bonds are indicated by ~. In the first step, an amino acid is attached to AMP with a high- 
energy bond. In the second step, the high-energy bond is transferred to the tRNA, which is then referred to as 
"charged." Depending on which class of aminoacyl-tRNA synthetase is involved, the amino acid will be attached to 
either the 2' or 3' carbon of the sugar of the 3' terminal adenosine. 



Recognition of the Aminoacyl-tRNA 
During Protein Synthesis 

Although amino acids enter the protein-synthesizing 
process attached to transfer RNAs, it was theoretically 
possible that the ribosome recognized the amino acid it- 
self during translation. A simple experiment was done to 
determine whether the amino acid or the transfer RNA 
was recognized. 

In 1962, F. Chapeville and colleagues isolated trans- 
fer RNA with cysteine attached. They chemically con- 



verted the cysteine to alanine by using Raney nickel, a 
catalytic form of nickel that removes the SH group of 
cysteine (fig. 11.7). When these transfer RNAs were 
used in protein synthesis, alanine was incorporated 
where cysteine should have been, demonstrating that 
the transfer RNA, not the amino acid, was recognized 
during protein synthesis. The synthetase puts a specific 
amino acid on a specific transfer RNA; then, during pro- 
tein synthesis, the anticodon on the transfer RNA — not 
the amino acid itself — determines which amino acid is 
incorporated. 



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Chapter Eleven Gene Expression: Translation 



NH 3 + 



Cysteine 

SH — CH 2 — CH 



NH 3 + 



CH 3 — CH 



Alanine 



H H 



HC = 



O 



tRNA c y s 




5' 



Remove SH 
Raney nickel 



KJ 





tRNA°y s 



Figure 11.7 Cysteine-tRNA Cys treated with Raney nickel 
becomes alanine-tRNA Cys by the removal of the SH group of 
cysteine. During protein synthesis, alanine is incorporated in 
place of cysteine in proteins, indicating that the specificity of 
amino acid incorporation into proteins resides with the tRNA. 




Initiation Complex 

Translation can be divided into three stages: initiation, 
elongation, and termination. Elongation is the repetitive 
process of adding amino acids to a growing peptide 
chain. However, added complexity enters the picture in 
the initiation and termination of protein synthesis. 

It is especially important that the translation process 
start precisely. Remember that the genetic code is trans- 
lated in groups of three nucleotides (codons). If the 
reading of the messenger RNA begins one base too early 
or too late, the reading frame is shifted so that an en- 
tirely different set of codons is read (fig. 11. 8). The pro- 
tein produced, if any, will probably bear no structural or 
functional resemblance to the protein the gene is 
coded for. 



CAUCAUCAUCAU 





mRNA 



(a) 




CAUCAUCAUCAU 





C 

_L 



mRNA 




(b) 



Figure 11.8 (a) In the normal reading of the messenger RNA, 
these codons are read as repeats of CAU, coding for histidine. 
(b) A shift in the reading frame of the messenger RNA causes 
the codons to be read as AUC repeats coding for isoleucine. 



H— N 



— C — C 



// 



O 



H H 



\ 



H CH, 



CH, 



O 



= C — N 

H 



— C — C 



// 



o 



\ 



CH, 



CH, 



O 



S— CH, 



Methionine 



S— CH, 



N-formyl methionine 



Figure 11.9 The structures of the amino acids methionine and 
N-formyl methionine. 



Role ofN-Formyl Methionine 

The synthesis of every protein in Escherichia coli begins 
with the modified amino acid N-formyl methionine 
(fig. 11.9). However, none of the completed proteins in 
E. coli contains N-formyl methionine. Many of these pro- 
teins do not even have methionine as their first amino 
acid. Obviously, before a protein becomes functional, the 
initial amino acid is modified or removed. In eukaryotes 
the initial amino acid, also methionine, does not have an 
N-formyl group. 

Methionine, with a codon of 5-AUG-3', known as the 
initiation codon, has two transfer RNAs with the same 
complementary anticodon (3'-UAC-5') but with different 
structures (fig. 11.10). One of these transfer RNAs 
(tRNA^ et ) serves as a part of the initiation complex. Be- 
fore the initiation of translation, this transfer RNA will 
have its methionine chemically modified to N-formyl me- 
thionine (fMet).The other transfer RNA will not have its 
methionine modified (tRNA^ et ).The translation machin- 
ery will use it to insert methionine into proteins, where 
called for, in all but the first position. The cell thus has a 
mechanism to make use of methionine in the normal way 
as well as to use a modified form of it to initiate protein 
synthesis. Because of the structure of the prokaryotic ini- 
tiation transfer RNA, it can recognize AUG, GUG, and, 
rarely, UUG as initiation codons. In eukaryotes, CUG as 
well as AUG can serve as an initiation codon. Since the 
initiation methionine is not formylated in eukaryotes, the 
eukaryotic transfer RNA is designated tRNAj Met ; there is a 
separate internal methionine transfer RNA, termed 
tRNAm^ et , in eukaryotes, as in prokaryotes. 

Translation Initiation 

The subunits of the ribosome (30S and 5 OS) usually dis- 
sociate from each other when not involved in translation. 
To begin translation, an initiation complex forms, con- 
sisting of the following components in prokaryotes: the 
30S subunit of the ribosome, a messenger RNA, the 
charged N-formyl methionine tRNA (fMET-tRNA™ et ), and 



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u 



C C C G U 



G 



¥ 



G G A C 



X 



G 



(a) tRNA M f et 



A 
C 



G 



3' 
A 
C 
C 
A 



5' 



C — G 

C — G 

G — C 

A — U 



U — A 
















3' 










G — C 
















A 










C — G 
















C 










U 
















C 










A 






G 










A 


5' 








G C 


U C 


A 


\mA 


U 
D 

G o 








A - 

c - 

G — 


- C 

- G 

- c 








C G 


A G 






G 


















A 




A 


D 


D 








c - 


- G 








G — C 
















c - 


- G 








U — A 


A 


A 


■ i 










c - 


- G 








A — U 


A 




U 


C C 








c - 


- G 








G — C 


A 






| | 


| 


| 


| 






U 






\j/ — A 








G G 












G 




A G 


A C 


C 


¥ 


T 






u 


C 








A 


G C 


A U 


























U A C 

■ i 










G 


'7 

G 








G 




C G 

A D 


1 


Anticodon 














A 


A 
G 
C 


— 


U 
C 
G 






Odd bases 
















C 


— 


G 






D = Dihydrouridine 
















C 


— 


G 






\\f = Pseudouridine 
















A 




Co 




(b) tRNA J? et 


T = Ribothymidine 






















\ / ill 


G 7 = 7-Methylguany 


ic acic 














A 




U 






G = 2'-0-methylguanylic acid 
C = 2'-0-methylcytidylic acid 












U 

i 


A 


i 


















1 








X = 3-(3-Amino-3-carboxypropyl) 


uridine 








Anticodon 







u 



G 



Figure 11.10 The two tRNAs for methionine in E. coli. (a) The initiator tRNA. (o) The interior tRNA. 



three initiation factors (IF1, IF2, IF3). Initiation fac- 
tors (as well as elongation and termination factors) are 
proteins loosely associated with the ribosome.They were 
discovered when ribosomes were isolated and then 
washed, losing the ability to perform protein synthesis. 

The components that form the initiation complex in- 
teract in a series of steps. It is known that IF3 binds to the 
30S ribosomal subunit, allowing the 30S subunit to bind 
to messenger RNA (fig. 11.11, step 7). Meanwhile, a com- 
plex forms with IF2, the charged N-formyl methionine 
tRNA (fMET-tRNA™ et ) and GTP (guanosine triphosphate; 
fig. 11.11, step 2). It is IF2 that brings the initiator transfer 
RNA to the ribosome. IF2 binds only to the charged ini- 
tiator transfer RNA, and, without IF2, the initiator transfer 
RNA cannot bind to the ribosome. The final step in 



initiation-complex formation is bringing together the 
first two components (fig. 11.11, step 3). 

The hydrolysis of GTP to GDP + J> { (inorganic phos- 
phate, PO4 3 — see fig. 9.8) produces conformational 
changes; these changes allow the initiation complex to 
join the 50S ribosomal subunit to form the complete ri- 
bosome and then allow the initiation factors and GDP to 
be released. Frequently, the hydrolysis of a nucleoside 
triphosphate (e.g., ATP, GTP) in a cell occurs to release 
the energy in the phosphate bonds for use in a metabolic 
process. However, in the process of translation, the hy- 
drolysis apparently changes the shape of the GTP so that 
it and the initiation factors can be released from the ri- 
bosome after the 70S particle has been formed. Thus, hy- 
drolysis of GTP in translation is for conformational 



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Chapter Eleven Gene Expression: Translation 



Step 1 MF3J 



+ (30s) 

Ribosome subunit 





+ 



5'- 



mRNA 



3' 




Step 2 ( |F2 j + (j3TPj) 



+ 



fMet 

o 








fMet-tRNA^ 



Step 3 Combine end products of step 1 and step 2 






+ 



Initiation complex 




GDP+ 
P, 



GTP 



t 



~\ 



IF2 
IF3 




Complete 70S ribosome 



Figure 11.11 The prokaryotic 70S ribosome forms in a three-step process. In the first step, the 30S ribosomal subunit 
and the mRNA combine. In the second step, the initiator tRNA combines with IF2. In the final step, the components from 
steps 1 and 2 combine to form the initiation complex, followed by the formation of the 70S ribosome. 



change rather than covalent bond formation. IF1 helps 
the other two initiation factors bind to the 30S ribosomal 
subunit or stabilizes the 30S initiation complex. 

The process in eukaryotes is generally similar, but 
more complex. The eukaryotic initiation factor abbrevia- 
tions are preceded by an "e" to denote that they are eu- 
karyotic (elFl, eIF2, etc.). At least eleven initiation fac- 
tors are involved, including a specific cap-binding 
protein, eIF4E. 



The ribosome apparently recognizes the prokaryotic 
messenger RNA through complementarity of a region at 
the 3' end of the 16S ribosomal RNA and a region slightly 
upstream from the initiation sequence (AUG) on the mes- 
senger RNA. This idea, the Shine-Dalgarno hypothesis 
(fig. 1 1 . 1 2), is named after the people who first suggested 
it. The sequence (AGGAGGU) of complementarity be- 
tween the messenger RNA and the 16S ribosomal RNA is 
referred to as the Shine-Dalgarno sequence. Although 



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5'... 



(a) 



"i — i — i — i — i — i — i — i — i — i — i — i — r 

GUACUAAGGAGGU 

AUUCCUCCA 

3' J I I I I I I I I 



"l — i — i — r 

U G A A 



fMet 

~~ r~ 



Phage X cm gene 
Glu Gin Arg 



1 — I — I — I — I — I — I — I — I — r 

UGGAACAACGC 




• ••3' 
mRNA 



5' 
16SrRNA 



Normal 



5' cap 



AUG 



Shunting [■ 



IRES 



(b) 



5' cap 



AUG 



AUG 



ORF 



I- 



5' cap 



AUG 



IRES 



AUG 



Figure 11.12 Translation initiation signals, (a) The Shine-Dalgarno hypothesis for prokaryotic translation. The Shine-Dalgarno 
sequence (AGGAGGU) is on the prokaryotic messenger RNA just upstream from the initiation codon AUG. Complementarity 
exists between this sequence and a complementary sequence (UCCUCCA) on the 3' end of the 1 6S ribosomal RNA. 
[b) Scanning, shunting, and internal ribosome entry in eukaryotic messenger RNA. The 5' untranslated region of a eukaryotic 
gene is shown in red; the beginning of the gene in blue. Normally, in the scanning model, the initiation codon of the gene is 
the first AUG encountered. In shunting, an open reading frame (ORF, green) may or may not be present to provide 
secondary structure in the messenger RNA to shunt scanning to the main gene. If the open reading frame is translated, 
reinitiation of translation at the same ribosome may occur at the main gene. Finally, an internal ribosome entry site (IRES, 
yellow) allows translation to begin within the messenger RNA without scanning. 



there is a good deal of homology between prokaryotic 
and eukaryotic small ribosomal RNAs, the Shine-Dalgarno 
region is absent in eukaryotes.The actual mechanism for 
recognizing the 5 ' end of eukaryotic messenger RNA ap- 
pears to be based on recognition of the 5' cap of the mes- 
senger RNA by the cap-binding protein with recruitment 
of other initiation factors and the small subunit of the ri- 
bosome. This is followed by the small subunit's move- 
ment down the messenger RNA. The ribosome scans the 
messenger RNA until it recognizes the initiation codon. 
This model is referred to as the scanning hypothesis. 
In several known cases in eukaryotes, a process called 
shunting occurs, in which the first AUG does not serve 
as the initiation codon; rather, scanning begins, but it by- 
passes a region of the messenger RNA upstream of the 
initiation codon, called the leader or 5' untranslated 
region (5' UTR), in favor of an AUG further down the 
messenger RNA. The cause of shunting seems to be sec- 
ondary structure in the messenger RNA, upstream from 



the AUG codon that actually serves as the initiation 
codon. In some cases, very small genes, called open 
reading frames (ORFs), are present in this region of 
the messenger RNA and play some role in shunting. It 
seems also that some ORFs are translated, and then the 
main gene is translated by the same ribosome in a 
process called reinitiation (fig. 11.12£>).We have seen 
this in the genes of some plant and animal viruses; it is a 
topic under current study 

Under some circumstances, eukaryotic ribosomes 
can initiate protein synthesis within the messenger RNA 
if that messenger RNA contains a sequence called an in- 
ternal ribosome entry site. These sequences were 
discovered in the poliovirus RNA and in several cellular 
messenger RNAs. They are at least four hundred nu- 
cleotides long. Thus, although scanning accounts for the 
initiation of most eukaryotic messenger RNAs at their 5' 
ends, some initiation can take place internally in mes- 
senger RNAs that have internal ribosome entry sites. 



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Chapter Eleven Gene Expression: Translation 



Aminoacyl and Peptidyl Sites in the Ribosome 

When the initiator transfer RNA joins the 30S subunit 
of the prokaryotic ribosome with its messenger RNA 
attached, it fits into one of three sites in the ribosome. 
These sites, or cavities in the ribosome, are referred to 
as the aminoacyl site (A site), the peptidyl site (P 
site), and the exit site (E site, fig. 11.13) Here, we con- 
centrate on the A and P sites, each of which contains a 
transfer RNA just before forming a peptide bond: the P 
site contains the transfer RNA with the growing pep- 
tide chain (peptidyl-tRNA); the A site contains a new 
transfer RNA with its single amino acid (aminoacyl- 
tRNA).The exit site helps eject depleted transfer RNAs 
after a peptide bond forms. When the complete 70S ri- 
bosome of figure 11.11 has formed, the initiation fMET- 
tRNAf et is placed directly into the P site (fig. 11.13), 
the only charged transfer RNA that can be placed di- 
rectly there. The association of transfer RNA and ribo- 
some is aided by a G-C base pairing between the 
3' -CCA terminus of all transfer RNAs and a guanine in 
the 23S ribosomal RNA. 




Elongation 

Positioning a Second Transfer RNA 

The next step in prokaryotic translation is to position 
the second transfer RNA, which is specified by the 
codon at the A site. The second transfer RNA is posi- 
tioned in the A site of the ribosome so it is able to form 
hydrogen bonds between its anticodon and the second 
codon on the messenger RNA.This step requires the cor- 
rect transfer RNA, another GTP, and two proteins called 



70S - 




l_l_l ... mRNA 



elongation factors (EF-Ts and EF-Tu). EF-Tu, bound to 
GTP, is required to position a transfer RNA into the A site 
of the ribosome (fig. 1 1 . 14). After the transfer RNA is po- 
sitioned, the GTP is hydrolyzed to GDP + Pj. Upon hy- 
drolysis of the GTP, the EF-Tu/GDP complex is released 
from the ribosome. EF-Ts is required to regenerate an EF- 
Tu/GTP complex. EF-Ts displaces the GDP on EF-Tu. 
Then a new GTP displaces EF-Ts, and now the EF- 
Tu/GTP complex can bind another transfer RNA. Here 
again, the hydrolysis of GTP changes the shape of the 
GTP so that the EF-Tu/GDP complex can depart from 
the ribosome after the transfer RNA is in place in the A 
site (fig. 11.15). Figure 11.16 shows the ribosome at the 



Figure 11.13 The 70S ribosome contains an A site, a P site, 
and an E site that can receive tRNAs. The messenger RNA 
runs through the bottom of the sites. 



end of this step. EF-Tu does not bind fMet-tRNA™ et , so 
this blocked (formylated) methionine cannot be in- 
serted into a growing peptide chain. 

It takes several milliseconds for the GTP to be hy- 
drolyzed, and another few milliseconds for the EF- 
Tu/GDP to actually leave the ribosome. During those two 
intervals of time, the codon-anticodon fit of the transfer 
RNA is scrutinized. If the correct transfer RNA is in place, 
a peptide bond forms. If not, the charged transfer RNA is 
released and a new cycle of EF-Tu/GTP-mediated testing 
of transfer RNAs begins. The error rate is only about one 
mistake in ten thousand amino acids incorporated into 
protein. The speed of amino acid incorporation is about 
fifteen amino acids per second in prokaryotes and about 
two to five per second in eukaryotes. 

Peptide Bond Formation 

The two amino acids on the two transfer RNAs are now 
in position to form a peptide bond between them; both 
amino acids are juxtaposed to an enzymatic center, 
peptidyl transferase, in the 5 OS subunit. This enzy- 
matic center, an integral part of the 50S subunit, was 
originally believed to be composed of parts of several of 
the 50S proteins. Now, however, it is believed to have ri- 
bozymic activity, enzymatic activity of the ribosomal 
RNA of the ribosome. The enzymatic activity involves a 
bond transfer from the carboxyl end of N-formyl me- 
thionine to the amino end of the second amino acid 
(phenylalanine in fig. 11.16). Every subsequent peptide 
bond is identical, regardless of the amino acids in- 
volved. The energy used is contained in the high-energy 
ester bond between the transfer RNA in the P site and 
its amino acid (fig. 11.17). Immediately after the forma- 
tion of the peptide bond, the transfer RNA with the 
dipeptide is in the A site, and a depleted transfer RNA is 
in the P site (box 11.2). 

Translocation 

The next stage in elongation is translocation of the ribo- 
some in relation to the transfer RNAs and the messenger 



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RNA. Elongation factor EF-G, earlier called translocase, 
catalyzes the translocation process. The ribosome must 
be converted from the pretranslocational state to the 
posttranslocational state by the action of EF-G, which 
physically moves the messenger RNA and its associated 
transfer RNAs (fig. 11.18). This movement is accom- 
plished by the hydrolysis of a GTP to GDP after EF-G en- 
ters the ribosome at the A site. After the first posttranslo- 
cational state is reached, the depleted transfer RNA in the 
E site is ejected, leaving the ribosome ready to accept a 
new charged transfer RNA in the A site. A computer- 
generated diagram of a ribosome with all three transfer 
RNA sites occupied is shown in figure 11.18&. In eukary- 



otes, three elongation factors perform the same tasks that 
EF-Tu, EF-Ts, and EF-G perform in prokaryotes.The factor 
eEFla replaces EF-Tu, eEFip7 replaces EF-Ts, and eEF2 
replaces EF-G. 

When translocation is complete, the situation is again 
as diagrammed in figure 11.13, except that instead of 
fMet-tRNAf et , the P site contains the second transfer RNA 
(tRNA phe ) with a dipeptide attached to it. The process of 
elongation is then repeated, with a third transfer RNA 
coming into the A site. The process repeats from here to 
the end (fig. 11.19), synthesizing a peptide starting from 
the amino (N-terminal) end and proceeding to the car- 
boxyl (C-terminal) end. During the repetitive aspect of 



mRNA ••• 



mRNA ... 




Figure 11.14 The EF-Ts/EF-Tu cycle. EF-Ts and EF-Tu are required for a transfer RNA to attach to the A site of the ribosome. At 
top center, we have EF-Tu attached to a GDR The GDP is then displaced by EF-Ts, which in turn is displaced by GTP. A transfer 
RNA attaches and is brought to the ribosome. If the codon-anticodon fit is correct, the transfer RNA attaches at the A site with the 
help of the hydrolysis of GTP to GDP + P i; allowing the EF-Tu to release. The EF-Tu is now back where we started. Since EF-Tu has 
a strong affinity for GDP, the role of EF-Ts is to displace the GDP, and later to be replaced by GTP. 



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Chapter Eleven Gene Expression: Translation 



BOX 11.2 



Antibiotics, substances living 
organisms produce that are 
toxic to other living organ- 
isms, are of interest to us for two rea- 
sons: They have been extremely im- 
portant in fighting the diseases that 
strike human beings and farm ani- 
mals, and many are useful tools for an- 
alyzing protein synthesis. Some antibi- 
otics impede the process of protein 
synthesis in a variety of ways, often 
poisoning bacteria selectively; the ef- 
fectiveness of antibiotics normally de- 
rives from the metabolic differences 
between prokaryotes and eukaryotes. 
For example, an antibiotic that blocks 
a 70S bacterial ribosome without af- 
fecting an 80S human ribosome could 
be an excellent antibiotic. About 160 
antibiotics are known. 

PUROMYCIN 

Puromycin resembles the 3' end of an 
aminoacyl-tRNA (fig. 1). It is bound to 
the A site of the bacterial ribosome, 
where peptidyl transferase creates a 
bond from the nascent peptide at- 
tached to the transfer RNA in the P site 
to puromycin. Elongation can then no 
longer occur. The peptide chain is re- 
leased prematurely, and protein syn- 
thesis at the ribosome terminates. 

Experiments with puromycin 
helped demonstrate the existence of 
the A and P sites of the ribosome. It 
was found that puromycin could 
not bind to the ribosome if transloca- 
tion factor EF-G were absent. With 
EF-G, translocation took place, and 
puromycin could then bind to the ri- 
bosome. Puromycin's ability to bind 
only after translocation indicates that 
a second site on the ribosome be- 
comes available after translocation. 

STREPTOMYCIN, TETRACYCLINE, 
AND CHLORAMPHENICOL 

Streptomycin, which binds to one of 
the proteins (protein SI 2) of the 30S 
subunit of the prokaryotic ribosome, 
inhibits initiation of protein synthe- 
sis. Streptomycin also causes misread- 



Biomedical 
Applications 



Antibiotics 



ing of codons if chain initiation has 
already begun, presumably by alter- 
ing the conformation of the ribosome 
so that transfer RNAs are less firmly 
bound to it. Bacterial mutants that are 
streptomycin resistant, as well as 
mutants that are streptomycin de- 
pendent (they cannot survive with- 
out the antibiotic), occur. Both types 
of mutants have altered 30S subunits, 
specifically changed in protein SI 2. 

Tetracycline blocks protein syn- 
thesis by preventing an aminoacyl- 
tRNA from binding to the A site on 
the ribosome. Chloramphenicol 
blocks protein synthesis by binding 
to the 50S subunit of the prokaryotic 
ribosome, where it blocks the pep- 
tidyl transfer reaction. Chlorampheni- 
col does not affect the eukaryotic 
ribosome. However, chlorampheni- 
col, as well as several other antibi- 
otics, is used cautiously because the 
mitochondrial ribosomes within eu- 
karyotic cells are very similar to 
prokaryotic ribosomes. Some of the 
antibiotics that affect prokaryotic ri- 
bosomes thus also affect mitochon- 
dria. As was mentioned, the similarity 
between bacteria and mitochondria 
implies that mitochondria have a 
prokaryotic origin. (Similarities be- 
tween cyanobacteria and chloroplasts 
also support the idea that chloro- 
plasts have a prokaryotic origin.) 

THE TROUBLE WITH 
ANTIBIOTICS 

Over the years, antibiotics have virtu- 
ally eliminated certain diseases from 
the industrialized world. They have 
also made modern surgery possible 
by preventing most serious infections 



that tend to follow operations. Antibi- 
otics have been so successful that, in 
the 1980s, many pharmaceutical 
companies drastically cut back the 
development of new antibiotics. 
However, a disaster was in the mak- 
ing as we overprescribed antibiotics 
to people and farm animals: bacteria 
are not prepared to take this on- 
slaught without fighting back. 

Mutation takes place all the time 
at a low but dependable rate. Thus, re- 
sistant bacteria are constantly arising 
from sensitive strains. We can select 
for penicillin- and streptomycin- 
resistant strains of bacteria in the lab- 
oratory by allowing the antibiotic to 
act as a selective agent, removing all 
but the resistant individuals. The 
same sort of artificial selection that 
we can apply in the lab applies every 
time a person or animal takes an an- 
tibiotic. We may be at a point now 
where the ability of bacteria to de- 
velop resistance, and to pass that re- 
sistance to other strains, has put us 
on the verge of disaster. The process 
of evolution works amazingly fast in 
bacteria because of their ubiquity, 
large population sizes, and ability to 
transfer genetic material between 
individuals. We may shortly find 
ourselves as we were before World 
War II, when simple infections in 
hospitals were often lethal. Right 
now, only one antibiotic can keep the 
common — and potentially deadly — 
infectious bacterium Staphylococcus 
under control: vancomycin. Several 
types of disease-causing bacteria 
have already evolved a tolerance to 
vancomycin. 

The answer to this potentially dis- 
astrous problem is to develop new 
antibiotics and reduce the irresponsi- 
ble use of antibiotics in people and 
animals. Hopefully, the warning bell 
has sounded. At least a dozen new an- 
tibiotics that show promise are in the 
early stages of development by phar- 
maceutical companies. 



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~v^ 




Psite 



yv 



CH 3 CH 3 



Puromycin 



A site 



Peptidyl 
puromycin 



A site 



Figure 1 Puromycin is bound to the A site of the ribosome. A peptide bond then forms. Further elongation is 
prevented, and the chain is terminated. 



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Chapter Eleven Gene Expression: Translation 




(a) 




(b) 



Figure 11.15 Space-filling model of EF-Tu bound with (a) GDP and {b) GTP, showing the change in the protein's structure. Yellow, 
blue, and red are domains of the protein. The GTP and GDP are in white, with a magnesium ion, Mg 2+ , in green. When EF-Tu is 
bound with GDP, there is a visible hole in the molecule. The hole disappears when GTP is bound. The am inoacy I -transfer RNA is 
believed to bind between the red and yellow domains. (Courtesy of Rolf Hiigenfeid.) 




...mRNA 
3' 



Figure 11.16 A ribosome with two transfer RNAs attached. In 
this case, the second codon (UUU) is for the amino acid 
phenylalanine. The two amino acids are next to each other. 



protein synthesis, two GTPs are hydrolyzed per peptide 
bond: one GTP in the release of EF-Tu from the A site, and 
one GTP in the translocational process of the ribosome 
after the peptide bond has formed. In addition, every 



charged transfer RNA has had an amino acid attached at 
the expense of the hydrolysis of an ATP to AMP +PP 
There is some evidence that the action of EF-Tu 
hydrolyzes two GTPs. 

Termination t^ 

Nonsense Codons 

Termination of protein synthesis in both prokaryotes and 
eukaryotes occurs when one of three nonsense 
codons appears in the A site of the ribosome. These 
codons are UAG (sometimes referred to as amber), UAA 
(ochre), and UGA (opal). ("Amber," or brown stone, is the 
English translation of the name Bernstein, a graduate stu- 
dent who took part in the discovery of UAG in R. H. Ep- 
stein's lab at the California Institute of Technology. 
"Ochre" and "opal" are tongue-in-cheek extensions of the 
first label.) In prokaryotes, three proteins called release 
factors (RF) are involved in termination, and a GTP is hy- 
drolyzed to GDP + P A . 

When a nonsense codon enters the A site on the ri- 
bosome, a release factor recognizes it. RF1 and RF2 are 
class 1 release factors: They recognize stop codons and 
then promote hydrolysis of the bond between the termi- 
nal amino acid and its tRNA in the P site. Class 2 release 
factors (RF3) do not recognize stop codons, but they 
stimulate class 1 release factors to act. RF1 recognizes the 
stop codons UAA and UAG, and RF2 recognizes UAA and 
UGA (fig. 11.20). Both do so because they have tripep- 



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fMet 



Phe 



^V 




Psite 



A site 



Figure 11.17 Peptide bond formation on the ribosome 
between N-formyl methionine and phenylalanine. The bond 
attaching the carboxyl end of the first amino acid to its tRNA is 
transferred to the amino end of the second amino acid. The 
first tRNA is now uncharged, whereas the second tRNA has a 
dipeptide attached. 



tides that mimic anticodons to recognize the stop 
codons: proline-alanine-threonine in RF1 and serine- 
proline-phenylalanine in RF2. In this molecular mim- 
icry, a protein mimics the shape of a nucleic acid in or- 
der to function properly 

The next base in the messenger RNA past the stop 
codon is usually an adenine, required for efficient termi- 
nation. After the release factors act, with the hydrolysis of 
a GTP, the ribosome has completed its task of translating 
mRNA into a polypeptide. Final release of all factors and 
dissociation of the two subunits of the ribosome take 



place with the help of IF3, which rebinds to the 30S sub- 
unit, and a ribosome recycling factor (RRF). Table 
11.1 compares prokaryotic and eukaryotic translation. 



Rate and Cost of Translation 

As mentioned, the average speed of protein synthesis is 
about fifteen peptide bonds per second in prokaryotes. 
Discounting the time for initiation and termination, an 
average protein of three hundred amino acids is synthe- 
sized in about twenty seconds (the released protein 
forms its final structure spontaneously or is modified 
with the aid of other proteins, as we shall see). An equiv- 
alent eukaryotic protein takes about 2.5 minutes to be 
synthesized. The energy cost is at least four high-energy 
phosphate bonds per peptide bond (two from an ATP 
during transfer RNA charging, and two from GTP hydro- 
lysis during transfer RNA binding at the A site and 
translocation), or about twelve hundred high-energy 
bonds per protein. This cost is very high — about 90% of 
the energy production of an E. coli cell goes into protein 
synthesis. A high energy cost is presumably the price a 
living system has to pay for the speed and accuracy of its 
protein synthesis. 



Coupling of Transcription and Translation 

In prokaryotes, such as E. coli, in which no nuclear en- 
velope exists, translation begins before transcription is 
completed. Figure 11.21 shows a length of an E. coli 
chromosome. An RNA polymerase is visible on the 
DNA, transcribing a gene. The messenger RNA, still be- 
ing synthesized, can be seen extending away from the 
DNA.Attached to the messenger RNA are about a dozen 
ribosomes. Since translation starts at the end of the 
messenger that is synthesized first (5'), an initiation 
complex can form and translation can begin shortly af- 
ter transcription begins. As translation proceeds along 
the messenger, its 5' end will again become exposed, 
and a new initiation complex can form. The occurrence 
of several ribosomes translating the same messenger is 
referred to as a polyribosome, or simply a polysome 
(fig. 11.22). 

In prokaryotes, most messenger RNAs contain the in- 
formation for several genes. These RNAs are said to be 
polycistronic (fig. 11.23). (Cistron, another term for 
gene, is defined in chapter 12.) Each gene on the messen- 
ger RNA is translated independently: each has a Shine- 
Dalgarno sequence for ribosome recognition (see fig. 11.12) 
and an initiation codon (AUG) for fMet.The ribosome that 
completes the translation of the first gene may or may not 
continue to the second gene after dissociation. The trans- 
lation of any gene follows all the steps we have outlined. 



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Molecular Genetics 



11. Gene Expression: 
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Chapter Eleven Gene Expression: Translation 




Binding of 
EF-G + GTP' 



•••nnRNA 



30S 



Pretranslocational state 



(a) 



3' 




UU/A 
••• i i i i i i i 



••• nnRNA 



First posttranslocational state 



Final posttranslocational state 




mRNA 



lNflyjH'pLMe 



(b) 



Figure 11.18 (a) EF-G's translocation of the ribosome converts 
it from a pretranslocational state (P and A sites occupied) to a 
posttranslocational state (E and P sites occupied). The 
uncharged transfer RNA in the E site is then ejected, (b) A see- 
through model of the 70S ribosome of E. coli with transfer 
RNAs in the A, P, and E sites. The structure was determined 
by cryoEM mapping, an electron microscopic technique using 
rapidly frozen specimens. The position of the messenger RNA 
is shown, as well as the stalk of the 50S subunit (St) and one 
of the polypeptides of the large subunit, L1 . ([£>] Courtesy of 
Joachim Frank, Howard Hughes Medical Institute.) 



Table 11.1 Some Comparisons Between Prokaryotic and Eukaryotic Translation 





Prokaryotes 




Eukaryotes 


Initiation codon 


AUG, occasion 


ally GUG, UUG 


AUG, occasionally GUG,CUG 


Initiation amino acid 


N-formyl methionine 


Methionine 


Initiation tRNA 


tRNAf et 




tRNA™ et 


Interior methionine tRNA 


tRNAlT 




tRNA™ et 


Initiation factors 


IF1,IF2,IF3 




elF factors 


Elongation factor 


EF-Tu 




eEFla 


Elongation factor 


EF-Ts 




eEFl(37 


Translocation factor 


EF-G 




eEF2 


Release factors 


RF1,RF2, RF3, 


RRF 


eRFl,eRF3 



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iLu_i»»»mRNA 




i_u •• 



(a) 




(b) 



Figure 11.19 Cycle of peptide bond formation and 
translocation on the ribosome. (a) After the peptide bond is 
transferred (fig. 11.17), the ribosome and messenger RNA 
move over one codon. Now the transfer RNA with the peptide 
is in the P site, and the A site is again open. In this example, 
the next transfer RNA that moves into the A site carries 
glutamic acid, {b) Three-dimensional model of the translocation 
process minus the mRNA and amino acids. The tRNA in the A 
site is pale blue, the tRNA in the P site is green, and the tRNA 
in the E site is yellow, then brown when ready to leave. Going 
clockwise from a, in which the A and P sites are occupied: 
EF-G translocates the ribosome after peptide bond formation 
and then evacuates the A site. Ef-Tu brings a new charged 
tRNA to the A site while the E site is emptied, {[b] Courtesy of 
Joachim Frank.) 



In eukaryotes, however, almost all messenger RNAs 
contain the information for only one gene (mono- 
cistronic). Since most ribosomal recognition of eu- 
karyotic genes depends on the 5' cap, and since each 
eukaryotic messenger RNA has only one cap, usually 
only one polypeptide can be translated for any given 
messenger RNA. Exceptions occur when the messen- 
ger RNAs contain internal ribosome entry sites. Al- 
though it is certainly not the rule, the translated pep- 
tide can be modified or cleaved into smaller functional 
peptides. For example, in mice, a single messenger 
RNA codes for a protein that is later cleaved into epi- 
dermal growth factor and at least seven other related 
peptides. In addition, the same sequence can, in some 
cases, give rise to alternative proteins through alterna- 



tive start codons, termination read-through, or alterna- 
tive splicing. 

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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Molecular Genetics 



11. Gene Expression: 
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.) 



o 






■ . 



£ 






I."** 






*3 



' * ' - n 

* >$"■ '-■: 



(b) 



Nascent polypeptide 
(early) 



mRNA 
Ribosome 



Nascent polypeptide 
(late) 



Gene 1 

I 



Gene 2 

I 



fMet •••Terminator 



Gene 3 

I 



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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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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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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11. Gene Expression: 
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 



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 


C 


A 


G 


C 


A 


G 


C 


A 


G 


C 


A 


G 


codon 1 


C 


A 


G 




















codon 2 






G 


C 


A 
















codon 3 










A 


G 


C 












codon 4 














C 


A 


G 








codon 5 


















G 


C 


A 




codon 6 






















A 


G C 


Overlapping, 
two bases no 
punctuation 


C 


A 


G 


C 


A 


G 


C 


A 


G 


C 


A 


G 


codon 1 


C 


A 


G 




















codon 2 




A 


G 


C 


















codon 3 






G 


C 


A 
















codon 4 








C 


A 


G 














codon 5 










A 


G 


C 












codon 6 












G 


C 


A 










codon 7 














C 


A 


G 








codon 8 
















A 


G 


C 






codon 9 


















G 


C 


A 




codon 10 




















C 


A 


G 



• • • 



Nonoverlapping, 
punctuation 

codon 



Punctuation 
CAGCAGCAGCAG 



1 



• • • 



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) 


U 


C A 


G 


Phe 


Ser 


Tyr 


Cys 






Phe 


Ser 


Tyr 


Cys 


C 






U 


















Leu 


Ser 


stop 


stop 


A 








Leu 


Ser 


stop 


Trp 


G 
U 




Leu 


Pro 


His 


Arg 






Leu 


Pro 


His 


Arg 


C 






C 


















Leu 


Pro 


Gin 


Arg 


A 








Leu 


Pro 


Gin 


Arg 


G 
U 




lie 


Thr 


Asn 


Ser 






He 


Thr 


Asn 


Ser 


C 






A 


















He 


Thr 


Lys 


Arg 


A 








Met (start) 


Thr 


Lys 


Arg 


G 
U 




Val 


Ala 


Asp 


Gly 






Val 


Ala 


Asp 


Gly 


C 






G 


















Val 


Ala 


Glu 


Gly 


A 








Val 


Ala 


Glu 


Gly 


G 





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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The Genetic Code 



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(Met J 
3' 



5' 




tRNA 




Anticodon 



5' 



1 2 3 



3' mRNA 



L 



J 



T 

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 


in 


Number-three Base in 


tRNA (5' End) 




mRNA (3' End) 


G 




U or C 


C 




G 


A 




U 


U 




A or G 


I 




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 



H 



H 



Guanine 



/ 



y 



y 



H /-\ C ^^ CH 



o 



M 



N 



HC 



// 



y 



H 



y 



y 



\ 



N 



J_ 



To ribose sugar 



•C N 



N N 

H 



/ 



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 



H 
,C 



Inosine 



/ 



/ 



y 



H ^\ c ^ ^ CH 



HC 






o 

-C. 



\ 

r 
Z 



N' 



"N 



S 



CH 



H 



y 



y 



y 



y 



N 



-N' 



Cytosine 



C 
O 



To ribose 
sugar 



To ribose sugar 



H H 



y 



H 



/ 



N 



S 



Inosine 



/ 



O 



HC 



// C N 



y 



y 



> 

y 



-N 

> 

>N Adenine 



To ribose 
sugar 



\ 



N- 



z 



V N 



S 



CH 



To ribose sugar 









H 
X ^CH 


Uracil 


Inosine 


y 
y 


L 




To ribose 
sugar 






J 
\ 

/ 


y 

^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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Molecular Genetics 



11. Gene Expression: 
Translation 



©TheMcGraw-Hil 
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The Genetic Code 



311 



Table 1 1 .6 The Genetic Code Dictionary of Yeast Mitochondria* 



First Position (5' 


End) 
U 

C 

A 

G 




Second Position 




Third Position (3' 

U 
C 
A 
G 
U 
C 
A 
G 
U 
C 
A 
G 
U 
C 
A 
G 


End) 


U 


C A 


G 


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 


He 


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. 



Tamarin: Principles of 
Genetics, Seventh Edition 



Molecular Genetics 



11. Gene Expression: 
Translation 



©TheMcGraw-Hil 
Companies, 2001 



312 



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