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Questions and Problems

Chapter 1

1.1 In a few sentences, what were Mendel’s key ideas about

inheritance?

ANS: Mendel postulated transmissible factors—genes—to

explain the inheritance of traits. He discovered that

genes exist in different forms, which we now call alleles.

Each organism carries two copies of each gene. During

reproduction, one of the gene copies is randomly incor-

porated into each gamete. When the male and female

gametes unite at fertilization, the gene copy number is

restored to two. Different alleles may coexist in an organ-

ism. During the production of gametes, they separate

from each other without having been altered by

coexistence.

1.2 Both DNA and RNA are composed of nucleotides. What

molecules combine to form a nucleotide?

ANS: Each nucleotide consists of a sugar, a nitrogen-containing

base, and a phosphate.

1.3 Which bases are present in DNA? Which bases are pres-

ent in RNA? Which sugars are present in each of these

nucleic acids?

ANS: The bases present in DNA are adenine, thymine, gua-

nine, and cytosine; the bases present in RNA are adenine,

uracil, guanine, and cytosine. The sugar in DNA is

deoxyribose; the sugar in RNA is ribose.

1.4 What is a genome?

ANS: A genome is the set of all the DNA molecules that are

characteristic of an organism. Each DNA molecule

forms one chromosome in a cell of the organism.

1.5 The sequence of a strand of DNA is ATTGCCGTC. If

this strand serves as the template for DNA synthesis,

what will be the sequence of the newly synthesized

strand?

ANS: TAACGGCAG

1.6 A gene contains 141 codons. How many nucleotides are

present in the gene’s coding sequence? How many amino

acids are expected to be present in the polypeptide

encoded by this gene?

ANS: There are 3  141  423 nucleotides in the gene’s cod-

ing sequence. Its polypeptide product will contain 141

amino acids.

1.7 The template strand of a gene being transcribed is CTT-

GCCAGT. What will be the sequence of the RNA made

from this template?

ANS: GAACGGUCT

1.8 What is the difference between transcription and

translation?

ANS: Transcription is the production of an RNA chain using a

DNA chain as a template. Translation is the production

of a chain of amino acids—that is, a polypeptide—using

an RNA chain as a template.

1.9 RNA is synthesized using DNA as a template. Is DNA

ever synthesized using RNA as a template? Explain.

ANS: Sometimes, DNA is synthesized from RNA in a process

called reverse transcription. This process plays an impor-

tant role in the life cycles of some viruses.

1.10 The gene for a-globin is present in all vertebrate species.

Over millions of years, the DNA sequence of this gene

has changed in the lineage of each species. Consequently,

the amino acid sequence of a-globin has also changed in

these lineages. Among the 141 amino acid positions in

this polypeptide, human a-globin differs from shark

a-globin in 79 positions; it differs from carp a-globin in

68 and from cow a-globin in 17. Do these data suggest

an evolutionary phylogeny for these vertebrate species?

ANS: The human and cow a-globins are least different; there-

fore, on the assumption that differences in a-globin

re ect the degree of phylogenetic relationship, the

human and the cow are the most closely related organ-

isms among those mentioned. The next closest “relative”

of humans is the carp, and the most distant relative is the

shark.

1.11 Sickle-cell anemia is caused by a mutation in one of the

codons in the gene for b-globin; because of this mutation,

the sixth amino acid in the b-globin polypeptide is a

valine instead of a glutamic acid. A less severe type of ane-

mia is caused by a mutation that changes this same codon

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to one specifying lysine as the sixth amino acid in the

b-globin polypeptide. What word is used to describe the

two mutant forms of this gene? Do you think that an indi-

vidual carrying these two mutant forms of the b-globin

gene would suffer from anemia? Explain.

ANS: The two mutant forms of the b-globin gene are properly

described as alleles. Because neither of the mutant alleles

can specify a “normal” polypeptide, an individual who

carries each of them would probably suffer from

anemia.

1.12 Hemophilia is an inherited disorder in which the blood-

clotting mechanism is defective. Because of this defect,

people with hemophilia may die from cuts or bruises,

especially if internal organs such as the liver, lungs, or

kidneys have been damaged. One method of treatment

involves injecting a blood-clotting factor that has been

puried from blood donations. This factor is a protein

encoded by a human gene. Suggest a way in which mod-

ern genetic technology could be used to produce this

factor on an industrial scale. Is there a way in which the

inborn error of hemophilia could be corrected by human

gene therapy?

ANS: The gene for the human clotting factor could be isolated

from the human genome and transferred into bacteria,

which could then be grown in vats to produce large

amounts of the gene’s protein product. This product could

be isolated from the bacteria, puried, and then injected

into patients to treat hemophilia. Another approach would

be to transfer a normal copy of the clotting factor gene

into the cells of people who have hemophilia. If expressed

properly, the transferred normal gene might be able to

compensate for the mutant allele these people naturally

carry. For this approach to succeed, the normal clotting

factor gene would have to be transferred into the cells that

produce clotting factor, or into their precursors.

Chapter 2

2.1 Carbohydrates and proteins are linear polymers. What

types of molecules combine to form these polymers?

ANS: Sugars combine to form carbohydrates; amino acids

combine to form proteins.

2.2 All cells are surrounded by a membrane; some cells are

surrounded by a wall. What are the differences between

cell membranes and cell walls?

ANS: Cell membranes are made of lipids and proteins; they

have a uid structure. Cell walls are made of more rigid

materials such as cellulose.

2.3 What are the principal differences between prokaryotic

and eukaryotic cells?

ANS: In a eukaryotic cell, the many chromosomes are con-

tained within a membrane-bounded structure called the

nucleus; the chromosomes of prokaryotic cells are not

contained within a special subcellular compartment.

Eukaryotic cells usually possess a well-developed inter-

nal system of membranes and they also have membrane-

bounded subcellular organelles such as mitochondria

and chloroplasts; prokaryotic cells do not typically have

a system of internal membranes (although some do), nor

do they possess membrane-bounded organelles.

2.4 Distinguish between the haploid and diploid states.

What types of cells are haploid? What types of cells are

diploid?

ANS: In the haploid state, each chromosome is represented

once; in the diploid state, each chromosome is repre-

sented twice. Among multicellular eukaryotes, gam-

etes are haploid and somatic cells are diploid.

2.5 Compare the sizes and structures of prokaryotic and

eukaryotic chromosomes.

ANS: Prokaryotic chromosomes are typically (but not always)

smaller than eukaryotic chromosomes; in addition, pro-

karyotic chromosomes are circular, whereas eukaryotic

chromosomes are linear. For example, the circular chro-

mosome of E. coli, a prokaryote, is about 1.4 mm in cir-

cumference. By contrast, a linear human chromosome

may be 10–30 cm long. Prokaryotic chromosomes also

have a comparatively simple composition: DNA, some

RNA, and some protein. Eukaryotic chromosomes are

more complex: DNA, some RNA, and a lot of protein.

2.6 With a focus on the chromosomes, what are the key

events during interphase and M phase in the eukaryotic

cell cycle?

ANS: During interphase, the chromosomes duplicate. During

M phase (mitosis), the duplicated chromosomes, each

consisting of two identical sister chromatids, condense

(a feature of prophase), migrate to the equatorial plane of

the cell (a feature of metaphase), and then split so that

their constituent sister chromatids are separated into dif-

ferent daughter cells (a feature of anaphase); this last

process is called sister chromatid disjunction.

2.7 Which typically lasts longer, interphase or M phase? Can

you explain why one of these phases lasts longer than the

other?

ANS: Interphase typically lasts longer than M phase. During

interphase, DNA must be synthesized to replicate all the

chromosomes. Other materials must also be synthesized

to prepare for the upcoming cell division.

2.8 In what way do the microtubule organizing centers of

plant and animal cells differ?

ANS: The microtubule organizing centers of animal cells have

distinct centrosomes, whereas the microtubule organiz-

ing centers of plant cells do not.

2.9 Match the stages of mitosis with the events they encom-

pass: Stages: (1) anaphase, (2) metaphase, (3) prophase,

and (4) telophase. Events: (a) reformation of the nucleo-

lus, (b) disappearance of the nuclear membrane,

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the mitotic spindle, (e) movement of chromosomes to

the equatorial plane, (f) movement of chromosomes to

the poles, (g) decondensation of the chromosomes, (h)

splitting of the centromere, and (i) attachment of micro-

tubules to the kinetochore.

ANS: (1) Anaphase: (f), (h); (2) metaphase: (e), (i); (3) prophase:

(b), (c), (d); (4) telophase: (a), (g).

2.10 Arrange the following events in the correct temporal

sequence during eukaryotic cell division, starting with

the earliest: (a) condensation of the chromosomes,

(b) movement of chromosomes to the poles, (c) duplica-

tion of the chromosomes, (d) formation of the nuclear

membrane, (e) attachment of microtubules to the kineto-

chores, and (f) migration of centrosomes to positions on

opposite sides of the nucleus.

ANS: (c), (f), (a), (e), (b), (d).

2.11 In human beings, the gene for b-globin is located on chro-

mosome 11, and the gene for a-globin, which is another

component of the hemoglobin protein, is located on chro-

mosome 16. Would these two chromosomes be expected to

pair with each other during meiosis? Explain your answer.

ANS: Chromosomes 11 and 16 would not be expected to pair

with each other during meiosis; these chromosomes are

heterologues, not homologues.

2.12 A sperm cell from the fruit y Drosophila melanogaster

contains four chromosomes. How many chromosomes

would be present in a spermatogonial cell about to enter

meiosis? How many chromatids would be present in a

spermatogonial cell at metaphase I of meiosis? How

many would be present at metaphase II?

ANS: There are eight chromosomes in a Drosophila spermato-

gonial cell about to enter meiosis. There are 16 chroma-

tids in a Drosophilia spermatogonial cell at metaphase I of

meiosis. There are eight chromatids in a Drosophilia cell

at metaphase II of meiosis.

2.13 Does crossing over occur before or after chromosome

duplication in cells going through meiosis?

ANS: Crossing over occurs after chromosomes have duplicated

in cells going through meiosis.

2.14 What visible characteristics of chromosomes indicate

that they have undergone crossing over during meiosis?

ANS: The chiasmata, which are visible late in prophase I of

meiosis, indicate that chromosomes have crossed over.

2.15 During meiosis, when does chromosome disjunction

occur? When does chromatid disjunction occur?

ANS: Chromosome disjunction occurs during anaphase I.

Chromatid disjunction occurs during anaphase II.

2.16 In Arabidopsis, is leaf tissue haploid or diploid? How

many nuclei are present in the female gametophyte?

How many are present in the male gametophyte? Are

these nuclei haploid or diploid?

ANS: Leaf tissue is diploid. The female gametophyte contains

eight identical haploid nuclei. The male gametophyte

contains three identical haploid nuclei.

2.17 From the information given in Table 2.1 in this chapter,

is there a relationship between genome size (measured in

base pairs of DNA) and gene number? Explain.

ANS: Among eukaryotes, there does not seem to be a clear

relationship between genome size and gene number. For

example, humans, with 3.2 billion base pairs of genomic

DNA, have about 20,500 genes, and Arabidopsis plants,

with about 150 million base pairs of genomic DNA, have

roughly the same number of genes as humans. However,

among prokaryotes, gene number is rather tightly cor-

related with genome size, probably because there is so

little nongenic DNA.

2.18 Are the synergid cells in an Arabidopsis female gameto-

phyte genetically identical to the egg cell nestled between

them?

ANS: Yes.

2.19 A cell of the bacterium Escherichia coli, a prokaryote, con-

tains one chromosome with about 4.6 million base pairs

of DNA comprising 4288 protein-encoding genes. A cell

of the yeast Saccharomyces cerevisiae, a eukaryote, contains

about 12 million base pairs of DNA comprising 6268

genes, and this DNA is distributed over 16 distinct chro-

mosomes. Are you surprised that the chromosome of a

prokaryote is larger than some of the chromosomes of a

eukaryote? Explain your answer.

ANS: It is a bit surprising that yeast chromosomes are, on aver-

age, smaller than E. coli chromosomes because, as a rule,

eukaryotic chromosomes are larger than prokaryotic

chromosomes. Yeast is an exception because its genome—

not quite three times the size of the E. coli genome—is

distributed over 16 separate chromosomes.

2.20 Given the way that chromosomes behave during meiosis,

is there any advantage for an organism to have an even

number of chromosome pairs (such as Drosophila does),

as opposed to an odd number of chromosome pairs (such

as human beings do)?

ANS: No, there is no advantage associated with an even num-

ber of chromosomes. As long as the chromosomes come

in pairs, they will be able to synapse during prophase I

and then disjoin during anaphase I to distribute the

genetic material properly to the two daughter cells.

2.21 In owering plants, two nuclei from the pollen grain par-

ticipate in the events of fertilization. With which nuclei

from the female gametophyte do these nuclei combine?

What tissues are formed from the fertilization events?

ANS: One of the pollen nuclei fuses with the egg nucleus in the

female gametophyte to form the zygote, which then

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develops into an embryo and ultimately into a sporo-

phyte. The other genetically functional pollen nucleus

fuses with two nuclei in the female gametophyte to form

a triploid nucleus, which then develops into a triploid

tissue, the endosperm; this tissue nourishes the develop-

ing plant embryo.

2.22 The mouse haploid genome contains about 2.9  10

nucleotide pairs of DNA. How many nucleotide pairs of

DNA are present in each of the following mouse cells:

(a) somatic cell, (b) sperm cell, (c) fertilized egg, (d) pri-

mary oocyte, (e) rst polar body, and (f) secondary

spermatocyte?

ANS: (a) 5.8  10

nucleotide pairs (np); (b) 2.9  10

np;

np; (d) 11.6  10

np; (e) 5.8  10

np;

and (f) 5.8  10

2.23 Arabidopsis plants have 10 chromosomes (ve pairs) in

their somatic cells. How many chromosomes are present

in each of the following: (a) egg cell nucleus in the female

gametophyte, (b) generative cell nucleus in a pollen

grain, (c) fertilized endosperm nucleus, and (d) fertilized

egg nucleus?

ANS: (a) 5, (b) 5, (c) 15, (d) 10.

Chapter 3

3.1 On the basis of Mendel’s observations, predict the results

from the following crosses with peas: (a) a tall (dominant

and homozygous) variety crossed with a dwarf variety;

(b) the progeny of (a) self-fertilized; (c) the progeny from

(a) crossed with the original tall parent; (d) the progeny

of (a) crossed with the original dwarf parent.

ANS: (a) All tall; (b) 3/4 tall, 1/4 dwarf; (c) all tall; (d) 1/2 tall,

1/2 dwarf.

3.2 Mendel crossed pea plants that produced round seeds

with those that produced wrinkled seeds and self-fertil-

ized the progeny. In the F

, he observed 5474 round

seeds and 1850 wrinkled seeds. Using the letters W and

w for the seed texture alleles, diagram Mendel’s crosses,

showing the genotypes of the plants in each generation.

Are the results consistent with the Principle of

Segregation?

ANS: Round (WW )  wrinkled (ww) → F

round (Ww); F

self-fertilized → F

3/4 round (2 WW; 1 Ww), 1/4 wrin-

kled (ww). The expected results in the F

are 5493 round,

1831 wrinkled. To compare the observed and expected

results, compute c

with one degree of freedom;

(5474  5493)

/5493  (1850  1831)

/1831  0.263,

which is not signicant at the 5% level. Thus, the results

are consistent with the Principle of Segregation.

3.3 A geneticist crossed wild, gray-colored mice with white

(albino) mice. All the progeny were gray. These progeny

were intercrossed to produce an F

, which consisted of

198 gray and 72 white mice. Propose a hypothesis to

explain these results, diagram the crosses, and compare

the results with the predictions of the hypothesis.

ANS: The data suggest that coat color is controlled by a single

gene with two alleles, C (gray) and c (albino), and that C

is dominant over c. On this hypothesis, the crosses are

gray (CC)  albino (cc) → F

gray (Cc); F

 F

→ 3/4

gray (2 CC: 1 Cc), 1/4 albino (cc). The expected results in

the F

are 203 gray and 67 albino. To compare the

observed and expected results, compute c

with one

degree of freedom: (198  203)

/203  (67  72)

/72 

0.470, which is not signicant at the 5% level. Thus, the

results are consistent with the hypothesis.

3.4 A woman has a rare abnormality of the eyelids called

ptosis, which prevents her from opening her eyes com-

pletely. This condition is caused by a dominant allele,

P. The woman’s father had ptosis, but her mother had

normal eyelids. Her father’s mother had normal

eyelids.

(a) What are the genotypes of the woman, her father, and

her mother?

(b) What proportion of the woman’s children will have

ptosis if she marries a man with normal eyelids?

ANS: (a) Woman’s genotype Pp, father’s genotype Pp, mother’s

genotype pp; (b) ½

3.5 In pigeons, a dominant allele C causes a checkered pat-

tern in the feathers; its recessive allele c produces a plain

pattern. Feather coloration is controlled by an indepen-

dently assorting gene; the dominant allele B produces

red feathers, and the recessive allele b produces brown

feathers. Birds from a true-breeding checkered, red vari-

ety are crossed to birds from a true-breeding plain,

brown variety.

(a) Predict the phenotype of their progeny.

(b) If these progeny are intercrossed, what phenotypes

will appear in the F

and in what proportions?

ANS: (a) Checkered, red (CC BB)  plain, brown (cc bb) → F

all checkered, red (Cc Bb); (b) F

progeny: 9/16 check-

ered, red (C- B-), 3/16 plain, red (cc B-), 3/16 checkered,

brown (C- bb), 1/16 plain, brown (cc bb).

3.6 In mice, the allele C for colored fur is dominant over the

allele c for white fur, and the allele V for normal behavior

is dominant over the allele v for waltzing behavior, a

form of dis-coordination. Given the genotypes of the

parents in each of the following crosses:

(a) Colored, normal mice mated with white, normal mice

produced 29 colored, normal, and 10 colored, waltzing

progeny

(b) Colored, normal mice mated with colored, normal

mice produced 38 colored, normal, 15 colored, waltzing,

11 white, normal, and 4 white, waltzing progeny

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mice produced 8 colored, normal, 7 colored, waltzing,

9 white, normal, and 6 white, waltzing progeny.

ANS: (a) colored, normal (CC Vv)  white, normal (cc Vv)

(b) colored, normal (Cc Vv)  colored, normal (Cc Vv);

3.7 In rabbits, the dominant allele B causes black fur and the

recessive allele b causes brown fur; for an independently

assorting gene, the dominant allele R causes long fur and

the recessive allele r (for rex) causes short fur. A homozy-

gous rabbit with long, black fur is crossed with a rabbit

with short, brown fur, and the offspring are intercrossed.

In the F

, what proportion of the rabbits with long, black

fur will be homozygous for both genes?

ANS: Among the F

progeny with long, black fur, the geno-

typic ratio is 1 BB RR: 2 BB Rr: 2 Bb RR: 4 Bb Rr; thus, 1/9

of the rabbits with long, black fur are homozygous for

both genes.

3.8 In shorthorn cattle, the genotype RR causes a red coat,

the genotype rr causes a white coat, and the genotype

Rr causes a roan coat. A breeder has red, white, and

roan cows and bulls. What phenotypes might be

expected from the following matings and in what

proportions?

(a) Red  red

(b) Red  roan

(d) Roan  roan.

ANS: (a) All red; (b) 1/2 red, 1/2 roan; (c) all roan; (d) 1/4 red,

1/2 roan, 1/4 white

3.9 How many different kinds of F

gametes, F

genotypes,

and F

phenotypes would be expected from the following

crosses:

(a) AA  aa;

(b) AA BB  aa bb;

(d) What general formulas are suggested by these

answers?

ANS:

Gametes F

Genotypes F

Phenotypes

(a) 2 3 2

(b) 2  2  4 3  3  9 2  2  4

(d) 2

, where n is the

number of genes

3.10 A researcher studied six independently assorting genes in

a plant. Each gene has a dominant and a recessive allele:

R black stem, r red stem; D tall plant, d dwarf plant; C full

pods, c constricted pods; O round fruit, o oval fruit; H

hairless leaves, h hairy leaves; W purple ower, w white

ower. From the cross (P1) Rr Dd cc Oo Hh Ww  (P2)

Rr dd Cc oo Hh ww,

(a) How many kinds of gametes can be formed by P1?

(b) How many genotypes are possible among the prog-

eny of this cross?

progeny?

(d) What is the probability of obtaining the Rr Dd cc Oo

hh ww genotype in the progeny?

(e) What is the probability of obtaining a black, dwarf,

constricted, oval, hairy, purple phenotype in the

progeny?

ANS: (a) 2  2  1  2  2  2  32; (b) 3  2  2  2  3 

2  144; (c) 2  2  2  2  2  2  64; (d) (1/2) 

(1/2)  (1/2)  (1/2)  (1/4)  (1/2)  1/128; (e) (3/4) 

(1/2)  (1/2)  (1/2)  (1/4)  (1/2)  3/256.

3.11 For each of the following situations, determine the

degrees of freedom associated with the c

statistic and

decide whether or not the observed c

value warrants

acceptance or rejection of the hypothesized genetic ratio.

Hypothesized Ratio Observed−c

(a) 3:1 7.0

(b) 1:2:1 7.0

(d) 9:3:3:1 5.0

ANS: (a) 1, reject; (b) 2, reject; (c) 3, accept; (d) 3, accept.

3.12 Mendel testcrossed pea plants grown from yellow, round

seeds to plants grown from green, wrinkled seeds and

obtained the following results: 31 yellow, round; 26

green, round; 27 yellow, wrinkled; and 26 green, wrin-

kled. Are these results consistent with the hypothesis that

seed color and seed texture are controlled by indepen-

dently assorting genes, each segregating two alleles?

ANS: On the hypothesis, the expected number in each class

is 27.5; c

with three degrees of freedom is calculated as

(31  27.5)

/27.5  (26  27.5)

/27.5  (27  27.5)

/27.5 

(26  27.5)

/27.5  0.618, which is not signicant at the

5% level. Thus, the results are consistent with the hypoth-

esis of two independently assorting genes, each segregating

two alleles.

3.13 Perform a chi-square test to determine if an observed

ratio of 30 tall to 20 dwarf pea plants is consistent with

an expected ratio of 1:1 from the cross Dd  dd.

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ANS: c

 (30  25)

/25  (20  25)

/25  2, which is less

than 3.84, the 5 percent critical value for a chi-square

statistic with one degree of freedom; consequently, the

observed segregation ratio is consistent with the expected

ratio of 1:1.

3.14 Seed capsules of the Shepherd’s purse are either triangu-

lar or ovoid. A cross between a plant with triangular seed

capsules and a plant with ovoid seed capsules yielded F

hybrids that all had triangular seed capsules. When these

hybrids were intercrossed, they produced 80 F

plants,

72 of which had triangular seed capsules and 8 of which

had ovoid seed capsules. Are these results consistent with

the hypothesis that capsule shape is determined by a sin-

gle gene with two alleles?

ANS: If capsule shape is determined by a single gene with two

alleles, the F

plants should segregate in a 3:1 ratio. To

test for agreement between the observed segregation

data and the expected ratio, compute the expected num-

ber of plants with either triangular or ovoid seed cap-

sules: (3/4)  80  60 triangular and (1/4)  80  20

ovoid; then compute a c

statistic with one degree of

freedom: c

 (72  60)

/60  (8  20)

/20  9.6, which

exceeds the critical value of 3.84. Consequently, the data

are inconsistent with the hypothesis that capsule shape is

determined by a single gene with two alleles.

3.15 Albinism in humans is caused by a recessive allele a.

From marriages between people known to be carriers

(Aa) and people with albinism (aa), what proportion of

the children would be expected to have albinism? Among

three children, what is the chance of one without albi-

nism and two with albinism?

ANS: Half the children from Aa  aa matings would have

albinism. In a family of three children, the chance that

one will be unaffected and two affected is 3  (1/2)



(1/2)

 3/8.

3.16 If both husband and wife are known to be carriers of the

allele for albinism, what is the chance of the following

combinations in a family of four children: (a) all four

unaffected; (b) three unaffected and one affected; (c) two

unaffected and two affected; (d) one unaffected and three

affected?

ANS: (a) (3/4)

 81/256; (b) 4  (3/4)

 (1/4)

 108/256; (c)

6  (3/4)

 (1/4)

 54/256; (d) 4  (3/4)

 (1/4)



12/256.

3.17 In humans, cataracts in the eyes and fragility of the bones

are caused by dominant alleles that assort independently.

A man with cataracts and normal bones marries a woman

without cataracts but with fragile bones. The man’s

father had normal eyes, and the woman’s father had nor-

mal bones. What is the probability that the rst child of

this couple will (a) be free from both abnormalities; (b)

have cataracts but not have fragile bones; (c) have fragile

bones but not have cataracts; (d) have both cataracts and

fragile bones?

ANS: Man (Cc ff )  woman (cc Ff ). (a) cc ff, (1/2)  (1/2)  1/4;

(b) Cc ff, (1/2)  (1/2)  1/4; (c) cc Ff, (1/2)  (1/2)  1/4;

(d) Cc Ff, (1/2)  (1/2)  1/4.

3.18 In generation V in the pedigree in Figure 3.15, what is

the probability of observing seven children without the

cancer-causing mutation and two children with this

mutation among a total of nine children?

ANS: 9!/(7! 2!)  (1/2)

 (1/2)

 0.07

3.19 If a man and a woman are heterozygous for a gene, and if

they have three children, what is the chance that all three

will also be heterozygous?

ANS: (1/2)

 1/8

3.20 If four babies are born on a given day: (a) What is the

chance that two will be boys and two will be girls?

(b) What is the chance that all four will be girls? (c) What

combination of boys and girls among four babies is most

likely? (d) What is the chance that at least one baby will

be a girl?

ANS: (a) 4  (1/2)

 (1/2)

 4/16; (b) (1/2)

 1/16; (c) 2

boys, girls; (d) 1  probability that all four are boys  1

 (1/2)

 15/16.

3.21 In a family of six children, what is the chance that at least

three are girls?

ANS: (20/64)  (15/64)  (6/64)  (1/64)  42/64

3.22 The following pedigree shows the inheritance of a domi-

nant trait. What is the chance that the offspring of the

following matings will show the trait: (a) III-1  III-3;

(b) III-2  III-4?

III

3421

ANS: (a) zero; (b) 1/2

3.23 The following pedigree shows the inheritance of a reces-

sive trait. Unless there is evidence to the contrary, assume

that the individuals who have married into the family do

not carry the recessive allele. What is the chance that the

offspring of the following matings will show the trait: (a)

III-1  III-12; (b) II-4  III-14; (c) III-6  III-13; (d)

IV-1  IV-2?

1234

III

5678

4 7

14 15 16 17

109

11 12 13

ANS: (a) (1/2)  (1/4)  1/8; (b) (1/2)  (1/2)  (1/4)  1/16;

 1/24

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3.24 In the following pedigrees, determine whether the trait is

more likely to be due to a dominant or a recessive allele.

Assume the trait is rare in the population.

III

(a)

2345

III

1 234

(b)

45123678

123

ANS: (a) Recessive; (b) dominant.

3.25 In pedigree (b) of Problem 3.24, what is the chance that

the couple III-1 and III-2 will have an affected child?

What is the chance that the couple IV-2 and IV-3 will

have an affected child?

ANS: For III-1  III-2, the chance of an affected child is 1/2.

For IV-2  IV-3, the chance is zero.

3.26 Peas heterozygous for three independently assorting

genes were intercrossed.

(a) What proportion of the offspring will be homozy-

gous for all three recessive alleles?

(b) What proportion of the offspring will be homozy-

gous for all three genes?

gous for one gene and heterozygous for the other two?

(d) What proportion of the offspring will be homozy-

gous for the recessive allele of at least one gene?

ANS: (a) (1/4)

 1/64; (b) (1/2)

 1/8; (c) 3  (1/2)



(1/2)

 3/8; (d) 1  probability that the offspring is not

homozygous for the recessive allele of any gene  1  (3/4)

 37/64.

3.27 The following pedigree shows the inheritance of a reces-

sive trait. What is the chance that the couple III-3 and

III-4 will have an affected child?

III

2413

ANS: 1/2

3.28 A geneticist crosses tall pea plants with short pea plants.

All the F

plants are tall. The F

plants are then allowed

to self-fertilize, and the F

plants are classied by height:

62 tall and 26 short. From these results, the geneticist

concludes that shortness in peas is due to a recessive

allele (s) and that tallness is due to a dominant allele (S).

On this hypothesis, 2/3 of the tall F

plants should be

heterozygous Ss. To test this prediction, the geneticist

uses pollen from each of the 62 tall plants to fertilize the

ovules of emasculated owers on short pea plants. The

next year, three seeds from each of the 62 crosses are

sown in the garden and the resulting plants are grown to

maturity. If none of the three plants from a cross is short,

the male parent is classied as having been homozygous

SS; if at least one of the three plants from a cross is short,

the male parent is classied as having been heterozygous

Ss. Using this system of progeny testing, the geneticist

concludes that 29 of the 62 tall F

plants were homozy-

gous SS and that 33 of these plants were heterozygous Ss.

(a) Using the chi-square procedure, evaluate these results

for goodness of t to the prediction that 2/3 of the tall F

plants should be heterozygous.

(b) Informed by what you read in A Milestone in Genet-

ics: Mendel’s 1866 Paper, which you can nd in the Stu-

dent Companion Site, explain why the geneticist’s

procedure for classifying tall F

plants by genotype is not

denitive.

cation procedure and calculate the expected frequencies

of homozygotes and heterozygotes among the tall

plants.

(d) Evaluate the predictions obtained in (c) using the chi-

square procedure.

ANS: (a) The observed numbers, expected numbers, and chi-

square calculation are laid out in the following table:

Observed Expected (Obs − Exp)

/Exp

Dominant

homozygotes

(SS)

62  1/3  20.7

3.33

Heterozygotes

(Ss)

62  2/3  41.3

1.67

Total 62 62 5.00

The total chi-square value is greater than the critical

value for a chi-square statistic with one degree of free-

dom (3.84). Therefore, we reject the hypothesis that the

expected proportions are 1/3 and 2/3.

(b) The problem with the geneticist’s classication

procedure is that it allows for a heterozygote to be

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misclassied as a homozygote if none of its three prog-

eny shows the recessive (short) phenotype. The proba-

bility of this event is 1/2 for any one offspring—therefore

(1/2)

 1/8 for all three offspring.

probability of misclassifying a heterozygote as a homo-

zygote. The frequency of heterozygotes expected a priori

(2/3) must be decreased by the probability of misclassi-

cation (1/8); thus, the predicted frequency of heterozy-

gotes is 62  (2/3)  (1  1/8)  62  (7/12)  36.2.

The predicted frequency of homozygotes is obtained by

subtraction: 62 – 36.2  25.8.

(d) The chi-square calculation is (29  25.8)

/25.8  (33

 36.2)

/36.2  0.68, which is much less than the critical

value for a chi-square statistic with one degree of free-

dom. Therefore, we tentatively accept the idea that

adjusting for the probability of misclassication explains

the observed data.

3.29 A researcher who has been studying albinism has identi-

ed a large group of families with four children in which

at least one child shows albinism. None of the parents in

this group of families shows albinism. Among the chil-

dren, the ratio of those without albinism to those with

albinism is 1.7:1. The researcher is surprised by this

result because he thought that a 3:1 ratio would be

expected on the basis of Mendel’s Principle of Segrega-

tion. Can you explain the apparently non-Mendelian

segregation ratio in the researcher’s data?

ANS: The researcher has obtained what appears to be a non-

Mendelian ratio because he has been studying only fami-

lies in which at least one child shows albinism. In these

families, both parents are heterozygous for the mutant

allele that causes albinism. However, other couples in the

population might also be heterozygous for this allele but,

simply due to chance, have failed to produce a child with

albinism. If a man and a woman are both heterozygous

carriers of the mutant allele, the chance that a child they

produce will not have albinism is 3/4. The chance that

four children they produce will not have albinism is

therefore (3/4)

 0.316. In the entire population of fami-

lies in which two heterozygous parents have produced a

total of four children, the average number of affected

children is 1. Among families in which two heterozygous

parents have produced at least one affected child among a

total of four children, the average must be greater than 1.

To calculate this conditional average, let us denote the

number of children with albinism by x, and the probabil-

ity that exactly x of the four children have albinism by

P(x). The average number of affected children among

families in which at least one of the four children is

affected—that is, the conditional average—is therefore

SxP(x)/(1  P(0)), where the sum starts at x  1 and ends

at x  4. We start the sum at x  1 because we must

exclude those cases in which none of the four children is

affected. The divisor (1  P(0)) is the probability that the

couple has had at least one affected child among their

four children. Now P(0)  0.316 and SxP(x)  1. There-

fore, the average we seek is simply 1/(1  0.316)  1.46.

If, in the subset of families with at least one affected child,

the average number of affected children is 1.46, then the

average number of unaffected children is 4 – 1.46  2.54.

Thus, the expected ratio of unaffected to affected chil-

dren in these families is 2.54:1.46, or 1.74:1, which is

what the researcher has observed.

Chapter 4

4.1 What blood types could be observed in children born to

a woman who has blood type M and a man who has

blood type MN?

ANS: M and MN.

4.2 In rabbits, coloration of the fur depends on alleles of the

gene c. From information given in the chapter, what phe-

notypes and proportions would be expected from the fol-

lowing crosses: (a) c



 cc; (b) c



c  c



c; (c) c



 c



;

(d) cc

 cc; (e) c



 c



c; (f) c

c  cc?

ANS: (a) All wild-type; (b) 3/4 wild-type, 1/4 albino; (c) 3/4

wild-type, 1/4 chinchilla; (d) 1/2 chinchilla, 1/2 albino;

(e) 3/4 wild-type, 1/4 Himalayan; (f) 1/2 Himalayan, 1/2

albino.

4.3 In mice, a series of ve alleles determines fur color. In

order of dominance, these alleles are as follows: A

, yel-

low fur but homozygous lethal; A

, agouti with light

belly; A



, agouti (wild-type); a

, black and tan; and a,

black. For each of the following crosses, give the coat

color of the parents and the phenotypic ratios expected

among the progeny: (a) A

 A

; (b) A

a  A

;

a  A

a; (d) A

 A

; (e) A

 A



; (f) A





a; (g) a

a  aa; (h) A

 A



; and (i) A

 A



ANS:

Parents Offspring

(a) Yellow  yellow 2 yellow: 1 light belly

(b) Yellow  light belly 2 yellow: 1 light belly:

1 black and tan

1 black

(d) Light belly  light belly All light belly

(e) Light belly  yellow 1 yellow: 1 light belly

(f) Agouti  black and tan 1 agouti: 1 black and tan

(g) Black and tan  black 1 black and tan: 1 black

(h) Yellow  agouti 1 yellow: 1 light belly

(i) Yellow  yellow 2 yellow: 1 light belly

4.4 In several plants, such as tobacco, primrose, and red clo-

ver, combinations of alleles in eggs and pollen have been

found to inuence the reproductive compatibility of the

plants. Homozygous combinations, such as S

, do not

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develop because S

pollen is not effective on S

stigmas.

However, S

pollen is effective on S

stigmas. What

progeny might be expected from the following crosses

(seed parent written rst): (a) S

 S

;

(b) S

 S

;

 S

; and (d) S

 S

? .

ANS: (a) S

, S

; (b) S

, S

; (c) S

;

(d) S

, S

4.5 From information in the chapter about the ABO blood

types, what phenotypes and ratios are expected from the

following matings: (a) I

 I

; (b) I

 ii; (c) I

i 

i; and (d) I

i  ii;

ANS: (a) All AB; (b) 1 A: 1 B; (c) 1 A: 1 B: 1 AB: 1 O; (d) 1 A: 1 O.

4.6 A woman with type O blood gave birth to a baby, also

with type O blood. The woman stated that a man with

type AB blood was the father of the baby. Is there any

merit to her statement?

ANS: No. The woman must be ii; if her mate is I

; they could

not have an ii child.

4.7 A woman with type AB blood gave birth to a baby with

type B blood. Two different men claim to be the father.

One has type A blood, the other has type B blood. Can

the genetic evidence decide in favor of either?

ANS: No. The woman is I

. One man could be either I

i; the other could be either I

or I

i. Given the uncer-

tainty in the genotype of each man, either could be the

father of the child.

4.8 The ower colors of plants in a particular population

may be blue, purple, turquoise, light blue, or white. A

series of crosses between different members of the popu-

lation produced the following results:

Cross Parents Progeny

Purple

 blue

All purple

Purple

 purple

76 purple, 25 turquoise

Blue

 blue

86 blue, 29 turquoise

Purple

 turquoise

49 purple, 52 turquoise

Purple

 purple

69 purple, 22 blue

Purple

 blue

50 purple, 51 blue

Purple

 blue

54 purple, 26 blue,

25 turquoise

Turquoise

 turquoise

All turquoise

Purple

 blue

49 purple, 25 blue,

23 light blue

Light blue

 light blue

60 light blue,

29 turquoise, 31 white

Turquoise

 white

All light blue

White

 white

All white

Purple

 white

All purple

How many genes and alleles are involved in the inheri-

tance of ower color? Indicate all possible genotypes for

the following phenotypes: (a) purple; (b) blue; (c) tur-

quoise; (d) light blue; (e) white.

ANS: One gene with four alleles. (a) purple: c

, c

; c

, c

;

(b) blue: c

, c

; (c) turquoise: c

, c

; (d) light

blue: c

; (e) white: c

4.9 A woman who has blood type O and blood type M mar-

ries a man who has blood type AB and blood type MN. If

we assume that the genes for the A-B-O and M-N blood-

typing systems assort independently, what blood types

might the children of this couple have, and in what

proportions?

ANS: The woman is ii L

; the man is I

; the blood

types of the children will be A and M, A and MN, B and

M, and B and MN, all equally likely.

4.10 A Japanese strain of mice has a peculiar, uncoordinated

gait called waltzing, which is due to a recessive allele, v.

The dominant allele V causes mice to move in a coordi-

nated manner. A mouse geneticist has recently isolated

another recessive mutation that causes uncoordinated

movement. This mutation, called tango, could be an allele

of the waltzing gene, or it could be a mutation in an

entirely different gene. Propose a test to determine

whether the waltzing and tango mutations are alleles, and

if they are, propose symbols to denote them.

ANS: Cross homozygous waltzing with homozygous tango. If

the mutations are alleles, all the offspring will have an

uncoordinated gait; if they are not alleles, all the off-

spring will be wild-type. If the two mutations are alleles,

they could be denoted with the symbols v (waltzing) and

(tango).

4.11 Congenital deafness in human beings is inherited as a

recessive condition. In the following pedigree, two deaf

individuals, each presumably homozygous for a recessive

mutation, have married and produced four children with

normal hearing. Propose an explanation.

III

ANS: The individuals III-4 and III-5 must be homozygous for

recessive mutations in different genes; that is, one is aa

BB and the other is AA bb; none of their children is deaf

because all of them are heterozygous for both genes

(Aa Bb).

4.12 In the fruit y, recessive mutations in either of two inde-

pendently assorting genes, brown and purple, prevent the

synthesis of red pigment in the eyes. Thus, homozygotes

for either of these mutations have brownish-purple eyes.

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However, heterozygotes for both of these mutations

have dark red, that is, wild-type eyes. If such double het-

erozygotes are intercrossed, what kinds of progeny will

be produced, and in what proportions?

ANS: 9/16 dark red, 7/16 brownish purple.

4.13 The dominant mutation Plum in the fruit y also causes

brownish-purple eyes. Is it possible to determine by

genetic experiments whether Plum is an allele of the

brown or purple genes?

ANS: No. The test for allelism cannot be performed with dom-

inant mutations.

4.14 From information given in the chapter, explain why mice

with yellow coat color are not true-breeding.

ANS: The allele for yellow fur is homozygous lethal.

4.15 A couple has four children. Neither the father nor the

mother is bald; one of the two sons is bald, but neither of

the daughters is bald.

(a) If one of the daughters marries a nonbald man and

they have a son, what is the chance that the son will

become bald as an adult?

(b) If the couple has a daughter, what is the chance that

she will become bald as an adult?

ANS: The mother is Bb and the father is bb. The chance that a

daughter is Bb is 1/2. (a) The chance that the daughter

will have a bald son is (1/2)  (1/2)  1/4. (b) The chance

that the daughter will have a bald daughter is zero.

4.16 The following pedigree shows the inheritance of ataxia,

a rare neurological disorder characterized by uncoordi-

nated movements. Is ataxia caused by a dominant or a

recessive allele? Explain.

III

ANS: Dominant. The condition appears in every generation

and nearly every affected individual has an affected par-

ent. The exception, IV-2, had a father who carried the

ataxia allele but did not manifest the trait—an example of

incomplete penetrance.

4.17 Chickens that carry both the alleles for rose comb (R)

and pea comb (P) have walnut combs, whereas chickens

that lack both of these alleles (i.e., they are genotypically

rr pp) have single combs. From the information about

interactions between these two genes given in the chap-

ter, determine the phenotypes and proportions expected

from the following crosses:

(a) RR Pp  rr Pp;

(b) rr PP  Rr Pp;

(d) Rr pp  rr pp.

ANS: (a) 3/4 walnut, 1/4 rose; (b) 1/2 walnut, 1/2 pea; (c) 3/8

walnut, 3/8 rose, 1/8 pea, 1/8 single; (d) 1/2 rose, 1/2

single.

4.18 Rose-comb chickens mated with walnut-comb chickens

produced 15 walnut-, 14 rose-, 5 pea-, and 6 single-comb

chicks. Determine the genotypes of the parents.

ANS: Rr pp  Rr Pp.

4.19 Summer squash plants with the dominant allele C bear

white fruit, whereas plants homozygous for the recessive

allele c bear colored fruit. When the fruit is colored, the

dominant allele G causes it to be yellow; in the absence

of this allele (i.e., with genotype gg), the fruit color is

green. What are the F

phenotypes and proportions

expected from intercrossing the progeny of CC GG and

cc gg plants? Assume that the C and G genes assort

independently.

ANS: 12/16 white, 3/16 yellow, 1/16 green.

4.20 The white Leghorn breed of chickens is homozygous for

the dominant allele C, which produces colored feathers.

However, this breed is also homozygous for the domi-

nant allele I of an independently assorting gene that

inhibits coloration of the feathers. Consequently, Leg-

horn chickens have white feathers. The white Wyandotte

breed of chickens has neither the allele for color nor the

inhibitor of color; it is therefore genotypically cc ii. What

are the F

phenotypes and proportions expected from

intercrossing the progeny of a white Leghorn hen and a

white Wyandotte rooster?

ANS: 13/16 white, 3/16 colored.

4.21 Fruit ies homozygous for the recessive mutation scarlet

have bright red eyes because they cannot synthesize

brown pigment. Fruit ies homozygous for the recessive

mutation brown have brownish-purple eyes because they

cannot synthesize red pigment. Fruit ies homozygous

for both of these mutations have white eyes because they

cannot synthesize either type of pigment. The brown and

scarlet mutations assort independently. If fruit ies that

are heterozygous for both of these mutations are inter-

crossed, what kinds of progeny will they produce, and in

what proportions?

ANS: 9/16 dark red (wild-type), 3/16 brownish purple, 3/16

bright red, 1/16 white.

4.22 Consider the following hypothetical scheme of determi-

nation of coat color in a mammal. Gene A controls the

conversion of a white pigment P

into a gray pigment

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; the dominant allele A produces the enzyme necessary

for this conversion, and the recessive allele a produces an

enzyme without biochemical activity. Gene B controls

the conversion of the gray pigment P

into a black pig-

ment P

; the dominant allele B produces the active

enzyme for this conversion, and the recessive allele b

produces an enzyme without activity. The dominant

allele C of a third gene produces a polypeptide that com-

pletely inhibits the activity of the enzyme produced by

gene A; that is, it prevents the reaction P

→P

. Allele c of

this gene produces a defective polypeptide that does not

inhibit the reaction P

→P

. Genes A, B, and C assort

independently, and no other genes are involved. In the F

of the cross AA bb CC  aa BB cc, what is the expected

phenotypic segregation ratio?

ANS: 9 black: 3 gray: 52 white.

4.23 What F

phenotypic segregation ratio would be expected

for the cross described in the preceding problem if the

dominant allele, C, of the third gene produced a product

that completely inhibited the activity of the enzyme pro-

duced by gene B—that is, prevented the reaction P

→P

rather than inhibiting the activity of the enzyme pro-

duced by gene A?

ANS: 9 black: 39 gray: 16 white.

4.24 The Micronesian Kingsher, Halcyon cinnamomina, has a

cinnamon-colored face. In some birds, the color contin-

ues onto the chest, producing one of three patterns: a

circle, a shield, or a triangle; in other birds, there is no

color on the chest. A male with a colored triangle was

crossed with a female that had no color on her chest, and

all their offspring had a colored shield on the chest.

When these offspring were intercrossed, they produced

an F

with a phenotypic ratio of 3 circle: 6 shield: 3 tri-

angle: 4 no color. (a) Determine the mode of inheritance

for this trait and indicate the genotypes of the birds in all

three generations. (b) If a male without color on his chest

is mated to a female with a colored shield on her chest

and the F

segregate in the ratio of 1 circle: 2 shield:

1 triangle, what are the genotypes of the parents and

their progeny?

(a) The simplest explanation for the inheritance of the

trait is recessive epistasis combined with incomplete

dominance, summarized in the following table:

Genotype Phenotype Frequency in F

AA B- Circle 3/16

Aa B- Shield 6/16

aa B- Triangular 3/16

A- bb No color 3/16

aa bb No color 1/16

(b) Father’s genotype: Aa bb; mother’s genotype: Aa BB

Circle Shield Triangle

Progeny

genotypes:

AA Bb Aa Bb aa Bb

4.25 In a species of tree, seed color is determined by four

independently assorting genes: A, B, C, and D. The

recessive alleles of each of these genes (a, b, c, and d) pro-

duce abnormal enzymes that cannot catalyze a reaction

in the biosynthetic pathway for seed pigment. This path-

way is diagrammed as follows:

White precursor

Yellow Orange

Red

Blue

When both red and blue pigments are present, the seeds

are purple. Trees with the genotypes Aa Bb Cc Dd and Aa

Bb Cc dd were crossed.

(a) What color are the seeds in these two parental

genotypes?

(b) What proportion of the offspring from the cross will

have white seeds?

blue offspring from the cross.

ANS: (a) Purple  red; (b) proportion white (aa)  1/4;

27/128, proportion white (aa)  1/4  32/128, propor-

tion blue (A- B- cc Dd)  (3/4)(3/4)(1/4)(1/2)  9/128.

4.26 Multiple crosses were made between true-breeding lines

of black and yellow Labrador retrievers. All the F

prog-

eny were black. When these progeny were intercrossed,

they produced an F

consisting of 91 black, 39 yellow,

and 30 chocolate. (a) Propose an explanation for the

inheritance of coat color in Labrador retrievers. (b) Pro-

pose a biochemical pathway for coat color determination

and indicate how the relevant genes control coat

coloration.

ANS: (a) Because the F

segregation is approximately 9 black:

3 chocolate: 4 yellow, coat color is determined by epista-

sis between two independently assorting genes: black 

B- E-; chocolate  bb E-; yellow  B- ee or bb ee.

(b) Yellow pigment—E brown pigment—B black

pigment.

4.27 Two plants with white owers, each from true-breeding

strains, were crossed. All the F

plants had red owers.

When these F

plants were intercrossed, they produced

an F

consisting of 177 plants with red owers and

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142 with white owers. (a) Propose an explanation for

the inheritance of ower color in this plant species.

(b) Propose a biochemical pathway for ower pigmenta-

tion and indicate which genes control which steps in this

pathway.

ANS: (a) Because the F

segregation is approximately 9 red: 7

white, ower color is due to epistasis between two inde-

pendently assorting genes: red  A- B- and white  aa

B-, A- bb, or aa bb. (b) Colorless precursor—A colorless

product—B red pigment.

4.28 Consider the following genetically controlled biosyn-

thetic pathway for pigments in the owers of a hypo-

thetical plant:

Gene A

Enzyme A

Gene B

Enzyme B

Gene C

Enzyme C

Assume that gene A controls the conversion of a white

pigment, P

, into another white pigment, P

; the

dominant allele A species an enzyme necessary for

this conversion, and the recessive allele a species a

defective enzyme without biochemical function. Gene

B controls the conversion of the white pigment, P

into a pink pigment, P

; the dominant allele, B, pro-

duces the enzyme necessary for this conversion, and

the recessive allele, b, produces a defective enzyme.

The dominant allele, C, of the third gene species an

enzyme that converts the pink pigment, P

, into a red

pigment, P

; its recessive allele, c, produces an altered

enzyme that cannot carry out this conversion. The

dominant allele, D, of a fourth gene produces a poly-

peptide that completely inhibits the function of

enzyme C; that is, it blocks the reaction P

→P

. Its

recessive allele, d, produces a defective polypeptide

that does not block this reaction. Assume that ower

color is determined solely by these four genes and that

they assort independently. In the F

of a cross between

plants of the genotype AA bb CC DD and plants of the

genotype aa BB cc dd, what proportion of the plants

will have (a) red owers? (b) pink owers? (c) white

owers?

ANS: (a) Proportion red  (3/4)

 (1/4)  27/256; (b) pro-

portion pink  (3/4)

 [(3/4)

 (1/4)]  117/256; (c)

proportion white  1  144/256  112/256.

4.29 In the following pedigrees, what are the inbreeding coef-

cients of A, B, and C?

Offspring of ﬁrst

cousins once

removed

Offspring of second

cousins

Offspring of half–ﬁrst

cousins

ANS: F

 (1/2)

 1/32; F

 2  (1/2)

 1/32; F

 2 

(1/2)

 1/64.

4.30 A, B, and C are inbred strains of mice, assumed to be

completely homozygous. A is mated to B and B to C.

Then the A  B hybrids are mated to C, and the off-

spring of this mating are mated to the B  C hybrids.

What is the inbreeding coefcient of the offspring of this

last mating?

ANS: From the following pedigree, the inbreeding coefcient

is (1/2)

(1  F

)  (1/2)

(1  F

)  3/8 because F



 1.

A ×

× C

(A × B) × C

4.31 Mabel and Frank are half siblings, as are Tina and Tim.

However, these two pairs of half siblings do not have any

common ancestors. If Mabel marries Tim and Frank

marries Tina and each couple has a child, what fraction

of their genes will these children share by virtue of

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common ancestry? Will the children be more or less

closely related than rst cousins?

ANS: The pedigree is shown below:

Frank

Mabel Tina

Tim

The coefcient of relationship between the offspring of

the two couples is obtained by calculating the inbreeding

coefcient of the imaginary child from a mating between

these offspring and multiplying by 2: [(1/2)

 2]  2 

1/8. This is the same degree of relatedness as rst

cousins.

4.32 Suppose that the inbreeding coefcient of I in the fol-

lowing pedigree is 0.25. What is the inbreeding coef-

cient of I’s common ancestor, C?

ANS: F

 (1/2)

(1  F

)  0.25; thus, F

 1.

4.33 A randomly pollinated strain of maize produces ears that

are 24 cm long, on average. After one generation of self-

fertilization, the ear length is reduced to 20 cm. Predict

the ear length if self-fertilization is continued for one

more generation.

ANS: The mean ear length for randomly mated maize is

24 cm and that for maize from one generation of self-

fertilization is 20 cm. The inbreeding coefcient of the

offspring of one generation of self-fertilization is 1/2,

and the inbreeding coefcient of the offspring of two

generations of self-fertilization is (1/2)(11/2)  3/4.

Mean ear length (Y ) is expected to decline linearly with

inbreeding according to the equation Y  24  b F

where b is the slope of the line. The value of b can be

determined from the two values of Y that are given. The

difference between these two values (4 cm) corresponds

to an increase in F from 0 to 1/2. Thus, b  4/(1/2) 

8 cm, and for F  3/4, the predicted mean ear length is

Y  24  8  (3/4)  18 cm.

Chapter 5

5.1 What are the genetic differences between male- and

female-determining sperm in animals with heteroga-

metic males?

ANS: The male-determining sperm carries a Y chro mo-

some; the female-determining sperm carries an X

chromosome.

5.2 A male with singed bristles appeared in a culture of Dro-

sophila. How would you determine if this unusual pheno-

type was due to an X-linked mutation?

ANS: Cross the singed male to wild-type females and then

intercross the offspring. If the singed bristle phenotype is

due to an X-linked mutation, approximately half the F

males, but none of the F

females, will show it.

5.3 In grasshoppers, rosy body color is caused by a recessive

mutation; the wild-type body color is green. If the gene

for body color is on the X chromosome, what kind of

progeny would be obtained from a mating between a

homozygous rosy female and a hemizygous wild-type

male? (In grasshoppers, females are XX and males are

XO.)

ANS: All the daughters will be green and all the sons will be

rosy.

5.4 In the mosquito Anopheles culicifacies, golden body (go) is a

recessive X-linked mutation, and brown eyes (bw) is a

recessive autosomal mutation. A homozygous XX female

with golden body is mated to a homozygous XY male

with brown eyes. Predict the phenotypes of their F

offspring. If the F

progeny are intercrossed, what kinds

of progeny will appear in the F

and in what

proportions?

ANS: The cross is go/go / female  /Y bw/bw male → F

go/ bw/ females (wild-type eyes and body) and go/Y

bw/ males (golden body, wild-type eyes). An intercross

of the F

offspring yields the following F

phenotypes in

both sexes.

Body Eyes Genotype Proportion

Golden Brown go/go or Y bw/bw

(1/2)  (1/4)

 1/8

Golden Wild-

type

go/go or Y

/bw

or 

(1/2)  (3/4)

 3/8

Wild-

type

Brown

/go or Y

bw/bw

(1/2)  (1/4)

 1/8

Wild-

type

Wild-

type

/go or Y /bw

or 

(1/2)  (3/4)

 3/8

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5.5 What are the sexual phenotypes of the following geno-

types in Drosophila: XX, XY, XXY, XXX, XO?

ANS: XX is female, XY is male, XXY is female, XXX is female

(but barely viable), XO is male (but sterile).

5.6 In human beings, a recessive X-linked mutation, g, causes

green-defective color vision; the wild-type allele, G,

causes normal color vision. A man (a) and a woman (b),

both with normal vision, have three children, all married

to people with normal vision: a color-defective son (c),

who has a daughter with normal vision (f); a daughter

with normal vision (d), who has one color-defective son

(g) and two normal sons (h); and a daughter with normal

vision (e), who has six normal sons (i). Give the most

likely genotypes for the individuals (a–i) in this family.

(a)

(c)

(f)

(i) 6

(d) (e)

(b)

(h)

(g)

G/Y

g/Y

G/Y G/Y

g/Y

G/g

g/G G/G

G/g

ANS: (a) X

Y; (b) X

; (c) X

Y; (d) X

; (e) X

; (f) X

;

(g) X

Y; (h) X

Y; (i) X

5.7 If both father and son have defective color vision, is it

likely that the son inherited the trait from his father?

ANS: No. Defective color vision is caused by an X-linked

mutation. The son’s X chromosome came from his

mother, not his father.

5.8 A normal woman, whose father had hemophilia, marries

a normal man. What is the chance that their rst child

will have hemophilia?

ANS: The risk for the child is P(woman transmits mutant

allele)  P(child is male)  (1/2)  (1/2)  1/4.

5.9 A man with X-linked color blindness marries a woman

with no history of color blindness in her family. The

daughter of this couple marries a normal man, and their

daughter also marries a normal man. What is the chance

that this last couple will have a child with color blindness?

If this couple has already had a child with color blindness,

what is the chance that their next child will be color blind?

ANS: The risk for the child is P(mother is C/c)  P(mother

transmits c)  P(child is male)  (1/2)  (1/2)  (1/2) 

1/8; if the couple has already had a child with color

blindness, P(mother is C/c)  1, and the risk for each

subsequent child is 1/4.

5.10 A man who has color blindness and type O blood has

children with a woman who has normal color vision and

type AB blood. The woman’s father had color blindness.

Color blindness is determined by an X-linked gene, and

blood type is determined by an autosomal gene.

(a) What are the genotypes of the man and the woman?

(b) What proportion of their children will have color

blindness and type B blood?

blindness and type A blood?

(d) What proportion of their children will be color blind

and have type AB blood?

ANS: (a) The man is X

Y ii; the woman is X



. (b) Prob-

ability color blind  1/2; probability type B blood  1/2;

combined probability  (1/2)  (1/2)  1/4. (c) Proba-

bility color blind  1/2; probability type A blood  1/2;

combined probability (1/2)  (1/2)  1/4. (d) 0.

5.11 A Drosophila female homozygous for a recessive X-linked

mutation that causes vermilion eyes is mated to a wild-

type male with red eyes. Among their progeny, all the

sons have vermilion eyes, and nearly all the daughters

have red eyes; however, a few daughters have vermilion

eyes. Explain the origin of these vermilion-eyed

daughters.

ANS: Each of the rare vermilion daughters must have resulted

from the union of an X(v) X(v) egg with a Y-bearing

sperm. The diplo-X eggs must have originated through

nondisjunction of the X chromosomes during oogenesis

in the mother. However, we cannot determine if the

nondisjunction occurred in the rst or the second mei-

otic division.

5.12 In Drosophila, vermilion eye color is due to a recessive

allele (v) located on the X chromosome. Curved wings

are due to a recessive allele (cu) located on one autosome,

and ebony body is due to a recessive allele (e) located on

another autosome. A vermilion male is mated to a curved,

ebony female, and the F

males are phenotypically wild-

type. If these males were backcrossed to curved, ebony

females, what proportion of the F

offspring will be wild-

type males?

ANS: P(male)  1/2; P(male transmits rst wild-type autosome)

 1/2; P(male transmits other wild-type autosome)  1/2;

therefore, combined proportion, P(wild-type male)  1/8

5.13 A Drosophila female heterozygous for the recessive

X-linked mutation w (for white eyes) and its wild-type

allele w

is mated to a wild-type male with red eyes.

Among the sons, half have white eyes and half have red

eyes. Among the daughters, nearly all have red eyes;

however, a few have white eyes. Explain the origin of

these white-eyed daughters.

ANS: Each of the rare white-eyed daughters must have resulted

from the union of an X(w) X(w) egg with a Y-bearing

sperm. The rare diplo-X eggs must have originated

through nondisjunction of the X chromosomes during

the second meiotic division in the mother.

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5.14 In Drosophila, a recessive mutation called chocolate (c)

causes the eyes to be darkly pigmented. The mutant phe-

notype is indistinguishable from that of an autosomal

recessive mutation called brown (bw). A cross of choco-

late-eyed females to homozygous brown males yielded

wild-type F

females and darkly pigmented F

males. If

the F

ies are intercrossed, what types of progeny are

expected, and in what proportions? (Assume that the

double mutant combination has the same phenotype as

either of the single mutants alone.)

ANS: 3/8 wild-type (red), 5/8 brown for both male and female

progeny.

5.15 Suppose that a mutation occurred in the SRY gene on the

human Y chromosome, knocking out its ability to pro-

duce the testis-determining factor. Predict the pheno-

type of an individual who carried this mutation and a

normal X chromosome.

ANS: Female.

5.16 A woman carries the androgen-insensitivity mutation

(ar) on one of her X chromosomes; the other X carries

the wild-type allele (AR). If the woman marries a normal

man, what fraction of her children will be phenotypically

female? Of these, what fraction will be fertile?

ANS: Three-fourths will be phenotypically female (genotypi-

cally ar/AR, AR/AR, or ar/Y). Among the females, 2/3

(ar/AR and AR/AR) will be fertile; the ar/Y females will

be sterile.

5.17 Would a human with two X chromosomes and a Y chro-

mosome be male or female?

ANS: Male.

5.18 In Drosophila, the gene for bobbed bristles (recessive allele

bb, bobbed bristles; wild-type allele



, normal bristles) is

located on the X chromosome and on a homologous seg-

ment of the Y chromosome. Give the genotypes and

phenotypes of the offspring from the following crosses:

(a) X

 X



;

(b) X

 X



;



 X



;

(d) X



 X



ANS: (a) 1/2 X

bobbed females, 1/2 X



wild-type

males; (b) 1/2 X



wild-type females, 1/2 X

bobbed males; (c) 1/4 X



wild-type females, 1/4 X



wild-type females, 1/4 X



wild-type males, 1/4 X

bobbed males; (d) 1/4 X



wild-type females, 1/4

bobbed females, 1/4 X



wild-type males, 1/4



wild-type males.

5.19 Predict the sex of Drosophila with the following chromo-

some compositions (A = haploid set of autosomes):

(a) 4X 4A;

(b) 3X 4A;

(d) 1X 3A;

(e) 2X 2A;

(f) 1X 2A.

ANS: (a) Female; (b) intersex; (c) intersex; (d) male: (e) female;

(f) male.

5.20 In chickens, the absence of barred feathers is due to a

recessive allele. A barred rooster was mated with a non-

barred hen, and all the offspring were barred. These F

chickens were intercrossed to produce F

progeny,

among which all the males were barred; half the females

were barred and half were nonbarred. Are these results

consistent with the hypothesis that the gene for barred

feathers is located on one of the sex chromosomes?

ANS: Yes. The gene for feather patterning is on the Z chromo-

some. If we denote the allele for barred feathers as B and

the allele for nonbarred feathers as b, the crosses are as

follows: B/B (barred) male  b/W (nonbarred) female →

: B/b (barred) males and B/W (barred) females. Inter-

crossing the F

produces B/B (barred) males, B/b (barred)

males, B/W (barred) females, and b/W (nonbarred)

females, all in equal proportions.

5.21 A Drosophila male carrying a recessive X-linked mutation

for yellow body is mated to a homozygous wild-type

female with gray body. The daughters of this mating all

have uniformly gray bodies. Why are not their bodies a

mosaic of yellow and gray patches?

ANS: Drosophila does not achieve dosage compensation by

inactivating one of the X chromosomes in females.

5.22 What is the maximum number of Barr bodies in the

nuclei of human cells with the following chromosome

compositions:

(a) XY;

(b) XX;

(d) XXX;

(e) XXXX;

(f) XYY?

ANS: a) Zero; (b) one; (c) one; (d) two; (e) three; (f) zero.

5.23 Males in a certain species of deer have two nonhomolo-

gous X chromosomes, denoted X

and X

, and a Y chro-

mosome. Each X chromosome is about half as large as

the Y chromosome, and its centromere is located near

one of the ends; the centromere of the Y chromosome is

located in the middle. Females in this species have two

copies of each of the X chromosomes and lack a Y chro-

mosome. How would you predict the X and Y chromo-

somes to pair and disjoin during spermatogenesis to

produce equal numbers of male- and female-determin-

ing sperm?

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ANS:

Metaphase Anaphase

Since the centromere is at the end of each small X chro-

mosome but in the middle of the larger Y, both X

and X

pair at the centromere of the Y chromosome during meta-

phase so that the two X chromosomes disjoin together

and segregate from the Y chromosome during anaphase.

5.24 A breeder of sun conures (a type of bird) has obtained

two true-breeding strains, A and B, which have red eyes

instead of the normal brown found in natural popula-

tions. In Cross 1, a male from strain A was mated to a

female from strain B, and the male and female offspring

all had brown eyes. In Cross 2, a female from strain A was

mated to a male from strain B, and the male offspring

had brown eyes and the female offspring had red eyes.

When the F

birds from each cross were mated brother

to sister, the breeder obtained the following results:

Phenotype

Proportion in

of Cross 1

Proportion in

of Cross 2

Brown male 6/16 3/16

Red male 3/16 5/16

Brown female 3/16 3/16

Red female 5/16 5/16

Provide a genetic explanation for these results.

ANS: Color is determined by an autosomal gene (alleles A

and a) and a sex-linked gene (alleles B and b) on the Z

chromosome (females are ZW and males are ZZ) and

the recessive alleles are mutually epistatic—that is, aa, bb,

or bW birds have red eyes, and A- B- or A- BW birds

have brown eyes.

Cross 1

P Strain A red male



Strain B red female

aa BB AA bW

Aa Bb

Brown males



Aa BW

Brown females

A- Bb

Brown males (6/16)

A- BW

Brown females (3/16)

aa Bb

Red males (2/16)

A- bW

Red females (4/16)

aa BW

Red females (1/16)

Cross 2

P Strain A red female



Strain B red male

aa BW AA bb

Aa Bb

Brown males



Aa bW

Red females

A- Bb

Brown males (3/16)

A- bW

Brown females (3/16)

A- bb

Red males (3/16)

A- bW

Red females (3/16)

aa b-

Red males 2/16

aa -W

Red females (2/16)

5.25 In 1908, F. M. Durham and D. C. E. Marryat reported

the results of breeding experiments with canaries. Cin-

namon canaries have pink eyes when they rst hatch,

whereas green canaries have black eyes. Durham and

Marryat crossed cinnamon females with green males

and observed that all the F

progeny had black eyes,

just like those of the green strain. When the F

males

were crossed to green females, all the male progeny

had black eyes, whereas all the female progeny had

either black or pink eyes, in about equal proportions.

When the F

males were crossed to cinnamon females,

four classes of progeny were obtained: females with

black eyes, females with pink eyes, males with black

eyes, and males with pink eyes—all in approximately

equal proportions. Propose an explanation for these

ndings.

ANS: Eye color in canaries is due to a gene on the Z chromo-

some, which is present in two copies in males and one

copy in females. The allele for pink color at hatching

(p) is recessive to the allele for black color at hatching

(P). There is no eye color gene on the other sex chro-

mosome (W), which is present in one copy in females

and absent in males. The parental birds were genotypi-

cally p/W (cinnamon females) and P/P (green males).

Their F

sons were genotypically p/P (with black eyes

at hatching). When these sons were crossed to green

females (genotype P/W), they produced F

progeny

that sorted into three categories: males with black eyes

at hatching (P/-, half the total progeny), females with

black eyes at hatching (P/W, a fourth of the total prog-

eny), and females with pink eyes at hatching (p/W, a

fourth of the total progeny). When these sons were

crossed to cinnamon females (genotype p/W), they

produced F

progeny that sorted into four equally fre-

quent categories: males with black eyes at hatching

(genotype P/p), males with pink eyes at hatching (gen-

otype p/p), females with black eyes at hatching

(genotype P/W), and females with pink eyes at hatch-

ing (genotype p/W).

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Chapter 6

6.1 In the human karyotype, the X chromosome is approxi-

mately the same size as seven of the autosomes (the

so-called C group of chromosomes). What procedure

could be used to distinguish the X chromosome from the

other members of this group?

ANS: Use one of the banding techniques.

6.2 In humans, a cytologically abnormal chromosome 22,

called the “Philadelphia” chromosome because of the

city in which it was discovered, is associated with

chronic leukemia. This chromosome is missing part of

its long arm. How would you denote the karyotype of

an individual who had 46 chromosomes in his somatic

cells, including one normal 22 and one Philadelphia

chromosome?

ANS: 46, XX, del(22)(q) or 46, XY, del(22)(q), depending on

the sex chromosome constitution.

6.3 During meiosis, why do some tetraploids behave more

regularly than triploids?

ANS: In allotetraploids, each member of the different sets

of chromosomes can pair with a homologous partner

during prophase I and then disjoin during anaphase I.

In triploids, disjunction is irregular because homol o-

gous chromosomes associate during prophase I either

by forming bivalents and univalents or by forming

trivalents.

6.4 The following table presents chromosome data on four

species of plants and their F

hybrids:

Meiosis I Metaphase

Species or

Hybrid

Root Tip

Chromosome

Number

Number of

Bivalents

Number of

Univalents

A 20 10 0

B 20 10 0

C 10 5 0

D 10 5 0

A  B

20 0 20

A  C

15 5 5

A  D

15 5 5

C  D

10 0 10

(a) Deduce the chromosomal origin of species A.

(b) How many bivalents and univalents would you expect

to observe at meiotic metaphase I in a hybrid between

species C and species B?

to observe at meiotic metaphase I in a hybrid between

species D and species B?

ANS: (a) Species A is an allotetraploid with a genome from

each of species C and species D; (b) 0 bivalents and

15univalents; (c) 0 bivalents and 15 univalents.

6.5 A plant species A, which has seven chromosomes in its

gametes, was crossed with a related species B, which has

nine. The hybrids were sterile, and microscopic observa-

tion of their pollen mother cells showed no chromosome

pairing. A section from one of the hybrids that grew vig-

orously was propagated vegetatively, producing a plant

with 32 chromosomes in its somatic cells. This plant was

fertile. Explain.

ANS: The fertile plant is an allotetraploid with 7 pairs of chro-

mosomes from species A and 9 pairs of chromosomes

from species B; the total number of chromosomes is

(2  7)  (2  9)  32.

6.6 A plant species X with n  5 was crossed with a related

species Y with n  7. The F

hybrid produced only a few

pollen grains, which were used to fertilize the ovules of

species Y. A few plants were produced from this cross,

and all had 19 chromosomes. Following self-fertilization,

the F

hybrids produced a few F

plants, each with

24chromosomes. These plants were phenotypically dif-

ferent from either of the original species and were highly

fertile. Explain the sequence of events that produced

these fertile F

hybrids.

ANS: The F

hybrid had 5 chromosomes from species X and 7

chromosomes from species Y, for a total of 12. When this

hybrid was backcrossed to species Y, the few progeny

that were produced had 5  7  12 chromosomes from

the hybrid and 7 from species Y, for a total of 19. This

hybrid was therefore a triploid. Upon self-fertilization,

a few F

plants were formed, each with 24 chromosomes.

Presumably the chromosomes in these plants consisted

of 2  5  10 from species X and 2  7  14 from spe-

cies Y. These vigorous and fertile F

plants were there-

fore allotetraploids.

6.7 Identify the sexual phenotypes of the following

genotypes in human beings: XX, XY, XO, XXX, XXY,

XYY.

ANS: XX is female, XY is male, XO is female (but sterile), XXX

is female, XXY is male (but sterile), and XYY is male.

6.8 If nondisjunction of chromosome 21 occurs in the divi-

sion of a secondary oocyte in a human female, what is the

chance that a mature egg derived from this division will

receive two number 21 chromosomes?

ANS: 1/2

6.9 A Drosophila female homozygous for a recessive X-linked

mutation causing yellow body was crossed to a wild-type

male. Among the progeny, one y had sectors of yellow

pigment in an otherwise gray body. These yellow sectors

were distinctly male, whereas the gray areas were female.

Explain the peculiar phenotype of this y.

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ANS: The y is a gynandromorph, that is, a sexual mosaic. The

yellow tissue is X(y)/O and the gray tissue is X(y)/X().

This mosaicism must have arisen through loss of the

X chromosome that carried the wild-type allele, pre-

sumably during one of the early embryonic cleavage

divisions.

6.10 The Drosophila fourth chromosome is so small that ies

monosomic or trisomic for it survive and are fertile. Sev-

eral genes, including eyeless (ey), have been located on this

chromosome. If a cytologically normal y homozygous

for a recessive eyeless mutation is crossed to a y mono-

somic for a wild-type fourth chromosome, what kinds of

progeny will be produced, and in what proportions?

ANS: Approximately half the progeny should be disomic ey/

and half should be monosomic ey/O. The disomic prog-

eny will be wild-type, and the monosomic progeny will

be eyeless.

6.11 A woman with X-linked color blindness and Turner syn-

drome had a color-blind father and a normal mother. In

which of her parents did nondisjunction of the sex chro-

mosomes occur?

ANS: Nondisjunction must have occurred in the mother. The

color blind woman with Turner syndrome was produced

by the union of an X-bearing sperm, which carried the

mutant allele for color blindness, and a nullo-X egg.

6.12 In humans, Hunter syndrome is known to be an X-linked

trait with complete penetrance. In family A, two pheno-

typically normal parents have produced a normal son,

a daughter with Hunter and Turner syndromes, and a son

with Hunter syndrome. In family B, two phenotypically

normal parents have produced two phenotypically nor-

mal daughters and a son with Hunter and Klinefelter syn-

dromes. In family C, two phenotypically normal parents

have produced a phenotypically normal daughter,

a daughter with Hunter syndrome, and a son with Hunter

syndrome. For each family, explain the origin of the child

indicated in italics.

ANS: The daughter with Turner and Hunter syndromes in

family A must have received her single X chromosome

from her mother, who is heterozygous for the mutation

causing Hunter syndrome. The daughter did not receive

a sex chromosome from her father because sex chromo-

some nondisjunction must have occurred during meiosis

in his germline. The son with Klinefelter syndrome in

family B is karyotypically XXY, and both of his X chro-

mosomes carry the mutant allele for Hunter syndrome.

This individual must have received two mutant

X chromosomes from his heterozygous mother due to X

chromosome nondisjunction during the second meiotic

division in her germline. The daughter with Hunter syn-

drome in family C is karyotypically XX, and both of her

X chromosomes carry the mutant allele for Hunter syn-

drome. This individual received the two mutant X

chromosomes from her heterozygous mother through

nondisjunction during the second meiotic division in the

mother’s germline. Furthermore, because the daughter

did not receive a sex chromosome from her father, sex

chromosome nondisjunction must have occurred during

meiosis in his germline too.

6.13 Although XYY men are phenotypically normal, would

they be expected to produce more children with sex

chromosome abnormalities than XY men? Explain.

ANS: XYY men would produce more children with sex chro-

mosome abnormalities because their three sex chromo-

somes will disjoin irregularly during meiosis. This

irregular disjunction will produce a variety of aneuploid

gametes, including the XY, YY, XYY, and nullo sex chro-

mosome constitutions.

6.14 In a Drosophila salivary chromosome, the bands have a

sequence of 1 2 3 4 5 6 7 8. The homologue with which

this chromosome is synapsed has a sequence of 1 2 3 6 5

4 7 8. What kind of chromosome change has occurred?

Draw the synapsed chromosomes.

ANS: The animal is heterozygous for an inversion:

66 44

6.15 Other chromosomes have sequences as follows: (a) 1 2 5

6 7 8; (b) 1 2 3 4 4 5 6 7 8; (c) 1 2 3 4 5 8 7 6. What kind

of chromosome change is present in each? Illustrate how

these chromosomes would pair with a chromosome

whose sequence is 1 2 3 4 5 6 7 8.

ANS: (a) Deletion:

12 5678

5678

(b) Duplication:

1 234 56 87

234

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12345

1234

6.16 In plants translocation heterozygotes display about 50 per-

cent pollen abortion. Why?

ANS: In translocation heterozygotes, only alternate segrega-

tion leads to euploid gametes, and the frequency of alter-

nate segregation is typically around 5 percent.

6.17 One chromosome in a plant has the sequence A B C D E

F, and another has the sequence M N O P Q R. A recip-

rocal translocation between these chromosomes pro-

duced the following arrangement: A B C P Q R on one

chromosome and M N O D E F on the other. Illustrate

how these translocated chromosomes would pair with

their normal counterparts in a heterozygous individual

during meiosis.

ANS:

CONM

6.18 In Drosophila, the genes bw and st are located on chromo-

somes 2 and 3, respectively. Flies homozygous for bw

mutations have brown eyes, ies homozygous for st

mutations have scarlet eyes, and ies homozygous for

bw and st mutations have white eyes. Doubly heterozy-

gous males were mated individually to homozygous bw;

st females. All but one of the matings produced four

classes of progeny: wild-type, and brown-, scarlet- and

white-eyed. The single exception produced only wild-

type and white-eyed progeny. Explain the nature of this

exception.

ANS: The exceptional male, whose genotype is bw/ st/, is

heterozygous for a translocation between chromosomes

2 and 3. It is not possible to determine whether the

translocation is between the two mutant chromosomes

or between the two wild-type chromosomes, that is,

whether it is T(bw; st) or T(; ); however, it clearly is

not between a mutant chromosome and a wild-type

chromosome, that is, T(bw; ) or T(; st). If it were, the

progeny would be either brown or scarlet, not either

wild-type or white.

6.19 A phenotypically normal boy has 45 chromosomes, but

his sister, who has Down syndrome, has 46. Suggest an

explanation for this paradox.

ANS: The boy carries a translocation between chromosome 21

and another chromosome, say chromosome 14. He also

carries a normal chromosome 21 and a normal chromo-

some 14. The boy’s sister carries the translocation, one

normal chromosome 14, and two normal copies of chro-

mosome 21.

6.20 Distinguish between a compound chromosome and a

Robertsonian translocation.

ANS: A compound chromosome is composed of segments

from the same pair of chromosomes, as when two X

chromosomes become attached to each other. A Rob-

ertsonian translocation involves a fusion of segments

from two different pairs of chromosomes. These seg-

ments fuse at or near the centromeres, usually with the

loss of the short arms of each of the par tic i pat ing

chromosomes.

6.21 A yellow-bodied Drosophila female with attached-X

chromosomes was crossed to a white-eyed male. Both

of the parental phenotypes are caused by X-linked

recessive mutations. Predict the phenotypes of the

progeny.

ANS: All the daughters will be yellow-bodied and all the sons

will be white-eyed.

6.22 A man has attached chromosomes 21. If his wife is cyto-

logically normal, what is the chance their rst child will

have Down syndrome?

ANS: Zygotes produced by this couple will be either trisomic

or monosomic for chromosome 21. Thus, 100 percent of

their viable children will develop Down syndrome.

6.23 Analysis of the polytene chromosomes of three popula-

tions of Drosophila has revealed three different banding

sequences in a region of the second chromosome:

Population Banding Sequence

P1 1 2 3 4 5 6 7 8 9 10

P2 1 2 3 9 8 7 6 5 4 10

P3 1 2 3 9 8 5 6 7 4 10

Explain the evolutionary relationships among these

populations.

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ANS: The three populations are related by a series of

inversions:

P1 1 2 3 4 5 6 7 8 9 10

P2 1 2 3 9 8 7 6 5 4 10

P3 1 2 3 9 8 5 6 7 4 10

6.24 Each of six populations of Drosophila in different geo-

graphic regions had a specic arrangement of bands in

one of the large autosomes:

(a) 12345678

(b) 12263478

(d) 14322678

(e) 16223478

(f) 154322678

Assume that arrangement (a) is the original one. In what

order did the other arrangements most likely arise, and

what type of chromosomal aberration is responsible for

each change?

ANS: Arrangement (a) produced (c) by inversion of segment

2345; (c) produced (f) by a duplication of band 2; (f) pro-

duced (d) by a deletion of band 5; (d) produced (e) by

inversion of segment 43226; (e) produced (b) by inver-

sion of segment 622.

6.25 The following diagram shows two pairs of chromosomes

in the karyotypes of a man, a woman, and their child.

The man and the woman are phenotypically normal, but

the child (a boy) suffers from a syndrome of abnormali-

ties, including poor motor control and severe mental

impairment. What is the genetic basis of the child’s

abnormal phenotype? Is the child hyperploid or hypo-

ploid for a segment in one of his chromosomes?

Mother Father Child

ANS: The mother is heterozygous for a reciprocal transloca-

tion between the long arms of the large and small chro-

mosomes; a piece from the long arm of the large

chromosome has been broken off and attached to the

long arm of the short chromosome. The child has inher-

ited the rearranged large chromosome and the normal

small chromosome from the mother. Thus, because the

rearranged large chromosome is decient for some of its

genes, the child is hypoploid.

6.26 A male mouse that is heterozygous for a reciprocal

translocation between the X chromosome and an auto-

some is crossed to a female mouse with a normal karyo-

type. The autosome involved in the translocation

carries a gene responsible for coloration of the fur. The

allele on the male’s translocated autosome is wild-type,

and the allele on its nontranslocated autosome is

mutant; however, because the wild-type allele is domi-

nant to the mutant allele, the male’s fur is wild-type

(dark in color). The female mouse has light color in her

fur because she is homozygous for the mutant allele of

the color-determining gene. When the offspring of the

cross are examined, all the males have light fur and all

the females have patches of light and dark fur. Explain

these peculiar results.

ANS: The phenotype in the female offspring is mosaic because

one of the X chromosomes is inactivated in each of their

cells. If the translocated X is inactivated, the autosome

attached to it could also be partially inactivated by a

spreading of the inactivation process across the translo-

cation breakpoint. This spreading could therefore inacti-

vate the color-determining gene on the translocated

autosome and cause patches of tissue to be phenotypi-

cally mutant.

6.27 In Drosophila, the autosomal genes cinnabar (cn) and

brown (bw) control the production of brown and red eye

pigments, respectively. Flies homozygous for cinnabar

mutations have bright red eyes, ies homozygous for

brown mutations have brown eyes, and ies homozy-

gous for mutations in both of these genes have white

eyes. A male homozygous for mutations in the cn and bw

genes has bright red eyes because a small duplication

that carries the wild-type allele of bw (bw



) is attached

to the Y chromosome. If this male is mated to a karyo-

typically normal female that is homozygous for the cn

and bw mutations, what types of progeny will be

produced?

ANS: The sons will have bright red eyes because they will

inherit the Y chromosome with the bw



allele from their

father. The daughters will have white eyes because they

will inherit an X chromosome from their father.

6.28 In Drosophila, vestigial wing (vg), hairy body (h), and eye-

less (ey) are recessive mutations on chromosomes 2, 3,

and 4, respectively. Wild-type males that had been irradi-

ated with X rays were crossed to triply homozygous

recessive females. The F

males (all phenotypically wild-

type) were then testcrossed to triply homozygous reces-

sive females. Most of the F

males produced eight classes

of progeny in approximately equal proportions, as would

be expected if the vg, h, and ey genes assort indepen-

dently. However, one F

male produced only four classes

of offspring, each approximately one-fourth of the total:

(1) wild-type, (2) eyeless, (3) vestigial, hairy, and (4) ves-

tigial, hairy, eyeless. What kind of chromosome aberra-

tion did the exceptional F

male carry, and which

chromosomes were involved?

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ANS: A reciprocal translocation between chromosomes 2

and 3. One translocated chromosome carries the wild-

type alleles of vg and h on chromosomes 2 and 3, respec-

tively, and the other carries the recessive mutant alleles

of these genes. The chromosome that carries the ey gene

(chromosome 4) is not involved in the rearrangement.

6.29 Cytological examination of the sex chromosomes in a man

has revealed that he carries an insertional translocation. A

small segment has been deleted from the Y chromosome

and inserted into the short arm of the X chromosome; this

segment contains the gene responsible for male differen-

tiation (SRY). If this man marries a karyotypically normal

woman, what types of progeny will the couple produce?

ANS: XX zygotes will develop into males because one of their X

chromosomes carries the SRY gene that was translocated

from the Y chromosome. XY zygotes will develop into

females because their Y chromosome has lost the SRY gene.

Chapter 7

7.1 Mendel did not know of the existence of chromosomes.

Had he known, what change might he have made in his

Principle of Independent Assortment?

ANS: If Mendel had known of the existence of chromosomes, he

would have realized that the number of factors determining

traits exceeds the number of chromosomes, and he would

have concluded that some factors must be linked on the

same chromosome. Thus, Mendel would have revised the

Principle of Independent Assortment to say that factors on

different chromosomes (or far apart on the same chromo-

some) are inherited independently.

7.2 From a cross between individuals with the genotypes Cc

Dd Ee  cc dd ee, 1000 offspring were produced. The class

that was C- D- ee included 351 individuals. Are the genes

c, d, and e on the same or different chromosomes? Explain.

ANS: The class represented by 351 offspring indicates that at

least two of the three genes are linked.

7.3 If a is linked to b, and b to c, and c to d, does it follow

that a recombination experiment would detect linkage

between a and d? Explain.

ANS: No. The genes a and d could be very far apart on the

same chromosome—so far apart that they recombine

freely, that is, 50 percent of the time.

7.4 Mice have 19 autosomes in their genome, each about

the same size. If two autosomal genes are chosen ran-

domly, what is the chance that they will be on the same

chromosome?

ANS: 1/19

7.5 Genes on different chromosomes recombine with a fre-

quency of 50 percent. Is it possible for two genes on the

same chromosome to recombine with this frequency?

ANS: Yes, if they are very far apart.

7.6 If two loci are 10 cM apart, what proportion of the cells

in prophase of the rst meiotic division will contain a

single crossover in the region between them?

ANS: 20%

7.7 Genes a and b are 20 cM apart. An a



individual

was mated with an a b/a b individual.

(a) Diagram the cross and show the gametes produced by

each parent and the genotype of the F

(b) What gametes can the F

produce, and in what

proportions?

was crossed to a b/a b individuals, what off-

spring would be expected, and in what proportions?

(d) Is this an example of the coupling or repulsion link-

age phase?

(e) If the F

were intercrossed, what offspring would be

expected, and in what proportions?

ANS: (a) Cross: a



 a b/a b. Gametes: a



from one

parent, a b from the other. F

: a



/a b

(b) 40% a



, 40% a b, 10% a



b, 10% a b



from testcross: 40% a



/a b, 40% a b/a b, 10% a



b/a b, 10% a b



/a b

(d) Coupling linkage phase

(e) F

from intercross:

Sperm

40% a

/a b

a b/a

ba b/a b

a b/a b

b/a

b/a b

a b

ba b

/a b

a b

/a b

10% a b

10% a

b 10% a b

40% a

10% a

16% 16% 4% 4%

40% a b

16% 16% 4% 4%

Eggs

4% 4% 1% 1%

Summary of phenotypes:



and b



66%



and b

a and b



a and b

16%

7.8 Answer questions (a)–(e) in the preceding problem under

the assumption that the original cross was a



b/a



b 

a b



/a b



ANS: (a) Cross: a



b/a



b  a b



/a b



Gametes: a



b from one

parent, a b



from the other F

: a



b/a b



(b) 40% a



b, 40% a b



, 10% a



, 10% a b

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from testcross: 40% a



b/a b, 40% a b



/a b, 10%



/a b, 10% a b/a b

(d) Repulsion linkage phase

(e) F

from intercross:

Sperm

40% a

b/a

b/a ba

b/a b

a b

ba b

a b

/a ba b

/a b

/a ba

/a b

a b/a

ba b/a

a b/a ba b/a b

10% a b

10% a

10% a b40% a

10% a

16% 16% 4% 4%

40% a b

16% 16% 4% 4%

Eggs

4% 4% 1% 1%

Summary of phenotypes:



and b



51%



and b

24%

a and b



24%

a and b

7.9 If the recombination frequency in the previous two

problems were 40 percent instead of 20 percent, what

change would occur in the proportions of gametes and

testcross progeny?

ANS: Coupling heterozygotes a



/a b would produce the

following gametes: 30% a



, 30% a b, 20% a



b, 20%

a b



; repulsion heterozygotes a



b/a b



would produce the

following gametes: 30% a



b, 30% a b



, 20% a



, 20%

a b. In each case, the frequencies of the testcross progeny

would correspond to the frequencies of the gametes.

7.10 A homozygous variety of maize with red leaves and nor-

mal seeds was crossed with another homozygous variety

with green leaves and tassel seeds. The hybrids were then

backcrossed to the green, tassel-seeded variety, and the

following offspring were obtained: red, normal 124; red,

tassel 126; green, normal 125; green, tassel 123. Are the

genes for plant color and seed type linked? Explain.

ANS: No. The leaf color and tassel seed traits assort

independently.

7.11 A phenotypically wild-type female fruit y that was

heterozygous for genes controlling body color and

wing length was crossed to a homozygous mutant male

with black body (allele b) and vestigial wings (allele

vg). The cross produced the following progeny: gray

body, normal wings 126; gray body, vestigial wings 24;

black body, normal wings 26; black body, vestigial

wings 124. Do these data indicate linkage between the

genes for body color and wing length? What is the

frequency of recombination? Diagram the cross,

showing the arrangement of the genetic markers on

the chromosomes.

ANS: Yes. Recombination frequency  (24  26)/(126  24 

26  124)  0.167. Cross:

b vg

female

male

126 124 24 26

b vg

7.12 Another phenotypically wild-type female fruit y het-

erozygous for the two genes mentioned in the previous

problem was crossed to a homozygous black, vestigial

male. The cross produced the following progeny: gray

body, normal wings 23; gray body, vestigial wings 127;

black body, normal wings 124; black body, vestigial wings

26. Do these data indicate linkage? What is the frequency

of recombination? Diagram the cross, showing the

arrangement of the genetic markers on the

chromosomes.

ANS: Yes. Recombination frequency  (23  26)/(23  127 

124  26)  0.163. Cross:

female male

23 26 127 124

b vg

7.13 In rabbits, the dominant allele C is required for colored

fur; the recessive allele c makes the fur colorless (albino).

In the presence of at least one C allele, another gene

determines whether the fur is black (B, dominant) or

brown (b, recessive). A homozygous strain of brown rab-

bits was crossed with a homozygous strain of albinos.

The F

were then crossed to homozygous double reces-

sive rabbits, yielding the following results: black 34;

brown 66; albino 100. Are the genes b and c linked? What

is the frequency of recombination? Diagram the crosses,

showing the arrangement of the genetic markers on the

chromosomes.

ANS: Yes. Recombination frequency is estimated by the fre-

quency of black offspring among the colored offspring:

34/(66  34)  0.34. Cross:

C b c B

C b

c b

C b

c bc bc

c B c b C B

brown

albino albino

black

66 100 34

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7.14 In tomatoes, tall vine (D) is dominant over dwarf (d),

and spherical fruit shape (P) is dominant over pear

shape (p). The genes for vine height and fruit shape are

linked with 20 percent recombination between them.

One tall plant (I) with spherical fruit was crossed with

a dwarf, pear-fruited plant. The cross produced the

following results: tall, spherical 81; dwarf, pear 79;

tall, pear 22; dwarf spherical 17. Another tall plant

with spherical fruit (II) was crossed with the dwarf,

pear-fruited plant, and the following results were

obtained: tall, pear 21; dwarf, spherical 18; tall, spheri-

cal 5; dwarf, pear 4. Diagram these two crosses, show-

ing the genetic markers on the chromosomes. If the

two tall plants with spherical fruit were crossed with

each other, that is, I  II, what phenotypic classes

would you expect from the cross, and in what

proportions?

ANS: Plant I has the genotype D P/d p, and when crossed to a

d p/d p plant produces four classes of progeny:

D P d p D p d P

d p d p d p d p

81 79 22 17

Plant II has the genotype D p/d P, and when crossed to a

d p/d p plant produces four classes of progeny:

D p d P D P d p

d p d p d p d p

21 18 5 4

If the two plants are crossed (D P/d p  D p/d P), the

phenotypes of the offspring can be predicted from the

following table.

Gametes from plant I

Gametes

from

plant II

D p

D p/D P

0.16

d p/d p

0.04

D P/d p

0.04

d P/d p

0.16

D p/d p

0.16

d p/d P

0.01

D P/d P

0.01

d P/d P

0.04

D p/d P

0.04

d p/D p

0.01

D P/D p

0.01

d P/D p

0.04

D p/D p

0.04

D P/D P

0.04

d p/D P

0.04

d P/D P

0.16

D p

D P

0.40

0.40 0.10

0.10

D pd

d P

d p

Summary of phenotypes:

Tall, spherical 0.54

Tall, pear

0.21

Dwarf, spherical

0.21

Dwarf, pear

0.04

7.15 In Drosophila, the genes sr (stripe thorax) and e (ebony

body) are located at 62 and 70 cM, respectively, from

the left end of chromosome 3. A striped female

homozygous for e



was mated with an ebony male

homozygous for sr



. All the offspring were phenotypi-

cally wild-type (gray body and unstriped).

(a) What kind of gametes will be produced by the F

females, and in what proportions?

(b) What kind of gametes will be produced by the F

males, and in what proportions?

females are mated with striped, ebony

males, what offspring are expected, and in what

proportions?

(d) If the F

males and females are intercrossed, what off-

spring would you expect from this intercross, and in

what proportions?

ANS: (a) The F

females, which are sr e



/sr



e, produce four

types of gametes: 46% sr e



, 46% sr



e, 4% sr e, 4% sr



(b) The F

males, which have the same genotype as the F

females, produce two types of gametes: 50% sr e



, 50%



e; remember, there is no crossing over in Drosophila

males. (c) 46% striped, gray; 46% unstriped, ebony; 4%

striped, ebony; 4% unstriped, gray. (d) The offspring

from the intercross can be obtained from the follow ing

table.

Sperm

sr e

0.46

0.23 0.23

0.002 0.002

0.04

sr e

0.04

sr e

/sr e

sr e

/sr

0.50 0.50

sr e

0.46

e/sr e

e/sr

Eggs

sr e/sr e

sr e/sr

/sr e

/sr

Summary of phenotypes:

Striped, gray 0.25

Unstriped, gray 0.50

Striped, ebony 0

Unstriped, ebony 0.25

7.16 In Drosophila, genes a and b are located at positions 22.0

and 42.0 on chromosome 2, and genes c and d are

located at positions 10.0 and 25.0 on chromosome 3.

A y homozygous for the wild-type alleles of these four

genes was crossed with a y homozygous for the reces-

sive alleles, and the F

daughters were backcrossed to

their quadruply recessive fathers. What offspring would

you expect from this backcross, and in what

proportions?

ANS: Because the two chromosomes assort independently, the

genetic makeup of the gametes (and, therefore, of the

backcross progeny) can be obtained from the following

table.

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24-WC Answers to All Questions and Problems

Chromosome 3 in gametes

Chr

omo-

some

gametes

c d

a ba b c da b c

c d

b c d

a b

c da b

a b

c d

a b

bcd

c d

b c

a b c

da b c d

0.425

0.40 0.17

0.17

0.0425

0.0075 0.0075

0.17

0.17 0.03 0.03

0.03 0.030.40

0.10

0.425 0.075 0.075

a b

dc d

7.17 The Drosophila genes vg (vestigial wings) and cn (cinnabar

eyes) are located at 67.0 and 57.0, respectively, on chro-

mosome 2. A female from a homozygous strain of vesti-

gial ies was crossed with a male from a homozygous

strain of cinnabar ies. The F

hybrids were phenotypi-

cally wild-type (long wings and dark red eyes).

(a) How many different kinds of gametes could the F

females produce, and in what proportions?

(b) If these females are mated with cinnabar, vestigial

males, what kinds of progeny would you expect, and in

what proportions?

ANS: (a) The F

females, which are cn vg



/cn



vg, produce four

types of gametes: 45% cn vg



, 45% cn



vg, 5% cn



5% cn vg. (b) 45% cinnabar eyes, normal wings; 45%

reddish-brown eyes, vestigial wings; 5% reddish-brown

eyes, normal wings; 5% cinnabar eyes, vestigial wings.

7.18 In Drosophila, the genes st (scarlet eyes), ss (spineless bris-

tles), and e (ebony body) are located on chromosome 3,

with map positions as indicated:

st ss e

44 58 70

Each of these mutations is recessive to its wild-type allele

(st



, dark red eyes; ss



, smooth bristles; e



, gray body).

Phenotypically wild-type females with the genotype st ss



/st



e were crossed with triply recessive males.

Predict the phenotypes of the progeny and the frequen-

cies with which they will occur assuming (a) no interfer-

ence and (b) complete interference.

ANS: In the following enumeration, classes 1 and 2 are paren-

tal types, classes 3 and 4 result from a single crossover

between st and ss, classes 5 and 6 result from a single

crossover between ss and e, and classes 7 and 8 result

from a double crossover, with one of the exchanges

between st and ss and the other between ss and e.

Class Phenotypes

(a) Frequency

with No

Interference

(b) Frequency

with Complete

Interference

1 Scarlet,

spineless

0.3784 0.37

2 Ebony 0.3784 0.37

3 Scarlet, ebony 0.0616 0.07

4 Spineless 0.0616 0.07

5 Scarlet,

spineless, ebony

0.0516 0.06

6 Wild-type 0.0516 0.06

7 Scarlet 0.0084 0

8 Spineless, ebony 0.0084 0

7.19 In maize, the genes Pl for purple leaves (dominant over Pl

for green leaves), sm for salmon silk (recessive to Sm for

yellow silk), and py for pigmy plant (recessive to Py for

normal-size plant) are on chromosome 6, with map posi-

tions as shown:

pl sm py

45 55 65

Hybrids from the cross Pl sm py / Pl sm py  pl Sm Py / pl

Sm Py were testcrossed with pl sm py / pl sm py plants. Pre-

dict the phenotypes of the offspring and their frequen-

cies assuming (a) no interference and (b) complete

interference.

ANS: In the enumeration below, classes 1 and 2 are parental

types, classes 3 and 4 result from a single crossover

between Pl and Sm, classes 5 and 6 result from a single

crossover between Sm and Py, and classes 7 and 8 result

from a double crossover, with one of the exchanges

between Pl and Sm and the other between Sm and Py.

Class Phenotypes

(a) Frequency

with No

Interference

(b) Frequency

with Complete

Interference

1 Purple, salmon,

pigmy

0.405 0.40

2 Green, yellow,

normal

0.405 0.40

3 Purple, yellow,

normal

0.045 0.05

4 Green, salmon,

pigmy

0.045 0.05

5 Purple, salmon,

normal

0.045 0.05

6 Green, yellow,

pigmy

0.045 0.05

7 Purple, yellow,

pigmy

0.005 0

8 Green, salmon,

normal

0.005 0

7.20 In maize, the genes Tu, j2, and gl3 are located on chro-

mosome 4 at map positions 101, 106, and 112, respec-

tively. If plants homozygous for the recessive alleles of

these genes are crossed with plants homozygous for the

dominant alleles, and the F

plants are testcrossed to tri-

ply recessive plants, what genotypes would you expect,

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Answers to All Questions and Problems WC-25

and in what proportions? Assume that interference is

complete over this map interval.

ANS: In the following enumeration, classes 1 and 2 are paren-

tal types, classes 3 and 4 result from crossing over

between Tu and j2, and classes 5 and 6 result from cross-

ing over between j2 and Gl3; only the chromosome from

the triply heterozygous F

plant is shown. Because inter-

ference is complete, there are no double crossover

progeny.

Class Genotype Frequency

1 tu j2 gl3 0.445

2 Tu J2 Gl3 0.445

3 tu J2 Gl3 0.025

4 Tu j2 gl3 0.025

5 tu j2 Gl3 0.030

6 Tu J2 gl3 0.030

7.21 A Drosophila geneticist made a cross between females

homozygous for three X-linked recessive mutations (y,

yellow body; ec, echinus eye shape; w, white eye color) and

wild-type males. He then mated the F

females to triply

mutant males and obtained the following results:

Females Males Number

   / y ec w   

475

y ec w / y ec w y ec w 469

y   / y ec w y  

 ec w / y ec w  ec w

y  w / y ec w y  w

 ec  / y ec w  ec 

  w / y ec w   w

y ec  / y ec w y ec 

Determine the order of the three loci y, ec, and w, and

estimate the distances between them on the linkage map

of the X chromosome.

ANS: The double crossover classes, which are the two that

were not observed, establish that the gene order is

y—w—ec. Thus, the F

females had the genotype y w ec/

 . The distance between y and w is estimated by the

frequency of recombination between these two genes:

(8  7)/1000  0.015; similarly, the distance between w

and ec is (18  23)/1000  0.041. Thus, the genetic map

for this segment of the X chromosome is y—1.5 cM—w—

4.1 cM—ec.

7.22 A Drosophila geneticist crossed females homozygous for

three X-linked mutations (y, yellow body; B, bar eye

shape; v, vermilion eye color) to wild-type males. The F

females, which had gray bodies and bar eyes with dark

red pigment, were then crossed to y B



v males, yielding

the following results:

Phenotype

546

244

160

ellow, bar

vermilion

}

ellow, vermilion

bar

}

ellow

bar

, vermilion

}

ellow, bar, vermilion

wild-type

}

Number

Determine the order of these three loci on the X chro-

mosome and estimate the distances between them.

ANS: The last two classes, consisting of yellow, bar ies and ver-

milion ies, with a total of 50 progeny, result from double

crossovers. Thus, the order of the genes is y—v—B, and

the F

females had the genotype y v B/  . The dis-

tance between y and v is the average number of crossovers

between them: (244  50)/1000  29.4 cM; likewise, the

distance between v and B is (160  50)/1000  21.0 cM.

Thus, the genetic map is y—29.4 cM—v—21.0 cM—B.

7.23 Female Drosophila heterozygous for three recessive

mutations e (ebony body), st (scarlet eyes), and ss (spineless

bristles) were testcrossed, and the following progeny

were obtained:

Phenotype Number

Wild-type 67

Ebony 8

Ebony, scarlet 68

Ebony, spineless 347

Ebony, scarlet, spineless 78

Scarlet 368

Scarlet, spineless 10

Spineless 54

(a) What indicates that the genes are linked?

(b) What was the genotype of the original heterozygous

females?

(d) What is the map distance between e and st?

(e) What is the map distance between e and ss?

(f) What is the coefcient of coincidence?

(g) Diagram the crosses in this experiment.

ANS: (a) Two of the classes (the parental types) vastly outnum-

ber the other six classes (recombinant types); (b) st 

/ ss e; (c) st—ss—e; (d) [(145  122)  1  (18) 

2]/1000  30.3 cM; (e) (122  18)/1000  14.0 cM;

(f)(0.018)/(0.163  0.140)  0.789. (g) st  / ss e

females  st ss e/st ss e males → two parental classes and

six recombinant classes.

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26-WC Answers to All Questions and Problems

7.24 Consider a female Drosophila with the following X chro-

mosome genotype:

w dor



dor

The recessive alleles w and dor cause mutant eye colors

(white and deep orange, respectively). However, w is epi-

static over dor; that is, the genotypes w dor / Y and w

dor/ w dor have white eyes. If there is 40 percent recom-

bination between w and dor, what proportion of the sons

from this heterozygous female will show a mutant phe-

notype? What proportion will have either red or deep

orange eyes?

ANS: The female will produce four kinds of gametes: 30% w ,

30%  dor, 20% w dor, and 20%  ; thus, 80% of the

progeny will be mutant (either white or deep orange)

and 50% will be pigmented (either red or deep orange).

7.25 In Drosophila, the X-linked recessive mutations prune (pn)

and garnet (g) recombine with a frequency of 0.4. Both of

these mutations cause the eyes to be brown instead of

dark red. Females homozygous for the pn mutation were

crossed to males hemizygous for the g mutation, and the

daughters, all with dark red eyes, were crossed with

their brown-eyed brothers. Predict the frequency of sons

from this last cross that will have dark red eyes.

ANS: The F

females are genotypically pn / g. Among their

sons, 40 percent will be recombinant for the two X-linked

genes, and half of the recombinants will have the wild-

type alleles of these genes. Thus, the frequency of sons

with dark red eyes will be 1/2  40%  20%.

7.26 Assume that in Drosophila there are three genes x, y, and

z, with each mutant allele recessive to the wild-type

allele. A cross between females heterozygous for these

three loci and wild-type males yielded the following

progeny:

Females

  

1010

Males

  

  z

430

 y z

x  

x y 

441

x y z 31

Total: 2010

Using these data, construct a linkage map of the three

genes and calculate the coefcient of coincidence.

ANS: Ignore the female progeny and base the map on the male

progeny. The parental types are   z and x y . The

two missing classes ( y  and x  z) must represent

double crossovers; thus, the gene order is y—x—z. The

distance between y and x is (32  27)/1000  5.9 cM and

that between x and z is (31  39)/1000  7.0 cM. Thus,

the map is y—5.9 cM—x—7.0 cM—z. The coefcient of

coincidence is zero.

7.27 In the nematode Caenorhabditis elegans, the linked genes

dpy (dumpy body) and unc (uncoordinated behavior) recom-

bine with a frequency P. If a repulsion heterozygote car-

rying recessive mutations in these genes is self-fertilized,

what fraction of the offspring will be both dumpy and

uncoordinated?

ANS: (P/2)

7.28 In the following testcross, genes a and b are 20 cM apart,

and genes b and c are 10 cM apart: a  c /  b   a b

c / abc. If the coefcient of coincidence is 0.5 over this

interval on the linkage map, how many triply homozy-

gous recessive individuals are expected among 1000

progeny?

ANS: 5.

7.29 Drosophila females heterozygous for three recessive

mutations, a, b, and c, were crossed to males homozygous

for all three mutations. The cross yielded the following

results:

Phenotype Number

  

  c

348

 b c

a  

110

a b 

306

a b c 65

Construct a linkage map showing the correct order of

these genes and estimate the distances between them.

ANS: From the parental classes,   c and a b , the heterozy-

gous females must have had the genotype   c/a b .

The missing classes,  b  and a  c, which would rep-

resent double crossovers, establish that the gene order

is b—a—c. The distance between b and a is (96  110)/

1000  20.6 cM and that between a and c is (65 

75)/1000  14.0 cM. Thus, the genetic map is b—20.6

cM—a—14.0 cM—c.

7.30 A Drosophila second chromosome that carried a recessive

lethal mutation, l(2)g14, was maintained in a stock with a

balancer chromosome marked with a dominant muta-

tion for curly wings. This latter mutation, denoted Cy, is

also associated with a recessive lethal effect—but this

effect is different from that of l(2)g14. Thus, l(2)g14 /Cy

ies survive, and they have curly wings. Flies without the

Cy mutation have straight wings. A researcher crossed

l(2)g14 / Cy females to males that carried second chromo-

somes with different deletions (all homozygous lethal)

balanced over the Cy chromosome (genotype Df /Cy).

Each cross was scored for the presence or absence of

progeny with straight wings.

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Answers to All Questions and Problems WC-27

Cross

Location of deletion Non-Curly

progeny

3 yes

4 yes

5no

In which band is the lethal mutation l(2)g14 located?

ANS: The lethal mutation resides in band 7.

7.31 The following pedigree, described in 1937 by C. L.

Birch, shows the inheritance of X-linked color blindness

and hemophilia in a family. What is the genotype of II-2?

Do any of her children provide evidence for recombina-

tion between the genes for color blindness and

hemophilia?

Key to phenotypes:

Normal

Hemophilic

Color blind

123

III

1 23

ANS: II-1 has the genotype C h/c H, that is, she is a repulsion

heterozygote for the alleles for color blindness (c) and

hemophilia (h). None of her children are recombinant

for these alleles.

7.32 The following pedigree, described in 1938 by B. Rath,

shows the inheritance of X-linked color blindness and

hemophilia in a family. What are the possible genotypes

of II-1? For each possible genotype, evaluate the chil-

dren of II-1 for evidence of recombination between the

color blindness and hemophilia genes.

III

12 3 4

Key to phenotypes:

Normal

Hemophilic

Color vision uncertain

Color blind and hemophilic

Color blind

ANS: II-1 is either (a) C h/c H or (b) c h/C H. Her four sons

have the genotypes c h (1), C h (2), c H (3), and C H (4). If

II-1 has the genotype C h/c H, sons 1 and 4 are recombi-

nant and sons 2 and 3 are nonrecombinant. If II-1 has

the genotype c h/C H, sons 2 and 3 are recombinant and

sons 1 and 4 are nonrecombinant. Either way, the fre-

quency of recombination is 0.5.

7.33 A normal woman with a color-blind father married a

normal man, and their rst child, a boy, had hemophilia.

Both color blindness and hemophilia are due to X-linked

recessive mutations, and the relevant genes are separated

by 10 cM. This couple plans to have a second child. What

is the probability that it will have hemophilia? Color

blindness? Both hemophilia and color blindness? Nei-

ther hemophilia nor color blindness?

ANS: The woman is a repulsion heterozygote for the alleles

for color blindness and hemophilia—that is, she is C h/c

H. If the woman has a boy, the chance that he will have

hemophilia is 0.5 and the chance that he will have color

blindness is 0.5. If we specify that the boy have only one

of these two conditions, then the chance that he will

have color blindness is 0.45. The reason is that the boy

will inherit a nonrecombinant X chromosome with a

probability of 0.9, and half the nonrecombinant X chro-

mosomes will carry the mutant allele for color blindness

and the other half will carry the mutant allele for hemo-

philia. The chance that the boy will have both conditions

is 0.05, and the chance that he will have neither condi-

tion is 0.05. The reason is that the boy will inherit a

recombinant X chromosome with a probability of 0.1,

and half the recombinant X chromosomes will carry

both mutant alleles and the other half will carry neither

mutant allele.

7.34 Two strains of maize, M1 and M2, are homozygous for

four recessive mutations, a, b, c, and d, on one of the

large chromosomes in the genome. Strain W1 is

homozygous for the dominant alleles of these muta-

tions. Hybrids produced by crossing M1 and W1 yield

many different classes of recombinants, whereas

hybrids produced by crossing M2 and W1 do not yield

any recombinants at all. What is the difference between

M1 and M2?

ANS: M2 carries an inversion that suppresses recombination in

the chromosome.

7.35 A Drosophila geneticist has identied a strain of ies with

a large inversion in the left arm of chromosome 3. This

inversion includes two mutations, e (ebony body) and cd

(cardinal eyes), and is anked by two other mutations, sr

(stripe thorax) on the right and ro (rough eyes) on the left.

The geneticist wishes to replace the e and cd mutations

inside the inversion with their wild-type alleles; he plans

to accomplish this by recombining the multiply mutant,

inverted chromosome with a wild-type, inversion-free

chromosome. What event is the geneticist counting on

to achieve his objective? Explain.

ANS: A two-strand double crossover within the inversion;

the exchange points of the double crossover must lie

between the genetic markers and the inversion

breakpoints.

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28-WC Answers to All Questions and Problems

Chapter 8

8.1 By what criteria are viruses living? nonliving?

ANS: Viruses reproduce and transmit their genes to progeny

viruses. They utilize energy provided by host cells and

respond to environmental and cellular signals like other

living organisms. However, viruses are obligate parasites;

they can reproduce only in appropriate host cells.

8.2 How do bacteriophages differ from other viruses?

ANS: Bacteriophages reproduce in bacteria; other viruses

reproduce in protists, plants, and animals.

8.3 In what ways do the life cycles of bacteriophages T4 and

l differ? In what aspects are they the same?

ANS: Bacteriophage T4 is a virulent phage. When it infects a

host cell, it reproduces and kills the host cell in the pro-

cess. Bacteriophage lambda can reproduce and kill the

host bacterium—the lytic response—just like phage T4,

or it can insert its chromosome into the chromosome of

the host and remain there in a dormant state—the lyso-

genic response.

8.4 How does the structure of the l prophage differ from the

structure of the l chromosome packaged in the l head?

ANS: The mature (packaged) lambda chromosome and the

lambda prophage are circular permutations of one

another (see Figure 8.5).

8.5 In what way does the integration of the l chromosome

into the host chromosome during a lysogenic infection

differ from crossing over between homologous

chromosomes?

ANS: The insertion of the phage l chromosome into the host

chromosome is a site-specic recombination process

catalyzed by an enzyme that recognizes specic sequences

in the l and E. coli chromosomes. Crossing over between

homologous chromosomes is not sequence specic. It

can occur at many sites along the two chromosomes.

8.6 Geneticists have used mutations that cause altered pheno-

types such as white eyes in Drosophila, white owers and

wrinkled seeds in peas, and altered coat color in rabbits to

determine the locations of genes on the chromosomes of

these eukaryotes. What kinds of mutant phenotypes have

been used map genes in bacteria?

ANS: Three main types of bacterial mutants have been used to

map genes in bacteria; these include mutants unable to

utilize specic sugars as energy sources (such as lactose),

mutants unable to synthesize essential metabolites (these

are called auxotrophs), and mutants resistant to drugs

and antibiotics. Wild-type bacteria can use almost any

sugar as an energy source, can grow on minimal media,

and are killed by antibiotics, whereas mutants in genes

controlling these processes result in different growth

characteristics. These growth phenotypes can be used to

map genes in bacteria.

8.7 You have identied three mutations—a, b, and c—in

Streptococcus pneumoniae. All three are recessive to their

wild-type alleles a



, b



, and c



. You prepare DNA from a

wild-type donor strain and use it to transform a strain

with genotype a b c. You observe a



transformants and



transformants, but no b



transformants. Are these

mutations closely linked? If so, what is their order on the

Streptococcus chromosome?

ANS: The a, b, and c mutations are closely linked and in the

order b—a—c on the chromosome.

8.8 A nutritionally defective E. coli strain grows only on a

medium containing thymine, whereas another nutrition-

ally defective strain grows only on a medium containing

leucine. When these two strains were grown together, a

few progeny were able to grow on a minimal medium

with neither thymine nor leucine. How can this result be

explained?

ANS: There are two possible explanations. One possibility is

that a spontaneous mutation caused reversion of either

auxotrophic strain to the prototrophic condition. Because

this requires only one mutation in one cell, this is a pos-

sibility, although rare. Another, more likely, possibility is

that conjugation occurred between the E. coli parental

auxotrophic strains. During conjugation, genes from the

parental strains recombined. Because each parent had a

wild-type gene copy for either thymine or leucine, recom-

binant progeny containing the wild-type copy of each

gene would be able to synthesize both nutrients and grow

on minimal medium.

8.9 Assume that you have just demonstrated genetic recom-

bination (e.g., when a strain of genotype a b



is present

with a strain of genotype a



b, some recombinant geno-

types, a



and a b, are formed) in a previously unstudied

species of bacteria. How would you determine whether

the observed recombination resulted from transforma-

tion, conjugation, or transduction?

ANS: Perform two experiments: (1) determine whether the

process is sensitive to DNase and (2) determine whether

cell contact is required for the process to take place. The

cell contact requirement can be tested by a U-tube

experiment (see Figure 8.9). If the process is sensitive to

DNase, it is similar to transformation. If cell contact is

required, it is similar to conjugation. If it is neither sensi-

tive to DNase nor requires cell contact, it is similar to

transduction.

8.10 (a) What are the genotypic differences between F

cells,



cells, and Hfr cells? (b) What are the phenotypic dif-

ferences? (c) By what mechanism are F



cells converted

into F



cells? F



cells to Hfr cells? Hfr cells to F



cells?

ANS: (a) F



cells, no F factor present; F



cells, autonomous F

factor; Hfr cells, integrated F factor (see Figure 8.14).

(b) F



and Hfr cells have F pili; F



cells do not. (c) F



cells are converted into F



cells by the conjugative trans-

fer of F factors from F



cells. Hfr cells are formed when

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F factors in F



cells become integrated into the chromo-

somes of these cells. Hfr cells become F



cells when the

integrated F factors exit the chromosome and become

autonomous (self-replicating) genetic elements.

8.11 (a) Of what use are F′ factors in genetic analysis? (b) How

are F′ factors formed? (c) By what mechanism does sex-

duction occur?

ANS: (a) F′ factors are useful for genetic analyses where two

copies of a gene must be present in the same cell, for

example, in determining dominance relationships. (b) F′

factors are formed by abnormal excision of F factors

from Hfr chromosomes (see Figure 8.21). (c) By the con-

jugative transfer of an F′ factor from a donor cell to a

recipient (F



) cell.

8.12 What are the basic differences between generalized

transduction and specialized transduction?

ANS: Generalized transduction: (1) transducing particles often

contain only host DNA; (2) transducing particles may

carry any segment of the host chromosome. Thus, all

host genes are transduced. Specialized transduction:

(1) transducing particles carry a recombinant chromo-

some, which contains both phage DNA and host DNA;

(2) only host genes that are adjacent to the prophage

integration site are transduced.

8.13 What roles do IS elements play in the integration of F

factors?

ANS: IS elements (or insertion sequences) are short (800–

1400 nucleotide pairs) DNA sequences that are trans-

posable—that is, capable of moving from one position

in a chromosome to another position or from one

chromosome to another chromosome. IS elements

mediate recombination between nonhomologous

DNA molecules—for example, between F factors and

bacterial chromosomes.

8.14 How can bacterial genes be mapped by interrupted mat-

ing experiments?

ANS: By interrupting conjugation at various times after the

donor and recipient cells are mixed (using a blender or

other form of agitation), one can determine the length of

time required to transfer a given genetic marker from an

Hfr cell to an F



8.15 What does the term cotransduction mean? How can

cotransduction frequencies be used to map genetic

markers?

ANS: Cotransduction refers to the simultaneous transduction

of two different genetic markers to a single recipient

cell. Since bacteriophage particles can package only

1/100 to 1/50 of the total bacterial chromosome, only

markers that are relatively closely linked can be cotrans-

duced. The frequency of cotransduction of any two

markers will be an inverse function of the distance

between them on the chromosome. As such, this

frequency can be used as an estimate of the linkage dis-

tance. Specic cotransduction-linkage functions must be

prepared for each phage–host system studied.

8.16 In E. coli, the ability to utilize lactose as a carbon source

requires the presence of the enzymes b-galactosidase and

b-galactoside permease. These enzymes are encoded by

two closely linked genes, lacZ and lacY, respectively.

Another gene, proC, controls, in part, the ability of E. coli

cells to synthesize the amino acid proline. The alleles str

and str

, respectively, control resistance and sensitivity to

streptomycin. Hfr H is known to transfer the two lac

genes, proC, and str, in that order, during conjugation. A

cross was made between Hfr H of genotype lacZ



lacY



proC



str

and an F



strain of genotype lacZ



lacY



proC



str

. After about 2 hours, the mixture was diluted and

plated out on medium containing streptomycin but no

proline. When the resulting proC



str

recombinant colo-

nies were checked for their ability to grow on medium

containing lactose as the sole carbon source, very few of

them were capable of fermenting lactose. When the

reciprocal cross (Hfr H lacZ



lacY



proC



str

X F



lacZ



lacY



proC



str

) was done, many of the proC



str

recom-

binants were able to grow on medium containing lactose

as the sole carbon source. What is the order of the lacZ

and lacY genes relative to proC?

ANS: lacY—lacZ—proC.

8.17 An F



strain, marked at 10 loci, gives rise spontaneously

to Hfr progeny whenever the F factor becomes incorpo-

rated into the chromosome of the F



strain. The F factor

can integrate into the circular chromosome at many

points, so that the resulting Hfr strains transfer the

genetic markers in different orders. For any Hfr strain,

the order of markers entering a recipient cell can be

determined by interrupted mating experiments. From

the following data for several Hfr strains derived from

the same F



, determine the order of markers in the F



strain.

Hfr Strain Markers Donated in order

1 —Z–H–E–R→

2 —O–K–S–R→

3 —K–O–W–I→

4 —Z–T–I–W→

5 —H–Z–T–I→

ANS:

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8.18 The data in the following table were obtained from

three-point transduction tests made to determine the

order of mutant sites in the A gene encoding the a sub-

unit of tryptophan synthetase in E. coli. Anth is a linked,

unselected marker. In each cross, trp



recombinants were

selected and then scored for the anth marker (anth



anth

). What is the linear order of anth and the three

mutant alleles of the A gene indicated by the data in

the table?

Cross

Donor

Markers

Recipient

Markers

anth Allele in trp



Recombinants % anth



anth



—

A34

anth



—

A223

72 anth



: 332 anth



anth



—

A46

anth



—

A223

196 anth



: 180

anth



anth



—

A223

anth



—

A34

380 anth



: 379

anth



anth



—

A223

anth



—

A46

60 anth



: 280 anth



ANS: anth—A34—A223—A46.

8.19 Bacteriophage P1 mediates generalized transduction

in E. coli. A P1 transducing lysate was prepared by

growing P1 phage on pur



pro

his

bacteria. Genes

pur, pro, and his encode enzymes required for the syn-

thesis of purines, proline, and histidine, respectively.

The phage and transducing particles in this lysate

were then allowed to infect pur



pro



his



cells. After

incubating the infected bacteria for a period of time

sufcient to allow transduction to occur, they were

plated on minimal medium supplemented with proline

and histidine, but no purines to select for pur



trans-

ductants. The pur



colonies were then transferred to

minimal medium with and without proline and with

and without histidine to determine the frequencies of

each of the outside markers. Given the following

results, what is the order of the three genes on the

E. coli chromosome?

Genotype Number Observed

pro



his



100

pro



his



pro



his

150

pro

his

ANS: pro—pur—his.

8.20 Two additional mutations in the trp A gene of E. coli, trp

A58 and trp A487, were ordered relative to trp A223 and

the outside marker anth by three-factor transduction

crosses as described in Problem 8.18. The results of these

crosses are summarized in the following table. What is

the linear order of anth and the three mutant sites in the

trp A gene?

Cross

Donor

Markers

Recipient

Markers

anth Allele in trp



Recombinants

anth



anth



—

A487

anth



—

A223

72 anth



332 anth



anth



—

A58

anth



—

A223

196 anth



180 anth



anth



—

A223

anth



—

A487

380 anth



379 anth



anth



—

A223

anth



—

A58

60 anth



280 anth



ANS: anth—A487—A223—A58.

8.21 You have identied a mutant E. coli strain that cannot

synthesize histidine (His



). To determine the location

of the his



mutation on the E. coli chromosome, you

perform interrupted mating experiments with ve dif-

ferent Hfr strains. The following chart shows the time

of entry (minutes, in parentheses) of the wild-type

alleles of the rst ve markers (mutant genes) into the

His



strain.

Hfr A -------- his (1) man (9) gal (28) lac (37) thr (45)

Hfr B -------- man (15) his (23) cys (38) ser (42) arg (49)

Hfr C -------- thr (3) lac (11) gal (20) man (39) his (47)

Hfr D -------- cys (3) his (18) man (26) gal (45) lac (54)

Hfr E -------- thr (6) rha (18) arg (36) ser (43) cys (47)

On the map below of the circular E. coli chromosome,

indicate (1) the relative location of each gene relative

to thr (located at 0/100 Min), (2) the position where

the sex factor is integrated in each of the ve Hfr’s,

and (3) the direction of chromosome transfer for each

Hfr (indicate direction with an arrow or arrowhead).

thr

lac

100/0

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ANS:

thr

lac

gal

man

his

cys

ser

rha

Min

0/100

8.22 Mutations nrd 11 (gene nrd B, encoding the beta subunit

of the enzyme ribonucleotide reductase), am M69 (gene

63, encoding a protein that aids tail-ber attachment),

and nd 28 (gene denA, encoding the enzyme endonucle-

ase II) are known to be located between gene 31 and

gene 32 on the bacteriophage T4 chromosome. Muta-

tions am N54 and am A453 are located in genes 31 and

32, respectively. Given the three-factor cross data in the

following table, what is the linear order of the ve

mutant sites?

Three-Factor Cross Data

Cross % Recombination

1. am A453—am M69  nrd 11

2.6

2. am A453—am M69  nrd 11

4.2

3. am A453—am M69  nd 28

2.5

4. am A453—nd 28  am M69

3.5

5. am A453—nrd 11  nd 28

2.9

6. am A453—nd 28  nrd 11

2.1

7. am N54—am M69  nrd 11

3.5

8. am N54—nrd 11  am M69

1.9

9. am N54—nd 28  am M69

1.7

10. am N54—am M69  nd 28

2.7

11. am N54—nd 28  nrd 11

2.9

12. am N54—nrd 11  nd 28

1.9

All recombination frequencies are given as

(wild type progeny)

 100.

total progeny

ANS: amA453—nrd11—nd28—amM69—amN54.

Chapter 9

9.1 (a) How did the transformation experiments of Grifth

differ from those of Avery and his associates? (b) What was

the signicant contribution of each? (c) Why was Grifth’s

work not evidence for DNA as the genetic material,

whereas the experiments of Avery and coworkers provided

direct proof that DNA carried the genetic information?

ANS: (a) Grifth’s in vivo experiments demonstrated the occur-

rence of transformation in pneumococcus. They pro-

vided no indication as to the molecular basis of the

transformation phenomenon. Avery and colleagues car-

ried out in vitro experiments, employing biochemical

analyses to demonstrate that transformation was medi-

ated by DNA. (b) Grifth showed that a transforming

substance existed; Avery et al. dened it as DNA. (c)

Grifth’s experiments did not include any attempt to

characterize the substance responsible for transforma-

tion. Avery et al. isolated DNA in “pure” form and dem-

onstrated that it could mediate transformation.

9.2 A cell-free extract is prepared from Type IIIS pneumo-

coccal cells. What effect will treatment of this extract

with (a) protease, (b) RNase, and (c) DNase have on its

subsequent capacity to transform recipient Type IIR cells

to Type IIIS? Why?

ANS: (a) No effect; (b) no effect; (c) DNase will destroy the

capacity of the extract to transform type IIR cells to Type

IIIS by degrading the DNA in the extract. Protease and

RNase will degrade the proteins and RNA, respectively,

in the extract. They will have no effect, since the proteins

and RNA are not involved in transformation.

9.3 How could it be demonstrated that the mixing of heat-

killed Type III pneumococcus with live Type II resulted in

a transfer of genetic material from Type III to Type II

rather than a restoration of viability to Type III by Type II?

ANS: Puried DNA from Type III cells was shown to be

sufcient to transform Type II cells. This occurred in the

absence of any dead Type III cells.

9.4 What is the macromolecular composition of a bacterial

virus or bacteriophage such as phage T2?

ANS: About 1/2 protein, 1/2 DNA. A single long molecule of

DNA is enclosed within a complex “coat” composed of

many proteins.

9.5 (a) What was the objective of the experiment carried out

by Hershey and Chase? (b) How was the objective

accomplished? (c) What is the signicance of this

experiment?

ANS: (a) The objective was to determine whether the genetic

material was DNA or protein. (b) By labeling phospho-

rus, a constituent of DNA, and sulfur, a constituent of

protein, in a virus, it was possible to demonstrate that

only the labeled phosphorus was introduced into the

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host cell during the viral reproductive cycle. The DNA

was enough to produce new phages. (c) Therefore DNA,

not protein, is the genetic material.

9.6 How did the reconstitution experiment of Fraenkel–

Conrat and colleagues show that the genetic information

of tobacco mosaic virus (TMV) is stored in its RNA

rather than its protein?

ANS: When tobacco leaves were infected with reconstituted

virus particles containing RNA from type A viruses and

protein from type B viruses, the progeny viruses were

type A, showing that RNA, not protein, carries the

genetic information in TMV.

9.7 (a) What background material did Watson and Crick

have available for developing a model of DNA? (b) What

was their contribution to building the model?

ANS: (a) The ladder-like pattern was known from X-ray dif-

fraction studies. Chemical analyses had shown that a 1:1

relationship existed between the organic bases adenine

and thymine and between cytosine and guanine. Physical

data concerning the length of each spiral and the stack-

ing of bases were also available. (b) Watson and Crick

developed the model of a double helix, with the rigid

strands of sugar and phosphorus forming spirals around

an axis, and hydrogen bonds connecting the complemen-

tary bases in nucleotide pairs.

9.8 (a) Why did Watson and Crick choose a double helix for

their model of DNA structure? (b) Why were hydrogen

bonds placed in the model to connect the bases?

ANS: (a) A multistranded, spiral structure was suggested by the

X-ray diffraction patterns. A double-stranded helix with

specic base-pairing nicely ts the 1:1 stoichiometry

observed for A:T and G:C in DNA. (b) Use of the known

hydrogen-bonding potential of the bases provided a

means of holding the two complementary strands in a

stable conguration in such a double helix.

9.9 (a) If a virus particle contained double-stranded DNA

with 200,000 base pairs, how many nucleotides would be

present? (b) How many complete spirals would occur on

each strand? (c) How many atoms of phosphorus would

be present? (d) What would be the length of the DNA

conguration in the virus?

ANS: (a) 400,000; (b) 20,000; (c) 400,000; (d) 68,000 nm.

9.10 What are the differences between DNA and RNA?

ANS: DNA has one atom less of oxygen than RNA in the sugar

part of the molecule; the sugar in DNA is 2-deoxyribose,

whereas the sugar in RNA is ribose. In DNA, thymine

replaces the uracil that is present in RNA. (In certain

bacteriophages, DNA-containing uracil is present.)

DNA is most frequently double-stranded, but bacterio-

phages such as ΦX174 contain single-stranded DNA.

RNA is most frequently single-stranded. Some viruses,

such as the Reoviruses, however, contain double-stranded

RNA chromosomes.

9.11 RNA was extracted from TMV (tobacco mosaic virus) par-

ticles and found to contain 20 percent cytosine (20 percent

of the bases were cytosine). With this information, is it

possible to predict what percentage of the bases in TMV

are adenine? If so, what percentage? If not, why not?

ANS: No. TMV RNA is single-stranded. Thus, the base-pair

stoichiometry of DNA does not apply.

9.12 DNA was extracted from cells of Staphylococcus afermen-

tans and analyzed for base composition. It was found that

37 percent of the bases are cytosine. With this informa-

tion, is it possible to predict what percentage of the bases

are adenine? If so, what percentage? If not, why not?

ANS: Yes. Because DNA in bacteria is double-stranded, the 1:1

base-pair stoichiometry applies. Therefore, if 37% of the

bases are cytosine, then 37% are guanine. This means

that the remaining 26% of the bases are adenine and thy-

mine. Thus, 26%/2  13% of the bases are adenine.

9.13 If one strand of DNA in the Watson–Crick double helix

has a base sequence of 5-GTCATGAC-3, what is the

base sequence of the complementary strand?

ANS: 3′-C A G T A C T G-5′

9.14 Indicate whether each of the following statements about

the structure of DNA is true or false. (Each letter is used

to refer to the concentration of that base in DNA.)

(a) A  T  G  C

(b) A  G; C  T

(d) T/A  C/G

(e) A  G  C  T

(f) G/C  1

(g) A  T within each single strand.

(h) Hydrogen bonding provides stability to the double

helix in aqueous cytoplasms.

(i) Hydrophobic bonding provides stability to the double

helix in aqueous cytoplasms.

(j) When separated, the two strands of a double helix are

identical.

(k) Once the base sequence of one strand of a DNA dou-

ble helix is known, the base sequence of the second

strand can be deduced.

(l) The structure of a DNA double helix is invariant. (m)

Each nucleotide pair contains two phosphate groups,

two deoxyribose molecules, and two bases.

ANS: (a) False (b) False (c) True (d) True (e) True

(f) True (g) False (h) True (i) True (j) False

(k) Frue (l) False (m) True

9.15 The nucleic acids from various viruses were extracted

and examined to determine their base composition.

Given the following results, what can you hypothesize

about the physical nature of the nucleic acids from these

viruses?

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(a) 35% A, 35% T, 15% G, and 15% C. (b) 35% A, 15% T,

25% G, and 25% C. (c) 35% A, 30% U, 30% G, and 5% C.

ANS: (a) Double-stranded DNA; (b) single-stranded DNA;

9.16 Compare and contrast the structures of the A, B, and Z

forms of DNA.

ANS: The B form of DNA helix is that proposed by Watson

and Crick and is the conformation that DNA takes under

physiological conditions. It is a right-handed double

helical coil with 10 bases per turn of the helix and a diam-

eter of 1.9 nm. It has a major and a minor groove. Z-DNA

is left-handed, has 12 bases per turn, a single deep groove,

and is 1.8 nm in diameter. Its sugar–phosphate backbone

takes a zigzagged path, and it is G:C rich. A-DNA is a

right-handed helix with 11 base pairs per turn. It is a

shorter, thicker double helix with a diameter of 0.23 nm

and has a narrow, deep major groove and a broad, shal-

low minor groove. A-DNA forms in vitro under high salt

concentrations or in a partially dehydrated state.

9.17 The temperature at which one-half of a double-stranded

DNA molecule has been denatured is called the melting

temperature, T

. Why does T

depend directly on the

GC content of the DNA?

ANS: The value of T

increases with the GC content because

GC base pairs, connected by three hydrogen bonds, are

stronger than AT base pairs connected by two hydrogen

bonds.

9.18 A diploid rye plant, Secale cereale, has 2n  14 chromo-

somes and approximately 1.6  10

bp of DNA. How

much DNA is in a nucleus of a rye cell at (a) mitotic

metaphase, (b) meiotic metaphase I, (c) mitotic telo-

phase, and (d) meiotic telophase II?

ANS: (a) 3.2  10

bp (b) 3.2  10

bp (c) 1.6  10

bp (d) 0.8

 10

9.19 The available evidence indicates that each eukaryotic

chromosome (excluding polytene chromosomes) con-

tains a single giant molecule of DNA. What different

levels of organization of this DNA molecule are appar-

ent in chromosomes of eukaryotes at various times dur-

ing the cell cycle?

ANS: DNA during interphase is not yet organized into indi-

vidual chromosomes but consists of a series of ellipsoidal

“beads on a string” that form an 11-nm ber. Here, 146

bp of DNA is wrapped 1.65 turns around the nucleo-

some core of eight histones. However, during metaphase

of meiosis and mitosis, DNA becomes organized into

chromosomes. The 11-nm ber is folded and super-

coiled to produce a 30-nm chromatin ber, the basic

structural unit of the metaphase chromosome. A third

and nal level of packaging involves nonhistone chro-

mosomal proteins that form a scaffold to condense the

30-nm bers into tightly packaged metaphase chromo-

somes, the highest level of DNA condensation observed.

9.20 A diploid nucleus of Drosophila melanogaster contains

about 3.4  10

nucleotide pairs. Assume (1) that all

nuclear DNA is packaged in nucleosomes and (2) that an

average internucleosome linker size is 60 nucleotide

pairs. How many nucleosomes would be present in a dip-

loid nucleus of D. melanogaster? How many molecules of

histone H2a, H2b, H3, and H4 would be required?

ANS: In the diploid nucleus of D. melanogaster, 1.65  10

nucleosomes would be present; these would contain 3.3 

molecules of each histone, H2a, H2b, H3, and H4.

9.21 The relationship between the melting T

and GC con-

tent can be expressed, in its much simplied form, by the

formula T

 69  0.41 (% GC). (a) Calculate the melt-

ing temperature of E. coli DNA that has about 50% GC.

(b) Estimate the %GC of DNA from a human kidney

cell where T

 85°C.

ANS: (a) 89.5°C. (b) About 39%

9.22 Experimental evidence indicates that most highly repeti-

tive DNA sequences in the chromosomes of eukaryotes

do not produce any RNA or protein products. What

does this indicate about the function of highly repetitive

DNA?

ANS: It indicates that most highly repetitive DNA sequences

do not contain structural genes specifying RNA and

polypeptide gene products.

9.23 The satellite DNAs of Drosophila virilis can be isolated,

essentially free of main-fraction DNA, by density-gradi-

ent centrifugation. If these satellite DNAs are sheared

into approximately 40-nucleotide-pair-long fragments

and are analyzed in denaturation–renaturation experi-

ments, how would you expect their hybridization kinet-

ics to compare with the renaturation kinetics observed

using similarly sheared main-fraction DNA under the

same conditions? Why?

ANS: The satellite DNA fragments would renature much

more rapidly than the main-fraction DNA fragments. In

D. virilus satellite DNAs, all three have repeating hepta-

nucleotide-pair sequences. Thus, essentially every 40

nucleotide-long (average) single-stranded fragment

from one strand will have a sequence complementary (in

part) with every single-stranded fragment from the com-

plementary strand. Many of the nucleotide-pair

sequences in main-fraction DNA will be unique

sequences (present only once in the genome).

9.24 (a) What functions do (1) centromeres and (2) telomeres

provide? (b) Do telomeres have any unique structural

features? (c) When chromosomes are broken by expo-

sure to high-energy radiation such as X rays, the result-

ing broken ends exhibit a pronounced tendency to stick

to each other and fuse. Why might this occur?

ANS: (a) (1) Centromeres function as spindle-ber attachment

sites on chromosomes; they are required for the separa-

tion of homologous chromosomes to opposite poles of

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the spindle during anaphase I of meiosis and for the

separation of sister chromatids during anaphase of mito-

sis and anaphase II of meiosis. (2) Telomeres provide at

least three important functions: (i) prevention of exonu-

cleolytic degradation of the ends of the linear DNA mol-

ecules in eukaryotic chromosomes, (ii) prevention of the

fusion of ends of DNA molecules of different chromo-

somes, and (iii) provision of a mechanism for replication

of the distal tips of linear DNA molecules in eukaryotic

chromosomes. (b) Yes. Most telomeres studied to date

contain DNA sequence repeat units (e.g., TTAGGG in

human chromosomes), and at least in some species, telo-

meres terminate with single-stranded 3′ overhangs that

form “hairpin” structures. The bases in these hairpins

exhibit unique patterns of methylation that presumably

contribute to the structure and stability of telomeres.

contain telomeres; as a result, the free ends of the DNA

molecules are apparently subject to the activities of

enzymes such as exonucleases, ligases, and so on, which

modify the ends. They can regain stability by fusing to

broken ends of other DNA molecules that contain ter-

minal telomere sequences.

9.25 Are eukaryotic chromosomes metabolically most active

during prophase, metaphase, anaphase, telophase, or

interphase?

ANS: Interphase. Chromosomes are for the most part meta-

bolically inactive (exhibiting little transcription) during

the various stages of condensation in mitosis and

meiosis.

9.26 Are the scaffolds of eukaryotic chromosomes composed

of histone or nonhistone chromosomal proteins? How

has this been determined experimentally?

ANS: Nonhistone chromosomal proteins. The “scaffold”

structures of metaphase chromosomes can be observed

by light microscopy after removal of the histones by dif-

ferential extraction procedures.

9.27 (a) Which class of chromosomal proteins, histones or

nonhistones, is the more highly conserved in different

eukaryotic species? Why might this difference be

expected? (b) If one compares the histone and nonhis-

tone chromosomal proteins of chromatin isolated from

different tissues or cell types of a given eukaryotic organ-

ism, which class of proteins will exhibit the greater het-

erogeneity? Why are both classes of proteins not

expected to be equally homogeneous in chromosomes

from different tissues or cell types?

ANS: (a) Histones have been highly conserved throughout the

evolution of eukaryotes. A major function of histones is

to package DNA into nucleosomes and chromatin bers.

Since DNA is composed of the same four nucleotides

and has the same basic structure in all eukaryotes, one

might expect that the proteins that play a structural role

in packaging this DNA would be similarly conserved. (b)

The nonhistone chromosomal proteins exhibit the

greater heterogeneity in chromatin from different tis-

sues and cell types of an organism. The histone composi-

tion is largely the same in all cell types within a given

species—consistent with the role of histones in packag-

ing DNA into nucleosomes. The nonhistone chromo-

somal proteins include proteins that regulate gene

expression. Because different sets of genes are tran-

scribed in different cell types, one would expect hetero-

geneity in some of the nonhistone chromosomal proteins

of different tissues.

9.28 (a) If the haploid human genome contains 3  10

nucle-

otide pairs and the average molecular weight of a nucleo-

tide pair is 660, how many copies of the human genome

are present, on average, in 1 mg of human DNA?

(b) What is the weight of one copy of the human genome?

thaliana contains 7.7  107 nucleotide pairs, how many

copies of the A. thaliana genome are present, on average,

in 1 mg of A. thaliana DNA? (d) What is the weight of

one copy of the A. thaliana genome? (e) Of what impor-

tance are calculations of the above type to geneticists?

ANS: (a) One microgram of human DNA will contain, on

average, 3.04  10

copies of the genome. Using an

average molecular weight per nucleotide pair of 660,

the molecular weight of the entire human genome is

1.98  10

(3  10

 660). Thus, 1.98  10

(1 “mole”  number of grams equivalent to the “molec-

ular” weight) of human DNA will contain, on average,

6.02  10

molecules [Avogadro’s number  number of

molecules (here, copies of the genome) present in one

“mole” of a substance]. One gram will contain on aver-

age (3.04  10

)(6.02  10

/1.98  10

) copies of the

genome; thus, 1 mg will contain, on average, 3.04  10

copies of the human genome. (b) One copy of the

human genome weighs approximately (3.3  10

12

(1.98  10

g per “mole”/6.02  10

molecules per

“mole”) or 3.3  10

6

mg. (c) By analogous calculations,

1 g of A. thaliana DNA contains, on average, 1.18  10

copies of the genome. (d) Similarly, one copy of the A.

thaliana genome weighs approximately 8.4  10

8

mg.

(e) In carrying out molecular analyses of the structures

of genomes, geneticists frequently need to know how

many copies of a genome are present, on average, in a

given quantity of DNA.

Chapter 10

10.1 DNA polymerase I of E. coli is a single polypeptide of

molecular weight 103,000.

(a) What enzymatic activities other than polymerase

activity does this polypeptide possess? (b) What are the

in vivo functions of these activities? (c) Are these activi-

ties of major importance to an E. coli cell? Why?

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ANS: (a) Both 3′ → 5′ and 5′ → 3′ exonuclease activities.

(b) The 3′ → 5′ exonuclease “proofreads” the nascent

DNA strand during its synthesis. If a mismatched base

pair occurs at the 3′-OH end of the primer, the 3′ → 5′

exonuclease removes the incorrect terminal nucleotide

before polymerization proceeds again. The 5′ → 3′ exo-

nuclease is responsible for the removal of RNA primers

during DNA replication and functions in pathways

involved in the repair of damaged DNA (see Chapter

13). (c) Yes, both exonuclease activities appear to be very

important. Without the 3′ → 5′ proofreading activity

during replication, an intolerable mutation frequency

would occur. The 5′ → 3′ exonuclease activity is essential

to the survival of the cell. Conditional mutations that

alter the 5′ → 3′ exonuclease activity of DNA polymerase

I are lethal to the cell under conditions where the exo-

nuclease is nonfunctional.

10.2 Escherichia coli cells are grown for many generations in a

medium in which the only available nitrogen is the heavy

isotope

N. They are then transferred to a medium con-

taining

N as the only source of nitrogen.

(a) What distribution of

N and

N would be expected

in the DNA molecules of cells that had grown for one

generation in the

N-containing medium assuming that

DNA replication was (i) conservative, (ii) semiconserva-

tive, or (iii) dispersive?

(b) What distribution would be expected after two gen-

erations of growth in the

N-containing medium

assuming (i) conservative, (ii) semiconservative, or (iii)

dispersive replication?

ANS: (a) (i) One-half of the DNA molecules with

N in both

strands and the other half with

N in both strands; (ii) all

DNA molecules with one strand containing

N and the

complementary strand containing

N; (iii) all DNA

molecules with both strands containing roughly equal

amounts of

N and

N. (b) (i) 1/4 of the DNA molecules

with

N in both strands and 3/4 with

N in both strands;

(ii) half of the DNA molecules with one strand contain-

ing

N and the complementary strand containing

and the other half with

N in both strands; (iii) all DNA

molecules with both strands containing about 1/4

and 3/4

10.3 Why do DNA molecules containing

N band at a differ-

ent position than DNA molecules containing

N when

centrifuged to equilibrium in 6M CsCl?

ANS:

Nitrogen contains eight neutrons instead of the seven

neutrons in the normal isotope of nitrogen,

N. There-

fore,

N has an atomic mass of about 15, whereas

N has

a mass of about 14. This difference means that purines

and pyrimidines containing

N have a greater density

(weight per unit volume) than those containing

Equilibrium density-gradient centrifugation in 6M CsCl

separates DNAs or other macromolecules based on their

densities, and E. coli DNA, for example, that contains

has a density of 1.724 g/cm

, whereas E. coli DNA that

contains

N has a density of 1.710 g/cm

10.4 A DNA template plus primer with the structure

3´ P —TGCGAATTAGCGACAT— P 5´

5´ P —ATCGGTACGACGCTTAAC—OH 3´

...................

(where P  a phosphate group) is placed in an in vitro

DNA synthesis system (Mg

2

, an excess of the four

deoxyribonucleoside triphosphates, etc.) containing a

mutant form of E. coli DNA polymerase I that lacks exo-

nuclease activity. The polymerase and exonuclease activ-

ities of this aberrant enzyme are identical to those of

normal E. coli DNA polymerase I. It simply has no exo-

nuclease activity.

(a) What will be the structure of the nal product?

(b) What will be the rst step in the reaction sequence?

ANS: (a)

3´ P -TGCGAATTAGCGACAT- P 5´

5´ P - ATCGGTACGACGCTTAATCGCTGTA-OH 3´;

........ ......... .............. ........

Note that DNA synthesis will not occur on the left end

since the 3′-terminus of the potential primer strand is

blocked with a phosphate group—all DNA polymerases

require a free 3′-OH terminus.

(b) The rst step will be the removal of the mismatched

C (exiting as dCMP) from the 3′-OH primer terminus

by the 3′ 5′ exonuclease (“proofreading”) activity.

10.5 How might continuous and discontinuous modes of

DNA replication be distinguished experimentally?

ANS: If nascent DNA is labeled by exposure to

H-thymidine

for very short periods of time, continuous replication

predicts that the label would be incorporated into chro-

mosome-sized DNA molecules, whereas discontinuous

replication predicts that the label would rst appear in

small pieces of nascent DNA (prior to covalent joining,

catalyzed by polynucleotide ligase).

10.6 E. coli cells contain ve different DNA polymerases—I,

II, III, IV, and V. Which of these enzymes catalyzes the

semiconservative replication of the bacterial chromo-

some during cell division? What are the functions of the

other four DNA polymerases in E. coli?

ANS: DNA polymerase III is the true replicase. DNA poly-

merase I removes the RNA primers and replaces them

with DNA. The other DNA polymerases play important

roles in DNA repair pathways (see Chapter 13).

10.7 The Boston barberry is an imaginary plant with a diploid

chromosome number of 4, and Boston barberry cells

are easily grown in suspended cell cultures.

H-thymi-

dine was added to the culture medium in which a

G1-stage cell of this plant was growing. After one cell

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generation of growth in

H-thymidine-containing

medium, colchicine was added to the culture medium.

The medium now contained both

H-thymidine and

colchicine. After two “generations” of growth in

H-thy-

midine-containing medium (the second “generation”

occurring in the presence of colchicine as well), the two

progeny cells (each now containing eight chromosomes)

were transferred to culture medium containing nonra-

dioactive thymidine (

H-thymidine) plus colchicine.

Note that a “generation” in the presence of colchicine

consists of a normal cell cycle’s chromosomal duplication

but no cell division. The two progeny cells were allowed

to continue to grow, proceeding through the “cell cycle,”

until each cell contained a set of metaphase chromo-

somes that looked like the following.

If autoradiography were carried out on these metaphase

chromosomes (four large plus four small), what pattern

of radioactivity (as indicated by silver grains on the auto-

radiograph) would be expected? (Assume no recombina-

tion between DNA molecules.)

ANS:

Two Plus two

For both the

lar

ge and small

chromosomes

10.8 Suppose that the experiment described in Problem 10.7

was carried out again, except this time replacing the

H-thymidine with nonradioactive thymidine at the

same time that the colchicine was added (after one cell

generation of growth in

H-thymidine-containing

medium). The cells were then maintained in colchicine

plus nonradioactive thymidine until the metaphase

shown in Problem 10.7 occurred. What would the auto-

radiographs of these chromosomes look like?

ANS:

Plus twoTwo For both the

lar

ge and small

chromosomes

10.9 Suppose that the DNA of cells (growing in a cell culture)

in a eukaryotic species was labeled for a short period of

time by the addition of

H-thymidine to the medium.

Next assume that the label was removed and the cells

were resuspended in nonradioactive medium. After a

short period of growth in nonradioactive medium, the

DNA was extracted from these cells, diluted, gently lay-

ered on lters, and autoradiographed. If autoradiographs

of the type

. . . . . . . . . . . .

were observed, what would this indicate about the nature

of DNA replication in these cells? Why?

ANS: The DNA replication was unidirectional rather than

bidirectional. As the intracellular pools of radioactive

H-thymidine are gradually diluted after transfer to non-

radioactive medium, less and less

H-thymidine will be

incorporated into DNA at each replicating fork. This

will produce autoradiograms with tails of decreasing

grain density at each growing point. Since such tails

appear at only one end of each track, replication must be

unidirectional. Bidirectional replication would produce

such tails at both ends of an autoradiographic track (see

Figure 10.31).

10.10 Arrange the following enzymes in the order of their

action during DNA replication in E. coli: (1) DNA poly-

merase I, (2) DNA polymerase III, (3) DNA primase,

(4) DNA gyrase, and (5) DNA helicase.

ANS: The correct sequence of action is 4, 5, 3, 2, 1.

10.11 Fifteen distinct DNA polymerases—a, b, g, d, e, k, z, h,

q, k, l, m, s, f, and Rev1—have been characterized in

mammals. What are the intracellular locations and func-

tions of these polymerases?

ANS: Current evidence suggests that polymerases a, d, and/or

e are required for the replication of nuclear DNA. Poly-

merase e is thought to catalyze the continuous synthesis

of the leading strand, and polymerases a and d are

thought to catalyze the replication of the lagging strand.

Polymerase a forms a complex with primase and initiates

the synthesis of Okazaki fragments during the discon-

tinuous replication of the lagging strand. Polymerase a

catalyzes the incorporation of approximately the rst 30

nucleotides in each Okazaki fragment before being

replaced by polymerase d, which then completes the syn-

thesis of the fragments. Polymerase g catalyzes replica-

tion of organellar chromosomes. Polymerases b, k, z, h,

q, k, l, m, s, f, and Rev1 function in various DNA repair

pathways (see Chapter 13).

10.12 The E. coli chromosome contains approximately 4  10

nucleotide pairs and replicates as a single bidirectional

replicon in approximately 40 minutes under a wide vari-

ety of growth conditions. The largest chromosome of

D. melanogaster contains about 6  10

nucleotide pairs.

(a) If this chromosome contains one giant molecule of

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DNA that replicates bidirectionally from a single origin

located precisely in the middle of the DNA molecule,

how long would it take to replicate the entire chromo-

some if replication in Drosophila occurred at the same

rate as replication in E. coli? (b) Actually, replication

rates are slower in eukaryotes than in prokaryotes. If

each replication bubble grows at a rate of 5000 nucleo-

tide pairs per minute in Drosophila and 100,000 nucleo-

tide pairs per minute in E. coli, how long will it take to

replicate the largest Drosophila chromosome if it contains

a single bidirectional replicon as described in (a) above?

minutes. Based on your calculations in (a) and (b) earlier,

what do these rapid nuclear divisions indicate about the

number of replicons per chromosome in Drosophila?

ANS: (a) Given bidirectional replication of a single replicon,

each replication fork must traverse 2 10

nucleotide

pairs in E. coli and 3  10

nucleotide pairs in the largest

Drosophila chromosome. If the rates were the same in

both species, it would take 15 times (3  10

/2  10

) as

long to replicate the Drosophila chromosome or 10 hours

(40 minutes  15  600 minutes).(b) If replication forks

in E. coli move 20 times as fast as replication forks in

Drosophila (100,000 nucleotide pairs per minute/5000

nucleotide pairs per minute), the largest Drosophila chromo-

some would require 8.3 days (10 hours  20  200 hours)

to complete one round of replication. (c) Each Drosophila

chromosome must contain many replicons in order to

complete replication in less than 10 minutes.

10.13 E. coli cells that have been growing in

N for many gen-

erations are transferred to medium containing only

and allowed to grow in this medium for four generations.

Their DNA is then extracted and analyzed by equilib-

rium CsCl density-gradient centrifugation. What pro-

portion of this DNA will band at the “light,” “hybrid,”

and “heavy” positions in the gradient?

ANS: No DNA will band at the “light” position; 12.5 percent

(2 of 16 DNA molecules) will band at the “hybrid” den-

sity; and 87.5 percent (14 of 16 DNA molecules) will

band at the “heavy” position.

10.14 The bacteriophage lambda chromosome has several

A:T-rich segments that denature when exposed to pH

11.05 for 10 minutes. After such partial denaturation,

the linear packaged form of the lambda DNA molecule

has the structure shown in Figure 10.9a. Following its

injection into an E. coli cell, the lambda DNA molecule

is converted into a covalently closed circular molecule by

hydrogen bonding between its complementary single-

stranded termini and the action of DNA ligase. It then

replicates as a q-shaped structure. The entire lambda

chromosome is 17.5 mm long. It has a unique origin of

replication located 14.3 mm from the left end of the lin-

ear form shown in Figure 10.9a. Draw the structure that

would be observed by electron microscopy after both

(1) replication of an approximately 6-mm-long segment

of the lambda chromosomal DNA molecule (in vivo) and

(2) exposure of this partially replicated DNA molecule to

pH 11.05 for 10 minutes (in vitro), (a) if replication had

proceeded bidirectionally from the origin, and (b) if rep-

lication had proceeded unidirectionally from the origin.

ANS:

(a) and (b)

10.15 What enzyme activity catalyzes each of the following

steps in the semiconservative replication of DNA in

prokaryotes?

(a) The formation of negative supercoils in progeny

DNA molecules. (b) The synthesis of RNA primers.

sion of DNA chains at the 3′-OH termini of primer

strands. (e) Proofreading of the nucleotides at the 3′-OH

termini of DNA primer strands?

ANS: (a) DNA gyrase; (b) primase; (c) the 5′ → 3′ exonuclease

activity of DNA polymerase I; (d) the 5′ → 3′ polymerase

activity of DNA polymerase III; (e) the 3′ → 5′ exonucle-

ase activity of DNA polymerase III.

10.16 One species of tree has a very large genome consisting of

2.0  10

base pairs of DNA.

(a) If this DNA was organized into a single linear mole-

cule, how long (meters) would this molecule be? (b) If

the DNA is evenly distributed among 10 chromosomes

and each chromosome has one origin of DNA replica-

tion, how long would it take to complete the S phase of

the cell cycle, assuming that DNA polymerase can syn-

thesize 2  10

bp of DNA per minute? (c) An actively

growing cell can complete the S phase of the cell cycle in

approximately 300 minutes. Assuming that the origins of

replication are evenly distributed, how many origins of

replication are present on each chromosome? (d) What

is the average number of base pairs between adjacent ori-

gins of replication?

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ANS: (a) (34 nm/100 bp)(2  10

bp)  6.8  10

 6.8 meters.

(b) 2  10

/10 chromosomes  2  10

bp; 4  10

bp/

min (bidirectional); 2  10

/4  10

 5  10

min 

50,000 min. (c) (50,000 min)(1 ori)  (300 min)

(X ori); X  50,000/300  167 ori. (d) (2  10

bp/

chrom.)/(167 ori/chrom.)  1.2  10

bp/ori.

10.17 Why must each of the giant DNA molecules in eukaryotic

chromosomes contain multiple origins of replication?

ANS: In eukaryotes, the rate of DNA synthesis at each replica-

tion fork is about 2500–3000 nucleotide pairs per min-

ute. Large eukaryotic chromosomes often contain

–10

nucleotide pairs. A single replication fork could

not replicate the giant DNA in one of these large chro-

mosomes fast enough to permit the observed cell gen-

eration times.

10.18 In E. coli, viable polA mutants have been isolated that

produce a defective gene product with little or no 5′→3′

polymerase activity, but normal 5′→3′ exonuclease activ-

ity. However, no polA mutant has been identied that

is completely decient in the 5′→3′ exonuclease activity,

while retaining 5′→3′ polymerase activity, of DNA poly-

merase I. How can these results be explained?

ANS: The 5′ → 3′ exonuclease activity of DNA polymerase I is

essential to the survival of the bacterium, whereas the

5′→3′ polymerase activity of the enzyme is not essential.

10.19 Other polA mutants of E. coli lack the 3′ → 5′ exonuclease

activity of DNA polymerase I. Will the rate of DNA syn-

thesis be altered in these mutants? What effect(s) will

these polA mutations have on the phenotype of the

organism?

ANS: No, the rate of DNA synthesis will not be altered. E. coli

strains carrying polA mutations that eliminate the 3′ → 5′

exonuclease activity of DNA polymerase I will exhibit

unusually high mutation rates.

10.20 Many of the origins of replication that have been charac-

terized contain AT-rich core sequences. Are these

AT-rich cores of any functional signicance? If so, what?

ANS: Because AT base pairs are held together by only two

hydrogen bonds instead of the three hydrogen bonds

present in GC base pairs, the two strands of AT-rich

regions of double helices are separated more easily, pro-

viding the single-stranded template regions required for

DNA replication.

10.21 (a) Why isn’t DNA primase activity required to initiate

rolling-circle replication?(b) DNA primase is required

for the discontinuous synthesis of the lagging strand,

which occurs on the single-stranded tail of the rolling

circle. Why?

ANS: Rolling-circle replication begins when an endonuclease

cleaves one strand of a circular DNA double helix. This

cleavage produces a free 3′-OH on one end of the cut

strand, allowing it to function as a primer. The discon-

tinuous synthesis of the lagging strand requires the de

novo initiation of each Okazaki fragment, which requires

DNA primase activity.

10.22 DNA polymerase I is needed to remove RNA primers

during chromosome replication in E. coli. However,

DNA polymerase III is the true replicase in E. coli. Why

does not DNA polymerase III remove the RNA

primers?

ANS: DNA polymerase III does not have a 5′ → 3′ exonuclease

activity that acts on double-stranded nucleic acids. Thus,

it cannot excise RNA primer strands from replicating

DNA molecules. DNA polymerase I is present in cells at

much higher concentrations and functions as a mono-

mer. Thus, DNA polymerase I is able to catalyze the

removal of RNA primers from the vast number of Oka-

zaki fragments formed during the discontinuous replica-

tion of the lagging strand.

10.23 In E. coli, three different proteins are required to unwind

the parental double helix and keep the unwound strands

in an extended template form. What are these proteins,

and what are their respective functions?

ANS: DNA helicase unwinds the DNA double helix, and sin-

gle-strand DNA-binding protein coats the unwound

strands, keeping them in an extended state. DNA gyrase

catalyzes the formation of negative supercoiling in E. coli

DNA, and this negative supercoiling behind the replica-

tion forks is thought to drive the unwinding process

because superhelical tension is reduced by unwinding

the complementary strands.

10.24 How similar are the structures of DNA polymerase I and

DNA polymerase III in E. coli? What is the structure of

the DNA polymerase III holoenzyme? What is the func-

tion of the dnaN gene product in E. coli?

ANS: DNA polymerase I is a single polypeptide of molecular

weight 109,000, whereas DNA polymerase III is a

complex multimeric protein. The DNA polymerase III

holoenzyme has a molecular mass of about 900,000

daltons and is composed of at least 20 different poly-

peptides. The dnaN gene product, the b subunit of

DNA polymerase III, forms a dimeric clamp that encir-

cles the DNA molecule and prevents the enzyme from

dissociating from the template DNA during replication.

10.25 The dnaA gene product of E. coli is required for the ini-

tiation of DNA synthesis at oriC. What is its function?

How do we know that the DnaA protein is essential to

the initiation process?

ANS: DnaA protein initiates the formation of the replication

bubble by binding to the 9-bp repeats of OriC. DnaA

protein is known to be required for the initiation process

because bacteria with temperature-sensitive mutations in

the dnaA gene cannot initiate DNA replication at restric-

tive temperatures.

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10.26 What is a primosome, and what are its functions? What

essential enzymes are present in the primosome? What

are the major components of the E. coli replisome? How

can geneticists determine whether these components are

required for DNA replication?

ANS: The primosome is a protein complex that initiates the

synthesis of Okazaki fragments during lagging strand

synthesis. The major components of the E. coli DNA

primosome are DNA primase and DNA helicase. Genet-

icists have been able to show that both DNA primase

and DNA helicase are required for DNA replication by

demonstrating that mutations in the genes encoding

these enzymes result in the arrest of DNA synthesis in

mutant cells under conditions where the altered proteins

are inactive.

10.27 The chromosomal DNA of eukaryotes is packaged into

nucleosomes during the S phase of the cell cycle. What

obstacles do the size and complexity of both the repli-

some and the nucleosome present during the semicon-

servative replication of eukaryotic DNA? How might

these obstacles be overcome?

ANS: Nucleosomes and replisomes are both large macromo-

lecular structures, and the packaging of eukaryotic

DNA into nucleosomes raises the question of how a

replisome can move past a nucleosome and replicate

the DNA in the nucleosome in the process. The most

obvious solution to this problem would be to com-

pletely or partially disassemble the nucleosome to

allow the replisome to pass. The nucleosome would

then reassemble after the replisome had passed. One

popular model has the nucleosome partially disassem-

bling, allowing the replisome to move past it (see

Figure 10.33b).

10.28 Two mutant strains of E. coli each have a temperature-

sensitive mutation in a gene that encodes a product

required for chromosome duplication. Both strains repli-

cate their DNA and divide normally at 25°C but are

unable to replicate their DNA or divide at 42°C. When

cells of one strain are shifted from growth at 25°C to

growth at 42°C, DNA synthesis stops immediately. When

cells of the other strain are subjected to the same tempera-

ture shift, DNA synthesis continues, albeit at a decreasing

rate, for about a half hour. What can you conclude about

the functions of the products of these two genes?

ANS: The product of the rst gene is required for DNA chain

extension, whereas the product of the second gene is

only required for the initiation of DNA synthesis.

10.29 In what ways does chromosomal DNA replication in

eukaryotes differ from DNA replication in prokaryotes?

ANS: (1) DNA replication usually occurs continuously in

rapidly growing prokaryotic cells but is restricted to

the S phase of the cell cycle in eukaryotes. (2) Most

eukaryotic chromosomes contain multiple origins of

replication, whereas most prokaryotic chromosomes

contain a single origin of replication. (3) Prokaryotes

utilize two catalytic complexes that contain the same

DNA polymerase to replicate the leading and lagging

strands, whereas eukaryotes utilize two or three distinct

DNA polymerases for leading and lagging strand syn-

thesis. (4) Replication of eukaryotic chromosomes

requires the partial disassembly and reassembly of

nucleosomes as replisomes move along parental DNA

molecules. In prokaryotes, replication probably involves

a similar partial disassembly/reassembly of nucleo-

some-like structures. (5) Most prokaryotic chromo-

somes are circular and thus have no ends. Most

eukaryotic chromosomes are linear and have unique

termini called telomeres that are added to replicating

DNA molecules by a unique, RNA-containing enzyme

called telomerase.

10.30 (a) The chromosome of the bacterium Salmonella

typhimurium contains about 4  10

nucleotide pairs.

Approximately how many Okazaki fragments are pro-

duced during one complete replication of the S.

typhimurium chromosome? (b) The largest chromosome

of D. melanogaster contains approximately 6  10

nucle-

otide pairs. About how many Okazaki fragments are pro-

duced during the replication of this chromosome?

ANS: (a) 2000–4000 Okazaki fragments. (b) 300,000–600,000

Okazaki fragments.

10.31 In the yeast S. cerevisiae, haploid cells carrying a mutation

called est1 (for ever-shorter telomeres) lose distal telo-

mere sequences during each cell division. Predict the

ultimate phenotypic effect of this mutation on the prog-

eny of these cells.

ANS: The chromosomes of haploid yeast cells that carry the

est1 mutation become shorter during each cell division.

Eventually, chromosome instability results from the

complete loss of telomeres, and cell death occurs because

of the deletion of essential genes near the ends of

chromosomes.

10.32 Assume that the sequence of a double-stranded DNA

shown below is present at one end of a large DNA mol-

ecule in a eukaryotic chromosome.

5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′

3′-(centromere sequence)-ctaaggggcccttcgaaccccccgggtagaagcatgagaaacgt-5′

You have reconstituted a eukaryotic replisome that is

active in vitro. However, it lacks telomerase activity. If

you isolate the DNA molecule shown above and repli-

cate it in your in vitro system, what products would you

expect?

ANS: Without telomerase, the 5′ end of the newly replicated

strand will be missing some bases. The exact number of

missing bases does not matter:

5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′

3′-(centromere sequence)-ctaaggggcccttcgaaccccccgcgggtagaagcatg-5′

5′-(centromere sequence)-gattccccgggaagcttggggggcccatcttcgtacgtctttgca-3′

3′-(centromere sequence)-ctaaggggcccttc-5′

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Chapter 11

11.1 Distinguish between DNA and RNA (a) chemically,

(b) functionally, and (c) by location in the cell.

ANS: (a) RNA contains the sugar ribose, which has a hydroxyl

(OH) group on the 2-carbon; DNA contains the sugar

2-deoxyribose, with only hydrogens on the 2-carbon.

RNA usually contains the base uracil at positions where

thymine is present in DNA. However, some DNAs con-

tain uracil, and some RNAs contain thymine. DNA exists

most frequently as a double helix (double-stranded mole-

cule); RNA exists more frequently as a single-stranded

molecule. But, some DNAs are single-stranded and some

RNAs are double-stranded. (b) The main function of

DNA is to store genetic information and to transmit that

information from cell to cell and from generation to gen-

eration. RNA stores and transmits genetic information in

some viruses that contain no DNA. In cells with both

DNA and RNA: (1) mRNA acts as an intermediary in pro-

tein synthesis, carrying the information from DNA in the

chromosomes to the ribosomes (sites at which proteins are

synthesized). (2) tRNAs carry amino acids to the ribo-

somes and function in codon recognition during the syn-

thesis of polypeptides. (3) rRNA molecules are essential

components of the ribosomes. (4) snRNAs are important

components of spliceosomes. (5) miRNAs play key roles

in regulating gene expression (see Chapter 18). (c) DNA is

located primarily in the chromosomes, which are found in

the nucleus of eukaryotic cells; however, some DNA is

also found in cytoplasmic organelles, such as mitochon-

dria and chloroplasts. RNA is located throughout cells.

11.2 What bases in the mRNA transcript would represent the

following DNA template sequence: 5′-TGCAGACA-3′?

ANS: 3′-ACGUCUGU-5′

11.3 What bases in the transcribed strand of DNA would give

rise to the following mRNA base sequence:

5′-CUGAU-3′?

ANS: 3′-GACTA-5′

11.4 On the basis of what evidence was the messenger RNA

hypothesis established?

ANS: The genetic information of cells is stored in DNA, which

is located predominantly in the chromosomes. The gene

products (polypeptides) are synthesized primarily in the

cytoplasm on ribosomes. Some intermediate must there-

fore carry the genetic information from the chromo-

somes to the ribosomes. RNA molecules (mRNAs) were

shown to perform this function by means of RNA pulse-

labeling and pulse-chase experiments combined with

autoradiography. The enzyme RNA polymerase was

subsequently shown to catalyze the synthesis of mRNA

using chromosomal DNA as a template. Finally, the

mRNA molecules synthesized by RNA polymerase were

shown to faithfully direct the synthesis of specic poly-

peptides when used in in vitro protein synthesis systems.

11.5 At what locations in a eukaryotic cell does protein syn-

thesis occur?

ANS: Protein synthesis occurs on ribosomes. In eukaryotes,

most of the ribosomes are located in the cytoplasm and

are attached to the extensive membranous network of

endoplasmic reticulum. Some protein synthesis also

occurs in cytoplasmic organelles such as chloroplasts and

mitochondria.

11.6 List three ways in which the mRNAs of eukaryotes differ

from the mRNAs of prokaryotes.

ANS: (1) Eukaryotes have a 5′ cap on their mRNAs; prokary-

otes do not. (2) Messenger RNAs of eukaryotes generally

have a 3′ poly-A tail, prokaryotic mRNAs do not. (3)

Messenger RNA formation in eukaryotes involves

removal of introns (when present) and the splicing

together of exons. Prokaryotic genes (with very rare

exceptions) do not have introns.

11.7 What different types of RNA molecules are present in

prokaryotic cells? in eukaryotic cells? What roles do

these different classes of RNA molecules play in the cell?

ANS: Both prokaryotic and eukaryotic organisms contain mes-

senger RNAs, transfer RNAs, and ribosomal RNAs. In

addition, eukaryotes contain small nuclear RNAs and

micro RNAs. Messenger RNA molecules carry genetic

information from the chromosomes (where the informa-

tion is stored) to the ribosomes in the cytoplasm (where

the information is expressed during protein synthesis).

The linear sequence of triplet codons in an mRNA mol-

ecule species the linear sequence of amino acids in the

polypeptides produced during translation of that mRNA.

Transfer RNA molecules are small (about 80 nucleotides

long) molecules that carry amino acids to the ribosomes

and provide the codon-recognition specicity during

translation. Ribosomal RNA molecules provide part of

the structure and function of ribosomes; they represent

an important part of the machinery required for the syn-

thesis of polypeptides. Small nuclear RNAs are struc-

tural components of spliceosomes, which excise

noncoding intron sequences from nuclear gene tran-

scripts. Micro RNAs are involved in the regulation of

gene expression.

11.8 Many eukaryotic genes contain noncoding introns that

separate the coding sequences or exons of these genes. At

what stage during the expression of these split genes are

the noncoding intron sequences removed?

ANS: The entire nucleotide-pair sequences—including the

introns—of the genes are transcribed by RNA poly-

merase to produce primary transcripts that still contain

the intron sequences. The intron sequences are then

spliced out of the primary transcripts to produce the

mature, functional RNA molecules. In the case of

protein-encoding nuclear genes of higher eukaryotes,

the introns are spliced out by complex macromolecular

structures called spliceosomes.

11.9 For several decades, the dogma in biology has been that

molecular reactions in living cells are catalyzed by

enzymes composed of polypeptides. We now know that

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the introns of some precursor RNA molecules such as the

rRNA precursors in Tetrahymena are removed autocata-

lytically (self-spliced) with no involvement of any cata-

lytic protein. What does the demonstration of autocatalytic

splicing indicate about the dogma that biological reac-

tions are always catalyzed by proteinaceous enzymes?

ANS: “Self-splicing” of RNA precursors demonstrates that

RNA molecules can also contain catalytic sites; this

property is not restricted to proteins.

11.10 What role(s) do spliceosomes play in pathways of gene

expression? What is their macromolecular structure?

ANS: Spliceosomes excise intron sequences from nuclear gene

transcripts to produce the mature mRNA molecules that

are translated on ribosomes in the cytoplasm. Spliceo-

somes are complex macromolecular structures composed

of snRNA and protein molecules (see Figure 11.22).

11.11 What components of the introns of nuclear genes that

encode proteins in higher eukaryotes are conserved and

required for the correct excision of intron sequences

from primary transcripts by spliceosomes?

ANS: The introns of protein-encoding nuclear genes of higher

eukaryotes almost invariably begin (5′) with GT and end

(3′) with AG. In addition, the 3′ subterminal A in the

“TACTAAC box” is completely conserved; this A is

involved in bond formation during intron excision.

11.12 Match one of the following terms with each of the

descriptions given below. Terms: (1) sigma (s) factor;

(2) poly(A) tail; (3) TATAAT; (4) exons; (5) TATAAAA;

(6) RNA polymerase III; (7) intron; (8) RNA poly-

merase II; (9) heterogeneous nuclear RNA (hnRNA);

(10) snRNA; (11) RNA polymerase I; (12) TTGACA;

(13) GGCCAATCT (CAAT box).

Descriptions:

(a) Intervening sequence found in many eukaryotic

genes.

(b) A conserved nucleotide sequence (-30) in eukaryotic

promoters involved in the initiation of transcription.

eukaryotic cells, most as components of the spliceosome,

that participate in the excision of introns from nuclear

gene transcripts.

(d) A sequence (-10) in the nontemplate strand of the

promoter of E. coli that facilitates the localized unwind-

ing of DNA when complexed with RNA polymerase.

(e) The RNA polymerase in the nucleus that catalyzes

the synthesis of all rRNAs except for the small 5S rRNA.

(f) The subunit of prokaryotic RNA polymerase that is

responsible for the initiation of transcription at

promoters.

(g) An E. coli promoter sequence located 35 nucleotides

upstream from the transcription-initiation site; it serves

as a recognition site for the sigma factor.

(h) The RNA polymerase in the nucleus that catalyzes

the synthesis of the transfer RNA molecules and small

nuclear RNAs.

(i) A polyadenosine tract 20–200 nucleotides long that is

added to the 3′ end of most eukaryotic messenger RNAs.

(j) The RNA polymerase that transcribes nuclear genes

that encode proteins.

(k) A conserved sequence in the nontemplate strand of

eukaryotic promoters that is located about 80 nucleo-

tides upstream from the transcription start site.

(l) Segments of an eukaryotic gene that correspond to

the sequences in the nal processed RNA transcript of

the gene.

(m) The population of primary transcripts in the nucleus

of a eukaryotic cell.

ANS: (a) 7; (b) 5; (c) 10; (d) 3; (e) 11; (f) 1; (g) 12; (h) 6; (i) 2;

(j) 8; (k) 13; (l) 4; (m) 9.

11.13 (a) Which of the following nuclear pre-mRNA nucleo-

tide sequences potentially contains an intron?

(1) 5′-UGACCAUGGCGCUAACACUGCCAAU

UGGCAAUACUGACCUGAUAGCAUCAGCCAA-3′

(2) 5′-UAGUCUCAUCUGUCCAUUGACUUCGAAA

CUGAAUCGUAACUCCUACGUCUAUGGA-3′

(3) 5′-UAGCUGUUUGUCAUGACUGACUGGUCA

CUAUCGUACUAACCUGUCAUGCAAUGUC-3′

(4) 5′-UAGCAGUUCUGUCGCCUCGUGGUGCUG

CUGGCCCUUCGUCGCUCGGGCUUAGCUA-3′

(5) 5′-UAGGUUCGCAUUGACGUACUUCUGAAAC

UACUAACUACUAACGCAUCGAGUCUCAA-3′

(b) One of the ve pre-mRNAs shown in (a) may undergo

RNA splicing to excise an intron sequence. What mRNA

nucleotide sequence would be expected to result from

this splicing event?

ANS: (a) Sequence 5. It contains the conserved intron sequences:

a 5′ GU, a 3′ AG, and a UACUAAC internal sequence

providing a potential bonding site for intron excision.

Sequence 4 has a 5′ GU and a 3′ AG but contains no

internal A for the bonding site during intron excision.

(b) 5′—UAGUCUCAA—3′; the putative intron from the

5′ GU through the 3′ AG has been removed.

11.14 What is the function of the introns in eukaryotic genes?

ANS: This is a wide-open question at present! There is much

speculation, but little hard evidence. One popular

hypothesis is that introns enhance exon shufing by

increasing recombination events between sequences

encoding adjacent domains of a polypeptide. Also, in one

yeast mitochondrial gene, the introns contain open read-

ing frames that encode “maturases” that splice out these

introns—a neat negative feedback control. Other introns

may be merely relics of evolution.

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11.15 A particular gene is inserted into the phage lambda chro-

mosome and is shown to contain three introns.

(a) The primary transcript of this gene is puried from

isolated nuclei. When this primary transcript is hybrid-

ized under R-loop conditions with the recombinant

lambda chromosome carrying the gene, what will the

R-loop structure(s) look like? Label your diagram.

(b) The mRNA produced from the primary transcript of

this gene is then isolated from cytoplasmic polyribosomes

and similarly examined by the R-loop hybridization

procedure using the recombinant lambda chromosome

carrying the gene. Diagram what the R-loop structure(s)

will look like when the cytoplasmic mRNA is used. Again,

label the components of your diagram.

ANS:

(a)

(b)

Displaced single-stranded DNA ("R-loop")

Primary transcript

mRNA

Exon1

Intron1

Intron2

Exon2

Intron3

Exon3

Exon4

λ DNA

Intron1

Intron2 Intron3

DNA

Displaced single-stranded exon DNA ("R-loops")

DNA

λDNA

11.16 A segment of DNA in E. coli has the following sequence

of nucleotide pairs:

3´-A

5´-T TTT

TTTTTTTTT

AAAAAAAAA

AAA

GGGGG

GGGG

CCCC

CCCCCCG-5´

C-

3´

When this segment of DNA is transcribed by RNA poly-

merase, what will be the sequence of nucleotides in the

RNA transcript if the promoter is located to the left of

the sequence shown?

ANS: If there is a promoter located upstream from this DNA

segment, the nucleotide sequence of this portion of the

RNA transcript will be

5′-UACGAUGACGAUAAGCGACAUAGC-3′. If there

is no upstream promoter, this segment of DNA will not

be transcribed.

11.17 A segment of DNA in E. coli has the following sequence

of nucleotide pairs:

3´-A

5´

-T TTT

TTTTGATTT

AAAACTAAA

AAA

GCGGA

GGGA

CCCT

CCGCCTG-

C-

TTT

TTTTTTTTTA

AAAAAAAAAT

AAA

GGGGG

GGGG

CCCC

CCCCCC G-5´

C-3´

When this segment of DNA is transcribed by RNA poly-

merase, what will be the sequence of nucleotides in the

RNA transcript?

ANS: Assuming that there is a 35 sequence upstream from

the consensus 10 sequence in this segment of the DNA

molecule, the nucleotide sequence of the transcript will

be 5′-ACCCGACAUAGCUACGAUGACGAUAAGC

GACAUAGC-3′.

11.18 A segment of DNA in E. coli has the following sequence

of nucleotide pairs:

3´-A

5´

-T AGA

GCTTAGCGA

CGAATCGCT

TCT

TAGATA

GCGC

CGCG

ATCTATT

T-

A-

AAC

ACTATGGTCG

TGATACCAGC

TTG

TGACT

GGCA

CCGT

AACTG

-5´

A-3´

When this segment of DNA is transcribed by RNA poly-

merase, what will be the sequence of nucleotides in the

RNA transcript?

ANS: Given the consensus 35 and 10 sequences in this seg-

ment of DNA and the fact that transcripts almost always

start with a purine, the predicted nucleotide sequence of

the transcript is 5′-ACCCGACAUAGCUACGAUGA

CGAUA-3′.

11.19 A segment of human DNA has the following sequence of

nucleotide pairs:

3´-A

5´

-T AGA

GCTTAGCTT

CGAATCGAA

TCT

TAGATA

GCGA

CGCT

ATCTATT

T-

A-

AAC

ACTATGGTCG

TGATACCAGC

TTG

TGACT

GGCA

CCGT

AACTG

-5´

A-3´

When this segment of DNA is transcribed by RNA poly-

merase, what will be the sequence of nucleotides in the

RNA transcript?

ANS: Assuming that there is a CAAT box located upstream

from the TATA box shown in this segment of DNA,

the nucleotide sequence of the transcript will be

5′-ACCCGACAUAGCUACGAUGACGAUA-3′.

11.20 The genome of a human must store a tremendous amount

of information using the four nucleotide pairs present in

DNA. What does the language of computers tell us about

the feasibility of storing large amounts of information

using an alphabet composed of just four letters?

ANS: Given the vast amount of information that can be stored

on a small computer chip by using a binary code, it is

clear that large quantities of genetic information can be

stored in the genomes of organisms by using the four-

letter alphabet of the genetic code.

11.21 What is the central dogma of molecular genetics? What

impact did the discovery of RNA tumor viruses have on

the central dogma?

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ANS: According to the central dogma, genetic information is

stored in DNA and is transferred from DNA to RNA to

protein during gene expression. RNA tumor viruses

store their genetic information in RNA, and that infor-

mation is copied into DNA by the enzyme reverse tran-

scriptase after a virus infects a host cell. Thus, the

discovery of RNA tumor viruses or retroviruses—retro

for backwards ow of genetic information—provided an

exception to the central dogma.

11.22 The biosynthesis of metabolite X occurs via six steps

catalyzed by six different enzymes. What is the minimal

number of genes required for the genetic control of this

metabolic pathway? Might more genes be involved?

Why?

ANS: If six different enzymes are required for the pathway,

then minimally six genes are necessary for genetic con-

trol of the pathway. However, because the expression of

these enzymes may rely on the product of other genes,

such as transcription factors, more than six genes could

be involved in genetic control of this pathway.

11.23 What do processes of DNA synthesis, RNA synthesis,

and polypeptide synthesis have in common?

ANS: DNA, RNA, and protein synthesis all involve the synthe-

sis of long chains of repeating subunits. All three pro-

cesses can be divided into three stages: chain initiation,

chain elongation, and chain termination.

11.24 What are the two stages of gene expression? Where do

they occur in a eukaryotic cell? a prokaryotic cell?

ANS: The two stages of gene expression are as follows:

(1) Transcription—the transfer of genetic information

from DNA to RNA.

(2) Translation—the transfer of genetic information from

RNA to protein. In eukaryotes, transcription occurs in

the nucleus and translation occurs in the cytoplasm

on complex macromolecular structures called ribosomes.

In prokaryotes, transcription and translation are often

coupled with mRNA molecules often being translated

by ribosomes while still being synthesized during

transcription.

11.25 Compare the structures of primary transcripts with those

of mRNAs in prokaryotes and eukaryotes. On average, in

which group of organisms do they differ the most?

ANS: The primary transcripts of eukaryotes undergo more

extensive posttranscriptional processing than those of

prokaryotes. Thus, the largest differences between

mRNAs and primary transcripts occur in eukaryotes.

Transcript processing is usually restricted to the excision

of terminal sequences in prokaryotes. In contrast, eukary-

otic transcripts are usually modied by (1) the excision of

intron sequences; (2) the addition of 7-methyl guanosine

caps to the 5′ termini; (3) the addition of poly(A) tails to

the 3′ termini. In addition, the sequences of some eukary-

otic transcripts are modied by RNA editing processes.

11.26 What ve types of RNA molecules participate in the

process of gene expression? What are the functions of

each type of RNA? Which types of RNA perform their

function(s) in (a) the nucleus and (b) the cytoplasm?

ANS: The ve types of RNA molecules that are involved in

gene expression are mRNAs, rRNAs, tRNAs, micro

RNAs, and snRNAs. mRNA molecules carry genetic

information from genes to the sites of protein synthesis

and specify the amino acid sequences of polypeptides.

rRNAs are major structural components of the ribo-

somes and provide functions required for translation.

tRNA molecules are the adapters that provide amino

acid-codon specicity during translation; each tRNA is

activated by a specic amino acid and contains an antico-

don sequence that is complementary or partially comple-

mentary to one, two, or three codons in mRNAs.

Micro-RNAs are involved in regulative gene expression

(see Chapter 18). snRNAs are structural components of

the spliceosomes that excise introns from gene transcripts

in eukaryotes. snRNAs perform their splicing functions

in the nucleus. mRNAs carry information from the

nucleus to the cytoplasm, so they function in both com-

partments of the cell. However, their most prominent

function is to direct the synthesis of polypeptides during

translation, which occurs in the cytoplasm. rRNAs and

tRNAs perform their functions during translation in the

cytoplasm. Micro-RNAs become incorporated into ribo-

nucleoprotein complexes in the cytoplasm.

11.27 Why was the need for an RNA intermediary in protein

synthesis most obvious in eukaryotes? How did research-

ers rst demonstrate that RNA synthesis occurred in the

nucleus and that protein synthesis occurred in the

cytoplasm?

ANS: In eukaryotes, the genetic information is stored in DNA

in the nucleus, whereas proteins are synthesized on ribo-

somes in the cytoplasm. How could the genes, which are

separated from the sites of protein synthesis by a double-

membrane—the nuclear envelope, direct the synthesis of

polypeptides without some kind of intermediary to carry

the specications for the polypeptides from the nucleus

to the cytoplasm? Researchers rst used labeled RNA

and protein precursors and autoradiography to demon-

strate that RNA synthesis and protein synthesis occurred

in the nucleus and the cytoplasm, respectively.

11.28 Two eukaryotic genes encode two different polypeptides,

each of which is 335 amino acids long. One gene con-

tains a single exon; the other gene contains an intron of

41,324 nucleotide pairs long. Which gene would you

expect to be transcribed in the least amount of time?

Why? When the mRNAs specied by these genes are

translated, which mRNA would you expect to be trans-

lated in the least time? Why?

ANS: Because transcription results in a primary transcript from

the DNA template, the DNA sequence for the single-

exon gene is the shortest, and so it will be transcribed in

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the least amount of time. However, because each poly-

peptide is the same length, the mature mRNAs for both

genes will be the same length and will be translated in

the same amount of time.

11.29 Design an experiment to demonstrate that RNA tran-

scripts are synthesized in the nucleus of eukaryotes and

are subsequently transported to the cytoplasm.

ANS: A simple pulse- and pulse/chase-labeling experiment will

demonstrate that RNA is synthesized in the nucleus and

is subsequently transported to the cytoplasm. This

experiment has two parts: (1) pulse label eukaryotic cul-

ture cells by growing them in

H-uridine for a few min-

utes and localize the incorporated radioactivity by

autoradiography. (2) Repeat the experiment, but this

time add a large excess of nonradioactive uridine to the

medium in which the cells are growing after the labeling

period and allow the cells to grow in the nonradioactive

medium for about an hour. Then localize the incorpo-

rated radioactivity by autoradiography.

11.30 Total RNA was isolated from human cells growing in cul-

ture. This RNA was mixed with nontemplate strands

(single strands) of the human gene encoding the enzyme

thymidine kinase, and the RNA–DNA mixture was incu-

bated for 12 hours under renaturation conditions. Would

you expect any RNA–DNA duplexes to be formed during

the incubation? If so, why? If not, why not? The same

experiment was then performed using the template strand

of the thymidine kinase gene. Would you expect any

RNA–DNA duplexes to be formed in this second

experiment? If so, why? If not, why not?

ANS: RNA–DNA duplexes will be formed when the template

strand is used, but not when the nontemplate strand is

used. (However, if some hybridization is observed with

the nontemplate strand, this is because total RNA was

used and is nonspecic to the thymidine kinase gene)

Only one strand—the template strand—of most genes is

transcribed. Thus, RNA will contain nucleotide sequences

complementary to the template strand but not to the

nontemplate strand.

11.31 Two preparations of RNA polymerase from E. coli are

used in separate experiments to catalyze RNA synthesis

in vitro using a puried fragment of DNA carrying the

argH gene as template DNA. One preparation catalyzes

the synthesis of RNA chains that are highly heteroge-

neous in size. The other preparation catalyzes the syn-

thesis of RNA chains that are all the same length. What

is the most likely difference in the composition of the

RNA polymerases in the two preparations?

ANS: The rst preparation of RNA polymerase is probably

lacking the sigma subunit and, as a result, initiates the

synthesis of RNA chains at random sites along both

strands of the argH DNA. The second preparation prob-

ably contains the sigma subunit and initiates RNA chains

only at the site used in vivo, which is governed by the

position of the 10 and 35 sequences of the promoter.

11.32 Transcription and translation are coupled in prokaryotes.

Why is this not the case in eukaryotes?

ANS: In eukaryotes, transcription occurs in the nucleus and

translation occurs in the cytoplasm. Because these pro-

cesses occur in different compartments of the cell, they

cannot be coupled as they are in prokaryotes.

11.33 What two elements are almost always present in the

promoters of eukaryotic genes that are transcribed by

RNA polymerase II? Where are these elements located

relative to the transcription start site? What are their

functions?

ANS: TATA and CAAT boxes. The TATA and CAAT boxes are

usually centered at positions 30 and 80, respectively,

relative to the startpoint (1) of transcription. The

TATA box is responsible for positioning the transcrip-

tion startpoint; it is the binding site for the rst basal

transcription factor that interacts with the promoter.

The CAAT box enhances the efciency of transcrip-

tional initiation.

11.34 In what ways are most eukaryotic gene transcripts modi-

ed? What are the functions of these posttranscriptional

modications?

ANS: (1) Intron sequences are spliced out of gene transcripts to

provide contiguous coding sequences for translation. (2)

The 7-methyl guanosine caps added to the 5′ termini of

most eukaryotic mRNAs help protect them from degra-

dation by nucleases and are recognized by proteins

involved in the initiation of translation. (3) The poly(A)

tails at the 3′ termini of mRNAs play an important role

in their transport from the nucleus to the cytoplasm and

enhance their stability.

11.35 How does RNA editing contribute to protein diversity in

eukaryotes?

ANS: RNA editing sometimes leads to the synthesis of two or

more distinct polypeptides from a single mRNA.

11.36 How do the mechanisms by which the introns of tRNA

precursors, Tetrahymena rRNA precursors, and nuclear

pre-mRNAs are excised differ? In which process are

snRNAs involved? What role(s) do these snRNAs

play?

ANS: The introns of tRNA precursors, Tetrahymena rRNA

precursors, and nuclear pre-mRNAs are excised by com-

pletely different mechanisms. (1) Introns in tRNA are

excised by cleavage and joining events catalyzed by splic-

ing nucleases and ligases, respectively. (2) Introns in Tet-

rahymena rRNA precursors are excised autocatalytically.

(3) Introns of nuclear pre-mRNAs are excised by spli-

ceosomes. snRNAs are involved in nuclear pre-mRNA

splicing as structural components of spliceosomes. In

addition, snRNA U1 is required for the cleavage events

at the 5′ termini of introns; U1 is thought to base-pair

with a partially complementary consensus sequence at

this position in pre-mRNAs.

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11.37 A mutation in an essential human gene changes the

5′ splice site of a large intron from GT to CC. Predict

the phenotype of an individual homozygous for this

mutation.

ANS: This zygote will probably be nonviable because the gene

product is essential, and the elimination of the 5′ splice

site will almost certainly result in the production of a

nonfunctional gene product.

11.38 Total RNA was isolated from nuclei of human cells

growing in culture. This RNA was mixed with a puried,

denatured DNA fragment that carried a large intron of a

housekeeping gene (a gene expressed in essentially all

cells), and the RNA–DNA mixture was incubated for

12 hours under renaturation conditions. Would you

expect any RNA–DNA duplexes to be formed during the

incubation? If so, why? If not, why not? The same exper-

iment was then performed using total cytoplasmic RNA

from these cells. Would you expect any RNA–DNA

duplexes to be formed in this second experiment? If so,

why? If not, why not?

ANS: In the rst experiment, yes, some hybridization would be

expected because not all of the RNA in the preparation

has been completely processed and will still contain the

intron sequence. However, if only cytoplasmic RNA is

used in the hybridization experiment, all of the mRNA in

the preparation will have been processed (because pro-

cessing occurs in the nucleus) and no hybridization will

be observed.

Chapter 12

12.1 In a general way, describe the molecular organization of

proteins and distinguish proteins from DNA, chemically

and functionally. Why is the synthesis of proteins of par-

ticular interest to geneticists?

ANS: Proteins are long chainlike molecules made up of amino

acids linked together by peptide bonds. Proteins are com-

posed of carbon, hydrogen, nitrogen, oxygen, and usually

sulfur. They provide the enzymatic capacity and much of

the structure of living organisms. DNA is composed

of phosphate, the pentose sugar 2-deoxyribose, and four

nitrogen-containing organic bases (adenine, cytosine,

guanine, and thymine). DNA stores and transmits the

genetic information in most living organisms. Protein

synthesis is of particular interest to geneticists because

proteins are the primary gene products—the key inter-

mediates through which genes control the phenotypes of

living organisms.

12.2 At what locations in the cell does protein synthesis occur?

ANS: Protein synthesis occurs on ribosomes. In eukaryotes,

most of the ribosomes are located in the cytoplasm and

are attached to the extensive membranous network of

endoplasmic reticulum. Some protein synthesis also

occurs in cytoplasmic organelles such as chloroplasts and

mitochondria.

12.3 Is the number of potential alleles of a gene directly

related to the number of nucleotide pairs in the gene? Is

such a relationship more likely to occur in prokaryotes or

in eukaryotes? Why?

ANS: It depends on how you dene alleles. If every variation in

nucleotide sequence is considered to be a different allele,

even if the gene product and the phenotype of the organ-

ism carrying the mutation are unchanged, then the

number of alleles will be directly related to gene size.

However, if the nucleotide sequence change must pro-

duce an altered gene product or phenotype before it is

considered a distinct allele, then there will be a positive

correlation, but not a direct relationship, between the

number of alleles of a gene and its size in nucleotide

pairs. The relationship is more likely to occur in pro-

karyotes where most genes lack introns. In eukaryotic

genes, nucleotide sequence changes within introns are

usually neutral; that is, they do not affect the activity of

the gene product or the phenotype of the organism.

Thus, in the case of eukaryotic genes with introns, there

may be no correlation between gene size and number of

alleles producing altered phenotypes.

12.4 Why was it necessary to modify Beadle and Tatum’s one

gene–one enzyme concept of the gene to one gene–one

polypeptide?

ANS: Several enzymes were shown to contain two or more dif-

ferent polypeptides, and these polypeptides were some-

times controlled by genes that mapped to different

chromosomes. Thus, the mutations clearly were not in

the same gene.

12.5 (a) Why is the genetic code a triplet code instead of a

singlet or doublet code? (b) How many different amino

acids are specied by the genetic code? (c) How many

different amino acid sequences are possible in a polypep-

tide 146 amino acids long?

ANS: (a) Singlet and doublet codes provide a maximum of

4 and (4)

or 16 codons, respectively. Thus, neither code

would be able to specify all 20 amino acids. (b) 20.

146

12.6 What types of experimental evidence were used to deci-

pher the genetic code?

ANS: Synthetic RNA molecules (polyuridylic acid molecules)

containing only the base uracil were prepared. When

these synthetic molecules were used to activate in vitro

protein synthesis systems, small polypeptide containing

only the amino acid phenylalanine (polyphenylalanine

molecules) was synthesized. Codons composed only of

uracil were thus shown to specify phenylalanine. Similar

experiments were carried out using synthetic RNA mol-

ecules with different base compositions. Later, in vitro

systems activated with synthetic RNA molecules with

known repeating base sequences were developed. Ulti-

mately, in vitro systems in which specic aminoacyl-

tRNAs where shown to bind to ribosomes activated with

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specic mini-mRNAs, which were trinucleotides of

known base sequence, were developed and used in codon

identication.

12.7 In what sense and to what extent is the genetic code

(a) degenerate, (b) ordered, and (c) universal?

ANS: (a) The genetic code is degenerate in that all but 2 of the

20 amino acids are specied by two or more codons.

Some amino acids are specied by six different codons.

The degeneracy occurs largely at the third or 3′ base of

the codons. “Partial degeneracy” occurs where the third

base of the codon may be either of the two purines or

either of the two pyrimidines and the codon still species

the same amino acid. “Complete degeneracy” occurs

where the third base of the codon may be any one of the

four bases and the codon still species the same amino

acid. (b) The code is ordered in the sense that related

codons (codons that differ by a single base change) spec-

ify chemically similar amino acids. For example, the

codons CUU, AUU, and GUU specify the structurally

related amino acids leucine, isoleucine, and valine,

respectively. (c) The code appears to be almost com-

pletely universal. Known exceptions to universality

include strains carrying suppressor mutations that alter

the reading of certain codons (with low efciencies in

most cases) and the use of UGA as a tryptophan codon in

yeast and human mitochondria.

12.8 The thymine analog 5-bromouracil is a chemical muta-

gen that induces single base-pair substitutions in DNA

called transitions (substitutions of one purine for another

purine and one pyrimidine for another pyrimidine).

Using the known nature of the genetic code (Table 12.1),

which of the following amino acid substitutions should

you expect to be induced by 5-bromouracil with the

highest frequency:

(a) Met → Val; (b) Met → Leu; (c) Lys → Thr; (d) Lys →

Gln; (e) Pro → Arg; or (f) Pro → Gln?

Why?

ANS: (a) Met → Val. This substitution occurs as a result of a

transition. All other amino acid substitutions listed would

require transversions.

12.9 Using the information given in Problem 12.8, would you

expect 5-bromouracil to induce a higher frequency of

His → Arg or His → Pro substitutions? Why?

ANS: Expect a higher frequency of His → Arg substitutions.

His → Arg results from a transition; His → Pro would

require a transversion (not induced by 5-bromouracil).

12.10 What is the minimum number of tRNAs required to rec-

ognize the six codons specifying the amino acid leucine?

ANS: Because of wobble (Table 12.2), one tRNA can recognize

both UUA and UUG codons for leucine. However, it

takes two more tRNAs to recognize CUU, CUC, CUA,

and CUG. Therefore, a minimum of three tRNAs are

required to recognize the six codons for leucine.

12.11 Characterize ribosomes in general as to size, location,

function, and macromolecular composition.

ANS: Ribosomes are from 10 to 20 nm in diameter. They are

located primarily in the cytoplasm of cells. In bacteria,

they are largely free in the cytoplasm. In eukaryotes,

many of the ribosomes are attached to the endoplasmic

reticulum in the cytoplasm. Ribosomes are complex

structures composed of over 50 different polypeptides

and three to ve different RNA molecules. In both pro-

karyotes and eukaryotes, ribosomes are the site of

translation.

12.12 (a) Where in the cells of higher organisms do ribosomes

originate? (b) Where in the cells are ribosomes most

active in protein synthesis?

ANS: (a) The nucleus, specically the nucleoli. (b) The cytoplasm.

12.13 Identify three different types of RNA that are involved in

translation and list the characteristics and functions of

each.

ANS: Messenger RNA (mRNA) molecules carry genetic

information from the chromosomes (where the informa-

tion is stored) to the ribosomes in the cytoplasm (where

the information is expressed during protein synthesis).

The linear sequence of triplet codons in an mRNA

molecule species the linear sequence of amino acids

in the polypeptide(s) produced during translation of

that mRNA. Transfer RNA (tRNA) molecules are

small (about 80 nucleotides long) molecules that carry

amino acids to the ribosomes and provide the codon-

recognition specicity during translation. Ribosomal

RNA (rRNA) molecules provide part of the structure

and function of ribosomes; they represent an impor-

tant part of the machinery required for the synthesis of

polypeptides.

12.14. (a) How is messenger RNA related to polysome forma-

tion? (b) How does rRNA differ from mRNA and tRNA

in specicity? (c) How does the tRNA molecule differ

from that of DNA and mRNA in size and helical

arrangement?

ANS: (a) Polysomes are formed when two or more ribosomes

are simultaneously translating the same mRNA mole-

cule. Ribosomes are usually spaced about 90 nucleotides

apart on an mRNA molecule. Thus, polysome size is

determined by mRNA size. (b) A ribosome, which con-

tains rRNA molecules, can participate in the synthesis of

any polypeptide specied by the ribosome-associated

mRNA. In that sense, rRNA is nonspecic. Messenger

RNAs and tRNAs, in contrast, are specic, in directing the

synthesis of a particular polypeptide or set of polypep-

tides (mRNA) or in attaching to a particular amino acid

(tRNA). (c) Transfer RNA molecules are much smaller

(about 80 nucleotides) than DNA or mRNA molecules.

They are single-stranded molecules but have complex

secondary structures because of the base pairing between

different segments of the molecules.

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12.15 Outline the process of aminoacyl-tRNA formation.

ANS: A specic aminoacyl-tRNA synthetase catalyzes the forma-

tion of an amino acid-AMP complex from the appropriate

amino acid and ATP (with the release of pyro pho spha te).

The same enzyme then catalyzes the formation of the

aminoacyl-tRNA complex, with the release of AMP. Both

the amino acid-AMP and aminoacyl-tRNA linkages are

high-energy phosphate bonds.

12.16 How is translation (a) initiated and (b) terminated?

ANS: (a) Translation is initiated by a complex reaction involv-

ing mRNA, ribosomes, initiation factors (IF-1, IF-2, and

IF-3), GTP, the initiator codon AUG, and a special

initiator tRNA (rRNA

Met

). It also appears to involve a

base-pairing interaction between a base sequence near

the 3′-end of the 16S rRNA and a base sequence in the

“leader sequence” of the mRNA. (b) Translation is

terminated by recognition of one or more of the chain-

termination codons (UAG, UAA, and UGA) by the

appropriate protein release factor (RF-1 or RF-2).

12.17 Of what signicance is the wobble hypothesis?

ANS: Crick’s wobble hypothesis explains how the anticodon of

a given tRNA can base-pair with two or three different

mRNA codons. Crick proposed that the base-pairing

between the 5′ base of the anticodon in tRNA and the 3′

base of the codon in mRNA was less stringent than

normal and thus allowed some “wobble” at this site. As a

result, a single tRNA often recognizes two or three of

the related codons specifying a given amino acid (see

Table 12.2).

12.18 If the average molecular mass of an amino acid in a par-

ticular polypeptide is 100 daltons, about how many

nucleotides will be present in an mRNA coding sequence

specifying this polypeptide, which has a molecular mass

of 27,000 daltons?

ANS: At least 813 nucleotides [ (270 aa  3)  3 nucleotides

for termination codon].

12.19 The bases A, G, U, C, I (inosine) all occur at the 5′ posi-

tions of anticodons in tRNAs.

(a) Which base can pair with three different bases at the

3′ positions of codons in mRNA? (b) What is the mini-

mum number of tRNAs required to recognize all codons

of amino acids specied by codons with complete

degeneracy?

ANS: (a) Inosine. (b) Two.

12.20 Assume that in the year 2025, the rst expedition of

humans to Mars discovers several Martian life forms

thriving in hydrothermal vents that exist below the plan-

et’s surface. Several teams of molecular biologists extract

proteins and nucleic acids from these organisms and

make some momentous discoveries. Their rst discovery

is that the proteins in Martian life forms contain only 14

different amino acids instead of the 20 present in life

forms on Earth. Their second discovery is that the DNA

and RNA in these organisms have only two different

nucleotides instead of the four nucleotides present in liv-

ing organisms on Earth. (a) Assuming that transcription

and translation work similarly in Martians and Earth-

lings, what is the minimum number of nucleotides that

must be present in the Martian codon to specify all the

amino acids in Martians? (b) Assuming that the Martian

code proposed above has translational start-and-stop

signals, would you expect the Martian genetic code to be

degenerate like the genetic code used on Earth?

ANS: (a) Two nucleotides in all combinations of four (2

) would

produce 16 codons. Therefore, the minimum number of

nucleotides comprising the Martian genetic code must

be four. (b) Sixteen codons would allow code words for

14 amino acids, one initiation codon, and a translational

termination codon. The Martian genetic code could not

be degenerate.

12.21 What are the basic differences between translation in

prokaryotes and in eukaryotes?

ANS: Translation occurs by very similar mechanisms in pro-

karyotes and eukaryotes; however, there are some differ-

ences. (1) In prokaryotes, the initiation of translation

involves base-pairing between a conserved sequence

(AGGAGG)—the Shine–Dalgarno box—in mRNA and

a complementary sequence near the 3′ end of the 16S

rRNA. In eukaryotes, the initiation complex forms at the

5′ end of the transcript when a cap-binding protein

interacts with the 7-methyl guanosine on the mRNA.

The complex then scans the mRNA processively and ini-

tiates translation (with a few exceptions) at the AUG

closest to the 5′ terminus. (2) In prokaryotes, the amino

group of the initiator methionyl-tRNA

Met

is formylated;

in eukaryotes, the amino group of methionyl-tRNA

Met

not formylated. (3) In prokaryotes, two soluble protein

release factors (RFs) are required for chain termination.

RF-1 terminates polypeptides in response to UAA and

UAG codons; RF-2 terminates chains in response to

UAA and UGA codons. In eukaryotes, one release factor

responds to all three termination codons.

12.22 What is the function of each of the following compo-

nents of the protein-synthesizing apparatus:

(a) Aminoacyl-tRNA synthetase. (b) Release factor 1.

tion factor G

ANS: (a) Attachment of an amino acid to the correct tRNA.

(b) Recognition of termination codons UAA and UAG and

release of the nascent polypeptide from the tRNA in the

P site of the ribosome. (c) Formation of a peptide bond

between the amino group of the aminoacyl-tRNA in the

A site and the carboxyl group of the growing polypeptide

on the tRNA in the P site. (d) Formation of the initiation

complex required for translation; all steps leading up to

peptide bond formation. (e) Translocation of the pepti-

dyl-tRNA from the A site on the ribosome to the P site.

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12.23 An E. coli gene has been isolated and shown to be 68 nm

long. What is the maximum number of amino acids that

this gene could encode?

ANS: Assuming 0.34 nm per nucleotide pair in B-DNA, a gene

68 nm long would contain 200 nucleotide pairs. Given

the triplet code, this gene would contain 200/3  66.7

triplets, one of which must specify chain termination.

Disregarding the partial triplet, this gene could encode a

maximum of 65 amino acids.

12.24 (a) What is the difference between a nonsense mutation

and a missense mutation? (b) Are nonsense or missense

mutations more frequent in living organisms? (c) Why?

ANS: (a) A nonsense mutation changes a codon specifying an

amino acid to a chain-termination codon, whereas a mis-

sense mutation changes a codon specifying one amino

acid to a codon specifying a different amino acid. (b)

Missense mutations are more frequent. (c) Of the 64

codons, only three specify chain termination. Thus, the

number of possible missense mutations is much larger

than the number of possible nonsense mutations. More-

over, nonsense mutations almost always produce non-

functional gene products. As a result, nonsense mutations

in essential genes are usually lethal in the homozygous

state.

12.25 The human a-globin chain is 141 amino acids long. How

many nucleotides in mRNA are required to encode

human a-globin?

ANS: 426 nucleotides—3  141  423 specifying amino acids

plus three (one codon) specifying chain termination.

12.26 What are the functions of the A, P, and E aminoacyl-

tRNA binding sites on the ribosome?

ANS: The incoming aminoacyl-tRNA enters the A site of

the ribosome, the nascent polypeptide-tRNA occupies

the P site, and the uncharged exiting tRNA occupies the

E site.

12.27 (a) In what ways does the order in the genetic code mini-

mize mutational lethality? (b) Why do base-pair changes

that cause the substitution of a leucine for a valine in the

polypeptide gene product seldom produce a mutant

phenotype?

ANS: (a) Related codons often specify the same or very similar

amino acids. As a result, single base-pair substitutions

frequently result in the synthesis of identical proteins

(degeneracy) or proteins with amino acid substitutions

involving very similar amino acids. (b) Leucine and valine

have very similar structures and chemical properties;

both have nonpolar side groups and fold into essentially

the same three-dimensional structures when present in

polypeptides. Thus, substitutions of leucine for valine or

valine for leucine seldom alter the function of a protein.

12.28 (a) What is the function of the Shine–Dalgarno sequence

in prokaryotic mRNAs? (b) What effect does the dele-

tion of the Shine–Dalgarno sequence from an mRNA

have on its translation?

ANS: (a) The Shine–Dalgarno sequence is a conserved

polypurine tract, consensus AGGAGG, that is located

about seven nucleotides upstream from the AUG initiation

codon in mRNAs of prokaryotes. It is complementary to,

and is believed to base-pair with, a sequence near the 5′

terminus of the 16S ribosomal RNA. (b) Prokaryotic

mRNAs with the Shine– Dalgarno sequence deleted are

either not translated or are translated inefciently.

12.29 (a) In what ways are ribosomes and spliceosomes similar?

(b) In what ways are they different?

ANS: (a) Both ribosomes and spliceosomes play essential roles

in gene expression, and both are complex macromolecu-

lar structures composed of RNA and protein molecules.

(b) Ribosomes are located in the cytoplasm; spliceosomes

in the nucleus. Ribosomes are larger and more complex

than spliceosomes.

12.30 The 5′ terminus of a human mRNA has the following

sequence:

5′-GAAGAGACAAGGTCAUGGCCAUAUGC

UUGUUCCAAUCGUUAGCUGCGCAGGAUC-

GCCCUGGG . . . . . . 3′

When this mRNA is translated, what amino acid sequence

will be specied by this mRNA sequence?

ANS: NH

-Met-Ala-Ile-Cys-Leu-Phe-Gln-Ser-Leu-Ala-

Ala-Gln-Asp-Arg-Pro-Gly-COOH.

12.31 A partial (5′ subterminal) nucleotide sequence of a pro-

karyotic mRNA is as follows:

5′-.....AGGAGGCUCGAACAUGUCAAUAUG CUU

GUUCCAAUCGUUAGCUGCGCAGGACCGUCC

CG GA. . . . . . 3′

When this mRNA is translated, what amino acid

sequence will be specied by this portion of the mRNA?

ANS: Met-Ser-lle-Cys-Leu-Phe-Gln-Ser-Leu-Ala-Ala-

Gln-Asp-Arg-Pro-Gly

12.32 The following DNA sequence occurs in the nontem plate

strand of a gene in a bacterium (the promoter sequence

is located to the left but is not shown):

5′-GAATGTCAGAACTGCCATGCTTCATATGAA-

TAGACCTCTAG-3′

(a) What is the ribonucleotide sequence of the mRNA

molecule that is transcribed from this piece of DNA?

(b) What is the amino acid sequence of the polypeptide

encoded by this mRNA? (c) If the nucleotide indicated

↓

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by the arrow undergoes a mutation that changes T to A,

what will be the resulting amino acid sequence following

transcription and translation?

ANS: (a) 5′-GAAUGUCAGAACUGCCAUGCUUCAUAUG

AAUAGACCUCUAG-3′ (b) NH

-fMet-Ser-Glu-Leu-

Pro-Cys-Phe-Ile-COOH (c) NH

-fMet-Ser-Glu-Leu-

Pro-Cys-Phe-Ile-Arg-Ile-Asp-Leu-COOH.

12.33 Alan Garen extensively studied a particular nonsense

(chain-termination) mutation in the alkaline phospha-

tase gene of E. coli. This mutation resulted in the termi-

nation of the alkaline phosphatase polypeptide chain at a

position where the amino acid tryptophan occurred in

the wild-type polypeptide. Garen induced revertants (in

this case, mutations altering the same codon) of this

mutant with chemical mutagens that induced single

base-pair substitutions and sequenced the polypeptides

in the revertants. Seven different types of revertants

were found, each with a different amino acid at the tryp-

tophan position of the wild-type polypeptide (termina-

tion position of the mutant polypeptide fragment). The

amino acids present at this position in the various rever-

tants included tryptophan, serine, tyrosine, leucine, glu-

tamic acid, glutamine, and lysine. Did the nonsense

mutation studied by Garen contain a UAG, a UAA, or a

UGA nonsense mutation? Explain the basis of your

deduction.

ANS: (UAG). This is the only nonsense codon that is related to

tryptophan, serine, tyrosine, leucine, glutamic acid, glu-

tamine, and lysine codons by a single base-pair substitu-

tion in each case.

12.34 The following DNA sequence occurs in a bacterium

(the promoter sequence is located to the left but is not

shown).

↓

5′-CAATCATGGACTGCCATGCTTCATATGAATAGTTGACAT-3′

3′-GTTAGTACCTGACGGTACGAAGTATACTTATCAACTGTA-5′

(a) What is the ribonucleotide sequence of the mRNA

molecule that is transcribed from the template strand of

this piece of DNA? Assume that both translational start

and termination codons are present.

(b) What is the amino acid sequence of the polypeptide

encoded by this mRNA?

mutation that causes this C:G base pair to be deleted,

what will be the polypeptide encoded by the mutant

gene?

ANS: (a) 5′-CAAUCAUGGACUGCCAUGCUUCAUAUG

AAUAGUUGACAU-3′ (b) NH2-fMet-Asp-Cys-His-Ala-

Ser-Tyr-Glu-COOH (c) NH2-fMet-Asp-Cys -Met-Leu-

His -Met-Asn-Ser- COOH.

Chapter 13

13.1 Identify the following point mutations represented in

DNA and in RNA as (1) transitions, (2) transversions, or

(3) reading frameshifts. (a) A to G; (b) C to T; (c) C to G;

(d) T to A; (e) UAU ACC UAU to UAU AAC CUA;

(f) UUG CUA AUA to UUG CUG AUA.

ANS: (a) Transition, (b) transition, (c) transversion, (d) trans-

version, (e) frameshift, (f) transition.

13.2 Of all possible missense mutations that can occur in a

segment of DNA encoding the amino acid tryptophan,

what is the ratio of transversions to transitions if

all single base-pair substitutions occur at the same

frequency?

ANS: 6:1. UGG transitions: UGA (nonsense), UAG (nonsense),

CGG (Arg). UGG transversions: UGC (Cys), UGU

(Cys), UCG (Ser), UUG (Leu), AGG (Arg), GGG (Gly).

13.3 Both lethal and visible mutations are expected to occur

in fruit ies that are subjected to irradiation. Outline a

method for detecting (a) X-linked lethals and (b) X-linked

visible mutations in irradiated Drosophila.

ANS: (a) ClB method, (b) attached X method (see Chapter 6).

13.4 H. J. Muller used the ClB technique to identify many

radiation-induced recessive lethal mutations on Dro-

sophila’s X chromosome, which is now known to con-

tain more than a thousand genes. These mutations

could be propagated in stock cultures by keeping them

in heterozygous condition with the ClB chromosome.

Would you expect all these lethal mutations to be

alleles of one essential X-linked gene, or to be alleles

of different essential X-linked genes? Why couldn’t

H. J. Muller determine the answer to this question

experimentally?

ANS: The radiation-induced recessive lethal mutations that

Muller recovered in his experiments were likely

scattered across the entire X chromosome. Thus, they

likely affected different genes. However, Muller could

not study this issue because he could not carry out a

complementation test to determine if any of the lethal

mutations were alleles of the same gene. The reason is

that to perform the complementation test, Muller would

have had to cross females that carried a particular lethal

mutation balanced with the ClB chromosome to males

that carried a different lethal mutation. Because such

males are not viable, the required cross cannot be

performed.

13.5 Published spontaneous mutation rates for humans are

generally higher than those for bacteria. Does this indi-

cate that individual genes of humans mutate more fre-

quently than those of bacteria? Explain.

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ANS: Probably not. A human is larger than a bacterium, with

more cells and a longer life span. If mutation frequencies

are calculated in terms of cell generations, the rates for

human cells and bacterial cells are similar.

13.6 A precancerous condition (intestinal polyposis) in a par-

ticular human family group is caused by a single domi-

nant gene. Among the descendants of one woman who

died with cancer of the colon, 10 people have died with

the same type of cancer and 6 now have intestinal pol-

yposis. All other branches of the large kindred have been

carefully examined, and no cases have been found. Sug-

gest an explanation for the origin of the defective gene.

ANS: A dominant mutation presumably occurred in the woman

in whom the condition was rst known.

13.7 Juvenile muscular dystrophy in humans depends on an

X-linked recessive gene. In an intensive study, 33 cases

were found in a population of some 800,000 people. The

investigators were condent that they had found all cases

that were well enough advanced to be detected at the

time the study was made. The symptoms of the disease

were expressed only in males. Most of those with the dis-

ease died at an early age, and none lived beyond 21 years

of age. Usually, only one case was detected in a family,

but sometimes two or three cases occurred in the same

family. Suggest an explanation for the sporadic occur-

rence of the disease and the tendency for the gene to

persist in the population.

ANS: The X-linked gene is carried by mothers, and the disease

is expressed in half of their sons. Such a disease is dif-

cult to follow in pedigree studies because of the recessive

nature of the gene, the tendency for the expression to

skip generations in a family line, and the loss of the males

who carry the gene. One explanation for the sporadic

occurrence and tendency for the gene to persist is that,

by mutation, new defective genes are constantly being

added to the load already present in the population.

13.8 Products resulting from somatic mutations, such as the

navel orange and the Delicious apple, have become

widespread in citrus groves and apple orchards. However,

traits resulting from somatic mutations are seldom main-

tained in animals. Why?

ANS: Plants can be propagated vegetatively, but no such meth-

ods are available for widespread use in animals.

13.9 If a single short-legged sheep should occur in a ock,

suggest experiments to determine whether the short legs

are the result of a mutation or an environmental effect. If

due to a mutation, how can one determine whether the

mutation is dominant or recessive?

ANS: The sheep with short legs could be mated to unrelated

animals with long legs. If the trait is expressed in the rst

generation, it could be presumed to be inherited and to

depend on a dominant gene. On the other hand, if it does

not appear in the rst generation, F

sheep could be

crossed back to the short-legged parent. If the trait is

expressed in one-half of the backcross progeny, it is prob-

ably inherited as a simple recessive. If two short-legged

sheep of different sex could be obtained, they could be

mated repeatedly to test the hypothesis of dominance. In

the event that the trait is not transmitted to the progeny

that result from these matings, it might be considered to

be environmental or dependent on some complex genetic

mechanism that could not be identied by the simple test

used in the experiments.

13.10 How might enzymes such as DNA polymerase be in volved

in the mode of action of both mutator and antimutator

genes (mutant genes that increase and decrease, respec-

tively, mutation rates)?

ANS: Enzymes may discriminate among the different nucleo-

tides that are being incorporated. Mutator enzymes may

utilize a higher proportion of incorrect nucleotides,

whereas antimutator enzymes may select fewer incorrect

bases in DNA replication. In the case of the phage T4

DNA polymerase, the relative efciencies of polymeriza-

tion and proofreading by the polymerase’s 3′ → 5′ exo-

nuclease activity play key roles in determining the

mutation rate.

13.11 How could spontaneous mutation rates be optimized by

natural selection?

ANS: If both mutators and antimutators operate in the same

living system, an optimum mutation rate for a particular

organism in a given environment may result from natu-

ral selection.

13.12 A mutator gene Dt in maize increases the rate at which

the gene for colorless aleurone (a) mutates to the domi-

nant allele (A), which yields colored aleurone. When

reciprocal crosses were made (i.e., seed parent dt/dt, a/a 

Dt/Dt, a/a and seed parent Dt/Dt, a/a  dt/dt, a/a), the

cross with Dt/Dt seed parents produced three times as

many dots per kernel as the reciprocal cross. Explain

these results.

ANS: Dt is a mutator gene that induces somatic mutations in

developing kernels.

13.13 The deciency Df(1)w

rJ1

removes 16 contiguous bands

from a region near the left end of the Drosophila X chro-

mosome. Females homozygous for this deciency die.

However, females heterozygous for it and a ClB chromo-

some are viable and fertile. If such females are mated to

males that carry wild-type X and Y chromosomes, what

kinds of progeny will appear and in what proportions?

ANS: The cross is Df(1)w

rJ1

/ClB females  /Y (wild-type)

males. The genotypes of the daughters are Df(1)w

rJ1

/

(phenotypically wild-type) and ClB/ (bar-eyed). These

two classes of daughters will occur in equal proportions.

The sons inherit a Y chromosome and either the Df(1)

rJ1

or ClB X chromosomes, both of which act as reces-

sive lethals. Thus, no sons will appear in the progeny.

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13.14 In Drosophila, the Y chromosome Y·w



has a small piece

of the X chromosome translocated to it; this piece con-

tains the wild-type alleles of all the genes missing in

Df(1)w

rJ1

mentioned in Problem 13.13. If males carrying

Y·w



and a wild-type X chromosome are crossed to Df(1)

rJ1

/ClB females, what kinds of progeny will appear, and

in what proportions? How would your answer change if

the wild-type X chromosome in the males carried a radi-

ation-induced recessive lethal mutation located within

the region that is missing in Df(1)w

rJ1

? How could these

unusual chromosomes be used to devise a scheme that

would allow you to carry out complementation tests

between two independently induced recessive lethal

mutations that map within this region?

ANS: In the cross Df(1)w

rJ1

/ClB females  /Y·w



(wild-type)

males, the daughters will be either Df(1)w

rJ1

/ (pheno-

typically wild-type) or ClB/ (bar-eyed); these two types

of daughters will appear in equal proportions. The sons

from the cross will be either Df(1)w

rJ1

/ Y·w



(viable and

phenotypically wild-type because the small piece of the

X chromosome translocated to the Y chromosome con-

tains the genes that are missing in the deciency Df(1)

rJ1

) or ClB/ Y·w



. This latter genotype will be viable if

the lethal mutation in the ClB chromosome resides in the

region dened by the deciency Df(1)w

rJ1

. In that case,

bar-eyed males would appear and be as frequent as wild-

type males. If the lethal mutation in the ClB chromo-

some resides outside the region dened by Df(1)w

rJ1

then no bar-eyed males will appear among the progeny.

If the males in the cross carry a radiation-induced lethal

(ril) mutation in the region dened by Df(1)w

rJ1

—that is,

if the cross is Df(1)w

rJ1

/ClB females  ril/Y·w



males

(viable because the small piece of the X chromosome

carried by the Y chromosome includes the wild-type

allele of ril)—there will not be any wild-type daughters

(genotype Df(1)w

rJ1

/ril) among the progeny. However,

bar-eyed daughters (genotypically ClB/ril should appear

unless the lethal mutation in the ClB chromosome is

allelic to ril.

An induced lethal mutation that lies within the region

dened by Df(1)w

rJ1

could be tested for complementation

with another such lethal mutation. Let’s call the rst

mutation lethal-1 and the second mutation lethal-2. The

required cross for the complementation test is ClB/lethal-1

females  lethal-2/Y·w



males. This cross is possible

because the small piece of the X chromosome carried by

the Y·w



chromosome contains the wild-type allele of

lethal-2; thus, the lethal-2/Y·w



males are viable. Among

the progeny, we would look for daughters with the geno-

type lethal-1/lethal-2, which, because they lack the ClB

chromosome, will have normal (non-bar) eyes. If these

females appear, we know that lethal-1 and lethal-2 comple-

ment one another; that is, they are mutations in different

genes. If these females do not appear, we can conclude that

lethal-1 and lethal-2 are alleles of the same gene.

13.15 If CTT is a DNA triplet (transcribed strand of DNA)

specifying glutamic acid, what DNA and mRNA base

triplet alterations could account for valine and lysine in

position 6 of the b-globin chain?

ANS:

Amino Acid

Glumatic acid

Valine

GUA

Lysine

AAA

TTT

CAT

GTA

CTT

Transcribed strand

Mutation

GAA

mRNA DNA

GAA

... .. ..

13.16 The bacteriophage T4 genome contains about 50

percent A:T base pairs and 50 percent G:C base pairs. The

base analog 2-aminopurine induces A:T → G:C and G:C

→ A:T base-pair substitutions by undergoing tautomeric

shifts. Hydroxylamine is a mutagenic chemical that reacts

specically with cytosine and induces only G:C → A:T

substitutions. If a large number of independent muta-

tions were produced in bacteriophage T4 by treatment

with 2-aminopurine, what percentage of these mutations

should you expect to be induced to mutate back to the

wild-type genotype by treatment with hydroxylamine?

ANS: About half of the induced mutations would be expected

to mutate back to the wild-type genotype.

13.17 Assuming that the b-globin chain and the a-globin chain

shared a common ancestor, what mechanisms might

explain the differences that now exist in these two chains?

What changes in DNA and mRNA codons would

account for the differences that have resulted in unlike

amino acids at corresponding positions?

ANS: Mutations: transitions, transversions, and frameshifts.

13.18 In a given strain of bacteria, all of the cells are usually

killed when a specic concentration of streptomycin is

present in the medium. Mutations that confer resistance

to streptomycin occur. The streptomycin-resistant

mutants are of two types: some can live with or without

streptomycin; others cannot survive unless this drug is

present in the medium. Given a streptomycin-sensitive

strain of this species, outline an experimental procedure

by which streptomycin-resistant strains of the two types

could be established.

ANS: Irradiate the nonresistant strain and plate the irradiated

organisms on a medium containing streptomycin. Those

that survive and produce colonies are resistant. They

could then be replicated to a medium without strepto-

mycin. Those that survive would be of the rst type;

those that can live with streptomycin but not without it

would be the second type.

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13.19 One stock of fruit ies was treated with 1000 roentgens

(r) of X-rays. The X-ray treatment increased the muta-

tion rate of a particular gene by 2 percent. What percent-

age increases in the mutation rate of this gene would be

expected if this stock of ies was treated with X-ray doses

of 1500 r, 2000 r, and 3000 r?

ANS: 3%; 4%; 6%.

13.20 Why does the frequency of chromosome breaks induced

by X-rays vary with the total dosage and not with the

rate at which it is delivered?

ANS: Each quantum of energy from the X-rays that is absorbed

in a cell has a certain probability of hitting and breaking

a chromosome. Hence, the greater the number of quanta

of energy or dosage, the more likely the breaks are to

occur. The rate at which this dosage is delivered does not

change the probability of each quantum inducing a

break.

13.21 A reactor overheats and produces radioactive tritium

), radioactive iodine (I

131

), and radioactive xenon

(Xn

133

). Why should we be more concerned about radio-

active iodine than the other two radioactive isotopes?

ANS: Radioactive iodine is concentrated by living organisms

and food chains.

13.22 One person was in an accident and received 50 roentgens

(r) of X-rays at one time. Another person received 5 r in

each of 20 treatments. Assuming no intensity effect, what

proportionate number of mutations would be expected

in each person?

ANS: The person receiving a total of 100 r would be expected

to have twice as many mutations as the one receiving

50 r.

13.23 A cross was performed in Neurospora crassa between a

strain of mating type A and genotype x



z and a strain

of mating type a and genotype x m z



. Genes x, m, and z

are closely linked and are present in the order x–m–z on

the chromosome. An ascus produced from this cross

contained two copies (“identical twins”) of each of the

four products of meiosis. If the genotypes of the four

products of meiosis showed that gene conversion had

occurred at the m locus and that reciprocal recombina-

tion had occurred at the x and z loci, what might the

genotypes of the four products look like? In the paren-

theses below, write the genotypes of the four haploid

products of meiosis in an ascus showing gene conversion

at the m locus and reciprocal recombination of the ank-

ing markers (at the x and z loci).

Ascus Spore Pairs

1–2 3–4 5–6 7–8

( ) ( ) ( ) ( )

ANS: ( x



z) (x



) (x m



z) (x m z



) or equivalent.

13.24 How does nitrous acid induce mutations? What specic

end results might be expected in DNA and mRNA from

the treatment of viruses with nitrous acid?

ANS: Nitrous acid brings about a substitution of an OH group

for an NH

group in those bases (A, C, and G) having

side groups. In so doing, adenine is converted into

hypoxanthine, which base-pairs with cytosine, and cyto-

sine is converted into uracil, which base-pairs with ade-

nine. The net effects are GC ↔ AT base-pair substitutions

(see Figure 13.16).

13.25 Are mutational changes induced by nitrous acid more

likely to be transitions or transversions?

ANS: Transitions.

13.26 You are screening three new pesticides for potential

mutagenicity by using the Ames test. Two his



strains

resulting from either a frameshift or a transition muta-

tion were used and produced the following results (num-

ber of revertant colonies):

Strain 1

Transition

Mutant

Control

(no chemical)

Transion

Mutant



Chemical

Transion Mutant

 Chemical

 Rat liver

Enzymes

Pesticide #1 21 180 19

Pesticide #2 18 19 17

Pesticide #3 25 265 270

Strain 2

Frameshift

Mutant

Control

(no chemical)

Frameshift

Mutant 

Chemical

Frameshift

Mutant 

Chemical  Rat

liver Enzymes

Pesticide #1 5 4 5

Pesticide #2 7 5 93

Pesticide #3 6 9 7

What type of mutations, if any, do the three pesticides

induce?

ANS: P #1—Causes transition mutation. Liver enzymes convert

it into nonmutagen. Does not cause frameshift mutations.

P #2—Does not cause transition mutations. Liver enzymes

convert it into a frameshift mutagen. P #3—Causes

transition mutations. Liver enzymes have no effect on

mutagenicity. Does not cause frameshift mutations.

13.27 How does the action and mutagenic effect of 5-bromo-

uracil differ from that of nitrous acid?

ANS: Nitrous acid acts as a mutagen on either replicating or

nonreplicating DNA and produces transitions from A to

G or C to T, whereas 5-bromouracil does not affect non-

replicating DNA but acts during the replication process

causing GC ↔ AT transitions. 5-Bromouracil must be

incorporated into DNA during the replication process in

order to induce mispairing of bases and thus mutations.

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13.28 Sydney Brenner and A. O. W. Stretton found that non-

sense mutations did not terminate polypeptide synthesis

in the rII gene of the bacteriophage T4 when these

mutations were located within a DNA sequence interval

in which a single nucleotide insertion had been made on

one end and a single nucleotide deletion had been made

on the other. How can this nding be explained?

ANS: The reading frame will be shifted between the two

frameshift mutations. This shift in reading frame does

not read the normal nonsense codon and termination

does not occur.

13.29 Seymour Benzer and Ernst Freese compared spontane-

ous and 5-bromouracil-induced mutants in the rII gene

of the bacteriophage T4; the mutagen increased the

mutation rate (rII



→ rII) several hundred times above

the spontaneous mutation rate. Almost all (98 percent) of

the 5-bromouracil-induced mutants could be induced to

revert to wild-type (rII → rII



) by 5-bromouracil treat-

ment, but only 14 percent of the spontaneous mutants

could be induced to revert to wild-type by this treatment.

Discuss the reason for this result.

ANS: 5-BU causes GC ↔ AT transitions. 5-BU can, therefore,

revert almost all of the mutations that it induces by

enhancing the transition event that is the reverse of the

one that produced the mutation. In contrast, the sponta-

neous mutations will include transversions, frameshifts,

deletions, and other types of mutations, including

transitions. Only the spontaneous transitions will show

enhanced reversion after treatment with 5-BU.

13.30 How do acridine-induced changes in DNA result in

inactive proteins?

ANS: Mutations induced by acridine dyes are primarily inser-

tions or deletions of single base-pairs. Such mutations

alter the reading frame (the in-phase triplets specifying

mRNA codons) for that portion of the gene distal (rela-

tive to the direction of transcription and translation) to

the mutation (see Figure 13.7b). This would be expected

to totally change the amino acid sequences of polypep-

tides distal to the mutation site and produce inactive

polypeptides. In addition, such frameshift mutations fre-

quently produce in-frame termination codons that result

in truncated proteins.

Use the known codon-amino acid assignments (Table

12.1) to work the following problems.

13.31 Mutations in the genes encoding the - and -subunits

of hemoglobin lead to blood diseases such as thalassemia

and sickle-cell anemia. You have found a family in China

in which some members suffer from a new genetic form

of anemia. The DNA sequences at the 5′ end of the non-

template strand of the normal and mutant DNA encod-

ing the  subunit of hemoglobin are as follows:

Normal 5-ACGTTATGCCGTACTGCCAGCTAAC

TGCTAAAGAACAATTA……-3

Mutant 5-ACGTTATGCCCGTACTGCCAGCTAA

CTGCTAAAGAACAATTA….-3

(a) What type of mutation is present in the mutant hemo-

globin gene? (b) What are the codons in the translated

portion of the mRNA transcribed from the normal and

mutant genes? (c) What are the amino acid sequences of

the normal and mutant polypeptides?

ANS: (a) Frameshift due to the insertion of C at the 9th, 10th, or

11th nucleotide from the 5′ end. (b) Normal: 5′-AUGCC-

GUACUGCCAGCUAACUGCUAAAGAACAAUUA-3′.

Mutant: 5′-AUGCCCGUACUGCCAGCUAACUGC-

UAAAGAACAAUUA-3′. (c) Normal: NH

-Met-Pro-Tyr-

Cys-Gln-Leu-Thr-Ala-Lys-Glu-Gln-Leu. Mutant: NH

Met-Pro-Val-Leu-Pro-Ala-Asn-Cys.

13.32 Bacteriophage MS2 carries its genetic information in

RNA. Its chromosome is analogous to a polygenic mol-

ecule of mRNA in organisms that store their genetic

information in DNA. The MS2 minichromosome

encodes four polypeptides (i.e., it has four genes). One of

these four genes encodes the MS2 coat protein, a poly-

peptide of 129 amino acids long. The entire nucleotide

sequence in the RNA of MS2 is known. Codon 112 of

the coat protein gene is CUA, which species the amino

acid leucine. If you were to treat a replicating population

of bacteriophage MS2 with the mutagen 5-bromouracil,

what amino acid substitutions would you expect to be

induced at position 112 of the MS2 coat protein (i.e.,

Leu → other amino acid)? (Note: Bacteriophage MS2

RNA replicates using a complementary strand of RNA

and base-pairing as DNA.)

ANS: Proline and serine.

13.33 Would the different amino acid substitutions induced by

5-bromouracil at position 112 of the coat polypeptide

that you indicated in Problem 13.32 be expected to occur

with equal frequency? If so, why? If not, why not? Which

one(s), if any, would occur more frequently?

ANS: No. Leucine → proline would occur more frequently.

Leu (CUA) — 5-BU → Pro (CCA) occurs by a single

base-pair transition, whereas Leu (CUA) — 5-BU → Ser

(UCA) requires two base-pair transitions. Recall that

5-bromouracil (5-BU) induces only transitions (see

Figure 13.15).

13.34 Would such mutations occur if a nonreplicating suspen-

sion of MS2 phage was treated with 5-bromouracil?

ANS: No. 5-Bromouracil is mutagenic only to replicating

nucleic acids.

13.35 Recall that nitrous acid deaminates adenine, cytosine, and

guanine (adenine → hypoxanthine, which base-pairs with

cytosine; cytosine → uracil, which base-pairs with adenine;

and guanine → xanthine, which base-pairs with cytosine).

Would you expect nitrous acid to induce any mutations

that result in the substitution of another amino acid for a

glycine residue in a wild-type polypeptide (i.e., glycine →

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another amino acid) if the mutagenesis were carried out

on a suspension of mature (nonreplicating) T4 bacterio-

phage? (Note: After the mutagenic treatment of the phage

suspension, the nitrous acid is removed. The treated phage

is then allowed to infect E. coli cells to express any induced

mutations.) If so, by what mechanism? If not, why not?

ANS: Yes:

DNA:

mRNA:

Polypeptide:

GGX

HNO

AGX

Gly Ser or Arg

(depending on X)

CCX'

GGX

UCX'

DNA:

mRNA:

Polypeptide:

GGX

HNO

GAX

Gly Asp or Glu

(depending on X)

CCX'

GGX

CUX'

DNA:

mRNA:

Polypeptide:

GGX

HNO

AAX

Gly Asn or Lys

(depending on X)

CCX'

GGX

UUX'

Note: The X at the third position in each codon in mRNA

and in each triplet of base pairs in DNA refers to the fact

that there is complete degeneracy at the third base in the

glycine codon. Any base may be present in the codon,

and it will still specify glycine.

13.36 Keeping in mind the known nature of the genetic code,

the information given about phage MS2 in Problem

13.32, and the information you have learned about

nitrous acid in Problem 13.35, would you expect

nitrous acid to induce any mutations that would result

in amino acid substitutions of the type glycine →

another amino acid if the mutagenesis were carried out

on a suspension of mature (nonreplicating) MS2 bacte-

riophage? If so, by what mechanism? If not, why not?

ANS: No. The glycine codon is GGX, where X can be any one

of the four bases. Because of this complete degeneracy

at the third position of the glycine codon, changing X

to any other base will have no effect (i.e., the codon will

still specify glycine). Nitrous acid deaminates guanine

(G) to xanthine, but xanthine still base-pairs with cyto-

sine. Thus, guanine is not a target for mutagenesis by

nitrous acid.

13.37 Would you expect nitrous acid to induce a higher fre-

quency of Tyr → Ser or Tyr → Cys substitutions? Why?

ANS: Tyr → Cys substitutions; Tyr to Cys requires a transition,

which is induced by nitrous acid. Tyr to Ser would

require a transversion, and nitrous acid is not expected to

induce transversions.

13.38 Which of the following amino acid substitutions should

you expect to be induced by 5-bromouracil with the

highest frequency? (a) Met → Leu; (b) Met → Thr;

→ Gln? Why?

ANS: (b) Met → Thr. 5-Bromouracil induces transitions, not

transversions. All other changes listed require transversions.

13.39 The wild-type sequence of part of a protein is

-Trp-Trp-Trp-Met-Arg-Glu-Trp-Thr-Met

Each mutant in the following table differs from wild-

type by a single point mutation. Using this information,

determine the mRNA sequence coding for the wild-type

polypeptide. If there is more than one possible nucleo-

tide, list all possibilities.

Mutant Amino Acid Sequence of Polypeptide

1 Trp-Trp-Trp Met

2 Trp-Trp-Trp-Met-Arg-Asp-Trp-Thr-Met

3 Trp-Trp-Trp-Met-Arg-Lys-Trp-Thr-Met

4 Trp-Trp-Trp-Met-Arg-Glu-Trp-Met-Met

ANS: 5′-UGG-UGG-UGG-AUG-CGA(or AGA)-GAA(or

GAG)-UGG-AUG-3′

13.40 Acridine dyes such as proavin are known to induce

primarily single base-pair additions and deletions. Sup-

pose that the wild-type nucleotide sequence in the

mRNA produced from a gene is

5′-AUGCCCUUUGGGAAAGGGUUUCCCUAA-3′

Also, assume that a mutation is induced within this gene by

proavin, and, subsequently, a revertant of this mutation is

similarly induced with proavin and shown to result from

a second-site suppressor mutation within the same gene. If

the amino acid sequence of the polypeptide encoded by

this gene in the revertant (double mutant) strain is

-Met-Pro-Phe-Gly-Glu-Arg-Phe-Pro-COOH

what would be the most likely nucleotide sequence in the

mRNA of this gene in the revertant (double mutant)?

ANS: 5′-AUGCCCUUUGGGGAAAGGUUUCCCUAA-3′

13.41 Eight independently isolated mutants of E. coli, all of

which are unable to grow in the absence of histidine (his

−

were examined in all possible cis and trans heterozygotes

(partial diploids). All of the cis heterozygotes were able to

grow in the absence of histidine. The trans heterozygotes

yielded two different responses: some of them grew in the

absence of histidine; others did not. The experimental

results, using “” to indicate growth and “0” to indicate

no growth, are given in the accompanying table. How

many genes are dened by these eight mutations? Which

mutant strains carry mutations in the same gene(s)?

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Growth of Trans Heterozygotes (without Histidine)

Mutant 1 2 3 4 5 6 7 8

8 0 0 0 0 0 0 1 0

     

6 0 0 0 0 0 0

5 0 0 0 0 0

4 0 0 0 0

3 0 0 0

2 0 0

1 0

ANS: Two genes; mutations 1, 2, 3, 4, 5, 6, and 8 are in one

gene; mutation 7 is in a second gene.

13.42 Assume that the mutants described in Problem 13.41

yielded the following results. How many genes would

they have dened? Which mutations would have been in

the same gene(s)?

Mutant 1 2 3 4 5 6 7 8

     

0 0

     

   

0 0

   

 

0 0

 

2 0 0

1 0

ANS: Four genes; mutations 1 and 2 in one gene; mutations 3

and 4 in a second gene; mutations 5 and 6 in a third gene;

and mutations 7 and 8 in a fourth gene.

13.43 In Drosophila, white, white cherry, and vermilion are all sex-

linked mutations affecting eye color. All three mutations

are recessive to their wild-type allele(s) for red eyes. A

white-eyed female crossed with a vermilion-eyed male

produces white-eyed male offspring and red-eyed (wild-

type) female offspring. A white-eyed female crossed with

a white cherry-eyed male produces white-eyed sons and

light cherry-eyed daughters. Do these results indicate

whether or not any of the three mutations affecting eye

color are located in the same gene? If so, which

mutations?

ANS: The complementation test for allelism involves placing

mutations pairwise in a common protoplasm in the trans

conguration and determining whether the resulting

trans heterozygotes have wild-type or mutant pheno-

types. If the two mutations are in different genes, the

two mutations will complement each other, because the

wild-type copies of each gene will produce functional

gene products (see Figure 13.21a). However, if the two

mutations are in the same gene, both copies of the gene in

the trans heterozygote will produce defective gene prod-

ucts, resulting in a mutant phenotype (see Figure 13.21b).

When complementation occurs, the trans heterozy-

gote will have the wild-type phenotype. Thus, the

complementation test allows one to determine whether

any two recessive mutations are located in the same gene

or in different genes. Because the mutations of interest

are sex-linked, all the male progeny will have the same

phenotype as the female parent. They are hemizygous,

with one X chromosome obtained from their mother. In

contrast, the female progeny are trans heterozygotes. In

the cross between the white-eyed female and the vermil-

ion-eyed male, the female progeny have red eyes, the

wild-type phenotype. Thus, the white and vermilion

mutations are in different genes, as illustrated in the

following diagram:

X chr

omosome

from parent

trans heterozygote

X chr

omosome

from parent

Complementation yields wild-type phenotype; both v

and w

gene products are produced in the trans heterozygote.

gene product

gene pr

oduct

In the cross between a white-eyed female and a white

cherry-eyed male, the female progeny have light cherry-

colored eyes (a mutant phenotype), not wild-type red

eyes as in the rst cross. Since the trans heterozygote has

a mutant phenotype, the two mutations, white and white

cherry, are in the same gene:

trans heterozygote

No w

gene product; therefore, mutant phenotype.

No active (w

)

gene product

X chr

omosome

from parent

X chr

omosome

from parent

13.44 The loz (lethal on Z) mutants of bacteriophage X are con-

ditional lethal mutants that can grow on E. coli strain

Y but cannot grow on E. coli strain Z. The results shown

in the following table were obtained when seven loz

mutants were analyzed for complementation by infect-

ing E. coli strain Z with each possible pair of mutants.

A “” indicates that progeny phage were produced in

the infected cells, and a “0” indicates that no progeny

phage were produced. All possible cis tests were also

done, and all cis heterozygotes produced wild-type yields

of progeny phage.

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Mutant 1 2 3 4 5 6 7

 



0 0 0

    

 



4 0 0



 

2 0 0

1 0

Propose three plausible explanations for the apparently

anomalous complementation behavior of loz mutant

number 7. (b) What simple genetic experiments can be

used to distinguish between the three possible explana-

tions? (c) Explain why specic outcomes of the proposed

experiments will distinguish between the three possible

explanations.

ANS: Mutations 1, 2, and 4 do not complement one another

and thus appear to be located in the same gene, as

do mutations 3 and 5. The anomaly is that mutation 7

does not complement mutations 3, 5, and 6, even

though mutation 6 does complement mutations 3 and

5. (a) There are three simple explanations of the seem-

ingly anomalous complementation behavior of muta-

tion 7. (1) It is a deletion spanning all or parts of two

genes. (2) It is a double mutation with defects in two

genes. (3) It is a polar mutation—a nonsense mutation

that interferes with the expression of downstream

genes—in the promoter-proximal gene of a multigenic

transcription unit. (b) Three simple genetic operations

will distinguish between these three possibilities.

(1) Reversion. Plate a large number of mutant 7 phage

on E. coli strain Z and look for wild-type revertants.

(2) Backcross mutant 7 to wild-type phage and test the

mutant progeny for the ability to complement muta-

tions 3 and 6. (3) Introduce F’s carrying tRNA non-

sense suppressor genes into E. coli strain Z and

determine whether any of them suppress the loz7

mutation. (c) If mutation 7 is a deletion, it will not

revert, and, if it is a double mutation, the reversion

rate will probably be below the level of detection in

your experiment. On the other hand, if it is a polar

nonsense mutation, loz



revertants will be obtained. If

mutation 7 is a deletion, no new genotypes will be pro-

duced in the backcross to wild-type. However, if it is a

double mutation, some recombinant single-mutant

progeny will be produced in the backcross to wild-

type, and these single mutations will complement

either mutation 3 or mutation 6. If mutation 7 is a

polar nonsense mutation, it should be suppressed by

one or more of the tRNA suppressor genes introduced

into E. coli strain Z.

Chapter 14

14.1 (a) In what ways is the introduction of recombinant DNA

molecules into host cells similar to mutation? (b) In what

ways is it different?

ANS: (a) Both introduce new genetic variability into the cell. In

both cases, only one gene or a small segment of DNA

representing a small fraction of the total genome is

changed or added to the genome. The vast majority of

the genes of the organism remain the same. (b) The

introduction of recombinant DNA molecules, if they

come from a very different species, is more likely to

result in a novel, functional gene product in the cell, if

the introduced gene (or genes) is capable of being

expressed in the foreign protoplasm. The introduction of

recombinant DNA molecules is more analogous to

duplication mutations (see Chapter 6) than to other

types of mutations.

14.2 Listed in this question are four different single strands of

DNA. Which of these, in their double-stranded form,

would you expect to be cleaved by a restriction

endonuclease?

(a) ACTCCAGAATTCACTCCG

(b) GCCTCATTCGAAGCCTGA

(d) ACTCCACTCCCGACTCCA

ANS: Restriction endonucleases cleave at palindromes in dou-

ble-stranded DNA. A palindrome (indicated in bold-

faced letters) can be found in all the double-stranded

sequences except d. Therefore, sequence d would not be

cleaved by a restriction endonuclease, whereas a, b,

and c could be cleaved by the appropriate enzyme.

(a)

ACTCCAGA ATTC ACTCCG

TGCGGTCTTA AGTGAGGC

(b)

GCCTCATTCGA AGCCTGA

CGGAGTAAGCTTCGGACT

(c)

CTCGCCA ATTGACTCGTC

GAGCGGTTAACTGAGCAG

(d)

ACTCCACTCCCGACTCCA

TGAGGAGAGGGCTGAGGT

14.3 If the sequence of base pairs along a DNA molecule

occurs strictly at random, what is the expected frequency

of a specic restriction enzyme recognition sequence of

length (a) four and (b) six base pairs?

ANS: (a) (1/4)

 1/256; (b) (1/4)

 1/4096

14.4 In what ways do restriction endonucleases differ from

other endonucleases?

ANS: Restriction endonucleases recognize and cut specic

nucleotide sequences in DNA. Most other endonucleases

are not sequence-specic; many cut DNA sequences at

random.

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14.5 Of what value are recombinant DNA and gene-cloning

technologies to geneticists?

ANS: Recombinant DNA and gene-cloning techniques allow

geneticists to isolate essentially any gene or DNA

sequence of interest and to characterize it structurally

and functionally. Large quantities of a given gene can be

obtained in pure form, which permits one to determine

its nucleotide-pair sequence (to “sequence it” in com-

mon lab jargon). From the nucleotide sequence and our

knowledge of the genetic code, geneticists can predict

the amino acid sequence of any polypeptide encoded by

the gene. By using an appropriate subclone of the gene as

a hybridization probe in northern blot analyses, geneti-

cists can identify the tissues in which the gene is

expressed. Based on the predicted amino acid sequence

of a polypeptide encoded by a gene, geneticists can syn-

thesize oligopeptides and use these to raise antibodies

that, in turn, can be used to identify the actual product of

the gene and localize it within cells or tissues of the

organism. Thus, recombinant DNA and gene-cloning

technologies provide very powerful tools with which to

study the genetic control of essentially all biological pro-

cesses. These tools have played major roles in the explo-

sive progress in the eld of biology during the last three

decades.

14.6 What determines the sites at which DNA molecules will

be cleaved by a restriction endonuclease?

ANS: The nucleotide-pair sequence. Restriction endonucle-

ases recognize a specic nucleotide-pair sequence in

DNA regardless of the source of the DNA. In most cases,

this is a 4 or 6 nucleotide-pair sequence; in a few cases,

the recognition sequence is longer (e.g., 8 nucleotide

pairs). Most restriction enzymes cleave the two strands

of the DNA at a specic position (between the same two

adjacent nucleotides in each strand) within the recogni-

tion sequence. A few restriction enzymes bind at a spe-

cic recognition sequence but cut the DNA at a nearby

site outside of the recognition sequences. Some restric-

tion endonucleases cut both strands between the same

two nucleotide pairs (“blunt end” cutters), whereas oth-

ers cut the two strands at different positions and yield

complementary single-stranded ends (“sticky or stag-

gered end” cutters). See Table 14.1 for examples.

14.7 Restriction endonucleases are invaluable tools for biolo-

gists. However, genes encoding restriction enzymes

obviously did not evolve to provide tools for scientists.

Of what possible value are restriction endonucleases to

the microorganisms that produce them?

ANS: Restriction endonucleases are believed to provide a kind

of primitive immune system to the microorganisms

that produce them—protecting their genetic material

from “invasion” by foreign DNAs from viruses or other

pathogens or just DNA in the environment that might

be taken up by the microorganism. Obviously, these

microorganisms do not have a sophisticated immune sys-

tem like that of higher animals (see Chapter 22).

14.8 Why is the DNA of a microorganism not degraded by a

restriction endonuclease that it produces, even though

its DNA contains recognition sequences normally

cleaved by the endonuclease?

ANS: Microorganisms that produce restriction endonucleases

also produce enzymes that modify one or more bases in

the recognition sequence for that endonuclease so that it

can no longer cleave the DNA at that site. In most cases,

the modifying enzyme is a methylase that attaches a

methyl group to one or more of the bases in the recogni-

tion sequence. For example, E. coli strains that produce

the restriction endonuclease EcoRI also produce EcoRI

methylase, an enzyme that transfers a methyl group from

S-adenosylmethionine to the second (most 3′) adenine

residue in each strand of the recognition sequence

(5-GAATCC-3) producing N

-methyladenines at these

positions. EcoRI cannot cleave DNA that contains

-methyladenine at these positions even if the EcoRI

recognition sequence is present in this DNA (see Figure

14.1). Thus, if one wishes to digest DNA with EcoRI, that

DNA must not be isolated from an E. coli strain that is

producing EcoRI methylase.

14.9 One of the procedures for cloning foreign DNA segments

takes advantage of restriction endonucleases such as Hin-

dIII (see Table 14.1) that produce complementary single-

stranded ends. These enzymes produce identical

complementary ends on cleaved foreign DNAs and on the

vector DNAs into which the foreign DNAs are inserted.

Assume that you have inserted your favorite gene into the

HindIII site in the polycloning region of the Bluescript

cloning vector with DNA ligase, have amplied the plas-

mid containing your gene in E. coli, and have isolated a

large quantity of gene/Bluescript DNA. How could you

excise your favorite gene from the Bluescript vector?

ANS: A foreign DNA cloned using an enzyme that produces

single-stranded complementary ends can always be

excised from the cloning vector by cleavage with the

same restriction enzyme that was originally used to clone

it. For example, the HindIII fragment carrying your

favorite can be excised from the Bluescript DNA by

cleavage with restriction endonuclease HindIII. The

human HindIII fragment will be anked in the recombi-

nant plasmid DNA clone by HindIII cleavage sites.

14.10 You are working as part of a research team studying the

structure and function of a particular gene. Your job is to

clone the gene. A restriction map is available for the

region of the chromosome in which the gene is located;

the map is as follows:

Xba

I PstI

HindIII

EcoRIEcoRI Hind

III

SalI

GENE

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Your rst task is to prepare a genomic DNA library that

contains clones carrying the entire gene. Describe how

you would prepare such a library in plasmid vector

pBluescript II (see Figure 14.3), indicating which restric-

tion enzymes, media, and host cells you would use.

ANS: Step 1: Digest genomic DNA isolated from your research

organism with EcoRI. Step 2: Treat pBluescript II DNA

with EcoRI. Step 3: Mix the EcoRI-digested genomic and

vector DNAs under annealing conditions and incubate

with DNA ligase. Step 4: Transform amp

E. coli cells car-

rying the lacZM15 gene with the resulting ligation prod-

ucts. Step 5: Plate the transformed cells on nutrient agar

medium containing Xgal and ampicillin. Only trans-

formed cells will produce colonies in the presence of

ampicillin. Step 6: Prepare your genomic DNA library

by using bacteria from white colonies; these bacteria will

contain pBluescript II DNA with genomic DNA inserts.

Bacteria harboring Bluescript II plasmids with no insert

will produce blue colonies (see Figure 14.4).

14.11 Compare the nucleotide-pair sequences of genomic

DNA clones and cDNA clones of specic genes of higher

plants and animals. What is the most frequent difference

that you would observe?

ANS: Most genes of higher plants and animals contain non-

coding intron sequences. These intron sequences will be

present in genomic clones, but not in cDNA clones,

because cDNAs are synthesized using mRNA templates

and intron sequences are removed during the processing

of the primary transcripts to produce mature mRNAs.

14.12 Most of the genes of plants and animals that were cloned

soon after the development of recombinant DNA tech-

nologies were genes encoding products that are synthe-

sized in large quantities in specialized cells. For example,

about 90 percent of the protein synthesized in mature

red blood cells of mammals consists of - and b-globin

chains, and the globin genes were among the rst mam-

malian genes cloned. Why were genes of this type so

prevalent among the rst eukaryotic genes that were

cloned?

ANS: Higher eukaryotes have very large genomes; for example,

the genomes of mammals contain approximately 3  10

nucleotide pairs. Thus, trying to identify a particular

single-copy gene from a clone library is like looking for

the proverbial “needle-in-a-haystack.” To accomplish

this, one needs a nucleic acid hybridization probe specic

for the gene or an antibody probe specic for the gene

product. Given a specic cell or tissue type producing the

mRNA and/or the protein gene product in large amounts,

it was relatively easy to obtain pure mRNA or pure pro-

tein to use in making a hybridization or antibody probe,

respectively, with which to screen a library for the gene or

cDNA of interest. These approaches are much more dif-

cult for the majority of the genes that encode products

that represent only a small proportion of the total gene

products in any given cell type.

14.13 Genomic clones of the chloroplastic glutamine synthe-

tase gene (gln2) of maize are cleaved into two fragments

by digestion with restriction endonuclease HindIII,

whereas full-length maize gln2 cDNA clones are not cut

by HindIII. Explain these results.

ANS: The maize gln2 gene contains many introns, and one of

the introns contains a HindIII cleavage site. The intron

sequences (and thus the HindIII cleavage site) are not

present in mRNA sequences and thus are also not pres-

ent in full-length gln2 cDNA clones.

14.14 In the following illustration, the upper line shows a gene

composed of segments A–D. The lower circle shows a

mutant version of this gene, consisting of two fused

pieces (A-B, C-D), carried on a plasmid. You attempt a

targeted mutagenesis of a diploid cell by transforming

cells with the cloned mutant gene. The following dia-

gram shows the desired pairing of the plasmid and chro-

mosome just prior to recombination.

ABC

A' B' C'

Probe

Plasmid

You prepare DNA from the cells, digest it with an enzyme

that cuts at x, and hybridize the cleaved DNA with the

probe shown above. The following diagram shows a

Southern blot of possible results.

1 2 3 4 5

____ ____ ____ ____

____ ____

____

(a) Which lane shows fragments produced from DNA in

the cell before transformation? (b) Which lane shows

fragments produced from DNA in the cell in which the

anticipated targeted mutagenesis occurred?

crossovers occurred, one between A and B, and the other

between C and D?

ANS: (a) 1 (b) 2 (c) 2

14.15 (a) What experimental procedure is carried out in South-

ern, northern, and western blot analyses? (b) What is the

major difference between Southern, northern, and west-

ern blot analyses?

ANS: (a) Southern, northern, and western blot procedures

all share one common step, namely, the transfer of

macromolecules (DNAs, RNAs, and proteins, respec-

tively) that have been separated by gel electrophoresis

to a solid support—usually a nitrocellulose or nylon

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membrane—for further analysis. (b) The major difference

between these techniques is the class of macromolecules

that are separated during the electrophoresis step: DNA

for Southern blots, RNA for northern blots, and protein

for western blots.

14.16 What major advantage does the polymerase chain reac-

tion (PCR) have over other methods for analyzing

nucleic acid structure and function?

ANS: The PCR technique has much greater sensitivity than

any other method available for analyzing nucleic acids.

Thus, PCR procedures permit analysis of nucleic acid

structure given extremely minute amounts of starting

material. DNA sequences can be amplied and structur-

ally analyzed from very small amounts of tissue like

blood or sperm in assault and rape cases. In addition,

PCR methods permit investigators to detect the pres-

ence of rare gene transcripts (e.g., in specic types of

cells) that could not be detected by less sensitive proce-

dures such as northern blot analyses or in situ hybridiza-

tion studies.

14.17 The cloning vectors in use today contain an origin of

replication, a selectable marker gene (usually an antibi-

otic-resistance gene), and one additional component.

What is this component, and what is its function?

ANS: All modern cloning vectors contain a “polycloning site”

or “multiple cloning site” (MCS)—a cluster of unique

cleavage sites for a number of different restriction endo-

nucleases in a nonessential region of the vector into

which the foreign DNA can be inserted. In general, the

greater the complexity of the MCS—that is, the more

restriction endonuclease cleavage sites that are present—

the greater the utility of the vector for cloning a wide

variety of different restriction fragments. For example,

see the MCS present in plasmid Bluescript II shown in

Figure 14.3.

14.18 The drawing in this problem shows a restriction map of

a segment of a DNA molecule. Eco refers to locations

where the restriction endonuclease EcoRI cuts the DNA,

and Pst refers to locations where the restriction enzyme

PstI cuts the DNA. Potential restriction sites are num-

bered 1–6. Distances between restriction sites are shown

on the bottom scale in base pairs (bp). The thick line

represents the part of the molecule that has homology

with a probe.

Eco

5000 bp 3000 bp 4000 bp 5000 bp2000 bp

Pst

Eco

Pst

Eco

Pst

(a) Assume that individual 1 has restriction sites 1 through

6. If this individual’s DNA is digested with PstI, what are

the expected sizes of the DNA fragments that will

hybridize with the probe? (b) Assume that individual 2

has a mutation that eliminates site 4. If this individual’s

DNA is digested with PstI, what are the expected sizes of

the DNA fragments that will hybridize with the probe?

nates site 5. If the DNA of this individual is digested with

PstI, what are the expected sizes of the DNA fragments

that will hybridize with the probe? (d) If the DNA of

individual 1 is digested with both PstI and EcoRI, what

are the expected sizes of the DNA fragments that will

hybridize with the probe? (e) If the DNA of individual 3

is digested with both PstI and EcoRI, what are the

expected sizes of the DNA fragments that will hybridize

with the probe?

ANS: (a) 7000 bp  7000 bp; (b) 14,000 bp; (c) 7000 bp  7000

bp; (d) 4000 bp  2000 bp  5000 bp; (e) 4000 bp 

7000 bp.

14.19 The cystic brosis (CF) gene (location: chromosome 7,

region q31) has been cloned and sequenced, and studies

of CF patients have shown that about 70 percent of them

are homozygous for a mutant CF allele that has a specic

three-nucleotide-pair deletion (equivalent to one codon).

This deletion results in the loss of a phenylalanine residue

at position 508 in the predicted CF gene product. Assume

that you are a genetic counselor responsible for advising

families with CF in their pedigrees regarding the risk of

CF among their offspring. How might you screen puta-

tive CF patients and their parents and relatives for the

presence of the CFDF508 mutant gene? What would the

detection of this mutant gene in a family allow you to say

about the chances that CF will occur again in the family?

ANS: Because the nucleotide-pair sequences of both the normal

CF gene and the CF∆508 mutant gene are known, labeled

oligonucleotides can be synthesized and used as hybridiza-

tion probes to detect the presence of each allele (normal

and ∆508). Under high-stringency hybridization condi-

tions, each probe will hybridize only with the CF allele that

exhibits perfect complementarity to itself. Since the

sequences of the CF gene anking the ∆508 site are known,

oligonucleotide PCR primers can be synthesized and used

to amplify this segment of the DNA obtained from small

tissue explants of putative CF patients and their relatives

by PCR. The amplied DNAs can then be separated

by agarose gel electrophoresis, transferred to nylon

membranes, and hybridized to the respective labeled

oligonucleotide probes, and the presence of each CF allele

can be detected by autoradiography. For a demonstration

of the utility of this procedure, see Focus on Detection of

a Mutant Gene Causing Cystic Fibrosis. In the procedure

described there, two synthetic oligonucleotide probes—

oligo-N  3′-CTTTTATAGTAGAAACCAC-5′ and

oligo-DF  3′-TTCTTTTATAGTA—ACCACAA-5′

(the dash indicates the deleted nucleotides in the CFD508

mutant allele) were used to analyze the DNA of CF

patients and their parents. For conrmed CF families, the

results of these Southern blot hybridizations with the

oligo-N (normal) and oligo-DF (CFD508) labeled probes

were often as follows:

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Oligo-

∆F probe:

Oligo-N probe:

Both parents were heterozygous for the normal CF allele

and the mutant CFD508 allele as would be expected for a

rare recessive trait, and the CF patient was homozygous for

the CFD508 allele. In such families, one-fourth of the chil-

dren would be expected to be homozygous for the D508

mutant allele and exhibit the symptoms of CF, whereas

three-fourths would be normal (not have CF). However,

two-thirds of these normal children would be expected to

be heterozygous and transmit the allele to their children.

Only one-fourth of the children of this family would be

homozygous for the normal CF allele and have no chance

of transmitting the mutant CF gene to their offspring.

Note that the screening procedure described here can be

used to determine which of the normal children are carri-

ers of the CFD508 allele: that is, the mutant gene can be

detected in heterozygotes as well as in homozygotes.

14.20 Cereal grains are major food sources for humans and

other animals in many regions of the world. However,

most cereal grains contain inadequate supplies of certain

of the amino acids that are essential for monogastric ani-

mals such as humans. For example, corn contains insuf-

cient amounts of lysine, tryptophan, and threonine.

Thus, a major goal of plant geneticists is to produce corn

varieties with increased kernel lysine content. As a pre-

requisite to the engineering of high-lysine corn, molecu-

lar biologists need more basic information about the

regulation of the biosynthesis and the activity of the

enzymes involved in the synthesis of lysine. The rst step

in the anabolic pathway unique to the biosynthesis of

lysine is catalyzed by the enzyme dihydrodipicolinate

synthase. Assume that you have recently been hired by a

major U.S. plant research institute and that you have

been asked to isolate a clone of the nucleic acid sequence

encoding dihydrodipicolinate synthase in maize. Briey

describe four different approaches you might take in

attempting to isolate such a clone and include at least

one genetic approach.

ANS: You could attempt to isolate either a dihydropicolinate

synthase (DHPS) cDNA clone or a DHPS genomic

clone. Once you have isolated either DHPS clone, it can

be used as a hybridization probe to isolate the other

(genomic or cDNA) by screening an appropriate library

by in situ colony hybridization. Four approaches that

have proven effective in isolating other eukaryotic cod-

ing sequences of interest are the following: (1) You could

obtain a clone of the DHPS gene of a lower eukaryote

(a clone of the DHPS gene of Saccharomyces cerevisae

is available) or even a prokaryote and use it as a heter-

ologous hybridization probe to screen a maize cDNA

library using low stringency conditions. Sometimes this

approach is successful; sometimes it is not successful.

Whether or not this approach works depends on how

similar the coding sequences of the specic gene of inter-

est are in the two species. (2) You could purify the DHPS

enzyme from corn and use the puried protein to pro-

duce an antibody to DHPS. This DHPS-specic anti-

body could then be used to screen a maize cDNA

expression library by a protocol analogous to the western

blot procedure. (An expression library contains the

cDNA coding sequences fused to appropriate transcrip-

tion and translation signals so that they are expressed in

E. coli or other host cells in which the cDNA library is

prepared.) (3) You could purify the DHPS enzyme from

maize and determine the amino acid sequence of its

-terminus by microsequencing techniques. From

the amino acid sequence and the known genetic code,

you could predict the possible nucleotide sequences

encoding this segment of the protein. Because of the

degeneracy in the code, there would be a set of nucleo-

tide sequences that would all specify the same amino acid

sequence, and you would not know which one was pres-

ent in the maize DHPS gene. However, the synthesis of

oligonucleotides is now routine and quite inexpensive.

Thus, you could synthesize a mixture of oligonucleotides

containing all possible coding sequences and use this

mixture as a set of hybridization probes to screen an

appropriate library by in situ colony hybridization.

(4) Finally, you might try a simple and very quick genetic

approach based on the ability of cDNAs in an expression

library to rescue DHPS mutants of E. coli or other species

that can be transformed at high frequencies. You would

obtain a DHPS-decient mutant of E. coli (available from

the E. coli Genetics Stock Center at Yale University),

transform it with your cDNA expression library, and

plate the transformed cells on medium lacking diami-

nopimelic acid (the product of DHPS). DHPS-decient

E. coli mutants cannot grow in the absence of diaminopi-

melic acid; thus, any colonies that grow on your selection

plates should be the result of rescue of the DHPS mutant

bacteria by corn DHPS encoded by cDNAs in the library.

This entire screening procedure can be carried out in 3

or 4 days; thus, it is much simpler than the preceding

approaches. In fact, David A. Frisch rst isolated a maize

DHPS cDNA by this simple, but powerful genetic

approach. However, that this approach would only be

expected to work in the case of enzymes that are active as

monomers or homomultimers; it is not applicable when

the active form of the enzyme is a heteromultimer.

14.21 You have just isolated a mutant of the bacterium Shigella

dysenteriae that is resistant to the antibiotic kanamycin,

and you want to characterize the gene responsible for

this resistance. Design a protocol using genetic selection

to identify the gene of interest.

ANS: Genetic selection is the most efcient approach to clon-

ing genes of this type. Prepare a genomic library in an

expression vector such as Bluescript (see Figure 14.3)

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using DNA from the kanamycin-resistant strain of Shi-

gella dysenteriae. Then, screen the library for the kanamy-

cin-resistance gene by transforming kanamycin-sensitive

E. coli cells with the clones in the library and plating the

transformed cells on medium containing kanamycin.

Only cells that are transformed with the kanamycin-

resistance gene will produce colonies in the presence of

kanamycin.

14.22 You have isolated a cDNA clone encoding a protein of

interest in a higher eukaryote. This cDNA clone is not

cleaved by restriction endonuclease EcoRI. When this

cDNA is used as a radioactive probe for blot hybridiza-

tion analysis of EcoRI-digested genomic DNA, three

radioactive bands are seen on the resulting Southern

blot. Does this result indicate that the genome of the

eukaryote in question contains three copies of the gene

encoding the protein of interest?

ANS: No. The genome of the species in question may contain

one, two, or three copies of the gene (or family of closely

related genes) encoding this protein. The possibilities

are as follows:

(1) One copy of the gene with two EcoRI cleavage sites

located within intron sequences. (2) Two copies of the

gene with one EcoRI cleavage site located within an

intron sequence of one of the copies. (3) Three copies of

the gene with no EcoRI cleavage site in any of the copies,

that is, each copy present on a single EcoRI restriction

fragment.

14.23 A linear DNA molecule is subjected to single and double

digestions with restriction endonucleases, and the fol-

lowing results are obtained:

Enzymes Fragment Sizes (in kb)

EcoRI 2.9, 4.5, 7.4, 8.0

HindIII 3.9, 6.0, 12.9

EcoRI and HindIII 1.0, 2.0, 2.9, 3.5, 6.0, 7.4

Draw the restriction map dened by these data.

ANS:

6.0 2.0 7.4 3.5

1.0

2.9

14.24 A DNA molecule is subjected to single and double diges-

tions with restriction enzymes, and the products are sep-

arated by gel electrophoresis. The results are as follows

(fragment sizes are in kb):

EcoRI

and

HindIII HindIII BamHI

EcoRI

and

BamHI

HindIII

and

BamHI

8 5 12 6 6 6

4 4 6 4 5

3 2 1

Draw the restriction map of this DNA molecule.

ANS:

14.25 You are studying a circular plasmid DNA molecule of

size 10.5 kilobase pairs (kb). When you digest this plas-

mid with restriction endonucleases BamHI, EcoRI, and

HindIII, singly and in all possible combinations, you

obtain linear restriction fragments of the following sizes:

Enzymes Fragment Sizes (in kb)

BamHI 7.3, 3.2

EcoRI 10.5

HindIII 5.1, 3.4, 2.0

BamHI  EcoRI

6.7, 3.2, 0.6

BamHI  HindIII

4.6, 2.7, 2.0, 0.7, 0.5

EcoRI  HindIII

4.0, 3.4, 2.0, 1.1

BamHI  EcoRI  HindIII

4.0, 2.7, 2.0, 0.7, 0.6, 0.5

Draw a restriction map for the plasmid that ts your

data.

ANS: There are two possible restriction maps for these data as

shown below:

4.0

2.0

0.7

2.7

0.5

0.6

4.0

2.7

0.7

2.0

0.5

0.6

Restriction enzyme cleavage sites for BamHI, EcoRI, and

HindIII are denoted by B, E, and H, respectively. The

numbers give distances in kilobase pairs.

14.26 The automated DNA sequencing machines utilize uo-

rescent dyes to detect the nascent DNA chains synthe-

sized in the presence of the four dideoxy (ddX) chain

terminators, each labeled with a different uorescent

dye. The dyes uoresce at different wavelengths, which

are recorded by a photocell as the products of the reac-

tions are separated based on length by capillary gel elec-

trophoresis (see Figure 14.17). In the standard sequencing

reaction, the chains terminating with ddG uoresce dark

blue (peaks appear black in computer printout), those

terminating with ddC uoresce light blue, those termi-

nating with ddA uoresce green, and those terminating

with ddT uoresce red. The computer printout for the

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62-WC Answers to All Questions and Problems

sequence of a short segment of DNA is shown as

follows.

First

nucleotide

Last

nucleotide

What is the nucleotide sequence of the nascent strand of

DNA?

What is the nucleotide sequence of the DNA template

strand?

ANS: Nascent strand:

5-AGTTCTAGAGCGGCCGCCACCGCGTGGAGCTCCAGCTTTTGTTCCCTTT-3

Template strand:

3-TCAAGATCTCGCCGGCGGTGGCGCACCTCGAGGTCGAAAACAAGGGAAA-5

14.27 Ten micrograms of a decanucleotide-pair HpaI restriction

fragment were isolated from the double-stranded DNA

chromosome of a small virus. Octanucleotide poly(A)

tails were then added to the 3′ ends of both strands using

terminal transferase and dATP; that is,

5′-X X X X X X X X X X-3′

3′-X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′

↓ terminal transferase, dATP

5′-X X X X X X X X X X A A A A A A A A-3′

3′-A A A A A A A A X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′

where X and X′ can be any of the four standard nucleo-

tides, but X′ is always complementary to X.

The two complementary strands (“Watson” strand and

“Crick” strand) were then separated and sequenced by

the 2′,3′-dideoxyribonucleoside triphosphate chain-ter-

mination method. The reactions were primed using a

synthetic poly(T) octamer; that is,

Watson strand

3′-A A A A A A A A X′ X′ X′ X′ X′ X′ X′ X′ X′ X′-5′

5′-T T T T T T T T-OH

Crick strand

5′-X X X X X X X X X X A A A A A A A A-3′

HO-T T T T T T T T-5′

Two DNA sequencing reactions were carried out.

Reaction 1 contained the Watson strand template/primer

shown above; reaction 2 contained the Crick strand

template/primer. Both sequencing reactions contained

DNA polymerase and all other substrates and compo-

nents required for DNA synthesis in vitro plus the

standard four 2′,3′-dideoxyribonucleoside triphosphate

chain-terminators—ddGTP, ddCTP, ddATP, and ddTTP—

each labeled with a different uorescent dye. The dyes

uoresce at different wavelengths, which are recorded by

a photocell as the products of the reactions are separated

by capillary gel electrophoresis (see Figure 14.17). In the

standard sequencing reaction, the chains terminating with

ddG uoresce dark blue (peaks appear black in the com-

puter printouts), those terminating with ddC uoresce

light blue, those terminating with ddA uoresce green,

and those terminating with ddT uoresce red. The com-

puter printout for sequencing reaction 1, which contained

the Watson strand as template, is shown as follows.

110Nucleotide:

Draw the predicted computer printout for reaction

2, which contained the Crick strand as template, in the

following box. Remember that all DNA synthesis occurs

in the 5′ → 3′ direction and that the sequence of

the nascent strand reads 5′ to 3′ from left to right in the

printout.

110Nucleotide:

ANS:

110Nucleotide:

Chapter 15

15.1 Distinguish between a genetic map, a cytogenetic map,

and a physical map. How can each of these types of maps

be used to identify a gene by positional cloning?

ANS: Genetic map distances are determined by crossover fre-

quencies. Cytogenetic maps are based on chromosome

morphology or physical features of chromosomes. Physi-

cal maps are based on actual physical distances—the

number of nucleotide pairs (0.34 nm per base pair)—

separating genetic markers. If a gene or other DNA

sequence of interest is shown to be located near a mutant

gene, a specic band on a chromosome, or a particular

DNA restriction fragment, that genetic or physical

marker (mutation, band, or restriction fragment) can be

used to initiate a chromosome walk to the gene of

interest.

15.2 In the technique of positional cloning, a researcher

begins with a DNA library and selects a clone that is

tightly linked to the gene of interest. That clone, or a

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Answers to All Questions and Problems WC-63

piece of it, is then used as a probe to isolate an overlap-

ping clone from a different DNA library. The second

clone is used to isolate a third overlapping clone from

the rst library, and so on, until the researcher has

“walked” along the chromosome to the desired locus. (a)

How can the researcher walk consistently in the same

direction along the chromosome during the cloning pro-

cess? (b) What could happen if a long repetitive DNA

sequence such as a transposon was situated between the

starting clone and the gene of interest?

ANS: (a) To ensure that the researcher “walks” in the same

direction along the chromosome to the desired locus, it is

necessary to check at each step in the process, usually by

blot hybridization with subclones, which end of the clone

used as a probe overlaps with the next recovered clone.

(b) A long repetitive sequence could present a problem in

chromosome walking because if that sequence is used as a

probe to isolate a new clone, it will hybridize with restric-

tion fragments from all over the genome, and thereby

thwart the effort to walk systematically in one direction

along a single chromosome. To overcome this problem, a

researcher would have to “jump over” the repetitive ele-

ment and use a piece of unique DNA beyond the element

as a probe in the next step of the walking procedure.

15.3 What is a contig? What is an RFLP? What is a VNTR?

What is an STS? What is an EST? How is each of these

used in the construction of chromosome maps?

ANS: A contig (contiguous clones) is a physical map of a chro-

mosome or part of a chromosome prepared from a set of

overlapping genomic DNA clones. An RFLP (restriction

fragment length polymorphism) is a variation in the

length of a specic restriction fragment excised from a

chromosome by digestion with one or more restriction

endonucleases. A VNTR (variable number tandem

repeat) is a short DNA sequence that is present in the

genome as tandem repeats and in highly variable copy

number. An STS (sequence tagged site) is a unique DNA

sequence that has been mapped to a specic site on a

chromosome. An EST (expressed sequence tag) is a cDNA

sequence—a genomic sequence that is transcribed. Con-

tig maps permit researchers to obtain clones harboring

genes of interest directly from DNA Stock Centers—to

“clone by phone.” RFLPs are used to construct the high-

density genetic maps that are needed for positional clon-

ing. VNTRs are especially valuable RFLPs that are used

to identify multiple sites in genomes. STSs and ESTs

provide molecular probes that can be used to initiate

chromosome walks to nearby genes of interest.

15.4 The following is a Southern blot of EcoRI-digested DNA

of rye plants from two different inbred lines, A and B.

Developed autoradiogram I shows the bands resulting

from probing the blot with

P-labeled cDNA1. Autora-

diogram II shows the same Southern blot after it was

stripped of probe and reprobed with

P-labeled cDNA2.

I II

A B A B

a3 b3

(a) Which bands would you expect to see in the autora-

diogram of a similarly probed Southern blot prepared

using EcoRI-digested DNA from F

hybrid plants pro-

duced by crossing the two inbred lines? (b) What can you

conclude about the gene(s) represented by band a1 on

blot I in the two inbreds? (c) The F

plants were crossed

to plants possessing only bands a1, a4, and b3. DNA was

isolated from several individual progeny and digested

with EcoRI. The resulting DNA fragments were sepa-

rated by gel electrophoresis, transferred to a nylon mem-

brane, and hybridized with radioactive cDNA1 and

cDNA2 probes. The following table summarizes the

bands present in autoradiograms obtained using DNA

from individual progeny.

Plant

No. Bands Present

a1 a2 a3 a4 b1 b2 b3

  



  



  



  



     



   



   



   



   



  

Interpret these data. Do the data provide evidence for

RFLPs? At how many loci? Are any of the RFLPs

linked? If so, what are the linkage distances dened by

the data?

ANS: (a) a1, a2, a3, a4, b1, b2, and b3. (b) Band a1 represents a

locus whose DNA is homologous to cDNA1. Since the

marker is not polymorphic in the parents used, it cannot

be mapped in this cross. (c) The cDNA1 probe detects

one RFLP locus with alleles that are visualized as band

a4 and bands a2/a3. The cDNA2 detects a second RFLP

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64-WC Answers to All Questions and Problems

locus with alleles that are visualized as band b3 and bands

b1/b2. The two loci are linked with 20% recombination

observed in this cross.

15.5 As part of the Human Genome Mapping Project, you are

trying to clone a gene involved in colon cancer. Your rst

step is to localize the gene using RFLP markers. In the

following table, RFLP loci are dened by STS number

(e.g., STS1), and the gene for colon cancer is designated C.

Loci

% Recombi-

nation Loci

% Recombi-

nation

C, STS1 50 STS1, STS5 10

C, STS2 15 STS2, STS3 30

C, STS3 15 STS2, STS4 14

C, STS4 1 STS2, STS5 50

C, STS5 40 STS3, STS4 16

STS1, STS2 50 STS3, STS5 25

STS1, STS3 35 STS4, STS5 41

STS1, STS4 50

(a) Given the percentage recombination between differ-

ent RFLP loci and the gene for colon cancer shown in the

table, draw a genetic map showing the order and genetic

distances between adjacent RFLP markers and the gene

for colon cancer. (b) Given that the human genome con-

tains approximately 3.3  10

base pairs of DNA and that

the human genetic map contains approximately 3300 cM,

approximately how many base pairs of DNA are located

along the stretch of chromosome dened by this RFLP

map? (Hint: First gure how many base pairs of DNA are

present per cM in the human genome.) (c) How many

base pairs of DNA are present in the region between the

colon cancer gene and the nearest STS?

ANS: (a)

10 cM 25 cM 15 cM 1 cM 14 cM

STS1 STS5 STS3 C STS4 STS2

(b) 3.3  10

bp/3.3  10

cM  1  10

bp/cM. The

total map length is 65 cM, which equates to about 65 

or 65 million bp.

or about 1 million base pairs.

15.6 What are STRs? Why are they sometimes called

microsatellites?

ANS: STRs are polymorphic tandem repeats of sequences of

only two to four nucleotide pairs. They are called micro-

satellites because they are short in length and are compo-

nents of the highly repetitive satellite DNAs of eukaryotes

(see Chapter 9).

15.7 You have cloned a previously unknown human gene.

What procedure will allow you to position this gene on

the cytological map of the human genome without per-

forming any pedigree analyses? Describe how you would

carry out this procedure.

ANS: With a clone of the gene available, uorescent in situ

hybridization (FISH) can be used to determine which

human chromosome carries the gene and to localize the

gene on the chromosome. Single-stranded copies of the

clone are coupled to a uorescent probe and hybridized

to denatured DNA in chromosomes spread on a slide.

After hybridization, free probe is removed by washing,

and the location of the uorescent probe is determined

by photography using a uorescence microscope (see

Focus on: In Situ Hybridization).

15.8 You have identied a previously unknown human EST.

What must be done before this new EST can be called

an STS?

ANS: An EST must be placed on the physical map of a chro-

mosome before it can be called an STS. If it is also posi-

tioned on the genetic map of the chromosome, then it is

called an anchor marker.

15.9 VNTRs and STRs are specic classes of polymorphisms.

What is the difference between a VNTR and an STR?

ANS: Variable number tandem repeats (VNTRs) are com-

posed of repeated sequences of 10–80 nucleotide pairs,

and short tandem repeats (STRs) are composed of

repeated sequences of 2–10 nucleotide pairs.

15.10 An RFLP and a mutant allele that causes albinism in

humans cannot be shown to be separated by recombina-

tion based on pedigree analysis or by radiation hybrid

mapping. Do these observations mean that the RFLP

occurs within or overlaps the gene harboring the muta-

tion that causes albinism? If so, why? If not, why not?

ANS: The resolution of genetic mapping in humans is quite

low—in the range of 1–10 million base pairs. Radiation

hybrid mapping provides higher resolution—to about

50 kb. However, even with 50-kb resolution, there could

be several genes separating the RFLP from the mutation

responsible for albinism.

15.11 A cloned 6-kb fragment of DNA from human chromo-

some 9 contains a single site recognized by the restric-

tion enzyme EcoRI. This cloned fragment is demarcated

by sites for the restriction enzyme BamHI. There are no

other BamHI recognition sites within the clone.

A researcher has collected DNA samples from 10 people.

He digests each sample with a combination of EcoRI and

BamHI enzymes. The doubly digested DNA is then frac-

tionated by gel electrophoresis and blotted to a mem-

brane. After xing the DNA to the membrane, the

researcher hybridizes it with a radioactive probe made

from the entire cloned BamHI fragment. The autoradio-

gram obtained by exposing an X-ray lm to this mem-

brane yielded the following results. Three of the DNA

samples contained a 4-kb fragment and a 2-kb fragment

that hybridize with the probe, three of the DNA samples

contained a 6-kb DNA fragment that hybridizes with the

probe, and four of the DNA samples contain 6-, 4-, and

2-kb DNA fragments that hybridize with the probe.

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What has this analysis revealed? What are the genotypes

of the three different types of DNA samples?

ANS: The results of this analysis reveal that the EcoRI cleavage

site within the original 6-kb BamHI clone is polymorphic—

that is, it is present in some chromosomes 9 but absent in

others. We can represent these two types of chromo-

somes 9 as BEB (with the EcoRI site between anking

BamHI sites) and B-B (without the EcoRI site). The rst

three samples came from people who were homozygous

for the BEB version of chromosome 9, the next three

samples came from people who were homozygous for

the B-B version, and the last four samples came from

people who where heterozygous for the two versions—

that is, they had the genotype BEB/B-B. In the human

population that was sampled, the EcoRI site is therefore

the basis for a restriction fragment length polymorphism

(RFLP).

15.12 Both an RFLP and a mutation that causes deafness in

humans map to the same location on the same chromo-

some. How can you determine whether or not the

RFLP overlaps with the gene containing the deafness

mutation?

ANS: You can start a chromosome walk using the hybridization

probe that detects the RFLP. However, if a physical map

of this region of the chromosome already exists (see

Figure 15.5), a chromosome walk might not be neces-

sary. cDNAs can be used to locate candidate genes in the

region covered by the chromosome walk, and the

sequences of genes in individuals with this form of inher-

ited deafness can be compared with the sequences of

homologous genes of individuals with normal hearing,

looking for changes that would be expected to cause a

loss of gene function. The overall process is illustrated in

Figure 15.6.

15.13 What were the goals of the Human Genome Project?

What impact has achieving these goals had on the prac-

tice of medicine to date? What are some of the predicted

future impacts? What are some of the possible misuses of

human genome data?

ANS: The goals of the Human Genome Project were to pre-

pare genetic and physical maps showing the locations of

all the genes in the human genome and to determine the

nucleotide sequences of all 24 chromosomes in the

human genome. These maps and nucleotide sequences

of the human chromosomes helped scientists identify

mutant genes that result in inherited diseases. Hopefully,

the identication of these mutant disease genes will lead

to successful treatments, including gene therapies, for at

least some of these diseases in the future. Potential mis-

uses of these data include invasions of privacy by govern-

ments and businesses—especially employment agencies

and insurance companies. Individuals must not be denied

educational opportunities, employment, or insurance

because of inherited diseases or mutant genes that result

in a predisposition to mental or physical abnormalities.

15.14 What difculty does repetitive DNA pose for the assem-

bly of whole genome shotgun sequences by computer

analysis?

ANS: Repetitive DNA can make it difcult to assemble long

stretches of DNA sequence from the sequences of DNA

fragments that contain these sequences. The reason is

that in the computer analysis, a repetitive sequence in

one DNA fragment will match with the same repetitive

sequence in another DNA fragment even if the two

DNA fragments are not contiguous in the genome. The

effort to determine which short DNA fragments are

authentic neighbors (contiguous to each other) can

therefore be thrown off by these spurious matches.

15.15 Which type of molecular marker, RFLP or EST, is most

likely to mark a disease-causing mutant gene in humans?

Why?

ANS: An EST is more likely than an RFLP to occur in a dis-

ease-causing human gene. All ESTs correspond to

expressed sequences in a genome. RFLPs occur through-

out a genome, in both expressed and unexpressed

sequences. Because less than 2 percent of the human

genome encodes proteins, most RFLPs occur in noncod-

ing DNA.

15.16 Bacteriophage ΦX174 contains 11 genes in a genome of

5386 bp; E. coli has a predicted 4288 genes in a genome

of about 4.639 kb; S. cerevisiae has about 6000 genes in a

genome of size 12.1 mb; C. elegans has about 19,000

genes present in a genome of about 100 mb; and H. sapi-

ens has an estimated 22,000 genes in its 3000-mb genome.

Which genome has the highest gene density? Which

genome has the lowest gene density? Does there appear

to be any correlation between gene density and develop-

mental complexity? If so, describe the correlation.

ANS: The bacteriophage ΦX174 genome has the highest gene

density—one gene per 490 nucleotide pairs. The human

genome has the lowest gene density—about one gene

per 100 kb. In the species mentioned in this question,

there is a striking correlation between genome size and

developmental complexity. With some exceptions,

including species with polyploid genomes, there does

appear to be a rough correlation between genome size

and developmental complexity.

15.17 A contig map of one segment of chromosome 3 of Arabi-

dopsis is as follows.

12345

Chromosome segment

678910

Genomic clones in

YAC vectors

(a) If an EST hybridizes with genomic clones C, D, and

E, but not with the other clones, in which segment of

chromosome 3 is the EST located? (b) If a clone of gene

ARA hybridizes only with genomic clones C and D, in

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which chromosome segment is the gene located? (c) If a

restriction fragment hybridizes with only one of the

genomic clones shown above, in which chromosome

segment(s) could the fragment be located?

ANS: (a) Segment 5; (b) segment 4; (c) segment 1, 6, or 10.

15.18 Eight human–Chinese hamster radiation hybrids were

tested for the presence of six human ESTs designated A

through F. The results are shown in the following table,

where a plus indicates that a marker was present and a

minus indicates that it was absent.

12345678

––

– –––

––

–

F–

+++ +

+ ++++–

Radiation hybrid

Marker

Based on these data, do any of the ESTs appear to be

closely linked? Which ones? What would be needed for

you to be more certain of your answer?

ANS: EST markers D and E appear to be closely linked. The

eight human–Chinese hamster radiation hybrids contain

either both D and E or neither marker. More radiation

hybrids would need to be tested for the presence of these

ESTs to obtain convincing evidence of this linkage.

15.19 What is the major advantage of gene chips as a microar-

ray hybridization tool?

ANS: The major advantage of gene chips as a microarray

hybridization tool is that a single gene chip can be used

to quantify thousands of distinct nucleotide sequences

simultaneously. The gene-chip technology allows

researchers to investigate the levels of expression of a

large number of genes more efciently than was possible

using earlier microarray procedures.

15.20 What major advantage does the green uorescent pro-

tein of the jellysh have over other methods for studying

protein synthesis and localization?

ANS: The green uorescent protein (GFP) can be used to study

protein localization and movement over time in living

cells. Most other procedures for studying protein localiza-

tion require that cells be permeabilized and/or xed and

exposed to antibodies coupled to radioactive or uores-

cent compounds prior to visualization. As a result, these

procedures only provide information about the location

of a protein at a single time point. In contrast, GFP-tagged

proteins can be used to study the synthesis and movement

of proteins in living cells over time (hours to days).

15.21 You are given chromosome-specic cDNA libraries for

all 24 human chromosomes. How might these libraries

be used to study chromosome evolution in primates?

ANS: The DNA sequences in human chromosome-specic

cDNA libraries can be coupled to uorescent dyes and

hybridized in situ to the chromosomes of other primates.

The hybridization patterns can be used to detect changes

in genome structure that have occurred during the evo-

lution of the various species of primates from common

ancestors (see Figure 6.4). Such comparisons are espe-

cially effective in detecting new linkage relationships

resulting from translocations and centric fusions.

15.22 Of the cereal grass species, only maize contains two cop-

ies of each block of linked genes. What does this duplica-

tion of sets of maize genes indicate about the origin of

this agronomically important species?

ANS: The presence of two copies of each block of genes in the

corn genome indicates that maize has evolved from a tet-

raploid ancestor. The presence of one set of genes pri-

marily in the large chromosomes and the second set

largely in the small chromosomes suggests that maize

has evolved from an allotetraploid (see Chapter 6) pro-

duced by combining the diploid genomes of two ances-

tral cereal grass species.

15.23 Five human genomic DNA clones present in PAC vec-

tors were tested by hybridization for the presence of six

sequence-tagged sites designated STS1 through STS6.

The results are given in the following table: a plus indi-

cates the presence of the STS, and a minus indicates the

absence of the STS.

123456

+++

––+

––

–

––+

––

–

––

– ++

–––

– + –

STS

YAC clone

(a) What is the order of the STS sites on the

chromosome?

(b) Draw the contig map dened by these data.

ANS: (a) Order of STS sites: 2-5-1-4-3-6.

(b)

251436

STS markers:

YAC

clones

15.24 The complete sequences of six mitochondrial genomes

of H. neanderthalensis have been available for some time;

the rst H. neanderthalensis mtDNA sequence was pub-

lished in 2008. How similar are the sequences of the

mtDNAs of H. neanderthalensis and H. sapiens? Are the

genomes similar in size? Is the amount of diversity

observed in the mtDNAs of Neanderthals and humans

the same? If not, what might this tell us about the sizes of

Neanderthal and human populations? How many genes

are present in the H. neanderthalensis mitochondrial

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genome? How many of these genes encode proteins?

How many specify structural RNA molecules? Are there

any pseudogenes in H. neanderthalensis mtDNA? All of

these questions can be answered by visiting the http://

www.ncbi.nlm.nih.gov/ web site.

ANS: The mtDNAs of Homo neanderthalensis and H. sapiens are

very similar, both in size and in nucleotide sequence. The

H. neanderthalensis mtDNA contains 16,565 nucleotide

pairs, whereas the H. sapiens mtDNA contains 16,569

nucleotide pairs, and the two DNAs exhibit 99 percent

sequence identity. There is less sequence diversity in the

mtDNAs of Neanderthals than in the mtDNAs of

humans, which is consistent with Neanderthal popula-

tions being smaller than human populations. There are

37 genes in the Neanderthal mtDNA: 13 encode pro-

teins and 24 specify structural RNAs (tRNAs and

rRNAs). The Neanderthal mitochondrial genome does

not contain any pseudogenes.

15.25 Assume that you have just sequenced a small fragment of

DNA that you had cloned. The nucleotide sequence of

this segment of DNA is as follows.

aagtagtcgaaaccgaattccgtagaaacaactcgcacgctccggtttc

gtgttgcaacaaaataggcattcccatcgcggcagttagaatcaccga

gtgcccagagtcacgttcgtaagcaggcgcagtttacaggcagca

gaaaaatcgattgaacagaaatggctggcggtaaagcaggcaagga

ttcgggcaaggccaaggcgaaggcggtatcgcgttccgcgcgcgcggg

In an attempt to learn something about the identity or

possible function of this DNA sequence, you decide to

perform a BLAST (nucleotide blast) search on the NCBI

web site (http://www.ncbi.nlm.nih.gov). Paste or type

this sequence into the query sequence box. Run the

search and examine the sequences most closely related to

your query sequence. Are they coding sequences? What

proteins do they encode? Repeat the BLAST search with

only half of your sequence as the query sequence. Do you

still identify the same sequences in the databases? If you

use one-fourth of your sequence as a query, do you still

retrieve the same sequences? What is the shortest DNA

sequence that you can use as a query and still identify the

same sequences in the databanks?

ANS: All of the sequences identied by the megablast search

encode histone H2a proteins. The query sequence is

identical to the coding sequence of the Drosophila

melanogaster histone H2aV gene (a member of the gene

family encoding histone H2a proteins). The query

sequence encodes a Drosophila histone H2a polypeptide

designated variant V. The same databank sequences are

identied when one-half or one-fourth of the given

nucleotide sequence is used as the query in the megablast

search. Query sequences as short as 15–20 nucleotides

can be used to identify the Drosophila gene encoding the

histone H2a variant. However, the results will vary

depending on the specic nucleotide sequence used as

the query sequence.

15.26 The NCBI web site (http://www.ncbi.nlm.nih.gov) can

also be used to search for protein sequences. Instead of

performing a BLAST search with a nucleic acid query,

one performs a protein blast with a polypeptide (amino

acid sequence) query. Assume that you have the follow-

ing partial sequence of a polypeptide:

GYDVEKNNSRIKLGLKSLVSKGILVQTKGT

GASGSFKLNKKAASGEAKPQAKKAGAAKA

Go to the NCBI web site and access the BLAST tool.

Then click on protein blast and enter the query sequence

in the box at the top. Then click BLAST. What is the

identity of your query sequence?

ANS: Your query sequence is a portion of histone H1.2 of the

mouse (Mus musculus). This portion of the mouse histone

H1.2 polypeptide differs by one amino acid from the

corresponding part of the human H1 histone.

15.27 The sequence of a gene in Drosophila melanogaster that

encodes a histone H2A polypeptide is as follows:

aagtagtcgaaaccgaattccgtagaaacaactcgcacgctccggtttc

gtgttgcaacaaaataggcattcccatcgcggcagttagaatcacc

gagtgcccagagtcacgttcgtaagcaggcgcagtttacaggcagcag

aaaaatcgattgaacagaaatggctggcggtaaagcaggcaagg

attcgggcaaggccaaggcgaaggcggtatcgcgttccgcgcg

cgcgggtcttcagttccccgtgggtcgcatccatcgtcatctcaag

agccgcactacgtcacatggacgcgtcggagccactgcagccgtg

tactccgctgccatattggaatacctgaccgccgaggtcctggagtt

ggcaggcaacgcatcgaaggacttgaaagtgaaacgtatcactcc

tcgccacttacagctcgccattcgcggagacgaggagctggacag

cctgatcaaggcaaccatcgctggtggcggtgtcattccgcacata

cacaagtcgctgatcggcaaaaaggaggaaacggtgcaggatccgc

agcggaagggcaacgtcattctgtcgcaggcctactaagccagtcgg

caatcggacgccttcgaaacatgcaacactaatgtttaattcagattt

cagcagagacaagctaaaacaccgacgagttgtaatcatttctgtgcg

ccagcatatatttcttatatacaacgtaatacataattatgtaattctagca

tctccccaacactcacatacatacaaacaaaaaatacaaacacacaaaac

gtatttacccgcacgcatccttggcgaggttgagtatgaaacaaa

aacaaaacttaatttagagcaaagtaattacacgaataaatttaataa

aaaaaactataataaaaacgcc.

Let’s use the translation software available on the Inter-

net at http://www.expasy.org/tools/dna.html to translate

this gene in all six possible reading frames and see which

reading frame species histone H2A. Just type or paste

the DNA sequence in the “ExPASy Translate” tool box

and click TRANSLATE SEQUENCE. The results will

show the products of translation in all six reading frames

with Met’s and Stop’s boldfaced to highlight potential

open-reading frames. Which reading frame species his-

tone H2A?

ANS: Reading frame 5′ → 3′ number 1 has a large open read-

ing frame with a methionine codon near the 5′ end. You

can verify that this is the correct reading frame by using

the predicted translation product as a query to search

one of the protein databases (see Problem 15.26).

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Chapter 16

16.1 What are CpG islands? Of what value are CpG islands in

positional cloning of human genes?

ANS: CpG islands are clusters of cytosines and guanines that

are often located just upstream (5′) from the coding

regions of human genes. Their presence in nucleotide

sequences can provide hints as to the location of genes in

human chromosomes.

16.2 Why is the mutant gene that causes Huntington’s disease

called huntingtin? Why might this gene be renamed in

the future?

ANS: The gene was named huntingtin after the disease that it

causes when defective. The gene will probably be

renamed after the function of its gene product has been

determined.

16.3 How was the nucleotide sequence of the CF gene used to

obtain information about the structure and function of

its gene product?

ANS: The CF gene was identied by map position-based clon-

ing, and the nucleotide sequences of CF cDNAs were

used to predict the amino acid sequence of the CF gene

product. A computer search of the protein data banks

revealed that the CF gene product was similar to several

ion channel proteins. This result focused the attention of

scientists studying cystic brosis on proteins involved in

the transport of salts between cells and led to the discov-

ery that the CF gene product was a transmembrane con-

ductance regulator—now called the CFTR protein.

16.4 How might the characterization of the CF gene and its

product lead to the treatment of cystic brosis by

somatic-cell gene therapy? What obstacles must be over-

come before cystic brosis can be treated successfully by

gene therapy?

ANS: Once the function of the CF gene product has been

established, scientists should be able to develop proce-

dures for introducing wild-type copies of the CF gene

into the appropriate cells of cystic brosis patients to

alleviate the devastating effects of the mutant gene. A

major obstacle to somatic-cell gene-therapy treatment of

cystic brosis is the size of the CF gene—about 250 kb,

which is too large to t in the standard gene transfer vec-

tors. Perhaps a shortened version of the gene constructed

from the CF cDNA—about 6.5 kb—can be used in place

of the wild-type gene. A second major obstacle is getting

the transgene into enough of the target cells of the cystic

brosis patient to alleviate the symptoms of the disease.

A third challenge is to develop an expression vector con-

taining the gene that will result in long-term expression

of the introduced gene in transgenic cells. Another con-

cern is how to avoid possible side effects caused by over-

expression or inappropriate expression of the transgene

in cystic brosis patients. Despite these obstacles, many

scientists are optimistic that cystic brosis will be effec-

tively treated by somatic-cell gene therapy in the future.

16.5 Myotonic dystrophy (MD), occurring in about 1 of 8000

individuals, is the most common form of muscular

dystrophy in adults. The disease, which is characterized

by progressive muscle degeneration, is caused by a domi-

nant mutant gene that contains an expanded CAG repeat

region. Wild-type alleles of the MD gene contain 5–30

copies of the trinucleotide. Mutant MD alleles contain

50 to over 2000 copies of the CAG repeat. The complete

nucleotide sequence of the MD gene is available. Design

a diagnostic test for the mutant gene responsible for

myotonic dystrophy that can be carried out using

genomic DNA from newborns, fetal cells obtained by

amniocentesis, and single cells from eight-cell pre-

embryos produced by in vitro fertilization.

ANS: Oligonucleotide primers complementary to DNA

sequences on both sides (upstream and downstream) of the

CAG repeat region in the MD gene can be synthesized and

used to amplify the repeat region by PCR. One primer

must be complementary to an upstream region of the tem-

plate strand, and the other primer must be complementary

to a downstream region of the nontemplate strand. After

amplication, the size(s) of the CAG repeat regions can be

determined by gel electrophoresis (see Figure 16.2). Tri-

nucleotide repeat lengths can be measured by including

repeat regions of known length on the gel. If fewer than 30

copies of the trinucleotide repeat are present on each chro-

mosome, the newborn, fetus, or pre-embryo is homozy-

gous for a wild-type MD allele or heterozygous for two

different wild-type MD alleles. If more than 50 copies of

the repeat are present on each of the homologous chromo-

somes, the individual, fetus, or cell is homozygous for a

dominant mutant MD allele or heterozygous for two dif-

ferent mutant alleles. If one chromosome contains less

than 30 copies of the CAG repeat and the homologous

chromosome contains more than 50 copies, the newborn,

fetus, or pre-embryo is heterozygous, carrying one

wild-type MD allele and one mutant MD allele.

16.6 In humans, the absence of an enzyme called purine

nucleoside phosphorylase (PNP) results in a severe

T-cell immunodeciency similar to that of severe com-

bined immunodeciency disease (SCID). PNP deciency

exhibits an autosomal recessive pattern of inheritance,

and the gene encoding human PNP has been cloned and

sequenced. Would PNP deciency be a good candidate

for treatment by gene therapy? Design a procedure for

the treatment of PNP deciency by somatic-cell gene

therapy.

ANS: Yes. A somatic-cell gene therapy procedure similar to

that used for X-linked SCID (see Figure 16.7) might be

effective in treating purine nucleoside phosphorylase

(PNP) deciency. White blood cells could be isolated

from the patient, transfected with a vector carrying a

wild-type PNP gene, grown in culture and assayed for

the expression of the PNP transgene, and then infused

back into the patient after the expression of the trans-

gene had been veried.

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16.7 Human proteins can now be produced in bacteria such as

E. coli. However, one cannot simply introduce a human

gene into E. coli and expect it to be expressed. What steps

must be taken to construct an E. coli strain that will

produce a mammalian protein such as human growth

hormone?

ANS: The transcription initiation and termination and transla-

tion initiation signals or eukaryotes differ from those of

prokaryotes such as E. coli. Therefore, to produce a

human protein in E. coli, the coding sequence of the

human gene must be joined to appropriate E. coli regula-

tory signals—promoter, transcription terminator, and

translation initiator sequences. Moreover, if the gene

contains introns, they must be removed or the coding

sequence of a cDNA must be used, because E. coli does

not possess the spliceosomes required for the excision of

introns from nuclear gene transcripts. In addition, many

eukaryotic proteins undergo posttranslational process-

ing events that are not carried out in prokaryotic cells.

Such proteins are more easily produced in transgenic

eukaryotic cells growing in culture.

16.8 You have constructed a synthetic gene that encodes an

enzyme that degrades the herbicide glyphosate. You wish

to introduce your synthetic gene into Arabidopsis plants

and test the transgenic plants for resistance to glypho-

sate. How could you produce a transgenic Arabidopsis

plant harboring your synthetic gene by A. tumefaciens-

mediated transformation?

ANS: You would rst construct a chimeric gene containing

your synthetic gene fused to a plant promoter such as the

35S promoter of cauliower mosaic virus and a plant

transcription termination and polyadenylation signal

such as the one from the nos gene of the Ti plasmid. This

chimeric gene would then be inserted into the T-DNA

of a Ti plasmid carrying a dominant selectable marker

gene (e.g., 35S/NPTII/nos, which confers resistance to

kanamycin to host cells) and introduced into Agrobacte-

rium tumefaciens cells by transformation. Tissue explants

from Arabidopsis plants would be co-cultivated with A.

tumefaciens cells harboring the recombinant Ti plasmid,

and plant cells that carry T-DNAs inserted into their

chromosomes would be selected by growth on medium

containing the appropriate selective agent (e.g., kanamy-

cin). Transgenic plants would then be regenerated from

the transformed cells and tested for resistance to

glyphosate.

16.9 A human STR locus contains a tandem repeat (TAGA)

where n may be any number between 5 and 15. How

many alleles of this locus would you expect to nd in the

human population?

ANS: Eleven, ranging in multiples of 3, from 15 to 45 nucleo-

tides long.

16.10 A group of bodies are found buried in a forest. The police

suspect that they may include the missing Jones family

(two parents and two children). They extract DNA from

bones and examine (using PCR) genes A and B, which

are known to contain tandem triplet repeats of variable

length. They also analyze DNA from two other men.

The results are shown below where the numbers indi-

cate the number of copies of a tandem repeat in a par-

ticular allele; for example, male 1 has one allele with 8

and another allele with 9 copies of a tandem repeat in

gene A.

Gene A Gene B

Male 1

8/9 5/7

Male 2 6/8 5/5

Male 3 7/10 7/7

Woman 8/8 3/5

Child 1 7/8 5/7

Child 2 8/8 3/7

Could the woman have been the mother of both chil-

dren? Why or why not? Which man, if any, could have

been the father of child 1?

ANS: The woman could be the mother of both children, hav-

ing passed alleles A 8 and B 5 to child 1 and A 8 and B 3

to child 2. Male 3 could have been the father of child 1.

If so, he contributed alleles A 7 and B 7.

16.11 DNA proles have played central roles in many rape and

murder trials. What is a DNA prole? What roles do

DNA proles play in these forensic cases? In some cases,

geneticists have been concerned that DNA prole data

were being used improperly. What were some of their

concerns, and how can these concerns be properly

addressed?

ANS: DNA proles are the specic patterns (1) of peaks pres-

ent in electropherograms of chromosomal STRs or

VNTRs amplied by PCR using primers tagged with

uorescent dyes and separated by capillary gel electro-

phoresis (see Figures 16.11 and 16.12) or (2) of bands on

Southern blots of genomic DNAs that have been digested

with specic restriction enzymes and hybridized to

appropriate STR or VNTR sequences (see Figure 16.10).

DNA proles, “like” epidermal ngerprints, are used as

evidence for identity or nonidentity in forensic cases.

Geneticists have expressed concerns about the statistical

uses of DNA prole data. In particular, they have ques-

tioned some of the methods used to calculate the proba-

bility that DNA from someone other than the suspect

could have produced an observed prole. These con-

cerns have been based in part on the lack of adequate

DNA prole databases for various human subpopula-

tions and the lack of precise information about the

amount of variability in DNA proles for individuals of

different ethnic backgrounds. These concerns have been

addressed by the acquisition of data on prole frequen-

cies in different populations and ethnic groups from

throughout the world.

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16.12 The DNA proles shown in this problem were prepared

using genomic DNA from blood cells obtained from a

woman, her daughter, and three men who all claim to be

the girl’s father.

MCF1 F2 F3

Based on the DNA proles, what can be determined

about paternity in this case?

ANS: Neither F1 nor F2 could be the girl’s biological father;

only individual F3 could be the child’s father.

16.13 Most forensic experts agree that proles of DNA from

blood samples obtained at crime scenes and on personal

items can provide convincing evidence for murder con-

victions. However, the defense attorneys sometimes

argue successfully that sloppiness in handling blood sam-

ples results in contamination of the samples. What prob-

lems would contamination of blood samples present in

the interpretation of DNA proles? Would you expect

such errors to lead to the conviction of an innocent per-

son or the acquittal of a guilty person?

ANS: Contamination of blood samples would introduce more

variability into DNA proles. This would lead to a lack

of allelic matching of proles obtained from the blood

samples and from the defendant. Mixing errors would be

expected to lead to the acquittal of a guilty person and

not to the conviction of an innocent person. Only the

mislabeling of samples could implicate someone who is

innocent.

16.14 The Ti plasmid contains a region referred to as T-DNA.

Why is this region called T-DNA, and what is its

signicance?

ANS: The T in T-DNA is an abbreviation for “transferred.”

The T-DNA region of the Ti plasmid is the segment

that is transferred from the Ti plasmid of the bacterium

to the chromosomes of the plant cells during Agrobacte-

rium tumefaciens-mediated transformation.

16.15 The generation of transgenic plants using A. tumefaciens-

mediated transformation often results in multiple sites of

insertion. These sites frequently vary in the level of

transgene expression. What approaches could you use to

determine whether or not transgenic plants carry more

than one transgene and, if so, where the transgenes are

inserted into chromosomes?

ANS: Probing Southern blots of restriction enzyme-digested

DNA of the transgenic plants with

P-labeled transgene

may provide evidence of multiple insertions but would

not reveal the genomic location of the inserts. Fluores-

cence in situ hybridization (FISH) is a powerful proce-

dure for determining the genomic location of gene

inserts. FISH is used to visualize the location of trans-

genes in chromosomes.

16.16 Disarmed retroviral vectors can be used to introduce

genes into higher animals including humans. What

advantages do retroviral vectors have over other kinds of

gene-transfer vectors? What disadvantages?

ANS: Disarmed retroviruses are lacking genes essential for

reproduction in host cells but can still integrate into the

DNA of the host cell in the proviral state. The retroviral

genomes are small enough to allow them to be manipu-

lated easily in vitro and yet will accept foreign DNA

inserts of average gene size. The retroviruses contain

strong promoters in their long terminal repeats that can

be used to drive high levels of transcription of the for-

eign gene insert.

16.17 Transgenic mice are now routinely produced and studied

in research laboratories throughout the world. How are

transgenic mice produced? What kinds of information

can be obtained from studies performed on transgenic

mice? Does this information have any importance to the

practice of medicine? If so, what?

ANS: Transgenic mice are usually produced by microinjecting

the genes of interest into pronuclei of fertilized eggs or

by infecting pre-implantation embryos with retroviral

vectors containing the genes of interest. Transgenic mice

provide invaluable tools for studies of gene expression,

mammalian development, and the immune system of

mammals. Transgenic mice are of major importance in

medicine; they provide the model system most closely

related to humans. They have been, and undoubtedly

will continue to be, of great value in developing the tools

and technology that will be used for human gene therapy

in the future.

16.18 Two men claim to be the father of baby Joyce Doe. Joyce’s

mother had her CODIS STR DNA prole analyzed and

was homozygous for allele 8 at the TPOX locus (allele 8

contains 8 repeats of the GAAT sequence at this poly-

morphic locus). Baby Joyce is heterozygous for alleles 8

and 11 at this locus. In an attempt to resolve the disputed

paternity, the two men were tested for their STR DNA

proles at the TPOX locus on chromosome 2. Putative

father 1 was heterozygous for alleles 8 and 11 at the

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TPOX locus, and putative father 2 was homozygous for

allele 11 at this locus. Can these results resolve this case

of disputed paternity. If so, who is the biological father? If

not, why not?

ANS: The results cannot resolve the case of disputed paternity.

Baby Joyce received allele 8 at the TPOX locus from her

mother. Given that baby Joyce is heterozygous for alleles

8 and 11 at the TPOX locus, she must have received

allele 11 from her father. However, both of the putative

fathers carry allele 11 and could have passed it on to baby

Joyce.

16.19 Many valuable human proteins contain carbohydrate or

lipid components that are added posttranslationally. Bac-

teria do not contain the enzymes needed to add these

components to primary translation products. How might

these proteins be produced using transgenic animals?

ANS: Posttranslationally modied proteins can be produced in

transgenic eukaryotic cells growing in culture or in

transgenic plants and animals. Indeed, transgenic sheep

have been produced that secrete human blood-clotting

factor IX and a1-antitrypsin in their milk. These sheep

were produced by fusing the coding sequences of the

respective genes to a DNA sequence that encodes the

signal peptide required for secretion and introducing

this chimeric gene into fertilized eggs that were then

implanted and allowed to develop into transgenic ani-

mals. In principle, this approach could be used to pro-

duce any protein of interest.

16.20 Richard Meagher and coworkers have cloned a family of

10 genes that encode actins (a major component of the

cytoskeleton) in Arabidopsis thaliana. The 10 actin gene

products are similar, often differing by just a few amino

acids. Thus, the coding sequences of the 10 genes are

also very similar, so that the coding region of one gene

will cross-hybridize with the coding regions of the other

nine genes. In contrast, the noncoding regions of the 10

genes are quite divergent. Meagher has hypothesized

that the 10 actin genes exhibit quite different temporal

and spatial patterns of expression. You have been hired

by Meagher to test this hypothesis. Design experiments

that will allow you to determine the temporal and spatial

pattern of expression of each of the 10 actin genes in

Arabidopsis.

ANS: You can subclone the 3′ noncoding regions (sequences

between the translation-termination codons and the 3′

termini of the transcripts) of the actin genes and use

these sequences as gene-specic hybridization probes,

after verifying their specicity (no cross-hybridization to

the other genes in the gene family). This procedure has

worked elegantly for the 15 tubulin genes and the 10

actin genes of Arabidopsis, because the transcript

sequences are very divergent in the 5′ and 3′ noncoding

regions. You cannot, of course, use intron sequences

because they are excised during pre-mRNA processing.

These 3′ noncoding gene-specic hybridization probes

are then used to measure individual gene transcript lev-

els in various organs and tissues of developing plants by

either northern blot or in situ hybridization experiments.

Alternatively, the 5′ and 3′ noncoding regions can be

used to design gene-specic PCR primers, and reverse

transcription PCR (RT-PCR) can be used to measure

individual gene transcript levels in organs and tissues.

Indeed, Meagher and colleagues have already used this

approach to document the striking temporal and tissue-

specic patterns of actin gene expression in Arabidopsis.

16.21 The rst transgenic mice resulted from microinjecting

fertilized eggs with vector DNA similar to that dia-

grammed in Figure 16.15 except that it contained a pro-

moter for the mammalian metallothionein gene linked to

the HGH gene. The resulting transgenic mice showed

elevated levels of HGH in tissues of organs other than the

pituitary gland—for example, in heart, lung, and liver—

and the pituitary gland underwent atrophy. How might

the production of HGH in transgenic animals be better

regulated, with expression restricted to the pituitary gland?

ANS: The vector described contains the HGH gene; however,

it does not contain a mammalian HGH promoter that

will regulate the expression of the transgene in the

appropriate tissues. Construction of vectors containing a

properly positioned mammalian HGH-promoter

sequence should result in transgenic mice in which

HGH synthesis is restricted to the pituitary gland.

16.22 How do the reverse genetic approaches used to dissect

biological processes differ from classical genetic

approaches?

ANS: Classical genetic approaches use mutational dissection to

probe the functions of genes. Mutant alleles that produce

altered phenotypes are identied and used to investigate

the functions of the wild-type alleles. Comparative

molecular analyses of mutant and wild-type organisms

sometimes allow researchers to determine the precise

function of a gene. Reverse genetic approaches use the

known nucleotide sequences of genes to design proce-

dures to either produce null mutations in them or inhibit

their expression. RNA interference protocols are used to

block the expression of specic genes. T-DNA and trans-

poson insertion protocols are used to produce null muta-

tions (to “knock out” the function) of specic genes.

Basically, in classical genetics, you start with mutant

alleles and hope to uncover the wild-type gene function;

in reverse genetics, you start with the wild-type gene and

generate mutant alleles.

16.23 How can RNAi gene silencing be used to determine the

function of genes?

ANS: RNAi involves the use of double-stranded RNAs, where

one strand is complementary to the mRNA and the

other strand is equivalent to the mRNA, to silence the

expression of target genes. RNAi makes use of the

RNA-induced silencing complex (RISC) to block gene

expression (see Figure 16.23).

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16.24 How do insertional mutagenesis approaches differ from

other reverse genetic approaches?

ANS: Insertional mutagenesis approaches produce mutant

alleles—usually null alleles, or “knockouts,”—by the

insertion of foreign DNA into genes. Other reverse

genetic approaches, for example, RNA interference,

inhibit the expression of the genes but leave them struc-

turally intact.

16.25 Insertional mutagenesis is a powerful tool in both plants

and animals. However, when performing large-scale

insertional mutagenesis, what major advantage do plants

have over animals?

ANS: Plants have an advantage over animals in that once inser-

tional mutations are induced they can be stored for long

periods of time and distributed to researchers as dormant

seeds.

16.26 We discussed the unfortunate effects of insertional muta-

genesis in the four boys who developed leukemia after

treatment of X-linked severe combined immunode-

ciency disease by gene therapy. How might this conse-

quence of gene therapy be avoided in the future? Do you

believe that the use of somatic-cell gene therapy to treat

human diseases can ever be made 100 percent risk-free?

Why? Why not?

ANS: The vectors used in somatic-cell gene therapy must not

insert themselves preferentially into important genes,

especially proto-oncogenes. Ideally, the vectors should

insert themselves into unessential regions of the human

genome. However, such vectors may not exist. At the

minimum, however, vectors should be used that insert

into the genome at random sites so that their chance of

inserting into an important regulatory gene such as a

proto-oncogene is very low. Gene therapy will probably

never be completely risk-free, because anomalous inser-

tion events will always occur at some low frequency. No

biological process is 100 percent accurate. However, the

potential benets of gene therapy must outweigh the

potential risks before the procedure will become an

accepted tool for treating inherited diseases.

16.27 One strand of a gene in Arabidopsis thaliana has the fol-

lowing nucleotide sequence:

atgagtgacgggaggaggaagaagagcgtgaacggaggt

gcaccggcgcaaacaatcttggatgatcggagatctagtcttccgga

agttgaagcttctccaccggctgggaaacgagctgttatcaagagtgc

cgatatgaaagatgatatgcaaaaggaagctatcgaaatcgccatct

ccgcgtttgagaagtacagtgtggagaaggatatagctgagaatataa

agaaggagtttgacaagaaacatggtgctacttggcattgcattgttgg

tcgcaactttggttcttatgtaacgcatgagacaaaccatttcgtttacttct

acctcgaccagaaagctgtgctgctcttcaagtcgggttaa

The function(s) of this gene is still uncertain. (a) How

might insertional mutagenesis be used to investigate the

function(s) of the gene? (b) Design an experiment using

RNA interference to probe the function(s) of the gene.

ANS: (a) You would rst want to check the Salk Institute’s

Genome Analysis Laboratory web site to see if a T-DNA

or transposon insertion has already been identied in

this gene (see Problem 16.28). If so, you can simply order

seeds of the transgenic line from the Arabidopsis Biologi-

cal Resource Center at Ohio State University. If no

insertion is available in the gene, you can determine

where it maps in the genome and use transposons that

preferentially jump to nearby sites to identify a new

insertional mutation (see http://www.arabidopsis.org/

abrc/ima/jsp). (b) You can construct a gene that has sense

and antisense sequences transcribed to a single mRNA

molecule (see Figure 16.23b), introduce it into Arabidop-

sis plants by A. tumefaciens-mediated transformation, and

study its effect(s) on the expression of the gene and the

phenotype of transgenic plants. The transcript will form

a partially base-paired hairpin that will enter the RISC

silencing pathway and block the expression of the gene

(see Figure 16.23b).

16.28 Let’s check the Salk Institute’s Genome Analysis Labora-

tory web site (http://signal.salk.edu/cgi-bin/tdnaexpress)

to see if any of their T-DNA lines have insertions in the

gene shown in the previous question. At the SIGnAL

web site, scroll down to “Blast” and paste or type the

sequence in the box. The resulting map will show the

location of mapped T-DNA insertions relative to

the location of the gene (green rectangle at the top). The

blue arrows at the top right will let you focus on just the

short region containing the gene or relatively long

regions of chromosome 4 of Arabidopsis. Are there any

T-DNA insertions in the gene in question? Near the

gene?

ANS: A search of the Salk Institute’s Genome Analysis Labora-

tory (SIGnAL) web site reveals that there are numerous

T-DNA insertions (Salk T-DNA insertions and several

others) located in the promoter region of this gene.

Moreover, there are two insertion lines (FLAG_177C02

and FLAG_216D01) in the collection at the Institute of

Agronomic Research in Versailles, France, with inserts in

the coding region of this gene. All of these insertion lines

in these collections are available to researchers on

request. Thus, the function of this gene can be studied by

using these insertion lines.

16.29 The CRISPR/Cas9 anti-phage immunity system in Strep-

tococcus pyogenes deploys a variety of crRNAs derived from

the spacer and repeat sequences in the CRISPR array in

the S. pyogenes genome. In combination with the transac-

tivating RNA (tracrRNA), these crRNAs guide the Cas9

endonuclease to complementary sequences in infecting

phage genomes, whereupon Cas9 cleaves the phage DNA.

A requirement for cleavage is that the targeted phage

DNA sequence be immediately upstream of a protospacer

adjacent motif (PAM), which in the S. pyogenes system is

5′-NGG-3′. Why is it important that the CRISPR array

in the S. pyogenes genome not contain this PAM?

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ANS: It is important that the CRISPR array in the S. pyogenes

genome does not contain the PAM 5′NGG-3′ because if

it did, crRNAs generated from the array could target the

Cas9 endonuclease to the array and cleave it, resulting in

breakage of the S. pyogenes chromosome.

16.30 The Streptococcus pyogenes Cas9 endonuclease can be tar-

geted to a specic genomic DNA sequence by an sgRNA

that at its 5′ end has 20 nucleotides complementary to the

target sequence. If this target sequence is immediately

upstream of the protospacer adjacent motif (PAM)

5′NGG-3′, Cas9 will cleave the target DNA. Suppose you

have chosen a 20-nucleotide target sequence in the genome

of Drosophila melanogaster and that this sequence is next to

the required PAM. How could you determine if Cas9 will

cleave only this sequence in the Drosophila genome?

ANS: You could use the 20-nucleotide sequence as a query in

BLAST to scan the sequenced portion of the Drosophila

genome to see if all or part of the target sequence is pres-

ent anywhere else. If this sequence is present elsewhere,

then you could check to see if the PAM 5′-NGG-3′ is

immediately downstream of the sequence. If it is, then

the Cas9 endonuclease will cleave at this site as well as at

the intended target site.

16.31 How could the CRISPR/Cas9 system be used to create a

translocation between two autosomes in cultured human

cells?

ANS: Create two sgRNAs, one to target a sequence on a par-

ticular autosome and the other to target a sequence

on a different autosome. Then introduce these sgRNAs

and the Cas9 endonuclease into cultured cells to induce

breakage at the two target sites. The broken DNA mol-

ecules may be repaired by the NHEJ pathway, and if they

are, the broken pieces of different autosomes could be

joined covalently, creating a reciprocal translocation.

Chapter 17

17.1 How can inducible and repressible enzymes of microor-

ganisms be distinguished?

ANS: By studying the synthesis or lack of synthesis of the

enzyme in cells grown on chemically dened media. If

the enzyme is synthesized only in the presence of a cer-

tain metabolite or a particular set of metabolites, it is

probably inducible. If it is synthesized in the absence but

not in the presence of a particular metabolite or group of

metabolites, it is probably repressible.

17.2 Distinguish between (a) repression and (b) feedback

inhibition caused by the end-product of a biosynthetic

pathway. How do these two regulatory phenomena com-

plement each other to provide for the efcient regula-

tion of metabolism?

ANS: Repression occurs at the level of transcription during

enzyme synthesis. The end-product, or a derivative of

the end-product, of a repressible system acts as an

effector molecule that usually, if not always, combines

with the product of one or more regulator genes to turn

off the synthesis of the enzymes in the biosynthetic path-

way. Feedback inhibition occurs at the level of enzyme

activity; it usually involved the rst enzyme of the bio-

synthetic pathway. Feedback inhibition thus brings about

an immediate arrest of the biosynthesis of the end-

product. All together, feedback inhibition and repression

rapidly and efciently turn off the synthesis of both the

enzymes and the end-products that no longer need to be

synthesized by the cell.

17.3 In the lactose operon of E. coli, what is the function of

each of the following genes or sites: (a) regulator, (b)

operator, (c) promoter, (d) structural gene Z, and (e)

structural gene Y?

ANS:

Gene or Regulatory

Element Function

(a) Regulator gene Encodes the repressor

(b) Operator Binding site of repressor

polymerase and CAP-cAMP

complex

(d) Structural gene Z Encodes b-galactosidase

(e) Structural gene Y Encodes b-galactoside

permease

17.4 What would be the result of inactivation by mutation of

the following genes or sites in the E. coli lactose operon:

(a) regulator, (b) operator, (c) promoter, (d) structural

gene Z, and (e) structural gene Y?

ANS: (a) Constitutive synthesis of the lac enzymes.

(b) Constitutive synthesis of the lac enzymes.

(d) No b-galactosidase activity.

(e) No b-galactoside permease activity.

17.5 Groups of alleles associated with the lactose operon are

as follows (in order of dominance for each allelic series):

repressor, I

(superrepressor), I



(inducible), and I



(con-

stitutive); operator, O

(constitutive, cis



dominant) and



(inducible, cis-dominant); structural, Z



and Y



. (a)

Which of the following genotypes will produce b-galac-

tosidase and b-galactoside permease if lactose is present:

(1) I



, (2) I





, (3) I



, (4) I



and (5) I





? (b) Which of the above genotypes

will produce b-galactosidase and b-galactoside permease

if lactose is absent? Why?

ANS: (a) 1, 2, 3, and 5. Genotype 1 is wild-type and inducible,

whereas genotypes 2, 3, and 5 are constitutive; all except 4

will produce b-galactosidase and b-galactoside permease in

the presence of lactose. However, genotype 4 has a super-

repressor mutation (I

) and is uninducible with normal

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levels of lactose. (b) 2, 3, and 5. In genotypes 2 and 3, the

repressor cannot bind to O

and in genotype 5, no repressor

is made; both situations render the operon constitutive.

17.6 Assume that you have discovered a new strain of E. coli

that has a mutation in the lac operator region that causes

the wild-type repressor protein to bind irreversibly

to the operator. You have named this operator mutant

for “superbinding” operator. (a) What phenotype

would a partial diploid of genotype I









have with respect to the synthesis of the enzymes

b-galactosidase and b-galactoside permease? (b) Does

your new O

mutation exhibit cis or trans dominance in

its effects on the regulation of the lac operon?

ANS: (a) b-Galactosidase will be produced only when lactose is

present. Permease will not be produced at all. (b) cis

dominance.

17.7 Why is the O

mutation in the E. coli lac operon epistatic

to the I

mutation?

ANS: The O

mutant prevents the repressor from binding to

the operator. The I

mutant repressor cannot bind to O

The I

mutant protein has a defect in the allosteric site

that binds allolactose but has a normal operator binding

site. Therefore, because the single O

mutant would have

the same phenotype as the O

double mutant, the O

mutation is, by denition, epistatic to I

17.8 For each of the following partial diploids indicate

whether enzyme synthesis is constitutive or inducible

(see Problem 17.5 for dominance relationships):

(a) I



, (b) I



, (d) I



/IO



(e) I





/IO



Why?

ANS: (a) Inducible, this is the wild-type genotype and

phenotype.

(b) Constitutive, the O

mutation produces an operator

that is not recognized by the lac repressor.

(d) Inducible, I



is dominant to I



(e) Constitutive, no active repressor is synthesized in this

bacterium.

17.9 Write the partial diploid genotype for a strain that will

(a) produce b-galactosidase constitutively and permease

inducibly and (b) produce b-galactosidase constitutively

but not permease either constitutively or inducibly, even

though a Y



gene is known to be present.

ANS: (a) I







(b) I







17.10 As a genetics historian, you are repeating some of the

classic experiments conducted by Jacob and Monod with

the lactose operon in E. coli. You use an F plasmid to

construct several E. coli strains that are partially diploid

for the lac operon. You construct strains with the follow-

ing genotypes: (1) I











, (2) I









, (3) I







, (4) I







, and (5) I







. (a) Which of

these strains will produce functional b-galactosidase in

both the presence and absence of lactose? (b) Which of

these strains will exhibit constitutive synthesis of func-

tional b-galactoside permease? (c) Which of these strains

will express both gene Z and gene Y constitutively and

will produce functional products (b-galactosidase and

b-galactoside permease) of both genes? (d) Which of

these strains will show cis dominance of lac operon regu-

latory elements? (e) Which of these strains will exhibit

trans dominance of lac operon regulatory elements?

ANS: (a) 1, 5.

(b) 2, 5.

(d) 1, 2, 5.

(e) 3, 4.

17.11 Constitutive mutations produce elevated enzyme levels

at all times; they may be of two types: O

or I



. Assume

that all other DNA present is wild-type. Outline how the

two constitutive mutants can be distinguished with

respect to (a) map position, (b) regulation of enzyme

levels in O



versus I





partial diploids, and (c) the

position of the structural genes affected by an O

muta-

tion versus the genes affected by an I



mutation in a par-

tial diploid.

ANS: (a) The O

mutations map very close to the Z structural

gene; I

−

mutations map slightly farther from the struc-

tural gene (but still very close by; see Figure 17.5).

(b) An I



partial diploid would exhibit

constitutive synthesis of b-galactosidase and b-galacto-

side permease, whereas an I







partial

diploid would be inducible for the synthesis of these

enzymes.

mutation is cis-dominant; the I



mutation is

trans-recessive.

17.12 How could the tryptophan operon in E. coli have devel-

oped and been maintained by evolution?

ANS: The system could have developed from a series of tan-

dem duplications of a single ancestral gene. Mutational

changes that make the system more efcient and, there-

fore, favored by natural selection could have brought the

system to its present level of efciency.

17.13 Of what biological signicance is the phenomenon of

catabolite repression?

ANS: Catabolite repression has apparently evolved to assure

the use of glucose as a carbon source when this carbohy-

drate is available, rather than less efcient energy sources.

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17.14 How might the concentration of glucose in the medium

in which an E. coli cell is growing regulate the intracel-

lular level of cyclic AMP?

ANS: Possibly by directly or indirectly inhibiting the enzyme

adenylcyclase, which catalyzes the synthesis of cyclic

AMP from ATP.

17.15 Is the CAP–cAMP effect on the transcription of the lac

operon an example of positive or negative regulation? Why?

ANS: Positive regulation; the CAP–cAMP complex has a posi-

tive effect on the expression of the lac operon. It func-

tions in turning on the transcription of the structural

genes in the operon.

17.16 Would it be possible to isolate E. coli mutants in which

the transcription of the lac operon is not sensitive to

catabolite repression? If so, in what genes might the

mutations be located?

ANS: Yes; in the gene encoding CAP. Some mutations in this

gene might result in a CAP that binds to the promoter in

the absence of cAMP. Also, mutations in the gene (or

genes) coding for the protein (or proteins) that regulate

the cAMP level as a function of glucose concentration.

17.17 Using examples, distinguish between negative regulatory

mechanisms and positive regulatory mechanisms.

ANS: Negative regulatory mechanisms, such as that involving

the repressor in the lactose operon, block the transcrip-

tion of the structural genes of the operon, whereas posi-

tive mechanisms, such as the CAP–cAMP complex in the

lac operon, promote the transcription of the structural

genes of the operon.

17.18 The following table gives the relative activities of the

enzymes b-galactosidase and b-galactoside permease in

cells with different genotypes at the lac locus in E. coli.

The level of activity of each enzyme in wild-type E. coli

not carrying F’s was arbitrarily set at 100; all other values

are relative to the observed levels of activity in these

wild-type bacteria. Based on the data given in the table

for genotypes 1 through 4, ll in the levels of enzyme

activity that would be expected for the fth genotype.

-Galactoside

-Galactosidase

Permease

Genotype - Inducer  Inducer - Inducer  Inducer

1. I



0.1 100 0.1 100

2. I





100 100 100 100

3. I



25 100 25 100

4. I







F I





200 200 100 100

5. I





F I



ANS: 0.1; 100; 25.1; 200.

17.19 The rate of transcription of the trp operon in E. coli

is controlled by both (1) repression/derepression and

(2) attenuation. By what mechanisms do these two regu-

latory processes modulate trp operon transcript levels?

ANS: Repression/derepression of the trp operon occurs at the

level of transcription initiation, modulating the fre-

quency at which RNA polymerase initiates transcription

from the trp operon promoters. Attenuation modulates

trp transcript levels by altering the frequency of termina-

tion of transcription within the trp operon leader region

(trpL).

17.20 What effect will deletion of the trpL region of the trp

operon have on the rates of synthesis of the enzymes

encoded by the ve genes in the trp operon in E. coli cells

growing in the presence of tryptophan?

ANS: Deletion of the trpL region would result in the levels of

the tryptophan biosynthetic enzymes in cells growing in

the presence of tryptophan being increased about 10-fold

because attenuation would no longer occur if this region

were absent.

17.21 By what mechanism does the presence of tryptophan in

the medium in which E. coli cells are growing result in

premature termination or attenuation of transcription of

the trp operon?

ANS: First, remember that transcription and translation are

coupled in prokaryotes. When tryptophan is present in

cells, tryptophan-charged tRNA

Trp

is produced. This

allows translation of the trp leader sequence through the

two UGG Trp codons to the trp leader sequence UGA

termination codon. This translation of the trp leader

region prevents base-pairing between the partially com-

plementary mRNA leader sequences 75–83 and 110–121

(see Figure 17.15b), which in turn permits formation of

the transcription–termination “hairpin” involving leader

sequences 110–121 and 126–134 (see Figure 17.15c).

17.22 Suppose that you used site-specic mutagenesis to mod-

ify the trpL sequence such that the two UGG Trp codons

at positions 54–56 and 57–60 (see Figure 17.14) in the

mRNA leader sequence were changed to GGG Gly

codons. Will attenuation of the trp operon still be regu-

lated by the presence or absence of tryptophan in the

medium in which the E. coli cells are growing?

ANS: No. Attenuation of the trp operon would now be con-

trolled by the presence or absence of Gly-tRNA

Gly

17.23 What do trp attenuation and the lysine riboswitch have

in common?

ANS: Both trp attenuation and the lysine riboswitch turn off

gene expression by terminating transcription upstream

from the coding regions of the regulated genes. Both

involve the formation of alternative mRNA secondary

structures—switching between the formation of antiter-

minator and transcription–terminator hairpins—in

response to the presence or absence of a specic

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metabolite (compare Figure 17.15 and Figure 2 in the

Focus On The Lysine Riboswitch).

17.24 Would attenuation of the type that regulates the level of

trp transcripts in E. coli be likely to occur in eukaryotic

organisms?

ANS: No. Since transcription (nucleus) and translation (cyto-

plasm) are no coupled in eukaryotes, attenuation of the

type occurring in prokaryotes would not be possible.

Chapter 18

18.1 Operons are common in bacteria but not in eukaryotes.

Suggest a reason why.

ANS: In multicellular eukaryotes, the environment of an indi-

vidual cell is relatively stable. There is no need to respond

quickly to changes in the external environment. In addi-

tion, the development of a multicellular organism

involves complex regulatory hierarchies composed of

hundreds of different genes. The expression of these

genes is regulated spatially and temporally, often through

intricate intercellular signaling processes.

18.2 In bacteria, translation of an mRNA begins before the

synthesis of that mRNA is completed. Why is this “cou-

pling” of transcription and translation not possible in

eukaryotes?

ANS: Coupling of transcription and translation is not possible

in eukaryotes because these two processes take place in

different cellular compartments—transcription in the

nucleus and translation in the cytoplasm.

18.3 Muscular dystrophy in humans is caused by mutations in

an X-linked gene that encodes a protein called dystro-

phin. What techniques could you use to determine if this

gene is active in different types of cells, say skin cells,

nerve cells, and muscle cells?

ANS: Activity of the dystrophin gene could be assessed by blot-

ting RNA extracted from the different types of cells and

hybridizing it with a probe from the gene (northern

blotting); or the RNA could be reverse transcribed into

cDNA using one or more primers specic to the dystro-

phin gene and the resulting cDNA could be amplied by

the polymerase chain reaction (RT-PCR). Another tech-

nique would be to hybridize dystrophin RNA in situ—that

is, in the cells themselves—with a probe from the gene.

It would also be possible to check each cell type for pro-

duction of dystrophin protein by using anti-dystrophin

antibodies to analyze proteins from the different cell

types on western blots or analyze the proteins in the cells

themselves—that is, in situ.

18.4 Why do steroid hormones interact with receptors inside

the cell, whereas peptide hormones interact with recep-

tors on the cell surface?

ANS: Steroid hormones are small, lipid-soluble molecules that

have little difculty passing through the cell membrane.

Peptide hormones are typically too large to pass through

the cell membrane freely; rather, their inuence must be

mediated by a signaling system that begins with a mem-

brane-bound receptor protein that binds the hormone.

18.5 In the polytene chromosomes of Drosophila larvae (Chap-

ter 6), some bands form large “puffs” when the larvae are

subjected to high temperatures. How could you show

that these puffs contain genes that are vigorously tran-

scribed in response to this heat shock treatment?

ANS: One procedure would be to provide larvae with radioac-

tively labeled UTP, a building block of RNA, under dif-

ferent conditions—with and without heat shock. Then

prepare samples of polytene cells from these larvae for

autoradiography. If the heat shock-induced puffs contain

genes that are vigorously transcribed, the radioactive sig-

nal should be abundant in the puffs.

18.6 How would you distinguish between an enhancer and a

promoter?

ANS: An enhancer can be located upstream, downstream, or

within a gene, and it functions independently of its ori-

entation. A promoter is almost always immediately

upstream of a gene and it functions only in one direction

with respect to the gene.

18.7 Tropomyosins are proteins that mediate the interaction

of actin and troponin, two proteins involved in muscle

contractions. In higher animals, tropomyosins exist as a

family of closely related proteins that share some amino

acid sequences but differ in others. Explain how these

proteins could be created from the transcript of a single

gene.

ANS: By alternate splicing of the transcript.

18.8 A polypeptide consists of three separate segments of

amino acids, A—B—C. Another polypeptide contains

segments A and C but not segment B. How might you

determine if these two polypeptides are produced by

translating alternately spliced versions of RNA from a

single gene or by translating mRNA from two different

genes?

ANS: Southern blotting of genomic DNA digested with an

appropriate restriction enzyme, followed by hybridiza-

tion of the blot with a probe containing the DNA encod-

ing segments A and B, or B and C, or at least parts of

these adjacent segments. If one DNA fragment is

detected on the blot, the two polypeptides are encoded

by a single gene whose RNA is alternately spliced to pro-

duce two mRNAs. If two DNA fragments are detected,

the two polypeptides are encoded by two different genes.

18.9 What techniques could be used to show that a plant gene

is transcribed when the plant is illuminated with light?

ANS: Northern blotting of RNA extracted from plants grown

with and without light, or PCR amplication of cDNA

made by reverse transcribing these same RNA extracts.

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18.10 When introns were rst discovered, they were thought

to be genetic “junk”—that is, sequences without any use-

ful function. In fact, they appeared to be worse than junk

because they actually interrupted the coding sequences

of genes. However, among eukaryotes, introns are perva-

sive and anything that is pervasive in biology usually has

a function. What function might introns have? What

benet might they confer on an organism?

ANS: Introns make it possible for genes to encode different—

but related—polypeptides by alternate splicing of their

RNA transcripts.

18.11 The GAL4 transcription factor in yeast regulates two

adjacent genes, GAL1 and GAL10, by binding to DNA

sequences between them. These two genes are tran-

scribed in opposite directions on the chromosome, one

to the left of the GAL4 protein’s binding site and the

other to the right of this site. What property of enhanc-

ers does this situation illustrate?

ANS: That enhancers can function in either orientation.

18.12 Using the techniques of genetic engineering, a researcher

has constructed a fusion gene containing the heat-shock

response elements from a Drosophila hsp70 gene and the

coding region of a jellysh gene (gfp) for green uores-

cent protein. This fusion gene has been inserted into the

chromosomes of living Drosophila by the technique of

transposon-mediated transformation (Chapter 21 on the

Instructor Companion site). Under what conditions will

the green uorescent protein be synthesized in these

genetically transformed ies? Explain.

ANS: The green uorescent protein will be made after the ies

are heat shocked.

18.13 Suppose that the segment of the hsp70 gene that was used

to make the hsp70/gfp fusion in the preceding problem

had mutations in each of its heat-shock response ele-

ments. Would the green uorescent protein encoded by

this fusion gene be synthesized in genetically trans-

formed ies?

ANS: Probably not unless the promoter of the gfp gene is recog-

nized and transcribed by the Drosophila RNA polymerase

independently of the heat-shock response elements.

18.14 The polypeptide products of two different genes, A and

B, each function as transcription factors. These polypep-

tides interact to form dimers: AA homodimers, BB

homodimers, and AB heterodimers. If the A and B poly-

peptides are equally abundant in cells, and if dimer for-

mation is random, what is the expected ratio of

homodimers to heterodimers in these cells?

ANS: With equal abundance of the A and B polypeptides, AA

homodimers should constitute 1/4 of the total dimers

formed, BB homodimers should constitute 1/4 of the

total, and AB heterodimers should constitute 1/2 of the

total. The expected ratio of homodimers to heterodimers

is therefore (1/4  1/4):(1/2)  1:1.

18.15 A particular transcription factor binds to enhancers in 40

different genes. Predict the phenotype of individuals homo-

zygous for a frameshift mutation in the coding sequence of

the gene that species this transcription factor.

ANS: The mutation is likely to be lethal in homozygous condi-

tion because the transcription factor controls so many

different genes, and a frameshift mutation in the coding

sequence will almost certainly destroy the transcription

factor’s function.

18.16 The alternately spliced forms of the RNA from the Dro-

sophila doublesex gene encode proteins that are needed to

block the development of one or the other set of sexual

characteristics. The protein that is made in female ani-

mals blocks the development of male characteristics, and

the protein that is made in male animals blocks the

development of female characteristics. Predict the phe-

notype of XX and XY animals homozygous for a null

mutation in the doublesex gene.

ANS: Both XX and XY animals would develop as intersexes

because neither of the forms of the doublesex protein

will be able to block sexual development. In these ani-

mals, both developmental pathways will be carried out,

leading to tissues that have both male and female

characteristics.

18.17 The RNA from the Drosophila Sex-lethal (Sxl) gene is

alternately spliced. In males, the sequence of the mRNA

derived from the primary transcript contains all eight

exons of the Sxl gene. In females, the mRNA contains

only seven of the exons because during splicing exon 3 is

removed from the primary transcript along with its

anking introns. The coding region in the female’s

mRNA is therefore shorter than it is in the male’s mRNA.

However, the protein encoded by the female’s mRNA is

longer than the one encoded by the male’s mRNA. How

might you explain this paradox?

ANS: Exon 3 contains an in-frame stop codon. Thus, the pro-

tein translated from the Sxl mRNA in males will be

shorter than the protein translated from the shorter Sxl

mRNA in females.

18.18 In Drosophila, expression of the yellow gene is needed for

the formation of dark pigment in many different tissues;

without this expression, a tissue appears yellow in color.

In the wings, the expression of the yellow gene is con-

trolled by an enhancer located upstream of the gene’s

transcription initiation site. In the tarsal claws, expres-

sion is controlled by an enhancer located within the

gene’s only intron. Suppose that by genetic engineering,

the wing enhancer is placed within the intron and the

claw enhancer is placed upstream of the transcription

initiation site. Would a y that carried this modied yel-

low gene in place of its natural yellow gene have darkly

pigmented wings and claws? Explain.

ANS: Yes. Enhancers are able to function in different positions

in and around a gene.

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18.19 A researcher suspects that a 550-bp-long intron contains

an enhancer that drives expression of an Arabidopsis gene

specically in root tip tissue. Outline an experiment to

test this hypothesis.

ANS: The intron could be placed in a GUS expression vector,

which could then be inserted into Arabidopsis plants. If

the intron contains an enhancer that drives gene expres-

sion in root tips, transgenic plants should show GUS

expression in their root tips. See the Problem-Solving

Skills feature in Chapter 18 for an example of this type of

analysis.

18.20 What is the nature of each of the following classes of

enzymes? What does each type of enzyme do to chroma-

tin? (a) HATs, (b) HDACs, (c) HMTs.

ANS: (a) HATs, histone acetyl transferases, transfer acetyl

groups to the amino acid lysine in histones; (b) HDACs,

histone deacetylases, remove acetyl groups from these

lysines; (c) HMTs, histone methyl transferases, transfer

methyl groups to lysine, arginine, and histidine in

histones.

18.21 In Drosophila larvae, the single X chromosome in males

appears diffuse and bloated in the polytene cells of the

salivary gland. Is this observation compatible with the

idea that X-linked genes are hyperactivated in Drosophila

males?

ANS: Yes. The diffuse, bloated appearance indicates that the

genes on this chromosome are being transcribed

vigorously—the chromatin is “open for business.”

18.22 Suppose that the LCR of the b-globin gene cluster was

deleted from one of the two chromosomes 11 in a man.

What disease might this deletion cause?

ANS: The LCR regulates the expression of all the genes linked

to it. Deletion of the LCR would ablate or impair globin

gene expression from one of the two chromosomes 11.

With less b-globin being produced, the individual would

likely suffer from anemia.

18.23 Would double-stranded RNA derived from an intron be

able to induce RNA interference?

ANS: Short interfering RNAs target messenger RNA mole-

cules, which are devoid of introns. Thus, if siRNA were

made from double-stranded RNA derived from an

intron, it would be ineffective against an mRNA target.

18.24 An RNA interference-like phenomenon has been impli-

cated in the regulation of transposable elements. In Dro-

sophila, two of the key proteins involved in this regulation

are encoded by the genes aubergine and piwi. Flies that

are homozygous for mutant alleles of these genes are

lethal or sterile, but ies that are heterozygous for them

are viable and fertile. Suppose that you have strains of

Drosophila that are heterozygous for aubergine or piwi

mutant alleles. Why might the genomic mutation rate in

these mutant strains be greater than the genomic muta-

tion rate in a wild-type strain?

ANS: The aubergine and piwi gene products mediate the RNAi-

like response. Reduction in the amount of aubergine or

piwi protein would likely impair the organism’s ability to

mount this response, and without a vigorous capacity for

regulation, transposable elements would be more likely

to move in the genome. This movement would likely

cause mutations because the transposons could insert

into genes and inactivate them. Thus, Drosophila that are

heterozygous for mutations in aubergine or piwi might

experience higher mutation rates than Drosophila lacking

these mutations.

18.25 Suppose that female mice homozygous for the a allele of

the Igf2 gene are crossed to male mice homozygous for

the b allele of this gene. Which of these two alleles will

be expressed in the F

progeny?

ANS: The paternally contributed allele (b) will be expressed in

the F

progeny.

18.26 Epigenetic states are transmitted clonally through cell

division. What kinds of observations indicate that these

states can be reversed or reset?

ANS: Here are two observations that show reversal or resetting

of an epigenetic state: (1) For imprinted genes in mam-

mals, the epigenetic state can be reset when the gene

passes through the germ line of the opposite sex.

(2) Genes on the inactive X chromosome in mammals

are reactivated in the female germ line.

18.27 A researcher hypothesizes that in mice gene A is actively

transcribed in liver cells, whereas gene B is actively tran-

scribed in brain cells. Describe procedures that would

allow the researcher to test this hypothesis.

ANS: RNA could be isolated from liver and brain tissue.

Northern blotting or RT-PCR with this RNA could

then establish which of the genes (A or B) is transcribed

in which tissue. For northern blotting, the RNA samples

would be fractionated in a denaturing gel and blotted to

a membrane and then the RNA on the membrane would

be hybridized with gene-specic probes, rst for one

gene, then for the other (or the researcher could prepare

two separate blots and hybridize each one with a differ-

ent probe). For RT-PCR, the RNA samples would be

reverse transcribed into cDNA using primers specic for

each gene; then the cDNA molecules would be amplied

by standard PCR and the products of the amplications

would be fractionated by gel electrophoresis to deter-

mine which gene’s RNA was present in the original

samples.

18.28 Suppose that the hypothesis mentioned in the previous

question is correct and that gene A is actively transcribed

in liver cells, whereas gene B is actively transcribed in

brain cells. The researcher now extracts equivalent

amounts of chromatin from liver and brain tissues and

treats these extracts separately with DNase I for a lim-

ited period of time. If the DNA that remains after the

treatments is then fractionated by gel electrophoresis,

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transferred to a membrane by Southern blotting, and

hybridized with a radioactively labeled probe specic for

gene A, which sample (liver or brain) will be expected to

show the greater signal on the autoradiogram? Explain

your answer.

ANS: The sample of chromatin from brain tissue would be

expected to show the greater signal on a Southern blot

hybridized with a probe specic for gene A. The reason

is that this gene is not so well transcribed in brain cells;

consequently, it will be more resistant to digestion with

DNase I in chromatin derived from brain cells than in

chromatin derived from liver cells in which it is actively

transcribed (and therefore more open to digestion with

DNase I).

18.29 Why do null mutations in the msl gene in Drosophila have

no effect in females?

ANS: The msl gene is not functional in females.

18.30 Suppose that a woman carries an X chromosome in

which the XIST locus has been deleted. The woman’s

other X chromosome has an intact XIST locus. What

pattern of X-inactivation would be observed throughout

the woman’s body?

ANS: The X chromosome containing the intact XIST locus

will be silenced in all cases because that locus is located

with the X inactivation center (XIC) and aids in silencing

the inactive X.

18.31 In Drosophila, the variegated phenotype of the white mot-

tled allele is suppressed by a dominant autosomal muta-

tion that knocks out the function of the gene for

heterochromatin protein 1 (HP1), an important factor in

heterochromatin formation. Flies with the white mottled

allele and the suppressor mutation have an almost uni-

form red color in their eyes; without the suppressor

mutation, the eyes are mosaics of red and white tissue.

Can you suggest an explanation for the effect of the sup-

pressor mutation?

ANS: HP1, the protein encoded by the wild-type allele of the

suppressor gene, is involved in chromatin organization.

Perhaps this heterochromatic protein spreads from the

region near the inversion breakpoint in the chromosome

that carries the white mottled allele and brings about the

“heterochromatization” of the white locus. When HP1 is

depleted by knocking out one copy of the gene encoding

it—that is, by putting the suppressor mutation into the

y’s genotype, the “heterochromatization” of the white

locus would be less likely to occur, and perhaps not occur

at all. The white locus would then function fully in all

eye cells, producing a uniform red eye color.

18.32 The sheep Dolly (Chapter 2) was the rst cloned mam-

mal. Dolly was created by implanting a nucleus from a

cell taken from the udder of a female sheep into an enu-

cleated egg. This nucleus had two X chromosomes, and

because it came from a differentiated cell, one of them

must have been inactivated. If the udder cell was

heterozygous for at least one X-linked gene whose

expression you could assay, how could you determine if

all of Dolly’s cells had the same X chromosome inacti-

vated? If, upon testing, Dolly’s cells prove to be mosaic

for X chromosome activity—that is, different X’s are

active in different clones of cells—what must have hap-

pened during her embryological development?

ANS: Take samples of cells from Dolly and determine if the

products of both alleles of the X-linked gene are present

in them. If the products of both alleles are present, then

Dolly must be a genetic mosaic for X chromosome

activity. Thus, during her development, the pattern of

X-inactivation that existed in the udder cell from which

she was derived must have been reset. If only one of the

gene’s products is detected—and if the sample of cells is

representative of all Dolly’s cells—then Dolly must have

maintained the pattern of X-inactivation that existed in

the udder cell from which she was derived.

Chapter 19

19.1 If heart disease is considered to be a threshold trait, what

genetic and environmental factors might contribute to

the underlying liability for a person to develop this

disease?

ANS: Some of the genes implicated in heart disease are listed

in Table 19.2. Environmental factors might include diet,

amount of exercise, and whether or not the person

smokes.

19.2 A wheat variety with red kernels (genotype A′A′ B′B′)

was crossed with a variety with white kernels (genotype

AA BB). The F

were intercrossed to produce an F

. If

each primed allele increases the amount of pigment in

the kernel by an equal amount, what phenotypes will be

expected in the F

? Assuming that the A and B loci assort

independently, what will the phenotypic frequencies be?

ANS: 1/16 red; 4/16  1/4 dark pink; 6/16 pink; 4/16  1/4

light pink; 1/16 white.

19.3 For alcoholism, the concordance rate for monozygotic

twins is 55 percent, whereas for dizygotic twins, it is

28 percent. Do these data suggest that alcoholism has a

genetic basis?

ANS: The concordance for monozygotic twins is almost twice

as great as that for dizygotic twins. Monozygotic twins

share twice as many genes as dizygotic twins. The data

strongly suggest that alcoholism has a genetic basis.

19.4 The height of the seed head in wheat at maturity is deter-

mined by several genes. In one variety, the head is just

9 inches above the ground; in another, it is 33 inches

above the ground. Plants from the 9-inch variety were

crossed to plants from the 33-inch variety. Among the F

the seed head was 21 inches above the ground. After self-

fertilization, the F

plants produced an F

population in

which 9-inch and 33-inch plants each appeared with a

frequency of 1/256. (a) How many genes are involved in

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the determination of seed head height in these strains of

wheat? (b) How much does each allele of these genes

contribute to seed head height? (c) If a 21-inch F

plant

were crossed to a 9-inch plant, how often would you

expect 18-inch wheat to occur in the progeny?

ANS: (a) 4; (b) 3 inches; (c) frequency of 1/4.

19.5 Assume that size in rabbits is determined by genes with

equal and additive effects. From a total of 2012 F

prog-

eny from crosses between true-breeding large and small

varieties, eight rabbits were as small as the small variety

and eight were as large as the large variety. How many

size-determining genes were segregating in these

crosses?

ANS: Because 8/2012 is approximately 1/256  (1/4)

, it appears

that four size-determining genes were segregating in the

crosses.

19.6 A sample of 20 plants from a population was measured in

inches as follows: 18, 21, 20, 23, 20, 21, 20, 22, 19, 20, 17,

21, 20, 22, 20, 21, 20, 22, 19, and 23. Calculate (a) the

mean, (b) the variance, and (c) the standard deviation.

ANS: (a) The mean is 20.45 inches. (b) The variance is 2.37

inches

. (c) The standard deviation is 1.54 inches.

19.7 Quantitative geneticists use the variance as a measure of

scatter in a sample of data; they calculate this statistic by

averaging the squared deviations between each measure-

ment and the sample mean. Why don’t they simply mea-

sure the scatter by computing the average of the

deviations without bothering to square them?

ANS: Because S(X

− mean)  0.

19.8 Two inbred strains of corn were crossed to produce an

, which was then intercrossed to produce an F

. Data

on ear length from a sample of F

and F

individuals gave

phenotypic variances of 15.2 cm

and 27.6 cm

, respec-

tively. Why was the phenotypic variance greater for the

than for the F

ANS: For the F

, V

 0 because they are all genetically identi-

cal and heterozygous; for the F

, V

> 0 because genetic

differences result from the segregation and independent

assortment of genes. Thus, in the F

, the phenotypic

variance has a pronounced genetic component.

19.9 A study of quantitative variation for abdominal bristle

number in female Drosophila yielded estimates of V



6.08, V

 3.17, and V

 2.91. What was the broad-sense

heritability?

ANS: 3.17/6.08  0.52

19.10 A researcher has been studying kernel number on ears of

corn. In one highly inbred strain, the variance for kernel

number is 426. Within this strain, what is the broad-

sense heritability for kernel number?

ANS: The broad-sense heritability within a highly inbred

strain is expected to be zero because there is no genetic

variability.

19.11 Measurements on ear length were obtained from three

populations of corn—two inbred varieties and a ran-

domly pollinated population derived from a cross

between the two inbred strains. The phenotypic vari-

ances were 9.2 cm

and 9.6 cm

for the two inbred variet-

ies and 26.4 cm

for the randomly pollinated population.

Estimate the broad-sense heritability of ear length for

these populations.

ANS: V

is estimated by the average of the variances of the

inbreds: 9.4 cm

. V

is estimated by the difference between

the variances of the randomly pollinated population and

the inbreds: (26.4  9.4)  17.0 cm

. The broad-sense

heritability is H

 V

 17.0/26.4  0.64.

19.12 Figure 19.4 summarizes data on maturation time in pop-

ulations of wheat. Do these data provide any insight as to

whether or not this trait is inuenced by dominance?

Explain.

ANS: Because the F

plants have maturation times midway

between those of the parental strains, there seems to be

little or no dominance for this trait.

19.13 A quantitative geneticist claims that the narrow-sense

heritability for body mass in human beings is 0.7, while

the broad-sense heritability is only 0.3. Why must there

be an error?

ANS: Broad-sense heritability must be greater than narrow-

sense heritability because H

 V

> V

 h

19.14 The mean value of a trait is 100 units, and the narrow-

sense heritability is 0.4. A male and a female measuring

124 and 126 units, respectively, mate and produce a large

number of offspring, which are reared in an average

environment. What is the expected value of the trait

among these offspring?

ANS: (125  100)(0.4)  100  110 units.

19.15 The narrow-sense heritability for abdominal bristle

number in a population of Drosophila is 0.3. The mean

bristle number is 12. A male with 10 bristles is mated to

a female with 20 bristles, and a large number of progeny

are scored for bristle number. What is the expected num-

ber of bristles among these progeny?

ANS: (15  12)(0.3)  12  12.9 bristles.

19.16 A breeder is trying to decrease the maturation time in a

population of sunowers. In this population, the mean

time to owering is 100 days. Plants with a mean ower-

ing time of only 90 days were used to produce the next

generation. If the narrow-sense heritability for owering

time is 0.2, what will the average time to owering be in

the next generation?

ANS: (90  100)(0.2)  100  98 days.

19.17 A sh breeder wishes to increase the rate of growth in a

stock by selecting for increased length at 6 weeks after

hatching. The mean length of 6-week-old ngerlings is

currently 10 cm. Adult sh that had a mean length of

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15cm at 6 weeks of age were used to produce a new gen-

eration of ngerlings. Among these, the mean length was

12.5 cm. Estimate the narrow-sense heritability of n-

gerling length at 6 weeks of age and advise the breeder

about the feasibility of the plan to increase growth rate.

ANS: h

 R/S  (12.5  10)/(15  10)  0.5; selection for

increased growth rate should be effective.

19.18 Leo’s IQ is 86 and Julie’s IQ is 110. The mean IQ in the

population is 100. Assume that the narrow-sense herita-

bility for IQ is 0.4. What is the expected IQ of Leo and

Julie’s rst child?

ANS: (98  100)(0.4)  100  99.2.

19.19 One way to estimate a maximum value for the narrow-

sense heritability is to calculate the correlation between

half-siblings that have been reared apart and divide it by

the fraction of genes that half-siblings share by virtue of

common ancestry. A study of human half-siblings found

that the correlation coefcient for height was 0.14. From

this result, what is the maximum value of the narrow-

sense heritability for height in this population?

ANS: Half-siblings share 25 percent of their genes. The maxi-

mum value for h

is therefore 0.14/0.25  0.56.

19.20 A selection differential of 40 mg per generation was used

in an experiment to select for increased pupa weight in

Tribolium. The narrow-sense heritability for pupa weight

was estimated to be 0.3. If the mean pupa weight was

initially 2000 mg and selection was practiced for 10 gen-

erations, what was the mean pupa weight expected to

become?

ANS: The response to selection in one generation is R  h

S 

(0.3)(40 mg)  12 mg. The cumulative effect over 10 gen-

erations is therefore 10  12 mg  120 mg. Thus, the

mean pupa weight should become 2120 mg.

19.21 On the basis of the observed correlations for personality

traits shown in Table 19.5, what can you say about the

value of the environmentality (C

in Table 19.3)?

ANS: The correlations for MZT are not much different from

those for MZA. Evidently, for these personality traits,

the environmentality (C

in Table 19.3) is negligible.

19.22 Correlations between relatives provide estimates of the

broad and narrow-sense heritabilities on the assumption

that the genetic and environmental factors inuencing

quantitative traits are independent of each other and that

they do not interact in some peculiar way. In Chapter 18,

we considered epigenetic modications of chromatin

that regulate genes and noted the possibility that some of

these modications might be induced by environmental

factors. How could epigenetic inuences on complex

traits be incorporated into the basic theory of quantita-

tive genetics?

ANS: We might represent the value of a quantitative trait, T, as

m + g + e + eg, where m is the mean of the population, g is

the deviation due to genetic factors, e is the deviation due

to environmental factors, and eg is the deviation due to

epigenetic factors arising from the interaction of genetic

and environmental factors.

Chapter 20

20.1 The following data for the M–N blood types were

obtained from native villages in Central and North

America:

Group Sample Size M MN N

Central American 86 53 29 4

North American 278 78 61 139

Calculate the frequencies of the L

and L

alleles for the

two groups.

ANS: Frequency of L

in Central American population: p  (2 

53  29)/(2  86)  0.78; q  0.22. Frequency of L

North American population: p  (2  78  61)/(2  278) 

0.39; q  0.61.

20.2 The frequency of an allele in a large randomly mating

population is 0.2. What is the frequency of heterozygous

carriers?

ANS: 2pq  2(0.2)(0.8)  0.32.

20.3 The incidence of recessive albinism is 0.0004 in a human

population. If mating for this trait is random in the pop-

ulation, what is the frequency of the recessive allele?

ANS: q

 0.0004; q  0.02.

20.4 In a sample from an African population, the frequencies

of the LM and LN alleles were 0.78 and 0.22, respec-

tively. If the population mates randomly with respect to

the M–N blood types, what are the expected frequencies

of the M, MN, and N phenotypes?

ANS:

Phenotype Hardy–Weinberg Frequency

(0.78)

 0.61

2(0.78)(0.22)  0.34

(0.22)

 0.05

20.5 Human beings carrying the dominant allele T can taste

the substance phenylthiocarbamide (PTC). In a popula-

tion in which the frequency of this allele is 0.4, what is

the probability that a particular taster is homozygous?

ANS: Frequency of tasters (genotypes TT and Tt): (0.4)

 2

(0.4)(0.6)  0.64. Frequency of TT tasters among all

tasters: (0.4)

/(0.64)  0.25.

20.6 A gene has three alleles, A

, A

, and A

, with frequencies

0.6, 0.3, and 0.1, respectively. If mating is random, pre-

dict the combined frequency of all the heterozygotes in

the population.

ANS: Frequency of heterozygotes combined  2[(0.6)(0.3) 

(0.3)(0.1)  ((0.6)(0.1)]  0.54.

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20.7 Hemophilia is caused by an X-linked recessive allele. In a

particular population, the frequency of males with hemo-

philia is 1/4000. What is the expected frequency of

females with hemophilia?

ANS: (0.00025)

 6.25  10

8

20.8 In Drosophila, the ruby eye phenotype is caused by a

recessive, X-linked mutant allele. The wild-type eye

color is red. A laboratory population of Drosophila is

started with 25 percent ruby-eyed females, 25 percent

homozygous red-eyed females, 5 percent ruby-eyed

males, and 45 percent red-eyed males. (a) If this popula-

tion mates randomly for one generation, what is the

expected frequency of ruby-eyed males and females? (b)

What is the frequency of the recessive allele in each of

the sexes?

ANS: (a) Half the males will be ruby-eyed and 5 percent (0.50 

0.10  100 percent) of the females will be ruby-eyed.

(b) Among males, the frequency of the recessive allele

will be 0.5, which was its frequency among females in the

previous generation; among females, the frequency of

the recessive allele will be (0.5  0.1)/2  0.3, which is

the average of the frequencies of this allele in males and

females in the previous generation.

20.9 A trait determined by an X-linked dominant allele shows

100 percent penetrance and is expressed in 36 percent of

the females in a population. Assuming that the popula-

tion is in Hardy–Weinberg equilibrium, what proportion

of the males in this population express the trait?

ANS: In females, the frequency of the dominant phenotype is

0.36. The frequency of the recessive phenotype is 0.64  q

;

thus, q  0.8 and p  0.2. The frequency of the dominant

phenotype in males is therefore p  0.2.

20.10 A phenotypically normal couple has had one normal

child and a child with cystic brosis, an autosomal reces-

sive disease. The incidence of cystic brosis in the popu-

lation from which this couple came is 1/500. If their

normal child eventually marries a phenotypically normal

person from the same population, what is the risk that

the newlyweds will produce a child with cystic brosis?

ANS: The probability that the unaffected child of the couple is

a carrier of the mutant allele for cystic brosis is 2/3. The

probability that the mate of this individual is a carrier can

be determined by using the population incidence of the

disease. The mutant allele frequency is the square root of

the incidence—0.045—and the frequency of heterozy-

gotes under the assumption of random mating is 2 

0.045  (1  0.045)  0.086. If both individuals are car-

riers, the chance that they will have an affected child is

1/4. Putting all this analysis together, the risk for the

child to have cystic brosis is therefore 2/3  0.086 

1/4  0.014, which is seven times the incidence in the

population at large.

20.11 What frequencies of alleles A and a in a randomly mating

population maximize the frequency of heterozygotes?

ANS: Frequency of heterozygotes  H  2pq  2p(1  p).

Using calculus, take the derivative of H and set the result

to zero to solve for the value of p that maximizes H: dH/

dp  2  4p  0 implies that p  2/4  0.5.

20.12 In an isolated population, the frequencies of the I

, I

and i alleles of the A–B–O blood type gene are, respec-

tively, 0.15, 0.25, and 0.60. If the genotypes of the A–B–O

blood type gene are in Hardy–Weinberg proportions,

what fraction of the people who have type A blood in this

population is expected to be homozygous for the I

allele?

ANS: In a Hardy–Weinberg population, the frequency of I

homozygotes is (0.15)

 0.0335 and the frequency of I

heterozygotes is 2  0.15  0.60  0.18. The sum of

these frequencies—0.2025—is the frequency of individ-

uals with type A blood. Thus, the frequency of I

homozygotes among all individuals with type A blood is

0.0225/0.2025  0.1111.

20.13 In a survey of moths collected from a natural population,

a researcher found 51 dark specimens and 49 light speci-

mens. The dark moths carry a dominant allele, and the

light moths are homozygous for a recessive allele. If the

population is in Hardy–Weinberg equilibrium, what is

the estimated frequency of the recessive allele in the

population? How many of the dark moths in the sample

are likely to be homozygous for the dominant allele?

ANS: Under the assumption that the population is in Hardy–

Weinberg equilibrium, the frequency of the allele for

light coloration is the square root of the frequency of

recessive homozygotes. Thus, q  √0.49  0.7, and the

frequency of the allele for dark color is 1  q  p  0.3.

From p

 0.09, we estimate that 0.09  100  9 of

the dark moths in the sample are homozygous for the

dominant allele.

20.14 A population of Hawaiian Drosophila is segregating two

alleles, P

and P

, of the phosphoglucose isomerase (PGI)

gene. In a sample of 100 ies from this population, 30

were P

homozygotes, 60 were P

heterozygotes, and

10 were P

homozygotes. (a) What are the frequencies

of the P

and P

alleles in this sample? (b) Perform

a chi-square test to determine if the genotypes in the

sample are in Hardy–Weinberg proportions. (c) Assum-

ing that the sample is representative of the population,

how many generations of random mating would be

required to establish Hardy–Weinberg proportions in

the population?

ANS: (a) P

 0.6 and P

 0.4. (b) Predict H–W proportions

by multiplying the expected genotype frequencies by the

sample size and compare these values with the observed

genotype frequencies using a chi-square test:

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Genotype H–W Frequency Predicted Number

0.6

 0.36 0.36  100  36

2(0.6)(0.4)  0.48 0.28  100  48

0.4

 0.16 0.16  100  16

 (30  36)

/36  (60  48)

/48  (10  16)

/16  6.25,

df  1; 6.25 is greater than 3.841; therefore, the observed

genotypes are not in agreement with H–W.

20.15 In a large population that reproduces by random mating,

the frequencies of the genotypes GG, Gg, and gg are 0.04,

0.32, and 0.64, respectively. Assume that a change in the

climate induces the population to reproduce exclusively

by self-fertilization. Predict the frequencies of the geno-

types in this population after many generations of

self-fertilization.

ANS: Ultimate frequency of GG is 0.2; ultimate frequency of

gg is 0.8.

20.16 The frequencies of the alleles A and a are 0.6 and 0.4,

respectively, in a particular plant population. After many

generations of random mating, the population goes

through one cycle of self-fertilization. What is the

expected frequency of heterozygotes in the progeny of

the self-fertilized plants?

ANS: 2pq(1  F)  2(0.6)(0.4)(1  0.5)  0.24.

20.17 Each of two isolated populations is in Hardy–Weinberg

equilibrium with the genotype frequencies shown below:

Genotype: AA Aa Aa

Frequency in

Population 1:

0.04 0.32 0.64

Frequency in

Population 2:

0.64 0.32 0.04

(a) If the populations are equal in size and they merge to

form a single large population, predict the allele and

genotype frequencies in the large population immedi-

ately after merger.

(b) If the merged population reproduces by random mat-

ing, predict the genotype frequencies in the next

generation.

random mating, will these genotype frequencies remain

constant?

ANS: (a) Frequency of A in merged population is 0.5 and that

of a is also 0.5; (b) 0.25 (AA), 0.50 (Aa), and 0.25 (aa);

20.18 A population consists of 25 percent tall individuals

(genotype TT), 25 percent short individuals (genotype

tt), and 50 percent individuals of intermediate height

(genotype Tt). Predict the ultimate phenotypic and

genotypic composition of the population if, generation

after generation, mating is strictly assortative (i.e., tall

individuals mate with tall individuals, short individuals

mate with short individuals, and intermediate individuals

mate with intermediate individuals).

ANS: Ultimately, all members of the population will either be

TT or tt, each 50% of the total.

20.19 In controlled experiments with different genotypes of

an insect, a researcher has measured the probability of

survival from fertilized eggs to mature, breeding

adults. The survival probabilities of the three geno-

types tested are 0.92 (for GG), 0.90 (for Gg), and 0.56

(for gg). If all breeding adults are equally fertile, what

are the relative tnesses of the three genotypes? What

are the selection coefcients for the two least t

genotypes?

ANS: The relative tnesses can be obtained by dividing each of

the survival probabilities by the largest probability (0.92).

Thus, the relative tnesses are 1 for GG, 0.98  1  0.02

for Gg, and 0.61  1  0.39 for gg. The selection coef-

cients are s

 0.02 for Gg and s

 0.39 for gg.

20.20 In a large randomly mating population, 0.84 of the indi-

viduals express the phenotype of the dominant allele A

and 0.16 express the phenotype of the recessive allele a.

(a) What is the frequency of the dominant allele? (b) If

the aa homozygotes are 5 percent less t than the other

two genotypes, what will the frequency of A be in the

next generation?

ANS: Frequency of a is q  0.4; (a) thus p  1  q  0.6;

(b) use the following scheme:

Genotype AA Aa Aa

Hardy–Weinberg frequency 0.36 0.48 0.16

Relative tness 1 1 0.95

Relative contribution to

next generation

0.36 0.48 0.152

Proportional contribution

to next generation

0.363 0.484 0.153

Thus, the frequency of the A allele in the next generation

will be (0.363  0.484/2)  0.605.

20.21 Because individuals with cystic brosis die before they

can reproduce, the coefcient of selection against them

is s  1. Assume that heterozygous carriers of the

recessive mutant allele responsible for this disease are as

t as wild-type homozygotes and that the population

frequency of the mutant allele is 0.02. (a) Predict the

incidence of cystic brosis in the population after one

generation of selection. (b) Explain why the incidence of

cystic brosis hardly changes even with s  1.

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ANS: (a) Use the following scheme:

Genotype CC Cc cc

Hardy–Weinberg (0.98)

 2(0.98)(0.02)  (0.02)



frequency 0.9604 0.0392 0.0004

Relative tness 1 1 0

Relative (0.9604)  1 (0.0392)  1 0

contribution

to next generation

Proportional 0.9604/0.9996 0.0392/0.9996 0

contribution  0.9608  0.0392

The new frequency of the allele for cystic brosis is (0.5)

(0.0392)  0.0196; thus, the incidence of the disease will

be (0.0196)

 0.00038, which is very slightly less than

the incidence in the previous generation. (b) The inci-

dence of cystic brosis does not change much because

selection can only act against the recessive allele when it

is in homozygotes, which are rare in the population.

20.22 For each set of relative tnesses for the genotypes AA,

Aa, and aa, explain how selection is operating. Assume

that 0 < t < s < 1.

AA Aa aa

Case 1 1 1 1 - s

Case 2 1 - s 1 - s 1

Case 3 1 1 - t 1 - s

Case 4 1 - s 1 1 - t

ANS: Case 1: selection is operating against a deleterious

recessive allele. Case 2: selection is operating against a

deleterious dominant allele. Case 3: selection is operat-

ing against a deleterious allele that has some expression

in heterozygotes, that is, it is partially dominant. Case 4:

selection is operating against both alleles in homozygous

condition; this is a case of balancing selection.

20.23 The frequency of newborn infants homozygous for a

recessive lethal allele is about 1 in 25,000. What is the

expected frequency of carriers of this allele in the

population?

ANS: q

 4  10

5

; thus q  6.3  10

3

and 2pq  0.0126.

20.24 A population of size 50 reproduces in such a way that

the population size remains constant. If mating is ran-

dom, how rapidly will genetic variability, as measured by

the frequency of heterozygotes, be lost from this

population?

ANS: The frequency of heterozygotes will decrease by a

1/(2N)  1/100  0.01 per generation.

20.25 A population is segregating three alleles, A

, A

, and A

with frequencies 0.2, 0.5, and 0.3, respectively. If these

alleles are selectively neutral, what is the probability that

will ultimately be xed by genetic drift? What is the

probability that A

will ultimately be lost by genetic drift?

ANS: Probability of ultimate xation of A

is 0.5; probability of

ultimate loss of A

is 1  0.3  0.7.

20.26 A small island population of mice consists of roughly

equal numbers of males and females. The Y chromo-

some in one-fourth of the males is twice as long as the

Y chromosome in the other males because of an expan-

sion of heterochromatin. If mice with the large Y chro-

mosome have the same tness as mice with the small

Y chromosome, what is the probability that the large

Y chromosome will ultimately be xed in the mouse

population?

ANS: 0.25.

20.27 In some regions of west Africa, the frequency of the

HBB

allele is 0.2. If this frequency is the result of a

dynamic equilibrium due to the superior tness of

HBB

heterozygotes, and if HBB

HBB

homozy-

gotes are essentially lethal, what is the intensity of selec-

tion against the HBB

HBB

homozygotes?

ANS: p  0.2; at equilibrium, p  t/(s  t). Because s  1, we

can solve for t; t  0.25.

20.28 Mice with the genotype Hh are twice as t as either of

the homozygotes HH and hh. With random mating, what

is the expected frequency of the h allele when the mouse

population reaches a dynamic equilibrium because of

balancing selection?

ANS: The relative tnesses of the genotypes HH, Hh, and hh

are 0.5, 1, and 0.5, respectively. At equilibrium, the fre-

quency of h will be s/(t  s)  (0.5)/(0.5  0.5)  0.5.

20.29 A completely recessive allele g is lethal in homozygous

condition. If the dominant allele G mutates to g at a rate

of 10

6

per generation, what is the expected frequency of

the lethal allele when the population reaches mutation–

selection equilibrium?

ANS: At mutation–selection equilibrium

qus

== =

10 10

.001.

−6

20.30 Individuals with the genotype bb are 20 percent less t

than individuals with the genotypes BB or Bb. If B

mutates to b at a rate of 10

6

per generation, what is the

expected frequency of the allele b when the population

reaches mutation–selection equilibrium?

ANS:

qus== = 2.2 ×//(0.2)10 10

−−63

Chapter 21

21.1 Which of the following pairs of DNA sequences could

qualify as the terminal repeats of a bacterial IS element.

Explain.

(a) 5′-GAATCCGCA-3′ and 5′-ACGCCTAAG-3′

(b) 5′-GAATCCGCA-3′ and 5′-CTTAGGCGT-3′

(d) 5′-GAATCCGCA-3′ and 5′-TGCGGATTC-3′.

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ANS: The pair in (d) are inverted repeats and could therefore

qualify.

21.2 Which of the following pairs of DNA sequences could

qualify as target site duplications at the point of an IS50

insertion? Explain.

(a) 5′-AATTCGCGT-3′ and 5′-AATTCGCGT-3′

(b) 5′-AATTCGCGT-3′ and 5′-TGCGCTTAA-3′

(d) 5′-AATTCGCGT-3′ and 5′-ACGCGAATT-3′.

ANS: The pair in (a) are direct repeats and could therefore

qualify.

21.3 One strain of E. coli is resistant to the antibiotic strepto-

mycin, and another strain is resistant to the antibiotic

ampicillin. The two strains were cultured together and

then plated on selective medium containing streptomy-

cin and ampicillin. Several colonies appeared, indicating

that cells had acquired resistance to both antibiotics.

Suggest a mechanism to explain the acquisition of dou-

ble resistance.

ANS: Resistance for the second antibiotic was acquired by con-

jugative gene transfer between the two types of cells.

21.4 What distinguishes IS and Tn3 elements in bacteria?

ANS: Tn3 elements carry a gene that is not essential for

transposition.

21.5 The circular order of genes on the E. coli chromosome is

*A B C D E F G H*, with the * indicating that the ends of

the chromosome are attached to each other. Two copies

of an IS element are located in this chromosome: one

between genes C and D, and the other between genes D

and E. A single copy of this element is also present in the

F plasmid. Two Hfr strains were obtained by selecting

for integration of the F plasmid into the chromosome.

During conjugation, one strain transfers the chromo-

somal genes in the order D E F G H A B C, whereas the

other transfers them in the order D C B A H G F E.

Explain the origin of these two Hfr strains. Why do they

transfer genes in different orders? Does the order of

transfer reveal anything about the orientation of the IS

elements in the E. coli chromosome?

ANS: In the rst strain, the F factor integrated into the chro-

mosome by recombination with the IS element between

genes C and D. In the second strain, it integrated by

recombination with the IS element between genes D and

E. The two strains transfer their genes in different orders

because the two chromosomal IS elements are in oppo-

site orientation.

21.6 The composite transposon Tn5 consists of two IS50 ele-

ments, one on either side of a group of three genes for

antibiotic resistance. The entire unit IS50L kan

ble

str

IS50R can transpose to a new location in the E. coli chro-

mosome. However, of the two IS50 elements in this

transposon, only IS50R produces the catalytically active

transposase. Would you expect IS50R to be able to be

excised from the Tn5 composite transposon and insert

elsewhere in the chromosome? Would you expect IS50L

to be able to do this?

ANS: Both IS50 elements should be able to excise from the

transposon and insert elsewhere in the chromosome,

because even though IS50L does not produce its own

transposase, IS50R provides a source of this enzyme.

21.7 By chance, an IS1 element has inserted near an IS2 ele-

ment in the E. coli chromosome. The gene between them,

sug



, confers the ability to metabolize certain sugars.

Will the unit IS1 sug



IS2 behave as a composite trans-

poson? Explain.

ANS: No. IS1 and IS2 are mobilized by different transposases.

21.8 A researcher has found a new Tn5 element with the

structure IS50L str

ble

kan

IS50L. What is the most

likely origin of this element?

ANS: IS50L inserted on each side of the cluster of antibiotic

resistance genes.

21.9 Would a Tn3 element with a frameshift mutation early

in the tnpA gene be able to form a cointegrate? Would a

Tn3 element with a frameshift mutation early in the

tnpR gene be able to form a cointegrate?

ANS: The tnpA mutation: no; the tnpR mutation: yes.

21.10 What enzymes are necessary for replicative transposition

of Tn3? What are their respective functions?

ANS: Two enzymes, transposase and resolvase, are needed for

replicative transposition. These enzymes are encoded by

genes of Tn3. Transposase catalyzes formation of a coin-

tegrate between donor and recipient plasmids. During

this process, Tn3 is replicated so that there is a copy of it

at each junction in the cointegrate. Resolvase catalyzes

the site-specic recombination between the two Tn3

elements and thereby resolves the cointegrate, generat-

ing two molecules each with a copy of the transposon.

Resolvase also represses the synthesis of both the trans-

posase and resolvase enzymes.

21.11 What is the medical signicance of bacterial transposons?

ANS: Many bacterial transposons carry genes for antibiotic

resistance, and it is relatively simple for these genes to

move from one DNA molecule to another. DNA mole-

cules that acquire resistance genes can be passed to other

cells in a bacterial population, both vertically (by descent)

and horizontally (by conjugative transfer). Over time,

continued exposure to an antibiotic will select for cells

that have acquired a gene for resistance to that antibiotic.

The antibiotic will therefore no longer be useful in com-

bating these bacteria.

21.12 Describe the structure of the Ac transposon in maize. In

what ways do the Ds transposons differ structurally and

functionally from the Ac transposon?

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ANS: The Ac element consists of 4563 nucleotide pairs

bounded by inverted repeats that are 11 nucleotide pairs

long. The Ac element is anked by direct repeats of eight

nucleotide pairs long; however, these repeats are created

at the time the element is inserted into a chromosome

(target site duplications) and are therefore not consid-

ered to be integral parts of the element itself. Ds ele-

ments possess the same terminal inverted repeats as Ac,

but their internal sequences vary. Some residue of the Ac

sequence may be present, or non-Ac sequences may be

present; sometimes, one Ds element is contained within

another Ds element.

21.13 In homozygous condition, a deletion mutation of the c

locus, c

, produces colorless (white) kernels in maize; the

dominant wild-type allele, C, causes the kernels to be

purple. A newly identied recessive mutation of the c

locus, c

, has the same phenotype as the deletion muta-

tion (white kernels), but when c

and c

plants are

crossed, they produce white kernels with purple stripes.

If it is known that the c

plants harbor Ac elements,

what is the most likely explanation for the c

mutation?

ANS: The c

mutation is due to a Ds or an Ac insertion.

21.14 In maize, the O2 gene, located on chromosome 7, con-

trols the texture of the endosperm, and the C gene,

located on chromosome 9, controls its color. The gene

on chromosome 7 has two alleles, a recessive, o2, which

causes the endosperm to be soft, and a dominant, O2,

which causes it to be hard. The gene on chromosome 9

also has two alleles, a recessive, c, which allows the endo-

sperm to be colored, and a dominant, C

, which inhibits

coloration. In one homozygous C

strain, a Ds element is

inserted on chromosome 9 between the C gene and the

centromere. This element can be activated by introduc-

ing an Ac element by appropriate crosses. Activation of

Ds causes the C

allele to be lost by chromosome break-

age. In C

/c/c kernels, such loss produces patches of col-

ored tissue in an otherwise colorless background.

A geneticist crosses a strain with the genotype o2/o2; C

Ds/C

Ds to a strain with the genotype O2/o2; c/c. The

latter strain also carries an Ac element somewhere in the

genome. Among the offspring, only those with hard

endosperm show patches of colored tissue. What does

this tell you about the location of the Ac element in the

O2/o2; c/c strain?

ANS: The Ac element must be tightly linked to the O2 allele.

21.15 In maize, the recessive allele bz (bronze) produces a lighter

color in the aleurone than does the dominant allele, Bz.

Ears on a homozygous bz/bz plant were fertilized by pol-

len from a homozygous Bz/Bz plant. The resulting cobs

contained kernels that were uniformly dark except for a

few on which light spots occurred. Suggest an

explanation.

ANS: The paternally inherited Bz allele was inactivated by a

transposable element insertion.

21.16 The X-linked singed locus is one of several in Drosophila

that controls the formation of bristles on the adult cuti-

cle. Males that are hemizygous for a mutant singed allele

have bent, twisted bristles that are often much reduced in

size. Several P element insertion mutations of the singed

locus have been characterized, and some have been

shown to revert to the wild-type allele by excision of the

inserted element. What conditions must be present to

allow such reversions to occur?

ANS: The P transposase to catalyze excision and the absence of

P-specic piRNAs that would repress excision.

21.17 Dysgenic hybrids in Drosophila have elevated mutation

rates as a result of P element transposition. How could

you take advantage of this situation to obtain P element

insertion mutations on the X chromosome?

ANS: Cross dysgenic (highly mutable) males carrying a wild-

type X chromosome to females homozygous for a bal-

ancer X chromosome; then cross the heterozygous F

daughters individually to their brothers and screen the

males that lack the balancer chromosome for mutant

phenotypes, including failure to survive (lethality). Muta-

tions identied in this screen are probably due to P ele-

ment insertions in X-linked genes.

21.18 If DNA from a P element insertion mutation of the Dro-

sophila white gene and DNA from a wild-type white gene

were puried, denatured, mixed with each other, rena-

tured, and then viewed with an electron microscope,

what would the hybrid DNA molecules look like?

ANS:

P element DNA

White gene DNA

(Should see a single-stranded DNA loop corresponding

to the insertion.)

21.19 When complete P elements are injected into embryos

from an M strain, they transpose into the chromosomes

of the germ line, and progeny reared from these embryos

can be used to establish new P strains. However, when

complete P elements are injected into embryos from

insects that lack these elements, such as mosquitoes, they

do not transpose into the chromosomes of the germ line.

What does this failure to insert in the chromosomes of

other insects indicate about the nature of P element

transposition?

ANS: Factors made by the y’s genome are required for trans-

position; other insects apparently lack the ability to pro-

vide these factors.

21.20 (a) What are retroviruslike elements? (b) Give examples

of retroviruslike elements in yeast and Drosophila.

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(d) After a retroviruslike element has been inserted into

a chromosome, is it ever expected to be excised?

ANS: (a) Retroviruslike elements resemble integrated retrovi-

ruses in overall structure and behavior. (b) Examples

include the Ty1 element in yeast and the copia element in

Drosophila. (c) Retroviruslike elements transpose using

an RNA intermediate. The element DNA is transcribed

into single-stranded RNA, which is reverse-transcribed

into double-stranded DNA (cDNA). The double-

stranded cDNA is then inserted into a site in the genome.

(d) No. However, the LTRs could pair and recombine to

excise all but one LTR.

21.21 Sometimes, solitary copies of the LTR of Ty1 elements

are found in yeast chromosomes. How might these soli-

tary LTRs originate?

ANS: Through crossing over between the LTRs of a Ty1

element.

21.22 Would you ever expect the genes in a retrotransposon to

possess introns? Explain.

ANS: No. The intron sequences would be removed by RNA

processing prior to reverse transcription into DNA.

21.23 Suggest a method to determine whether the TART

retroposon is situated at the telomeres of each of the

chromosomes in the Drosophila genome.

ANS: In situ hybridization to polytene chromosomes using a

TART probe.

21.24 It has been proposed that the hobo transposable elements in

Drosophila mediate intrachromosomal recombination—

that is, two hobo elements on the same chromosome pair

and recombine with each other. What would such a

recombination event produce if the hobo elements were

oriented in the same direction on the chromosome?

What if they were oriented in opposite directions?

ANS: Same orientation: a deletion; opposite orientation: an

inversion.

21.25 What evidence suggests that some transposable elements

are not simply genetic parasites?

ANS: TART and HeT-A replenish the ends of Drosophila

chromosomes.

21.26 Approximately half of all spontaneous mutations in Dro-

sophila are caused by transposable element insertions. In

human beings, however, the accumulated evidence sug-

gests that the vast majority of spontaneous mutations are

not caused by transposon insertions. Propose a hypothe-

sis to explain this difference.

ANS: The transposition rate in humans may be very much less

than it is in Drosophila.

21.27 Z. Ivics, Z. Izsvák, and P. B. Hackett have “resurrected” a

nonmobile member of the Tc1/mariner family of trans-

posable elements isolated from the DNA of salmon.

These researchers altered 12 codons within the coding

sequence of the transposase gene of the salmon element

to restore the catalytic function of its transposase. The

altered element, called Sleeping Beauty, is being tested as

an agent for the genetic transformation of vertebrates

such as mice and zebra sh (and possibly humans).

Suppose that you have a bacterial plasmid containing the

gene for green uorescent protein (gfp) inserted between

the ends of a Sleeping Beauty element. How would you

go about obtaining mice or zebra sh that express the

gfp gene?

ANS: The Sleeping Beauty element could be used as a transfor-

mation vector in vertebrates much like the P element has

been used in Drosophila. The gfp gene could be inserted

between the ends of the Sleeping Beauty element and

injected into eggs or embryos along with an intact Sleep-

ing Beauty element capable of encoding the element’s

transposase. If the transposase that is produced in the

injected egg or embryo acts on the element that contains

the gfp gene, it might cause the latter to be inserted into

genomic DNA. Then, if the egg or embryo develops into

an adult, that adult can be bred to determine if a Sleeping

Beauty/gfp transgene is transmitted to the next genera-

tion. In this way, it would be possible to obtain strains of

mice or zebra sh that express the gfp gene.

21.28 The human genome contains about 5000 “processed

pseudogenes,” which are derived from the insertion of

DNA copies of mRNA molecules derived from many

different genes. Predict the structure of these pseudo-

genes. Would each type of processed pseudogene be

expected to found a new family of retrotransposons

within the human genome? Would the copy number of

each type of processed pseudogene be expected to

increase signicantly over evolutionary time, as the copy

number of the Alu family has? Explain your answers.

ANS: The processed pseudogenes will, in the best of cases,

contain sequences from the transcription start site to the

poly-A tail of the transcript (if there is one); however,

they will not contain the gene’s promoter or any of its

introns. Because these pseudogenes will not have a pro-

moter, they are not likely to found new retrotransposon

families. Without a promoter, they will not be tran-

scribed; hence, they will not produce RNA to be reverse

transcribed into DNA for insertion into other sites in the

genome. Most likely, the copy number of each of these

processed pseudogenes will not increase as the copy

number of the Alu element has. The Alu element con-

tains an “internal” promoter recognized by RNA poly-

merase III. Each insertion of the Alu element contains

this promoter and can therefore be transcribed into

RNA, which can subsequently be reverse transcribed

into DNA. The L1 element also contains an “internal”

promoter, but this promoter is recognized by RNA poly-

merase II. Most protein-coding genes contain an “exter-

nal” promoter—that is, one that is not transcribed—and

this promoter is recognized by RNA polymerase II.

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Chapter 22

22.1 During oogenesis, what mechanisms enrich the cyto-

plasm of animal eggs with nutritive and determinative

materials?

ANS: Unequal division of the cytoplasm during the meiotic

divisions; transport of substances into the oocyte from

surrounding cells such as the nurse cells in Drosophila.

22.2 Predict the phenotype of a fruit y that develops from an

embryo in which the posterior pole cells had been

destroyed by a laser beam.

ANS: The y will be sterile because the posterior pole cells

form the germ line in adults of both sexes.

22.3 Outline the main steps in the genetic analysis of develop-

ment in a model organism such as Drosophila.

ANS: Collect mutations with diagnostic phenotypes; map the

mutations and test them for allelism with one another;

perform epistasis tests with mutations in different genes;

clone individual genes and analyze their function at the

molecular level.

22.4 Why is the early Drosophila embryo a syncytium?

ANS: Mitotic division is so rapid that there is not enough time

for membranes to form between cells.

22.5 In Drosophila, what larval tissues produce the external

organs of the adult?

ANS: Imaginal discs.

22.6 Like dorsal, bicoid is a strict maternal-effect gene in Dro-

sophila; that is, it has no zygotic expression. Recessive

mutations in bicoid (bcd) cause embryonic death by pre-

venting the formation of anterior structures. Predict the

phenotypes of (a) bcd/bcd animals produced by mating

heterozygous males and females; (b) bcd/bcd animals pro-

duced by mating bcd/bcd females with bcd/ males;

with bcd/ males; (d) bcd/bcd animals produced by mating

bcd/ females with bcd/bcd males; (e) bcd/ animals pro-

duced by mating bcd/ females with bcd/bcd males.

ANS: (a) Wild-type; (b) embryonic lethal; (c) embryonic lethal;

(d) wild-type; (e) wild-type.

22.7 Why do women, but not men, who are homozygous for

the mutant allele that causes phenylketonuria produce

children that are physically and mentally retarded?

ANS: In homozygous condition, the mutation that causes phe-

nylketonuria has a maternal effect. Women homozygous

for this mutation inuence the development of their

children in utero.

22.8 In Drosophila, recessive mutations in the dorsal–ventral

axis gene dorsal (dl) cause a dorsalized phenotype in

embryos produced by dl/dl mothers; that is, no ventral

structures develop. Predict the phenotype of embryos

produced by females homozygous for a recessive muta-

tion in the anterior–posterior axis gene nanos.

ANS: Some structures fail to develop in the posterior portion

of the embryo.

22.9 A researcher is planning to collect mutations in mater-

nal-effect genes that control the earliest events in

Drosophila development. What phenotype should the

researcher look for in this search for maternal-effect

mutations?

ANS: Female sterility. Females affected by these mutations will

lay abnormal eggs that will not develop into viable

embryos.

22.10 A researcher is planning to collect mutations in the gap

genes, which control the rst steps in the segmentation

of Drosophila embryos. What phenotype should the

researcher look for in this search for gap gene

mutations?

ANS: Screen for lethal mutations that prevent regions of the

embryo from developing normally.

22.11 How do the somatic cells that surround a developing

Drosophila egg in the ovary inuence the formation of

the dorsal–ventral axis in the embryo that will be pro-

duced after the egg is fertilized?

ANS: The somatic cells surrounding a developing Drosophila

egg in the ovary determine where the spätzle protein,

which is the ligand for the Toll receptor protein, will be

cleaved. This cleavage will eventually occur on the ven-

tral side of the developing embryo.

22.12 What events lead to a high concentration of hunchback

protein in the anterior of Drosophila embryos?

ANS: The hunchback mRNA is translated into protein only in

the anterior region of the developing embryo. This

RNA is supplied to the egg by the nurse cells and it is

also synthesized after fertilization by transcription of

the hunchback gene. This zygotic transcription is stimu-

lated by a transcription factor encoded by maternally

supplied bicoid mRNA, which is located in the anterior

of the egg. Thus, hunchback mRNA is concentrated in

the anterior of the embryo. In addition, the hunchback

mRNA that is located in the posterior of the embryo is

bound by nanos protein and then degraded. The nanos

protein is concentrated in the posterior of the embryo

because maternally supplied nanos mRNA is preferen-

tially localized there.

22.13 Diagram a pathway that shows the contributions of the

sevenless (sev) and bride of sevenless (boss) genes to the dif-

ferentiation of the R7 photoreceptor in the ommatidia of

Drosophila eyes. Where would eyeless (ey) t in this

pathway?

ANS: ey → boss → sev → R7 differentiation

22.14 The sev

allele is temperature sensitive; at 22.7°C, ies

that are homozygous for it develop normal R7 photore-

ceptors, but at 24.3°C, they fail to develop these photo-

receptors. sos

is a recessive, loss-of-function mutation

in the son of sevenless (sos) gene. Flies with the genotype

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sev

/sev

; sos

/ fail to develop R7 photoreceptors

if they are raised at 22.7°C. Therefore, sos2A acts as

a dominant enhancer of the sev

mutant phenotype

at this temperature. Based on this observation, where

is the protein product of the wild-type sos gene—

called SOS—likely to act in the pathway for R7

differentiation?

ANS: If the SEV protein is activated—either by the BOSS

ligand or by a gain-of-function mutation in the sev gene,

a faulty effector protein could stop it from inducing the

R7 cell to differentiate. The SOS protein is likely to be a

downstream effector in the pathway for R7 differentia-

tion because when it is depleted by mutating one copy of

the sos gene, ies that have a partially functional SEV

protein show a mutant phenotype—that is, transmission

of the developmental signal through SEV and its down-

stream effector proteins is weakened.

20.15 When the mouse Pax6 gene, which is homologous to the

Drosophila eyeless gene, is expressed in Drosophila, it pro-

duces extra compound eyes with ommatidia, just like

normal Drosophila eyes. If the Drosophila eyeless gene were

introduced into mice and expressed there, what effect

would you expect? Explain.

ANS: Because the Pax6 gave the same phenotype in ies as

overexpression of the eyeless gene, the genes must be

functionally homologous, as well as structurally homolo-

gous. Therefore, expect extra mouse eyes or eye pri-

morida when expressing eyeless in the mouse.

22.16 Would you expect to nd homologues of Drosophila’s

BX-C and ANT-C genes in animals with radial symme-

try such as sea urchins and starsh? How could you

address this question experimentally?

ANS: Maybe these organisms would not have homologues of

the BX-C and ANT-C genes because they do not have

segmented bodies with bilateral symmetry as Drosophila

does. To see if homologues to these genes are present,

use Drosophila BX-C and ANT-C DNA as probes to

hybridize with starsh or sea urchin genomic DNA on a

Southern blot. The hybridization would have to be done

under conditions that allow DNA that is not a perfect

match to form a duplex—that is, under conditions of low

stringency. Usually, hybridizations of this type are car-

ried out at lower temperatures than typical Southern

hybridizations. If the probes stick to the DNA on the

blot, there is evidence for homologues to the BX-C and

ANT-C genes in the genomic DNA. Follow-up experi-

ments might endeavor to clone this DNA and, ultimately,

to sequence it to determine just how close a match it is to

the Drosophila probe DNA.

22.17 How might you show that two mouse Hox genes are

expressed in different tissues and at different times dur-

ing development?

ANS: Northern blotting of RNA extracted from the tissues at

different times during development. Hybridize the blot

with gene-specic probes.

22.18 Distinguish between therapeutic and reproductive

cloning.

ANS: Therapeutic cloning involves the creation of an embryo

by implanting the nucleus of a somatic cell into an enu-

cleated egg and stimulating the egg to divide. Stem cells

are then taken from the embryo to differentiate into spe-

cic tissues in the individual from which the somatic cell

was taken. These tissues will be genetically identical to

the other tissues of the individual—thus, they are unlikely

to be rejected by the individual’s immune system. Repro-

ductive cloning involves the creation of an embryo by

implanting the nucleus of a somatic cell into an enucle-

ated egg and then allowing the egg to develop into an

entire individual.

22.19 What is the scientic signicance of reproductive

cloning?

ANS: Reproductive cloning of mammals such as sheep, mice,

and cats indicates that somatic cell nuclei have all the

genetic information to direct the development of a com-

plete, viable organism. It also shows that epigenetic

modications of chromatin, such as X chromosome inac-

tivation, can be reset.

22.20 The methylation of DNA, the acetylation of histones,

and the packaging of DNA into chromatin by certain

kinds of proteins are sometimes referred to as epigenetic

modications of the DNA. These modications portend

difculties for reproductive cloning. Do they also por-

tend difculties for therapeutic cloning and for the use of

stem cells to treat diseases or injuries that involve the loss

of specic cell types?

ANS: Methylation of DNA, acetylation of histones, and pack-

aging of DNA into chromatin by certain kinds of pro-

teins all portend difculties for therapeutic cloning as

well as for reproductive cloning. These epigenetic modi-

cations of somatic cell DNA would have to be “repro-

grammed” in the oocyte or they could affect how the

stem cells derived from the oocyte would develop.

22.21 Assume that an animal is capable of producing 100 mil-

lion different antibodies and that each antibody contains

a light chain of 220 amino acids long and a heavy chain

of 450 amino acids. How much genomic DNA would be

needed to accommodate the coding sequences of these

genes?

ANS: If each antibody consists of one kind of light chain and

one kind of heavy chain, and if light and heavy chains can

combine freely, the potential to produce 100 million dif-

ferent antibodies implies the existence of 10,000 light

chain genes and 10,000 heavy chain genes (10,000 

10,000  100 million). If each light chain is 220 amino

acids long, each light chain gene must comprise 3  220 

660 nucleotides because each amino acid is specied by a

triplet of nucleotides; similarly, each heavy chain gene

must comprise 3  450  1350 nucleotides. Therefore,

the genome must contain 10,000  660  6.6 million

nucleotides devoted to light chain production and

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10,000  1,350  13.5 million nucleotides devoted to

heavy chain production. Altogether, then, the genome

must contain 19.5 million nucleotides dedicated to

encoding the amino acids of the various antibody chains.

22.22 Each L

gene segment in the kappa light chain locus on

chromosome 2 consists of two coding exons, one for the

leader peptide and one for the variable portion of the

kappa light chain. Would you expect to nd a stop codon

at the end of the coding sequence in the second

) exon?

ANS: No, because the V

coding sequence must be joined to

the coding sequence of the constant region to encode a

complete kappa light chain.

Chapter 23

23.1 Many cancers seem to involve environmental factors.

Why, then, is cancer called a genetic disease?

ANS: Cancer has been called a genetic disease because it results

from mutations of genes that regulate cell growth and

division. Nonhereditary forms of cancer result from

mutations in somatic cells. These mutations, however,

can be induced by environmental factors including

tobacco smoke, chemical pollutants, ionizing radiation,

and UV light. Hereditary forms of cancer also frequently

involve the occurrence of environmentally induced

somatic mutations.

23.2 Both embryonic cells and cancer cells divide quickly.

How can these two types of cells be distinguished from

each other?

ANS: Cancer cells do not display contact inhibition—they pile

up on top of each other—whereas embryonic cells spread

out in at sheets. Cancer cells are frequently aneuploid;

embryonic cells are euploid.

23.3 Most cancer cells are aneuploid. Suggest how aneuploidy

might contribute to deregulation of the cell cycle.

ANS: Aneuploidy might involve the loss of functional copies of

tumor suppressor genes, or it might involve the inappro-

priate duplication of proto-oncogenes. Loss of tumor sup-

pressor genes would remove natural brakes on cell division,

and duplication of proto-oncogenes would increase the

abundance of factors that promote cell division.

23.4 Would you ever expect to nd a tumor-inducing retrovi-

rus that carried a processed cellular tumor suppressor

gene in its genome?

ANS: No. A virus that carried a processed copy of a tumor sup-

pressor gene would not be expected to induce tumor for-

mation because the product of the tumor suppressor

gene would help to restrain cell growth and division.

23.5 How do we know that normal cellular oncogenes are not

simply integrated retroviral oncogenes that have acquired

the proper regulation?

ANS: They possess introns.

23.6 How might the absence of introns in a retroviral onco-

gene explain that gene’s overexpression in the tissues of

an infected animal?

ANS: The absence of introns might speed up the expression of

the gene’s protein product because there would be no

need for splicing. In addition, some introns contain

sequences called silencers that negatively regulate tran-

scription. Removal of these sequences might cause tran-

scription to occur when it otherwise would not.

23.7 When cellular oncogenes are isolated from different ani-

mals and compared, the amino acid sequences of the

polypeptides they encode are found to be very similar.

What does this suggest about the functions of these

polypeptides?

ANS: The products of these genes play important roles in cell

activities.

23.8 The majority of the c-ras oncogenes obtained from can-

cerous tissues have mutations in codon 12, 59, or 61 in

the coding sequence. Suggest an explanation.

ANS: Mutations in these codons cause amino acid changes that

activate the Ras protein.

23.9 When a mutant c-H-ras oncogene with a valine for gly-

cine substitution in codon 12 is transfected into cultured

NIH 3T3 cells, it transforms those cells into cancer cells.

When the same mutant oncogene is transfected into cul-

tured embryonic cells, it does not transform them. Why?

ANS: The cultured NIH 3T3 cells probably carry other muta-

tions that predispose them to become cancerous; trans-

fection of such cells with a mutant c-H-ras oncogene may

be the last step in the process of transforming the cells

into cancer cells. Cultured embryonic cells probably do

not carry the predisposing mutations needed for them to

become cancerous; thus, when they are transfected with

the mutant c-H-ras oncogene, they continue to divide

normally.

23.10 A mutation in the ras cellular oncogene can cause cancer

when it is in heterozygous condition, but a mutation in

the RB tumor suppressor gene can cause cancer only

when it is in homozygous condition. What does this dif-

ference between dominant and recessive mutations

imply about the roles that the ras and RB gene products

play in normal cellular activities?

ANS: Ras protein is an activator of cell division, whereas RB

protein is a suppressor of cell division.

23.11 Explain why individuals who develop nonhereditary reti-

noblastoma usually have tumors in only one eye, whereas

individuals with hereditary retinoblastoma usually

develop tumors in both eyes.

ANS: Retinoblastoma results from homozygosity for a loss-of-

function (recessive) allele. The sporadic occurrence of

retinoblastoma requires two mutations of this gene in

the same cell or cell lineage. Therefore, retinoblastoma

is rare among individuals who, at conception, are

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homozygous for the wild-type allele of the RB gene. For

such individuals, we would expect the frequency of

tumors in both eyes to be the square of the frequency of

tumors in one eye. Individuals who are heterozygous for

a mutant RB allele require only one somatic mutation to

occur for them to develop retinoblastoma. Because there

are millions of cells in each retina, there is a high prob-

ability that this somatic mutation will occur in at least

one cell in each eye, causing both eyes to develop tumors.

23.12 Approximately 5 percent of the individuals who inherit

an inactivated RB gene do not develop retinoblastoma.

Use this statistic to estimate the number of cell divisions

that form the retinal tissues of the eye. Assume that the

rate at which somatic mutations inactivate the RB gene is

one mutation per 10

cell divisions.

ANS: The probability that a carrier does not develop retino-

blastoma is 0.05, which is equal to the probability that

the wild-type RB allele is not mutationally inactivated

during the cell divisions that form the retinas of the eyes.

If the rate of mutational inactivation is u  10

6

per cell

division, and n is the number of cell divisions, then 0.05 

(1  u)

. If we take logarithms of both sides, log(0.05) 

nlog(1  u), which to a good approximation, is equal to

nlog (u). After substituting 10

6

for u and solving, we nd

that n  1.3  10

23.13 Inherited cancers like retinoblastoma show a dominant

pattern of inheritance. However, the underlying genetic

defect is a recessive loss-of-function mutation—often

the result of a deletion. How can the dominant pattern of

inheritance be reconciled with the recessive nature of the

mutation?

ANS: At the cellular level, loss-of-function mutations in the RB

gene are recessive; a cell that is heterozygous for such a

mutation divides normally. However, when a second

mutation occurs, that cell becomes cancerous. If the rst

RB mutation was inherited, there is a high probability

that the individual carrying this mutation will develop

retinoblastoma because a second mutation can occur any

time during the formation of the retinas in either eye.

Thus, the individual is predisposed to develop retino-

blastoma, and it is this predisposition that shows a domi-

nant pattern of inheritance.

23.14 The following pedigree shows the inheritance of familial

ovarian cancer caused by a mutation in the BRCA1 gene.

Should II-1 be tested for the presence of the predispos-

ing mutation? Discuss the advantages and disadvantages

of testing.

Ovarian cancer

Normal

III

ANS: II-1 should be tested for the BRCA1 mutation that appar-

ently was involved in the ovarian cancer that developed

in her mother and sister. If she is found to carry this

mutation, a prophylactic oophorectomy can be pre-

scribed to reduce the chance that she will develop

cancer. If she is found to be free of the mutation carried

by her mother and sister, then she is not more likely to

develop ovarian cancer than a woman in the general

population.

23.15 In what sense is pRB a negative regulator of E2F tran-

scription factors?

ANS: By binding to E2F transcription factors, pRB prevents

those transcription factors from activating their target

genes—which encode proteins involved in progression

of the cell cycle; pRB is therefore a negative regulator of

transcription factors that stimulate cell division.

23.16 A particular E2F transcription factor recognizes the

sequence TTTCGCGC in the promoter of its target

gene. A temperature-sensitive mutation in the gene

encoding this E2F transcription factor alters the ability

of its protein product to activate transcription; at 25°C,

the mutant protein activates transcription normally, but

at 35°C, it fails to activate transcription at all. However,

the ability of the protein to recognize its target DNA

sequence is not impaired at either temperature. Would

cells heterozygous for this temperature-sensitive muta-

tion be expected to divide normally at 25°C? at 35°C?

Would your answers change if the E2F protein functions

as a homodimer?

ANS: At 25°, cell division should be normal—the same as for

cells homozygous for a wild-type allele of the E2F gene.

At 35°, division would be expected to be impaired either

because the mutant E2F protein binds unproductively to

the sequence in its target gene or because a mutant E2F

polypeptide dimerizes with a wild-type E2F polypeptide

and abolishes the activation function of the wild-type

polypeptide.

23.17 During the cell cycle, the p16 protein is an inhibitor of

cyclin/CDK activity. Predict the phenotype of cells

homozygous for a loss-of-function mutation in the gene

that encodes p16. Would this gene be classied as a

proto-oncogene or as a tumor suppressor gene?

ANS: Cells homozygous for a loss-of-function mutation in the

p16 gene might be expected to divide in an uncontrolled

manner because the p16 protein would not be able to

inhibit cyclin-CDK activity during the cell cycle. The

p16 gene would therefore be classied as a tumor sup-

pressor gene.

23.18 The BCL-2 gene encodes a protein that represses the

pathway for programmed cell death. Predict the pheno-

type of cells heterozygous for a dominant activating

mutation in this gene. Would the BCL-2 gene be classi-

ed as a proto-oncogene or as a tumor suppressor gene?

ANS: Cells heterozygous for a dominant activating mutation

in the BCL-2 gene would be expected to be unable to

execute the programmed cell death pathway in response

to DNA damage induced by radiation treatment.

Such cells would continue to divide and accumulate

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mutations; ultimately they would have a good chance of

becoming cancerous. The BCL-2 gene would therefore

be classied as a proto-oncogene.

23.19 The protein product of the BAX gene negatively regu-

lates the protein product of the BCL-2 gene—that is,

BAX protein interferes with the function of the BCL-2

protein. Predict the phenotype of cells homozygous for a

loss-of-function mutation in the BAX gene. Would this

gene be classied as a proto-oncogene or as a tumor sup-

pressor gene?

ANS: Cells homozygous for a loss-of-function mutation in the

BAX gene would be unable to prevent repression of the

programmed cell death pathway by the BCL-2 gene

product. Consequently, these cells would be unable to

execute that pathway in response to DNA damage

induced by radiation treatment. Such cells would

continue to divide and accumulate mutations; ultimately,

they would have a good chance of becoming cancerous.

The BAX gene would therefore be classied as a tumor

suppressor gene.

23.20 Cancer cells frequently are homozygous for loss-of-

function mutations in the TP53 gene, and many of these

mutations map in the portion of TP53 that encodes the

DNA-binding domain of p53. Explain how these muta-

tions contribute to the cancerous phenotype of the cells.

ANS: Loss-of-function mutations in the DNA-binding

domain of p53 abolish the ability of that protein to acti-

vate transcription of target genes whose products are

involved in the restraint of cell division or in the promo-

tion of programmed cell death. Without restraint of cell

division or promotion of programmed cell death, cells

accumulate damage to their DNA and ultimately become

cancerous.

23.21 Suppose that a cell is heterozygous for a mutation that

caused p53 to bind tightly and constitutively to the DNA

of its target genes. How would this mutation affect the

cell cycle? Would such a cell be expected to be more or

less sensitive to the effects of ionizing radiation?

ANS: If a cell were heterozygous for a mutation that caused

p53 to bind tightly and constitutively to the DNA of its

target genes, its growth and division might be retarded,

or it might be induced to undergo apoptosis. Such a cell

would be expected to be more sensitive to the effects of

ionizing radiation because radiation increases the expres-

sion of p53, and in this case, the p53 would be predis-

posed to activate its target genes, causing the cell to

respond vigorously to the radiation treatment.

23.22 Mice homozygous for a knockout mutation of the TP53

gene are viable. Would they be expected to be more or

less sensitive to the killing effects of ionizing radiation?

ANS: Homozygous TP53 knockout mice might actually be less

sensitive to the killing effects of ionizing radiation

because p53 would be unable to mediate the apoptotic

response to the radiation treatment.

23.23 Would cancer-causing mutations of the APC gene be

expected to increase or decrease the ability of pAPC to

bind b-catenin?

ANS: They would probably decrease the ability of pAPC to

bind b-catenin.

23.24 Mice that are heterozygous for a knockout mutation in

the RB gene develop pituitary and thyroid tumors. Mice

that are homozygous for this mutation die during embry-

onic development. Mice that are homozygous for a

knockout mutation in the gene encoding the p130

homologue of RB and heterozygous for a knockout

mutation in the gene encoding the p107 homologue of

RB do not have a tendency to develop tumors. However,

homozygotes for knockout mutations in both of these

genes die during embryonic development. What do

these ndings suggest about the roles of the RB, p139,

and p107 genes in embryos and adults?

ANS: All three genes (RB, p130, and p107) are essential for

embryonic development, although by themselves, p130

and p107 are dispensable, possibly because their products

are functionally redundant. (Both gene products must be

inactivated before any deleterious effect is seen.) In

adults, only pRB appears to play a role in suppressing

tumor formation.

23.25 It has been demonstrated that individuals with diets poor

in ber and rich in fatty foods have an increased risk to

develop colorectal cancer. Fiber-poor, fat-rich diets may

irritate the epithelial lining of the large intestine. How

could such irritation contribute to the increased risk for

colorectal cancer?

ANS: The increased irritation to the intestinal epithelium

caused by a ber-poor, fat-rich diet would be expected to

increase the need for cell division in this tissue (to replace

the cells that were lost because of the irritation), with a

corresponding increase in the opportunity for the occur-

rence of cancer-causing mutations.

23.26 Messenger RNA from the KAI1 gene is strongly

expressed in normal prostate tissues but weakly expressed

in cell lines derived from metastatic prostate cancers.

What does this nding suggest about the role of the

KAI1 gene product in the etiology of prostate cancer?

ANS: The KAI1 gene is a prostate tumor suppressor gene.

Functional inactivation of this gene allows prostate

tumors to develop.

23.27 The p21 protein is strongly expressed in cells that have

been irradiated. Researchers have thought that this strong

expression is elicited by transcriptional activation of the

p21 gene by the p53 protein acting as a transcription factor.

Does this hypothesis t with the observation that p21

expression is induced by radiation treatment in mice homo-

zygous for a knockout mutation in the TP53 gene? Explain.

ANS: No. Apparently there is another pathway—one not medi-

ated by p53—that leads to the activation of the p21 gene.

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Chapter 24

24.1 What was some of the evidence that led Charles Darwin

to argue that species change over time?

ANS: Among other things, Darwin observed species on islands

that were different from each other and from continental

species but were still similar enough to indicate that they

were related. He also observed variation within species,

especially within domesticated breeds, and saw how the

characteristics of an organism could be changed by selec-

tive breeding. His observations of fossilized organisms

indicated that some species have become extinct.

24.2 Darwin stressed that species evolve by natural selection.

What was the main gap in his theory?

ANS: Darwin did not understand the mechanism of inheri-

tance; he did not know of Mendel’s principles.

24.3 Using the data in Table 24.1, and assuming that mating is

random with respect to the blood type, predict the fre-

quencies of the three genotypes of the Duffy blood-type

locus in a South African and an English population.

ANS: The frequency of the a allele is 0.06 in the South African

population and 0.42 in the English population. The pre-

dicted genotype frequencies under the assumption of

random mating are as follows:

Genotype South Africa England

(0.06)

 0.004 (0.42)

 0.18

2(0.06)(0.94)  0.11 2(0.42)(0.58)  0.49

(0.94)

 0.88 (0.58)

 0.33

24.4 Theodosius Dobzhansky and his collaborators studied

chromosomal polymorphisms in Drosophila pseudoobscura

and its sister species in the western United States. In one

study of polymorphisms in chromosome III of D. pseu-

doobscura sampled from populations at different locations

in the Yosemite region of the Sierra Nevada, Dobzhan-

sky (1948, Genetics 33: 158–176) recorded the following

frequencies of the Standard (ST) banding pattern:

Location Frequency ST

Elevation

(in feet)

Jacksonville 0.46 850

Lost Claim 0.41 3,000

Mather 0.32 4,600

Aspen 0.26 6,200

Porcupine 0.14 8,000

Tuolumne 0.11 8,600

Timberline 0.10 9,900

Lyell Base 0.10 10,500

What is interesting about these data?

ANS: The frequency of the ST banding pattern declines with

increasing altitude. Thus, the data indicate that this

chromosomal polymorphism exhibits an altitudinal

cline.

24.5 In a survey of electrophoretically detectable genetic vari-

ation in the alcohol dehydrogenase gene of Drosophila

melanogaster, a researcher found two allozymes, denoted

F (fast) and S (slow) in a population; 32 individuals were

homozygous for the F allele of the gene, 22 were homo-

zygous for the S allele, and 46 were heterozygous for the

F and S alleles. Are the observed frequencies of the three

genotypes consistent with the assumption that the popu-

lation is in Hardy–Weinberg equilibrium?

ANS: In the sample, the frequency of the F allele is (2  32 

46)/(2  100)  0.55 and the frequency of the S allele is

1  0.55  0.45. The predicted and observed genotype

frequencies are as follows:

Genotype Observed Hardy–Weinberg Predicted

FF 32

100  (0.55)

 30.25

FS 46

100  2(0.55)(0.45)  49.5

SS 22

100  (0.45)

 20.25

To test for agreement between the observed and pre-

dicted values, we compute a chi-square statistic with

1 degree of freedom: c

 S(obs.  pred.)

/pred.  0.50,

which is not signicant at the 5 percent level. Thus, the

population appears to be in Hardy–Weinberg equilib-

rium for the alcohol dehydrogenase locus.

24.6 A researcher has been studying genetic variation in sh

populations by using PCR to amplify microsatellite

repeats at a particular site on a chromosome (see Chap-

ter 16). The diagram below shows the gel-fractionated

products of amplications with DNA samples from 10

different sh. How many distinct alleles of this microsat-

ellite locus are evident in the gel?

ANS: There are four alleles.

24.7 Within the coding region of a gene, where would you

most likely nd silent polymorphisms?

ANS: In the third position of some of the codons. Due to the

degeneracy of the genetic code, different codons can

specify the same amino acid. The degeneracy is most

pronounced in the third position of many codons, where

different nucleotides can be present without changing

the amino acid that is specied.

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24.8 Why are the nucleotide sequences of introns more poly-

morphic than the nucleotide sequences of exons?

ANS: Introns do not encode amino acids; most exons do.

Many—perhaps most—of the nucleotides within an

intron can be changed without impairing the expression

of the gene or the integrity of its polypeptide product.

By contrast, many of the nucleotides within exons—

especially the rst and second positions of codons—are

functionally constrained by the amino acids specied.

24.9 DNA and protein molecules are “documents of evolu-

tionary history.” Why aren’t complex carbohydrate mol-

ecules such as starch, cellulose, and glycogen considered

“documents of evolutionary history”?

ANS: Complex carbohydrates are not “documents of evolu-

tionary history” because, although they are polymers,

they are typically made of one subunit incorporated rep-

etitiously into a chain. Such a polymer has little or no

“information content.” Thus, there is little or no oppor-

tunity to distinguish a complex carbohydrate obtained

from two different organisms. Moreover, complex carbo-

hydrates are not part of the genetic machinery; their for-

mation is ultimately specied by the action of enzymes,

which are gene products, but they themselves are not

genetic material or the products of the genetic material.

24.10 A geneticist analyzed the sequences of a gene cloned

from four different individuals. The four clones were

identical except for a few base pair differences, a deletion

(gap), and a transposable element (TE) insertion:

Sequences

Using this information, compute the minimum number

of mutations required to explain the derivation of the

four sequences (1, 2, 3, and 4) in the following phyloge-

netic trees:

12431234

Tree ATree BTree C

3421

Which of these trees provides the most parsimonious

explanation for the evolutionary history of the four DNA

sequences?

ANS: The tree on the right would require nine mutations.

These mutations are a deletion in the branch leading to

the common ancestor of sequences 1, 2, and 3; a TE

insertion in the branch leading to the common ancestor

of sequences 1 and 2; and seven base-pair changes: one

leading to sequence 1, another leading to sequence 2, two

leading to sequence 3 and three leading to sequence 4.

The tree in the middle would also require nine mutations.

In this tree, we regard the gap as ancestral—which means

that sequence 4 has acquired an “insertion” at the posi-

tion of the gap. The other mutations are a TE insertion

in the branch leading to the common ancestor of

sequences 1 and 2, and seven base-pair changes: one

leading to sequence 1, another to sequence 2, two lead-

ing to sequence 3, and three leading to sequence 4.

The tree on the right would also require nine mutations.

Here again we regard the gap as ancestral; evidently, an

insertion occurred at the position of the gap in the

branch leading to sequence 4. The TE insertion can also

be regarded as ancestral, with loss occurring in the

branch leading to the common ancestor of sequences

3 and 4. There are also seven base-pair changes in this

tree: one leading to sequence 1, another leading to

sequence 2, two leading to sequence 3 and three leading

to sequence 4. These interpretations of the data assume

that insertions and deletions (gaps) are reversible and

that there is no way of telling which way the sequence

evolved—that is, through insertion or deletion. How-

ever, if we know, for example, that the TE is a retrotrans-

poson incapable of excision, then we would need more

than nine mutations to explain the tree on the right. The

TE insertions must have occurred independently in the

branches leading to sequences 1 and 2, and there is no

need to postulate excision of the TE in the branch lead-

ing to the common ancestor of sequences 3 and 4. Thus,

on the assumption that the TE is not excisable, 10 muta-

tions are needed to explain the tree on the right. Given

this reservation, the left and middle trees provide the

most parsimonious explanations for the evolution of the

four sequences.

24.11 The heme group in hemoglobin is held in place by histi-

dines in the globin polypeptides. All vertebrate globins

possess these histidines. Explain this observation in terms

of the Neutral Theory of Molecular Evolution.

ANS: The histidines are rigorously conserved because they

perform an important function—anchoring the heme

group in hemoglobin. Because these amino acids are

strongly constrained by natural selection, they do not

evolve by mutation and random genetic drift.

24.12 During the early evolutionary history of the vertebrates,

a primordial globin gene was duplicated to form the

α- and b-globin genes. The rate of evolution of the poly-

peptides encoded by these duplicate genes has been

estimated to be about 0.9 amino acid substitutions per

site every billion years. By comparing the human α- and

b-globins, the average number of amino acid substitu-

tions per site has been estimated to be 0.800. From this

estimate, calculate when the duplication event that

produced the α- and b-globin genes must have occurred.

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ANS: The total elapsed evolutionary time is 0.800/0.9  880

million years, which must be apportioned equally to the

 and  gene lineages by dividing by 2; thus, the time

since the duplication event is estimated to be 440 million

years.

24.13 Ribonuclease, a protein that degrades RNA, is 124 amino

acids long. A comparison between the amino acid

sequences of cow and rat ribonucleases reveals 40 differ-

ences. What is the average number of amino acid substi-

tutions that have occurred per site in these two

evolutionary lineages? If the cow and the rat lineages

diverged from a common ancestor 80 million years ago,

what is the rate of ribonuclease evolution?

ANS: Estimate the average number of substitutions per site in

the ribonuclease molecule as ln(S), where S  (124 

40)/124  0.68, the proportion of amino acids that are

the same in the rat and cow molecules. The average

number of substitutions per site since the cow and rat

lineages diverged from a common ancestor is therefore

0.39. The evolutionary rate in the cow and rat lineages is

0.39/(2  80 million years)  2.4 substitutions per site

every billion years.

24.14 If a randomly mating population is segregating n selec-

tively neutral alleles of a gene and each allele has the

same frequency, what is the frequency of all the homozy-

gotes in the population?

ANS: With n alleles having equal frequency, the frequency of

any one allele is 1/n. Under random mating, the fre-

quency of homozygotes for a particular allele is (1/n)

The frequency of all the homozygotes is therefore

∑(1/n)

 n(1/n)

 1/n.

24.15 If the evolutionary rate of amino acid substitution in a

protein is K, what is the average length of time between

successive amino acid substitutions in this protein?

ANS: The reciprocal of the rate, that is, 1/K.

24.16 The coding sequence of the alcohol dehydrogenase

(Adh) gene of D. melanogaster consists of 765 nucleo-

tides (255 codons); 192 of these nucleotides are func-

tionally silent—that is, they can be changed without

changing an amino acid in the Adh polypeptide. In a

study of genetic variation in the Adh gene, Martin

Kreitman observed that 13 of the 192 silent nucleo-

tides were polymorphic. If the same level of polymor-

phisms existed among the nonsilent nucleotides of the

Adh gene, how many amino acid polymorphisms

would Kreitman have observed in the populations he

studied?

ANS: The fraction of polymorphic sites among the silent

nucleotides is 13/192  0.068. If the same level of poly-

morphism existed among the nonsilent sites, the number

of amino acid polymorphisms would be 225  0.067 

17.2. Kreitman actually observed only one amino acid

polymorphism—evidence that amino acid changes in

alcohol dehydrogenase are deleterious.

24.17 How might you explain the 1000-fold difference in the

evolutionary rates of brinopeptide and histone 3?

ANS: The protein with the higher evolutionary rate is not as

constrained by natural selection as the protein with the

lower evolutionary rate.

24.18 A geneticist has studied the sequence of a gene in each of

three species, A, B, and C. Species A and species B are

sister species; species C is more distantly related. The

geneticist has calculated the ratio of nonsynonymous

(NS) to synonymous (S) nucleotide substitutions in the

coding region of the gene in two ways—rst, by compar-

ing the gene sequences of species A and C, and second,

by comparing the gene sequences of species B and C.

The NS:S ratio for the comparison of species B and C is

ve times greater than it is for the comparison of species

A and C. What might this difference in the NS:S ratios

suggest?

ANS: The difference in the NS:S ratios suggests that in at least

one lineage, positive selection has been operating to

change nucleotides in the gene.

24.19 Dispersed, repetitive sequences such as transposable ele-

ments may have played a role in duplicating short regions

in a genome. Can you suggest a mechanism? (Hint: See

Chapter 21 on the Instructor Companion site.)

ANS: Repetitive sequences that are near each other can medi-

ate displaced pairing during meiosis. Exchange involving

the displaced sequences can duplicate the region between

them.

24.20 Exon shufing is a mechanism that combines exons from

different sources into a coherent sequence that can

encode a composite protein—one that contains peptides

from each of the contributing exons. Alternate splicing is

a mechanism that allows exons to be deleted during the

expression of a gene; the mRNAs produced by alternate

splicing may encode different, but related, polypeptides

(see Chapter 18). What bearing do these two mecha-

nisms have on the number of genes in a eukaryotic

genome? Do these mechanisms help to explain why the

gene number in the nematode Caenorabditis elegans is not

too different from the gene number in Homo sapiens?

ANS: Exon shufing is a way of creating genes that have pieces

from disparate sources. Through exon shufing, the

number of genes in a genome could be increased without

having to evolve the genes “from scratch.” However, the

genome sequencing projects indicate that the gene num-

ber in complex, multicellular vertebrates is not much dif-

ferent from the gene number in simple, multicellular

invertebrates or in plants. If, judging from their pheno-

types, multicellular vertebrates need more gene products

than phenotypically simpler invertebrates such as C. ele-

gans, alternate splicing could provide some of these gene

products without increasing the number of genes.

24.21 Drosophila mauritiana inhabits the island of Mauritius in

the Indian Ocean. Drosophila simulans, a close relative, is

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widely distributed throughout the world. What experi-

mental tests would you perform to determine if D. mau-

ritiana and D. simulans are genetically different species?

ANS: Cross D. mauritiana with D. simulans and determine if

these two species are reproductively isolated. For

instance, can they produce offspring? If they can, are the

offspring fertile?

24.22 Distinguish between allopatric and sympatric modes of

speciation.

ANS: Allopatric speciation occurs when populations diverge

genetically while they are geographically separated.

Sympatric speciation occurs when populations diverge

genetically while they inhabit the same territory.

24.23 The prune gene (symbol pn) is X-linked in Drosophila

melanogaster. Mutant alleles of this gene cause the eyes to

be brown instead of red. A dominant mutant allele of

another gene located on a large autosome causes

hemizygous or homozygous pn ies to die; this dominant

mutant allele is therefore called Killer of prune (symbol

Kpn). How could mutants such as these play a role in

the evolution of reproductive isolation between

populations?

ANS: The Kpn–pn interaction is an example of the kind of neg-

ative epistasis that might prevent populations that have

evolved separately from merging into one panmictic

population. The Kpn mutation would have evolved in

one population and the pn mutation in another, geo-

graphically separate population. When the populations

merge, the two mutations can be brought into the same

y by interbreeding. If the combination of these muta-

tions is lethal, then the previously separate populations

will not be able to exchange genes, that is, they will be

reproductively isolated.

24.24 A segment of DNA in an individual may differ at several

nucleotide positions from a corresponding DNA seg-

ment in another individual. For instance, one individual

may have the sequence …A…G…C… and another indi-

vidual may have the sequence …T…A…A…. These two

DNA segments differ in three nucleotide positions.

Because the nucleotides within each segment are tightly

linked, they will tend to be inherited together as a unit,

that is, without being scrambled by recombination.

We call such heritable units DNA haplotypes. Through

sampling and DNA sequencing, researchers can deter-

mine which DNA haplotypes are present in a particular

population. When this kind of analysis is performed on

human populations by sequencing, for example, a seg-

ment of mitochondrial DNA, it is found that samples

from Africa exhibit more haplotype diversity than sam-

ples from other continents. What does this observation

tell us about human evolution?

ANS: More haplotype diversity refers to the number of differ-

ent haplotypes that are found in a population. If African

populations of humans have the greatest haplotype

diversity, then these populations appear to have had a

longer time to accumulate different haplotypes—that is,

they are older than other populations. Greater haplotype

diversity in African populations of humans is therefore

evidence that African populations were at the root of the

modern human evolutionary tree.

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