Answers to
Questions and Probl
Chapter 1 ANS: There are 3 × 141 = 423 nucleotides in
1.1 In a few sentences, what were Mendel’s key ideas about ing sequence. Its polypeptide product
inheritance? amino acids.
ANS: Mendel postulated transmissible factors—genes—to 1.7 The template strand of a gene being tran
explain the inheritance of traits. He discovered that GCCAGT. What will be the sequence o
genes exist in different forms, which we now call alleles. from this template?
Each organism carries two copies of each gene. During ANS: GAACGGUCT
reproduction, one of the gene copies is randomly incor-
porated into each gamete. When the male and female 1.8 What is the difference between tr
gametes unite at fertilization, the gene copy number is translation?
restored to two. Different alleles may coexist in an organ- ANS: Transcription is the production of an RN
ism. During the production of gametes, they separate DNA chain as a template. Translation i
from each other without having been altered by of a chain of amino acids—that is, a pol
coexistence. an RNA chain as a template.
1.2 Both DNA and RNA are composed of nucleotides. What 1.9 RNA is synthesized using DNA as a te
molecules combine to form a nucleotide? ever synthesized using RNA as a templa
ANS: Each nucleotide consists of a sugar, a nitrogen-containing ANS: Sometimes, DNA is synthesized from R
base, and a phosphate. called reverse transcription. This proces
1.3 Which bases are present in DNA? Which bases are pres- tant role in the life cycles of some viruse
ent in RNA? Which sugars are present in each of these 1.10 The gene for a-globin is present in all ve
nucleic acids? Over millions of years, the DNA seque
ANS: The bases present in DNA are adenine, thymine, gua- has changed in the lineage of each specie
nine, and cytosine; the bases present in RNA are adenine, the amino acid sequence of a-globin has
uracil, guanine, and cytosine. The sugar in DNA is these lineages. Among the 141 amino
deoxyribose; the sugar in RNA is ribose. this polypeptide, human a-globin dif
a-globin in 79 positions; it differs from
1.4 What is a genome? 68 and from cow a-globin in 17. Do th
an evolutionary phylogeny for these ver
ANS: A genome is the set of all the DNA molecules that are
characteristic of an organism. Each DNA molecule ANS: The human and cow a-globins are least
forms one chromosome in a cell of the organism. fore, on the assumption that differen
reflect the degree of phylogenetic r
1.5 The sequence of a strand of DNA is ATTGCCGTC. If
human and the cow are the most closel
this strand serves as the template for DNA synthesis,
isms among those mentioned. The next
what will be the sequence of the newly synthesized
of humans is the carp, and the most dista
strand?
shark.
ANS: TAACGGCAG
1.11 Sickle-cell anemia is caused by a mutati
1.6 A gene contains 141 codons. How many nucleotides are codons in the gene for b-globin; because
present in the gene’s coding sequence? How many amino the sixth amino acid in the b-globin
acids are expected to be present in the polypeptide valine instead of a glutamic acid. A less se
encoded by this gene? mia is caused by a mutation that changes
ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 1
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to one specifying lysine as the sixth amino acid in the Eukaryotic cells usually possess a well-develope
b-globin polypeptide. What word is used to describe the nal system of membranes and they also have mem
two mutant forms of this gene? Do you think that an indi- bounded subcellular organelles such as mitoc
vidual carrying these two mutant forms of the b-globin and chloroplasts; prokaryotic cells do not typica
gene would suffer from anemia? Explain. a system of internal membranes (although some
do they possess membrane-bounded organelles.
ANS: The two mutant forms of the b-globin gene are properly
described as alleles. Because neither of the mutant alleles 2.4 Distinguish between the haploid and diploid
can specify a “normal” polypeptide, an individual who What types of cells are haploid? What types of
carries each of them would probably suffer from diploid?
anemia.
ANS: In the haploid state, each chromosome is repr
1.12 Hemophilia is an inherited disorder in which the blood- once; in the diploid state, each chromosome i
clotting mechanism is defective. Because of this defect, sented twice. Among multicellular eukaryote
people with hemophilia may die from cuts or bruises, etes are haploid and somatic cells are diploid.
especially if internal organs such as the liver, lungs, or
2.5 Compare the sizes and structures of prokaryo
kidneys have been damaged. One method of treatment
eukaryotic chromosomes.
involves injecting a blood-clotting factor that has been
purified from blood donations. This factor is a protein ANS: Prokaryotic chromosomes are typically (but not
encoded by a human gene. Suggest a way in which mod- smaller than eukaryotic chromosomes; in additi
ern genetic technology could be used to produce this karyotic chromosomes are circular, whereas eu
factor on an industrial scale. Is there a way in which the chromosomes are linear. For example, the circul
inborn error of hemophilia could be corrected by human mosome of E. coli, a prokaryote, is about 1.4 mm
gene therapy? cumference. By contrast, a linear human chrom
may be 10–30 cm long. Prokaryotic chromosom
ANS: The gene for the human clotting factor could be isolated
have a comparatively simple composition: DNA
from the human genome and transferred into bacteria,
RNA, and some protein. Eukaryotic chromoso
which could then be grown in vats to produce large
more complex: DNA, some RNA, and a lot of pr
amounts of the gene’s protein product. This product could
be isolated from the bacteria, purified, and then injected 2.6 With a focus on the chromosomes, what are
into patients to treat hemophilia. Another approach would events during interphase and M phase in the eu
be to transfer a normal copy of the clotting factor gene cell cycle?
into the cells of people who have hemophilia. If expressed
ANS: During interphase, the chromosomes duplicate.
properly, the transferred normal gene might be able to
M phase (mitosis), the duplicated chromosom
compensate for the mutant allele these people naturally
consisting of two identical sister chromatids, co
carry. For this approach to succeed, the normal clotting
(a feature of prophase), migrate to the equatorial
factor gene would have to be transferred into the cells that
the cell (a feature of metaphase), and then split
produce clotting factor, or into their precursors.
their constituent sister chromatids are separated
ferent daughter cells (a feature of anaphase);
Chapter 2
process is called sister chromatid disjunction.
2.1 Carbohydrates and proteins are linear polymers. What
2.7 Which typically lasts longer, interphase or M pha
types of molecules combine to form these polymers?
you explain why one of these phases lasts longer t
ANS: Sugars combine to form carbohydrates; amino acids other?
combine to form proteins.
ANS: Interphase typically lasts longer than M phase.
2.2 All cells are surrounded by a membrane; some cells are interphase, DNA must be synthesized to replicat
surrounded by a wall. What are the differences between chromosomes. Other materials must also be syn
cell membranes and cell walls? to prepare for the upcoming cell division.
ANS: Cell membranes are made of lipids and proteins; they 2.8 In what way do the microtubule organizing ce
have a fluid structure. Cell walls are made of more rigid plant and animal cells differ?
materials such as cellulose.
ANS: The microtubule organizing centers of animal ce
2.3 What are the principal differences between prokaryotic distinct centrosomes, whereas the microtubule o
and eukaryotic cells? ing centers of plant cells do not.
ANS: In a eukaryotic cell, the many chromosomes are con- 2.9 Match the stages of mitosis with the events they
tained within a membrane-bounded structure called the pass: Stages: (1) anaphase, (2) metaphase, (3) pr
nucleus; the chromosomes of prokaryotic cells are not and (4) telophase. Events: (a) reformation of the
contained within a special subcellular compartment. lus, (b) disappearance of the nuclear me
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(c) condensation of the chromosomes, (d) formation of How many are present in the male ga
the mitotic spindle, (e) movement of chromosomes to these nuclei haploid or diploid?
the equatorial plane, (f) movement of chromosomes to
ANS: Leaf tissue is diploid. The female game
the poles, (g) decondensation of the chromosomes, (h)
eight identical haploid nuclei. The ma
splitting of the centromere, and (i) attachment of micro-
contains three identical haploid nuclei.
tubules to the kinetochore.
2.17 From the information given in Table 2.
ANS: (1) Anaphase: (f), (h); (2) metaphase: (e), (i); (3) prophase:
is there a relationship between genome s
(b), (c), (d); (4) telophase: (a), (g).
base pairs of DNA) and gene number? E
2.10 Arrange the following events in the correct temporal ANS: Among eukaryotes, there does not see
sequence during eukaryotic cell division, starting with relationship between genome size and g
the earliest: (a) condensation of the chromosomes, example, humans, with 3.2 billion base
(b) movement of chromosomes to the poles, (c) duplica- DNA, have about 20,500 genes, and A
tion of the chromosomes, (d) formation of the nuclear with about 150 million base pairs of gen
membrane, (e) attachment of microtubules to the kineto- roughly the same number of genes as hu
chores, and (f) migration of centrosomes to positions on among prokaryotes, gene number is ra
opposite sides of the nucleus. related with genome size, probably bec
ANS: (c), (f), (a), (e), (b), (d). little nongenic DNA.
2.11 In human beings, the gene for b-globin is located on chro- 2.18 Are the synergid cells in an Arabidopsis
mosome 11, and the gene for a-globin, which is another phyte genetically identical to the egg cell
component of the hemoglobin protein, is located on chro- them?
mosome 16. Would these two chromosomes be expected to ANS: Yes.
pair with each other during meiosis? Explain your answer.
2.19 A cell of the bacterium Escherichia coli, a
ANS: Chromosomes 11 and 16 would not be expected to pair tains one chromosome with about 4.6 m
with each other during meiosis; these chromosomes are of DNA comprising 4288 protein-encod
heterologues, not homologues. of the yeast Saccharomyces cerevisiae, a euk
2.12 A sperm cell from the fruit fly Drosophila melanogaster about 12 million base pairs of DNA c
contains four chromosomes. How many chromosomes genes, and this DNA is distributed over
would be present in a spermatogonial cell about to enter mosomes. Are you surprised that the c
meiosis? How many chromatids would be present in a prokaryote is larger than some of the ch
spermatogonial cell at metaphase I of meiosis? How eukaryote? Explain your answer.
many would be present at metaphase II? ANS: It is a bit surprising that yeast chromoso
ANS: There are eight chromosomes in a Drosophila spermato- age, smaller than E. coli chromosomes b
gonial cell about to enter meiosis. There are 16 chroma- eukaryotic chromosomes are larger t
tids in a Drosophilia spermatogonial cell at metaphase I of chromosomes. Yeast is an exception beca
meiosis. There are eight chromatids in a Drosophilia cell not quite three times the size of the E.
at metaphase II of meiosis. distributed over 16 separate chromosom
2.13 Does crossing over occur before or after chromosome 2.20 Given the way that chromosomes behav
duplication in cells going through meiosis? is there any advantage for an organism
number of chromosome pairs (such as
ANS: Crossing over occurs after chromosomes have duplicated as opposed to an odd number of chromo
in cells going through meiosis. as human beings do)?
2.14 What visible characteristics of chromosomes indicate ANS: No, there is no advantage associated wi
that they have undergone crossing over during meiosis? ber of chromosomes. As long as the chr
ANS: The chiasmata, which are visible late in prophase I of in pairs, they will be able to synapse du
meiosis, indicate that chromosomes have crossed over. and then disjoin during anaphase I t
genetic material properly to the two dau
2.15 During meiosis, when does chromosome disjunction
occur? When does chromatid disjunction occur? 2.21 In flowering plants, two nuclei from the
ticipate in the events of fertilization. W
ANS: Chromosome disjunction occurs during anaphase I. from the female gametophyte do these
Chromatid disjunction occurs during anaphase II. What tissues are formed from the fertili
2.16 In Arabidopsis, is leaf tissue haploid or diploid? How ANS: One of the pollen nuclei fuses with the e
many nuclei are present in the female gametophyte? female gametophyte to form the zyg
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develops into an embryo and ultimately into a sporo- explain these results, diagram the crosses, and c
phyte. The other genetically functional pollen nucleus the results with the predictions of the hypothesis
fuses with two nuclei in the female gametophyte to form
ANS: The data suggest that coat color is controlled by
a triploid nucleus, which then develops into a triploid
gene with two alleles, C (gray) and c (albino), an
tissue, the endosperm; this tissue nourishes the develop-
is dominant over c. On this hypothesis, the cro
ing plant embryo.
gray (CC) × albino (cc) → F1 gray (Cc); F1 × F
2.22 The mouse haploid genome contains about 2.9 × 109 gray (2 CC: 1 Cc), 1/4 albino (cc). The expected r
nucleotide pairs of DNA. How many nucleotide pairs of the F2 are 203 gray and 67 albino. To comp
DNA are present in each of the following mouse cells: observed and expected results, compute c2 w
(a) somatic cell, (b) sperm cell, (c) fertilized egg, (d) pri- degree of freedom: (198 − 203)2/203 + (67 − 72
mary oocyte, (e) first polar body, and (f) secondary 0.470, which is not significant at the 5% level. T
spermatocyte? results are consistent with the hypothesis.
ANS: (a) 5.8 × 109 nucleotide pairs (np); (b) 2.9 × 109 np; 3.4 A woman has a rare abnormality of the eyelid
(c) 5.8 × 109 np; (d) 11.6 × 109 np; (e) 5.8 × 109 np; ptosis, which prevents her from opening her ey
and (f) 5.8 × 109 np pletely. This condition is caused by a dominan
2.23 Arabidopsis plants have 10 chromosomes (five pairs) in P. The woman’s father had ptosis, but her mot
their somatic cells. How many chromosomes are present normal eyelids. Her father’s mother had
in each of the following: (a) egg cell nucleus in the female eyelids.
gametophyte, (b) generative cell nucleus in a pollen (a) What are the genotypes of the woman, her fat
grain, (c) fertilized endosperm nucleus, and (d) fertilized her mother?
egg nucleus?
(b) What proportion of the woman’s children w
ANS: (a) 5, (b) 5, (c) 15, (d) 10. ptosis if she marries a man with normal eyelids?
Chapter 3 ANS: (a) Woman’s genotype Pp, father’s genotype Pp, m
genotype pp; (b) ½
3.1 On the basis of Mendel’s observations, predict the results
from the following crosses with peas: (a) a tall (dominant 3.5 In pigeons, a dominant allele C causes a checke
and homozygous) variety crossed with a dwarf variety; tern in the feathers; its recessive allele c produce
(b) the progeny of (a) self-fertilized; (c) the progeny from pattern. Feather coloration is controlled by an in
(a) crossed with the original tall parent; (d) the progeny dently assorting gene; the dominant allele B p
of (a) crossed with the original dwarf parent. red feathers, and the recessive allele b produce
feathers. Birds from a true-breeding checkered, r
ANS: (a) All tall; (b) 3/4 tall, 1/4 dwarf; (c) all tall; (d) 1/2 tall, ety are crossed to birds from a true-breedin
1/2 dwarf. brown variety.
3.2 Mendel crossed pea plants that produced round seeds (a) Predict the phenotype of their progeny.
with those that produced wrinkled seeds and self-fertil-
ized the progeny. In the F2, he observed 5474 round (b) If these progeny are intercrossed, what phe
seeds and 1850 wrinkled seeds. Using the letters W and will appear in the F2 and in what proportions?
w for the seed texture alleles, diagram Mendel’s crosses, ANS: (a) Checkered, red (CC BB) × plain, brown (cc b
showing the genotypes of the plants in each generation. all checkered, red (Cc Bb); (b) F2 progeny: 9/16
Are the results consistent with the Principle of ered, red (C- B-), 3/16 plain, red (cc B-), 3/16 che
Segregation? brown (C- bb), 1/16 plain, brown (cc bb).
ANS: Round (WW ) × wrinkled (ww) → F1 round (Ww); F1
3.6 In mice, the allele C for colored fur is dominant
self-fertilized → F2 3/4 round (2 WW; 1 Ww), 1/4 wrin-
allele c for white fur, and the allele V for normal b
kled (ww). The expected results in the F2 are 5493 round,
is dominant over the allele v for waltzing beh
1831 wrinkled. To compare the observed and expected
form of dis-coordination. Given the genotypes
results, compute c2 with one degree of freedom;
parents in each of the following crosses:
(5474 − 5493)2/5493 = (1850 − 1831)2/1831 = 0.263,
which is not significant at the 5% level. Thus, the results (a) Colored, normal mice mated with white, norm
are consistent with the Principle of Segregation. produced 29 colored, normal, and 10 colored, w
progeny
3.3 A geneticist crossed wild, gray-colored mice with white
(albino) mice. All the progeny were gray. These progeny (b) Colored, normal mice mated with colored,
were intercrossed to produce an F2, which consisted of mice produced 38 colored, normal, 15 colored, w
198 gray and 72 white mice. Propose a hypothesis to 11 white, normal, and 4 white, waltzing progeny
ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 4
Questions and Probl
Chapter 1 ANS: There are 3 × 141 = 423 nucleotides in
1.1 In a few sentences, what were Mendel’s key ideas about ing sequence. Its polypeptide product
inheritance? amino acids.
ANS: Mendel postulated transmissible factors—genes—to 1.7 The template strand of a gene being tran
explain the inheritance of traits. He discovered that GCCAGT. What will be the sequence o
genes exist in different forms, which we now call alleles. from this template?
Each organism carries two copies of each gene. During ANS: GAACGGUCT
reproduction, one of the gene copies is randomly incor-
porated into each gamete. When the male and female 1.8 What is the difference between tr
gametes unite at fertilization, the gene copy number is translation?
restored to two. Different alleles may coexist in an organ- ANS: Transcription is the production of an RN
ism. During the production of gametes, they separate DNA chain as a template. Translation i
from each other without having been altered by of a chain of amino acids—that is, a pol
coexistence. an RNA chain as a template.
1.2 Both DNA and RNA are composed of nucleotides. What 1.9 RNA is synthesized using DNA as a te
molecules combine to form a nucleotide? ever synthesized using RNA as a templa
ANS: Each nucleotide consists of a sugar, a nitrogen-containing ANS: Sometimes, DNA is synthesized from R
base, and a phosphate. called reverse transcription. This proces
1.3 Which bases are present in DNA? Which bases are pres- tant role in the life cycles of some viruse
ent in RNA? Which sugars are present in each of these 1.10 The gene for a-globin is present in all ve
nucleic acids? Over millions of years, the DNA seque
ANS: The bases present in DNA are adenine, thymine, gua- has changed in the lineage of each specie
nine, and cytosine; the bases present in RNA are adenine, the amino acid sequence of a-globin has
uracil, guanine, and cytosine. The sugar in DNA is these lineages. Among the 141 amino
deoxyribose; the sugar in RNA is ribose. this polypeptide, human a-globin dif
a-globin in 79 positions; it differs from
1.4 What is a genome? 68 and from cow a-globin in 17. Do th
an evolutionary phylogeny for these ver
ANS: A genome is the set of all the DNA molecules that are
characteristic of an organism. Each DNA molecule ANS: The human and cow a-globins are least
forms one chromosome in a cell of the organism. fore, on the assumption that differen
reflect the degree of phylogenetic r
1.5 The sequence of a strand of DNA is ATTGCCGTC. If
human and the cow are the most closel
this strand serves as the template for DNA synthesis,
isms among those mentioned. The next
what will be the sequence of the newly synthesized
of humans is the carp, and the most dista
strand?
shark.
ANS: TAACGGCAG
1.11 Sickle-cell anemia is caused by a mutati
1.6 A gene contains 141 codons. How many nucleotides are codons in the gene for b-globin; because
present in the gene’s coding sequence? How many amino the sixth amino acid in the b-globin
acids are expected to be present in the polypeptide valine instead of a glutamic acid. A less se
encoded by this gene? mia is caused by a mutation that changes
ONLINE_AnswerstoOddNumberedQuestionsandProblems.indd 1
, 2-WC Answers to All Questions and Problems
to one specifying lysine as the sixth amino acid in the Eukaryotic cells usually possess a well-develope
b-globin polypeptide. What word is used to describe the nal system of membranes and they also have mem
two mutant forms of this gene? Do you think that an indi- bounded subcellular organelles such as mitoc
vidual carrying these two mutant forms of the b-globin and chloroplasts; prokaryotic cells do not typica
gene would suffer from anemia? Explain. a system of internal membranes (although some
do they possess membrane-bounded organelles.
ANS: The two mutant forms of the b-globin gene are properly
described as alleles. Because neither of the mutant alleles 2.4 Distinguish between the haploid and diploid
can specify a “normal” polypeptide, an individual who What types of cells are haploid? What types of
carries each of them would probably suffer from diploid?
anemia.
ANS: In the haploid state, each chromosome is repr
1.12 Hemophilia is an inherited disorder in which the blood- once; in the diploid state, each chromosome i
clotting mechanism is defective. Because of this defect, sented twice. Among multicellular eukaryote
people with hemophilia may die from cuts or bruises, etes are haploid and somatic cells are diploid.
especially if internal organs such as the liver, lungs, or
2.5 Compare the sizes and structures of prokaryo
kidneys have been damaged. One method of treatment
eukaryotic chromosomes.
involves injecting a blood-clotting factor that has been
purified from blood donations. This factor is a protein ANS: Prokaryotic chromosomes are typically (but not
encoded by a human gene. Suggest a way in which mod- smaller than eukaryotic chromosomes; in additi
ern genetic technology could be used to produce this karyotic chromosomes are circular, whereas eu
factor on an industrial scale. Is there a way in which the chromosomes are linear. For example, the circul
inborn error of hemophilia could be corrected by human mosome of E. coli, a prokaryote, is about 1.4 mm
gene therapy? cumference. By contrast, a linear human chrom
may be 10–30 cm long. Prokaryotic chromosom
ANS: The gene for the human clotting factor could be isolated
have a comparatively simple composition: DNA
from the human genome and transferred into bacteria,
RNA, and some protein. Eukaryotic chromoso
which could then be grown in vats to produce large
more complex: DNA, some RNA, and a lot of pr
amounts of the gene’s protein product. This product could
be isolated from the bacteria, purified, and then injected 2.6 With a focus on the chromosomes, what are
into patients to treat hemophilia. Another approach would events during interphase and M phase in the eu
be to transfer a normal copy of the clotting factor gene cell cycle?
into the cells of people who have hemophilia. If expressed
ANS: During interphase, the chromosomes duplicate.
properly, the transferred normal gene might be able to
M phase (mitosis), the duplicated chromosom
compensate for the mutant allele these people naturally
consisting of two identical sister chromatids, co
carry. For this approach to succeed, the normal clotting
(a feature of prophase), migrate to the equatorial
factor gene would have to be transferred into the cells that
the cell (a feature of metaphase), and then split
produce clotting factor, or into their precursors.
their constituent sister chromatids are separated
ferent daughter cells (a feature of anaphase);
Chapter 2
process is called sister chromatid disjunction.
2.1 Carbohydrates and proteins are linear polymers. What
2.7 Which typically lasts longer, interphase or M pha
types of molecules combine to form these polymers?
you explain why one of these phases lasts longer t
ANS: Sugars combine to form carbohydrates; amino acids other?
combine to form proteins.
ANS: Interphase typically lasts longer than M phase.
2.2 All cells are surrounded by a membrane; some cells are interphase, DNA must be synthesized to replicat
surrounded by a wall. What are the differences between chromosomes. Other materials must also be syn
cell membranes and cell walls? to prepare for the upcoming cell division.
ANS: Cell membranes are made of lipids and proteins; they 2.8 In what way do the microtubule organizing ce
have a fluid structure. Cell walls are made of more rigid plant and animal cells differ?
materials such as cellulose.
ANS: The microtubule organizing centers of animal ce
2.3 What are the principal differences between prokaryotic distinct centrosomes, whereas the microtubule o
and eukaryotic cells? ing centers of plant cells do not.
ANS: In a eukaryotic cell, the many chromosomes are con- 2.9 Match the stages of mitosis with the events they
tained within a membrane-bounded structure called the pass: Stages: (1) anaphase, (2) metaphase, (3) pr
nucleus; the chromosomes of prokaryotic cells are not and (4) telophase. Events: (a) reformation of the
contained within a special subcellular compartment. lus, (b) disappearance of the nuclear me
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(c) condensation of the chromosomes, (d) formation of How many are present in the male ga
the mitotic spindle, (e) movement of chromosomes to these nuclei haploid or diploid?
the equatorial plane, (f) movement of chromosomes to
ANS: Leaf tissue is diploid. The female game
the poles, (g) decondensation of the chromosomes, (h)
eight identical haploid nuclei. The ma
splitting of the centromere, and (i) attachment of micro-
contains three identical haploid nuclei.
tubules to the kinetochore.
2.17 From the information given in Table 2.
ANS: (1) Anaphase: (f), (h); (2) metaphase: (e), (i); (3) prophase:
is there a relationship between genome s
(b), (c), (d); (4) telophase: (a), (g).
base pairs of DNA) and gene number? E
2.10 Arrange the following events in the correct temporal ANS: Among eukaryotes, there does not see
sequence during eukaryotic cell division, starting with relationship between genome size and g
the earliest: (a) condensation of the chromosomes, example, humans, with 3.2 billion base
(b) movement of chromosomes to the poles, (c) duplica- DNA, have about 20,500 genes, and A
tion of the chromosomes, (d) formation of the nuclear with about 150 million base pairs of gen
membrane, (e) attachment of microtubules to the kineto- roughly the same number of genes as hu
chores, and (f) migration of centrosomes to positions on among prokaryotes, gene number is ra
opposite sides of the nucleus. related with genome size, probably bec
ANS: (c), (f), (a), (e), (b), (d). little nongenic DNA.
2.11 In human beings, the gene for b-globin is located on chro- 2.18 Are the synergid cells in an Arabidopsis
mosome 11, and the gene for a-globin, which is another phyte genetically identical to the egg cell
component of the hemoglobin protein, is located on chro- them?
mosome 16. Would these two chromosomes be expected to ANS: Yes.
pair with each other during meiosis? Explain your answer.
2.19 A cell of the bacterium Escherichia coli, a
ANS: Chromosomes 11 and 16 would not be expected to pair tains one chromosome with about 4.6 m
with each other during meiosis; these chromosomes are of DNA comprising 4288 protein-encod
heterologues, not homologues. of the yeast Saccharomyces cerevisiae, a euk
2.12 A sperm cell from the fruit fly Drosophila melanogaster about 12 million base pairs of DNA c
contains four chromosomes. How many chromosomes genes, and this DNA is distributed over
would be present in a spermatogonial cell about to enter mosomes. Are you surprised that the c
meiosis? How many chromatids would be present in a prokaryote is larger than some of the ch
spermatogonial cell at metaphase I of meiosis? How eukaryote? Explain your answer.
many would be present at metaphase II? ANS: It is a bit surprising that yeast chromoso
ANS: There are eight chromosomes in a Drosophila spermato- age, smaller than E. coli chromosomes b
gonial cell about to enter meiosis. There are 16 chroma- eukaryotic chromosomes are larger t
tids in a Drosophilia spermatogonial cell at metaphase I of chromosomes. Yeast is an exception beca
meiosis. There are eight chromatids in a Drosophilia cell not quite three times the size of the E.
at metaphase II of meiosis. distributed over 16 separate chromosom
2.13 Does crossing over occur before or after chromosome 2.20 Given the way that chromosomes behav
duplication in cells going through meiosis? is there any advantage for an organism
number of chromosome pairs (such as
ANS: Crossing over occurs after chromosomes have duplicated as opposed to an odd number of chromo
in cells going through meiosis. as human beings do)?
2.14 What visible characteristics of chromosomes indicate ANS: No, there is no advantage associated wi
that they have undergone crossing over during meiosis? ber of chromosomes. As long as the chr
ANS: The chiasmata, which are visible late in prophase I of in pairs, they will be able to synapse du
meiosis, indicate that chromosomes have crossed over. and then disjoin during anaphase I t
genetic material properly to the two dau
2.15 During meiosis, when does chromosome disjunction
occur? When does chromatid disjunction occur? 2.21 In flowering plants, two nuclei from the
ticipate in the events of fertilization. W
ANS: Chromosome disjunction occurs during anaphase I. from the female gametophyte do these
Chromatid disjunction occurs during anaphase II. What tissues are formed from the fertili
2.16 In Arabidopsis, is leaf tissue haploid or diploid? How ANS: One of the pollen nuclei fuses with the e
many nuclei are present in the female gametophyte? female gametophyte to form the zyg
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, 4-WC Answers to All Questions and Problems
develops into an embryo and ultimately into a sporo- explain these results, diagram the crosses, and c
phyte. The other genetically functional pollen nucleus the results with the predictions of the hypothesis
fuses with two nuclei in the female gametophyte to form
ANS: The data suggest that coat color is controlled by
a triploid nucleus, which then develops into a triploid
gene with two alleles, C (gray) and c (albino), an
tissue, the endosperm; this tissue nourishes the develop-
is dominant over c. On this hypothesis, the cro
ing plant embryo.
gray (CC) × albino (cc) → F1 gray (Cc); F1 × F
2.22 The mouse haploid genome contains about 2.9 × 109 gray (2 CC: 1 Cc), 1/4 albino (cc). The expected r
nucleotide pairs of DNA. How many nucleotide pairs of the F2 are 203 gray and 67 albino. To comp
DNA are present in each of the following mouse cells: observed and expected results, compute c2 w
(a) somatic cell, (b) sperm cell, (c) fertilized egg, (d) pri- degree of freedom: (198 − 203)2/203 + (67 − 72
mary oocyte, (e) first polar body, and (f) secondary 0.470, which is not significant at the 5% level. T
spermatocyte? results are consistent with the hypothesis.
ANS: (a) 5.8 × 109 nucleotide pairs (np); (b) 2.9 × 109 np; 3.4 A woman has a rare abnormality of the eyelid
(c) 5.8 × 109 np; (d) 11.6 × 109 np; (e) 5.8 × 109 np; ptosis, which prevents her from opening her ey
and (f) 5.8 × 109 np pletely. This condition is caused by a dominan
2.23 Arabidopsis plants have 10 chromosomes (five pairs) in P. The woman’s father had ptosis, but her mot
their somatic cells. How many chromosomes are present normal eyelids. Her father’s mother had
in each of the following: (a) egg cell nucleus in the female eyelids.
gametophyte, (b) generative cell nucleus in a pollen (a) What are the genotypes of the woman, her fat
grain, (c) fertilized endosperm nucleus, and (d) fertilized her mother?
egg nucleus?
(b) What proportion of the woman’s children w
ANS: (a) 5, (b) 5, (c) 15, (d) 10. ptosis if she marries a man with normal eyelids?
Chapter 3 ANS: (a) Woman’s genotype Pp, father’s genotype Pp, m
genotype pp; (b) ½
3.1 On the basis of Mendel’s observations, predict the results
from the following crosses with peas: (a) a tall (dominant 3.5 In pigeons, a dominant allele C causes a checke
and homozygous) variety crossed with a dwarf variety; tern in the feathers; its recessive allele c produce
(b) the progeny of (a) self-fertilized; (c) the progeny from pattern. Feather coloration is controlled by an in
(a) crossed with the original tall parent; (d) the progeny dently assorting gene; the dominant allele B p
of (a) crossed with the original dwarf parent. red feathers, and the recessive allele b produce
feathers. Birds from a true-breeding checkered, r
ANS: (a) All tall; (b) 3/4 tall, 1/4 dwarf; (c) all tall; (d) 1/2 tall, ety are crossed to birds from a true-breedin
1/2 dwarf. brown variety.
3.2 Mendel crossed pea plants that produced round seeds (a) Predict the phenotype of their progeny.
with those that produced wrinkled seeds and self-fertil-
ized the progeny. In the F2, he observed 5474 round (b) If these progeny are intercrossed, what phe
seeds and 1850 wrinkled seeds. Using the letters W and will appear in the F2 and in what proportions?
w for the seed texture alleles, diagram Mendel’s crosses, ANS: (a) Checkered, red (CC BB) × plain, brown (cc b
showing the genotypes of the plants in each generation. all checkered, red (Cc Bb); (b) F2 progeny: 9/16
Are the results consistent with the Principle of ered, red (C- B-), 3/16 plain, red (cc B-), 3/16 che
Segregation? brown (C- bb), 1/16 plain, brown (cc bb).
ANS: Round (WW ) × wrinkled (ww) → F1 round (Ww); F1
3.6 In mice, the allele C for colored fur is dominant
self-fertilized → F2 3/4 round (2 WW; 1 Ww), 1/4 wrin-
allele c for white fur, and the allele V for normal b
kled (ww). The expected results in the F2 are 5493 round,
is dominant over the allele v for waltzing beh
1831 wrinkled. To compare the observed and expected
form of dis-coordination. Given the genotypes
results, compute c2 with one degree of freedom;
parents in each of the following crosses:
(5474 − 5493)2/5493 = (1850 − 1831)2/1831 = 0.263,
which is not significant at the 5% level. Thus, the results (a) Colored, normal mice mated with white, norm
are consistent with the Principle of Segregation. produced 29 colored, normal, and 10 colored, w
progeny
3.3 A geneticist crossed wild, gray-colored mice with white
(albino) mice. All the progeny were gray. These progeny (b) Colored, normal mice mated with colored,
were intercrossed to produce an F2, which consisted of mice produced 38 colored, normal, 15 colored, w
198 gray and 72 white mice. Propose a hypothesis to 11 white, normal, and 4 white, waltzing progeny
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