Hyde Chapter 3 – Solutions
9
3
MITOSIS AND MEIOSIS
CHAPTER SUMMARY QUESTIONS
2.
See table 3.1 for a summary answer.
4.
When a eukaryotic chromosome replicates during the S phase of the cell cycle, one
chromosome becomes two chromatids, attached near the centromere. These are sister
chromatids. Chromatids of different chromosomes are nonsister chromatids.
6.
The terms reductional and equational refer to the segregation of chromosomes
during meiosis. The first division is termed reductional because the homologous
chromosomes separate from each other, reducing the number of chromosomes in half.
The second meiotic division is called equational because the number of chromosomes
remains the same although the number of chromatids is halved.
8.
After the S phase, the sister chromatids are kept together by a multiprotein complex
termed cohesin. The anaphase-promoting complex (APC) triggers the destruction of
cohesin. Activated APC catalyzes the ubiquination and degradation of the protein
securin, an inhibitory protein that binds and inactivates a protease called separase.
Free from securin, separase can now degrade cohesin and permit the separation of
the sister chromatids from each other.
10. The two meiotic mechanisms that generate genetic diversity are (1) an independent
assortment of chromosomes, which starts in metaphase I and ends in anaphase I; and
(2) crossing over between nonsister chromatids of homologous chromosomes, which
takes place in prophase I.
12. In diploid organisms, chromosomes occur in pairs. The haploid number, or number
of pairs of chromosomes, may be even (4 in the fruit fly, for example) or odd (23 in
humans). However, the total diploid number is given by 2 the number of
chromosome pairs, and therefore it will be an even number.
14. 1.
In spermatogenesis, both meiotic divisions are followed by equal cytoplasmic
divisions, yielding four functional gametes. In oogenesis, unequal cytoplasmic
divisions usually produce only one functional gamete and up to three polar
bodies.
10
Hyde Chapter 3—Solutions
2.
3.
4.
Once spermatogenesis begins, it occurs continuously. Oogenesis, on the other
hand, starts during embryonic development and then pauses until puberty, where
it resumes for only one oocyte, usually, per month.
Spermatogenesis occurs throughout adult life, while oogenesis ends at
menopause.
Spermatogenesis results in the production of hundreds of millions of sperm per
day, whereas on average a human female produces about 400 ova in her
lifetime.
EXERCISES AND PROBLEMS
16.
18. a.
Mitosis, prophase, 2n = 6 (the two parallel threads represent sister chromatids);
or meiosis II, early prophase, 2n = 12 (each thread represents a chromosome,
and sister chromatids are not evident yet)
b. Mitosis, metaphase, 2n = 6, or meiosis II, metaphase, 2n = 12
c. Mitosis, anaphase, 2n = 6, or meiosis II, anaphase, 2n = 12
d. Meiosis I, early prophase, 2n = 6 (each thread represents a chromosome, and
sister chromatids are not evident yet)
e. Meiosis I, anaphase, 2n = 6
f. Meiosis II, anaphase, 2n = 6
20.
Human being
Garden pea
Fruit fly
House mouse
2n
(= Dyads)
46
14
8
40
Bivalent
(= Tetrads)
23
7
4
20
Hyde Chapter 3 – Solutions
Roundworm
Pigeon
Boa constrictor
Cricket
Lily
Indian fern
11
2
80
36
22
24
1260
1
40
18
11
12
630
22. The number of combinations is 2n where n = the number of different chromosome
pairs. In this case, n = 6, so we expect 26 = 64 different combinations.
24.
DNA
(Number of
Chromatids)
Spermatogonium or oogonium
2
Primary spermatocyte or primary oocyte
4
Secondary spermatocyte or secondary oocyte
2
Spermatid or ovum
1
Sperm
1
Ploidy
2n
2n
n
n
n
26. a.
The primary oocyte is diploid and will undergo meiosis, but only one functional
ovum results from each primary oocyte: 1 50 = 50 eggs.
b. Each secondary oocyte will undergo meiosis II to produce an ovum and a polar
body: 1 50 = 50 eggs.
28. Homologous chromosomes will pair during meiosis. Each gamete gets one of each
chromosome, A, B, C, D, and E. Fertilization fuses two cells with the chromosome
complement given. Since root cells are somatic tissue, these cells will be diploid.
Therefore, the answer is b. AA BB CC DD EE.
30. a. 21 chromosomes (n)
b. 42 chromosomes (2n)
c. 63 chromosomes (3n)
32. The sperm cells produced by the pollen parent have the A genotype. The female
parent contains the egg cell and two polar nuclei, all of which have genotype a. One
sperm cell will fertilize the egg cell resulting in a diploid zygote with genotype Aa.
The two polar nuclei are fertilized by a second sperm cell, producing a triploid
endosperm with genotype Aaa.
34. The number of different gametes produced by each parent is 220. Using the “and
rule” (refer back to chapter 2), we can calculate the number of different types of
offspring that can be produced from the mating: 220 220 = 240 possibilities!
36. AaBb queen ab drone yields for sons (or gametes): AB, Ab, aB, ab and for
daughters: AaBb, Aabb, aaBb, aabb.
12
Hyde Chapter 3—Solutions
38. a. 250 or about 1.1 1015
b. 22 = 4
The number of gametes produced is 2n, where n = number of independently behaving
entities. If the genes are completely independent, we expect 250, and if they are
completely linked, then the number of different gametes will be determined by the
number of different chromosomes, so we expect 22. In reality, the number falls
between these two extremes.
40.
Mitosis
Chromosome no.
Chromatid no.
Picograms DNA
Mitosis
4
8
0.6
Meiosis I
Meiosis I
4
8
0.6
Meiosis II
Meiosis II
2
4
0.3
CHAPTER INTEGRATION PROBLEM
a.
The process of mitosis does not relate directly to Mendel’s rules. The behavior of
chromosomes during meiosis, however, explains both equal segregation (first law)
and independent assortment (second law). Mendel’s law of segregation can be
explained by the homologous pairing and segregation of chromosomes during
meiosis. Only one chromosome from each homologous pair goes into a gamete. So,
for example, the A and a alleles will each segregate to a different gamete. This is also
true for the B/b, C/c, and D/d alleles. Mendel’s law of independent assortment can be
explained by the relative behavior of different (nonhomologous) chromosomes
during meiosis I. The pairing arrangement, separation, and migration of one
homologous pair do not influence the orientation of an adjacent homologous pair. In
our example, the assortment of gene A is independent of that of genes B, C, or D. In
other words, a gamete is equally likely to inherit the A and B alleles or the A and b
Hyde Chapter 3 – Solutions
13
alleles. Similarly, it can obtain the c and D alleles or the c and d alleles, with equal
probabilities.
b. Gene C is found on the p arm of an acrocentric chromosome, while genes B and D
are found on the q arm of different metacentric chromosomes.
c. The cell is in prophase of mitosis. The chromosomes are condensed but not yet lined
up in the center of the cell. In prophase I of meiosis there would be a synapsis
between maternal and paternal homologs. In prophase II of meiosis there would only
be four chromosomes (the haploid number) in the cell.
d. AaBbCcDd.
e. Four tetrads will form during prophase I of meiosis. These can align in eight
different ways in metaphase I. To draw them all, keep one tetrad fixed and change
the other three tetrads in systematic fashion.
The following eight possible arrangements correspond to Mendel’s second law
of independent assortment. Each of these arrangements will occur with equal
probability and each will produce two different types of gametes.
A
A a
a
B
B b
b
C
C c
A
A a
a
B
B b
b
C
C c
c
c
D
D d
d
d
d D
D
A
A a
a
A
A a
a
B
B b
b
B
B b
b
c
c C
C
c
c C
C
D
D d
d
d
d D
D
14
f.
Hyde Chapter 3—Solutions
A
A a
a
A
A a
a
b
b B
B
b
b B
B
C
C c
c
C
C c
c
D
D d
d
d
d D
D
A
A a
a
A
A a
a
b
b B
B
b
b B
B
c
c C
C
c
c C
C
D
D d
d
d
d D
D
The number of chromosomal combinations is 2n, where n = the number of different
chromosome pairs. Thus, there are 24 = 16 possible combinations. The chance of
getting one member of a set is 1/2; so the chance of getting ABCD, or all four from
the maternal side, is (1/2)4 = 1/16.
g. To solve this part use an empirical approach that lists all possible combinations with
two chromosomes from one parent. These are ABcd, abCD, AbCd, aBcD, AbcD, and
aBCd. So the answer is 6/16 = 3/8.
h. The set of gametes with a mixture of maternal and paternal chromosomes consists of
all gametes except those with only maternal or only paternal chromosomes. The
chance of a gamete getting only maternal chromosomes is 1/16 (see part f).
Similarly, the chance of a gamete getting only paternal chromosomes will be 1/16.
Therefore, the chance of getting a mixture of maternal and paternal chromosomes is
1 – (1/16 + 1/16) = 14/16 = 7/8.