Hyde Chapter 4 – Solutions
15
4
SEX LINKAGE AND PEDIGREE
ANALYSIS
CHAPTER SUMMARY QUESTIONS
2.
The differences are in terminology only, not in shape or size of the chromosomes. In
species in which females have a homomorphic pair of sex chromosomes, the
members of the pair are called X chromosomes. In species in which males have a
homomorphic sex chromosome pair, the members of the pair are called
Z chromosomes.
4.
Because a human male has only one X chromosome, he would be neither
homozygous nor heterozygous for Duchenne muscular dystrophy. These terms only
make sense in the female, where two X chromosomes are present. An affected male
is said to be hemizygous.
6.
Both Drosophila melanogaster and Caenorhabditis elegans use a genic balance
mechanism, where sex is determined by the ratio of X chromosomes to sets of
autosomes (X:A). An X:A ratio of 0.5 produces males in both species, whereas an
X:A ratio of 1.0 produces a female in Drosophila and a hermaphrodite in
Caenorhabditis.
8.
ESD refers to environmental sex determination, where the sex of the organism is
determined by the environment and not by sex chromosomes of the individual. For
example, sex determination in many reptilian species is determined by the
environmental temperature during egg development. The mechanisms of ESD likely
involve pathways for steroid biosynthesis. Two enzymes appear to play a major role:
reductase, which converts testosterone into dihydro-testosterone, is found in higher
levels at male-producing temperatures; aromatase, which converts testosterone into
oestradiol, is found in higher levels at female-producing temperatures.
10. Y-linked traits share two main characteristics: (1) they are only found in males and
(2) they are transmitted from an affected father to all his sons and none of his daughters.
16
Hyde Chapter 4—Solutions
EXERCISES AND PROBLEMS
12. Because male honeybees are haploid, they produce sperm via mitosis. Meiosis would
not be a successful process because there is only one copy of each chromosome and
therefore no homologs to pair with. Mitosis, on the other hand, would be successful
because it does not require pairing (synapsis) of homologous chromosomes.
14. In Drosophila, sex is determined by the ratio of X chromosomes to sets of autosomes
(X:A). Males have an X:A ratio of 0.5 (if the ratio is less than 0.5, metamales are
produced); females have an X:A ratio of 1.0 (if the ratio is more than 1.0,
metafemales are produced). A ratio that is between 0.5 and 1.0 yields intersex
individuals.
Chromosome Composition
Sex
Chromosomes
Autosomes
a.
X
Diploid
b.
XX
Diploid
c.
XY
Diploid
d.
XXX
Diploid
e.
XXY
Diploid
f.
X
Triploid
g.
XX
Triploid
h.
XY
Triploid
i.
XXX
Triploid
j.
XXY
Triploid
k.
XX
Tetraploid
l.
XXX
Tetraploid
m.
XXYY
Tetraploid
Sex of Fly
Male (X:A = 0.5)
Female (X:A = 1.0)
Male (X:A = 0.5)
Metafemale (X:A = 1.5)
Female (X:A = 1.0)
Metamale (X:A = 0.33)
Intersex (X:A = 0.67)
Metamale (X:A = 0.33)
Female (X:A = 1.0)
Intersex (X:A = 0.67)
Male (X:A = 0.5)
Intersex (X:A = 0.75)
Male (X:A = 0.5)
16. A grandson does not inherit a sex chromosome from his father’s mother. A
granddaughter, on the other hand, does not inherit any sex chromosomes from her
father’s father.
18. a. Apparent ratio of offspring is 3:1, male to female. Fertile males are 50% tra tra
and 50% tra+tra. Fertile daughters are all tra+tra; daughters counted as sterile
males are tra tra = 1/4 of total.
b. If the progeny in part (a) are mated among themselves, all females will be
tra+tra (tra tra females are sterile). Males will be tra tra and tra+tra. The mating
of tra+tra females with tra tra males produces 3/4 males and 1/4 females. The
mating of tra+tra females with tra+tra males produces 5/8 males (1/2 normal
males plus 1/4 1/2 = 1/8 transformed females) and 3/8 females (1/2 females
minus the 1/8 transformed). Since these matings are in equal frequencies, there
are 11/16 males [(3/4 1/2) + (5/8 1/2)] and 5/16 females [(1/4 1/2) +
(3/8 1/2)].
Hyde Chapter 4 – Solutions
20. a.
17
In chicken, the male is homogametic (ZZ) and the female heterogametic (ZW).
The cross can be depicted as such:
P
ZBW
Barred hen
F1
ZbW
ZBZb
Nonbarred hen Barred rooster
ZbZb
Nonbarred rooster
These would occur in equal proportions, so 50% of the F1 offspring would be
nonbarred hens, and 50% of the F1 offspring would be barred roosters.
F2
ZBW
;
ZbW
;
ZBZb
;
ZbZb
Barred hen
Nonbarred hen Barred rooster Nonbarred rooster
These would occur in equal proportions, so the F2 would consist of 1/4 barred
hens, 1/4 nonbarred hens, 1/4 barred roosters, and 1/4 nonbarred roosters.
b. The cross must be ZBW ZBZb. The only nonbarred offspring would be ZbW,
and thus female.
22. Nothing can be inferred from the first cross. The second cross indicates that black is
dominant to pink. The third cross alerts us to sex linkage, as we see a difference
between males and females. Therefore, eye color in canaries is sex-linked, with
black eye dominant to pink eye. Recall that in birds, the male is the homogametic
(ZZ) sex. If we let B = black and b = pink, the crosses can be diagrammed as
follows:
a.
ZbW
ZbZb
b. ZbW
b
ZBZB
c.
ZBW
b b
B
ZW
ZZ
All pink-eyed
ZbZb
B b
Z W
Z Z
All black-eyed
b
ZW
Pink-eyed
female
24.
Cross
Reciprocal
P
Female
Male
ZpW
Z+Z+
Z+W
ZpZp
F1
Female
Male
Z+W
Z+Zp
ZpW
Z+Zp
F2
Females
Males
Z+W, ZpW
Z+Z+, Z+Zp
Z+W, ZpW
Z+Zp, ZpZp
ZBZb
Black-eyed
male
18
Hyde Chapter 4—Solutions
26. Yes. Begin by determining genotypes of the two individuals. The woman must be
heterozygous XCXc. A man with normal vision must be XCY, and all his daughters
must receive his X chromosome and should be normal, either XCXC or XCXc. Since
color blindness is recessive, the daughter must have two Xc chromosomes. Note: A
very rare possibility is that the man is the father and nondisjunction occurred in both
parents: at meiosis II in the female and either meiosis I or II in the male (see chapter
8).
28. All the offspring of the abnormal-1 cross are normal, indicating that the abnormal-1
trait must be recessive. If the trait were X-linked, the male offspring of this mating
would be expected to have abnormal eyes. Since all offspring have normal eyes, the
abnormal-1 trait must be autosomal and the abnormal-1 flies are homozygous. In the
abnormal-2 cross, the 1:1 ratio within each sex indicates a mating between a
heterozygote and a homozygote. The cross could be XaY XAXa or aa Aa.
Therefore, abnormal-2 flies are heterozygous for a dominant gene that could be
either autosomal or X-linked. Finally, the abnormal-3 trait must be dominant, but we
cannot determine if it is X-linked or autosomal. We can tell, however, that the
abnormal-3 flies are homozygous for that dominant gene.
30. a.
Let X+ = wild-type eye, Xv = vermillion eye.
P
XvXv
F1
+
X+Y
v
XvY
X X
Therefore, a ratio of 1 wild-type female:1 vermillion male.
b. F1
X+Xv XvY
F2
+
v
X X
XvXv
X+Y
XvY
The F2 progeny are expected in a ratio of 1 wild-type female:1 vermillion female:
1 wild-type male:1 vermillion male.
Observed numbers (O)
Expected ratio
Expected numbers (E)
O–E
(O – E)2
(O – E)2/E
X+Xv
58
1/4
61.25
–3.25
10.56
0.172
Offspring
XvXv
X+Y
62
60
1/4
1/4
61.25 61.25
0.75
–1.25
0.56
1.56
0.009 0.025
XvY
65
1/4
61.25
3.75
14.06
0.229
Total
245
245
0.435 = x2
Critical chi-square at 0.05, 3 df, = 7.815. The chi-square, 0.435, is less than the
critical chi-square. Therefore, we fail to reject (that is, accept) the null hypothesis
Hyde Chapter 4 – Solutions
19
and conclude that the observed numbers are consistent with the expectations of Xlinked gene inheritance.
32. Assuming 100% penetrance:
a.
b.
c.
d.
Note: Other pedigrees are possible.
34. a.
Autosomal recessive inheritance is a possibility. However, X-linked recessive
is more likely because all affected individuals are males.
b.
Autosomal dominant and autosomal recessive modes of inheritance are
possible. However, the most likely mode of inheritance is X-linked dominant
(affected males transmit the trait to all of their daughters but none of their sons).
c.
Autosomal dominant inheritance is possible. However, Y-linked inheritance is
the most likely mode of inheritance. Only males are affected, and affected
males transmit the trait to all of their sons but none of their daughters.
36. a.
The phenotype is the propensity to have twin offspring. It could be caused
by a recessive or dominant, sex-linked or autosomal allele.
b. X-linked and Y-linked modes of inheritance are impossible. Both autosomal
dominant and autosomal recessive inheritance are possible. However, autosomal
dominance is more probable because autosomal recessive inheritance requires
four different individuals (I-1, II-3, III-1 and IV-5) to be carriers of this rare
trait.
c. Y-linked, X-linked dominant, and autosomal dominant modes of inheritance are
impossible. Autosomal or X-linked recessive inheritance are possible. However,
X-linked recessive is more probable because it requires less people to be carriers
than autosomal recessive (namely, individuals II-5 and II-7).
d. Autosomal recessive inheritance is the only possibility.
20
Hyde Chapter 4—Solutions
38. The genotype/phenotype designations are as follows:
Genotype
BB
Bb
bb
Bb
Phenotype in Males
Bald
Bald
Normal
Bb
–b
bb
Phenotype in Females
Bald
Normal
Normal
B–
Bb
Bb
40. Because one of their daughters is albino (aa), both parents must be carriers for the
disease (Aa). The second daughter is hemophilic (XhXh). Therefore, the father must
have the trait (XhY), and the mother must have at least one Xh allele. The fact that her
father had normal blood clotting means that she must be normal for hemophilia
(XHXh). Therefore, the cross can be designated as Aa XHXh Aa XhY.
Because the two traits assort independently, we can treat the cross as two separate
crosses and then use the product rule to determine the combined probabilities.
a. For albinism: Aa Aa 3/4 A–
For hemophilia: XHXh XhY 1/4 XHY
Therefore, the probability of genotype A– XHY = 3/4 1/4 = 3/16.
b. For albinism: Aa Aa 1/4 aa
For hemophilia: XHXh XhY 1/4 XhXh
Therefore, the probability of genotype aa XhXh = 1/4 1/4 = 1/16.
c. Only a daughter can be a carrier for both traits
For albinism: Aa Aa 1/2 Aa
For hemophilia: XHXh XhY 1/4 XHXh
Therefore, the probability of genotype Aa XHXh = 1/2 1/4 = 1/8.
Hyde Chapter 4 – Solutions
21
CHAPTER INTEGRATION PROBLEM
a.
I
1
II
1
3
2
2
3
4
Mork
Mindy
5
4
7
6
3
III
1–3
6
b. The only possibility for the Nanunanu syndrome is autosomal recessive. The other
four modes of inheritance are excluded.
c.
Y-linked inheritance is excluded because a female (II-1) is affected. The trait cannot
be autosomal dominant or X-linked dominant because it appears in the offspring (II1 and III-6) of unaffected parents (I-1, I-2 and II-4, II-5, respectively). X-linked
recessive inheritance is excluded because the father (I-1) is not affected, yet his
daughter (II-1) is.
d.
I
Nn
Nn
II
nn
N
–
N
N
–
–
Nn
Nn
–
nn
3
II
N
–
N
–
Mork
N
Mindy
N
N
–
–
N
–
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Hyde Chapter 4—Solutions
e.
For Mork and Mindy’s first child to have Nanunanu syndrome, both would have to be
carriers. This means Mork’s father would have to be a carrier as well. We have to
calculate each of these probabilities and then use the product rule. Mork’s grandparents
(Nn Nn) produce offspring in a ratio of 1/4 NN:2/4 Nn:1/4 nn. However, we know that
Mork’s father is not affected, and so we are left with three possibilities: 2 Nn and 1 nn.
Therefore, the probability of Mork’s father being heterozygous is 2/3. Mork’s father
would then have a 50:50 chance of transmitting the n allele to Mork. Therefore, the
probability of Mork being a carrier is 2/3 1/2 = 1/3. Mindy has a 2/3 probability of
being a carrier (same reasoning as Mork’s father). If Mork and Mindy are carriers, their
offspring have a 1/4 chance of inheriting the Nanunanu syndrome. Thus, the overall
probability is 1/4 1/3 2/3 = 1/18.
f.
Two scenarios are possible: (1) Mork and Mindy are heterozygous and both children
are unaffected, or (2) at least one parent is homozygous. Calculate each of these
probabilities, and then use the sum (“or”) rule to get to the overall probability.
Probability of scenario 1 = P(Mork heterozygous) P(Mindy heterozygous) P(two unaffected children) = 1/3 2/3 3/4 3/4 = 1/8.
The probability of both parents being heterozygous is 1/3 2/3 = 2/9. Therefore, the
probability of at least one of them being homozygous (scenario 2) is 1 – 2/9 = 7/9.
The overall probability is thus = 1/8 + 7/9 = 65/72 or 90.28%.
g.
Because their first child had Nanunanu syndrome, both Mork and Mindy must be
carriers. Therefore, their next child has a 25% chance of developing the syndrome.
h. The only scenario that does not satisfy the “at least one normal child out of three”
proposition is if all three children are affected. P(three affected children) = 1/4 1/4
1/4 = 1/64. Therefore, P(at least one normal child out of three) = 1 – 1/64 = 63/64.