Study Questions for Chapter 12
1. Describe in detail the structure of a chromosome
2. What are homologous chromosomes?
3. What are autosomes? What are sex chromosomes?
4. What are karyotypes?
5. Define metacentric, submetacentric, acrocentric, telocentric
6. Explain the cell life cycle and its stages.
7. Define Mitosis. Name the phases involved in Mitosis.
8. Explain the mitotic process.
9. In detail explain the location of the chromosomes at the mitotic plate.
10. What is the function of mitosis?
11. What type of cells are produced in mitosis?
12. Genetically speaking, what are the differences between mother cells and daughter cells?
13. What is ploidy?
14. What are haploids, diploids, tetraploids and hexaploids?
15. Define Meiosis.
16. Name the stages and steps involved in Meiosis.
17. Where does meiosis take place in animals and plants?
18. Explain in detail the process of prophase I. Name the different steps and events that take place
during Prophase. Define Synaptonemal complex, crossing over, genetic recombination, chiasma
and bivalents.
19. Explain in detail the process of metaphase I.
20. Explain in detail the process of anaphase I.
21. Explain in detail the process of telophase I.
22. Explain the process of Meiosis II.
23. Genetically speaking, after meiosis, what are the differences between mother cells and daughter
cells? Explain.
24. Briefly explain spermatogenesis.
25. Briefly explain oogenesis
26. Briefly explain pollen grain production in plants.
27. Briefly explain embryo sac production in plants.
28. Briefly explain the process of double fertilization in plants.
29. What does it mean that plants have “alternation of generations”?
30. Explain the chromosomal theory of inheritance.
31. 2. Who proclaimed the above theory?
32. What was McClung, Stevens, and Wilson contribution?
33. What are hetero and homogametic sexes? Give examples.
34. Explain Morgan’s experiments with sex linkage.
35. What were Morgan’s results?
36. What is Non-disjunction? Explain in detail non-disjunction of chromosomes
37. What was Bridges hypothesis?
38. What would be the results of non-disjunction of the X chromosomes?
39. Name the mechanisms of sex determination.
40. 11.Explain Genotypic sex determination.
41. What is dosage compensation? Examples.
42. What is sex-inactivation? Examples
43. Explain in detail the genetics of calico cats.
44. What is the sex chromosome composition in birds, butterflies and moths?
45. Do plants have sex chromosomes? Explain
46. Explain genic sex determination
47. Discuss human traits involving recessive alleles on the X chromosome.
48. Discuss human traits involving dominant alleles on the X chromosome.
49. Discuss human traits linked to the Y chromosome.
Selected book problems.
(modified from book). Depict each of the crosses that follow, first using Mendelian or using
Drosophila notation, give the genotype and phenotype of the F1 progeny that can be produced.
a. In humans, a mating between two individuals, each heterozygous for the recessive trait
phenylketonuria, whose locus is on chromosome 12.
b. In humans, a mating between a female heterozygous for both phenylketonuria and X-linked color blindness
and a male with normal color vision
who is heterozygous for phenylketonuria.
Answer:
a.
Allele symbols:
P= normal
P= phenylketonuria
Cross: Pp ×Pp
F1 Genotypes :1 PP:2 Pp:1 pp
F1 phenotypes: 3⁄4 normal, 1⁄4 phenylketonuria
b.
Allele symbols: C =normal c = color blind Cross:
Cc Pp ×CY Pp
F1 Genotypes:
1⁄16 CCPP
1⁄8 CC Pp
1⁄16 CC pp
1⁄16 CcPP
1⁄8 CcPp
1⁄16 Cc pp
1⁄16 CY PP
1⁄8 CY Pp
1⁄16 CY pp
1⁄16 cY PP
1⁄8 cY Pp
1⁄16 cY pp
F1
3⁄8 normal females
1⁄16 phenylketonuria males
1⁄8 phenylketonuria
3⁄16 color-blind males
phenotypes:
females
3⁄16 normal males
1⁄16 color-blind
phenylketonuria males
In Drosophila, white eyes are a sex-linked character. The mutant allele for white eyes (w) is recessive
to the wild-type allele for brick-red eye color (w+). A white-eyed female is crossed with a red-eyed
male. An F1 female from this cross is mated with her father, and an F1 male is mated with his mother.
What will be the eye color of the offspring of these last two crosses?
Answer: The initial cross is ww´w+Y, so that the F1 females are ww+ and the F1 males are wY. The
second set of crosses are therefore w+w´w+Y and wY´ww. The former will give all brick-red females
(w+ – ) and half white (wY) and half brick-red (w+Y) males. The latter will give only white-eyed males
and females (wY and ww).
One form of color blindness in humans is caused by a sex-linked recessive mutant gene (c). A woman
with normal color vision (c+) and whose father was color-blind marries a man of normal vision whose
father was also color-blind. What proportion of their offspring will be color-blind? (Give your answer
separately for males and females.)
Answer: Since fathers always give their X chromosome to their daughters, the woman must be
heterozygous for the color-blind trait and is c+c. As her husband received his X chromosome from his
mother and has normal color vision, he is c+Y. The cross is therefore c+c´c+Y. All daughters will
receive the paternal X bearing the c+ allele and have normal color vision. As sons will receive the
maternal X half will be cY and be color-blind and half will be c+Y and have normal color vision.
12.24
In humans, red-green color blindness is recessive and X-linked, whereas albinism is recessive and
autosomal. What types of children can be produced as the result of marriages between two
homozygous parents — a normal-visioned albino woman and a color-blind, normally pigmented man?
Answer: Let c and c+ be the color-blind and normal vision alleles, respectively, and let a and a+ be the
albino and normal pigmentation alleles, respectively. Then the cross can be represented as c+c+ aa´cY
a+a+. As all the offspring will be a+a, all will have normal pigmentation. The offspring will be either
c+c or c+Y and have normal color vision. The daughters will, however, be carriers for the color-blind
trait.
In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+), and
the gene for this trait is autosomal. The gene for the white-eye trait is on the X chromosome. Suppose
a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigialwinged male.
What will be the genotypes and phenotypes of the F1 flies?
What will be the genotypes and phenotypes of the F2 flies?
What will be the genotypes and phenotypes of the offspring of a cross of the F1 flies back to each
parent?
Answer:
The initial cross is ww vg+vg+´w+Y vgvg. The F1 consists of wY vg+vg (white, normal-winged) males
and ww+ vg+vg (red, normal-winged) females.
The F2 would be produced by crossing wY vg+vg males and w+w vg+vg females. In both the male and the
female progeny, 1⁄8 will be white and vestigial, 1⁄8 will be red and vestigial, 3⁄8 will be white and normal
winged, and 3⁄8 will be red and normal winged.
If the F1 males are crossed back to the female parent, the cross is wY vg+vg´ww vg+vg+. All the
progeny would be white and normal winged. If the F1 females are crossed back to the male parent, the
cross is ww+ vg+vg´w+Y vgvg. Among the male progeny, there would be 1⁄4 white, vestigial; 1⁄4 red,
vestigial; 1⁄4 white, normal winged; and 1⁄4 red, normal winged. Among the female progeny, half
would be red and normal winged and half would be red and vestigial.
In Drosophila, two red-eyed, long-winged flies are bred together and produce the offspring listed in
the following table:
Females
Males
red-eyed, long-winged
3⁄4
3⁄8
red-eyed, vestigial-winged
1⁄4
1⁄8
white-eyed, long-winged
—
3⁄8
white-eyed, vestigial-winged
—
1⁄8
What are the genotypes of the parents?
Answer: From problem 12.25, we know that w is X-linked while vg is autosomal. This can also be
determined by considering just one trait at a time and examining the frequency of progeny
phenotypes. The ratio of long-winged to vestigial-winged progeny is 3:1 (3⁄4 to 1⁄4) in both sexes,
while the ratio of red-eyed to white-eyed progeny is all to none in females and 1:1 in males. This is
consistent with vg being autosomal and w being X-linked. The 3:1 ratio of long-winged to vestigialwinged progeny indicates that each parent was heterozygous at the vg locus. Since both parents had
red eyes, both had (at least) one w allele. Since half of the sons are white eyed, the mother must have
been heterozygous. Therefore, the parents were ww vgvg and wY vgvg.
In chickens, a dominant sex-linked gene (B) produces barred feathers, and the recessive allele (b),
when homozygous, produces nonbarred (solid-color) feathers. Suppose a nonbarred cock is crossed
with a barred hen.
What will be the appearance of the F1 birds?
If an F1 female is mated with her father, what will be the appearance of the offspring?
If an F1 male is mated with his mother, what will be the appearance of the offspring?
Answer:
In poultry, sex type is determined by a ZZ (male) and ZW (female) system. The cross can be depicted
as bb (nonbarred cock) x BW (barred hen). The F1 progeny will be bW (nonbarred) hens and Bb
(barred) cocks.
The cross can be represented as bW x bb. All the progeny will be nonbarred.
The cross can be represented as Bb x BW. The progeny will be 1⁄2 barred cocks (1⁄4 BB, 1⁄4 Bb), 1⁄4
barred hens (BW), and 1⁄4 nonbarred hens (bW).
A man (A) suffering from defective tooth enamel, which results in brown-colored teeth, marries a
normal woman. All their daughters have brown teeth, but the sons are normal. The sons of man A
marry normal women, and all their children are normal. The daughters of man A marry normal men,
and 50 percent of their children have brown teeth. Explain these facts genetically.
Answer: Notice that the trait is transmitted from the father to his daughters, indicating crisscross
inheritance. This is typical of an X-linked trait. Since the man marries a normal woman and all of their
daughters have the trait, the trait must be dominant. The man’s X chromosome bearing the defective
tooth enamel allele is inherited by all of his daughters and none of his sons. All of his daughters would
therefore have defective tooth enamel and be heterozygous for the defective enamel allele. These
daughters would transmit the defective enamel allele half of the time, giving rise to 50 percent of their
children being affected.
In humans, differences in the ability to taste phenylthiourea are due to a pair of autosomal alleles.
Inability to taste is recessive to ability to taste. A child who is a nontaster is born to a couple who can
both taste the substance. What is the probability that their next child will be a taster?
Answer: Since the inability to taste the substance is recessive, the nontaster child must be homozygous
for the recessive allele, and each of his parents must have given the child a recessive allele. Since both
parents can taste, they must also bear a dominant allele. Let T represent the dominant (taster) allele,
and t represent the recessive (nontaster) allele [Note: Mendelian notation is used here for convenience,
but also because there is no value in assigning a normal (+) and abnormal allele.] Then the cross can
be written as Tt ´Tt. The chance that their next child will be a taster is the chance that the child will be
TT or Tt, or 3⁄4.
An individual with Turner syndrome would be expected to have how many Barr bodies in the
majority of cells?
Answer: None. Turner syndrome individuals are XO. They have only one X, so no X is inactivated.
An XXY individual with Klinefelter syndrome would be expected to have how many Barr bodies in
the majority of cells?
Answer: All but one X chromosome is inactivated. An XXY individual, having two X chromosomes,
has one inactivated.
Which of the following statements is not true for a disease that is inherited as a rare X-linked
dominant trait?
All daughters of an affected male will inherit the disease.
Sons will inherit the disease only if their mothers have it.
Both affected males and affected females will pass the trait to half the children.
Daughters will inherit the disease only if their fathers have it.
Answer: The only untrue statement is (d). Since daughters receive an X chromosome from each of
their parents, they can inherit an X-linked dominant disease from either their mother or father.
Women who were known to be carriers of the X-linked recessive hemophilia gene were studied to
determine the amount of time required for blood to clot. It was found that the time required for
clotting was extremely variable from individual to individual. The values obtained ranged from
normal clotting time at one extreme to clinical hemophilia at the other. What is the most probable
explanation for these findings?
Answer: Since hemophilia is an X-linked trait, the most likely explanation is that random inactivation of X
chromosomes (lyonization) produces individuals with different proportions of cells with a functioning
allele. Normal clotting times would be expected in females with a functional h+ allele, i.e., females whose hbearing X chromosome was very frequently inactivated. Clinical hemophilia would be expected in females
without a functional h+ allele, i.e., females whose h+-bearing X chromosome was very frequently
inactivated. Intermediate clotting times would be expected to be proportional to the amount of h+ function,
which is related to the frequency of inactivation of the h+-bearing X chromosome.
Hurler syndrome is a genetically transmitted disorder of mucopolysaccharide metabolism resulting in
short stature, mental retardation, and various bony malformations. Two specific types are described
with extensive pedigrees in the medical genetics literature:
Type I: recessive autosomal
Type II: recessive X linked
You are a consultant in a hospital ward with several patients with Hurler syndrome who have asked
you for advice about their relatives’ offspring. Being aware that both types are extremely rare and that
afflicted individuals almost never reproduce, what counsel would you give to a woman with Type I
Hurler syndrome, whose normal brother’s daughter is planning marriage, about the offspring of that
proposed marriage? In your answer, state the probabilities that the offspring will be affected and
whether male and female offspring have an equal probability of being affected.
Answer: Draw out the pedigree of the patient and try to assign genotypes to the relevant individuals.
Let h represent the Hurler syndrome Type I allele. Then the patient is hh. Since we are told that Hurler
syndrome patients virtually never reproduce, neither of her parents is expected to be hh. Still, in order
for the patient to be hh, her parents must have both been Hh. (The cross was Hh´Hh.) Since her
brother is normal, he is H –, with a 2⁄3 chance of being Hh. The brother’s daughter has a half chance
of receiving either of the brother’s (H or h) alleles. Thus, the chance that the brother’s daughter has an
h allele is 2⁄3 ´1⁄2=1⁄3. Since the trait is extremely rare, it is likely she will marry an HH individual,
and have H– children. Therefore there is no chance the brother’s daughter will have affected children,
as her husband most likely will provide a dominant, normal H allele. Since Type I is autosomal, there
will be no sex differences.