Study Questions For Chapter 12 (Chromosomes,Karyotypes,Cell,Mitotic Process...)

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

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 Xchromosome 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 aand 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 Xchromosome. Suppose a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigial-winged 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 (3⁄4to 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 Xlinked. The 3:1 ratio of long-winged to vestigial-winged 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 Trepresent 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 Xlinked 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 h-bearing 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.