52

Hyde Chapter 8—Solutions

8 CHANGES IN CHROMOSOME STRUCTURE AND NUMBER CHAPTER SUMMARY QUESTIONS 2. A pericentric inversion is one in which the inverted DNA region contains the centromere, while in a paracentric inversion, the centromere lies outside the inverted region. Figure 8.10 provides a schematic. Both types of inversion suppress recombination in an inversion heterozygote in two ways: (1) A real mechanism due to the difficulty in complete pairing between the two homologs near the inversion breakpoints (figure 8.11), and (2) An apparent mechanism due to crossovers in the inversion loop producing gametes that will generate nonviable zygotes because of duplications and deficiencies (figures 8.13 and 8.14). 4. Animals use chromosomal sex-determining mechanisms that are disrupted by polyploidy. Plants typically do not use those same mechanisms. Many plants can exist vegetatively, allowing more time for the rare somatic doubling of genome sets to occur. In addition, many plants rely on insect pollinators or wind to fertilize them, and therefore are more likely to undergo hybridization. 6. An acentric chromosome has no centromere, while a dicentric chromosome has two. An isochromosome is a chromosome that has a centromere and two genetically and morphologically identical arms (two long arms for example, instead of a short and a long arm). It undergoes normal segregation during cell division, whereas acentric and dicentric chromosomes do not. 8. All chromosomes form linear bivalents. The cross-shaped figure is seen only in heterozygotes. 10. No, there are no inversion loops formed in homozygotes. Therefore, crossover suppression is not observed in an inversion homozygote. 12. A diagram will show that a crossover between a centromere and the center of the cross can change the consequences of the pattern of centromere separation. For example, the following figure diagrams a crossover between loci 4 and 5. Refer also to figures 8.17 and 8.18.

Hyde Chapter 8—Solutions

53

14. Problems occur during meiosis, not mitosis. Problems are worse in the odd-ploid organisms such as triploids because of the production of aneuploid gametes (see figure 8.38). 16. Sporadic Down syndrome, also known as trisomy 21, is caused by a spontaneous, nondisjunction of chromosome 21. This form is random and comprises about 95% of all Down syndrome individuals. Familial Down syndrome (approximately 5% of cases) typically results from a Robertsonian translocation involving chromosome 21 and chromosome 14 (or 15). Translocation carriers do not exhibit Down syndrome; however, their children have an increased incidence of Down syndrome. (Please

54

Hyde Chapter 8—Solutions refer to figure 8.29). The incidence of sporadic Down syndrome increases among children born to older mothers (because older eggs are more likely to undergo nondisjunction of chromosome 21). Familial Down syndrome does not exhibit this age-dependent expression.

EXERCISES AND PROBLEMS 18. In humans, 2n = 46. a. 69 (3n). b. 47 (2n + 1). c. 44 (2n – 1– 1). d. 44 (2n – 2). 20. Reciprocal translocation (some effects occur only in the heterozygous condition). Look for the cross-shaped figure at meiosis or in salivary gland chromosomes. 22. a. b. c. d. e.

47, XY, + 11. 44, XX, – 5 – 7. 46, XY, 20q+. 46, XX, t(10p–;15q+). 45, Y, – 3 + 18.

24. a.

The male and female have the same genotype, so the discussion applies to both. Start by giving a different designation to each of the two homologs that are carrying the same allele: ey+–1 and ey+–2. Therefore, the genotype of the individual is now ey+–1 ey+–2 ey. Three different types of segregation can occur in this trisomic during gamete formation in meiosis: ey+–1 ey+–2 to one pole and ey to the other pole; ey+–1 ey to one pole and ey+–2 to the other; and finally, ey+–2 ey to one pole and ey+–1 to the other. Thus, each trisomic parent can produce four different types of gametes: 2/6 ey+ ey , 2/6 ey+, 1/6 ey+ ey+, and 1/6 ey. The various possible fertilizations can be obtained by using the forkedline method for gametes.

2/6 ey+ ey

2/6 ey+

2/6 ey+ ey

4/36 ey+ ey+ ey ey

normal-eyed, tetrasomic

2/6 ey+

4/36 ey+ ey+ ey

normal-eyed, trisomic

1/6 ey+ ey+

2/36 ey+ ey+ ey+ ey

normal-eyed, tetrasomic

1/6 ey

2/36 ey+ ey ey

normal-eyed, trisomic

2/6 ey+ ey

4/36 ey+ ey+ ey

normal-eyed, trisomic

2/6 ey+

4/36 ey+ ey+

normal-eyed, disomic

1/6 ey+ ey+

2/36 ey+ ey+ ey+

normal-eyed, trisomic

1/6 ey

2/36 ey+ ey

normal-eyed, disomic

Hyde Chapter 8—Solutions

1/6 ey+ ey+

1/6 ey

55

2/6 ey+ ey

2/36 ey+ ey+ ey+ ey

normal-eyed, tetrasomic

2/6 ey+

2/36 ey+ ey+ ey+

normal-eyed, trisomic

1/6 ey+ ey+

1/36 ey+ ey+ ey+ ey+ normal-eyed, tetrasomic

1/6 ey

1/36 ey+ ey+ ey

normal-eyed, trisomic

2/6 ey+ ey

2/36 ey+ ey ey

normal-eyed, trisomic

2/6 ey+

2/36 ey+ ey

normal-eyed, disomic

1/6 ey+ ey+

1/36 ey+ ey+ ey

normal-eyed, trisomic

1/6 ey

1/36 ey ey

eyeless, disomic

Therefore, the phenotypic ratio in the offspring is 9/36 normal-eyed, tetrasomic:18/36 normal-eyed, trisomic:8/36 normal-eyed, disomic:1/36 eyeless, disomic. b. The male is trisomic and can produce four types of gametes, as discussed in part (a). The female is disomic and can produce two typics of gametes in equal proportions: 1/2 ey+ and 1/2 ey.

+

1/2 ey

1/2 ey

2/6 ey+ ey

2/12 ey+ ey+ ey

normal-eyed, trisomic

2/6 ey+

2/12 ey+ ey+

normal-eyed, disomic

1/6 ey+ ey+

1/12 ey+ ey+ ey+

normal-eyed, trisomic

1/6 ey

1/12 ey+ ey

normal-eyed, disomic

2/6 ey+ ey

2/12 ey+ ey ey

normal-eyed, trisomic

2/6 ey+

2/12 ey+ ey

normal-eyed, disomic

1/6 ey+ ey+

1/12 ey+ ey+ ey

normal-eyed, trisomic

1/6 ey

1/12 ey ey

eyeless, disomic

Therefore, the phenotypic ratio in the offspring is 6/12 normal-eyed, trisomic:5/12 normal-eyed, disomic:1/12 eyeless, disomic.

56

Hyde Chapter 8—Solutions

26. a. Nonreciprocal translocation of GH from chromosome 1 to chromosome 2. b. Robertsonian translocation and the loss of the acentric chromosomes. c. Deletion of B from chromosome 1; pericentric inversion of C•DE in chromosome 1; then a reciprocal translocation of GH and TUV between the two chromosomes. d. Deletion of E from chromosome 1; reciprocal translocation of FGH and UV between the two chromosomes; duplication of UV in chromosome 1, followed by an inversion of the duplicated segment. e. Deletion of U from chromosome 2; reciprocal translocation of FGH and TV between the two chromosomes; then a pericentric inversion of C•DETV in chromosome 1. 28. We have five heterozygous genes, so we expect to see 25 = 32 genotypes, but we see only six genotypes. We see no exchanges between genes B, C, and D. These three genes could be so tightly linked that no recombination occurs between them (unlikely). Genes B, C, and D could be within an inversion that is heterozygous in the heterozygous parent. The recombination that does occur within this inversion results in inviable zygotes. 30. We see about half the enzyme activity in crosses with strains B and C. Therefore, the gene must be located in the region that is common to both strains, approximately in the region located 25 to 35 map units from the left end. 32.

v //////////////////////1.1

m m+

s s+

The use of X-rays alerts us to chromosomal aberrations. The fact that a vermilion female appears when we expect all wild-type females indicates that we have a deletion. The deletion must end between v and m. If the deletion included miniature, we should have seen a vermilion, miniature female. We can draw the chromosomes of the flies in the second cross as v /////////////////

m m+

s s+

v




m

s

The question is, What is the distance between the deletion and the gene for miniature? We must look at the males from the cross. Note that we see only half as many males compared with females; those males that received the deletion X chromosome must have died. The vermilion male must have resulted from a recombination between the end of the deletion and miniature: v //////////////

m m+

s s+

Hyde Chapter 8—Solutions

57

yielding the following chromosomes: m+

v

s+

and

///////////////// m

s

Therefore, map distance = (vermilion males/total males)
100 = 1/90
100 = 1.1 map units. 34. 3  4  1  2. Inversion of u p o will yield (4). If this arrangement is followed by an inversion of t s o p, (1) results. Finally, an inversion of r q p o will yield (2). 36. A translocation from the tip of the normal X in the male to the Y. We expect all males to receive an X chromosome with the white-eye allele from the female. For the male to be wild type, he still must have part of the wild-type X chromosome. To test, cross this wild-type male with white-eyed females. All the female progeny should be white-eyed and all the male progeny red-eyed. Cytological examination of the chromosomes should reveal the translocation. 38. a.

If the autotetraploid has 64 chromosomes, each set of chromosomes would have 64/4 = 16 chromosomes. Therefore, there will be 16 linkage groups. b. An allotetraploid is in essence a double diploid (also termed amphidiploid). This means that all 64 chromosomes exist as pairs, and so the number of chromosomes per set is 64/2 = 32. Therefore, the number of linkage groups is 32. This is true regardless of the number of chromosome pairs contributed by the original parents. For example, if one species is 2n = 20 and the other species is 2n = 44, the linkage group contributions of the species will be 10 and 22, respectively. If both species are 2n = 32, each will contribute 16 linkage groups. So the total number of linkage groups will be 32 in all such cases.

40. Nondisjunction in one of the cells in meiosis II in the female leads to the following gametes: X, X, XX, and O (lacking a sex chromosome). The male’s gametes will be X and Y. Gametes X Y 42. a.

X

X

XX

O

XX normal female XY normal male

XX normal female XY normal male

XXX triple-X female XXY Klinefelter male

XO Turner female YO lethal

These mosaics arise as a result of nondisjunction of sister chromatids in mitosis in embryonic development. b. The phenotypic variance can be explained by the timing of the mitotic nondisjunction. If it occurs early in embryonic development, the aneuploidy is likely to be spread throughout the body, and the phenotype may be severe; if it occurs late in embryonic development, the aneuploidy may be limited to one or a few organs or tissues, and the phenotypic effect will be milder.

58

Hyde Chapter 8—Solutions

44. 32. The gamete from P will have 9 chromosomes, and the gamete from U will have 7 chromosomes. The original zygote will have 16 chromosomes, but none of these will pair. To be fertile, each chromosome must be duplicated to yield 32 chromosomes. 46. The father. The allele for color blindness can only come from the mother. If meiosis in her is normal, an egg could get the X chromosome carrying the mutant allele. The daughter has only one X chromosome, so the sex chromosomes failed to separate in the father, and a sperm with neither X nor Y fertilized the egg. The mother. Let XE = Faulty tooth enamel, and Xe = normal tooth enamel. The child has Klinefelter syndrome and normal teeth, so he has to be XeXeY. The father has faulty teeth, XEY, and must have donated the Y chromosome to this child. Therefore, the child must have received both of his X chromosomes from his mother, who must be XEXe. b. Meiosis II, because the child received two copies of the same X chromosome from his mother.

48. a.

50. Since this karyotype is a tetraploid, it is likely that after fertilization all the chromosomes duplicated. An alternative, if quite unlikely, explanation is that the fetus may have resulted from the fertilization of a diploid egg by a diploid sperm. 52. The key is to remember that a deletion can only cause pseudodominance of adjacent genes. For example, del 1 leads to pseudodominance of genes e and j. Therefore, genes e and j must be adjacent to each other. Using a systematic logical approach, you should come up with the following gene order: j e l • r a d y. 54. If the plants were closely enough related, both in physiology and chromosome number, it is conceivable that a successful hybrid could be created. Perhaps chromosome doubling would be necessary to achieve successful meiosis. Although the odds of this are low, the odds are even lower that the hybrid would combine the desired attributes of each species. Similar experiments done in the past have rarely been successful even when viable hybrids were produced.

CHAPTER INTEGRATION PROBLEM a.

The P female has normal wings, so she must be homozygous for wild-type wing shape: N+N+. She is red-eyed, so she must have at least one w+ allele. Her F1 offspring consist of approximately equal numbers of red-eyed and white-eyed males, implying that she is heterozygous for eye color: w+w. Therefore, her genotype is N+ w+/N+ w.

b. The use of X-rays alerts us to chromosomal aberrations. Indeed, this is justified by the presence of 20 Notched females in the F1 offspring. These must be heterozygous for Notch: N+N. Therefore, X-ray irradiation of the P female caused mutations (most likely deletions) in its gametes, so that some now carry the N allele. These mutant

Hyde Chapter 8—Solutions

59

gametes, when fertilized by sperm with the N+-carrying X chromosome, will produce zygotes that are N+N. The Notched females produced by these zygotes have red eyes. So they could be w+w+ or w+w. These females inherit the w+ allele from their father, so they would get the w+ or w allele from their mother. Therefore, either maternal X chromosome could have incurred the mutation. c.

Given that 4% of the F1 offspring are Notched females, the absence of Notched males is surprising. This could be explained by the fact that Notched is lethal in the hemizygous state. Males that inherit the N allele from their mother die during embryonic development, and therefore no Notched males will survive to birth.

d.

For wild-type gametes: P

N+ w+/N+ w

F1

1/4 N+ w+/N+ w+ 1/4 N+ w/N+ w+ 1/4 N+ w+/Y 1/4 N+ w/Y

N+ w+/Y




and phenotypically F1

1/2 normal-winged, red-eyed females 1/4 normal-winged, red-eyed males 1/4 normal-winged, white-eyed males

For mutant gametes: Scenario 1 (chromosomal aberration in X chromosome carrying the w+ allele): P

N w+/N+ w

F1

1/4 N w+/N+ w+ 1/4 N+ w/N+ w+ 1/4 N w+/Y Die in embryonic development 1/4 N+ w/Y




N+ w+/Y

and phenotypically F1

1/3 Notched, red-eyed females 1/3 normal-winged, red-eyed females 1/3 normal-winged, white-eyed males

60

Hyde Chapter 8—Solutions Scenario 2 (chromosomal aberration in X chromosome carrying the w allele): P

N+ w+/N w

F1

1/4 N+ w+/N+ w+ 1/4 N w/N+ w+ 1/4 N+ w+/Y 1/4 N w/Y Die in embryonic development




N+ w+/Y

and phenotypically F1

e.

i.

White-eyed, Notched females P N+ w/N w
F1

F1

ii.

1/3 normal-winged, red-eyed females 1/3 Notched, red-eyed females 1/3 normal-winged, red-eyed males

1/4 N+ w/N+ w+ 1/4 N w/N+ w+ 1/4 N+ w/Y 1/4 N w/Y

Die in embryonic development

and phenotypically 1/3 normal-winged, red-eyed females 1/3 Notched, red-eyed females 1/3 normal-winged, white-eyed males

White-eyed, Notched females
P N+ w/N w F1

Red-eyed, normal-winged males N+ w+/Y

1/4 N+ w/N+ w 1/4 N w/N+ w 1/4 N+ w/Y 1/4 N w/Y

White-eyed, normal-winged males N+ w/Y

Die in embryonic development

and phenotypically F1

1/3 normal-winged, white-eyed females 1/3 Notched, white-eyed females 1/3 normal-winged, white-eyed males

Hyde Chapter 8—Solutions

61

f.

For a cross to yield females that are homozygous for Notch, it would have to be between a heterozygous female (N+N) and a hemizygous male (N/Y). However, those males are not viable. Therefore, this cross is not feasible.

g.

The cross would be between a white-eyed male and a Notched female that is homozygous for red eyes: P

N+ w+/N w+




N+ w/Y

One-third of the offspring should be Notched females. If those have red eyes, their genotype would have to be w+w, implying that the Notch mutation left the w+ allele intact. Therefore, the mutation involved only the N gene. If on the other hand the Notched females have white eyes, this indicates that the wild-type w+ allele is not being expressed in these flies. Therefore, the cause of the Notched phenotype is a mutation that spanned the white locus.