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Tamarin: Principles of Genetics, Seventh Edition
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Back Matter
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App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
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The McGraw Hill Companies, 2001
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Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
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7. a. 0; human female, male y b. 1; human female, female y c. 0; human male, male y d. 1; human male, female y e. 2; human female, female y f. 4; human female, female y g. 1/0 mosaic; human male-female mosaic, male-female y mosaic 9.
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Now diagram the second cross: XyXy X Xy gray X Xy gray XyXy yellow XyY yellow X Y gray XyY yellow X Y
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Cross P1 female male F1 female male F2 females males X X XlzY X Xlz X Y X X , X Xlz X Y, XlzY
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Reciprocal XlzXlz X Y X Xlz XlzY X Xlz, XlzXlz X Y, XlzY
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17. 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. A very rare possibility is that the man is the father and nondisjunction occurred in both parents: at meiosis I in the male and at meiosis II in the female (see chapter 8). 19. a. F1: wild-type females, white-eyed males. b. F2: 3 wild-type:3 white-eyed:1 ebony:1 ebony, white-eyed in both sexes. c. Reciprocal F2: 3 wild-type females:1 ebony female:3 wild-type males:3 white-eyed males:1 ebony male:1 ebony, white-eyed male. Let X red, Xw white, e wild-type, e ebony. XwXw e e X Xw e e (wild-type) XwY e e (white-eyed) XwY ee P1 F1
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11. Exemptions should be made minimally for hemophilia in brother, sister s son, mother s brother, mother s sister s son, mother s father, and others more distantly related. 13. P1 F1 fyfy X X (female) fy fy X X (female)
fy fy XctY (male) fy fy X Y (male)
Male fy X Female fy X fy Xct fy X fy Xct fy X fy Y fy Y
Use probability for the F2 generation rather than the Punnett square: (1/2)X (1/2)Xw (1/2)Y (1/2)Xw (1/2)Xw (1/2)Y (3/4)e (1/4)ee (3/4)e (1/4)ee (3/4)e (1/4)ee (3/4)e (1/4)ee 3/16 wild-type females 1/16 ebony females 3/16 wild-type males 1/16 ebony males 3/16 white-eyed females 1/16 white-eyed, ebony females 3/16 white-eyed males 1/16 white-eyed, ebony males
fy fy fy fy
fy X X fy X Xct fy X X fy X Xct
fy fy X X fy fy X Xct fyfy X X fyfy X Xct
fy fy fy fy
fy X Y fy XctY fy X Y fy XctY
fy fyX Y fy fy XctY fyfy X Y fyfy XctY
F2: females, 3/4 wild-type, 1/4 fuzzy; males, 3/8 wild-type, 3/8 cut, 1/8 fuzzy, and 1/8 cut and fuzzy. 15. a. X linked b. gray c. 1/2 gray:1/2 yellow in both sexes In both crosses, we see a difference in the phenotypes of the sexes, suggesting sex linkage. The F1 offspring from the rst cross indicate that gray is dominant to yellow. The F1 females from this cross must be heterozygous, and the two phenotypes in the F2 males result from each of the X chromosomes in the F1 female being hemizygous in the F2 males. The rst cross is therefore (calling gray the wild-type): X X X Xy gray X X gray X Xy gray X Y gray X Y gray XyY yellow XyY
For the reciprocal cross, X X ee X Xw e e (wild-type) X Ye e (wild-type) XwY e e
All F2 females will get X ; e :e will be 3:1. The males will be as in the males shown. 21. The female is heterozygous for an X-linked color gene (one of the X chromosomes in the cells of female cats is inactivated, leading to the black and yellow spots). We immediately deduce X-linkage because of the different phenotypes in the sexes. Finding two types of males indicates that the female was heterozygous. The patches of yellow and black come from X-inactivation. The cross is: XbXo XbXb black XbXo black and yellow XbY black XoY yellow XbY
23. Penetrance is the proportion of individuals of a particular genotype that shows the appropriate phenotype; expressivity is the degree to which a trait is expressed.
Tamarin: Principles of Genetics, Seventh Edition
Back Matter
App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
The McGraw Hill Companies, 2001
Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
25. 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. Autosomal dominant or possibly autosomal recessive inheritance. c. Autosomal, or sex-linked, recessive inheritance. d. Autosomal recessive inheritance. 27. Assuming 100% penetrance:
the offspring. Phenotypic classes will be equally distributed between the two sexes. The same results will be found for an autosomal locus if the dihybrids are females (no crossing over in males). A reciprocal cross cannot be done for X-linked genes because males cannot be dihybrid. Males dihybrid for an autosomal gene produce only two classes of offspring when testcrossed parentals. 5. a. The hotfoot locus is in the middle (compare, for example, hotfoot, a double crossover, with the wild-type, a parental); there are 16.0 map units from hotfoot to either end locus: 74 66 11 9 recombinants between hotfoot and waved and 79 61 11 9 recombinants between hotfoot and obese. b. The trihybrid parent was o h wa/o h wa . c. The coef cient of coincidence is 20/25.6 (20/[0.16 1,000]); interference is 1 (20/25.6) 0.22, or 22%. 0.16
Critical Thinking Question: 1. The immediate effect of a null allele is to make heterozygous genotypes appear to be homozygotes.That is, in the simplest system, we expect one band in a homozygote and two bands in a heterozygote. If we see only one band, we assume that the individual is homozygous for that allele, when in fact that individual might be heterozygous for the allele that produces the particular band and the allele that produces no band (null allele).The null allele can be veri ed by both the absence of any bands in the null homozygote and the results of breeding experiments when the null allele is suspected. 6 1. a. P1 groucho rough grogro ro ro gro gro roro F1 female gro gro ro ro male grogro roro F2 grogro ro ro 518 471 gro gro roro grogro roro 6 gro gro ro ro 5 (6 5)/1,000 0.011 1.1% recombination 1.1 map units apart b. Given the map units, F1 gametes are produced on the average 0.4945 (98.9%/2); by females as follows: gro ro , 49.45% gro ro, 49.45% 0.4945; gro ro, 0.55% (1.1%/2) 0.0055; and gro ro , 0.55% 0.0055. Males, lacking crossing over, produce only two gamete types: gro ro and gro ro, each 50% 0.50. Summing from the Punnett square following, the phenotypes of the offspring would be as follows: wild-type, 50%; groucho, rough, 0%; groucho, 25%; and rough, 25%. Linkage and Mapping in Eukaryotes
7. a. Work backward from the 0.61% double recombinants (0.100 0.061 100). Thus, there would be 6 of 1,000 double recombinants. In the an-sple region, we need the total of single double recombinants 100 of 1,000 (10 map units). Thus, 100 6 94; divided by 2 (two phenotypes) is 47 each. For the spleat region, the total of single and double recombinants 61 (6.1 map units). Thus, 61 6 55; divided by 2 is 27 and 28. The parentals make up the remainder for a total of 1,000. b. With a coef cient of coincidence of 0.60, only 0.366% (0.61 0.60) of the expected double recombinants will occur, that is, 4 instead of 6. Thus:
Coef cient of Coincidence 1.0 ancon, spiny, arctus oculus wild-type ancon, spiny arctus oculus ancon spiny, arctus oculus ancon, arctus oculus spiny Total 422 423 27 28 47 47 3 3 1,000 0.6 421 422 28 29 48 48 2 2 1,000
Male Female gro ro (0.4945) gro ro (0.4945) gro ro (0.0055) gro ro (0.0055) gro ro (0.5) groucho 0.24725 wild-type 0.24725 groucho 0.00275 wild-type 0.00275 gro ro (0.5) wild-type 0.24725 rough 0.24725 rough 0.00275 wild-type 0.00275
9. a. linked; b. trans; c. 28.7%. The cross is a testcross. If the genes were not linked, we would expect a 1:1:1:1 ratio of offspring; we don t see that. The alleles that are linked will appear as the majority classes, which are Trembling, long-haired and normal, Rex. Therefore, Trembling and Rex are in the trans position. If we let T Trembling, and R Rex, the cross is Tr tR tr tr
Recombinants are Trembling, Rex and normal, long-haired; 42 44 300 11. a. k e cd k e cd b. k k e cd k e cd 100 28.7%
e 5.1
3. A dihybrid female is testcrossed (with a hemizygous male having both recessive alleles). Each recombinant class will make up about 5% of the offspring. Each parental class will make up about 45% of
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