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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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Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
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29. From the F1 progeny, we can see that long and tan must be dominant, and the F2 result con rms this assumption. We see a 3:1 ratio for both tan:dark and long:short. The total number of ies is 80. An ideal 9:3:3:1 ratio would be 45:15:15:5. Our results are very close to this.Therefore, we conclude that tan and long are dominant, and that the F1 ies were heterozygous. 31. First list all possible genotypes for colored plants: AACCRR AACCRr AACcRR AACcRr AaCCRR AaCCRr AaCcRR AaCcRr The rst genotype can be eliminated because all progeny should be colored, regardless of the tester strain. AACCRr can be eliminated because the progeny of the rst cross would have all been colored. AACcRR can be eliminated because the progeny of the second cross would have all been colored. AaCCRR can also be eliminated because the progeny of the third cross would have all been colored. We are now left with AACcRr, AaCCRr, AaCcRR, and AaCcRr. Try AACcRr aaccRR (cross 1), which will give 1/2 colored (A-C-R-): 1/2 colorless (A-ccR-) progeny. This could be the genotype, so try it in the second cross: AACcRr aaCCrr. This too will give 1/2 colored and 1/2 colorless offspring (A-C-rr). Since this does not t the observed result, the unknown genotype is not AACcRr. Now try AaCCRr aaccRR (cross 1). This ts the results. Try AaCcRR aaccRR. This will give 3/4 colorless (aaccRR, A-ccRR, or aaC-RR), which does not t the results; therefore, the genotype is not AaCcRR. Now try AaCcRr aaccRR. This, too, will give 3/4 colorless (aaccR-, A-ccR-, or aaC-R-), which is not seen. Therefore, the genotype must be AaCCRr. Con rm this with the other two crosses: AaCCRr AaCCRr aaCCrr 1/4 colored : 3/4 colorless, which ts. AAccrr 1/2 colored : 1/2 colorless, which ts.
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35. 9 wild-type:3 orange-1:3 orange-2:1 pink. The rst two crosses indicate that wild-type is dominant to both oranges, and the fourth indicates that orange-2 is dominant to pink. The fth cross produces four phenotypes, indicating we are dealing with at least two genes. The presence of two genes is also suggested by orange-1 orange-2. If these two traits were allelic, all pink produces the progeny should have been orange. The F1 progeny that resemble those from a double testcross. If A-Bwild-type, A-bb orange-1, aaBorange-2, aabb is probably pink. The crosses in question are then: AAbb (orange-1) aaBB (orange-2)
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AaBb (wild-type) self 9/16 A-B3/16 A-bb 3/16 aaB1/16 aabb wild-type orange-1 orange-2 pink
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37. Two loci with epistasis. AaBb self yields 9/16 A-B-:6/16 A-bb aaB-:1/16 aabb. Verify by testcrossing the various classes. 39. See for example gure 2.25. Other ratios are 10:3:3; 10:6; 12:3:1; 12:4. 41. 9 nonworkers: 7 hard workers.The F1 indicates nonworker is dominant; therefore, a worker must be a recessive homozygote. If one gene is involved, the cross of the F1 female worker male (Ww ww) should produce 1 worker:1 nonworker, for this is a testcross. This result is not seen, so we must have more than one gene involved. Perhaps a worker can result from more than one gene. Let A-B- nonworker, and A-bb, aaB-, or aabb workers. The original worker is aabb, and the cross is: aabb b AaBb (nonworker) 1/4 A-B-: nonworker 1/4 A-bb: worker aabb (worker) 1/4 aaB-: worker 1/4 aabb worker AABB
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33. a. all normal b. 9 normal: 7 dark c. 1/2 ebony:1/2 normal; 1/2 black:1/2 normal If we let e ebony, e type, the rst cross is: ee b b wild-type, b black, and b wild-
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If AaBb
AaBb 9/16 A-B- nonworkers 3/16 aaB3/16 A-bb 1/16 aabb workers
e e bb 43.
all e e b b sel ng 9/16 e b wild-type 3/16 ee b ebony 3/16 e bb black 1/16 ee bb ebony, black Since it is dif cult to distinguish black and ebony, 7/16 will be darkbodied. In c, the crosses are: 1. e e b b 2. e e b b ee b b e e bb
2 1 3 4 indole tryptophan 3-hydroxyanthranilic niacin or kynurenine acid Accumulation: 1, indole; 2, ; 3, tryptophan or kynurenine; 4, 3hydroxyanthranilic acid. Without serine, the pathway would be blocked before the point of tryptophan production, after indole.
45. a. Maple sugar urine disease is recessive. If two individuals with the same phenotype produce offspring, some of whom have a different phenotype, both parents must be heterozygous. Since they are heterozygous, their phenotype must be dominant. b. 3/4. Let M normal and m maple sugar urine alleles. The cross is Mm Mm. At each conception, the probability is threefourths that each child is of the M- genotype.
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