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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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Critical Thinking Question: 1. Both meiosis and mitosis are processes that initiate under certain circumstances of cell cycle and place. Neither is actually dependent on the chromosomal content of the cell. Thus, meiosis could begin in a haploid cell, but it would not be a successful process because there is no homologue for any chromosome to pair with. Mitosis would, however, be successful because there is no pairing (synapsis) required for successful completion of the process. 4 b. (1/2) Probability and Statistics 0.3125 son, D daughter) DSDSD) all daughters) (1/2)5
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17. Hypothesis: RrYy RrYy produces R-Y-:R-yy:rrY-:rryy in a 9:3:3:1 ratio. The critical chi-square, three degrees of freedom at probability of 0.05, 7.815.
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R-YObserved Expected O (O (O E E )2 E )2/E 315 9/16 312.75 2.25 5.06 0.016
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R-yy 108 3/16 104.25 3.75 14.06 0.135
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rrY101 3/16 104.25 3.25 10.56 0.101
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rryy
1. a. (5!/3!2!)(1/2)3(1/2)2 c. 2(1/2)5 d. (1/2)5 e. 2(1/2)5
0.03125 (SDSDS, in which S 0.0625 (SDSDS 0.03125 0.0625 (all sons
32 556 1/16 34.75 556 2.75 7.56 0.218 0.470
f. 4 daughters, 1 son 0.1875 g. (1/2)2 with p
5 daughters: (5!/4!1!)(1/2)4(1/2)
Since this chi-square, 0.470, is less than the critical chi-square, we fail to reject our hypothesis of two-locus genetic control with dominant alleles at each locus. 19. We reject the 3:1 ratio as an appropriate null hypothesis. If we calculate the chi-square using 3:1 as the expected ratio, we expect 72 and 24 (3/4 and 1/4 of 96, respectively):
0.25 (DXXXS, in which X is either a daughter or a son, 1; P [1/2][1][1][1][1/2])
3. Remember that albinos have blue eyes. Therefore, 7/16 of the offspring will have blue eyes. If we let B brown, b blue, C normal color expression, and c albinism, the following genotypes are blue-eyed: C-bb and cc . a. (1/4)5 b. (1/8)5 0.0009765 0.0000305 0.00322 0.0014648
O Curly-winged ies Straight-winged ies 61 35
E 72 24
O E 11 11
(O E)2 121 121
(O E)2/E 121/72 121/24 1.681 5.042
c. (5!/4!1!)(7/32)4(9/32) d. (4!/2!2!)(1/8)2(1/8)2 5. a. 2(1/2)4 0.125
b. (1/2)4(1/2)4 0.0039063 (Probability that sperm and egg creating the zygote each had only paternal centromeres.) 7. One-half. 9. a. 81/256 b. 108/256 c. 9/256. In (a), since all children have the same phenotype, each child will have the same probability of having no molars. Therefore, (3/4)4 81/256. In (b), P 4! (3 4)3(1 4) 3!1! 108 256.
Chi-square 6.723 (1.681 5.042). With one degree of freedom, p 0.05 (critical chi-square 3.841). Critical Thinking Question: 1. You should change your choice because the box you chose originally has a 1/3 chance of containing the prize, whereas the remaining box has a 2/3 chance of containing the prize. The 1/3 chance of your choice is set by the fact that there were three equally likely choices at the beginning. When your friend eliminated an empty box, she left two choices: your original box and the third box. Since the probability of your original choice has not changed, the probability of the remaining box must be 2/3 to give a combined probability of 1.0 that a box contains a prize. 5 Sex Determination, Sex Linkage, and Pedigree Analysis 1. The differences are in terminology only, not in shape or size of the chromosomes. In species in which females have a homomorphic pair of sex chromosomes, the members of the pair are called X chromosomes. In species in which males have a homomorphic sex chromosome pair, the members of the pair are called Z chromosomes. 3. 3/8 males, 3/8 females, 2/8 intersexes (dsx dsx homozygotes). The 1/4 dsx dsx ies will be intersexes, whereas half of the 3/4 will be normal males and half will be normal females. 5. The protein is probably a dimer, which, in the heterozygote, can be of fast-fast, fast-slow, or slow-slow subunit combinations. A female heterozygous for a sex-linked gene controlling a dimeric enzyme should show the pattern of lane 3 in whole blood (mixture of slowslow and fast-fast dimers) and lanes 1 or 2 in individual cells.
When order is given, we multiply the chance of each event, 1/4 1/4 3/4 3/4 9/256. 11. One-eighth. B must be heterozygous (Gg), as must A s father. We assume A s mother is GG, since there is no mention of the disease in her family. Therefore, A has a one-half chance of getting g from his father. If two heterozygotes mate, the chance of a recessive child is one-fourth, so P 1/4 1/2. 13. 1/512. The F1 progeny are Aa Bb Cc Dd Ee. The chance of getting any individual with a particular homozygous genotype is (1/4)5. Since we are looking for two different possibilities, we have 2(1/4)5 2/1024 1/512. 15. 0.049. Since the order is not speci ed, we use the multinomial formula. We have six mice, so n 6. If p chance of agouti, p 3/4 1/2 3/8; q (black coat color) 1/4 1/2 1/8, and r (albino) 1/2. The equation becomes: p 6! (3 8)2(1 8)2(1 2)2 2!2!2! 6 5 4 3 2 1 9 1 2 2 2 64 64 4 1 0.049
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