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MM Observed Expected Chi-square 41 p2 100 36 0.694
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MN 38 2pq 100 48 2.083
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NN 21 q2 100 16 1.563
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Total 100 100 4.340
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Thus (HN) (selection differential) selective differential 9, (0.5)(9.0)
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The critical chi-square (0.05, one degree of freedom) 3.841. We thus reject the null hypothesis that this population is in HardyWeinberg proportions. 3. Here we must assume Hardy-Weinberg equilibrium because of dominance. The f(tt) 65/215 0.302. Thus q f(t) 0.302 0.55; and p f(T ) 1 0.55 0.45. Since f(tt) there are zero degrees of freedom (number of phenotypes number of alleles 2 2 0), we cannot do a chi-square test
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App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
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A-22
Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
to determine if the population is mating at random (in HardyWeinberg proportions). 5. a. Most human populations should be in Hardy-Weinberg proportions at the taster locus regardless of what the allelic frequencies are. 3:1 is a family ratio when the parents are heterozygotes or a population ratio when p q 0.5, given dominance. b. With no other information, it is probably safest to assume p q 0.5. At equal frequencies of alleles, the dominant phenotype (tasting) occurs in 75% of people (p2 2pq). 7. p 0.99, q 0.01. If the population is in equilibrium, there should be p2 of AA 2pq of Aa q2 of aa individuals. Since 1/10,000 shows the recessive trait, this is q2. Therefore, q Since p q 1, p 1 10,000 1 0.01 0.0001 0.99. 0.01
19. Inbreeding is disadvantageous when it causes recessive deleterious alleles to become homozygous. This occurs in normally outbred populations of diploids that have built up these harmful alleles. In species that normally inbreed, these deleterious alleles are probably no longer present; they were either removed by selection long ago or cannot build up in the population because of the regular pattern of inbreeding. 21. There are four paths passing through A, the only common ancestor (FA 0.01). All have ABCI as one side of the path. The second legs of the four other paths are ADEHI, ADGHI, AFGHI, and AFEHI. (A path such as IHEDAFGHI is invalid, passing through H twice.) Since each path has six ancestors, the inbreeding coef cient is F1 4(1/2)6(1.01) 0.063. 23. Using the formula F (2pq H)/2pq, we calculate that 2pq 0.48, and H 38/100 0.38.Therefore, F (0.48 0.38)/0.48 0.208. (0.32 0.20) 0.12 25. 0.375. F (2 pq H ) 2 pq 0.375 0.32 0.32 Critical Thinking Question: 1. Let us use the superscripts m and p for male and female, respectively. Then, we can construct the following Punnett square creating the next generation:
9. The expected frequency of brown-eyed individuals will depend on the allelic frequencies of the original population. If we assume that mating is random with respect to eye color, and there is no selection, allelic frequencies will not change with time. We can, for example, calculate the frequencies of brown-eyed individuals for two populations at equilibrium. Let p frequency of the brown-eye allele and q frequency of the blue-eye allele.
Population 1. 2.
p 0.7 0.5
q 0.3 0.5
Frequency of brown ( p2 0.91 0.75
2pq) Females, A; f(A) Females, a; f(a) pf qf
Males, A; f(A) f(AA) f(Aa) pmp f pmq f
pm Males, a; f(a) f(Aa) f(aa) p fq m q mq f
We see that the original premise will be met only if the alleles are equally frequent. 11. 0.187 MM, 0.491 MN, 0.321 NN. The next generation will achieve equilibrium and there will be (0.43)2 MM 2(0.43)(0.57)MN (0.57)2MN. 13. f(A) 0.792; f(B) 0.208. The easiest way to calculate frequencies is to do it empirically. We have three hundred people, so we have six hundred alleles. f(A) f(B) 2 2 200(AA) 600 75(AB) 475 600 125 600 0.792 0.208
Since the distribution of offspring in the table is independent of sex, it is the same in both sexes. The frequency of the A allele, p (in both sexes), will be the sum of the frequencies of the homozygotes and half the heterozygotes, or: p pmp f (1/2)( pmq f p) for all q s: p f) (1/2) p f (1 pm) p fq m )
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