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CHAPTER 14
Population Genetics and Evolution
14-1. Neither AA nor aa can reproduce, the first because of lethality and the
second because of sterility. The only matings that produce progeny are Aa x Aa,
and the zygotes are 1/4 AA, 1/2 Aa, and 1/4 aa. Because the AA zygotes do not
survive, the adult population consists of 2/3 Aa and 1/3 aa. The assumption of
random mating is not necessary.
14-2. (a) Each genomic DNA sample shows two bands because the person is
heterozygous. (b) It would be possible if a person were homozygous. (c)
Codominant. (d) 2 and 7 are the parents of A, 1 and 4 are the parents of B, 6
and 8 the parents of C, and 3 and 5 the parents of D.
14-3. Set q = 1/50 = 0.02. Then the frequency of carrier females equals 2pq = 2
 0.02  0.98 = 0.0392, or about two times the frequency of affected males. The
expected frequency of affected females equals q2 = 0.0004, or 1 in 2500.
14-4. The 10 AA genotypes contribute 20 A alleles to the sample, the 15 Aa
genotypes contribute 15 A and 15 a alleles to the sample, and the 4 aa
genotypes contribute 8 a alleles to the sample. Altogether, there are 35 A and
23 a alleles in the sample, for a total of 58. In this sample the allele frequency
of A is 35/58 = 0.603 and that of a is 23/58 = 0.397.
14-5. Total number of individuals 5 100, total number of alleles 5 200. Allele
frequencies: “15.7,” (6 + 7 + 5 + 2 x 1)/200 = 0.10; “6.0,” (15 + 18 + 2 x 6 +
5)/200 = 0.25; “6.2,” (21 + 2 x 12 + 18 + 7)/200 = 0.35; and “6.5,” (2 x 9 + 21 +
15 + 6)/200 = 0.30.
14-6. Because the frequency of homozygous recessives is 0.10 the frequency of
the recessive allele q is (0.10)1/2 = 0.316. This means p = (1-0.316) = 0.684, so
the frequency of heterozygotes is 2 X 0.316 x 0.684 = 0.43. Because 0.90 of the
individuals are nondwarfs, the frequency of heterozygotes among nondwarfs
is 0.43/0/90 = 0.48.
14-7. IAIA, (0.16)2 = 0.026; IAIO, 2(0.16)(0.74) = 0.236; IBIB, (0.10)2 = 0.01; IBIO,
2(0.10)(0.74) = 0.148; IOIO, (0.74)2 = 0.548; IAIB, 2(0.16)(0.10) = 0.032. The
phenotype frequencies are O, 0.548; A, 0.026 + 0.236 = 0.262; B, 0.01 + 0.148 =
0.158; AB, 0.032.
14-8. (a) (1/n)2 = 1/n2; (b) 2(1/n)(1/n) = 2/n2; (c) n(1/n)2 = 1/n (d) [n (n-1)/2]
x (2/n2) = 1 – (1-n).
14-9. If the genotype frequencies satisfy the Hardy-Weinberg principle, they
should be in the proportions p2, 2pq, and q2. In each case, p equals the
frequency of AA plus one-half the frequency of Aa, and q = 1 p. The allele
frequencies and expected genotype frequencies with random mating are (a) p
= 0.50, expected: 0.25, 0.50, 0.25; ( b) p = 0.635, expected: 0.403, 0.464, 0.133; (c)
p = 0.7, expected : 0.49, 0.42, 0.09; (d) p = 0.775, expected: 0.601, 0.349, 0.051;
and (e) p = 0.5, expected: 0.25, 0.50, 0.25. Therefore, only (a) and (c) fit the
Hardy-Weinberg principle.
14-10.(a) The probability that the wife is a carrier is the probability that a
phenotypically
normal person is a carrier. This probability is 2pq / (p2 + 2pq), where q =
√(1/1700) = 0.024 and p = 0.976. The numerator 2pq = 0.047 and the
denominator p 2 + 2pq = 0.999 (close enough to 1 that it can be ignored).
Hence the probability that the wife is a carrier equals 0.047. (b) If she is a
carrier, the probability that the first child will be affected is 1/4, so the overall
probability that the first child will be affected equals 0.047  1/4 = 0.012, or
about 1 in 84.
14-11. The frequency of heterozygotes is smaller by the amount 2pqF, in
which F is the inbreeding coefficient.
14-12. The allele frequency is the square root of the frequency of affected
offspring among unrelated individuals, or (8.5 x 10 –6)1/2 = 2.9 x 10—3. With
inbreeding the expected frequency of homozygous recessives is q2 (1-F) + qF,
in which F is the inbreeding coefficient. When F = 1/16, the expected
frequency of homozygous recessives is (2.9 x 10—3)2 (1-1/16) + (2.9 x 10—3) =
1.9 x 10—4; when F = 1/64, the value is 5.3 x 10—5.
14-13. (a) 2/3; (b) 2pq /(p2 + 2pq) = 0.0198 because q = √(1/10,000) = 0.01; (c)
2/3  0.0198  1/4 = 0.0033, or roughly 1 in 300.
14-14. Set q = 2  2pq; hence p = 1/4 and q = 3/4.
14-15. The numbers of alleles are as follows: A, 2 + 2 + 5 + 1 = 10; B, 5 + 12 +
12 + 10 = 39; and C , 10 + 5 + 5 + 1 = 21. The total number of alleles is 70, so
the allele frequencies in the sample are as follows: A, 10/70 = 0.14; B, 39/70 =
0.56; and C, 21/70 = 0.30. Among 35 plants, the expected numbers are
Genotype AA
AB
Expected
5.57 10.86 11.70
0.71
BB
BC
CC
AC
3.15
3.00
2
14-16. (a) (0.01)  1,000,000 = 100 seedlings. (b) The lethal recessive allele will
persist for many generations at a low frequency, because it is present
primarily in heterozygotes and is shielded from selection by the dominant
allele.
14-17. The frequency of the recessive allele is q = √(1/14,000) = 0.0085. Hence
p = 1  0.0085 = 0.9915, and the frequency of heterozygotes is 2pq = 0.017
(about 1 in 60).
14-18. (a) Here, F = 0.66, p = 0.43, and q = 0.57. The expected genotype
2
frequencies are as follows: AA, (0.43) (1  0.66) + (0.43)(0.66) = 0.347; AB,
2
2(0.43)(0.57)(1  0.66) = 0.167; and BB, (0.57) (1  0.66) + (0.57)(0.66) = 0.487.
(b) With random mating, the expected genotype frequencies are as follows:
2
2
AA, (0.43) = 0.185; AB, 2(0.43)(0.57) = 0.490; and BB, (0.57) = 0.325.
14-19. To obtain the relative fitness of B relative to A use, the equation pn/qn =
(p0/q0)(1/w)n. Where n = 100, p is the frequency of A, and q is the frequency of
B. The selection coefficient s = 1 – w. (a) The result implies that pn/qn = 1.10 
p0/q0, so the equation to solve is 1.10 = (1/w)100. Rearranging and taking
logarithms, we find that w = 0.99905, so s = 1 – w = 0.00095. (b) In this case, w
= 0.9936 and s = 0.0064. (c) Here w = 0.99309 and s = 0.00691.
14-20. Let pn be the allele frequency of A in generation n, and for
convenience, set  = 10–5. The probability of a particular A allele not mutating
in any one generation is 1 – , hence p1 = (1 – )p0, but because p0 = 1, p1 = 1 –
 = 0.99999. The relationship between p2 and p1 is the same as that between p1
and p0, and so p2 = (1 – )p1 = (1 – )(1 – )p0 = (1 – )2p0 = (1 – )2 = 0.99998.
In general, the rule is pn = (1 – )np0, which, in this example, is (0.99999)n.