Evolution__3rd_Edition

126 PART 2 / Evolutionary Genetics
We deduce selection coefficients of
0.12 and 0.86
Table 5.9
Estimates of selection coefficients for sickle cell anemia, using genotype frequencies in adults.
The sickle cell hemoglobin allele is S, and the normal hemoglobin (which actually consists of
more than one allele) is A. The genotype frequencies are for the Yorubas of Ibadan, Nigeria.
One small detail is not explained in the text. The observed : expected ratio for the heterozygote
may not be equal to 1. Here it turned out to be 1.12. All the observed : expected ratios are
therefore divided by 1.12 to make them fit the standard fitness regime for heterozygote
advantage. From Bodmer & Cavalli-Sforza (1976).
Observed Expected
adult Hardy–Weinberg Ratio
Genotype frequency (O) frequency (E) O : E Fitness
SS 29 187.4 0.155 0.155/1.12 = 0.14 = 1 − t
SA 2,993 2,672.4 1.12 1.12/1.12 = 1.00
AA 9,365 9,527.2 0.983 0.983/1.12 = 0.88 = 1 − s
Total 12,387 12,387
Calculation of expected frequencies: gene frequency of S = frequency of SS + 1 /2 (frequency of SA) = (29 +
2,993/2)/12,387 = 0.123. Therefore the frequency of A allele = 1 − 0.123 = 0.877. From the Hardy–Weinberg
theorem, the expected genotype frequencies are (0.123) 2 × 12,387, 2(0.877)(0.123) × 12,387, and (0.877) 2 ×
12,387, for AA, AS, and SS, respectively.
Genotype AA AS SS
Fitness 1 − s 1 1 − t
If the frequency of gene A = p and of gene S = q, then the relative genotype frequencies
among adults will be p 2 (1 − s):2pq : q 2 (1 − t). If there were no selection (s = t = 0), the
three genotypes would have Hardy–Weinberg frequencies of p 2 :2pq : q 2 .
Selection causes deviations from the Hardy–Weinberg frequencies. Take the genotype
AA as an example. The ratio of the observed frequency in adults to that predicted
from the Hardy–Weinberg ratio will be (1 − s)/1. The frequency expected from the
Hardy–Weinberg principle is found by the usual method: the expected frequency is p 2 ,
where p is the observed proportion of AA plus half the observed proportion of AS. Table
5.9 illustrates the method for a Nigerian population, where s = 0.12 (1 − s = 0.88) and
t = 0.86 (1 − t = 0.14).
The method is only valid if the deviation from Hardy–Weinberg proportions is
caused by heterozygous advantage and the genotypes differ only in their chance of
survival (not their fertility). If heterozygotes are found to be in excess frequency in a
natural population, it may indeed be because the heterozygote has a higher fitness.
However, it could also be for other reasons. Disassortative mating, for instance,
can produce the same result (in this case, disassortative mating would mean that aa
individuals preferentially mate with AA individuals). But for sickle cell anemia, the
physiological observations showed that the heterozygote is fitter and the procedure is
well justified. Indeed, in this case, although it has not been checked whether mating is
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