Evolution__3rd_Edition

102 PART 2 / Evolutionary Genetics
Natural populations may or may not
fit the Hardy–Weinberg equilibrium
5.4 We can test, by simple observation, whether genotypes
in a population are at the Hardy–Weinberg equilibrium
The Hardy–Weinberg theorem depends on three main assumptions: no selection,
random mating, and large population size. In a natural population, any of these could
be false; we cannot assume that natural populations will be at the Hardy–Weinberg
equilibrium. In practice, we can find out whether a population is at the Hardy–
Weinberg equilibrium for a locus simply by counting the genotype frequencies.
From those frequencies, we first calculate the gene frequencies; then, if the observed
homozygote frequencies equal the square of their gene frequencies, the population is in
Hardy–Weinberg equilibrium. If they do not, it is not.
The MN blood group system in humans is a good example, because the three
genotypes are distinct and the genes have reasonably high frequencies in human
populations. Three phenotypes, M, MN, and N are produced by three genotypes (MM,
MN, NN) and two alleles at one locus. The phenotypes of the MN group, like the better
known ABO group, are recognized by reactions with antisera. The antisera are made
by injecting blood into a rabbit, which then makes an antiserum to the type of blood
that was injected. If the rabbit has been injected with M-type human blood, it produces
Table 5.2
The frequencies of the MM, MN, and NN blood groups in three American populations. The figures for expected proportions and
numbers have been rounded.
Population MM MN NN Total Frequency M Frequency N
African Americans Observed number 79 138 61 278
Expected proportion 0.283 0.499 0.219 0.532 0.468
Expected number 78.8 138.7 60.8
European Americans Observed number 1,787 3,039 1,303 6,129
Expected proportion 0.292 0.497 0.211 0.54 0.46
Expected number 1,787.2 3,044.9 1,296.9
Native Americans Observed number 123 72 10 205
Expected proportion 0.602 0.348 0.05 0.776 0.224
Expected number 123.3 71.4 10.3
Specimen calculation for African Americans: Frequency of M allele = 79 + ( 1 /2 × 138) = 0.532 = p
Frequency of N allele = 61 + ( 1 /2 × 138) = 0.468 = q
Expected proportion of MM = p 2 = (0.532) 2 = 0.283
Expected proportion of MN = 2pq = 2(0.532) (0.468) = 0.499
Expected proportion of NN = q 2 = (0.468) 2 = 0.219
Expected numbers = expected proportion × total number (n) Expected number of MM = p 2 n = 0.283 × 278 = 78.8
Expected number of MN = 2pqn = 0.499 × 278 = 138.7
Expected number of NN = q 2 n = 0.219 × 278 = 60.8
..