In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. The frequency of individuals that are heterozygous for this allele is 0.32.
The Hardy-Weinberg equation dictates that the sum of the allele frequencies for each allele at the locus must equal 1, therefore p + q = 1. The Hardy-Weinberg equation is also written as p2 + 2pq + q2 = 1, where p is the population's frequency of the "A" allele and q is that of the "a" allele.
The homozygous genotype AA frequency is represented by p2, the homozygous genotype aa frequency by q2, and the heterozygous genotype aa frequency by 2pq in the equation.
Here, q = 0.2. Because of this, p = '1- q = 1 - 0.2 = 0.8.
The population of individuals who are heterozygous will now be 2pq, or 2 * 0.8 * 0.2 = 0.32.
Hence, This indicates that the population is heterozygous with a frequency of 0.32.
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