Question

at a particular locus, the frequency of allele a is 0.8 and that of allele a is 0.2. what would be the frequency of heterozygotes in a random mating population at equilibrium?

Answer 1

The frequency of heterozygotes in a population at Hardy-Weinberg equilibrium with allele frequencies p (A) = 0.8 and q (a) = 0.2 is 0.32 or 32%.

Step-by-step explanation:

The frequency of heterozygotes in a random mating population at equilibrium can be calculated using the Hardy-Weinberg equation, which is p² + 2pq + q² = 1.

Given that the frequency of allele a (q) is 0.2 and the frequency of allele A (p) is 0.8, we can calculate the frequency of heterozygotes (2pq).

This is:

2pq = 2(0.8)(0.2)

= 0.32

So, the frequency of the heterozygote genotype (Aa) in this population is 0.32 or 32%.