Question

Consider a population in Hardy-Weinberg equilibrium. The frequency of recessive homozygotes is 0.16 what is the frequency of heterozygotes in this population? (show work)

Answer 1

According to Hardy-Weinberg equilibrium-

                                                    p² + 2pq + q² = 1 & p + q = 1

Where, p = frequency of dominant allele, q = frequency of recessive allele, p² = frequency of homozygous dominant genotype, q² = frequency of homozygous recessive genotype & 2pq = frequency of heterozygous genotype.

Given, frequency of recessive homozygotes (q²) = 0.16

So, frequency of recessive allele (q) = = = 0.4

So, frequency of dominant allele (p) = 1 - q = 1 - 0.4 = 0.6

So, frequency of heterozygotes (2pq) = 2 x p x q = 2 x 0.6 x 0.4 = 0.48