Question

The frequencies of two alleles in a gene pool are .42 (A) and .58 (a). Assume the population is in Hardy-Weinberg equilibrium. Calculate the percentage of heterozygous and homozygous recessive individuals in the population.

Answer 1

I believe the correct answer to be:

Since frequency of A allele is 0.42, the frequency of homozygous AA alele = AA or A2 = 0.422 = 0.176 or 17.6%.

And same for the frequency of homozygous aa allele = a2 = 0.582 = 0.336 or 33.6%.

So the percentage of homozygous individuals is 33.6+17.6 = 51.2%.

And the remaining that are heterozygous are 100-51.2 = 48.8%

feel free to leave a comment down below for any further query. good rating would be appreciated if you find my answer helpful. thank you.