Question

In a Hardy-Weinberg population, with two alleles (A and a) that are in equilibrium, the frequency of the allele A is 0.2. What is the percentage of the population that is heterozygous for this allele?

• A. 0.32
• B. 4
• C. 16
• D. 32

Answer 1

The correct option is D 32

According to the Hardy-Weinberg equation, the sum of the allele frequencies for all the alleles at the locus must be 1, so p + q = 1. Also, the Hardy-Weinberg equation is expressed as: p${}^2$ + 2pq +q${}^2$ = 1 where p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. In the equation, p${}^2$ represents the frequency of the homozygous genotype AA, q${}^2$ represents the frequency of the homozygous genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. Here, q = 0.2.Hence, p = `1- q = 1 - 0.2 = 0.8. Now, population of heterozygous individual will be 2pq as mentioned, that is 2 * 0.8 * 0.2 = 0.32. It means, there is 32% of the heterozygous population.

Thus, the correct answer is option D.