Question

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

• A. $0.32$
• B. $4$
• C. $16$
• D. $32$

Answer 1

The correct option is D $32$

Answer is option D i.e. "32"

According to Hardy-Weinberg equation, the sum of 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.