Question

In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele A is 0.35. What is the percentage of the population that is homozygous for this allele?

Answer 1

Introduction: An allele is a genetic unit that controls a specific entity of an organism. The frequency that appears for a particular allele in a population is known as allele frequency. The Hardy–Weinberg law is a basic principle in population genetics. It states that the genetic frequency of an allele in a population is same for generations and is not affected by other evolutionary influencing factors such as mutation and selection. This principle works with only the genetics of diploid organisms having larger population size, no mutation, no random mating, and no natural selection.

Explanation:

The Hardy-Weinberg population has two alleles, A and a, that are in equilibrium then, the equation is p²+2pq+q² = 1 where,

p – frequency of dominant allele "A";

q – frequency of recessive allele "a";

p² – frequency of dominant homozygous genotype "AA";

q² – frequency of recessive homozygous genotype "aa"; and

2pq – frequency of heterozygous genotype "Aa";

Also, the sum of frequency of both alleles is p + q = 1.

Based on the above mentioned problem, the frequency of an allele A is 0.35.

Therefore, p = 0.35.

p² represents the frequency of homozygous genotype "AA"

Thus, p² = A*A

               = 0.35*0.35

               = 0.12

Therefore, p² = 0.12

The percentage of population for this homozygous allele "A" is 12%.