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 frequency of individuals that are heterozygous for this allele

Answer 1

In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. The frequency of individuals that are heterozygous for this allele is 0.32.

Hardy Weinberg Equilibrium Principle

• A stable, idealized population's constant frequency of alleles and genotypes is described by the Hardy-Weinberg equilibrium principle.
• In a sizable population, random mating, or spontaneous mutations are necessary for Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation dictates that the sum of the allele frequencies for each allele at the locus must equal 1, therefore p + q = 1. The Hardy-Weinberg equation is also written as p2 + 2pq + q2 = 1, where p is the population's frequency of the "A" allele and q is that of the "a" allele.

The homozygous genotype AA frequency is represented by p2, the homozygous genotype aa frequency by q2, and the heterozygous genotype aa frequency by 2pq in the equation.

Here, q = 0.2. Because of this, p = '1- q = 1 - 0.2 = 0.8.

The population of individuals who are heterozygous will now be 2pq, or 2 * 0.8 * 0.2 = 0.32.

Hence, This indicates that the population is heterozygous with a frequency of 0.32.

To learn more about Hardy Weinberg Principle refer to:

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