at a particular locus, the frequency of allele a is 0.8 and that of allele a is 0.2. what would be the frequency of heterozygotes in a random mating population at equilibrium?
What would be the frequency of heterozygotes in a random mating population at equilibrium?
Answer: In a population at genetic equilibrium, under Hardy-Weinberg equilibrium, the frequency of heterozygotes can be calculated using the Hardy-Weinberg equation. This equation states that in a given population, the square of the frequency of the homozygous dominant genotype (AA), the frequency of the heterozygous genotype (Aa), and the square of the frequency of the homozygous recessive genotype (aa) will add up to 1.
Given that the frequency of allele ‘a’ (recessive allele) is 0.8 and that of allele ‘A’ (dominant allele) is 0.2, we can calculate the frequency of the heterozygotes using the Hardy-Weinberg equation as follows:
Let p = frequency of allele ‘A’ = 0.2
Let q = frequency of allele ‘a’ = 0.8
According to the Hardy-Weinberg equation:
- p^2 + 2pq + q^2 = 1
Substitute the values of p and q:
- (0.2)^2 + 2(0.2)(0.8) + (0.8)^2 = 1
- 0.04 + 0.32 + 0.64 = 1
- 1 = 1
Therefore, the frequency of heterozygotes (Aa) in the population is 0.32 or 32%.