Question

In a population that is in Hardy-Weinberg equilibrium, 49% of the population show the homozygous recessive genotype. The frequency of homozygous dominant individuals is:

Answer 1

The frequency of homozygous dominant individuals is: 42%

Hardy–Weinberg equilibrium (HWE) is a null model of the relationship between allele and genotype frequencies, both within and between generations, assuming no mutation, no migration, no selection, random mating, and infinite population size. To find the frequency of the recessive allele, we must first find the frequency of the dominant allele (p). According to the Hardy-Weinberg principle, the square root of the homozygous genotype frequency is equal to the allele frequency. The Hardy-Weinberg equation used to determine genotype frequencies is: p2 + 2pq + q2 = 1. Where 'p2' represents the frequency of the homozygous dominant genotype (AA), '2pq' the frequency of the heterozygous genotype (Aa) and 'q2' the frequency of the homozygous recessive genotype (aa).

Calculation:

q2 = 49

q= 0.7

According to Hardy–Weinberg equilibrium p + q= 1

p = 1 - q

1-0.7= 0.3

p2+ 2pq+q2=1

One can substitute the values

2pq= 2(0.3) (0.7) = 0.42

42%  is the answer.

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