Question

The frequency of two alleles in a gene pool is 0.19 (A) and 0.81 (a). Assume that the size of the population is 100 and the popultaion is in Hardy-Weinberg equilibrium. [3]
a) Calculate the number of heterozygous individuals in the population.
b) Calculate the number of homozygous recessive individual in the population.

Answer 1

Given,
Frequency of allele (A) in a gene pool = 0.19
Frequency of allele (a) in a gene pool = 0.81
According to Hardy-Weinberg equilibrium equation,
p²+2pq+q² = 1
where,
p² = frequency of homozygous dominant genotype
2pq = frequency of heterozygous genotype
q² = frequency of homozygous recessive genotype
a) According to Hardy-Weinberg equilibrium equation, heterozygotes are represented by the 2pq term. Therefore, the number of heterozygous individuals
(Aa) is equal to 2pq which equals to
2 x 0.19 x 0.81 = 0.3078
Therefore in a population of 100, the number of heterozygotes are = 30.78 = 31 [1.5]
b) The homozygous recessive individuals (aa) are represented by the q² term in the Hardy-Weinberg equilibrium equation which equals to
0.81 x 0.81 = 0.6561
Therefore in a population of 100, the number of homozygous recessive are = 65.61 = 66 [1.5]