Question

Calculate the allelic frequencies if they give us the following phenotypes:

A1A1 = 132, A1A2 = 69 y A2A2 = 14

Answer 1

Here total population is 132 + 69 + 14 = 215

According to Hardy Weinberg equation:

p² + 2pq + q² =1 and p + q = 1

where

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

p² = percentage of homozygous dominant individuals

q² = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Here two allele is A1 and A2. The frequency of homozygous dominant genotype (A1A1) = 132/ 215 = 0.6139 which means p² = 0.6139 therefore p = 0.78.

Since value of p = 0.78 therefore value of q = 1 - 0.78 = 0.22

Thus the frequency of both allele p (A1) and q (A2) are 0.78 and 0.22 respectively.