Question

In rabbit species , a disease condition called albinism develops due to recessive alleles. let A be the allele for normal condition and "a" be the allele for albinism,such that AA and Aa represent normal phenotypes and "aa" represents the disease. assume a large population in genetic equilibrium in which 16 % of the individuals are affected. calculate the frequency of the "a" allele and that of A allele.

Answer 1

According to Hardy-Weinberg equilibrium
p² + 2pq + q² = 1
The frequency of aa is 16% which means q² = 0.16
Hence, q = 0.4
Since, p + q = 1
Hence, p = 1 - 0.4 = 0.6
Hence, frequency of allele A = 60% and that of allele a 40%.