Answer:
980,100
Explanation:
let the recessive condition (aa) be represented as q² which is the genotypic frequency = 100/1,000,000 = 0.0001
allelic frequency of a = q = √q² = √0.0001 = 0.01. thus q = 0.01
From the formula p + q = 1 where p is the allelic frequency of A.
since q = 0.01
p = 1 - 0.01 = 0.99. The allelic frequency of A (p) = 0.99
p² = 0.99² = 0.9801. Genotypic frequency of AA= 0.9801
= 0.9801 x 1,000,000 = 980,100 individuals with AA (homozygous normal)
For 2pq genotypic frequency for hetrozygous (Aa).
Using the formula p² + 2pq + q² = 1. Since p² = 0.9801 and q² = 0.0001
2pq = 1 - (p² + q²)
= 1 - (0.9801 + 0.0001)
= 1 - (0.9802)
= 0.0198 = 0.0198 x 1,000,000 = 19,800 individuals with Aa
