Question

y individuals would be expected to be homozygous normal (AA) under equilibrium conditions? Albinism is an autosomal recessive trait in humans. Assume that there are 100 albinos (aa) in a population of 1 million. How many individuals would be expected to be homozygous normal (AA) under equilibrium conditions? 100 10,000 19,800 980,100 999,900

Answer 1

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