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? 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

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

the answer is lactic acid from anaerobic respiration. the acid in lactic acid is what creates sore and burning muscles. if you think about how acid burns that will give you a clue for questions like this.