Question

A population of horses has individuals that are either gray or chestnut colored. The gray color allele, G, is dominant over the chestnut color allele, g. If the frequency of the gray allele in the population is 0.8, what percent of horses are gray? 64 percent 96 percent 32 percent 80 percent

Answer 1

We solve the problem using the Hardy-Wineberg equation.

We have the frequency for gray trait as p =  = 0.8

∴ p² = 0.64

64 % of the horses are homozygous for the gray trait.

Solving for the frequency of white chestnut trait by using the fact that,

p + q = 1

∴ q = 1 - p = 1 - 0.8 = 0.2

Now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous gray trait,

∴ 2pq = 2 x 0.8 x 0.2 = 0.32

Therefore, in a population of 100 horses, 32% are heterozygous for gray trait.

Hence the total number of gray horses in a population of 100 will be,

32 (heterozygous gray horses) + 64 (homozygous gray horses) = 96 (gray horses)

Hence, 96% of the horses are gray in color.