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.
