If 64% of the people in a population are blue-eyed and the population is in Hardy-Weinberg equilibrium,what is the percentage of heterozygotes in the population?

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the frequency of homozygous individuals for a particular trait is given by the square of the frequency of one of the alleles, while the frequency of heterozygous individuals is given by 2 times the frequency of one allele times the frequency of the other allele.

Let P be the frequency of the blue-eyed allele. Then, q = 1-P is the frequency of the non-blue-eyed allele. In a population in Hardy-Weinberg equilibrium, we have:

P^2 + 2Pq + q^2 = 1

Plugging in P = 0.64 and q = 0.36, we get:

0.64^2 + 2(0.64)(0.36) + 0.36^2 = 0.4096 + 0.4608 + 0.1296 = 0.999

Therefore, the frequency of heterozygotes is 2Pq = 2(0.64)(0.36) = 0.4608, or 46.08%. Rounded to the nearest integer, this gives 46%, the closest answer choice being C (32%).

If 64% of the People in a Population Are Blue-Eyed