Question

You are studying a single-gene locus with two alleles in a population that is in Hardy-Weinberg equilibrium. Examination of a large sample of individuals from the population reveals there are six times as many heterozygote as there are homozygote recessive individuals in this population. What is the frequency of the recessive allele?

Answer 1

To determine the frequency of the recessive allele in a population in Hardy-Weinberg equilibrium, use the equation p² + 2pq + q² = 1. Given that there are six times as many heterozygotes as there are homozygous recessive individuals, the frequency of the recessive allele can be found to be 0.25.

Step-by-step explanation:

To determine the frequency of the recessive allele in a population in Hardy-Weinberg equilibrium, we can use the equation p² + 2pq + q² = 1. In this equation, p represents the frequency of the dominant allele and q represents the frequency of the recessive allele.

Given that there are six times as many heterozygotes as there are homozygous recessive individuals, we can set up the equation:

2pq = 6q²

By rearranging the equation, we get:

2p = 6q

Dividing both sides of the equation by 6, we find:

q = 2p/6 = p/3

Since p + q = 1, we can substitute p/3 for q in the equation and solve for p:

p + p/3 = 1

Multiplying both sides of the equation by 3, we find:

3p + p = 3

Combining like terms, we get:

4p = 3

Dividing both sides of the equation by 4, we find:

p = 3/4 = 0.75

Therefore, the frequency of the recessive allele is q = p/3 = 0.75/3 = 0.25.