Final answer:
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.