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 heterozygotes as there are homozygote recessive individuals in this population. What is the frequency of the recessive allele?

To find the frequency of the recessive allele in a population, we can use the information that there are six times as many heterozygotes as there are homozygote recessive individuals.

Let's define the following variables:
- p : frequency of the dominant allele (A)
- q : frequency of the recessive allele (a)

According to the Hardy-Weinberg principle, in equilibrium:
- p^2 : represents the frequency of homozygote dominant individuals (AA)
- 2pq : represents the frequency of heterozygotes (Aa)
- q^2 : represents the frequency of homozygote recessive individuals (aa)

Given the information that there are six times as many heterozygotes as there are homozygote recessive individuals, we can write the following equation:

2pq = 6 * q^2

Now, to calculate the frequency of the recessive allele (q), we need to substitute the value of 2pq with 6q^2:

6q^2 = 2pq

Simplifying this equation, we get:

6q^2 = 2q^2

Dividing both sides by 2q^2, we get:

3 = q

Therefore, the frequency of the recessive allele (q) is 3/10 or 0.3.

Keep in mind that in the Hardy-Weinberg equilibrium, the sum of the frequencies of the two alleles (p + q) is equal to 1. So, if q = 0.3, then p = 1 - q = 1 - 0.3 = 0.7.