Question

Assume a population is in Hardy-Weinberg equilibrium. If there are two alleles at a locus, R and r, and the frequency of heterozygotes is known to be 0.18. What could be the frequency of the R allele?

a) 0.9
b) 0.81
c) 0.1
d) a & c are correct
e) Cannot tell from information provided

Answer 1

The frequency of the R allele can be calculated using the Hardy-Weinberg equation. Given that the frequency of heterozygotes is 0.18, the possible frequency of the R allele is approximately 0.81.

Step-by-step explanation:

The frequency of the R allele can be calculated using the Hardy-Weinberg equation, which states that p² + 2pq + q² = 1, where p is the frequency of the dominant allele (R) and q is the frequency of the recessive allele (r).

Given that the frequency of heterozygotes (Rr) is 0.18, we can substitute this value into the equation:

0.18 = 2pq

Since p + q = 1, we can substitute 1 - p for q:

0.18 = 2p(1 - p)

Simplifying this equation, we get:

0.18 = 2p - 2p²

Rearranging the equation, we get:

2p² - 2p + 0.18 = 0

Solving this quadratic equation, we find that p ≈ 0.81.

Therefore, the possible frequency of the R allele is approximately 0.81, so the correct answer is b) 0.81.