Question

Assume fur color is controlled by one gene with two alleles in squirrels, and gray is dominant to brown. if a population of 360 gray and 640 brown squirrels is in hardy-weinberg equilibrium, what is the frequency of the dominant allele?

A. 0.80
B. 0.36
C. 0.64
D. 0.40
E. 0.20
F. 0.60

Answer 1

The frequency of the dominant allele in a Hardy-Weinberg equilibrium squirrel population with a majority of brown squirrels is 0.20.

Step-by-step explanation:

Hardy-Weinberg equilibrium:

The frequency of the dominant allele in a squirrel population under Hardy-Weinberg equilibrium with 360 gray and 640 brown squirrels can be calculated using allele frequencies based on established principles. Since we know the gray fur color is dominant (A) and brown is recessive (a), and that in Hardy-Weinberg equilibrium p² + 2pq + q² = 1 where p is the frequency of the dominant allele and q is the frequency of the recessive allele, we first need to find q.

In this population, all 640 brown squirrels are homozygous recessive (aa), thus q² equals 640/1000 (since the total number of squirrels is 1000). Hence, q equals the square root of 0.64, which is 0.8. According to the Hardy-Weinberg principle, p + q = 1, we can subtract 0.8 from 1 to find p. Therefore, p equals 1 - 0.8, which is 0.2. The correct answer is E. 0.20, as this represents the frequency of the dominant allele.