Question

Consider a locus with two alleles - red and green. In a population, all assumptions of Hardy-Weinberg are true, except for no mutation. In this population, the red allele mutates to the green allele with a probability of 0.0005, while the green allele mutates to the red red with probability of 0.01. What is the equilibrium frequency of the red allele in this population?

A. 0.01
B. 0
C. 0.05
D. 0.95
E. 0.5

Answer 1

The equilibrium frequency of the red allele in a population, considering mutation rates but no other violations of the Hardy-Weinberg equilibrium conditions, is approximately 5%, corresponding to option C.

Step-by-step explanation:

To determine the equilibrium frequency of the red allele in a population where the Hardy-Weinberg equilibrium is disturbed only by mutations, we can derive the frequency using the mutation rates provided. The equilibrium condition is reached when the rate of mutation from red to green is equal to the rate of mutation from green to red. This can be expressed with the formula:

p(mutation rate from green to red) = q(mutation rate from red to green)

Where p is the frequency of the red allele and q is the frequency of the green allele, with p + q = 1. Using the provided mutation rates (red to green is 0.0005, and green to red is 0.01), the equation can be set up as follows:

p(0.01) = (1-p)(0.0005)

Solving for p gives us:

p = 0.0005 / (0.01 + 0.0005) = 0.0005 / 0.0105 ≈ 0.0476

So, the equilibrium frequency of the red allele is approximately 0.05 or 5%, which corresponds to option C.

Answer: C. 0.05