Question

Use the Hardy-Weinberg equations: p + q = 1 and

p2 + 2pq + q2 = 1. If the dominant allele frequency is 0.8, then what percent of the population will be homozygous recessive?
A. 0.64
B. 0.32
C. 0.04
D. 0.16
E. 0.40

Answer 1

The percent of the population that will be homozygous recessive is 40%.

Step-by-step explanation:

To calculate the percent of the population that will be homozygous recessive, we need to determine the value of q (frequency of the recessive allele). From the given information, we know that p (frequency of the dominant allele) is 0.8. Since p + q = 1, we can subtract p from 1 to find the value of q. Therefore, q = 1 - 0.8 = 0.2.

The Hardy-Weinberg equation p² + 2pq + q² = 1 allows us to calculate the frequencies of different genotypes in a population. In this case, we are interested in the frequency of the homozygous recessive genotype (aa). Since q = 0.2, we can substitute this value into the equation:

(0.2)² + 2(0.8)(0.2) + (0.2)² = 0.04 + 0.32 + 0.04 = 0.4.

Therefore, 0.4 or 40% of the population will be homozygous recessive.