In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of the allele a is 0.70. What is the percentage of the population that is homozygous for this allele? (Your answer should have two decimal places following an initial 0. For example, if your answer was 56%, you should enter 0.56)

Question

Social Studies

Tristin Mckay

24 September, 23:47

In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of the allele a is 0.70. What is the percentage of the population that is homozygous for this allele? (Your answer should have two decimal places following an initial 0. For example, if your answer was 56%, you should enter 0.56)

+4

Answers (1)

Know the Answer?

Elyse Blankenship · Answer 1

Answer: 0.49

Explanation: In Hardy-weinberg, the population is in equilibrium, which means the expected proportions for the genotype is expressed as:

(p+q) ² = 1 or p²+2pq+q² = 1, where

p is the frequency of the dominat allele;

q is the frequency of the recessive allele;

If the frequency of the allele a is 0.70, it means q=0.70.

To find the percentage who is recessive homozygous:

q² = (0.70) ² = 0.49

Therefore, the percentage of population homozygous for allele a is 0.49