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)

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