Hardy-Weinberg Equilibrium - ScienceAid
Assume that a population is in Hardy-Weinburg equilibrium for a particular gene with two alleles, A and a. The frequency of AA is p, and the frequency of a is q. Because these are the only two alleles for this gene, p+q=1.0. If the frequency of homozygous recessive individuals (aa) is 0.04, what is the value of q? What is the value of p?
a
q=0.2; p=0.8
b
q=0.8; p=0.2
c
q=0.04; p=0.16
d
q=0.1; p=1.0
The frequency of homozygous recessive individuals (aa) is given as 0.04, which means q^2 = 0.04.
Therefore, q = √0.04 = 0.2
Since p + q = 1, we can calculate the value of p:
p = 1 - q = 1 - 0.2 = 0.8
Thus, the correct answer is:
a) q = 0.2; p = 0.8