Question

Suppose a population is carrying a condition controlled by two alleles: A (dominant) and a (recessive). Only homozygous individuals that have two copies of the recessive allele have the condition. If the a allele has a frequency of 20%, and the A allele had a frequency of 80%, what percentage of the population will have the condition?
A. 20%
B. 4%
C. 80%
D. 64%

Answer 1

Make a = q, aa (homozygous recessive) = q^2,
A = p, AA (homozygous dominant) = p^2, and
2pq = heterozygous
This was derived from p+q = 1
Therefore all a in pop (q) = 20% = .20
And all A in pop (p) = 80% = .80
Since the disease is homozygous recessive (affected), then aa = qq or q×q = q^2
${q}^{2} = {(.20)}^{2} = .040 = 4\%$
Therefore [B. 4%] is the correct answer.

Answer 2

The answer is B. 4% of the population will have the condition