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
[tex] {q}^{2} = {(.20)}^{2} = .040 = 4\%[/tex]
Therefore [B. 4%] is the correct answer.
