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 would have the condition

Answer 1

Hi! :)

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%

Answer: 4%