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

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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%

4%

A

Step-by-step explanation:

Since the theta is in second quadrant, x will be negative and y positive. cos(theta)=-2/sqrt(29)=B/H, the base is 2 and H=sqrt(29). The perpendicular is 5.