Question

Given a population of 100 individuals where 15 are AA, 25 are Aa, and 60 are aa, what is

Answer 1

Answer:

frequency of allele “A” is .39

frequency of allele “a” is .77

Explanation:

hardy weinberg equation

p^2 + 2pq + q^2 = 1 or (A+a)^2

p^2 =  (AA)

2pq =  (Aa)

q^2 = (aa)

AA -> 15/100 = .15

Aa -> 25/100 = .25

aa -> 60/100 = .60

if p^2 = AA = .15 then A = √.15 = 0.38729833462 = .39

if q^2 = aa = .60 then a = √.60 = 0.77459666924 = .77

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