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Consider a sample of 100 individuals sampled from a population in Hardy-Weinberg equilibrium that are genotyped at a single l
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Answer #1

Total individuals: 100

A1A1: 25

A1A2: 45

A2A2: 30

Observed genotypic frequency:

A1A1: 25/100 = 0.25

A1A2: 45/100 = 0.45

A2A2: 30/100 = 0.30

Expected genotypic frequency:

p = freq (A1A1) + 1/2 freq (A1A2) = 0.25 + 1/2 X 0.45 = 0.25 + 0.225 = 0.475

q = 1-p = 1-0.475 = 0.525

freq (A1A1) = p^2 = 0.475^2 = 0.225

freq (A1A2) = 2pq = 2 x 0.475 X 0.525 = 0.498

freq (A2A2) = q^2 = 0.525^2 = 0.275

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