Fe Sci 7-2
1 answer:
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Answer:
population = 600
heterozygotes = 2pq
The homozygous recessive percentage =
q = 0.1
Solve for q:
∴ q =
∴ q =
∴ q = 0.316
solve for p:
for p + q = 1,
thus p = 1 – q = 1 – 0.316 = 0.684
The heterozygotes = 2pq
2pq = 2 x 0.316 x 0.684 = 0.432
600 x 0.432 = 259.2 = 259
so, the population reaches Hardy Weinberg equilibrium, 259 of indviduals should be heterozygous for this trait.
Hence, the correct answer is 259 .