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100 individuals whose genetic makeup is composed of 20% recessive alleles join an established population. The established population was composed of 600 individuals whose frequency of their recessive allele was 10%. When the population reaches Hardy Weinberg equilibrium, _____________ of indviduals should be heterozygous for this trait.

User Jaryd
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1 Answer

5 votes

Answer:

population = 600

heterozygotes = 2pq

The homozygous recessive percentage =
q^(2)

q = 0.1

Solve for q:

∴ q =
√(q)

∴ q =
√(0.10)

∴ 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 .

User Vizllx
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