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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.

Respuesta :

Answer:

population = 600

heterozygotes = 2pq

The homozygous recessive percentage = [tex]q^{2}[/tex]

q = 0.1

Solve for q:

∴ q = [tex]\sqrt{q}[/tex]

∴ q = [tex]\sqrt{0.10}[/tex]

∴ 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 .