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By Suman Patra
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Frequencies of alleles A and a in a population at hardy weinberg equilibrium are 0.7 and 0.3 respectively. In a random sample of 250 individuals taken from the population how many are expected to be heterozygous?
Home » Queries & Discussions » Frequencies of alleles A and a in a population at hardy weinberg equilibrium are 0.7 and 0.3 respectively. In a random sample of 250 individuals taken from the population how many are expected to be heterozygous?
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Frequency of allele A (p) = 0.7 or 70%
Frequency of allele a (q) = 0.3 or 30%
Total number of individuals(p2 + 2pq + q2) =250
First we have to found out, 2pq (Heterozygous genotype) = 2 x 0.7 x 0.3 = 0.42 or 42%
in 250 populations, 42% individuals are heterozygous i.e.,
250 x (42/100) or 250 x 0.42= 105