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11 December, 03:47

Assume a population is in Hardy-Weinberg equilibrium for the B gene, which has two alleles, B and b. In this population of 100,000 individuals, if the frequency of the b allele (q) is 4%, how many homozygous bb individuals would you expect to find in the population?

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  1. 11 December, 03:50
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    160

    Explanation:

    If p = frequency of dominant allele and q = frequency of recessive allele according to Hardy-Weinberg equilibrium,

    p + q = 1

    p² + 2pq + q² = 1

    where p² = frequency of dominant homozygous genotype

    q² = frequency of recessive homozygous genotype

    2pq = frequency of heterozygous genotype

    Here,

    b = 4% = 0.04

    bb = 0.04 * 0.04 = 0.0016

    Number of individuals in population = 100000

    Number of homozygous bb individuals = 0.0016 * 100000 = 160
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