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18 July, 17:41

A fruit fly population has a gene with two alleles, A1 and A2. Molecular genetic tests show that 40% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies are homozygous for the A1 allele?

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  1. 18 July, 18:02
    0
    = 0.48

    Explanation:

    the genetic frequency can be calculated using hardy - Weinberg equation

    p + q = 1

    here p represent the dominant trait and q represent the recessive trait.

    the proportion A1 or p = 40/100

    = 0.4

    now, the proportion of A2 or q will be

    p + q = 1

    0.4 + q = 1

    q = 1 - 0.4

    q = 0.6

    proportion of the flies are homozygous for the A1 allele

    2pq

    2 x 0.4 x 0.6

    = 0.48
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