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28 June, 01:45

Eight percent of a population shows a recessive trait. What would be the frequency of the heterozygotes for that trait if the population is in hardy=w

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  1. 28 June, 01:55
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    Population that show recessive trait should be only homozygous recessive. If the population in Hardy Weinberg equilibrium, then the recessive trait gene frequency would be:

    homozygous recessive = aa = 8%=0.08

    a = √0.08=0.283

    a+A=1

    A = 1-a = 1 - 0.283 = 0.717

    The percentage of heterozygote in population would be:

    heterozygote = 2 Aa = (2) (0.717) (0.283) = 0.4058 = 40.58%
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