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30 December, 20:42

Suppose that in a randomly mating population of mammals, 160 of its 1,000 members exhibit a specific recessive trait that does not affect viability of the individual. How many individuals in this population are heterozygous carriers of the gene that causes this trait?

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  1. 30 December, 22:56
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    480

    Explanation:

    To calculate the gene frequency, q, of the recessive trait we will take the square root of 16/100. The square root of 16/100 will be. 4.

    The frequency, p, will be 1 - 0.4. Hence, p will be 0.6.

    Now,

    According to the Hardy-Weinberg law, the frequency of the carriers can be calculated by

    2pq

    2 x 0.4 x 0.6

    0.48

    0.48 means that 48% or 480 out of 1000 individuals in this population are heterozygous carriers of the gene that causes this trait.
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