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21 July, 21:19

The incidence of cystic fibrosis, a recessive genetic disorder in the Caucasian population of United States, is

in every 2,500 individuals. Find the number of heterozygous carriers. (p + q = 1, p2 + 2pq + q2 = 1) A. 0.0392 B. 0.0254 C. 0.045

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  1. 21 July, 22:13
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    Cystic fibrosis occurs 1 in every 2,500 individuals.

    Using Hardy Weinberg principle,

    freq. of recessive allele / q2 = 1/2500 = 0.0004

    Take the square root,

    q = 0.02

    freq. of dominant allele / p = q - 1

    p = 0.98

    freq. of heterozygous individuals = 2pq = 0.04

    Therefore, the answer is 0.04, which is C.
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