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The incidence of cystic fibrosis, a recessive genetic disorder in the Caucasian population of United States, is 1 in every 2,500 individuals. Find the number of heterozygous carriers. (p + q = 1, p2 + 2pq + q2 = 1)

Sagot :

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Answer:

The no. of heterozygous carriers = 0.0392

Explanation:

From the given information:

The incidence of this recessive disorder i.e. q² = 1/2500

q² = 0.0004

q = 0.02

From Hardy Weinberg's Equilibrium.

p + q = 1; &

p² + 2pq + q² = 1

∴

p + 0.02 = 1

p = 1 - 0.02

p = 0.98

So, the numbers of heterozygous carrier 2pq is:

= 2 × 0.98 × 0.02

= 0.0392

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