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Cystic fibrosis is caused by a recessive allele. Roughly one out of every 500 individuals (0.20%) have this disease. Using the Hardy-Weinberg equation
Question

Cystic fibrosis is caused by a recessive allele. Roughly one out of every 500 individuals (0.20%) have this disease. Using the Hardy-Weinberg equation, the percentage of individuals who are carriers of the recessive allele for the disease is:​

A.

10.2

B.

1.0

C.

15.2

D.

7.6

Correct option is D

The Hardy-Weinberg equation states:

p+2pq + q 2 = 1   and  p + q = 1

where:

  • p2 = Frequency of homozygous dominant (normal) individuals
  • 2pq2pq2pq = Frequency of heterozygous carriers
  • q2 = Frequency of homozygous recessive individuals (affected by cystic fibrosis)
  • ppp and qqq are the allele frequencies of the dominant and recessive alleles, respectively.

Cystic fibrosis - f(aa) = 0.002

q2= 0.002

q = 0.04

p = 1 - 0.04 

p = 0.96

(carrier )  2pq = 2×0.9553×0.04472 \times 0.9553 \times 0.0447

0.0768×100 0.0768 \times 100

=  7.68  answer

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