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Consider a disease caused by a recessive allele. In a study population, one out of every 500 individuals (0.20%) has the disease. Based on the Hardy-W
Question

Consider a disease caused by a recessive allele. In a study population, one out of every 500 individuals (0.20%) has the disease.
Based on the Hardy-Weinberg equation, what is the percentage of individuals who are carriers of the recessive allele for the disease?

A.

7.6%

B.

20.2%

C.

1.5%

D.

30.5%

Correct option is A

The Hardy-Weinberg equation states:

p2+2pq+q2=1andp+q=1p^2 + 2pq + q^2 = 1 \quad \text{and} \quad p + q = 1​​

where:

  • p2= Frequency of homozygous dominant (normal) individuals
  • 2pq2pq2pq= Frequency of heterozygous carriers
  • q2= Frequency of homozygous recessive individuals (affected)
  • ppandqq q are the allele frequencies of the dominant and recessive alleles, respectively.

Diseased - f(aa) = 0.002

q2= 0.002

q = 0.04

p = 1 - 0.04 

p = 0.96

(carrier)2pq=2×0.9553×0.0447=0.0768×100(\text{carrier}) \quad 2pq = 2 \times 0.9553 \times 0.0447 \\= 0.0768 \times 100​​

Answer  is  7.68% 

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