Pa1uMareendra
Answered

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)


A: 0.0392

B: 0.0254

C: 0.045

Verified Answer :

As hardy  Weinberg theorem formula is given below
freq. of recessive allele/ q2
 applying the values
 = 1/2500
 q2= 0.0004
now take the square root
so
q=0.02
now
q=p-1
q=0.98
As the frequency. of heterozygous individuals = 2pq
putting values we get the answer
 = 0.04
so correct option is C 
hope it helps

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