Question

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

Answer 1

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.

Answer 2

Answer:

A.) 0.0392

Explanation:

1/2500 =.0004=q^2 so we have to sqrt to get just q

$\sqrt{.0004} = .02$

q=0.2

then we use the first equation of the Hardy-Weinberg equation to get p

p + q= 1

1-.02=.98

use the second equation

p^2 +2pq +q^2

2(.98)(.02)=0.0392

This is correct on Plato/ Edmentum