Question

Cystic fibrosis is a homozygous recessive condition that affects 1 in 10,000 of the Hispanic population in the United States. Calculate the frequency of the dominant allele, the frequency of the recessive allele, and the percentage of heterozygous individuals (carriers) in the Hispanic population.

Answer 1

Assuming Hardy-Weinberg equilibrium, we can use the following equations:

p + q = 1 (where p = frequency of dominant allele, q = frequency of recessive allele)

p2 + 2pq + q2 = 1 (where p2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, q2 = frequency of homozygous recessive individuals)

We know that the frequency of individuals with cystic fibrosis (homozygous recessive) is 1 in 10,000 or 0.0001:

q2 = 0.0001

q = √0.0001 = 0.01

Using the first equation, we can solve for p:

p + 0.01 = 1

p = 0.99

Therefore, the frequency of the dominant allele is 0.99 and the frequency of the recessive allele is 0.01.

To calculate the percentage of carriers (heterozygous individuals), we can use the frequency of q (0.01) and plug it into the 2pq term:

2pq = 2(0.99)(0.01) = 0.0198

So, approximately 2% of the Hispanic population would be carriers of the cystic fibrosis gene.

Okay yeeee