Question

Cystic fibrosis is an autosomal recessive disease. In a population of one hundred individuals, twenty-five are found to have the disease. Assuming Hardy - Weinberg equilibrium, what is the percent of the population that are carriers for cystic fibrosis?

• A. 10%
• B. 75%
• C. 25%
• D. 50%

Answer 1

The correct option is D 50%

According to Hardy - Weinberg equilibrium

$p^2+2\mathrm{pq}+q^2=1$ where,

p = frequency of the dominant allele in the population

q = frequency of the recessive allele in the population

$p^2$ = percentage of homozygous dominant individuals

$q^2$ = percentage of homozygous recessive individuals

2pq = percentage of heterozygous individuals

Out of 100, 25 people are the patients, means $q^2$ = 0.25 (becasue Cystic fibrosis is an autosomal recessive disorder)

If p +q = 1 and q = 0.5, then p = 0.50

So, the percent of carrier individuals is :

2 x 0.5 x 0.5 = 0.5

Therefore, 50% of the population are carriers.