Cystic Fibrosis is a homzygous recessive disease that affects 1 in 3400 people. What percent of the population are carriers?

The correct answer is = 0.02 or 2.5 percent approx.

Explanation:

Given:

CF is a recessive disease that affects 1 in 3400 people

The frequency of the recessive allele in the population (q2) -

= 1/3400

= 0.0002. Therefore, q is the square root, or 0.014.

The frequency of the dominant allele in the population (p)

= 1 - 0.014 = 0.986 (or 98.6%).

The percentage of s carriers in the population.

2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows:

2pq = (2)(.986)(.014) = 0.02 or 2.5 percent approx.

it can be both

Explanation: