Answer:
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.
