## To solve this problem, we can use the Hardy-Weinberg equilibrium equation and the information provided.

The Hardy-Weinberg equilibrium equation is:

p^2 + 2pq + q^2 = 1

In this equation:

- p represents the frequency of the dominant allele (normal allele)

- q represents the frequency of the recessive allele (CF allele)

- p^2 represents the frequency of individuals homozygous for the dominant allele (NN)

- 2pq represents the frequency of individuals heterozygous for both alleles (NCF)

- q^2 represents the frequency of individuals homozygous for the recessive allele (CC)

Given:

- The incidence of cystic fibrosis (CC) is 1 in 2500, which means q^2 = 1/2500 or q^2 = 0.0004

- The population is in Hardy-Weinberg equilibrium, which means the frequencies of alleles will not change from generation to generation

Now, let's use the information provided to determine the values of q and p.

From the equation p + q = 1, we can rewrite it as q = 1 - p.

Substituting q = 1 - p in the equation q^2 = 0.0004, we get:

(1 - p)^2 = 0.0004

Solving for p, we can take the square root of both sides:

1 - p = âˆš(0.0004)

Rearranging, we get:

p = 1 - âˆš(0.0004)

Now that we have the value for p, we can find the value of q:

q = 1 - p

Using these values of p and q, we can calculate the frequency of carriers for the CF alleles, which is 2pq.

Let me calculate that for you.