if the incidence of cystic fibrosis is 1 in 2500, what is the expected frequency of carriers for the CF allels? assume that this population is in Hardy-Weinberg equilibrium.

I appreciate a hint to start this answering this problem.

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

p + q = 1

q^2 is the percentage of cystic fibrosis cases (cc) cases.

That is how you start it. Repost if you get stuck!

8 months ago

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.

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