Question

10. Write the 2 Hardy-Weinberg equations. Use the equation to calculate the percentage of carriers of cystic fibrosis (a recessive disease, you must have two recessive alleles to have the disease) in a population if the frequency of cystic fibrosis births in that population is 9 in 1000.

Answer 1

Using the Hardy-Weinberg equations, the percentage of carriers of cystic fibrosis is 8.19%.

What are the Hardy-Weinberg equations?

The Hardy-Weinberg equations describe the relationship between alleles in a population that is fairly constant.

The Hardy-Weinberg equations are given below as;

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

$q^{2} = 0.009\\q = 0.09\\p = 1 - 0.09\\p = 0.91$ Hardy-Weinberg equations

Percentage of carriers = 2pq

Percentage of carriers = 2 * 0.09 * 0.91 * 100 = 8.19%

Therefore, the percentage of carriers of cystic fibrosis is 8.19%.

Learn more about Hardy-Weinberg equations at: brainly.com/question/16039271

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