Question

Cystic fibrosis is caused by a recessive allele. The frequency of this allele is 0.1 in a population of 2,500.
What is the frequency of the dominant allele?
How many people in this population will have cystic fibrosis?
How many people in this population are cystic fibrosis carriers? (Heterozygous)

Answer 1

Answer:

For example, the allele frequency of the mutant cystic fibrosis allele among Caucasians is 0.025, while the frequency of the normal allele is 0.975.

Explanation: hope fully this helps

Answer 2

The frequency of the dominant allele is 0.975, and the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%). 42 % is the people in this population are cystic fibrosis carriers.

What is the Hardy-Weinberg equation?

Hardy-Weinberg equation states that there will be no change in the genotype of a population unless, there is any evolutionary change happens.

This is the Hardy-Weinberg equation

p2 + 2pq + q2 = 1 and p + q = 1

The  people in this population are cystic fibrosis carriers

2pq = 2 (0.7 x 0.3) = 0.42 = 42%

Thus, the correct option is 0.975, 0.02 (or 2%), and 42 %

Learn more about Hardy-Weinberg equation

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