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Multiple Choice
Approximately 1 out of every 2,500 Caucasians in the United States is born with the recessive disease cystic fibrosis. According to the Hardy-Weinberg equilibrium equation, approximately what percentage of people are carriers?
A
About 2%
B
About 96%
C
About 10%
D
About 0.08%
E
About 4%
Verified step by step guidance
1
Understand that cystic fibrosis is a recessive genetic disorder, which means it is expressed in individuals who are homozygous recessive (q^2).
According to the problem, 1 out of every 2,500 individuals has cystic fibrosis, so q^2 = 1/2500.
Calculate q by taking the square root of q^2. This will give you the frequency of the recessive allele in the population.
Use the Hardy-Weinberg equation p + q = 1 to find p, the frequency of the dominant allele.
Calculate the carrier frequency (2pq) using the values of p and q. This will give you the percentage of people who are carriers of the cystic fibrosis allele.