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You are studying a rare disease, which affects 1 out of 10,000 individuals. The disease is caused by a gene with two alleles: D and d. Individuals with a dd genotype are affected, while individuals with a Dd phenotype are unaffected carriers. What is the expected frequency of unaffected carriers in the population, assuming this gene is in Hardy-Weinberg equilibrium?Multiple choice question.approximately 0.02%approximately 1%approximately 2%approximately 98%

Question

You are studying a rare disease, which affects 1 out of 10,000 individuals. The disease is caused by a gene with two alleles: D and d. Individuals with a dd genotype are affected, while individuals with a Dd phenotype are unaffected carriers. What is the expected frequency of unaffected carriers in the population, assuming this gene is in Hardy-Weinberg equilibrium?Multiple choice question.approximately 0.02%approximately 1%approximately 2%approximately 98%

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Solution

To solve this problem, we need to use the Hardy-Weinberg equilibrium equation, which is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele (D in this case), q is the frequency of the other allele (d in this case), p^2 is the frequency of homozygous dominant individuals (DD), 2pq is the frequency of heterozygous individuals (Dd), and q^2 is the frequency of homozygous recessive individuals (dd).

We know that the frequency of individuals with the disease (dd) is 1/10,000 = 0.0001. This is equal to q^2. So, we can solve for q (the frequency of the d allele) by taking the square root of 0.0001, which is 0.01.

The frequency of the D allele (p) is 1 - q, which is 1 - 0.01 = 0.99.

The frequency of unaffected carriers (Dd) is 2pq, which is 2 * 0.99 * 0.01 = 0.0198, or approximately 2%.

So, the expected frequency of unaffected carriers in the population is approximately 2%.

This problem has been solved

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