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A rare disease which is due to a recessive allele (a) that is lethal when homozygous (aa), occurs with a frequency of one in a million. How many individuals in a town of 14,000 can be expected to carry this allele?

A. 30
B. 15
C. 28
D. 40

User Jelder
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1 Answer

4 votes

Final answer:

To find the number of individuals in a town of 14,000 who carry a rare recessive allele, the Hardy-Weinberg principle is utilized to estimate allele frequency and calculate carrier frequency, resulting in an expected 28 carriers.

Step-by-step explanation:

The student's question pertains to the genetics behind a rare recessive disease and how many individuals might carry the allele in a given population. To answer the question, we use the Hardy-Weinberg principle to estimate allele frequency and then calculate carrier frequency. Since the disease occurs with a frequency of one in a million, or 1/1,000,000, which is the frequency of homozygous recessives (aa), we denote this as q².

The frequency of the allele q is the square root of 1/1,000,000, which is 1/1000 or 0.001. The frequency of heterozygotes (carriers) is 2pq; assuming p ~ 1 because q is very small, 2pq ~ 2q. Multiplying the carrier frequency by the number of individuals in the town, 2(0.001)(14,000), we get an expected value of 28 carriers.

User Natros
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