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A certain disease in humans follows a simple inheritance pattern through a gene locus with only two alleles: D (the dominant non-disease allele) and d (the recessive disease allele). If the frequency of the recessive allele is 10% in a population of 100,000, how many people would you expect to be healthy carriers of the recessive allele? (Assume Hardy-Weinberg equilibrium.)
  • a)
    36,000
  • b)
    24,000
  • c)
    9,000
  • d)
    18,000
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A certain disease in humans follows a simple inheritance pattern throu...
To calculate the number of healthy carriers of the recessive allele, we need to use the Hardy-Weinberg equilibrium equation. According to the equation, in a population in equilibrium, the frequency of the homozygous dominant genotype (DD) is given by the square of the frequency of the dominant allele (D), and the frequency of the heterozygous genotype (Dd) is given by 2 times the frequency of the dominant allele (D) multiplied by the frequency of the recessive allele (d).
Given that the frequency of the recessive allele (d) is 10% or 0.10, the frequency of the dominant allele (D) would be 1 - 0.10 = 0.90.
The frequency of the DD genotype would be (0.90)2 = 0.81.
The frequency of the Dd genotype would be 2 × 0.90 × 0.10 = 0.18.
To calculate the number of people who are healthy carriers (Dd genotype), we multiply the frequency of the Dd genotype by the total population size:
Number of healthy carriers = Frequency of Dd genotype × Total population size
= 0.18 × 100,000
= 18,000
Therefore, the expected number of healthy carriers of the recessive allele in the population is 18,000.
The correct answer is D. 18,000.
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A certain disease in humans follows a simple inheritance pattern through a gene locus with only two alleles: D (the dominant non-disease allele) and d (the recessive disease allele). If the frequency of the recessive allele is 10% in a population of 100,000, how many people would you expect to be healthy carriers of the recessive allele? (Assume Hardy-Weinberg equilibrium.)a)36,000b)24,000c)9,000d)18,000Correct answer is option 'D'. Can you explain this answer? for MCAT 2025 is part of MCAT preparation. The Question and answers have been prepared according to the MCAT exam syllabus. Information about A certain disease in humans follows a simple inheritance pattern through a gene locus with only two alleles: D (the dominant non-disease allele) and d (the recessive disease allele). If the frequency of the recessive allele is 10% in a population of 100,000, how many people would you expect to be healthy carriers of the recessive allele? (Assume Hardy-Weinberg equilibrium.)a)36,000b)24,000c)9,000d)18,000Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for MCAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A certain disease in humans follows a simple inheritance pattern through a gene locus with only two alleles: D (the dominant non-disease allele) and d (the recessive disease allele). If the frequency of the recessive allele is 10% in a population of 100,000, how many people would you expect to be healthy carriers of the recessive allele? (Assume Hardy-Weinberg equilibrium.)a)36,000b)24,000c)9,000d)18,000Correct answer is option 'D'. Can you explain this answer?.
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