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  • In a human population of 100,000 individuals, where a certain disease follows a simple inheritance pattern through a gene locus with two alleles (D and d), and the freque

In a human population of 100,000 individuals, where a certain disease follows a simple inheritance pattern through a gene locus with two alleles (D and d), and the frequency of the recessive allele (d) is 10%, what is the estimated number of individuals who are expected to be healthy carriers of the recessive allele? Assume the population is in Hardy-Weinberg equilibrium.

 

Option: 1

36,000

 


Option: 2

 24,000


Option: 3

9,000

 


Option: 4

18,000


Answers (1)

best_answer

As per Hardy-Weinberg law,

In the given scenario, the frequency of the recessive allele (d) = 10% or 0.1,
The total population size = is 1,00,000 individuals.

According to the Hardy-Weinberg equation,
The frequency of homozygous recessive individuals (dd) = q2
The frequency of heterozygous carriers (Dd) = 2pq
The frequency of homozygous recessive individuals (dd) = p2

Therefore, p+q = 1 , p = 1 - q = 1 - 0.1 = 0. 9
Therefore, p = 0.9

Using these formulas, we can calculate the expected number of individuals who are healthy carriers of the recessive allele:
Expected number of carriers (Dd) = 2pq x population size
= 2 * 0.1 x 0.9 x 100,000
= 0.18 x 100,000
= 18,000 

Hence, the correct answer is option 4.

Posted by

Ajit Kumar Dubey

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