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The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium and
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessives in the population.
Options:
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessives in the population.
Solution:
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Verified Answer
The correct answer is:
31% and 66%
According to the Hardy-Weinberg Equilibrium equation, the percentage of heterozygous individuals in the population is represented by the 2pq. Therefore, the number of heterozygous individuals (Aa) is equal to 2pq which equals 2 × 0.19 × 0.81 = 0.31 or 31%.
The percentage of homozygous recessives individuals in the population is represented by the q2, which equals to 0.81 × 0.81 = 0.66 or 66%.
The percentage of homozygous recessives individuals in the population is represented by the q2, which equals to 0.81 × 0.81 = 0.66 or 66%.
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