In a population of 3,000 individuals in Hardy-Weinberg equilibrium, the frequency of dominant allele of an autosomal gene is 0.6. The gene has only two alleles. What is the number of homozygous dominant, heterozygous dominant and recessive individuals in that population, with reference to that gene?
AA Aa aa
1. 1440 1080 480
2. 1470 1260 270
3. 1080 1440 480
4. 1260 1470 270

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answer is C.

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Detailed Solution

p = 0.6; q = 0.4
Frequency of ‘AA’ = p²= 0.6 x 0.6 = 0.36
Number of ‘AA’ = 0.36 x 3,000 = 1080
Frequency of ‘Aa’ = 2pq = 2 x 0.6 x 0.4 = 0.48
Number of ‘Aa’ = 0.48 x 3,000 = 1440
Frequency of ‘aa’ = q² = 0.4 x 0.4 = 0.16
Number of ‘aa’ = 0.16 x 3,000 = 480

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