Question

0.9 points Consider a population of 100 individuals and the frequency of the Aalele is 0.6 and the frequency of the alle is 0
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Answer #1

The answer will be option a (0.81).

Explanation:

Now, frequency of AA = (frequency of A)2 = (0.6)2 = 0.36

Now, frequency of aa = (frequency of a)2 = (0.4)2 = 0.16

Now, frequency of Aa = 2 x frequency of A x frequency of a = 2 x 0.6 x 0.4 = 0.48

Also, mean fitness = (Frequency of AA x Fitness of AA) + (Frequency of Aa x Fitness of Aa) + (Frequency of aa x Fitness of aa) = (0.36 x 1) + (0.48 x 0.25) + (0.16 x 0.25) = 0.36 + 0.12 + 0.04 = 0.52

So, after selection frequency of AA = (Frequency of AA x Fitness of AA) / mean fitness = 0.36 / 0.52 = 0.692 (Up to 3 decimals)

So, after selection frequency of Aa = (Frequency of Aa x Fitness of Aa) / mean fitness = 0.12 / 0.52 = 0.231 (Up to 3 decimals)

Thus, frequency of A allele in the next generation = Frequency of AA + 1/2 of Frequency of Aa = 0.692 + (0.5 x 0.231) = 0.8075 = 0.81 (Up to 2 decimals)

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