The question is incomplete as the frequency of alleles is not given, however, the frequency and population are given below :
Frequency of a = 0.506
total population = 500, Number of aa = 128
Answer:
The correct answer is - option C. 0.336.
Explanation:
Let, A = Normal allele, a = Defective allele
So, AA & Aa will develop normal phenotype & aa will develop defective phenotype.
Frequency of a = 0.506
So, Frequency of A = 1 - 0.506 = 0.494
So, frequency of AA = (0.494)2 = 0.244036
So, frequency of Aa = 2 x 0.494 x 0.506 = 0.499928
so, frequency of aa = (0.506)2 = 0.256036
total population = 500, Number of aa = 128
So, Number of AA = (0.494)2 x 500
= 122.018 which is almost 122 so considering it 122
So, Number of Aa = (2 x 0.494 x 0.506) x 500
= 249.964 which is almost 250 so considering it 250
It is given that, Relative fitness of AA (W11) & Aa (W12) is 1, and the relative fitness of aa (W22) is 0.
Now, mean fitness = (Frequency of AA x Fitness of AA) + (Frequency of Aa x Fitness of Aa) + (Frequency of aa x Fitness of aa)
= (0.244036 x 1) + (0.499928 x 1) + (0.256036 x 0) = 0.244036 + 0.499928 = 0.743964
So, after selection frequency of Aa
= (Frequency of Aa x Fitness of Aa) / mean fitness
= 0.499928 / 0.743964 = 0.67198 (Up to 5 decimal)
So, after the selection frequency of aa
= (Frequency of aa x Fitness of aa) / mean fitness
= 0 / 0.743964 = 0
So, frequency of a
= 1/2 of frequency of Aa + Frequency of aa
= 1/2 x 0.67198 + 0
= 0.33599 + 0
= 0.33599 which is almost 0.336
