A) DD = 0.04; Dd = 0.32; dd = 0.64
B) DD = 0.64; Dd = 0.32; dd = 0.64
C) DD = 0.04; Dd= 0.64; dd= 0.32
D) DD = 0.64; Dd = 0.32; dd = 0.04
Correct Answer: D
Solution :
[d] We are told that the allele frequency for D = 0.8; therefore, the frequency of d= 0.2. These values are the p and q that we need to calculate the genotype frequencies in the next generation. Using the Hardy- Weinberg equation, \[{{P}^{2}}\] (DD) = 0.64, 2pq (Dd) = 0.32, and \[{{q}^{2}}\] (dd) = 0.04.You need to login to perform this action.
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