Suppose genotypes AA, Aa, and aa have frequencies in zygotes of 0.16, 0.48, and
ID: 60940 • Letter: S
Question
Suppose genotypes AA, Aa, and aa have frequencies in zygotes of 0.16, 0.48, and 0.36, respectively (before mortality), and relative viabilities of w11= 1.0, w12 = 0.8, and w22= 0.6, respectively.
Calculate the allele frequencies in the gene pool that formed the zygotes. Calculate the genotype frequencies in the adults in the next generation. Remember that the new frequency of AA genotypes will be p62 w_11/ the new frequency of Aa genotypes will be 2pqw_12/ and the new frequency of Aa genotypes will be q^2 w_22/ Calculate the allele frequencies in the gamete pool of the next generation. Remember that Delta p = p/w (pw_11 +qw_12 - w) Calculate the genotype frequencies of the zygotes (pre-selection) in the next generation.Explanation / Answer
The relative fitness represents in terms of survival.
By multipying genotype frequencies with viabilities
0.16X 1.0 = 0.16
0.48 X 0.8 = 0.384
0.36 X 0.6 = 0.216
The sum of these converted into frequencies are
AA = 0.211,
Aa = 0.505,
aa = 0.284.
The zygotic frequencies are nothin but the allelic frequencies
Hence , the frequency of a A gamete is 0.211 + 0.505/2 = 0.464
the frequency of a a gamete is 0.284 + 0.505/2 = 0.536.
According to the Hardy Weinberg Law the alleles for the next generation are
AA = (0.464)2 = 0.215
Aa = 2 (0.464) (0.536) = 0.497
aa = (0.536) = 0.287