Posted by Josh on Thursday, February 7, 2013 at 11:37pm.
TT+Tt= Phenotypic frequencies for tan
tt=Phenotypic frequency for pink
557/(557+396)=Phenotypic frequency for tan=0.58
1-0.58=Phenotypic/genotypic frequency for pink=0.42
(T+t)^2=TT+2Tt+tt=1
Plugging in 0.42 for the genotypic frequency for pink in the above equation,
TT+2Tt+0.42=1 rearrangement gives,
TT+2Tt=0.58
Since the genotypes for Tt outnumber the genotype for TT 2:1, let TT=x and Tt=2x and solve for x:
x+2x=0.58
3x=0.58
x=0.58/3=0.19
Since x=0.19, the genotypic frequency for TT is 0.19, and the genotypic frequency for Tt is 0.38
Since the genotypic frequency for Tt is 0.38, the % of clams heterozygous for the Tt in the population is 0.38*(953)=362
For the new generation, calculate the number of clams with either phenotypes by multiplying the phenotypic frequencies by the new population number
1245 claims*(0.42)=523 pink clams
1245 claims-523 pink clams=722 tan clams
The fourth line from the bottom should say
Since the genotypic frequency for Tt is 0.38, the NUMBER of clams heterozygous for the Tt in the population is 0.38*(953)=362
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