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January 30, 2015

January 30, 2015

Posted by **Josh** on Thursday, February 7, 2013 at 11:37pm.

1. The frequencies of each allele.

2. The expected genotypic frequencies.

3. The number of heterozygous clams that you would predict to be in this population.

4. The expected phenotypic frequencies.

5. Assume that conditions were just right and over the course of the season, 1245 new baby clams were born. If you assume that all of the Hardy-Weinberg conditions are met (no evoution is or has occurred), how many of these baby clams would you expect to be pink and how would you many would you expect to be tan?

- Biology -
**Devron**, Friday, February 8, 2013 at 2:32amTT+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

- Biology-typo -
**Devron**, Friday, February 8, 2013 at 6:18amThe 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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