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Population Genetics (Hardy Weinberg) problem

Genetics as it applies to evolution, molecular biology, and medical aspects.

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Population Genetics (Hardy Weinberg) problem

Postby amdugg » Sat Feb 13, 2010 9:56 pm

Hello, this is my first post.

I am having trouble with a homework problem that should be really easy, and I can't figure out what I'm doing wrong.

Here it is:
In eggplants, fruit color is determined by a gene with two alleles: an allele that codes for purple fruit (Z) and an allele that codes for white fruit (W). Heterozygotes have neither purpler nor white fruit; instead ,their fruit is an intermediate color, light violet. A population of 10,000 eggplants is in Hardy-Weinberg equilibrium. In this population, the frequency of individuals with a genotype homozygous for the white allele is .49.

Here is what I did:

Calculate the Allelic frequencies:
p= the frequency of the allele Z = .3 (1-q)
q= the frequency of the allele W = (square root of .49) =.7

Calculate the genotypic frequencies:
genotype ZZ = p^2=.09
genotype WW = q^2 = .49
genotype ZW = 2pq = .0882

This should all add up to 1, but it doesn't and I can't figure out what I'm doing wrong.
I'm sure I'm missing something so basic that I'm going to be embarrassed for even posting this problem. Regardless, any help is much appreciated.
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Postby JackBean » Sat Feb 13, 2010 10:14 pm

for me 2*0.3*0.7 = 0.42 not 0.0882
and that makes up 1 exactly ;)
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Re:

Postby amdugg » Sun Feb 14, 2010 12:16 am

JackBean wrote:for me 2*0.3*0.7 = 0.42 not 0.0882
and that makes up 1 exactly ;)


That would be fine, except that it's not 2*0.3*0.7

Its p^2+q^2+2pq

.03^2+.07^2+2[(.03^2)*(.07^2)]

...which does not equal 1.

Am I making this more complicated than I have to?
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Postby JackBean » Sun Feb 14, 2010 10:40 am

How that it's not? It is 2*p*q, not 2*p^2*q^2! And p and q are 0.3 and 0.7 respectively.
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Postby Haidarh » Wed Feb 17, 2010 11:38 am

p = 0.3 q = 0.7 2pq = 2 x 0.3 x 0.7 = 0.42

p^2 + q^2 + 2pq = 1

0.9 + 0.49 + 0.42 = 1
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