Hardy-Weinberg equilibrium ?

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LiL Homie
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Hardy-Weinberg equilibrium ?

Post by LiL Homie » Sun Feb 04, 2007 10:48 pm

1- Assuming we have a Hardy-Weinberg equilibrium population & the frequency of the dominant allele "B" is 90% & that "b", the recessive allele is the only other allele at this gene locus in the population, what would be the expected frequency of Heterozygous individuals in this population?

Is this correct?...

p +q= 1

0.9 +0.1= 1.00

B= 90%
b=10%

Idk what I'm doing... but if this by some chance is correct, would my final answer be a percentage to express frequency?


:? I don't understand how to figure this out. Can someone please help explain this to me? I would really appreciate it! Thanks so much! :)
Last edited by LiL Homie on Mon Feb 05, 2007 1:11 am, edited 1 time in total.

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mith
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Post by mith » Sun Feb 04, 2007 11:24 pm

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clumpymold
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Re: Hardy-Weinberg equilibrium ?

Post by clumpymold » Sat Feb 17, 2007 6:48 am

LiL Homie wrote:1- Assuming we have a Hardy-Weinberg equilibrium population & the frequency of the dominant allele "B" is 90% & that "b", the recessive allele is the only other allele at this gene locus in the population, what would be the expected frequency of Heterozygous individuals in this population?

Is this correct?...

p +q= 1

0.9 +0.1= 1.00

B= 90%
b=10%

Idk what I'm doing... but if this by some chance is correct, would my final answer be a percentage to express frequency?


:? I don't understand how to figure this out. Can someone please help explain this to me? I would really appreciate it! Thanks so much! :)


You probably figured this out already since this post is over a week old.

Anyway, when a population is achieving Hardy-Weinberg equilibrium, there are two equations that will be satisfied (using your variables):

1.) B + b = 1
2.) B^2 + 2Bb + b^2 = 1

Equation 1 is used for ALLELES.

Equation 2 represents individuals according to the following:

B^2 = homozygous dominant individuals
2Bb = heterozygous individuals
b^2 = homozygous recessive individuals & recessive PHENOTYPE individuals
B^2 + 2Bb = dominant PHENOTYPE individuals

For your problem, B = 0.9, b = 0.1 (since B + b = 1). Therefore, to solve for heterozygous individuals, simply solve for 2Bb = 2(0.9)(0.1) = 0.18 = 18% of the population.

Hope that helps.

:)

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