## Hardy-Weinberg equilibrium ?

Discussion of everything related to the Theory of Evolution.

LiL Homie
Garter
Posts: 10
Joined: Mon Nov 27, 2006 7:48 pm

### Hardy-Weinberg equilibrium ?

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.

mith
Inland Taipan
Posts: 5345
Joined: Thu Jan 20, 2005 8:14 pm
Location: Nashville, TN
Contact:
search.php

Search for hardy weinberg
Living one day at a time;
Enjoying one moment at a time;
Accepting hardships as the pathway to peace;
~Niebuhr

clumpymold
Garter
Posts: 2
Joined: Sat Feb 17, 2007 6:29 am

### Re: Hardy-Weinberg equilibrium ?

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.

### Who is online

Users browsing this forum: No registered users and 0 guests