Hardy Weinberg Equilibrium Question

[booji]

Hi Guys,

I was hoping that someone could clarify when to use one of the two Hardy-Weinberg equilibrium equations. I am not sure when to use the p2 + 2pq + q2 =1 over the p+q=1 equation and I find myself making mistakes on this when taking practice exams.

Your help is always appreciated,

Booji

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[killinsound]

p itself is the frequency of that one allele.
q itself is the freq of one recessive allele.

p^2 is the frequency of homozygous for that allele.
2pq = freq of heterozygous
q^2 = freq of homozyg recessive

regardless p + q is always equal to 1... (assuming there are only 2 alleles)

i might have p and q backwards but its the same either way... just see what you need to solve for

for instance: if you're given the freq of one allele 'p', and someone asked you to find the freq of heterzygous individuals.

from p + q = 1, you would figure out the freq of q

then from p2 +2pq + q2 = 1, you could solve for the freq of heteryzogus invidiuals.

 

[corbis11]

Hi Guys,

I was hoping that someone could clarify when to use one of the two Hardy-Weinberg equilibrium equations. I am not sure when to use the p2 + 2pq + q2 =1 over the p+q=1 equation and I find myself making mistakes on this when taking practice exams.

Your help is always appreciated,

Booji

know both. if you are given allelic frequency, say p=30% then you know q=70% from p+q=1. You can take those percentages and plug them in as decimels in to p^2 + 2pq + q^2 ----> (0.3)^2 + 2(0.7)(0.3) + (0.7)^2 for a genotypic frequency of 9% pp, 42% pq, and 49% qq.

You could also be give the genotypic frequency of pp as, say, 64% so you would just take square root of it (0.64) to give you the allelic frequency of p=0.8 so then you know q=0.2

 

[booji]

Thank you very much!