Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Background Info

Hardy-Weinberg equilibrium is a mathematical model of the frequencies of alleles (i.e., versions of a gene) in a population. The model states that the frequency of the 2 alleles in a population are described by:

$p + q = 1$

where $p$ = the frequency of one allele in the population , and $q$ = the frequency of the other allele in the population.

Since individual organisms each have any combination of 2 alleles, the frequency of individuals with the different combinations of the alleles are described as:

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

where $p^2$ and $q^2$ = the frequencies of individuals with two copies of the same allele and $2pq$ is the frequency of individuals with one copy of each allele.

Question

If I begin with a hypothetical population where $p^2 = 0.49$, $2pq = 0.42$ and $q^2 = 0.09$ I can calculate $p$ as $\sqrt{0.49} = 0.7$ and $q = 1- 0.7 =0.3$. I can then use these $p$ and $q$ values to calculate the original population values (e.g., $q^2 = 0.3^2 = 0.09$).

However if I begin with a hypothetical population where $p^2 = 0.21$, $2pq = 0.66$ and $q^2 = 0.13$ (which satisfies the second equation above) I calculate $p = \sqrt{0.21} = 0.45$ and $q = 1 - 0.46 = 0.54$. These frequencies do not allow me to recalculate the original population frequencies as $0.54^2 = 0.29$ not $0.13$.

Is there a general explanation for why these calculations only work under certain circumstances?

share|improve this question
    
I am not sure what tags would be most appropriate. Please feel free to edit the tags for a better fit - thanks –  KennyPeanuts May 2 '12 at 2:36
add comment

1 Answer 1

up vote 2 down vote accepted

Hardy-Weinberg equilibrium is reached after one generation. In your second example, the initial population is not at equilibrium.

To wit, $4\times0.21\times0.13\ne0.66^2$ hence there exists no parameters $(p,q)$ such that $p^2 = 0.21$, $2pq = 0.66$ and $q^2 = 0.13$. The proportions $(0.21,0.66,0.13)$ describe a population where the Hardy-Weinberg proportions of the alleles A and a are $0.21+\frac120.66=0.54$ and $0.13+\frac120.66=0.46$ respectively. After one generation, and forever after, the alleles AA, Aa and aa are present in proportions $0.54^2\approx0.29$, $2\times0.54\times0.46\approx0.50$ and $0.46^2\approx0.21$ respectively.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.