# Why does this example of Hardy-Weinberg equilibrium not work?

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?

-
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

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.