The Hardy-Weinberg Model was put forth to explain why the recessive genes in a population would not be pushed out by the dominant genes. While it does not give an accurate representation of how gene flow works in real population (it doesn't take into account selective breeding, the founder effect, emigration and immigration, random genetic drift, etc.), it can be used to figure out the gene frequencies in a population at a specific time.

I have a hard time doing these, and I know that I could not do one of these with only the information given here in the previous WUs. So here's more information and a demonstration.

**The Hardy-Weinberg Equations**

* These apply only to a two allelic system. *

p = dominant allele

q = recessive allele

p + q = 1 (or 100% of the alleles)

p^{2} + 2pq + q^{2} = 1

So... Say you have a sample of 100 people. 36% have mid-digital hair. You know that mid-digital hair is caused by a dominant gene. This means that the 64% of the population that *doesn't* have mid-digital hair has two recessive alleles (qq). Lets plug that in.

q^{2} (qq) = .64 (64%)

Therefore, q = .8 (the square root of q = the square root of .64)

Remember that p + q = 1

So, p + .8 = 1

Therefore, p = .2

p^{2} = .04 (4%)

Remember, p^{2} + 2pq + q^{2} = 1

So, .04 + 2pq + .64 = 1

Therefore, 2pq = .32 (this is the number of heterozygous individuals)

And what, pray tell, does this mean? Well...

*The allele or gene frequency:* 80% for p, 20% for q.

*Genotype frequencies:* pq = 32%, pp = 4%, qq = 64%

*Phenotype frequencies:* well, this is what we started with. M-D hair 36%, no M-D hair 64%. We just add pq and pp together, since p is dominant, both will be expressed the same.