try this
math refresher (just substitute p for a, and q for b)
To understand H-W equilibrium and to be able to solve the equations,
you have to understand the difference between
allelic and genotypic frequency. Alleles are variants of genes.
In a sexually reproducing species, each gamete must carry
only one allele [think of sperm carrying an X or a Y sex chromosome,
never both]. Homozygous individuals produce gametes
carrying only one type of allele, either dominant or recessive.
Heterozygous individuals produce two kinds of gametes in equal
proportion - half carry the dominant, half the recessive. The frequency of the alleles in the gene pool are determined by the frequencies of the genotypes.
We symbolize the dominant allele as 'p' and the recessive as 'q'
We assume there is no difference between the sexes.
The frequencies of p and q must total 1.
So the first part of H-W equation is
p + q = 1
When gametes fuse during fertilization, a genotype is produced.
There are three genotypes.
Homozygous dominant
This zygote contains two p alleles. The odds of getting
two p alleles is the product of the frequency of p in the
gametes p X p = p^2. If p is present in the gametes at
a frequency of 0.6 then the odds of producing a
homozygous dominant zygote in the next generation is 0.6^2 or 0.36.
Heterozygous
Contains p and q. The odds of getting p and q is the product
of the frequencies of p and q in the gene pool. If p = 0.6 and q = 0.4
then pq = 0.24. But there are two ways to make a heterozygote.
There are two ways to make a heterozygote so we need to double
the last equation. The frequency of heterozygotes in a population
is given by
2pq = 2(0.6)(0.4) = 0.48
Homozygous recessive
contains two q alleles. The frequency of homozygous
recessive genotypes can be expressed as
q X q = q^2
If q = 0.4, the q^2 = (0.4)^2 or 0.16
The formation of zygotes comes from the union of the male and female gametes.
(p + q) (p + q) or (p + q) ^2 Since p + q = 1, (p+q)^2 =1
The genotypic frequencies add to 1
(p + q)^2 = p^2 + 2pq + q^2 = 1