Hardy Weinberg problems make up

Here are some things to remember when solving Hardy Weinberg problems:

• p² + 2pq + q² = 1 and p + q = 1
• p = frequency of the dominant allele in the population
• q = frequency of the recessive allele in the population
• p² = percentage of homozygous dominant individuals
• q² = percentage of homozygous recessive individuals
• 2pq = percentage of heterozygous individuals
• remember to check what the problem really wants you to solve for

Example Problem from : http://www.k-state.edu/parasitology/biology198/hardwein.html

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
----The frequency of the "aa" genotype.
So we know the "aa" must be the recessive allele so that means q^2 is 36% because q^2 is the population

----The frequency of the "a" allele.
Now to find the allele frequency, we first have to take the square root of .36 (36%). This means that the allele frequency is 0.6 or 60%   

----The frequency of the "A" allele.  
This allele frequency can be calculated by using the equation:     p + q = 1  Then we plug in the information. q=0.6 which means p must be 0.4 or 40%

----The frequencies of the genotypes "AA" and "Aa."
We must use the  equation:p² + 2pq + q² 
Then we plug in what we know  P=.4 Q= .6 p^2 = .16  q^2 = .36. So now we know that the frequency of the genotype "AA" is 16%. Now we plug in P and Q in order to solve for 2PQ = 2(.4)(.6)= .48
This means that the allele frequency for "Aa" is 48%

If you are still confused, watch this video: https://www.youtube.com/watch?v=xPkOAnK20kw