H-WEEZY

                                        Solving a Hardy-Weinberg Equation
                                                            q² given: .36   
                                                       Population: 1000         


1. q can be found by √.36 = .6
2. find the p value by using p+q=1.. subtract 1 from .6 = .4
3. square the p value to find p² = .16 
4. to find 2pq, multiply p & q.. (.4)(.6)=.24.. then multiply 2(.24) = .48
5. check if p²+2pq+q²=1

Frequency of dominant allele (p) = .4 .. 40%
Frequency of recessive allele (q) = .6 .. 60 %
Homozygous dominant individual (p²) = .16
Homozygous recessive individual (q²) = .36
Heterozygous individual (2pq) = .48

Find the number of individuals for the 3 possible genotypes by multiplying it with the population:
p² (.16)(1000) = 160 homozygous dominant individuals
q² (.36)(1000) = 360 homozygous recessive individuals
2pq (.48)(1000) = 480 heterozygous individuals