Population Genetics and Evolution

Hardy-Weinberg Equation 

The information that was given to me for the hardy-Weinberg problem is,
 the population is 1000 and my q2 is 0.28
 

 My q², homozygous recessive individual, is 0.28. With this I can find the dominant allele, q. I just need to squared root the 0.28 and I will receive q, which is 0.53. Then I can find P, the recessive allele, by using the equation P+q=1. The equation results me to subtract the 1 with q to get P, which becomes 0.47. To find P², homozygous dominant individual, you will squared P, P² will be 0.22. Then with P and q in hand, you can be able to find 2pq, heterozygous ind, by multiplying 2 with p and q. 2Pq will be 0.50.

In equation format it will be:
q² = 0.28    q= 0.53    P= 0.47    P² = 0.22    2pq= 0.50

√ 0.28 = 0.53
1- 0.53= 0.47
0.47² =0.22
 2*0.53*0.47= 0.50

With the population of 1000, I can find the number of alleles and the number of individuals with the genotype. All that is needed is to multiply the P²,p, q²,q and 2pq by 1000.

The results for the individual of the alleles and the genotype are:
q= 530     P= 470     q²= 280     P²= 220     2pq= 500

0.28*1000=280
 0.53*1000=530
0.47*1000=470
 0.22*1000=220
0.50*1000=500