Hardy Wienberg

      When doing the Hardy Weinberg the equation that is used is p²+ 2pq+ q²=1. In the equation p² is the frequency of the homozygous dominant individual, q² is the frequency of the homozygous recessive individual, and 2pq is heterozygous individual. In the problem It gave q²=.43 with a population of 1000. The steps for solving this equation would be to square root .43 and that will give you q=.65. After that you subtract .65 by 1 and you get .35. Then you have to do .35x.35 to get p²=.12. To get 2pq you do 2(.35)(.65) and get .45. After getting all that you multiply q², p² and 2pq by 1000. That should give you: 

q²= 430 people with a recessive trait

p²= 120 people with homozygous dominant trait

2pq= 450 people with heterozygous trait