The q2 I was given was .03 that goes to to 3% of the population carry the recessive trait. The population size is 1000.
1. We are given q2 that makes it easy to find q and to find that you take the square root of .03 and it gives you .17 so q is .17
2. Next is to find p by using q-1=p you plug it in .17-1=.83, p=.83
3. Now you are able to find p2 by squaring .83 and that gives you .69
4. Next you can find 2pq to find this you plug in 2(.83)(.17)=.28
5. You can now check the answer by plugging everything into the hardy Weinberg equation 2pq+q2=1.
Finding # of populations
1. q2 is the homozygous recessive individual so it's 1000x03=.30 so this means 30 people out of 1000 are homozygous recessive individual
2. p2 is the homozygous dominant individual followed by step one again all you do is 1000x.69=690 so 690 of the people in the population are homozygous dominant individual
3. You'll be doing the same thing for 2pq this is the heterozygous individual so it'll be 1000x.28=280
4. To check your answer all you have to do is add it all up and make sure it's equal to 1000