Hardy-Weinberg

 Step #1.) I was given the homozygous recessive (q2) which was .61. 

Step #2.) In order to get the recessive allele (q) I had to square-root .61 and the answer I got was .78 so applied it into the list below.

Step #3.) Now that I have the recessive allele (q) I can get the dominant allele (p) by subtracting it by 1. My answer for the dominant allele (p) was .22, *1-.78= .22*

Step #4.) The next step is the find the Homozygous dominant (p2) I got this by squaring the p (dominant allele) ((.22*.22= .05))

Step #5.) The last step is to get the heterozygous (2pq). To do that I multiplied q (recessive allele) and the p (dominant allele) ((2 (.22)(.78) = .34))

q2 (Homozygous recessive)- .61

q (Recessive allele)- .78

p (Dominant allele)- .22

p2 (Homozygous dominant)- .05

2pq (Heterozygous)- .34 
population: 1,000

~The genotypes are q2, p2, and 2pq.~ 

In the total of 1,000 individuals, I was asked to find the individuals of each genotype. Soooo the next thing I did was multiply each genotype by 1,000.

 For the q2 (homozygous recessive) I did this > .61*1,000 = 610
              p2 (homozygous dominant) > .05*1,000 = 50
             2pq (heterozygous) > .34*1,000 = 340 
                All together the individuals should add up to be 1,000. I checked, they do (:

 In conclusion there will be 610 that have the homozygous recessive allele in which they are not affected, 50 individuals who have the homozygous dominant allele, and 340 individuals who are heterozygous and carry it.