1000 population Question
q2 = .40
q2 (homozygous recessive individuals) = .40 400/1000
q (frequency of recessive allele) = .63
p2 (homozygous dominant individuals) = .14 140/1000
p (frequency of dominant allele) = .37
2pq (heterozygous individuals) = .46 460/1000
First of all, the q2 is given and in this case, it is .40. Next, you are supposed to square root the q2 value and this will give you .63. After that, since q and p is always equal to 1, and q is .63, therefore p has to be .37. Furthermore, you square .37 and this will give you .14. To find the heterozygous individuals, there is actually different kinds of methods; the main one being 2(p2+q2). However, the method in which I happen to find easiest for me, would be to add the q2 and p2 values, then subtracting that sum to 100. So, in this case it would be .46.
