08-26-2012, 01:48 PM
There are two equations when dealing with Hardy-Weinberg Equilibrium. p^2+2pq+q^2=1 is one, and the other is p+q=1 (that is, the percentage present for each allele must equal 1).
As you said, p^2 = .49 which implies that p = .7 (since p can't be a negative number). Therefore, .7 + q = 1 ... which means q = .3
So what's the percentage for heterozygous?
2pq = 2(.7)(.3) = .42 = 42%
Between this and clep3705's post, I hope you understand.
As you said, p^2 = .49 which implies that p = .7 (since p can't be a negative number). Therefore, .7 + q = 1 ... which means q = .3
So what's the percentage for heterozygous?
2pq = 2(.7)(.3) = .42 = 42%
Between this and clep3705's post, I hope you understand.
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