# Genotype probabilities

• Oct 11th 2010, 08:12 PM
root45
Genotype probabilities
I'm having difficulty with this problem, although it seems easy.

The possible genotypes of an organism are AA, Aa and aa (Aa and aA are equivalent). When two organisms mate, each independently contributes one of its two genes; either one of the pair is transmitted with probability .5.

Suppose that the probabilities of the genotypes AA, Aa and aa are p, 2q and r respectively, in the first generation. Find the probabilities in the second and third generations, and show that theses are the same.

I started by saying that each offspring has two genes, either A or a. An offspring has probability p that a parent has genotype AA, and probability 2q that a parent has genotype Aa. So the probability that it has an A is 1*p + .5*2q = p+q. Similarly, the probability that it has a for it's first gene is q+r. So the probability that it has genotype AA is (p+q)^2, genotype aa is (q+r)^2 and genotype Aa is 2(p+q)(q+r) (since Aa and aA are the same).

This seems fine, but when I try to calculate things for the third generation, it doesn't come out to the same as the second as it's apparently supposed to do. The calculation should be exactly the same, but with (p+q)^2 in place of p, 2(p+q)(q+r) in place of q and (q+r)^2 in place of r. So the probability having an A is (p+q)^2 + (p+q)(q+r) and the probability of having a is (q+r)^2 + (p+q)(q+r). But then the probability of being AA is

((p+q)^2 + (p+q)(q+r))^2

which obviously isn't equal to (p+q)^2. So I'm missing something, but I don't know what. Thanks for any help.
• Oct 11th 2010, 10:32 PM
harish21
You must have heard of/studied Hardy-Weinberg Principle .Referring to this link should clear things out for you.
• Oct 12th 2010, 04:27 AM
root45
Arg, I swear I looked at that page but didn't see the derivation. Thank you for your help.