I have this equation: Corresponds to HW in equilibria with three alleles:
$(p+q+r)^2=1$
Expanding the square results:
$p^2+2pq+r^2+2pr+q^2+2qr = 1$
I need to separate homozygous and heterozygous, that means: $2pq+2pr+2qr=1-p^2-q^2-r^2$
How I can use a equivalence similar to $(p=1-q) or (q=1-p)$ in two alleles to resolve the last equation?
Answer
Everybody said it already, but there is none. The original HWE equation ($(p+q)^2=1$) works because you've got two variables and two equations ($p+q=1$) to work with (in reality, these are just one equation and one variable, since $q=1-p$ so $p+(1-p))^2=1$). Now you have three variables and still only the one equation ($p+q+r=1$) which is, mathematically, impossible.
You need go out and measure some frequencies. In particular, you need to measure two of them - I'd choose two of the homozygous types for convenience's sake.
This basically summarizes everything nicely for ya: http://www.bio.miami.edu/dana/dox/trinomial.html
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