Hardy-Weinberg law makes a series of assumptions. One of them is the absence of selective effects. As you talk about disease, this assumption of neutrality is obviously not met.
At the moment of fecundation
Imagine for example that at the moment of the fecundation, the genotypes AA
, Aa
and aa
are at Hardy-Weinberg equilibrium. Let x
be the frequency of the allele A
at the moment of fecundation, the frequencies of the genotypes AA
, Aa
and aa
are $x^2$, $2x(1-x)$ and $(1-x)^2$, respectively.
After selection
As all genotypes aa
die off, you're left with the A
allele being at frequency $y=2x^2 + x(1-x)=x(1+x)$ and the frequencies of the genotypes AA
, Aa
and aa
are $\frac{x^2}{x^2 + 2x(1-x)}=\frac{2x-2}{x-2}$, $\frac{2x(1-x)}{x^2 + 2x(1-x)}=\frac{x}{2-x}$, and $0$, respectively.
Expressing the genotype frequencies in terms of the new allele frequency
You can as well plug $y$ into those genotype frequencies to get the expression in terms of the genotype frequencies at birth ($y$; after selection). It gives you $\frac{1-\sqrt{4 y+1}}{\sqrt{4 y+1}-5}$, $\frac{2 \left(\sqrt{4 y+1}-3\right)}{\sqrt{4 y+1}-5}$ and $0$ for the genotypes AA
, Aa
and aa
respectively. So, obviously, it is not at Hardy-Weinberg equilibrium.
Assumptions
You can make some more interesting calculations by leaving out the assumptions that all aa
genotypes die at birth and you could use a Leslie matrix to describe the probability for an individual of each genotype to make to the next age. I also assumed that the a
is completely recessive. You might want to release this assumption as well by computing the death of a fraction of the Aa
genotypes.