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Is it true that many disease causing variants/mutations do not follow Hardy Weinberg Equilibrium? If so, then please elaborate on why this may be true (or not) and provide examples.

I am interested in single-gene disorders or those linked with single nucleotide variants. I was told that many disease causing variants/mutations do not follow Hardy Weinberg Equilibrium, but I cannot find any documentation showing this (so I am looking for a reference on this matter). In particular, I have considered Huntington's disease, Sickle cell anaemia and Hemophilia, but have not found any special relationships with these diseases and HWE.

Thank you.

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  • $\begingroup$ Welcome to Biology.SE. Homework questions are considered off-topic unless the OP shows attempts of answering its own question. $\endgroup$ – Remi.b Sep 17 '15 at 21:27
  • $\begingroup$ This is not a homework question. I am a research data analyst proposing a bioinformatics model and I imagine the answer to this question will be helpful to my work. I have done much searching, even considering HWE along with single-gene disorders such as Huntington's disease, Sickle cell anaemia and Hemophilia. I turned to StackExchange because my searches have proved unfruitful thus far. Thank you. $\endgroup$ – user918804 Sep 17 '15 at 21:38
  • $\begingroup$ Ok, I retracted my close vote! $\endgroup$ – Remi.b Sep 17 '15 at 21:41
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    $\begingroup$ @user918804 in that case, please edit your question and add more details as to what you've already looked at, what you've learned, and what exactly you're still having problems with. Citations would also be helpful. The more detailed a question, the better the chances of getting a good detailed answer. $\endgroup$ – MattDMo Sep 17 '15 at 21:51
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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.

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  • $\begingroup$ I welcome the readers to double check my calculations. $\endgroup$ – Remi.b Sep 17 '15 at 22:33
  • $\begingroup$ I think the real surprising effect is the rather high frequency of the sickness causing alleles. For sickle cell anemia there is a well-known explanation, but I am wondering about Huntington Chorea. $\endgroup$ – jknappen - Reinstate Monica Mar 6 '18 at 12:54
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In the case of sickle cell anemia, there is a positive selection effect of the illness causing gene variant: Heterozygous gene carriers have some immunity against malaria. Therefore, sickle cell anemia is quite frequent in malaria affected regions of the world.

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