Hi, I have two SNPs that give rise to five possible genotype combinations (for example AA for the SNPA and BB for the SNPB). If I want to test this pair for association can I do a chi-square on a 5x2 table (five genotype combinations, two status case/control)?
Thanks a lot in advance Rossella
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Rossella •
I think you are looking for epistasis, which traditionally is identified through a general linear model, and partition the variances to those explained by locus A only, locus B only and AxB interactions. It gets a bit messier in case-control studies though, here is a method paper.
That really depends of what you want to do.
What you seem to try here is to see a whole effect (main + interaction) of the two SNPs on some disease. The test you are carrying out if you make a table is a very general test which states that each classs of double genotypes can have any effect.
Such a test could detect atypical genetic effects. However, on most plausible biological models, the cost of this "freedom" of model (degrees of freedom) is very high in terms of power. You can afford it if you have a very very large dataset (this depends of the true effect but I'd say at least 1000 cases and 1000 controls).
IF THERE IS NO LD between the SNPs A more standard way would be to run 2 logistic models. First recode you data AA = 0 AB = 1 BB = 2
then Y = a + bSNP1+cSNP2+dSNP1SNP2 store the likelihood L1
then Y = a store L2
Test -2*(L1-L2) ~ chi square with 3 degrees of freedom. SNP1 and SNP2 take values 0, 1 and 2.
Obviously, here again you have a strong underlying hypothesis (linear effect for main effects b and c) and something taking value 0,1,2 and 4.
Using alternative codings, you can specify a model corresponding to the full tables that you are proposing.
IF THERE IS LD - (might be the case here with missing genotypes), I'd advise using algorithms liek the one implemented in THESIAS which allows haplotype analysis to test the combined effect of two (up to 5,6) with a lot of nested models that can be tested by liklihood tests.
Finally, it is important to know the sampel size because it may happen that the test is useless if you have 10 cases and 10 controls.
Thanks a lot, the approach you are proposing seems very reasonable. I was using the 5x2 table just as a quick way to screen my data, for the final publication I will definitely have to use something more accurate
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What did you count for table, cases? And why 5 genotypes? I think there would be six 6 genotypes possible, 3 per SNP as: AA, AB, BB. In general I don't think your approach makes much sense, it sounds a bit as if you are looking at linkage-disequilibrium. Further, this approach seems very reductive to me, given that there are millions of SNPs genotypes, what does knowing that the count distributions of two arbitrary chosen are independent tell you?
I am not sure I understand. Wouldn't you expect 9 (3*3) possible genotypes?
Sure Chris, 9 genotypes. But, I understood he wanted to make a table of counts for each observed allele frequency for the snps, like so: SNP1-AA: Control 10, Case 22, SNP1-AB: Case 20, Control 24, etc. Something which I think is not sound. Anyways, Rossella, we are in need for an example table.
Normally I would have 9 genotypes but only five have been observed. Do Do I need to do a 3x3 table with 0 counts?