How do I construct this 3x3 matrix? I have 3 genes and two states
Good Afternoon,
My name is Gabriel, I'm doing an analysis if there is increase or decrease in dependence on the mutated genes, using 3 or more genes using the fisher exact test. I performed with success an analysis for two genes using fisher.test( ). example of the 2x2 contigency table:
Gene A mutated | Gene A normal
Gene B mutated| 26 | 12
--------------------------------------------------------------------------------------------
Gene B normal | 10 | 50
Now I'm wondering how can I perform the analysis for 3 genes (and construct the contigency table), as follows: Gene A mutated, Gene A normal, Gene B mutated, Gene B normal, Gene C mutated and Gene C normal. How do I perform a fisher test using fisher.test( ) function using this data (3x3 contigency table)? Can someone help me ?
3 answers
The problem here is that Fisher test is meant for categorical data. That means that each of your patients must belong to exactly one category. In your case, you have 3 variables, each with 2 categories. This implies a 2x2x2 "cube" matrix, so that e.g. in one cell, there will be the number of patients that have mutated gene A & normal gene B & mutated gene C.
A 3x2 table would be for problems like: man - woman - alien vs. 10 fingers - more fingers i.e. two variables, three and two categories, respectively.
Common fisher test is not meant for such problems (see nice short definition here: http://mathworld.wolfram.com/FishersExactTest.html) I'm not sure, if there is a program that could compute this 2x2x2 Fisher test modification, but googling gives me some results.
If you have a 3x3 as a matrix, you just use fisher.test(yourTable).
I think you should use a 3x2 (2x3) contingency table because you have three genes and two states (mutated vs. normal).
but if do like this : Columns : Gene A, Gene B and Gene C , rows : mutated , normal . And if I assume that we are analyzing 3 patients.... one patients can have Gene A and Gene B mutated but not C.... another patient Only Gene C and the other patient all three genes mutated, at the end will have an contingency table with :
gene A | gene B | gene C
mutated 2 2 2
normal 1 1 1
a total of 6 mutated and 3 normal....and our n of patients is three ....not 9!
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