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So first, sorry if I use non-standard vocabulary, since I am not an expert. Please feel free to correct.

I would like to know if a set of genes in a patient cohort are mutated by a mutually exclusive manner. While one can observe this easily by eye, it is unclear to me how to test this in a statistical rigorous way for a gene set more than two genes. All the patients suffer the same disease)

The scenario is this: 4 genes (g1, ..., g4), 15 patients (p1, ..., p15). If a gene is mutated in a patient, its denoted as 1, otherwise 0 for no observation.

\  | p1 | p2 | p3 | p4 | p5 | p6 | p7 | p8 | p9 | p10|
-----------------------------------------------------
g1 | 1  | 1  | 1  | 1  | 1  | 0  | 0  | 0  | 0  | 0  |
g2 | 0  | 0  | 0  | 0  | 0  | 1  | 1  | 1  | 0  | 0  |
g3 | 0  | 0  | 0  | 0  | 0  | 0  | 0  | 0  | 1  | 0  |
g4 | 0  | 0  | 0  | 0  | 0  | 0  | 0  | 0  | 0  | 1  |

While it may be obvious to the observer that this group of genes are mutated in a mutually exclusive manner.

  • How would I test if this distribution of events has not happened by chance? The underlying hypothesis is, that there is a selective pressure on that any of the four genes must have a mutation to in order to have the disease.

Could it be done by Chi Square testing?

Thank you for any help

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  • $\begingroup$ Can you explain what exactly constitutes 'mutually exclusive mutation' in the table? Is it than there's no more than a single '1' in a column? Exactly one '1' in each column? Some other criterion? Is it actually a logical notion or a probabilistic one? $\endgroup$ – Glen_b Mar 20 '13 at 9:53
  • $\begingroup$ I take you to mean by mutually exclusive that each person has a mutation in one gene and none of the other 3. What are you trying to assess significance of, exactly? The significance of finding this cohort of people? The number of people with a mutation of a given gene? $\endgroup$ – learner Mar 20 '13 at 11:32
  • $\begingroup$ GlenB: yes, it is a binary information - mutation observed, yes/no. learner: you are right. I want to know if this observation could also be observed by chance. So if it only were two genes I would perform a Fisher´s test. $\endgroup$ – dmeu Mar 20 '13 at 19:48
  • $\begingroup$ I have added some more detail to the original question. I hope it makes things clearer $\endgroup$ – dmeu Mar 21 '13 at 7:22
  • $\begingroup$ @dmeu your response there was almost completely unhelpful. I already understood the data to be binary, indicating presence or absence of a mutation. My questions related to the way you were defining mutually exclusive in respect of that table. Please address the specific questions in my original comment. $\endgroup$ – Glen_b Mar 21 '13 at 7:41

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