Timeline for Bioinformatical problem - specific word enrichment in a given sequence
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2013 at 19:08 | history | bounty ended | CommunityBot | ||
| Jun 13, 2013 at 19:54 | comment | added | gui11aume | Yep. This makes sense. I would also use the second resampling scheme that I proposed (it is usually a good idea to buttress your p-values by resampling them). | |
| Jun 13, 2013 at 10:58 | comment | added | pogibas | Just to make it clear - I want to test if one set is enriched in tandem repeat compared to the other set. So it's probably test "is the coverage by the motif equal in sets A and B"? And I should use two proportions (how much of the sequence is tandem repeat) for t-test? | |
| Jun 13, 2013 at 9:33 | comment | added | gui11aume | There is a difference. Either you test "is the coverage by the motif equal in sets A and B?" or you test "is the distribution of coverage among proteins of set A the same as the distribution of coverage among proteins of set B?" | |
| Jun 13, 2013 at 9:21 | vote | accept | pogibas | ||
| Jun 13, 2013 at 9:18 | comment | added | pogibas | "you have only two numbers" - is there a difference if I compute average coverage for all Set_A and Set_B or use coverage for every sequence independently? r-bloggers example use non-averaged input (input for t-test is a string). | |
| Jun 13, 2013 at 9:16 | comment | added | gui11aume | @Poe Permutation is a very good option. You have to think about the way you are going to permute the data and there are several options, which test different null hypotheses. You can shuffle the coordinates, but I guess you scores are going to be identical. You can Randomly pick with replacement sequences from set A and set B to form set C and set D, and compare set C and D 1000 times. You can also regenerate the sequences of set A and B with the same Markov model (the will have the transitions between k-nucleotides) and recompute the occurrence of the motifs 1000 times. | |
| Jun 13, 2013 at 9:12 | comment | added | gui11aume | @Poe Regarding your concern about normality, the justification for the approach is the Central Limit Theorem. The score above is a mean computed with large $n$, so its distribution is asymptotically Gaussian. I don't see how you could do a Wilcoxon test in these conditions because you have only two numbers. | |
| Jun 12, 2013 at 22:45 | comment | added | pogibas | I was suggested to do permutations and it seems to be a reasonable solution. Just shuffle coordinates genome wide, calculate hits in simulated set , repeat *1000, calculate empirical p value and then I should be able to compare such enrichment between my original datasets. (I added figure just to for example how I would like my data to look like). | |
| Jun 12, 2013 at 22:31 | comment | added | pogibas | Thank you for the answer! I got the point and it's a little bit more clear now, but I still have few questions: 1) r-bloggers.com/two-sample-students-t-test-1 says that t-test should be used for normal distribution data and is #3 solution (percentage of coverage) distributed normally? Shouldn't I use wilcox for such data? 2) Do I have to do normality test as mentioned in the link above? 3) Let's assume that I did permutations and have datasets that contain equal size items (no length bias). How could I test such equally sized datasets difference with #1 & #2 solutions? | |
| Jun 9, 2013 at 21:14 | history | answered | gui11aume | CC BY-SA 3.0 |