Hey,
I did bisulfite sequencing of DNA from cells I treated with a demethylating substance at different concentrations before. I now want to show that the difference between control, negative control and the cells treated with the demethylating substance is significant. I have about 4 sequences each.
Which statistical test would fit best here?
Thanks in advance :)
The probabilistic framework here is not so obvious to me. How does the randomness come into play? Is it always the same DNA, for which you want to account for possible "measurement" errors? Or the measurement procedure is accurate, but you have various DNAs? Or both?
Besides, what kind of index do you use (or want to use) to measure the "difference"?
"... now want to show that the difference between ..."
The difference of what, precisely?
I want to show the difference in methylation. So that when I add the demethylating substance you can see a decrease in methylation.
And no, it's not always the same DNA. I isolated DNA from cells and sequenced them. So each sequence I got equals the DNA of one cell. The measurement procedure should be accurate.
The index I use is "% methylation". Let's say from 14 CpG's in the amplicon I sequenced, I get sequences from the untreated cells with 8, 6, 8 and 11 methylated CpG's. For the cells treated with the demethylating substance I get sequences with 2, 4, 6 and 4 methylated CpG's. My question is now: Is there a test which allows me to tell if the difference in methylation between these to groups is significant or not.
Currently, I'm trying to get more replicates but this is quite difficult actually.
I hope this makes it a bit clearer to you.
Okay, I get it better now.
With the sparse data you have, there are few options you could try. You could model the number of methylated CpG as a Poisson random variable. Then you can test whether your two or three samples come from the same Poisson population. Check out these papers:
http://repository.hkbu.edu.hk/cgi/viewcontent.cgi?article=1000&context=math_ja
http://repository.hkbu.edu.hk/cgi/viewcontent.cgi?article=1001&context=math_ja
http://www.ucs.louisiana.edu/~kxk4695/JSPI-04.pdf
Btw, make sure you make the "correct" H_0 hypothesis!
You can conduct an ANOVA test first to identify whether they come from the same population. If they come from the same population, you can then add demethylating substance to the experimental group. Then you conduct another ANOVA test to affirm the significance. Statistically speaking, if the addition of new substance cause the experimental group to change significantly, then the ANOVA test will fail on the experimental group. We identify the change in variations and mean difference to follow F distribution and T distribution respectively, with only 5% of chance in making type I error - you reject the null hypothesis wrongly. The F-test in the ANOVA procedure is important to observe for your research.
Hi Sven,
So the t-test works best when the variable you are measuring follows a Normal distribution with some mean and variance (http://en.wikipedia.org/wiki/Student's_t-test, see assumptions). You have count data, and quite a small sample (4 sequences, 2groups), so you cannot check if this assumption holds.
You can use what Michael suggested, which is a glm assuming poisson distribution, to test whether the treatment has a significant effect.
Another way is to use a binomial distribution followed by a likelihood test. The null hypothesis is that both groups of samples have the same probability of being methylated. Alternative is that the probability is different. you can use (http://www.vassarstats.net/textbook/ch5apx.html) to calculate the probabilities or basically use R (which is what I use). I'll show you with the numbers you have provided above:
x = 8, 6, 8, 11 , y = 2, 4, 6, 4
overall mean = 6.125
mean x = 8.25 ; mean y = 4
for 1 sample, e.g 8 out 14 in x
P(8 out 14, given mean = 6.125) = 0.1277 (use N=14,k=8,p=6.125/14=0.4375)
P(all samples have same mean) = P(N=14,k=8,p=0.4375) * P(N=14,k=6,p=0.4375) .. P(N=14,k=4,p=0.4375) = 1.249903e-09
P(all samples have different mean) = P(N=14,k=8,p=) * P(N=14,k=6,p=0.4375) .. *P(N=14,k=4,p=0.2857143) = 2.589363e-07
Normally I do the steps above in log (so u can just add), but to illustrate I keep it in decimals
Next you log both probabilities and take the difference, to get a Xsq value
Xsq = 2*(log(2.589363e-07) - log(1.249903e-09)) = 10.67703
You can then check the significance of the chi-square value assuming 1 degree of freedom, which gives you p = 0.001090621
Hope this helps, and I can also provided the R code if you would be able to use that.
Thanks!
So Gen, I can follow your idea best. If I understood correctly, you would reject the Null Hypothesis that both groups have the same possibility of being methylated at a significance level of 0.05? So one could say, that in fact the difference in methylation between both groups is significant?
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