Question: Can we crosscheck log2 fold change in gene expression data?
gravatar for madkitty
5.9 years ago by
madkitty610 wrote:

So I have a very simple experimental design. 

A (control) ------- treatment -------> B

C (control) ------- treatment -------> D 


A and C are controls for different phenotype. We did multiple gene expression analysis, including these two:

  • B vs A
  • D vs. B 

 IF B vs. A in certain gene gives log2 = -2 and the same gene in D vs. B has log2 = -4 .. then can we estimate that the difference D vs B is (-2 * -4) ==> log2 = -8?

rna-seq • 2.1k views
ADD COMMENTlink modified 5.9 years ago by seidel7.1k • written 5.9 years ago by madkitty610

so you compared B(treatment) vs. A(control) and D(treatment) vs. B(treatment)? Why did you compare two treatments? Why din't you compare D(treatment) vs. C(control)? What difference are you looking at?

I am sorry I am a little confused, are you saying that you have log2FC values for B-A & D-B and you want to find D-B again?

Also, it would be helpful to show us some data (what A,B,C,D values look like), are they log2 values as well as how did you calculate your log2(Fold changes).

And honestly, even without knowing the above answers, I "sort of know" that you cannot extrapolate the log2FC values like you are trying to do.

ADD REPLYlink modified 5.9 years ago • written 5.9 years ago by komal.rathi3.6k
gravatar for Devon Ryan
5.9 years ago by
Devon Ryan96k
Freiburg, Germany
Devon Ryan96k wrote:

On the log2 scale they add, so -6. On the non-log2 scale they multiply, so so 0.015625 (aka 2^-6). BTW, you'd really need to run the test itself since the parameter maximum likelihood estimation could end up being a bit different (I'm assuming that you have replicates).

ADD COMMENTlink written 5.9 years ago by Devon Ryan96k
gravatar for seidel
5.9 years ago by
United States
seidel7.1k wrote:

You might have a typo (as @komal.rathi hints at):

"[if] the same gene in D vs. B has log2 = -4 .. then can we estimate that the difference D vs B is..."

You don't have to estimate the difference of D vs. B, you just told us what it is.

On the other hand, if you've done B/A and D/C, and you want to use that data to estimate D/B, you can't do it directly without knowing something about the controls.
You don't mention what technology you're using (Northern Blot? qPCR? RNA Seq?), thus it's hard to assess what assumptions come into play. However if we assume RNA Seq data, and you have large populations of normalized values, you can't simply estimate the D/B difference reliably without examining the A/C difference to ensure that the gene is unchanged in both controls. Otherwise we can assume the data sets are all normalized to each other and the following numbers would satisfy the criteria log2(B(10)/A(40)) = -2, log2(D(10)/C(160)) = -4, yet log2(D/B) != -6 because the gene is changing in each control. But again, this assumes a typo, and we don't know what normalization assumptions you are employing.

ADD COMMENTlink written 5.9 years ago by seidel7.1k
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