Question: Quick question about analysis of log2 RMA data
0
mforde841.2k wrote:

Say I have some probe level and RMA normalized array data. The data is supposedly log2 normalized already. How do I calculate the relative difference in intensity between two probes?

If I understand correctly, it should just be the 2^residual.

For example

``````probe1 = 1
probe2 = 2

2^(2-1) = probe2 has 2x as much signal as probe1
``````

However for values less than 1, I'm not sure this makes sense:

``````probe1 = 0.25
probe2 = 0.5
2^(0.5-0.25) = probe2 has 1.19x as much signal as probe2?
``````

Shouldn't it be 2x as well in this instance?

microarray • 1.6k views
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modified 2.3 years ago • written 2.3 years ago by mforde841.2k

If your values are already log2 of the intensities, then you get the difference by subtracting the log2 values, because log(A/B) = log(A) - log(B). This is equivalent to dividing the initial (base 10) intensities, A/B, so essentially it gives you a fold-change.

ADD REPLYlink written 2.3 years ago by mastal5112.0k

If I understand correctly the resulting fold change is still in log2, correct?

So 0.5 is not actually double the intensity of 0.25, where 2 = 0.5 / 0.25. But instead 1.19x because 2^(0.5-0.25).

And 2^( (0.5-0.25) / 0.25) doesn't make sense because log2 isn't linear.

ADD REPLYlink modified 2.3 years ago • written 2.3 years ago by mforde841.2k

Yes, you're right, you get a log2FoldChange rather than a Fold-Change. 0.5 is not double the intensity of 0.25 because those were the log2 values of the intensities.

ADD REPLYlink modified 2.3 years ago • written 2.3 years ago by mastal5112.0k
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