![]() It just shows that the z-score is a crude measure, because obviously the score does not take into account the functional level and activation dependencies of the single genes within a process. At least not as long as it is significantly enriched. I got very few differential expressed genes, with 1.5 Foldchange very few, can I consider 1.41 foldchange is this is acceptable ? suggestions please. I'd like to compare the normal FC values among these groups, how do I say this particular gene has expressed more in this particular group ? Logfc = 0.58 2^(0.58) = 1.5 Foldchange Using the above function I converted logFC to normal FC values for three groups(MC,SC,OC). #take 2^(absolute(log2FoldChange)) > res.new$FC res.new$FC<-ifelse(res.new$logFC<0,res.new$FC*(-1),res.new$FC*1)īut I am not sure if I am correct. How do I convert the values (positive and negative) to normal fold changes? I calculated differential expressed genes using limma's topTable() function. I have a microarray set of 26933 genes and 6 samples (3 controls and 3 mutants).
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