Question: how to calculate the log FC for -1.5 fold change?
0
2.5 years ago by
Calangoa30
Calangoa30 wrote:

Hi there, I have 6 microarray data sets that I think FDR<0.05 and fold change >1.5 and <-1.5 for up- and dw-regulated genes respectively are the best (based on Mark R Dalman et al paper) but after normalization with R packages, we have log transformation of fold changes and I know if log FC is 2 it means 4 fold change for gene expression. and logFC -1= FC 0.5. it means, we don't have negative measure for fold change in statistic but have in biology. what is the similar value of FC -1.5 for log FC? (?logFC=FC-1.5)

next-gen • 6.0k views
modified 2.5 years ago by h.mon32k • written 2.5 years ago by Calangoa30
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So if log2FC = log2(FC), to get the FC you have to do FC = 2^log2(FC). So if your log2FC is -1.5 you have to compute 2^-1.5

1
2.5 years ago by
h.mon32k
Brazil
h.mon32k wrote:

When we use fold-change as ratios of expression, values are bound between [0, +Inf], hence a value of -1.5 fold-change is impossible. Also, the scale is centered at 1, meaning up-regulation are values greater than 1, and down-regulation are values smaller than one. A fold-change of 2 and a fold-change of 0.5 have the same magnitude: both means one sample has double the expression of the other. This way of interpreting fold-changes have two drawbacks: fold-changes are centered around one, and they are not symmetrical.

By transforming to log2( fold-change ), values are now centered at zero, and they are symmetrical: -1 and 1 indicate the same magnitude of change, down or up.

So both ways are showing the same thing, but log2( fold-change ) are more convenient.

Many thanks h.mon so you mean log2 (fold change) for cut-off is more better and it is symmetrical. so, I suppose FC=0.667(1/1.5) show down-regulation of gene about 1.5 fold since log 0.667 and log 1.5 show near - and +0.58 (symmetrically).