Question: % of change and log2 fold change (L2FC)?
2
Wox340 wrote:

I am trying to understand Gene expression data analysis. Could someone the percentage of change of the GE calculating from log2 fold change (L2FC)?

FoldChange = μ_Treatment / μ_Control

log2FC=Log2(μ_Treatment)-Log2(μ_Control)

rna-seq gene • 1.4k views
modified 19 months ago • written 19 months ago by Wox340
5
geek_y11k wrote:

seems correct to me. log2FC is more informative as you will see the direction (sign) and the effect.

``````Treatment = 10
Control = 100
``````

`FoldChange = 10/100 = 0.1` which is not informative

`log2FC = log2(10)-log2(100) = -3.3` which tells the effect and direction.

Thank you @geek_y.

After getting log2FC; e.g. -3.3, how can we calculate the % of change? is that %change = (Change/Original) x 100 ?

2

If you do `2^log2FC`, which tells how many folds the change is. Here `2^3.3=9.8` folds. The sign gives information about the up/down regulation.

2
Wox340 wrote:

calculate fold change (FC) When comparing these log transformed values, we use the quotient rule of logarithms:

log(A/B) = log(A) - log(B)

log(A) = 4

log(B) = 1

Therefore:

log(A/B) = 4 - 1

log(A/B) = 3 This gives a 3-fold change.

Please note that in this case we are reporting the log(fold change). Biologists often use the log(fold change) because without taking the log, down regulated genes would have values between 0 and 1, whereas up regulated genes would have any value between 1 and infinity. This distribution makes graphing and further statistical analysis difficult. Taking the log typically makes the resulting values more normally distributed, which is better for further analysis.