Question

% of change and log2 fold change (L2FC)?

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5.7 years ago

WUSCHEL ▴ 810

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 • 8.9k views

ADD COMMENT • link 5.7 years ago by WUSCHEL ▴ 810

score 7 · Accepted Answer · 2019-03-17

7

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5.7 years ago

GouthamAtla 12k

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.

ADD COMMENT • link 5.7 years ago by GouthamAtla 12k

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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 ?

ADD REPLY • link 5.7 years ago by WUSCHEL ▴ 810

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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.

ADD REPLY • link 5.7 years ago by GouthamAtla 12k

score 3 · Accepted Answer · 2019-03-18

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.