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# Confusion over how to calculate fold change error

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### #1 sialic acid

sialic acid

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Posted 22 April 2015 - 06:40 PM

Some confusion is going around in the lab over how to specifically calculate the error bars for a control group when you present data as 'fold changes'

Let's say I run the same experiment with drug doses 0, 5, and 10 uM on the same set of cells and the instrument I use outputs its readings as arbitrary numbers.  The output readings depend heavily on simply what day it is.  I get the the following data for 3 experimental replicates

0              5              10    uM

Replicate 1          0.5            1                2

Replicate 2            2             3              7

Replicate 3           5              10            12

If I simply plot that data as a bar graph for each treatment and do ANOVA with Dunnett's post test to test for significance vs. the control group (0), the data isn't significant because of the high variability in the control group as well as the other groups.  However if I normalize the data that was obtain for each assay on the days that they were run the data is transformed into the following:

0              5              10    uM

Replicate 1          1             2                4

Replicate 2           1            1.5             3.5

Replicate 3          1              2            2.4

After normalizing and running ANOVA with Dunnett's post test, the data is significant now with 10 uM statistically significant over the control.  The only problem is that since everything is normalized based on the day that the expeirments were run, the are no error bars for the control group since it is always 1.  In some publications you see control bars with values listed as 1.0 with no error bars while in many other publications you see control bars normalized to 1.0 with error bars.  I don't really understand why normalizing first each day and then tabulating the results would be wrong (which would result in a control group with no error bars).  Is this wrong?  Can someone explain?

### #2 DRT

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Posted 23 April 2015 - 12:35 PM

You are correct to highlight the assumption of equal variability as a problem for normalised data.

Either use a non parametric ANOVA or a two way ANOVA treating replicates as a separate factor.