--Hi,
a reviewer ask me to add error bar on my qPCR control condition. But this is what i think:
Noteably, the error bar for the control group has a different statistical meaning as the error bars of the other groups, in the representation in fold induction we admit that there is an uncertainty for the control group but we try to quantify the target genes so we make the assumption that the control normalizer does not vary in relation to itself.
To simplify the calculation phase of the fold change, we normalize as follows: target ddCt - control ddCt.
To express target genes as a function of Control, Control fold induction is like doing:
ddCt Control - ddCt Control.
So we admit that Control fold induction does not require error bars.
We admit that it exists on dCt and ddCt but not on 2^(-ddCt) (or fold induction).
Someone can tell me his opinion ?
thank you --
Hi Laurent Manchon,
You can try this:
1) Take the mean of all 2-dCq of control group.
2-Cq interest gene control 1 / 2-Cq reference gene control 1 = 2-dCq control 1
2-Cq interest gene control 2 / 2-Cq reference gene control 2 = 2-dCq control 2
2-Cq interest gene control 3 / 2-Cq reference gene control 3 = 2-dCq sample 3
Take the mean of all control 2-dCq.
In this case: (2-dCq control 1 + 2-dCq control 2 + 2-dCq sample 3)/3 = 2-dCq mean control
2) Divide all samples 2-dCq (control and treated) by the mean 2-dCq mean control of control group.
2-dCq of control 1 sample/2-dCq mean control;
2-dCq of control 2 sample/2-dCq mean control;
2-dCq of treated 1 sample/2-dCq mean control;
2-dCq of treated 2 sample/2-dCq mean control;
.....
3) If you did right, your new control mean will be 1 and you can calculate the standard deviation of control group. The remaining groups will be a "percentage" of control group.
Best
Jacó
it makes no sense, I have to represent a standard deviation and not an average. For the control I have 3 values of dCq: 9.72 ; 8.37 ; 8.26 and my plot represent log2(FC)
Hi Laurent Manchon ,
If you are showing your results as log2(Fold-Change)to linearize the scale, i agree it does not make sense talk about standard deviation. I think you need to explain this for the referee. It depend how you are presenting the results.
Best,
Jacó
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