Having a standard deviation of DDCt values is possible only when you do a paired analysis, that is, for each "treated" sample (DCt) you have a paired "control" sample (DCt) of the same individual, so that you calculate one DDCt per individual. These values will scatter around their mean with some standard deviation. To give a summary of the calculated DDCt values, one might provide the mean and the standard deviation, typically like mean(DDCt) ± SD(DDCt). To simplify the notation, set m = mean(DDCt) and s = SD(DDCt). Under the (absolutely reasonable) assumption that the DDCt values will be approximately normal distributed, this implies that this range (from m-s to m+s) is expected to contain about 68% of the values (and the range from m-2s to m+2s contains about 95% of the values).

If people prefer to show/report fold-changes, these DDCt values need to be exponentiated. Noteably, it makes no sense to exponentiate the standard deviation! The standard deviation of the 2^-DDCt is not equal to 2^s. On could calculate the standard deviation based on the individual 2^-DDCt values, but the distribution of 2^-DDCt is not normal (it is skewed; approximately log-normal), so there is no simple interpretation of what this standard deviation value means, practically. Thus, instead one likes to provide a range for the 2^-DDCt values that is expected to contain 68% (or 95%) of the 2^-DDCt values, and this range is obtained by exponentiation the limits of the corresponding range for the DDCt values: the lower limit is 2^(-(m-s)) and the upper is 2^(-(m+s)).

If you don't have paired samples, you can't calculate meaningful individual DDCt values. Instead you can "only" calculate the mean DDCt-value. This mean value has a standard error (not standard deviation!), that can be obtained with the usual methods from the DCt values. If fold-changes should be reported, it is then adviseable to present the confidence interval for the mean fold-change. This is obtained by first calculating the limits of the confidence interval for the mean DDCt (as usual, because we can assume the mean DDCt has a normal distribution), and then exponentiate these limits.

I attach these two file.

I hope that I will help you.

Having a standard deviation of DDCt values is possible only when you do a paired analysis, that is, for each "treated" sample (DCt) you have a paired "control" sample (DCt) of the same individual, so that you calculate one DDCt per individual. These values will scatter around their mean with some standard deviation. To give a summary of the calculated DDCt values, one might provide the mean and the standard deviation, typically like mean(DDCt) ± SD(DDCt). To simplify the notation, set m = mean(DDCt) and s = SD(DDCt). Under the (absolutely reasonable) assumption that the DDCt values will be approximately normal distributed, this implies that this range (from m-s to m+s) is expected to contain about 68% of the values (and the range from m-2s to m+2s contains about 95% of the values).

If people prefer to show/report fold-changes, these DDCt values need to be exponentiated. Noteably, it makes no sense to exponentiate the standard deviation! The standard deviation of the 2^-DDCt is not equal to 2^s. On could calculate the standard deviation based on the individual 2^-DDCt values, but the distribution of 2^-DDCt is not normal (it is skewed; approximately log-normal), so there is no simple interpretation of what this standard deviation value means, practically. Thus, instead one likes to provide a range for the 2^-DDCt values that is expected to contain 68% (or 95%) of the 2^-DDCt values, and this range is obtained by exponentiation the limits of the corresponding range for the DDCt values: the lower limit is 2^(-(m-s)) and the upper is 2^(-(m+s)).

If you don't have paired samples, you can't calculate meaningful individual DDCt values. Instead you can "only" calculate the mean DDCt-value. This mean value has a standard error (not standard deviation!), that can be obtained with the usual methods from the DCt values. If fold-changes should be reported, it is then adviseable to present the confidence interval for the mean fold-change. This is obtained by first calculating the limits of the confidence interval for the mean DDCt (as usual, because we can assume the mean DDCt has a normal distribution), and then exponentiate these limits.

Ok thank you for your answers!

I also have question regarding my experiment and how I should calculate the ddCts.

I have 2 experiments.

1st experiment:

I have 3 cell lines (WT, carrier and patient). I have stimulation or no stimulation for each cell type.

When calculating ddCt, is it more appropriate to compare the stimulated to unstimulated WITHIN each cell type (dCt Wt stimulated - dCt Wt unstimulated) or compare everything against the same sample, say patient unstimulated?

2nd experiment.

I do transfection rescue of the patient cells only, with the plasmid (containing GOI) and empty vector as -ve control. So I have 2 groups. For each group, I have stimulation or no stimulation and do qPCR.

When calculating ddCt, is it more appropriate to compare the stimulated to unstimulated WITHIN each group or compare everything against same sample, say empty vector unstimulated?

Thank you!

Thank you Dr. Kubista for your kind answer!

Also, when is it better to graph ddCt values over fold change?

For example, I want to report the results like the following with 2^ddCt values:

- After stimulation, sample A had 40-fold upregulation whereas sample B had only 1.3-fold upregulation

- Here, I want to emphasize that the degree of upregulation is huge in sample A, but very small in B and I think this is more apparent in fold change.

Do you think it's still more appropriate to graph the ddCt values?