• Usually, ∆Ct are calculated as the difference between the Ct values of the target gene and the reference gene within the same sample, so that each ∆Ct value is a measure of a "normalized expression" (normalized to a loading control; the expression of the reference gene serves as the loading control).

Thus: ∆Ct (treated) = Ct (treated, reference gene) – Ct (treated, target gene) ∆Ct (untreated) = Ct (untreated, reference gene) – Ct (untreated, target gene)

• Usually, gene expression changes are shown as log2(FC) (base is 2, not 10). It's not wrong to take the base 10, but it is uncommon. If you calculate the ∆Ct values as shown above, ∆∆Ct = mean(∆Ct (treated)) - mean(∆Ct (untreated)) already *is* the mean log2(FC).

• You can use the ∆Ct values as calculated above to run t-tests or ANOVAs.

• In your presentation you should ideally show the individual ∆Ct values. If you have +20 ∆Ct values per group, it may be better to show them as box-and-whisker plots. Showing means and 95%CIs makes sense for mean ∆∆Ct values.

• It is simplest to run the stats on the ∆Ct values. If, for instance, you do a t-test of ∆Cts from a treated sample against ∆Cts from an untreated sample, any usual software will give you the mean ∆∆Ct (the mean difference of the ∆Cts between the groups) together with its standard error, its 95%CI, and a p-values testing the hypothesis that the mean ∆∆Ct = 0.

• Usually, ∆Ct are calculated as the difference between the Ct values of the target gene and the reference gene within the same sample, so that each ∆Ct value is a measure of a "normalized expression" (normalized to a loading control; the expression of the reference gene serves as the loading control).

Thus: ∆Ct (treated) = Ct (treated, reference gene) – Ct (treated, target gene) ∆Ct (untreated) = Ct (untreated, reference gene) – Ct (untreated, target gene)

• Usually, gene expression changes are shown as log2(FC) (base is 2, not 10). It's not wrong to take the base 10, but it is uncommon. If you calculate the ∆Ct values as shown above, ∆∆Ct = mean(∆Ct (treated)) - mean(∆Ct (untreated)) already *is* the mean log2(FC).

• You can use the ∆Ct values as calculated above to run t-tests or ANOVAs.

• In your presentation you should ideally show the individual ∆Ct values. If you have +20 ∆Ct values per group, it may be better to show them as box-and-whisker plots. Showing means and 95%CIs makes sense for mean ∆∆Ct values.

• It is simplest to run the stats on the ∆Ct values. If, for instance, you do a t-test of ∆Cts from a treated sample against ∆Cts from an untreated sample, any usual software will give you the mean ∆∆Ct (the mean difference of the ∆Cts between the groups) together with its standard error, its 95%CI, and a p-values testing the hypothesis that the mean ∆∆Ct = 0.

Honestly, I would always advice not using ddCt methods at all, because you run into these problems constantly, and keeping everything as log values makes it incredibly easy to miss outliers, very easy to make a mistake that isn't immediately obvious, and incredibly easy to lose track of what your data is telling you. If you get "-0.3, 1.7, -1.2" at the end of it, are you going to know what those values represent?

Plus it assumes all your qPCR efficiencies are exactly 2 (i.e. 100% efficient). They may be, but if not, this method is invalid.

Still, bonus points for using multiple reference genes.

Anyway, things to remember: the reference gene values are PER SAMPLE, and should be used to normalise THAT SAMPLE. The ref gene values for a treated sample have literally nothing to do with the ref gene values for an untreated sample other than as a straight comparator, and should not be combined (or at least, not without extreme caution).

I would convert all my data to relative quantities (RQ), such that for each gene, all the sample values are between 0 and 1, and are linear (so a sample with 0.5 has exactly half the cDNA of a sample with 1).

The equation for this is

efficiency (lowest Cq -sample Cq)

Which allows you to factor efficiency into the comparison, and also allows you to immediately spot outliers.

You can then use the RQ values for the reference genes to generate a normalisation factor (geometric mean of ref values), so:

NFsample1 = cuberoot(RQsample1(ref1)*RQsample1(ref2)*RQsample1(ref3))

And then simply calculate your final values by dividing your RQ for the target gene by the normalisation factor.

These final (normalised) values are the relative quantities of target gene in your treated and untreated samples, and can be compared using normal statistical methods, and presented using normal presentation methods exactly as you would present any linear comparison.

Remember: the numbers you get are arbitrary. qPCR isn't telling you HOW MANY transcripts you have, it's telling you the difference in transcript content between samples. Statistical tests don't care either: an ANOVA using 0.1, 0.2, 0.14, 0.5 vs 0.7, 1.2, 0.4, 0.7 will give you the same result as one using 1000, 2000, 1400, 5000 vs 7000, 12000, 4000, 7000.

I've attached a sample excel file taking the same data and using dCt methods and RQ methods: the results are the same, but the latter method is simply less prone to confusion.

Always use means of data plus SD not converted means.

You can simply average the Ct values of the two reference genes to get the "reference Ct": ∆Ct = mean(Ct of all ref. genes) - Ct(gene of interest).

The authors write: "the Ct values of each sample were normalized against the geometric mean Ct values of housekeeping genes (RPS11 and RPL13)."

So if you want to do it that way, just exactly do that. However, I don't recommend following this because this is simply wrong. It makes no sense to take a geometric mean of Ct values. (Apart from that I don't know what that CFX Manager 3.1 does, and it is not described, and the software is a proprietary black box).