Hello,

I skip all theoretical aspects behind.

Assuming that you want to calculate relative expression ratio between two samples (i.e. treated vs control), this formula includes a correction for different PCR efficiency of the individual amplicons.

Expression Ratio = [(E tar) ^deltaCp tar ] / [(E ref) ^deltaCp ref]

with

E = PCR efficiency

tar = target gene, ref = reference gene (ex. GAPDH)

delta Cp = difference in Cp between two samples (i.e. treated vs control)

The algorithm is implemented in REST software, which holds multiple reference genes and performs a randomisation test for statistical significance of differential expression.

Link to REST

http://www.gene-quantification.de/rest.html

Literature references

Pfaffl et al. 2002. Nucleic Acid Research, vol 30 (9): e36.

A deep theoretical and practical explanation of quantification strategies in real-time RT-PCR can be find here

http://www.gene-quantification.de

When the reaction with efficiency a reaches the threshold T after c cycles, the initial concentration N0 is given from the exponential equation

N0*a^c = T

Simpler to log this:

log(N0) + c*log(a) = log(T)

The threshold cycles is then

c = (log(T)-log(N0)) / log(a)

Now consider a reaction 1 that gives a ct-value = c1 for an efficiency of a1:

c1 = (log(T)-log(N0)) / log(a1)

For the efficiency a2, the reaction 2 gives a ct-value of

c2 = (log(T)-log(N0)) / log(a2)

For your question, you know a1 and a2 and you measured c1, and you wand to claculate the "corrected" c2. Example: Your target gene was amplified with a1 and gave ct=c1. Your reference gene was amplfied with a2. In order to compare the ct values of the target and the reference gene, ct-values from both genes must be obtained with the same efficiency a. Thus you want to get the corrected value for c1 that would have been obtained if the same sample would have been amplified with a2. This is obtained by combining tha last two equations, that both share (log(T)-log(N0)):

(log(T)-log(N0)) = c1*log(a1) and (log(T)-log(N0))= c2*log(a2)

Setting equal:

c1*log(a1) = c2*log(a2)

and thus

c2 = c1*log(a1)/log(a2)

That's it. c2 is the "corrected" ct-value, that would have been measured if the sample would have been amplified with the efficiency a2.

The problem is the error propagation. Ich you estimate a1 and a2 from the data you have (so do obviousely do have reasons to assume that a < 2.0 in the first cycles where c < ct), you will have some uncertainty about the correct values of a1 and a2.

Let s be the standard errors: sc for c1, sa1 for a1 and sa2 for a2. We can derive the stadard error for c2 fron Gauss error propagation. The partial derivatives are (if I did not make some mistakes!):

d/d(c1) = log(a1)/log(a2)

d/d(a1) = c1/(log(a1)log(a2))

d/d(a2) = c1*log(a1)/(log(a2)³)

and s2 will be

s2 = sqrt( sc²*log(a1)²/log(a2)² + sa1²*c1²/(log(a1)²log(a2)²) + sa2²*c1²*log(a1)²/(log(a2)³)² )

To give an example:

sc = 0.1 (quite good, I think)

a1 = 2.0

a2 = 1.8

sa1 = sa2 = 0.01 (also not so bad)

c1 = 25

s2 = sqrt(...) = 1.06

So although all parameters have been determined quite precisely, the standard deviation of the corrected ct value so large that no reasonable inference can be made. When the standard error of a1 and a2 were just 0.02, the standard error of c2 would already be more than 2 cycles!

All given that I did not make any mistakes (feel free to correct me).

Hello,

I skip all theoretical aspects behind.

Assuming that you want to calculate relative expression ratio between two samples (i.e. treated vs control), this formula includes a correction for different PCR efficiency of the individual amplicons.

Expression Ratio = [(E tar) ^deltaCp tar ] / [(E ref) ^deltaCp ref]

with

E = PCR efficiency

tar = target gene, ref = reference gene (ex. GAPDH)

delta Cp = difference in Cp between two samples (i.e. treated vs control)

The algorithm is implemented in REST software, which holds multiple reference genes and performs a randomisation test for statistical significance of differential expression.

Link to REST

http://www.gene-quantification.de/rest.html

Literature references

Pfaffl et al. 2002. Nucleic Acid Research, vol 30 (9): e36.

A deep theoretical and practical explanation of quantification strategies in real-time RT-PCR can be find here

http://www.gene-quantification.de

Noteably, since two pairs of errors are perfectly correlated and subtracted, the resulting error in this formula is considerably lower. I have no derivation of a proper estimation but only a simulation. For a standard deviation of >0.2 of replicate cts and a differenec of efficiencies of 2.0 vs. 1.8, the correction is not critical. For a standard deviation of 0.1 cycles, the log fold-change is over-(under-)estimated by approx. 0.1 log2's (what still is not quite much) when no correction is done. The standard errors for the slope of the standard curves was approx. 0.15 (7 dilutions in duplicates); ct-values of target and reference genes in treated and control samples were also taken from duplicates.

Perhaps, we, at times, become lost in the truth and beauty of these calculations...?

Although I believe them full-heartedly - I also feel that math and mother nature become arch-nemeses on occasion.

My attempt at reconciling the possibly irreconcilable is attached.

Lars, Issaco, Jochen: Great post!!!

Jochen - I am a fan.

My inclination is to pursue droplet digital PCR... but, qPCR/RT-qPCR math is too beautiful to leave behind. There are so many excellent things about it. It can make a non-mathematical person accidentally love math. So, qPCR and RT-qPCR it still is for me. Until a windfall draws the curtains otherwise.

ddPCR might also have problems we do not yet foresee. I remember the good old days of qPCR where things like the efficiency issues and the background problem (for instance) were still unknown, unrecognized, or just ignored :)

And, Jack, you can enter the beauty of Poisson regression models with ddPCR, possibly leading towards negative binomials and zero-inflated models... ;)