All you need is to notice that

• F(c)= p*N*Ec.

F = fluorescence intensity (fixed at threshold)

p = (unknown) proportionality factor (will cancel out)

N = concentration of the amplicon at the start of the PCR (needs to be known only relatively, see below)

E = amplification efficiency (to be determined)

c = cycle number (measured as cT at the threshold fluorescence FT)

At FT the cycle is cT. Solved for cT:

• log(FT) = log(p*N)+cT*log(E)
• cT =(log(FT)-log(p*N)) / log(E) = (log(FT/(p*N)) / log(E)

A dilution of factor D of a standard with initial concentaration N gives

• cT(D) = (log(FT/(p*N*D)) / log(E) =  (log(FT/(p*N) - log(D)) / log(E)

Now recognize that log(FT/(p*N) = constant = A:

• cT(D) =(A - log(D)) / log(E) = -1/log(E) * log(D) + A/log(E)

Now you should see how cT depends on the dilution factor D: the cTplotted against log(D) is a straight line with slope -1/log(E) and intercept A/log(E). Since the intercept contains A and A contains (several) unknowns (at least p, possibly also N), the efficiency can be best estimated using the slope. When m is the slope determined from the linear regression line, the efficiency is obtained by

• m = -1/log(E)  ---> log(E) = -1/m  ---> E = exp(-1/m)

("exp" being the inverse of the log. When the log was taken to the base b then E = b^(-1/m))

Note that, since E is positive (1

Hi. When you do it manually you just have to put your Ct values on the y-Axis and the log value of the template concentration on the x-axis. The slope of the resulting standard curve can be put in this calculator for efficiency http://www.thermoscientificbio.com/webtools/qpcrefficiency/

All you need is to notice that

• F(c)= p*N*Ec.

F = fluorescence intensity (fixed at threshold)

p = (unknown) proportionality factor (will cancel out)

N = concentration of the amplicon at the start of the PCR (needs to be known only relatively, see below)

E = amplification efficiency (to be determined)

c = cycle number (measured as cT at the threshold fluorescence FT)

At FT the cycle is cT. Solved for cT:

• log(FT) = log(p*N)+cT*log(E)
• cT =(log(FT)-log(p*N)) / log(E) = (log(FT/(p*N)) / log(E)

A dilution of factor D of a standard with initial concentaration N gives

• cT(D) = (log(FT/(p*N*D)) / log(E) =  (log(FT/(p*N) - log(D)) / log(E)

Now recognize that log(FT/(p*N) = constant = A:

• cT(D) =(A - log(D)) / log(E) = -1/log(E) * log(D) + A/log(E)

Now you should see how cT depends on the dilution factor D: the cTplotted against log(D) is a straight line with slope -1/log(E) and intercept A/log(E). Since the intercept contains A and A contains (several) unknowns (at least p, possibly also N), the efficiency can be best estimated using the slope. When m is the slope determined from the linear regression line, the efficiency is obtained by

• m = -1/log(E)  ---> log(E) = -1/m  ---> E = exp(-1/m)

("exp" being the inverse of the log. When the log was taken to the base b then E = b^(-1/m))

Note that, since E is positive (1

Hi,

You can get slope value(m) from your diluted cDNA curve. you have to make the standard curve by diluting the cDNA like 10-1, 10-2, 10-3, 10-4, 10-5. 

Slope value -3.3 is considered as 100% primer efficiency.

Suppose, you get the slope value 3.1 that means primer has 93.9% effiiency. Above 90% primer efficiency is acceptable.

Best wishes

Abhay

@Norbert Bittner  and Jochen Wilhelm

Thanks to both  of you for valuable answers

I am doing reverse transcription qPCR. I have 4 rat groups, 1st negative control, 2nd positive control, experimental drug treated group and reference drug treated group having 10 rats each. I have extracted RNA and synthesized cDNA for each sample. I wish to determine expression levels of Interleukin-1 cytokine in different groups. So, for efficiency testing, should I take the cDNA from one sample of each group or can i take cDNA from any sample regardless of group? would there be any effect on results if i do not calculate the efficiency and assume that it would be 2? Is it right if i assume that what ever the efficiency, it would not effect the results in this scenario? 

Arham,

the efficiencies for a given gene should be similar (actually identical) in all samples. Otherwise you will observe a lot of "noise" and it won't be simple to draw reliable concusions.(there are methods to account for sample-wise differenmces in efficiencies, but I am not convinced that they are improbing the results). Typically the efficiencies are not explicitely determined for each sample (it would be too much work). However, at least the parallelity of the log-linear parts of the amplification curves must be visually checked/confirmed. Another "trick" is to quantify each sample in different dilutions (instead of doing identical technical replicates). Then there is at least some data to estimate the efficiency for the individual sample based on ct-values and not on the slope of the log-linear phase of the amplification curve (what is difficult to determine and noisy and error-prone).

The efficiencies for different genes can be different. If this is the case, you have to account for this by the analysis method. Using calibration curves is a good option here. The ddCt-Method as describes by Livak et al. is applicable only when the efficiencies of all genes are similar (identical). The common efficiency should then be used to calculate the fold-changes (instead of simply using 2 as in Livals paper). So having an efficiency that is identical for all genes will not affect the conclusions but it will affect the estimated effect size ("fold change"). Unequal efficiencies will give completely erroneous results with the ddCt method. The "calibration curve method" will always give correct results but at the cost of performing all these calibration curves.

To check the efficiency using cDNA is often not a very good approach since you wil need relatively large amounts of the cDNA in the highest concentration, what is not always easy to obtain. I prefer generating a standard by a conventional PCR, possibly with outer primers, and use a (rather high) dilution as the standard to determine the efficiency. If you do this you must take greatest care not to get carry-over contaminations of PCR products!

To check if the sample (the RT mixture used as sample in the qPCR) itself does inhibit the PCR you can spike the standard to such samples.

Lot of information in a short text, I hope you understand everything.

PS: Apart from efficiency-considerations the selection of appropriate reference genes is often cruicial. Consider to use several reference genes right away.