Newbie question about theoretical PCR yield equation

Go to the profile of zlikowski
Apr 06, 2018

Hello everyone,

First of all, my apologies if this is the wrong thread. I am a total newbie (both theory and practice) in qPCR, but I have started reading several books/articles about the method. Of course, I have immediately encountered a simple equation that I do not understand, so I would greatly appreciate if any of you guys could help me out. Let's start, and sorry for an extremely long post. Ok, so if I understood correctly, a theoretical yield of a general PCR reaction (Np=number of product molecules) after n cycles, and starting with No molecules of the template, is given by:

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(1)      Np=No x 2^n

assuming that the template length = product length (for simplicity not going here with cases in which the template is larger that the desired product (the 2^n-2n equation). This means that, starting with a single molecule of template, after 1st cycle we have 2 molecules of the product, after 4 cycle 4 molecules, etc. Another important assumption in this equation is that the efficiency of each PCR cycle equals 100% or 1. Ok, so far so good. Now, let's incorporate the efficiency less than 100% into this story. When one looks up how the efficiency of chemical reactions (E) (even enzymatically catalyzed reactions, such as PCR) is expressed, usually one finds that the E is expressed as a percentage yield:

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(2)   E=(actual yield)/theoretical yield

Applying this to PCR, if we start a PCR with 1 molecule of the template and after one perfectly performed cycle we obtain 2 molecules of the product, efficiency of our reaction, according to (2), is 1:

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(3)  E=(actual yield)/(theoretical yield) = 2 / 2 = 1

But what happens if the efficiency of each cycle is, say, 0.9. According to (2) this means that after first PCR cycle we obtain 1.8 molecules of the product:

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(4)   Np (after first cycle) = 1 x 2^1 x 0.9 = 2 x 0.9 = 1.8 DNA molecules

In the second cycle, we again double the amount of present DNA molecules, again with the efficiency of 0.9:

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(5)   Np (after second cycle) = No. of DNA molecules after 1st cycle x 2 x efficiency = 1.8 x 2 x 0.9 = 3.24 DNA molecules

In the third cycle, we again double the amount of DNA molecules in the reaction mixture, again with the efficiency of 0.9:

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(6)   Np (after third cycle) = No. of DNA molecules after 2nd cycle x 2 x efficiency = 3.24 x 2 x 0.9 = 5.832 DNA molecules

More generally, if we want to incorporate efficiency of each cycle into the equation (2), we obtain (without proof by mathematical induction, but you can perform the exercise for 30-40 cycles):
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(7)  Np=No x (2E)^n

My question here is the following. Why is this wrong? Namely, in a large number of papers and books about PCR, qPCR, real-time PCR, you name it, the PCR is described by the following equation:

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(8)  Np = No(1 + E)^n

which is quite different from the equation (7). Obviously, I have made a mistake in my reasoning somewhere, and I would greatly appreciate you input here. I went through at least a two dozen RT-PCR and regular PCR papers which only cite this equation, and I have been unable to find an original source in which equation 8 was derived. I have also been quite confused with the definition of PCR efficiency in the current literature. For example, Svec et al. (How good is a PCR efficiency estimate: Recommendations for precise and robust qPCR efficiency assessments, 2015) write this sentence in their paper:

"For example, let say a test tube contains 100 target molecules and after one amplification cycle it contains 180 molecules, E = 80%, since 80% of the target molecules present were amplified."

But, if the tube contains 180 molecules after one round of amplification, this means that 90 molecules were amplified, not 80, right? And E should then be 90%, and not 80%, as the authors state. What am I doing wrong here? Is efficiency defined differently in PCR? Does someone knows the reference in which equation (8) has been derived? Is my reasoning for theoretical yield of a PCR reaction, which incorporates efficency of each cycle (eq 7) wrong? I mean, I know it is wrong for late stages of PCR, but so is eq (8). Please help, and once again thanks for reading a long post!


Go to the profile of Roberto Rosati
Roberto Rosati 9 months ago

I think you can fix the apparent divergence if you consider "yield" not the total number of molecules at the end of the cycle, but the number of molecules *produced* during the cycle (as it should be).
So if you had 100 molecules, and you get 180 total at the end of the cycle, it means that 80 of those 100 molecules were amplified. Hence, 80% efficiency.
If as you say, you had amplified 90 molecules, you'd have 190 in total at the end of the cycle, because you already started with 100.

Go to the profile of zlikowski
zlikowski 9 months ago

Thanks, that solved it, and makes the derivation of eq 8 a breeze. Much obliged!