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!