It is a cliche of freshman biology labs to point out that "every cycle of PCR doubles the DNA, so the yield will be $2^{cycles}$ times the template amount". However, if this were true, 1 ng of template would generate about 35 billion ng after 35 cycles, or 35 grams of DNA. This is clearly absurd and not the case.

Of course, the power-of-2 claim is a gross oversimplification (if anything, it is an upper bound - but even so, a very uninformative one), and in practice, yields will fall far short of it because:

  • Every single duplex of DNA does not denature at each cycle
  • Primers do not bind to every single molecule of DNA at each cycle
  • Not every DNA strand gets bound by a polymerase at every cycle
  • Not every polymerase that binds manages to complete the entire product in time in every cycle
  • The reaction inhibits itself by depleting dNTPs
  • The heat denatures the reaction by degrading enzyme

In fact, cursory examination of qPCR output often follows saturation kinetics:

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Mathematical methods for modeling qPCR are obviously well developed.

My question is about ordinary PCR: Is it possible get a reasonable expectation of nanogram yield for an ordinary PCR done in a tabletop cycler, with typical PCR reagents?

For instance, when amplifying from a plasmid, I would like to calculate how many cycles to do, how much template to use, and how much product to load on the agarose gel to ensure that I will be able to clearly distinguish exponential amplification (both primers anneal), linear amplification (only one primer anneals), and no amplification (neither primer can anneal or the reaction did not work).


2 Answers 2


An expected efficiency for a typical PCR is 80%, meaning each cycle multiplies the copy number of the targeted DNA sequence 1.58 times.

Firstly, it makes more sense to refer to the amount of DNA in a polymerase chain reaction in terms of copy number or in terms of moles; the number of DNA molecules of interest is what the reaction is operating on, and the mass of product generated is a function of the length of the product (and, to a lesser degree, on the composition of the product).

The following discussion is sourced from this URL: https://www.csun.edu/~hcbio027/biotechnology/lec3/pcr/p.htm

According to Perkin-Elemer, copy-number amplification of 100,000 fold of the targeted sequence of DNA can be expected from a PCR with 0.1 ng of Lambda phage DNA (a well-characterized and standard DNA isolate) in a 100 µL reaction with > 25 cycles of denaturation, annealing, and extension.

In the above 100,000-fold amplification example, if the targeted amplicon were to be 500 bp in length, the estimated molecular-weight of duplex DNA of 500bp is 325,000 g/mol (based an average base-pair having a molecular mass of 650 g/mol).

The Lamdba Phage genome is 42,502 base-pairs in length. 42,502 bp × 650 grams/mol/bp = 2.762×10^7 grams/mol.

0.1 ng Lambda DNA -> 0.1×10^-9 grams. 0.1×10^-9 g ÷ 2.762×10^7 g/mol = 3.619 × 10^-18 moles. 3.619 × 10^-18 moles × NA (Avogradro's Number) = 2.179×10^6 copies, or 2,179,000 copies.

2.179×10^6 copies × 100,000 = 2.179×10^11 copies. 2.179×10^11 copies ÷ NA × 325,000 g/mol = 1.176×10^-7 grams of sequence. 1.176×10^-7 grams is equal to 0.117 µg or 117 ng.

An amplification yield of 100,000x after 25 cycles would mean at each cycle 1 template would yield 1.58 templates for the next round of synthesis.

How was this calculated? If c is the number of copies made per round of synthesis, then:

c^25 = 100,000 = 10^5
so c^5 = 10
and so 5(log c) = log 10 = 1
so log c = 0.2 and c = 1.58 (approximately)

(Or you could calculate the 25th root of 100,000 on a calculator, if you prefer.) 

If we obtain 1.58 copies instead of the theoretical maximum of 2 copies, then the efficiency of the reaction could be said to be 79% (because 1.58/2.00 = 0.79).

One reason this calculation is important is that a slight loss of efficiency is magnified through the amplification. A reaction may appear to have not worked if the efficiency drops (in each cycle) by just a few percent. Optimization is critically important in the polymerase chain reaction.

  • $\begingroup$ Here is a useful website for calculating the molecular mass and number of copies of double-stranded DNA molecules based on their length in base-pairs. cels.uri.edu/gsc/cndna.html $\endgroup$
    – Patrick
    Jan 12, 2016 at 20:08

The equation is correct, but there's an additional asymptotic limit to a maximum concentration of product depending on the starting concentration of NTPs, template and primer pairs in solution too.


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