Equation for accurate prediction of PCR yield
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.