= PCR efficiency
{c}
{title2=$E_n$}
For a stipulated population of amplifiable copies, let $N_n$ be the amount after cycle $n$ and define <PCR efficiency> by the fractional new yield
$$
N_{n+1}=N_n(1+E_n),\qquad 0\leq E_n\leq1.
$$
Here $E_n=1$ means one additional copy per existing copy, hence ideal doubling. If every cycle has the same <PCR efficiency> $E$, the <exponential growth> model gives $N_n=N_0(1+E)^n$. Starting with $N_0=100$ copies, twenty ideal cycles give $104\,857\,600$ copies; the stipulated value $E=0.8$ gives about $12\,748\,236$. An efficiency of 80 percent therefore means multiplication by 1.8 each cycle, not retention of 80 percent of the final ideal yield.
\Image[/pcr-stipulated-efficiency-cycle-curves.png]
{title=Stipulated <PCR efficiency> models: ideal doubling, slower <exponential growth> and a <PCR plateau>}
{description=Original calculated curves, not laboratory observations or measured <fluorescence>. All begin at $N_0=100$. Blue has $E_n=1$; orange has $E_n=0.8$; green stipulates $E_n=0.9(1-N_n/K)$ with $K=10^8$. The left panel uses a <logarithmic scale> for copy amount; the right shows the assigned cycle efficiencies. Only the green model includes a finite ceiling. The illustrative ceiling and its equation are assumptions, not fitted chemical kinetics or a universal explanation of a plateau.}
In the ideal strand-counting example, all duplex-equivalent material doubles, but exact-length <amplicons> lag initially. Starting from one longer duplex and retaining its two original strands, cycle $n\geq1$ contains $2(2^n-n-1)$ exact-length strands out of $2^{n+1}$ total strands. Thus their fraction is $1-(n+1)/2^n$: zero after cycle 1, one quarter after cycle 2 and one half after cycle 3. This counts strands rather than assuming every strand immediately finds an exact-length partner. The generic yield model above ignores this early distinction.
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