PCR Amplification Simulator

Watch DNA copies grow exponentially through PCR cycles. Find the Ct threshold cycle.

Worked example — PCR Amplification Simulator

Final copies	1.07 × 10⁹
Fold amplification	1.07 × 10⁹×
Ct	27
Formula	1 × (1+1)n

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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C_t turns multiplication into subtraction 🖖

The chart's vertical axis is logarithmic, so this tool's pure exponential draws a straight line — and that is what makes the C_t readout worth anything. C_t is the cycle where the curve crosses your threshold, so multiplying the starting material slides the whole line sideways without changing its slope. Each tenfold change in C₀ moves C_t by ln 10 / ln(1+e) cycles: 3.32 at perfect efficiency, 3.59 at e = 0.9. Load the clinical preset and step C₀ from 10 to 100 to 1000 — C_t falls 18, 15, 11. Reading concentration backwards off that shift is the whole basis of quantitative PCR.

Not quite doubling every cycle 🖖

PCR works like a molecular photocopier: every cycle tries to duplicate each DNA strand in the tube. But no enzyme copies every template perfectly, so the efficiency e sets the true per-cycle multiplier (1 + e) — 2.0 when perfect, maybe 1.9 at 90%. Lower the efficiency in the tool and watch that small per-round shortfall snowball: over 30 cycles the gap between 1.9× and 2.0× becomes millions of missing copies.

Why PCR needed a Yellowstone microbe 🖖

Each cycle heats the tube to about 95°C to peel the two DNA strands apart — hot enough to cook an ordinary polymerase. Early PCR was miserable: technicians pipetted in fresh enzyme by hand after every single cycle. The rescue came from Thermus aquaticus, a bacterium thriving in Yellowstone's near-boiling springs, whose Taq polymerase shrugs off the heat — letting the entire reaction run untouched in one machine.

Problem solved in full

Thirty PCR cycles at perfect efficiency turning one molecule into a billion 5 steps
Thirty PCR cycles at perfect efficiency turn one molecule into a billion. Derive that, then find what a 5% drop in efficiency costs — the answer is not 5%.
1. Each cycle copies every strand present, so the population multiplies by 1 + e. Perfect efficiency means e = 1 and a straight doubling.
  
  $C_{n} = C_{0} (1 + e)^{n} = 1 \times 2^{30} = 1.07 \times 1 0^{9}$
2. Thirty doublings is 2³⁰, about 1.07 billion, from a single starting molecule. This is the whole reason PCR works: a quantity too small to detect becomes one too large to miss, in an afternoon.
3. The threshold cycle inverts the same relation — the cycle at which the count first crosses a detection threshold. It is a logarithm, which is why C_t values are compared by subtraction and why one cycle means a factor of two.
  
  $C_{0} (1 + e)^{C_{t}} = 1 0^{8} ⟹ C_{t} = lo g_{2} 1 0^{8} = 26.58 \to 27$
4. Now drop efficiency to 0.95, so each cycle multiplies by 1.95 instead of 2. After thirty cycles that is 5.02 × 10⁸.
  
  $e = 0.95 : 1.9 5^{30} = 5.02 \times 1 0^{8}$
5. The ratio is (2/1.95)³⁰ — a per-cycle shortfall of 2.5%, compounded thirty times.
  
  $\frac{2 ^{30}}{1.9 5 ^{30}} = (\frac{2}{1.95})^{30} = 2.14$
Answer

The tool prints 1.07 × 10⁹ copies and C_t = 27. A 5% efficiency loss costs a factor of 2.14, more than half the yield, because the loss is multiplied thirty times and not added once. That is the central difficulty of quantitative PCR: the readout is exponential in a quantity — efficiency — that varies between reactions with primer design, template purity and the machine's own thermal uniformity. It is why qPCR is always run against a standard curve rather than trusted as an absolute count, and why a single inhibitor in the sample can look exactly like less starting material.

$C_{n} = C_{0} (1 + e)^{n} : e = 1 \Rightarrow 1.07 \times 1 0^{9}, e = 0.95 \Rightarrow 2.14 \times fewer$

References (3)

Cₜ, amplification efficiency and how the two are reported: S. A. Bustin et al., "The MIQE Guidelines: Minimum Information for Publication of Quantitative Real-Time PCR Experiments." Clinical Chemistry 55(4), 611–622, 2009.
The real-time kinetic measurement the Cₜ readout models: R. Higuchi, C. Fockler, G. Dollinger and R. Watson, "Kinetic PCR Analysis: Real-time Monitoring of DNA Amplification Reactions." Nature Biotechnology 11(9), 1026–1030, 1993.
The thermostable polymerase named in the third block: R. K. Saiki et al., "Primer-directed enzymatic amplification of DNA with a thermostable DNA polymerase." Science 239(4839), 487–491, 1988.

By the 42Math team · Updated 2026-06-24

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Perfect (e=1) - Thirty doublings from a single molecule: 1.07 × 10⁹ copies, with the 10⁸ threshold crossed at cycle 27. That is the whole of PCR on this preset — no chemistry in the number at all, just 2³⁰.
95% efficient - Drop the efficiency from 1 to 0.95 and the same thirty cycles give 5.02 × 10⁸ instead of 1.07 × 10⁹. Five percent per cycle, compounded thirty times, costs you half the product. The threshold moves by a single cycle, 27 to 28 — which is why qPCR reports Ct rather than yield.
Clinical - Ten starting copies at 90% efficiency over forty cycles: 1.41 × 10¹² copies, Ct 18. Note that the fold amplification reads 1.41 × 10¹¹, ten times smaller than the copy count — it is measured against the ten molecules you began with, not against one.
Low efficiency - A hundred starting copies, but only 85% efficiency: 2.24 × 10¹¹ after thirty-five cycles, Ct 19. Ten times the template of the clinical preset and the threshold still arrives a cycle later, because that threshold is ten times higher and every cycle multiplies by 1.85 instead of 1.9.