A PCR amplification starts with 500 copies of a DNA fragment. Each cycle has an efficiency of 88 % (i.e., each cycle multiplies the number of copies by 1.88). How many copies will be present after 15 cycles? Give the answer in scientific notation rounded to two significant figures.

Step-by-step solution

1. Write the general formula for PCR copy number: $N=N_0\times (E{)}^n$, where $N_0$ is initial copies, $E$ is the efficiency factor, and $n$ is the number of cycles.
2. Substitute the given values: $N_0=500$, $E=1.88$, $n=15$.
3. Calculate the amplification factor: $(1.88{)}^{15}$.
4. Using a calculator, $(1.88{)}^{15}\approx 1.29\times 10^4$.
5. Multiply by the initial copies: $500\times 1.29\times 10^4=6.45\times 10^6$.
6. Round to two significant figures: $6.5\times 10^6$.

Concept tested: PCR efficiency calculation

Published 1 October 2026 · How this solution was prepared and checked