Biology · Introductory biology · Worked example
Calculate DNA copies after PCR cycles
A PCR starts with 500 copies of the target. How many copies would 25 cycles make at 100% efficiency, and at 90%? How many ideal cycles take the count past one billion?
Perfect efficiency
At E = 1 every copy is duplicated each cycle, so 25 cycles multiply 500 by 2²⁵ = 33,554,432.
90% efficiency
At E = 0.9 the factor per cycle is 1.9, so 25 cycles give 500 × 1.9²⁵ ≈ 4.65 × 10⁹ copies.
Compare the two
A 10% shortfall in each cycle compounds: (2/1.9)²⁵ ≈ 3.6, so the less efficient reaction makes under a third as much product.
Cycles to pass one billion
Starting from 500 copies, 2ⁿ must pass 2 million. Twenty cycles fall short and 21 are enough.
State the assumption
These counts assume the same efficiency in every cycle. Real reactions plateau, so the numbers describe the exponential phase only, which is why quantitative PCR measures the product during every cycle.
Result
About 1.68 × 10¹⁰ copies at 100% efficiency and 4.65 × 10⁹ at 90%; 21 ideal cycles take 500 copies past one billion.
Your turn
How many ideal cycles take a single copy past one million?
Show the answer and explanation
20 cycles.
2¹⁹ = 524,288 is short of a million, and 2²⁰ = 1,048,576 passes it.
Keep exploring
The PCR tool opens with 500 starting copies, 25 cycles and an efficiency of 0.9, and projects about 4.65 × 10⁹ copies. Set the efficiency to 1 to compare.
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