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During a PCR amplification, the efficiency of each cycle is 85 %. Starting with 100 copies of a target DNA fragment, how many copies will be present after 20 cycles? Express the answer in scientific notation.

medium 3 marks Class 12 · Biology · Biotechnology Principles and Processes
Answer
8.28×10^8

The correct answer is 8.28×10^8 — Step 5: Final answer is 8.28×10^8 copies.

Step-by-step solution

  1. Step 1: Write the growth factor per cycle: 1 + 0.85 = 1.85.
  2. Step 2: Express the number of copies after n cycles as N = N0 × (growth factor)^n, where N0 = 100 and n = 20.
  3. Step 3: Compute (1.85)^20 using logarithms or a calculator: (1.85)^20 ≈ 8.28×10^6.
  4. Step 4: Multiply by the initial 100 copies: 100 × 8.28×10^6 = 8.28×10^8 copies.
  5. Step 5: Final answer is 8.28×10^8 copies.
Common mistake: Assuming 100 % efficiency or using 0.85 instead of 1.85 as the growth factor

Concept tested: PCR efficiency calculation

Published · How this solution was prepared and checked

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