A 200 µl of polymerase chain reaction has 100 template DNA molecules and the reaction was performed for 10 cycles.
Q.55 How many molecules of amplicons will be present in 0.1 µl of reaction?
✅ Correct Answer: 51.2 molecules (Option C)
Quick Summary: PCR starts with 100 template DNA molecules in 200 µl, doubling per cycle over 10 cycles to yield 102,400 amplicons total. In 0.1 µl, that’s 51.2 molecules.
🔬 PCR Calculation Breakdown
The standard PCR amplification formula is: N = N₀ × 2ⁿ
2¹⁰ = 1024
N = 100 × 1024 = 102,400 amplicons in 200 µl
Volume fraction: 0.1 µl / 200 µl = 0.0005
Final calculation: 102,400 × 0.0005 = 51.2 molecules
❌ Option Analysis Table
| Option | Value | Common Mistake |
|---|---|---|
| (A) | 102.4 | Divides total by 1000 (wrong volume ratio) |
| (B) | 1024 | Only 2¹⁰, forgets × 100 initial templates |
| (C) | 51.2 | ✅ CORRECT: Full formula + volume adjustment |
| (D) | 512 | Uses 2⁹ (assumes only 9 cycles) |
📚 SEO Optimized Article
In polymerase chain reaction (PCR), DNA templates double exponentially each cycle, making precise yield calculations essential for biotechnology applications, GATE exams, and molecular biology research. This detailed guide solves the classic question: “A 200 µl PCR reaction starts with 100 template DNA molecules run for 10 cycles—how many amplicons in 0.1 µl?”
Step-by-Step Solution
- Total amplicons: N = 100 × 2¹⁰ = 102,400 in 200 µl
- Volume ratio: 0.1/200 = 5 × 10⁻⁴
- Final answer: 102,400 × 0.0005 = 51.2 molecules
Real-World PCR Considerations
- Theoretical doubling assumes 100% efficiency
- Actual yields plateau after 25-35 cycles
- Low inputs (100 molecules) work well for 10 cycles
- Essential for qPCR, cloning, and diagnostics
GATE Exam Tips
Biotechnology students preparing for GATE BT must master PCR calculations. Common traps include forgetting initial template count or incorrect volume scaling. Practice with different cycle numbers and volumes for exam success.
