Direction: A 200 µl of polymerase chain reaction has 100 template DNA molecules and the reaction was performed for 10 cycles.
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How many molecules of amplicons will be present in 0.1 µl of reaction?
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- 102.4
- 1024
- 51.2
- 512
Correct Option: C
From previous question, we have
Number of amplicons in 200 μL of the solution after PCR run = 1.024 × 105
Therefore, number of amplicons in 0.1 μL solution = 1.024 × 105 × 0.1 / 200 = 51.2
Hence, the required number of amplicons present in 0.1 μL of the solution = 51.2
