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 generated?
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- 1.024 × 104
- 1.024 × 105
- 2.048 × 104
- 2.048 × 105
Correct Option: B
Let N0 be the initial number of molecules present in the aliquot and let N be the end product of PCR amplification. Let n be the number of cycles performed by the PCR.
We, know N = N0 × 2n for n number of cycles since in each cycle the number of amplicons double the number of parent strands in an exponential manner.
Here, N0 = 100 for a 200 μL solution (given), n = 10
Therefore, N = 100 × 210
= 100 × 1024 = 1.024 × 105
Hence, the required amplicons generated after 10 cycles of PCR run = 1.024 × 105
