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Say I have one dsDNA that undergoes normal PCR (where amplification is exponential). If there is a mistake, say a G is swapped for an A during the 2nd round of replication, what percentage of the final DNA will have the mistake if there are 12 more rounds?

I was talking to my teacher about this today. I say 33%, my teacher said 12.5%.

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I will format it to .ppt as soon as I have more time! If something is not readable, please let me know!

I have to point out several things:

  1. The fraction of incorrect DNA molecules does not depend on the number of cycles (as long as the number of cycles is higher than 2).

  2. There is one exception: if the mutation occurs outside of the region to be amplified (for example at the 4th molecule, downstream of the region of interest), the total number of wrong DNA molecules will be 0.

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Your teacher is indeed correct.

In the first round you would get two identical molecules of the dsDNA.

In the second round you would get 3 identical molecules and one molecular with an A substituted for a G in one of the strands. ie.

No error (3 of the 4 molecules):

------G-------
------C-------

One mismatch (1 of the 4 molecules):

------A-------
------C-------

So there are a total of 8 strands of DNA after the second round and one of those strands has the mismatch. 1 / 8 = 0.125 = 12.5%

In round 3 you would have 8 dsDNA molecules and only one of those 8 dsDNA molecules would have the mismatch.

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