pfu and Taq PCR mutation percentage
Gys de Jongh
GysdeJongh at compuserve.com
Tue Jan 18 18:34:26 EST 2000
Frank O. Fackelmayer <Frank.Fackelmayer at uni-konstanz.de> wrote in message
news:388445FA.C7D1D73E at uni-konstanz.de...
> Hi Gys,
> I like your discussion with Duncan, but I have some questions. When you
> calculate P(k) in the last post, you find that P(0)=0.92396. If I
> that right, it means that there is roughly a 92% chance to get error-free
> amplification, or, in other words, only 8% of all fragments will contain
> (or more) mistakes. Apart from a gut feeling that this percentage is way
> low, I don´t understand how your calculation via poisson distribution
> on the number of cycles. It is clear that mistakes accumulate, so the
> percentage of "wrong" fragments MUST depend on the cycle number.
> Be it as it may, can anyone of you answer the following practical
> If I have a 3000 bp fragment, how many doubling (cycles) may I perform to
> less than, say 5%, of error-containing fragments with Pfu or Taq. This
> allow us to think of a rule-of-thumb to estimate the amount of template to
> and the number of cycles to perform.
sorry for being not clear enough. The P(0) is the chance that the polymerase
copies *one_whole_strand* of the target just*once* with *no_errors_at_all*.
This chance can be calculated in the two different ways as I did in that
post . It is than the basis for calculating the error after each PCR cycle.
(I did not yet figure out how ; it just looks related to F via the poisson
distribution for me) This chance can not be far from 1 because at each cycle
it increases as some sort of a power function. To solve your problem
howerver , you can use the formula or the references Duncan mentioned.
f=-lnF/( d*b) Where f is also an error rate. It is the chance that the
polymerase copies *one_base* of the target *wrong* . F is the error fraction
in the final PCR product . d= the number of duplications which will allways
be a bit less than the number of PCR cycles. b is the target size. So in
your case 2.6 x 10E-5 = - ln (0.95) / (d x 3000) For Taq . If you want 5%
error containing fragments than you want 95% fragments with no error which
is a fraction of 0.95 As explained above there is only a fraction of 0.92
after just 1 error free duplication After solving : d=0.6576 ; less than 1 ;
as expected. The number of 0.6576 duplications can be maybe 2 or 5 cycles
depending on the actual PCR. Just like a Km determination : if you take
longer (extension) times or higher substrate concentrations the
chartrecorder would produce a longer straight line further above the zero. I
did not yet look up Duncan's references because I like to have some fun with
the problem myself first. So don't just believe me ; have a look at the
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