pfu and Taq PCR mutation percentage
Gys de Jongh
GysdeJongh at compuserve.com
Wed Jan 19 18:17:20 EST 2000
Gys de Jongh <GysdeJongh at compuserve.com> wrote in message
news:862t5t$4mi$1 at ssauraab-i-1.production.compuserve.com...
Hi,
I think I found an analytical solution for the error propagation in a
symetrical PCR.
Symbols,etc
right : does not contain any error.
wrong : contains more than zero errors ; contains 1 or *more* errors.
Right molecules can only be copied from other right molecules ; a copy of a
wrong molecule will allways be be wrong.
A,B : to distinguish the forward from the reverse strand.
n = PCR cycle number.
b=number of bases in the ss DNA molecule ; the number of base pairs in the
ds DNA molecule
N(n) = The number of ds DNA molecules at the completion of PCR cycle number
n ; equals the number of ss DNA molecules of strand A. Equals the number of
ss DNA molecules of strand B. Because we only consider symetrical PCR.
p1w = the polymerase error rate ; the chance for incorporating 1 base wrong.
p1r = the chance for incorporating 1 base right.
pssMr = the chance that the ss DNA molecule of b bases is copied ( *only 1
time* ) right.
FssMr(n) = the chance that the ss DNA molecule of b bases is copied right
after the completion of n PCR cycles.
FdsMr(n) = the chance that the ds DNA molecule of b bases is copied right
after the completion of n PCR cycles.
CF = Copied Fraction , is between 0 and one. The fraction of the template
molecules that is copied.
First calculate pssMr :
The chance for incorporating 1 base right = p1r = (1 - p1w ) , by
definition. If a ss DNA molecule of b bases must be copied (*only 1 time*)
right than the first base must be right ; p1r = ( 1 - p1w ) AND the second
base must be right ; p1r = ( 1 - p1w ) and so on upto the last b base. So b
events with chance (1 - p1w ) must *all* happen. From chance theory it
follows that : pssMr = ( 1- p1w ) ^ b
In a symetrical PCR pssMr is the same for the A strand and the B strand.
At the start of cycle 1 there are N(0) ds DNA molecules of b base pairs with
no errors at all. After melting there are N(0) ss DNA molecules of the A
strand and N(0) ss DNA molecules of b bases of the B strand ; both with no
errors at all.
The amplification of the A strand :
The N(0) ss DNA molecules of the B strand will be copied to ss DNA molecules
of the A strand. However only a fraction , CF , of the B strand molecules
will be copied given the finite extension time and reaction velocity of the
enzyme. Of this fraction only a second fraction , pssMr , is copied right.
So there where N(0) right ss DNA molecules at the start of cycle 1. After
cycle 1 there will be the original N(0) right ss DNA molecules plus pssMr *
CF * N(0) new right ones synthesized from the right B strand as template.
Or after the completion of cycle 1 there are :
N(0) + pssMr * CF * N(0) = N(0) * (1 + pssMr * CF) right A strand
molecules. This number will be the input for the next cycle where the
process is repeated. Thus the number of right ss DNA molecules of b bases of
the A strand will be :
N(0) * (1 + pssMr * CF) ^ n after completion of cycle number n.
In a similar way we find the total number of ss DNA molecules :
N(0) * (1 + CF) ^ n ( right or wrong ) after completion of cycle number n.
Division gives the fraction of right ss DNA molecules of b bases of the A
strand after completion of cycle number n :
FssMr(n) = [ (1 + pssMr * CF) / (1 + CF) ] ^ n
which is , of course , also the chance that we pick 1 right ss DNA molecule
of b bases of the A strand from the reaction mix after completion of cyle
number n.
Observe that :
1) if the reaction is driven to completion , thus CF = 1 , than the total
number will increase as
N(0) * (1 + 1) ^ n = N(0) * 2 ^ n
2) if also the duplication is perfect , thus pssMr = 1 than the number of
right molecules will increase as :
N(0) * (1 + 1 *1) ^ n = N(0) * 2 ^ n
3) The fraction of right molecules is *not* a linear function of the cycle
number.
As the PCR is symetrical the same is true for the B strand.
The amplification of the ds DNA molecule :
If we are interested in right ds DNA molecules than at the condensation of
the ss DNA molecules we must first pick a right A strand AND than a right B
strand. The chance for this event will be the product of these , equal ,
chances. This will of course also be the fraction of right ds DNA molecules
after the completion of cycle number n.
So : FdsMr(n) = FssMr(n) ^ 2
Observe that in this model a combination of 1 right A strand and 1 B strand
with only 1 error base will be considered wrong.
If we put all this in a spreadsheet the influence of the various parameters
can be seen. FssMr(n) can be very good approximated by a linear function of
n with some "real life" PCR parameters. So there are usefull approximations
of the problem.
--
Gys
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