pfu and Taq PCR mutation percentage

[Gys de Jongh]

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