# Basic question about multiplex PCR

Let's say I have a DNA sequence with the following structure:

$$5' - N_n - S_1 - N_{1000} - S_2 - N_{1000} - S_3 - N_n - 3'$$

Here, the $N$s represent stretches of arbitrary sequence of the indicated length. $S_1$, $S_2$ and $S_3$ are all unique sequences 20 bp in length.

I design a forward primer PF that anneals to the positive strand at $S_1$. I design two reverse primers PR1 and PR2 that anneal to the negative strand at $S_2$ and $S_3$ respectively. Their annealing temperatures are reasonably close.

Clearly, PF and PR1 together will produce a 1040 bp PCR product, and PF with PR2 will produce a 2060 bp product.

Now let's say I add 0.2 μM PF (standard concentration), 0.1 μM PR1 and 0.1 μM PR2 all in the same reaction. I then run a PCR, with extension time sufficient for the 2 kb product.

I can see the following possible outcomes:

• I get a 50/50 mix of 1 kb and 2 kb products.
• PR1 competes aggressively and I get hardly any 2 kb product and a lot of 1 kb product.
• After a few cycles, the 1 kb product that has accumulated serves as an efficient forward primer for PR2, and I end up getting mostly 2 kb product and little 1 kb product.

Which one will happen? If it depends, what are the major factors on which it depends? Primer concentrations, extension time, annealing temperature, dNTP amount, number of cycles?

This question is basically a very simplified multiplex PCR. While much has been written about practical applications of multiplex PCR, I am hoping that answers will help me understand the considerations that go into deducing optimal multiplex PCR conditions without doing any empirical calibration. This isn't to circumvent the calibration, but to understand the process conceptually.

I haven't done this exact experiment. I am just deducing from the known facts about PCR.

The product of PF and PR2 will serve as a template for both 1kb product (P1) and 2kb product (P2).

Lets assume that after 2nd cycle there is 1 copy each of P1 and P2 (Delay of 1 cycle to make a smaller length product. See here).

Lets assume that the primer binding affinity is equal (and for just preliminaries consider that only one primer can bind a template — such as in overlapping binding sites). So half of P2 will bind to PR2 and half to PR1.

Number of copies at nth cycle is:

P2(n) = 1.5 × P2(n-1)

P1(n) = 2 × P1(n-1) + 0.5 × P2(n-2)

However this is not really the scenario; the primers are not competing for the template. Which means that both primers can bind to the substrate simultaneously. DNA polymerases (except Klenow) will remove any DNA strand bound to the template, ahead of it. In that case P1 will not form. But when PR2 runs out, the unused PR1 can produce P1.

Now there are additional factors, such as the distance between PR1 and PR2 binding sites. If the distance between the primer binding sites is less than the smaller product then the abovementioned case occurs. If the inter-primer distance is larger than smaller product then both products will form (and just like the competing primers case, P2 will serve as a template for both products). In this case the copy numbers will be:

P2(n) = 2 × P2(n-1)

P1(n) = 2 × P1(n-1) + 1 × P2(n-2)

This means that P1 will dominate.

Other assumptions:

• No limiting concentrations of dNTP, primers etc
• PCR efficiencies are equal for both primers and is equal to 2 (in non-competing cases)
• Primer annealing temperatures are same i.e binding affinities are same.
• you say DNA polymerases (except Klenow) will remove any DNA strand bound to the template do you mean DNA Polymerases have 5'-3' exonuclease activity? or helicase activity? if it'S the second, then both products should be preserved, right? – TheChymera Jun 18 '15 at 12:39