Theoretical mass yield

The idea behind PCR is amplification. In every cycle, the amount of the PCR product theoretically doubles. So, for example, if you start with 1 fmole of template: after 1 cycle, there will be (1 fmole)*(2^1) = 2 fmoles of product; after 2 cycles, there will be (1 fmole)*(2^2) = 4 fmoles; after 3 cycles, there will be (1 fmole)*(2^3) = 8 fmoles of product, etc. [exponential growth] But be careful! This is assuming that you don't run out of any of the other ingredients necessary for the reaction! The key to your problem is figuring out which ingredient runs out first (i.e., what is the limiting reagent). To begin thinking about this, let's take a look at what is necessary in a PCR reaction.

So in a standard PCR, there is:

- Template DNA
- Forward and reverse primers
- dNTPs
- DNA polymerase
- Buffer containing Mg2+ to keep the polymerase happy

For instance, if we have:

- 0.10ng of a 3000bp template
   -> can be converted to moles using the approximation of 650g/mole per bp
- 1.20uM of each primer
   -> can be converted to moles using the fact that the total volume is 100uL
- 100uM of each dNTP (means 400uM dNTPs total)
   -> can be converted to moles using the fact that the total volume is 100uL

Now think about how much of each reactant gets used up in each cycle. Don't forget that the amount of template doubles in each cycle, and so the amount of reactant used up doubles for each cycle as well! This will allow you to figure out what the limiting reagent is. How many moles of product is synthesized for each mole of that reagent? Do you run out of reagent before or after the 30 cycles is complete?

Your initial answer will probably be in MOLES of product. To convert this back into grams (or MICROGRAMS, as the problem indicates), remember that your product is 600bp, and that each bp is 650g/mole.

Here is some excellent information on this website (Go three quarters of the way down to: Actual yield is less than the theoretical maximum)

http://escience.ws/b572/L3/L3.htm