There are some other unknowns here... and subtleties... but I, like Jochen (and perhaps the late Jerry Garcia in the song "Ripple" by the Grateful Dead: wherein the sentiment occurs: "[these things] are for, your steps alone") merely think you should endeavor alone into solving these important, basic questions.

To explain further:

An obvious subtlety/detail that is not clarified in your brief question is:

If you are loading human gDNA and asking this question, about 6493 copies (of a single copy gene) exist in about 20 ng gDNA (assuming 3.08 pg human DNA per haploid genome). A 500-1000 bp amplicon would be a mere fraction of that. So, realize your question is in dire need of this other specific clarification - and only you can provide the answers yourself (e.g., try to arrive at the answers by yourself; and therein learn. The math associated is actually fun).

Without speaking for Jochen (whom I respect highly), I would say simply, try this thing out for yourself and see where you end up. Herein is your riddle/quest. 'You will learn much by trying to answer this for yourself' - I think is Jochen's inferred/hidden point. Stay up at night by all means to get to your answer? Yes. These are the basic beautiful things that get us all hooked on science in the first place. Asking to be taught about something, though eternally noble, is still easier than rolling up your own sleeves and getting dirty to the core. Remembering Avagadro's number is the only clue I will give you other than that PCR doubles template every cycle (when 100% efficient). Also recall that 660 g/mole per bp is an estimate, and if you know the actual sequence of your amplicon, you can fine tune this value accordingly. Hey - it's a huge interesting world out there!! ;]

A product of less than 1kb can be amplified to about 1E11 copies/µl. The molecular weight of a bp is about 660 g/mol. The rest is just math :)

E.g. in theory, if your dsDNA amplicon is 500 bp, it would take ~15.6 cycles (at 100% amplification efficiency) to obtain 1 mg of product if your initial rxn contains 1 ng amplicon/uL in a 20 uL PCR reaction.

In other words, a total of 20 ng of pure 500 bp dsDNA amplicon will become 1 mg of amplicon product after about 15.6 cycles... but, PCR is not continuously efficient - so it may take more cycles than this. If the PCR is only an average of 80% efficient, it would take about 18.4 cycles to get 1 mg (if dNTPs hold out and the reaction is unimpeded by any other unforeseen factors, etc.).

If it is the aim to get this amount precisely you can run the PCR into the plateu phase and add just enough primers to generate the required amount of PCR product (best use only one of the primers in a limiting concentration and take an excess of the other primer). The precondition is that the primers are specific and no primer-dimers are produced.

PS: much easier than determining the effciency and the number of cycles required to get the same result. Plus that it might be difficult to stop the PCR after 18.4 cycles, for instance ;) - yes, I know, this was likely not the problem (to produce a defined amount more or less precisely; rather to just produce a sufficient amount), but it is a nice mental exercise :)

dr jochen could you learn me how to calculate the required conc of primers and the no of cycles in this case?

and also how I could do successive runs precisely (whats the protocol ) that allow max efficiency?

many thanks

There are some other unknowns here... and subtleties... but I, like Jochen (and perhaps the late Jerry Garcia in the song "Ripple" by the Grateful Dead: wherein the sentiment occurs: "[these things] are for, your steps alone") merely think you should endeavor alone into solving these important, basic questions.

To explain further:

An obvious subtlety/detail that is not clarified in your brief question is:

If you are loading human gDNA and asking this question, about 6493 copies (of a single copy gene) exist in about 20 ng gDNA (assuming 3.08 pg human DNA per haploid genome). A 500-1000 bp amplicon would be a mere fraction of that. So, realize your question is in dire need of this other specific clarification - and only you can provide the answers yourself (e.g., try to arrive at the answers by yourself; and therein learn. The math associated is actually fun).

Without speaking for Jochen (whom I respect highly), I would say simply, try this thing out for yourself and see where you end up. Herein is your riddle/quest. 'You will learn much by trying to answer this for yourself' - I think is Jochen's inferred/hidden point. Stay up at night by all means to get to your answer? Yes. These are the basic beautiful things that get us all hooked on science in the first place. Asking to be taught about something, though eternally noble, is still easier than rolling up your own sleeves and getting dirty to the core. Remembering Avagadro's number is the only clue I will give you other than that PCR doubles template every cycle (when 100% efficient). Also recall that 660 g/mole per bp is an estimate, and if you know the actual sequence of your amplicon, you can fine tune this value accordingly. Hey - it's a huge interesting world out there!! ;]

Dear Jack Gallup, Is there any reference on which you are based to calculate the amount of PCR product (DNA) as you did?

The universal Handbook of Physics and Chemistry is a good place to start.

Google yawns at the finger-tips of all of us otherwise. I have used it often.

I think the limiting factor in most PCR reactions are the amount of nucleotides available. But you can do the calculations, just determine how many molecules you need for 1mg of your product, determine how much dNTPS you need, be sure you have excess primers and the amount of template will dictate the number of cycles (assuming high efficiency).

@Michael: Usually the dNTPs are not a limiting factor. Typically one has a mM conc of dNTPs and a sub-µM conc of primers in the reaction. dNTPs may become limiting only for very long PCR products. However, then the accumulation of pyrophosphate is limiting instead, before the dNTPs are used up. And it would still be easier to adjust the primer conc to fix the final amount of PCR product.

@Shaimaa: Apart from that I fully agree with Jack: find it out yourself! It is quite simple stuff you should have learned in school. Use your brains, think! It is much more beneficial than reading and following a given solution. You can then surely ask more specific questions in RG again (if you then still have problems).

Very nice discussion with high-quality answers!

However, PCR is not always predictable and the principal manthematical approximation is nothing but speculation when it comes to the real experiment:

So far, nobody mentioned secondary structure formation, GC-content of the template/amplicons or any of the nice things that make certain primer combinations useless. This, too, will affect the hypothetical mathematical calculation (@Jack).

So, first make a test-run, quantify your desired product by e.g. photometry and then calculate again!

@ Jochen - I am sure the limiting dNTPs will most likely stop most PCR reactions. I have never been able to reactivate a saturated PCR, except by adding additional dNTPs. Thus, the accumulation of PP seems not to affect amplicon production per se. Of course, with high-fidelity enzymes and perfect amplicon polymerization, the buffer capacity might play a role.

Still, you get product with some rounds of PCR even in 'water' instead of complete buffer, unless you have sufficient Mg and K in the reaction to stabilize the dsDNA-backbone and to keep the enzyme active. I can not recomend to do this, but it sort of works. - But the more amplicon you get, the more is hydrolysed by the acidic pH!

Jochen, a typical PCR mix has about 200uM each dNTP. You would expect to need about 1000-fold less primer than dNTPs (lets say you were making a product 1000nt in length) so that works out.

Since the students don't seem to want to crunch the numbers I just did a really quick calculation and your typical 50ul PCR reaction has enough dNTPs to generate about 15ug of product ASSUMING sufficient primer, no inhibitory products like pyrophosphate, and the polymerase could use all the dNTPs even as you fall well below the Km. Since we know there is inhibition and polymerase is very concentration dependent, I would expect the actual yield to be half or less.

So if you get 10ug of product from a 50ul rxn you would need 5ml PCR rxn to get 1mg of product. And in the real world your yield would be much less.

Feel free to check my math, I just had a beer with lunch and rounded very liberally.

It's awesome to see the Lions come out of their cages on this. A good example of a seemingly simple question exposing the underlying complexities and considerations that need to be explored and/or weighed in order to formulate a series of possible good answers. Cat's out the bag now. ;]