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# PCR solution calculus

Best Answer Trof, 22 August 2014 - 11:53 AM

You meant why is 10uM "equal" to 10 pmol (that's in your original example).

It isn't actually.

But, "molar" concentrations (moles/liter) are kind of standard way how to describe concentration. It's also shorter to write etc. Instead of "whatevermoles/whatevervolume".

So, that's why everybody is talking about uM primers and so.

As you now know, 10 uM means "10 micromoles in a liter". As you may already noticed, liter quantities are not exactly what molecular biology is about. The more sufficient volume would be actually ul (microliter). In PCR you add microliters, so it helps to know how many moles you have in a microliter.

As a matter of fact "molar" concentration is easily transformed to any "per microliter" concentration by reducing the moles six orders of magnitude. So 10uM (10 umoles/liter) becomes 10 pmol/ul (you can calculate it youself).

And of course you shouldn't mistake amount with a concentration.

And as I said before, in your example, you are required to put a particular amount of primers (not concentration) into a reaction. And since the concentration of primer 1 is 10 pmol/ul, it's easy to calculate what volume you need to put in reaction.

Sometimes you need a final concentration, sometimes you need a particular amount. If it's "molar" (ends with a big M), it's always a concentration. If it's just .. moles,..grams, without slash, it's just an amount and not a concentration.
So 10 umol/l can be never the same as 10 pmol, since the first is concentration, the second is amount.

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### #1 andra

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Posted 22 August 2014 - 01:58 AM

Hi

Let's take as example the following situation.

We want to prepare a 50 ul PCR product. Below there are the stock concentrations and the concentrations we need.

DNA polymerase: 5000 units/ml --> 1.25 units

Buffer: 10x --> 1x

dNTP mix: 2.5 mM each --> 200 uM

Primer 1: 10 uM --> 10 pmol

Primer 2: 25 uM --> 10 pmol

Template: 785 ug/ml --> 500 ng

Water: up to 50 ul

This is an exercise I got in a Molecular Biology class where the solution of the problem was like that:

DNA polymerase: 1.25 units / 5000 units/ml = 0.00025 ml = 0.25 ul

Buffer: 1 / 10 * 50 ul = 5 ul

dNTP mix: 2.5 mM each => dATP = 2.5 mM; dTTP = 2.5 mM; dGTP = 2.5 mM; dCTP = 2.5 mM => 2.5 mM * 4 (dATP + dTTP + dGTP + dCTP) = 10 mM = 10000 uM

C1V1 = C2V2 => V1 = C2V2 / C1 => V1 = 50 ul * 200 uM / 10000 uM = 1 ul

Primer 1:

10 uM/ml = 10 pmol/ul => 10 uM/10 uM = 1 ul

Primer 2: 10 uM/25 uM = 0.4 ul

Template:

ug/ml = ng/ul => 500 ug / 785 ug/ml = 0.63 ul

Water: 50 ul - (0.25 ul + 5 ul + 1 ul + 1 ul + 0.4 ul + 0.63 ul) = 41.72 ul

I don't understand two things:

1. Why in some cases we care about the final volume and in others we don't? For example, the dNTP mix is expressed in mM and we count it according to the final volume. On the other hand, the primers are also expressed in uM but we don't care about the final volume.

2. Why 10 uM/ml = 10 pmol/ul when 1 uM = 1000000 pmol but 1 ml = 1000 ul?

### #2 Trof

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Posted 22 August 2014 - 05:25 AM

Hello, welcome to the forums.
I moved your post, because it is a homework question and not a "methods" one. It belongs into Student forums section.

Question 1. depends a lot on your exercise setting.

Sometimes you just want a defined amount of primers and DNA and  polymerase in your reaction regradless of overal volume. I would think this is common for polymerase and DNA, less commmon for primers. By using larger volume, you may for example use bigger volume of DNA if needed (if you have a low concentration), and at the same time keep the polymerase amount stable, so you save on it.
As for primers, they are one of the limiting reagents in amplification, but mostly, primers are kept on the same final concentration in reaction. So using a set amount of DNA and polymerase is probably usual, in real work, set amount of primers usually not, but I can't tell you why you get it in your exercise.
It can be, that you just say "ah, I will put there 10 pmol of each primer" and then you calculate with amount. I for example sometimes reduce the PCR volume but keep the same pmol of primers there and the same DNA amount, have a comparable amplification as with the larger volume and save on the reagents (polymerase and others).

Why to care for a constant concentration instead in buffers and dNTPS is that these create the "enviroment" of reaction, polymerase need a constant concentration, so amount of buffer increases with volume, the same with MgCl ions and dNTPs needed for reaction.

And honestly, its not very good exercise you got... when making PCR in reality, you mix several things beforehand and this should be part of the calculation, stock primers are not 10uM, that's a "working" concentration made from the more concentrated stock (usually 100uM), I didn't get the dNTP calculation at all, since in reaction you need a concentration of each nucleotide, so no summation of the concentrations, usually "working concentration" (bough as a mix or prepared from aech separate nucleotide.. that would be nice calculation insted of what you have) is 10mM each and the final is 200uM each, what you got is really confusing..

Question 2.
You got it wrong. "uM" and any other concentration expressed with a big "M" at the end means "molar"  (e.g. 10 micromolar primers) and that means concentration (aka amount/volume). The term "molar" is defined as "equal to 1 mole per litre". Not a mililitre.

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### #3 andra

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Posted 22 August 2014 - 08:32 AM

Thanks for the nice explanations and for putting the topic to the right place.

I got it why we care about the final volume for dNTPs and buffer.

Anyway, I still don't understand why 10 mM (which is 10 mmol/l?) is equal to 10 pmol and why we don't have to transform the numbers and bring the equation to the same unit (mmol or pmol). Even if doing that for primers is not common in practice, I would like to get the principle.

About the dNTPs, if each comes in 10 mM concentration and we bring each to 200 uM, does it mean that we finally add 1 ul of each dNTP so it would be totally 4 ul and it would decrease the volume of the water?

I am very interested in having a good handling of these things. I just started my PhD in Molecular Biology and I will have to do a lot of PCR reactions (I know, maybe it's too late to get these things clear).

### #4 Trof

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Posted 22 August 2014 - 11:53 AM   Best Answer

You meant why is 10uM "equal" to 10 pmol (that's in your original example).

It isn't actually.

But, "molar" concentrations (moles/liter) are kind of standard way how to describe concentration. It's also shorter to write etc. Instead of "whatevermoles/whatevervolume".

So, that's why everybody is talking about uM primers and so.

As you now know, 10 uM means "10 micromoles in a liter". As you may already noticed, liter quantities are not exactly what molecular biology is about. The more sufficient volume would be actually ul (microliter). In PCR you add microliters, so it helps to know how many moles you have in a microliter.

As a matter of fact "molar" concentration is easily transformed to any "per microliter" concentration by reducing the moles six orders of magnitude. So 10uM (10 umoles/liter) becomes 10 pmol/ul (you can calculate it youself).

And of course you shouldn't mistake amount with a concentration.

And as I said before, in your example, you are required to put a particular amount of primers (not concentration) into a reaction. And since the concentration of primer 1 is 10 pmol/ul, it's easy to calculate what volume you need to put in reaction.

Sometimes you need a final concentration, sometimes you need a particular amount. If it's "molar" (ends with a big M), it's always a concentration. If it's just .. moles,..grams, without slash, it's just an amount and not a concentration.
So 10 umol/l can be never the same as 10 pmol, since the first is concentration, the second is amount.

Our country has a serious deficiency in lighthouses. I assume the main reason is that we have no sea.

I never trust anything that can't be doubted.

'Normal' is a dryer setting. - Elizabeth Moon

### #5 andra

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Posted 24 August 2014 - 03:21 AM

Fully got it now.

Thank you for the patience and explanations.

Lots of luck with your PCRs.