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# DNA amount calculation for PCR

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### #1 Nephrite

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Posted 17 February 2011 - 03:07 AM

Hello everybody.

I measured my DNA amount as 48 microgrammes/mL, and I accept it as 4.8 microgr/100 microliters.

The recommended DNA template/reaction is up to 1 microg/100 microliters. The volume of reaction is 30 microliters. I tend to apply 0.2 microgr/reaction, which is 30 microliters.

This is how I calculate it:

V1C1 = V2C2, where:

V1 - the required volume (X);
C1 - the total amount of DNA (4.8 microgr/100 microliters)
V2 - volume of reaction (30 microliters)
C2 - DNA amount in final volume (0.2 microg.

In this way: 30*0.2/4.8 = 1.25 microliters template DNA/ 30 microliters reaction.

So far this way of calculation worked nice, but yesterday my PCR failed and I need to check all possible reasons.

I am not confident with calculations, so could anybody have a look if this is a correct way?

Thank you.

### #2 ElHo

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Posted 17 February 2011 - 05:18 AM

C stands for concentration, not amount. So in your case C1 would be 4.8 microgr/100 microliters and C2 0.2 microgr/30µl. If you do the calculation you will get a volume of 4.2 µl to have 200 ng of DNA in your 30 µl reaction volume. But in my opinion that´s way too much for a pcr as too much DNA can inhibit the reaction. We usually have 50 ng of genomic DNA in a reaction volume of 50 µl and it works fine. Your original volume of 1.25 µl which means 60 ng of DNA, although calculated incorrectly, is far more suitable in my opinion.

### #3 ivanbio

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Posted 17 February 2011 - 08:14 AM

ElHO's explanation is correct = you need to add 4.2 ul to have 200 ng of DNA in your reaction. What I wanted to add is that while lower amounts of gDNA tend to be better for a PCR reaction, 60 ng may not be enough. If you have a very good PCR reaction then the amount of DNA you added (1.25 ul or 60 ng) would be fine, but it may not have worked because there was not enough gDNA (if your PCR is not that good). Adding 4.2 ul (200 ng) or even more may actually be better, or at least give you an amplified product.

Ivan

### #4 Nephrite

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Posted 17 February 2011 - 08:59 AM

C stands for concentration, not amount. So in your case C1 would be 4.8 microgr/100 microliters and C2 0.2 microgr/30µl. If you do the calculation you will get a volume of 4.2 µl to have 200 ng of DNA in your 30 µl reaction volume. But in my opinion that´s way too much for a pcr as too much DNA can inhibit the reaction. We usually have 50 ng of genomic DNA in a reaction volume of 50 µl and it works fine. Your original volume of 1.25 µl which means 60 ng of DNA, although calculated incorrectly, is far more suitable in my opinion.

Thank you very much for the reply.
I tried to achieve your numbers (volume 4.2 microl for 200 ng and 60 ng in 1.25 microl), in order to understand the philosophy of calculation.
This is how I did it:
my original concentration is 4.8 micrograms/100 microliters, and in 30 microliters (the volume of my reaction) it will be 1.44 micrograms - I used the simple triple rule to calculate the ratio.

And then, to obtain 60 ng:
V1 - the volume I take, which is 1.25 microl;
C1 - concentration of DNA in volume of 30 microl, which is 1.44 micrograms;
V2 - volume of reaction, which is 30 microl;
C1 - concentration of DNA in my reaction volume of 30 microl.

1.25 * 1.44 / 30 = 0.06 microgr = 60 ng.

And also, to obtain 4.2 microliters:
V1 - the requested volume;
C1 - concentration of DNA in volume of 30 microl, which is 1.44 micrograms;
V2 - volume of reaction, which is 30 microl;
C1 - concentration of DNA in my reaction volume of 30 microl, which is 0.2 microgr (200 ng).

30 * 0.2 / 1.44 = 4.16 microliters.

As you can see, I got your numbers, but do not think this is a correct way to do the calculations, or...?
It is a shame I do not know such basic things, but nobody teached me. I will appreciate if you explain me more about this.

### #5 Nephrite

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Posted 17 February 2011 - 09:21 AM

ElHO's explanation is correct = you need to add 4.2 ul to have 200 ng of DNA in your reaction. What I wanted to add is that while lower amounts of gDNA tend to be better for a PCR reaction, 60 ng may not be enough. If you have a very good PCR reaction then the amount of DNA you added (1.25 ul or 60 ng) would be fine, but it may not have worked because there was not enough gDNA (if your PCR is not that good). Adding 4.2 ul (200 ng) or even more may actually be better, or at least give you an amplified product.

Thank you, Ivan :-) I just replied ElHO - one of my problems are the calculations. You can see my reply. I will aprreciate if you put some light in my chaos.
I try to perform two step RT-PCR on material from parafin embedded sections. So far I tested several primer pairs (one R and several F) amplifying different fragments - it worked fine with lower amounts of DNA. I have tested higher amount - no signal.
Yesterday I got signal on my gel, but it was for a product, whic does not span any introns - so, it means that the reaction works, but can not say if the amplified product is from cDNA or gDNA. The other pair, which spans intron did not amplified anything.
I just repeated the RT reaction - the cDNA values are very low. I started to think either the RNA isolation went wrong, or one of the F primers has stopped working. Is the second possible?

### #6 pito

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Posted 17 February 2011 - 09:25 AM

About the calculations: maybe this will help: http://www.ruf.rice..../dilutions.html

And C is a concentration, not Xmoles.. Alwyas write down what units you use, it can help you understand the formula.

Check the link and if its not clear, then ask what exactly you dont understand.

And then, to obtain 60 ng:
V1 - the volume I take, which is 1.25 microl;
C1 - concentration of DNA in volume of 30 microl, which is 1.44 micrograms;
V2 - volume of reaction, which is 30 microl;
C1 - concentration of DNA in my reaction volume of 30 microl.

1.25 * 1.44 / 30 = 0.06 microgr = 60 ng.
...

It should be:

end volume x end concentration = Startvolume (that you dont know) x startconcentration.
With end concentration: 0,2 microgram/30microliter
end volume: 30 microliter
Startconcentration: 4.8 microgr/100 microliters

and startvolume =?

If you calculate it, it is:
4,2 microliter.

C is a concentration, its 4.8 micrograms/100 microliters and not 1,44 micrograms.

I hope you see where your "mistake" is?

At least thats how I see it.... to me it looks that you dont use C (concentration) but that you use a mass(micrograms) in stead.
You are using the formules wrong.

Its just 1 formula you use, you dont use it twice... check the units!

Edited by pito, 17 February 2011 - 09:49 AM.

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.

### #7 Nephrite

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Posted 18 February 2011 - 12:00 AM

It should be:

end volume x end concentration = Startvolume (that you dont know) x startconcentration.
With end concentration: 0,2 microgram/30microliter
end volume: 30 microliter
Startconcentration: 4.8 microgr/100 microliters

and startvolume =?

If you calculate it, it is:
4,2 microliter.

Thank you, Pito. So far I never had this problem - I guess it is because the different volumes and still do not understand how do you get the exact number 4.2 microliters.

Ok, let me start from the begging:

Start volume - V1
Start concentration - 4.8 ug/100 ul
End volume - 30 ul
End concentration - 0.2ug/30 ul

V1 = 30ul * 0.2ug/30ul / 4.8ug/100ul = 1.25 ul - this is what I obtain using the formula.

Then, because of the discussion I payed attention that I have different volumes but with the same units, i.e. ul. If the units were different - easy. In my logic, the end concentration 0.2ug/30ul should be transformed in volume of 100ul - in order to unify the volumes. Then 0.2ug/30ul will be 0.666666ug/100ul. Then if I use the above calculation I get volume 4.16 ul, which I should dilute in 30 ul. But is does not corresponds to 200 ng - this is not the answer. I asked several friends of mine - they all have the same logic as mine. Apparently we all are missing something obvious - please, let us know what it is.

### #8 pito

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Posted 18 February 2011 - 12:41 AM

It should be:

end volume x end concentration = Startvolume (that you dont know) x startconcentration.
With end concentration: 0,2 microgram/30microliter
end volume: 30 microliter
Startconcentration: 4.8 microgr/100 microliters

and startvolume =?

If you calculate it, it is:
4,2 microliter.

Thank you, Pito. So far I never had this problem - I guess it is because the different volumes and still do not understand how do you get the exact number 4.2 microliters.

Ok, let me start from the begging:

Start volume - V1
Start concentration - 4.8 ug/100 ul
End volume - 30 ul
End concentration - 0.2ug/30 ul

V1 = 30ul * 0.2ug/30ul / 4.8ug/100ul = 1.25 ul - this is what I obtain using the formula.

Then, because of the discussion I payed attention that I have different volumes but with the same units, i.e. ul. If the units were different - easy. In my logic, the end concentration 0.2ug/30ul should be transformed in volume of 100ul - in order to unify the volumes. Then 0.2ug/30ul will be 0.666666ug/100ul. Then if I use the above calculation I get volume 4.16 ul, which I should dilute in 30 ul. But is does not corresponds to 200 ng - this is not the answer. I asked several friends of mine - they all have the same logic as mine. Apparently we all are missing something obvious - please, let us know what it is.

I really dont see how you can make a mistake here.

V1 = (30ul * 0.2ug/30ul) / (4.8ug/100ul) or V1= 0,2µl / (4,8/100) = (0,2*100)/4,8 = 4,2 (its 4,16666666)

Or from the start:
end volume x end concentration = Startvolume (that you dont know) x startconcentration

30µL * 0,2µg/30µL = ? * 4.8µg/100µL
Wich is: 0,2µg = 4,8µg/100µl * ?
(30µL divided by 30µL is 1... they balance each other out.. so only the 0,2µg stays there.
And then:

0,2µg = 4,8µg/100µl * ?
or= 0,2µg*100µL = 4,8µg * ?
Wich is: 20µl*µg = 4,8µg * ?
Or: 20µl*µg / 4,8µg = ?µl
thus ?µl = 4,2µL.

So when you got this 4,2µl, what do you do then? You say it doesnt correspondend to the 200ng? I do not understand this?

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.

### #9 Nephrite

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Posted 18 February 2011 - 02:05 AM

It should be:

end volume x end concentration = Startvolume (that you dont know) x startconcentration.
With end concentration: 0,2 microgram/30microliter
end volume: 30 microliter
Startconcentration: 4.8 microgr/100 microliters

and startvolume =?

If you calculate it, it is:
4,2 microliter.

Thank you, Pito. So far I never had this problem - I guess it is because the different volumes and still do not understand how do you get the exact number 4.2 microliters.

Ok, let me start from the begging:

Start volume - V1
Start concentration - 4.8 ug/100 ul
End volume - 30 ul
End concentration - 0.2ug/30 ul

V1 = 30ul * 0.2ug/30ul / 4.8ug/100ul = 1.25 ul - this is what I obtain using the formula.

Then, because of the discussion I payed attention that I have different volumes but with the same units, i.e. ul. If the units were different - easy. In my logic, the end concentration 0.2ug/30ul should be transformed in volume of 100ul - in order to unify the volumes. Then 0.2ug/30ul will be 0.666666ug/100ul. Then if I use the above calculation I get volume 4.16 ul, which I should dilute in 30 ul. But is does not corresponds to 200 ng - this is not the answer. I asked several friends of mine - they all have the same logic as mine. Apparently we all are missing something obvious - please, let us know what it is.

I really dont see how you can make a mistake here.

V1 = (30ul * 0.2ug/30ul) / (4.8ug/100ul) or V1= 0,2µl / (4,8/100) = (0,2*100)/4,8 = 4,2 (its 4,16666666)

Or from the start:
end volume x end concentration = Startvolume (that you dont know) x startconcentration

30µL * 0,2µg/30µL = ? * 4.8µg/100µL
Wich is: 0,2µg = 4,8µg/100µl * ?
(30µL divided by 30µL is 1... they balance each other out.. so only the 0,2µg stays there.
And then:

0,2µg = 4,8µg/100µl * ?
or= 0,2µg*100µL = 4,8µg * ?
Wich is: 20µl*µg = 4,8µg * ?
Or: 20µl*µg / 4,8µg = ?µl
thus ?µl = 4,2µL.

So when you got this 4,2µl, what do you do then? You say it doesnt correspondend to the 200ng? I do not understand this?

''But is does not corresponds to 200 ng - this is not the answer.'' - please, forget about this sentence. My brain already blocked and I started writing nonsence.
Thank you very much for explanation - now I understand it and I will learn it once forever. The problem is that I miss basic mathematics - that`s why I did not understand how do you obtain this 4.2 (which is actually 4.16).
Once again - thank you very much.

### #10 pito

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Posted 18 February 2011 - 03:26 AM

It should be:

end volume x end concentration = Startvolume (that you dont know) x startconcentration.
With end concentration: 0,2 microgram/30microliter
end volume: 30 microliter
Startconcentration: 4.8 microgr/100 microliters

and startvolume =?

If you calculate it, it is:
4,2 microliter.

Thank you, Pito. So far I never had this problem - I guess it is because the different volumes and still do not understand how do you get the exact number 4.2 microliters.

Ok, let me start from the begging:

Start volume - V1
Start concentration - 4.8 ug/100 ul
End volume - 30 ul
End concentration - 0.2ug/30 ul

V1 = 30ul * 0.2ug/30ul / 4.8ug/100ul = 1.25 ul - this is what I obtain using the formula.

Then, because of the discussion I payed attention that I have different volumes but with the same units, i.e. ul. If the units were different - easy. In my logic, the end concentration 0.2ug/30ul should be transformed in volume of 100ul - in order to unify the volumes. Then 0.2ug/30ul will be 0.666666ug/100ul. Then if I use the above calculation I get volume 4.16 ul, which I should dilute in 30 ul. But is does not corresponds to 200 ng - this is not the answer. I asked several friends of mine - they all have the same logic as mine. Apparently we all are missing something obvious - please, let us know what it is.

I really dont see how you can make a mistake here.

V1 = (30ul * 0.2ug/30ul) / (4.8ug/100ul) or V1= 0,2µl / (4,8/100) = (0,2*100)/4,8 = 4,2 (its 4,16666666)

Or from the start:
end volume x end concentration = Startvolume (that you dont know) x startconcentration

30µL * 0,2µg/30µL = ? * 4.8µg/100µL
Wich is: 0,2µg = 4,8µg/100µl * ?
(30µL divided by 30µL is 1... they balance each other out.. so only the 0,2µg stays there.
And then:

0,2µg = 4,8µg/100µl * ?
or= 0,2µg*100µL = 4,8µg * ?
Wich is: 20µl*µg = 4,8µg * ?
Or: 20µl*µg / 4,8µg = ?µl
thus ?µl = 4,2µL.

So when you got this 4,2µl, what do you do then? You say it doesnt correspondend to the 200ng? I do not understand this?

''But is does not corresponds to 200 ng - this is not the answer.'' - please, forget about this sentence. My brain already blocked and I started writing nonsence.
Thank you very much for explanation - now I understand it and I will learn it once forever. The problem is that I miss basic mathematics - that`s why I did not understand how do you obtain this 4.2 (which is actually 4.16).
Once again - thank you very much.

Dont forget the website I posted, it shows the basics and explains the concept.

And basic maths? Eum, this is really very basic. Its nothing more then multuplying and dividing. Check the formula, check the website and always remember to use units! If you use units you cant make (or make less) mistakes!

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.

### #11 Trof

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Posted 21 February 2011 - 03:28 AM

I was never able to use the eqations like V1C1 something, seems to me too alien, and complicated. I, instead, imagine amounts od stuff flowing in the different volumes like this:

In PCR you work with microliters and nanograms, so let's transform it to them first.

48 microgrammes/mL = 48 nanograms/microliter
that's easy, you lower the volume 1000 times, so you lower the weight 1000 times, that means only to change the units on both sides.

Now what you need in your PCR reaction.. 1 microgram/100 microliters, that's 1000 nanograms/100 microliters -> 300ng/30 microliters, but you choose to use 200ng in reaction insted (personaly I don't think it's that important to calculate the "right" amount for a reaction volume, I just put 100 ng per reaction in normal volumes (10-50 ul) and don't calculate anything)

So you need 200ng in reaction and got 48ng in one microliter, so how many microliters you need to put in reaction to have 200ng flowing in there? You just divide 200 by 48, 200/48 = 4.17 microliters.

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### #12 ElHo

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Posted 21 February 2011 - 06:09 AM

Thank you, Ivan :-) I just replied ElHO - one of my problems are the calculations. You can see my reply. I will aprreciate if you put some light in my chaos.
I try to perform two step RT-PCR on material from parafin embedded sections. So far I tested several primer pairs (one R and several F) amplifying different fragments - it worked fine with lower amounts of DNA. I have tested higher amount - no signal.
Yesterday I got signal on my gel, but it was for a product, whic does not span any introns - so, it means that the reaction works, but can not say if the amplified product is from cDNA or gDNA. The other pair, which spans intron did not amplified anything.
I just repeated the RT reaction - the cDNA values are very low. I started to think either the RNA isolation went wrong, or one of the F primers has stopped working. Is the second possible?
[/quote]

DNA/RNA from parraffin embedded tissue is usually quite fragmented and difficult to amplify, especially with longer amplicons. Maybe that´s the reason you are not getting any pcr product with the longer, intron-spanning pcr product. What are the sizes of you different amplicons? Did you make a quality check of your isolated RNA? You could also use cDNA that has already worked in a previous RT-PCR run as a positive control in parallel.

### #13 Nephrite

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Posted 23 February 2011 - 02:41 PM

DNA/RNA from parraffin embedded tissue is usually quite fragmented and difficult to amplify, especially with longer amplicons. Maybe that´s the reason you are not getting any pcr product with the longer, intron-spanning pcr product. What are the sizes of you different amplicons? Did you make a quality check of your isolated RNA? You could also use cDNA that has already worked in a previous RT-PCR run as a positive control in parallel.
[/quote]

Thank you very much for reply :-)
Well, I am very new in molecular biology, so far I was working other things and I have no experience in this technique. I thought the problem was my wrong calculations (which I know how to do now :-)))thanks to explanations of Pito and Trof) but it was the gel. I use 2% agarose with Gold View in dilution 1 : 100 000. It seems it doesn`t work more than twice, even if I add new amount of Gold View.
Yes, I know about the fragmentation of NA. From my materials I was able to amplify fragments long up to 577 bp and spanning 4 introns. But most frequent results I have with fragment 316 bp long and spanning 1 intron. I use this primer to control the RT reaction. What do you mean as a quality check of RNA? When I measure my RNAs I always measure also the possible gDNA contamination but this is all.

### #14 pito

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Posted 27 February 2011 - 05:11 AM

DNA/RNA from parraffin embedded tissue is usually quite fragmented and difficult to amplify, especially with longer amplicons. Maybe that´s the reason you are not getting any pcr product with the longer, intron-spanning pcr product. What are the sizes of you different amplicons? Did you make a quality check of your isolated RNA? You could also use cDNA that has already worked in a previous RT-PCR run as a positive control in parallel.

Thank you very much for reply :-)
Well, I am very new in molecular biology, so far I was working other things and I have no experience in this technique. I thought the problem was my wrong calculations (which I know how to do now :-)))thanks to explanations of Pito and Trof) but it was the gel. I use 2% agarose with Gold View in dilution 1 : 100 000. It seems it doesn`t work more than twice, even if I add new amount of Gold View.
Yes, I know about the fragmentation of NA. From my materials I was able to amplify fragments long up to 577 bp and spanning 4 introns. But most frequent results I have with fragment 316 bp long and spanning 1 intron. I use this primer to control the RT reaction. What do you mean as a quality check of RNA? When I measure my RNAs I always measure also the possible gDNA contamination but this is all.

Check the file.

He is talking about a general check (I think).
http://www.flychip.o...sion/rna_qc.pdf

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.

### #15 Nephrite

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Posted 27 February 2011 - 07:13 AM

Check the file.
He is talking about a general check (I think).
http://www.flychip.o...sion/rna_qc.pdf
[/quote]

Many thanks, Pito :-)