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qpcr and sample dilution


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11 replies to this topic

#1 Labrat18

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Posted 21 September 2013 - 05:21 AM

Dear friends and colleagues,

 

I have some questions, it might be stupid so forgive me in advance. I have a protocol that is used to quantify a virus, that is being extracted from plasma or serum. Before adding the extracted sample to the PCR master mix, it is asking to dilute the sample to 5ng/ul or if the concentration is not known just dilute it to 1:20. Now my questions are:

 

1- since we are extracting sample from serum and plasma which is considered as cell-free. How much usually the is the DNA quantity in those type of samples?

 

2- is it really critical to dilute the sample. I know too much DNA might inhibit PCR reaction, but given we are extracting from plasma or serum, is this critical?

 

3- If we did dilute the sample, dont we have to multiply the result by the dilution factor which is 20?. No such thing is mentioned in the protocol? I just need to double check.

 

Thanks in advance :)



#2 theo22

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Posted 23 September 2013 - 03:27 AM

1: serum contains cell-free DNA, but the total concentration will be low.

 

2: if ypou would like to perform quantitative PCR, the amount of starting material is very important. So yes, in most cases dilution is important to stay within the linear range

 

#; For absolute measurement, you will need a standard which you have to dilute in the same way as your sample



#3 Labrat18

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Posted 28 September 2013 - 01:03 AM

Hello Theo,

 

Regarding the linear range of the assay, as far as I know the linear range corrosponds to the concentration of your standards. The standards concentration range from 10 copies/ul - 10E6 copies/ul (6 standard points). For that reason I said is it really important to dilute my sample, since I have vast linear range and it is most probably will fall within that range, maybe I should have clarified that in my initial post.  

 

Ok, generally speaking, if you diluted the sample for quantitative assay to fit in that range, you do have to multiply it by the dilution factor, which is in my case 20, right?



#4 theo22

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Posted 28 September 2013 - 02:17 AM

ofcourse you will have to take the dilution into account if you are trying to measure absoluut concentration. Most people compare qPCR results with their control sample (e.g. untreated vs treated). Then you only have to make sure that both samples are diluted equally and that you include a proper non-inducible gene (as a 'loading control'.



#5 Labrat18

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Posted 28 September 2013 - 03:15 AM

Hey Theo, thanks again for your input. I can guess from your replies that you are talking more about the considerations of a gene expression expirement. My expirement is about absolute quantification of a virus.

 

So, what I am planning to do, just to amplify the unknown sample directly, without dilution, if the sample fall within the linear range then  I will take that result, if somehow the sample is out of the linear range then I will dilute the sample, rerun the qPCR then mulitply my result with the dilution factor to get the absolute quantification.


Edited by Labrat18, 28 September 2013 - 03:15 AM.


#6 theo22

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Posted 28 September 2013 - 03:23 AM

Just one remark: what is the solution that contains your virus? Is it serum? Or is the virus isolated and now present in a buffer? Since you are not going to dilute your sample, make sure that your standard control samples are in the same solvent.



#7 Labrat18

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Posted 28 September 2013 - 04:43 AM

Hey Theo, Yes, the virus will be extracted from serum. And both the samples and the standard will be in the same solvent.

 

Thanks again for your input :)



#8 Labrat18

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Posted 29 September 2013 - 05:36 AM

Another thing came into my mind. The standards concentrations are provided 2.0E5 - 2.0E1 copies/ul. And the input of the sample is 5ul, so in total we are introducing 1000 000 copies/ul to 10 copies/ul, right?

 

I assume the unknown results will be in copies/ul, same unit of the standard.

 

Now if I want to convert this result to IU/ml, I have to first convert my result to copies/ml (multiply by 1000) then divide it by a certain number to convert to IU/ml... what is that number? some articles I found says its 5.7 some articles are 2.7...



#9 theo22

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Posted 29 September 2013 - 11:40 PM

Another thing came into my mind. The standards concentrations are provided 2.0E5 - 2.0E1 copies/ul. And the input of the sample is 5ul, so in total we are introducing 1000 000 copies/ul to 10 copies/ul, right?

No, 5 ul should contain 1E6 - 100 copies

 

What is IU? Infection units? multiplicity of infection?



#10 Labrat18

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Posted 30 September 2013 - 03:47 AM

 

Another thing came into my mind. The standards concentrations are provided 2.0E5 - 2.0E1 copies/ul. And the input of the sample is 5ul, so in total we are introducing 1000 000 copies/ul to 10 copies/ul, right?

No, 5 ul should contain 1E6 - 100 copies

 

What is IU? Infection units? multiplicity of infection?

 

 

IU is International Unit.

 

Ignore my question above, this is a new question

 

Since I will enter the standard values into the software as absolute quantities...

 

that is 1000 000 copies

100 000 copies

...

..

..

100 copies..

 

My unknown samples result will be in copies as well, right? So lets assume I got a result of 2000 copies, and I want to note my result as concentration of copies/ul.. 

So,  the concentratopm of 2000 copies; does it equal 2000/20 (the total volume of the reaction)= 100 copies/ul. Or is it 2000/5 (the input volume of the sample)= 400 copies/ul?

 

To sum it up, do we divide by the total volume of the reaction (which is 20ul) or just the volume of the sample input (which is 5ul)?



#11 theo22

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Posted 30 September 2013 - 03:54 AM

number of copies divided by the ul input will give you copies/ul



#12 Labrat18

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Posted 30 September 2013 - 04:00 AM

Thanks again Theo, Yeah, that make more sense to me as well.






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