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How to quantify virus by PCR

virus quantification pcr

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12 replies to this topic

#1 Mohamed 1984

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Posted 29 August 2012 - 04:43 PM

i want to correlate the quantity of microRNA and influenza virus , so i isolated the total RNA from experimentally infected lung. now i convert all the isolated RNA to cDNA. iam preparing to measure both the amount of miRNA 223 and miRNA21 and one of influenza genes

So my question: How can i measure the quantity of influnza virus in my sample so that i can correlate it to the quantity of my miRNAs? please in details.

#2 Curtis

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Posted 29 August 2012 - 06:27 PM

I would go for Real-Time PCR. I have a feeling you already know your answer but you are afraid to do it ;) Don't be. I used to run Real-Time PCR with patient samples in a pathobiology lab for a variety of viruses (HBV,HCV,EBV, HIV etc.) but I used commercial kits (Arthus, Qiagen, Roche etc) to do that. Then I gave the results to the doctors and they told the patients at what stage of infection they are....but in your case you must study qPCR well. Many people use SYBR green method, but I would recommend Taqman/probe method. you must design primers and probes specific for genes or areas of your virus that are unique. I used RNeasy kit from qiagen to extract total RNA. It extracted any RNA in the sample, so your virus RNA will be extracted too.

#3 Mohamed 1984

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Posted 30 August 2012 - 09:33 AM

Thanks for reply , now i will start the QPCR. i have another important question:
i synthesized cDNA with a concentration of 40ng/µl in a volum of 3 µl. i need for the the QPCR a conc of 0.15 ng/µl in a volum of 20 µl. i calculated it in this way
0.15/40*20µ so i will take a volum of 0.075 µl from the 3 µl. but i need much higher amount for QPCR ) 2 µl? what should i make. So the question how can i form a solution of 2µl from a solution of 0,057 µl but with conc remain fixed ( 0.15 ng/µl)

#4 leelee

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Posted 30 August 2012 - 08:21 PM

Just remember that the amount of viral DNA or RNA (depending on your virus) does not always correlate with infectivity in a linear way.

This may not be a concern for your application.

#5 Trof

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Posted 31 August 2012 - 04:57 AM

Thanks for reply , now i will start the QPCR. i have another important question:
i synthesized cDNA with a concentration of 40ng/µl in a volum of 3 µl. i need for the the QPCR a conc of 0.15 ng/µl in a volum of 20 µl. i calculated it in this way
0.15/40*20µ so i will take a volum of 0.075 µl from the 3 µl. but i need much higher amount for QPCR ) 2 µl? what should i make. So the question how can i form a solution of 2µl from a solution of 0,057 µl but with conc remain fixed ( 0.15 ng/µl)


First question, how do you know the concentration of your cDNA? The cDNA is difficult to measure, usually you transcribe defined amnout of RNA and for with that, you can only measure single stranded cDNA (i.e. after Rnase H digest the RNA strand) after you purify it from RT reaction and calculate concentration with the formula for ssDNA. Otherwise it has no use.
I'm bit confused by your overal amount of cDNA also, you say you have only 3 ul of it? Most RT reactions have amount at least 20 ul. How exactly did you do your reverse transcription?


Second question, how you got to the amount needed in reaction? You usually don't know the amount needed, it depends on many things and can't be calculated. You put maximum of 10 % of qPCR reaction volume from the RT, that is usually 2 ul in 20ul reaction. If this amount is not sufficient, you must transcribe more RNA in the first place.



Other thing.. miRNAs can't be reverse-transcribed the way normal RNA is (e.i. with random hexamers or oligo dT primers), you need special kits for that that have affinity for specific miRNAs.


And more, I'm thinking if it's really possible to just correlate virus RNA and miRNA together, since their RT efficiencies may vary. but probably you can, it depends on what you want to know it for. Maybe consider using a housekeeping gene from the lung tissue in addition for normalization. Because I think the miRNA function depends on the amount relative to the amount of transcript they regulate, that means the lung tissue, so you need to know their base level. Virus concentration is completely independent on this, it's used only as a second correlate for increase or decrease of miRNA level. And if you don't put there the housekeeping gene, you wouldn't know it they actually increased od recreased. Does that sound clear?

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.

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#6 Mohamed 1984

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Posted 01 September 2012 - 08:14 AM

I find your answer right . but for the question no. one, i suppose that if I have 40 ng /µl RNA templet , them make RT, knowing that each RNA strand will produce one cDNA strand, so i suuposed that the cDNA amount will be also 40 ng/µl. i want to know is this difficult to measure the amount of cDNA as a general or u mean after RT reaction? would u like to give me this formula , by which i can confirm the conc of cDNA before going to QPCR.
in my experiment i will take whatever the volum of sample contain 40ng/µl of cDNA then i will take a defined volum from this to form a concentration of 3 ng/µl per qPCR reaction ) i.e, 0.15 ng/µl if the reaction volum is 20 µl), Sure i had a special kits for miRNA with two primers ) dT oligo primer for miRNA and an Universal primer for mRNA . anyway i did this experiment on a lung homogenate so i will be unable to detect the growing level of the virus with or without miRNA , i will only know the amount of virus with the amount of special miRNA. would you like to give me your E-mail address or facebook so that we can share or ideas easily.

#7 Trof

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Posted 01 September 2012 - 02:58 PM

40 ng/ul is a concentration, it doesn't say how much RNA you took for RT, unless you say how many ul you used. Then you will have this amount of RNA and in theory you will have cDNA for each RNA molecule, but DNA has different weight, so the copy number of cDNA will be same, but not the weight and not the concentration. Then also you usually take some ul of RNA and put it into mixture, usually final volume of RT is 20 ul. Concentration of yet a different weight of cDNA would be much lower, because it's in bigger volume. And that is the theory, actually the efficiency of RT is never 100%, so there will be less cDNA created.

To practically measure the exact cDNA concentration you would need to get rid of the original RNA (if the RT anzyme doesn't have Rnase H activity) and then purify the cDNA, only then you can measure it's concentration on spectrophotometer as a ssDNA using this calculator for example.

But all of this doesn't really matter, because cDNA is not measured for RT-PCR. You just work with a RNA amount equivalent. You decide how much RNA use in RT and then just use portion of RT reaction (usually maximum of 10 % of PCR reaction volume).
You still didn't specify what amount of RNA you use for RT, so I can't tell you what portion of RT reaction you should use. For example we usually put 1 ug of RNA into 20ul RT reaction and use 1-2 ul to each PCR. That would make equivalent of 50-100 ng of RNA in each reaction. But that doesn't mean there is this actual amount of cDNA. but that doesn't matter. What matter are Cts, and since relative abundance of your target gene can vary a lot, it's no use to count with fixed amouts of cDNA, which in addition is even difficult to measure.

I'm still getting a bit lost in your RT procedure. Do you transcribe both miRNA and mRNA together or separately? miRNA RT uses either stem-loop primers for specific miRNAs or oligo dT after poly A tail is added for all of them, and you can either use oligo dT primer for mRNA (different than the miRNA one) or random hexamers for all RNA species, which may be probably called universal, but I'm not sure.

Also it's still not clear how you want to compare virus and miRNA levels, when you don't have normalisation for cell amount or a housekeeping gene. If you RT mRNA from your lung tissue including cells, you probably have cDNA containing not only virus, but all other RNAs, so you just need to select a housekeeping gene to run with your samples.

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.

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#8 Mohamed 1984

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Posted 02 September 2012 - 04:54 AM

Very simply, I isolated RNA from Lung homogenate , then i fixed its conc in all sample to 40ng/µl in a total volum of 10 µl RNA, RNAse free water mix. then i went to do cDNA.synthesis. in this case i will take 3 µl volum from 10 µl. and their conc still 40ng/µl. i want now to know how much conc of RNA should i take to start cDNA synthesis and then qpcr reaction. So i look for the qPCR kit and find that for 20µl reaction voulm i need 3 ng/20µl of cDNA this implies that i need 0.15 ng/µl of cDNA , Ok. So i suuposed that 40ng/µl RNA will produce around 40 ng/µl cDNA. here it dose not matter which weight of cDNA. SO now the question turned to be how can i convert this high conc of cDNA (40) to a very low one (0.15 ng) then i calculated it by making firstly a 1.5 ng solution and then make 0.15 solution. because i have no 1 µl pipet in the lab. Is this ok or not from ur opinion. also i checked this website but from where can i obtain this OD to measure the ssDNA conc؟

#9 Trof

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Posted 02 September 2012 - 02:52 PM

You have 40 ng /ul of RNA (is it DNase treated?) in 10 ul, thats 400 ng overal.
You take 3 ul from that to RT, right? Which is 120 ng overal. Let's assume your RT has a volume of 20ul.

For some reason your manual (what kit exactly?) specifies using 3 ng in 20 ul reaction, that implies you need 3 ng. Not a concentration. An amount. Concentration 3 ng/20 ul is what you get at the end, what you should focus on is getting there 3 ng.

If we take RNA-cDNA amount as equal, you now have like 120 ng of cDNA in your 20ul RT reaction. That means 6 ng/ul. You need 3 ng, SO you take 0.5 ul from RT reaction. If you don't have 1 ul pipette you can dilute the RT accordingly to get 3 ng in larger volume, but you need to make space for other PCR reagents too. Generaly doing PCR without 1 ul pipette is pretty difficult, most reagents require such low volumes.

Since I don't know your kit, I don't know why do you need such en exact amount, when the sufficent amounts vary a lot depending on the target gen abundance.

However, I'm really not confident in advising anything to you. From my point you seem to have deep problems with absolutelly basic things like concentration, measurement of nucleic acid concentration (OD260 is optical density, value measured at 260 nm on spectrophotometer) and so, how do you know concentration of your RNA in the first place?

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


#10 Mohamed 1984

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Posted 05 September 2012 - 02:38 PM

I can say that i knew the conc of RNA firstly by nanoquant plat. it is proved to be various conc in ng/µl in my samples so i actually did dilution for theses samples to be similar around 200 ng/µl. so that i can start the cDNA easily. now i made again a conc of 40 ng RNA after diultion to start cDNA. I used the qiagen kits for two step qRT-PCR and the qPCR require a volum of 20 µl with a conc of 3ng/20µl which means 0,15 ng/ µl that is right? since i used no 1µl pipets, i did a diltution of 40 ng/µl to 0.15 µl to obtain the 2 µl volum used in the Qpcr. that is all.

another question from ur experience. i recieved the primers for QPCR in lypholized state. do u know how can i dilute it ؟

#11 Trof

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Posted 06 September 2012 - 12:12 AM

You got 40 ng/ul RNA. You needed 3 ng, that would mean to pipett 0.075 ul. That's not possible, right. How exactly did you diluted?
You wrote "i did a diltution of 40 ng/µl to 0.15 µl to obtain the 2 µl volum used", that's kind of confusing, what concentration of RNA did you diluted to? (not the final concentration in the PCR reaction, that is absolutely unimportant value right now)
If you diluted RNA to 0.15ng/ul before you pipett to PCR, that would be like 266 times, and that's too many. You say that you used 2 ul to PCR, that would mean you should have your RNA concentration 1.5 ng/ul, that means diluting 1 ul of RNA [40ng/ul] in 25.7 ul of water. Any other dilution would be wrong.

As for lyophilised primers, you just add water or buffer you use. Common storage concentration is 100uM (100 pmol/ul). There should be come information from your primer provider, nmol, ug or OD values of the synthetised primer. Mostly there is already an information how much water/buffer to add to get certain concentrations.

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


#12 Mohamed 1984

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Posted 10 September 2012 - 05:23 PM

My dear i diluted 40ng/µl to 1.5 ng/µl then to 0.15ng/µl

that is right they say i should dilute it in 550 µl TE buffer with PH 8. The question if i dilute it in RNAse free water . will it affects the PH of storage as the water is not similar to TE buffer in a term of PH .

#13 Trof

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Posted 17 September 2012 - 01:30 PM

Primers are usually diluted in TE or 10mM Tris pH 8. Higher pH is better because it prevents degradation. Unbuffered water can vary in pH in time. Technically your buffers should be made with nuclease free water, so it would be sterile and have the righ pH.

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






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