I am conducting the qPCR to check the gene expression. After reverse-transcribing the RNA, I am going to perform the qPCR of cDNA. In my laboratory, to make a standard curve for qPCR, we have to make a serial dilution by mixing every sample to one tube. For example, there are 4 samples that will be used for qPCR. we will transfer 5ul of each sample to first tube (totally 20ul), then transfer half of it (10ul) to the second tube which will be diluted with water (10ul). After that, transfer 10ul of the second tube to the third tube which will be diluted with 10ul of water as well. By continuing doing this for 6 times, we will get the 6 serial dilution series of concentration.
I would like to know if this is the normal procedure in order to make a series of dilution for standard curve in qPCR. If so, why do we have to mix the cDNA of every sample to make a dilution series?
Thanks
Dear Jojo,
My first question is if the 4 samples you mentioned are just repilcates of the same experiment. If not, why you merge them?
In addition, the dilution you perform are not the ideal for qPCR, since you make 1:2 dilutions, therefore you do not cover a wide spectrum. It is recommended for dilutions 1:10 and at least 6 different if it is possible. Please give more info about the 4 different samples. In addition, do you know if the gene you study is high or low expressed? In case that you have a low expression, more dilutions will give you more accurate results.
@Jojo: yes, that's fine. You should consider more dilution steps. You should reach a dilution at which you don't get a (valid) Ct value anymore.
@Panagiotis: It's about esimating the primer efficiency. Therefor it is irrelevant where the standard comes from. And the 1:2 dilutions are ideal, because it minimizes any bias due to pipetting errors (no matter what volume you pipet - as long as you always pipet the same volume you will get an exact 1:2 dilution), and you can span the dynamic range with many dilution steps, what makes is possible to detemine the "linear range" of the method. You were right noting that six 1:2 dilutions will not span a wide range of concentrations, but the answer is to use more steps, not larger steps.
Hi,
I use serial dilutions 1:10 starting from 10ng/ul and finish in 0,00001 ng/ul to do the standard curve.
You can see more information here:
http://www.gene-quantification.com/real-time-pcr-guide-bio-rad.pdf
Good luck!
@Jojo: I meant that you should do as many dilutions (likely "more than 6 times") so that the highest dilution you measure will give you a Ct value above 30. If your undiluted standard gives you a Ct of, say, 20, then you shoul measure at least 10 dilutions of a 1:2 dilution-series (each step will increase the Ct by approximately 1 cycle, so after 10 steps you will reach a Ct-value of 30).
Be careful with the dilution series. Many have written that it's important to use 1:10, I agree. About the expression level of the gene, too, was asked. I'm just going to clarify a bit. The latter dilutions may have a different efficiency (the slope of the curve). It seems to me that this knowledge can be useful.
@Andrei: "The latter dilutions may have a different efficiency (the slope of the curve)" - why/how that?If you have an inhibitor in the buffer of the standard, one will find the highest concentrations being inhibited. As the inhibitor gets diluted in subsequent dilution steps, the further diluted standards will have a higher (hopefully ideal) efficiency. If you dilute with the same buffer that contains the inhibitor, all dilutions will be inhibited to the same degree and show the similar (low) efficiency. Just using less template should not affect the efficiency. It may be that a bad/inappropriiate background subtraction of the amplification curves can make the log-linear phases look less steep with increasing Ct, but that indicates a problem with data processing (that should be solved, for sure), but is not a sign of a lower efficiency.
Jochen, Thank You for You answer. You could see my results with 10x dilutions series in triplicates with pDNA. File name Eff.jpg. But if I did the dilution further, I saw that the slope was different. Please, see the untitled.png (it is the same dilutions series with pDNA, but not included in standard curve because slope in last sample is different). Thank You a lot or you attention. It is important for me to know your opinion.
It seems you mean the slope of the points in the right plot. I meant the slope of the individual amplification curves (as log(signal) vs. cycle; which is also an indicator of the amplification efficiency). The dynamic range of the method is from about ct=18 to about ct=32. The curves / diluion series you show covers this range quite fully. The highest concentrated sample is possibly a bit too concentrated, but it seems still ok. Dilutions that will given Ct values higher than 32 are not recommended. These PCRs contain only a hand full of molecules, what makes a precise quantification difficult, for mainly two reasons: 1) the "Poisson error" becomes relevant and 2) there might be considerable unspecific amplification (like of primer dimers).
You would be able to recognise the "linear part" of the points in the calibration plot better if you used 1:2 dilutions.
Thank You. I understand You, I think. But what can I do when expression near 35 cycle and higher? It is difficult to estimate, but some times it is nesessary.
Higher than 35 cycles is very likely some false-positive signal, as it would mean that you have less than a single molecule in your PCR.
High Ct values do not allow a precise quantification. When the effect is large, low precision is not a problem. But small effects simply won't be resoluted (at least with the usualy rather small sample sizes).
What you describe is not a STD curve. It is a serial dilution that tests the reliability-reproducibility of your PCR. Conventionally a STD curve has to give the concentration of your product by interpolation in a curve derived by serial dilutions of known concentrations. For DNA , we usually generate STD curves by using a series of concentrations of a plasmid-clone of the PCR product in question. This will give an absolute mass measurement. The starting amount of cDNA is not that critical unless is expressed in extremely low amounts or in a fraction of the cells/tissue of origin.
1.Find the length of E.coli, for example.
2. Mass of the genome M=genome size*(1.096*10^-21g/bp)
3. Convert the value from step 2 to picogram [M*(10^12 pg)=X]
4. Identify how many gene of interest copy is there.
5. Division the value pf step 3 by that of step 4, the unit will be a pg.
6. The weight of step 5 should be multiplied in 300000, 30000, 3000, 300, 30 copy)
*In step six, you provide exactly copy number of gene of interest.
7. the copy number values provided from step six should divided by 5 ul because
We basically use 5ul of template (gDNA) for each reaction in qPCR assay.
8. Measure your gDNA (or plasmid) concentration extracted (ug/ul).
9. Now dilution step
*The common formula used in dilution is C1*V1 = C2*V2
C1 is gDNA concentration. You have to convert ug/ul to pg.
V1= ?
C2= the values from step 7
V2= Desired volume (e.g. 100ul)
gDNA concentration(pg)*V1 = Step 7 values*100ul
V1= standard one
Now, we reached your answer:
After the formula mathematical function, you have four the same dilutions as standard 2 to 5. If you want to prepare standard 2, you have to take standard one. And if you want to prepare standard three, you have to take from standard 2 and so on.
Good Luck
hello this is Kyaw from TMDU. I would like to know how can I run qPCR with standard. In my research, I would like to 4 primers in transfected group and control group in human. They also said to use house keeping gene as reference. My question is how can I get standard template and it would be 2 standard templates these are one for control and one for experimental group?
and how can I compare gene expression of interest I am going to perform/
Thank you everyone
- Badges
- Science topic
- Similar topics
- Biological Science
- Molecular Biology
- Biomolecules
- Nucleic Acids
- RNA