In the preexperiment, reverse transcription should be done separately for the sample and the competitor, and then the unchanged amount of this sample reverse transcription product is combined with the amount of a logarithmic multiple dilution of the competitor reverse transcription product 2 for PCR, a method that determines the optimal copy number of the sample. This experiment was derived from PCR Laboratory Guide (Second Edition) by Seed Kang and Qu Lijia.
Operation method
Competitive RT-PCR: Evaluation of copy number
Materials and Instruments
Double-distilled water RT-PCR buffer MMLV Reverse transcriptase SUPERaseIN Taq DNA polymerase Total RNA Random primers dNTP mixture Gene-specific forward primers Gene-specific reverse primers Move Part I: Synthesis of single-stranded cDNA For more product details, please visit Aladdin Scientific website.
PCR Instrument PCR Tubes
I. Materials
1. Buffers, solutions and reagents
Double-distilled water to remove RNAase
10X RT-PCR buffer (100 mmol/L Tris-HCl, pH 8.3, 500 mmol/L KC1, 15 mmol/L MgCl2)
2. Enzyme and enzyme buffer
MMLV reverse transcriptase (100~200U/ul)
SUPERaseIN(20U/ul,Ambion)
3. Nucleic acids and oligonucleotides
dNTP mixture (10 mmol/L, contains all four dNTPs)
Random primers (50umol/L)
Total RNA (1~5ug)
4. Radioactive complex
Competitive RNA transcript, 109 copies (see scheme 4)
5. Experimental equipment
Thin-walled PCR tubes
II.
1. Add 2ul of 50umol/L random primers to the thin-walled PCR tubes on ice, one tube for RNA samples and one tube for competitive RNA transcripts.
2. Add 1-5ug of total RNA to the sample tube (maximum volume 10ul).
3. Add 109 copies of the competitive RNA transcript to the appropriate tube (10ul maximum volume).
4. Add 4ul of 10mmol/L dNTP to each tube.
5. Increase volume to 16ul with RNAase-removed double-distilled water.
6. Heat at 70°C for 10 min to denature the secondary structure, then immediately place on ice.
7. Add 2ul of 10XRT-PCR buffer.
8. Add 1ul of 20U/ul SUPERaseIN.
9. Add 1ul of 100~200U/ul MMLV reverse transcriptase.
10. Incubate at 42°C for 1h.
11. Heat at 95°C for 10 min to inactivate reverse transcription.
12. Continue amplification reaction or store at -20°C.
Part II: PCR Amplification
The purpose of PCR amplification in this protocol is the same as that depicted in Relative RT-PCR (see Protocol 1). The protocol design consists of 6 reactions and an additional 10% fraction. This is an adequate amount for a sample with 4 dilutions of competitor, 1 control without competitor and 1 control without template.
I. Materials
1. Buffers, solutions and reagents
Double-distilled water for RNAase removal
10X RT-PCR buffer (100 mmol/LTris-HCl, pH 8.3, 500 mmol/LKC1, 15 mmol/LMgCl2 )
2. Enzyme and enzyme buffer
TaqDNA polymerase (5U/pl)
3. nucleic acids and oligonucleotides
dNTP mixture (10 mmol/L, contains all four dNTPs)
Gene-specific forward primers (5 umol/L)
Gene-specific reverse primer (5umol/L)
cDNA from the above reverse transcription reaction
4. Radiocomplexes
Competitive RNA reverse transcription reactants (undiluted = 5X107 copies/ul )
Thin-walled PCR tubes
6. Specialized instruments
Programmable PCR instrument
II. Methods
1. Prepare the PCR reaction mixture.
10XRT-PCR buffer 35ul
10 mmol/LdNTP mixture 28ul
5umol/L gene-specific forward primer 14ul
5umol/L gene-specific reverse primer 14ul
5U/ul Taq DNA Polymerase 1.75ul
Double-distilled water for RNAase removal 236.25ul
2. Add an average of 47ul of PCR reaction mixture to each PCR tube, 6 tubes in total, labeled 1~6, and operate on ice.
3. Add 2ul of prepared template cDNA to tubes 1-5 (see Single-stranded cDNA Synthesis).
4. Gradient dilution of the reverse transcription reaction contains the following competitive RNA transcripts.
Undiluted: 1ul reverse transcription mixture = 5X107 copies/ul
102-fold dilution: 1ul of reverse transcription mixture added to 99ul of TE buffer = 5X105 copies/ul
104-fold dilution: add 1ul of 102-fold diluted reverse transcription mixture to 99ul of TE buffer = 5X103 copies/ul
Dilution 106x: add 1ul of 104x reverse transcription mixture to 99ul of TE buffer = 5X101/ul
5. Add 1ul of undiluted competitor to tube #1, corresponding to 5X107 copies/ul.
6. Add 1ul of 102x diluted competitor to tube 2, corresponding to 5X105 copies/ul.
7. Add 1ul of diluted 104 competitor into tube 3, which is 5X103 copies/ul .
8. Add 1ul of the competitor of dilution 106 into tube 4, which corresponds to 5Xl01 copies/ul.
9. Add 1ul of double-distilled water with RNAase removed to tube 5 as a control without competitor.
10. Add 3ul of double-distilled water with RNAase removed to tube 6 as no template control.
11. Run PCR amplification on an appropriate PCR instrument with a hot cap (e.g. GeneAmpPCRSystem9700, AppliedBioSystems). The amplification procedure is as follows.
94°C 30s
Annealing temperature 30s
72°C 30s
30 cycles
12. Check the results on a 2%~2.5% agarose gel containing 1ul/ml ethidium bromide. Sample 5ul of each reaction and the results are expressed in terms of the amount and copy number of transcripts. Because the competitor is designed to have the same reaction kinetics as the target gene, it can be co-amplified with the target gene and the amount of PCR product from the sample and the competitor can be directly compared in each reaction. The intensity of the PCR product from the competitor is consistent with amplification from an endogenous RNA transcript, with the same intensity representing the endogenous information present in the RNA sample. An example of a competitive quantitative RT-PCR experiment is shown in Figure 13-6. 
13. In order to control for variables that affect the efficiency of reverse transcription of competitor and exogenous transcripts, a final RT-PCR experiment is performed in which the sample RNA and competitor RNA are reverse transcribed in the same tube. The amount of competitor RNA used should be close to the amount presented in the initial RT-PCR experiment, and reverse transcription and amplification should be done as described previously. The amount of competitor RNA that produces the same signal intensity as the endogenous message represents the amount of endogenous message in the sample, as described previously.