I understand the reasons why too much starting template can inhibit the formation of PCR amplicons, and I am aware of the guideline amounts of DNA in weight/volume measurements. But obviously the same amount of copy numbers will not be present in a 50ug/ml DNA sample extracted from a virus that has a genome size of 100,000 base pairs compared with the same concentration of DNA extracted from a plant that has a genome size of 150Gb. So what is the optimal starting DNA template copy number desired for a PCR reaction (assuming of course that the primers will only bind the targeted region of the genome?)
this is a recommed dosage from my pfu pCR kit (50ul)
genome DNA: 100-200ng
plasmid DNA: 1-10ng
cDNA: 50-100ng
λDNA: 1-10ng. hope this will help you.
Well, as per our experience it can be around 25 to 50 copies. Real time PCR can detect these copies in a 40 cycles amplification program, Ct would be around 38-39.
Approximately 104 copies of target DNA are required to detect product in 25-30 PCR cycles
I would first try to obtain the PCR product and clone it, it will be beneficial too as it will be your positive control. Next, get the concenration of the plasmid and convert to copy number.
Finally, use the primers and try to amplify 20 copy number to up to 2 x10^6 copy number. This will be a test to indicate the sensitivity of the primer. If it cannot detect to as low a 20 copy number, you can further optimize other components in the PCR reaction to be able to detect to 20 copy number.
Once you have reached an optimized PCR mix and cylce to detect up to 20 copy numbers, you can rest assure as you have a positive control and also proof of the ability of your PCR to ampilfy your target DNA.
- Badges
- Science topic
- Similar topics
- Biological Science
- Molecular Biology
- Biomolecules
- Nucleic Acids
- DNA