Hello,
I did the PCR (long-range, target ~5 Kb) that is shown in the attached file. The lanes are:
1. ladder
2. target (plasmid purified from E. coli after transfection), 50 ng/uL corresponding to 10^9 plasmids.
3. 1:1000 dilution of previous (~10^6 plasmids)
4. 1:10 dilution of previous (~10^5 plasmids)
My question is: can the PCR have failed in lane 2 because there were too many plasmids?
The failure in lane 4 can be explained either in poor dilution or poor sensitivity of the assay; lane 3 worked as expected (the target is the high mol. band); I am not sure about lane 2.
The fact that there is a smear suggests that there is a lot of DNA, thus I guess the reaction got saturated. Would that be right to assume?
Also, I did not linearize the plasmid to save time and money...
Thank you
I think that you are correct. The smear is concatenated pcr product caused by over amplification so that in later cycles there is much product and less primer. The product anneals with itself at wrong positions so starts to form longer templates which have primer binding sites at the ends so these also amplify . The logic is that every 10 cycles there is 1000 times more product so after 30 cycles theorstically 10exp9 times as much product. This is impossible so something has to fail in the pcr and one of these things is template binding
You may use the DNA concentration for PCR less than 5 ng to get a better amplification.
If you use 1 ul of the plasmid (50ng/ul) as a template in the PCR for lane2, why I cannot see the plasmid band in lane2? The plasmid should be there after PCR.
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