Assume you want to generate the 2ug PCR product of a 1-kb single-copy gene of tomato 950 Mb/1C) from 6 ng of genomic DNA template, what is the minimal amount (ng) of primers you should use, if both primers are 24 bp long. (IC-haploid)
Only calculating primer is not solution u have to calculate actual templets also
But if u calculate only actual templets and rest fallow slandered PCR concentrations than also u can stop your reaction a cycle where u reach desired quantity,
Because primer and dNTPs always used in abundant in slandered PCR reaction.
Templet u can calculate with help of this
1 µg of 1000 bp DNA = 1.52 pmol = 9.1 X 1011 molecules,
So first find out how much actual templets are there for tomato, 6 ng of genomic DNA,
With that u can get how much fragment u going to get in second cycle of PCR, from where u going to double in each cycle.
Still u want to go to calculate primer so your primer should have HPLC grade primer not salt precipitated, than only u can get actual primer concentration of your interest...
Thank you
Praveen Rajvanshi
Hi Praveen,
Thanks for your answer.
So 6ng DNA comes to 0.009 pmol.
Now can you elaborate how much ng primer I need?
you should start with the standard primer concentration of 0,2 uM in the final PCR mix.
You can't predict final PCR product amount from primer concentration. In theory, PCR reactions are assumed to be 100% efficient. In reality, they are not. Standard practice is to add primers to 0.2 microMolar final concentration of each (e.g. 0.5 microliters of a 10 microMolar primer in a total volume of 25 microliters).
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