Copies per reaction-clarification needed - (Jul/21/2009 )
I wondering this one thing.
For real time pcr (detection of virus), one wants to measure the sensitivity of the assay by determining the minumun amount of viral copies per reaction that the assay is able to detect.
So, letīs say I have 8 ng/ul of target RNA in the sample (measured by nanospec). By using this formula:
Transcript length: 75nucleotides
Concentration: 8 ng/ul = 8 x 10-9 g/ul
Calculation: (8 x 10-9 g/ul / <75x 340>) x 6.022 x 10e23 =about 1,8e11 molecules per ul
I dilute this by factor 10-9 and I get 188 molecules per ul.
My PCR mix is 25ul that includes 2ul of this RNA sample. My assay is able to detect this dilution so can I say that the detection limit is 376 copies PER REACTION (I put 2 microliters and 2X 188/ul=376) or is the limit this 376 divided by the reaction volume of 25??
Thank you for your help!!
I think your calculation is probably dependent on the size of the viral genome rather than the size of your transcript (unless the transcript is the only RNA in your sample).
bob1 on Jul 23 2009, 11:11 AM said:
Well, itīs the trancript lenght that matters. See the following formula :http://www1.qiagen.com/resources/info/Guidelines_RTPCR/Quantifying.aspx
Check the RNA standards.
Iīm just wondering what the "copies per reaction" actually means. Cos if I now have this,letīs say, 10 000 molecules/ul and I dilute it with the factor 10 to 3, I get 10 molecules per ul. When I add 2 ul of this solution to my 25ul reaction, should I take into account the diluting effect of the reaction volume itself? So would it be 20molecules divided by 25= 0,8 molecules per reaction or?
bob1 on Jul 23 2009, 10:11 AM said:
Or actually youīre absolutely right...Cos if I measure how many molecules or RNA there are in the sample before any PCR reaction, OF COURSE the lenghts should be the entire genome