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Enzyme Units RNaseA


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#1 Axolotl9250

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Posted 25 July 2012 - 03:54 AM

Hi,

I'm following a protocol which requires 20U of RNAse A to be added to a DNA extraction. I can't get my head around working out how much I need to add. The RNaseA I have is this one http://products.invi...roduct/12091021 and apparently it has an activity of not less than 70 U⁄mg material.
Searching the web all I've discovered is "1 enzyme unit (U) = 1 μmol min−1. 1 U corresponds to 16.67 nanokatals.[1]"

I'm thinking if I need 20 units absolute, I need to work that out as a concentration of my volume in microlitres, and then use C1xV1=C2xV2:

2ml of DNA+TE Buffer Solution
20U = 20/2000 = 0.01 Units per microlitre.

I'm assuming if 1ml more or less equals 1g then 1 microlitre more or less = 1 milligram.

70U/mg x V1 = 0.01U/ul x 2000

V1 = (0.01 / 70) x 2000 = 0.285ul

This may be completely wrong but I'm not sure how else to work this out.

Thanks,
Ben W.


#2 mdfenko

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Posted 25 July 2012 - 05:35 AM

this is not a c1v1=c2v2 problem. you are not looking for a final concentration. you are looking for an absolute amount.

you have a 20mg/ml solution of rnase and the specific activity is 70 units/mg (1400 units/ml). you want 20 units to put into your dna solution. 20/70=0.2857.

you, therefore, want approximately 0.29mg (20.3 units). 20mg/ml=1mg/50ul. 0.29x50=14.5ul.

you want to add ~14.5ul of your 20mg/ml rnase solution to your dna solution.

or you can calculate this way:

20units/1400units/ml=0.014286ml=~14.3ul

Edited by mdfenko, 25 July 2012 - 05:42 AM.

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#3 Axolotl9250

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Posted 26 July 2012 - 01:12 AM

Thanks! That's much easier than I was playing it up in my head I think.




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