1 reply to this topic

[hesaguy]

Hi-

I'm trying to make up a nuclear lysis buffer off of a paper's protocol.  They call for using the following:

400mM   EDTA
2% (p/v)  N Lauryl sarkosyl
1 mg/ml  Proteinase K

As the title was alluding, my question is in regards to the proteinase K (sigma 39450-01-6).  Sigma's directions said the enzyme is soluble in water (1mg/ml) and I have 100mg of it.  I am confused by how much to add.  Assuming I add 100 ml of H2O to my bottle (make some aliquots out of it), would I just add 1ml, at 1mg/ml, of the enzyme to the rest of my buffer?

Help would be appreciated.  Thanks!




BTW: http://www.amjbot.or...3.full.pdf+html the link to the paper...

[phage434]

No, the final concentration of proteinase-K is supposed to be 1 mg/ml.  If you make 10 ml of buffer, you need 10 mg of proteinase-K.  You could get that by, for example, dissolving your 100 mg sample in 10 ml of water (a solution of 10 mg/ml) and using 1 ml of that in 9 ml containing the remainder of your buffer components.  Given that the EDTA is normally at 500 mM stock, the buffer was probably intended to be made this way:

(for 10 ml)
8 ml 500 mM EDTA stock
1 ml proteinase-K at 10 mg/ml
0.2 g N lauryl sarkosyl
water to bring the solution to 10 ml