Jump to content

  • Log in with Facebook Log in with Twitter Log in with Windows Live Log In with Google      Sign In   
  • Create Account

Submit your paper to J Biol Methods today!
- - - - -

Proteinase K, anybody?

DNA extraction proteinase proteinase k

  • Please log in to reply
1 reply to this topic

#1 hesaguy



  • Members
  • Pip
  • 1 posts

Posted 18 May 2012 - 03:58 PM


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...

#2 phage434



  • Global Moderators
  • PipPipPipPipPipPipPipPipPipPip
  • 2,747 posts

Posted 18 May 2012 - 04:56 PM

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

Home - About - Terms of Service - Privacy - Contact Us

©1999-2013 Protocol Online, All rights reserved.