Help with Enzymatic Calculations

Darren Natale camdna at ubvmsb.cc.buffalo.edu
Mon Jun 7 09:22:00 EST 1993


In article <1993Jun5.060506.6712 at news.weeg.uiowa.edu>, rcuevas at news.weeg.uiowa.edu (Ramon 'Jammin' Cuevas) writes...
> 
> 
>I hope somebody can help me out with this (potentially easy) problem.  :)
>I'm currently a second year biology student, and I'm gradually learning the
>laboratory techniques involved with molecular biology (ie. electrophoresis,
>blotting, uses of enzymes/buffer/DNA solutions, etc.)  Right now, I'm learning
>the methods involved with restriction enzyme digestion, and I must admit that
>I'm having a difficult time understanding how to determine the "correct"
>amount of restriction enzyme to use for a given amount of DNA sample - to be
>more specific, I don't understand how the concept of "units" fits into the
>overall schema (I haven't had any formal classes in molecular biology or
>genetics, so I don't know if this is an "elementary"-type of question.) For
>example, given that I have 500ng=0.5ug of DNA diluted in 20uL of solution, and
>also given that the enzyme (stock) solution contains 20,000 "units" per mL,
>how do I determine the volume amount of enzyme solution to use (based on the
>"units" concept).  Any help is most greatly appreciated. Thanks!
> 
> 
>Ramon F. Cuevas
>rcuevas at umaxc.weeg.uiowa.edu
> 

As another poster said, a unit of restriction enzyme is the amount that will
digest to completion some reference DNA (usually lambda phage) in a specified
amount of time. Unit definitions can be found in the catalog of the enzyme
supplier.

I was taught to account for differences in the number of cutting sites between
the reference DNA (e.g. lambda) and the DNA of interest. For example, lambda
has 121 TaqI sites, while pUC19 has only four. Intuitively, one would expect
that an pUC19 would require much less enzyme than lambda based on number of
sites. Of course, the total number of sites in the reaction will depend on
the total number of molecules, too. So, this would need to be taken into
account also. Back when both money, enzyme, and time was very precious, I
derived the following simple formula to do just the calculation you need:


the number of units needed per ug DNA= 

    (number of plasmid cuts/Kb size of plasmid) * (Kb lambda/# of lambda cuts)

kb lambda = 48.5

This formula will give a good estimate of the amount of enzyme needed to 
digest your DNA of interest. If the reference DNA differs from lambda, make
the appropriate changes. Of course, you will need to know the number of 
sites in your molecule of interest to use the formula. If you don't, use an
estimate based on the size of DNA and the number of bases in the recognition
sequence for the enzyme of interest.

D. Natale
camdna at ubvms..cc.buffalo.edu

P.S. Thank goodness I don't do this anymore :)



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