How Much Ligase Do I Need?

htert2020 at yahoo.com htert2020 at yahoo.com
Fri Aug 18 20:11:20 EST 2006


Hello,

I was wondering how much DNA ligase I will need for the following
situation.

I will be circularizing a 36-base oligonucleotide.  To accomplish the
circularization, the two endpoints of the oligonucleotide will be
brought together by an 18-base oligonucleotide containing bases
complementary to the endpoints of the 36-base oligo.

So, what makes this situation different from a typical vector/insert
ligation is that the DNA is extremely light -- only 36 bases per
circularization.  Therefore, for equivalent masses of DNA at equal
volume, the concentration of DNA endpoints for these tiny pieces of DNA
will be hundreds of times greater than the concentration of DNA
endpoints for a typical vector/insert ligation.

It is the concentration of DNA endpoints that concerns me.  If the
concentration of DNA endpoints is hundreds of times greater than a
typical vector/insert ligation, does that mean I'll need hundreds of
times the amount of ligase?  (My situation requires ligating 200ug of
the oligonucleotide for an in-vivo DNA transfection application)

To me, it seems that the same amount of ligase should work, regardless
of the number of DNA endpoints that need to be ligated at equal volume.
 The reason for this is that at higher concentration of DNA endpoints,
ligase molecules will coincide (thermodynamically) with DNA endpoints
at higher frequency.  This higher frequency means that the ligase
molecules will be finished with the entire job within roughly the same
amount of time as an environment of low DNA endpoint concentration.
And ligase enzymes, being enzymes, don't theoretically break down after
a certain number of ligations.

I guess my main concern is cost.  Ligase can be very expensive.  One of
my requirements is that I must ligate approximately 1 x 10(16) of these
oligos (approx. 200ug).  And it must be affordable, because I may
potentially need to repeat the experiment many times.

Does my above theory seem correct?  Perhaps I will only need 2-10 times
more ligase than is used in a typical reaction, because the ligase will
need an extra bit of time to perform the ligation on each DNA endpoint.
 Financially, that is acceptable.  But it's hard for me to believe that
I will need hundreds of times more ligase than is used in a typical
reaction.

Can someone comment on this?  Your thoughts would be very much
appreciated, and my wallet will thank you as well.

Regards,
Andrew



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