How Much Ligase Do I Need?
hzhen at freeuk.com
Sat Aug 19 03:58:54 EST 2006
htert2020 at yahoo.com wrote:
> I was wondering how much DNA ligase I will need for the following
> 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
> Can someone comment on this? Your thoughts would be very much
> appreciated, and my wallet will thank you as well.
I'm not sure if I understand the problem. It will be the
concentration of the DNA that is more likely to give you trouble, no?
If it is too concentrated, you will get more intermolecular reaction
than intramolecular reaction so you get concatemers rather than
Anyway, you just need to leave the reactions for longer at 25 deg C if
you want to use less ligase. I find the NEB one to be very good, I
think they give far more ligase for the money you pay than say
Promega. In fact the difference is so great that it makes me wonder
if the unit they use are the same. Anyway it doesn't look expensive
More information about the Methods