How Much Ligase Do I Need?

ChenHA hzhen at
Sat Aug 19 19:20:08 EST 2006

htert2020 at wrote:

> Thanks for your response.  The concentration of DNA will not give me
> any undesirable concatemers because the ratio between the two different
> oligonucleotides will be, more or less, one to one.  Moreover, an oligo
> will have a very difficult time forming concatemers because the only
> way to join two oligos in that manner is through blunt end ligation...
> there are no "sticky ends" because an oligo is one-sided.  Here is a
> text diagram to illustrate.  For best viewing, copy and paste it into
> Windows Notepad or other word processor that displays text in fixed
> character width font:
>        OOOOO
>       O     O
>      O       O
>      O       O
>      O       O
>       O  .  O
>        OOOOO        <== 36mer oligo circularizes where the "." is
>  TTTTTTTTTTTTTTTTT  <== template 18mer oligo
>                     The "." is where the ligase acts

Umm, I don't see how you can't not get concatemers -

    oooo         ooooooo        ooooo
   o    0   .   o       o   .   o
         ooooooo         ooooooo

Although perhaps if your molecules are small, then perhaps it might not 
be too bad.

Off the tangent, but drawing picture like suddenly reminded me of a 
recent paper in Nature using oligos to make tiny picture (lots of 
hairpin DNA structures all annealed together to form a pattern).  Have 
your read that?

> My issue is, 2ug of these little miniature DNA oligos has roughly
> hundreds of times the molar quantity of DNA endpoints than 2ug of DNA
> plasmid/vector of much larger size.  This is just mathematics because
> we're assuming constant DNA mass (2ug).  The typical recommended amount
> of ligase to use, for a typical plasmid/vector application at a 10ul or
> 20ul volume, is anywhere from 1 unit of ligase to maybe 10 or 20 units.
>  But my experiment is not a "typical" plasmid/vector operation that
> consists of large DNA (2kbp-6kbp), because for the same mass of DNA
> (say, 2ug), there will be hundreds of times more DNA endpoints to
> ligate in comparison to the "typical" experiment due to the fact that
> each piece of DNA oligo is very small.  And at the same volume (say,
> 20ul), this means a super-high concentration of DNA endpoints to ligate
> -- by a factor of many hundreds.
> Thus, the initial concerns would be, would a "typical" amount of ligase
> be sufficient, and if not, how much ligase would be effective?  Would I
> have to double, triple, or quadruple the amount of ligase "typically"
> used in a "typical" experiment, or would I have to multiply the amount
> of ligase by a factor of many hundreds?  If hundreds, that can be
> extremely expensive.

If the ligase is stable, and your DNA is likely to be stable as well (no 
contaminant that is likely to degraded it), and there are sufficient ATP 
in the mixture, then presumably you can leave the ligation around for a 
  long time until the reaction complete.  Of course I don't know if any 
of the above conditions are true, since they are always likely to be 
contaminants of some sort, and protein can't be eternally stable, so 
they must be an optimum time period for your reaction, and only you can 
determine that.

I'm just saying that there is no fixed amount of ligase as such, rather 
how long you can leave your reaction for (just make sure there is enough 
ATP).  Is there a method for determining if your reaction is effective?

> But, as mentioned in my previous message, I suspect that roughly the
> same (or slightly more) ligase would be needed than the "typical"
> amount because, although the concentration of DNA endpoints is super
> high, the frequency at which the ligase molecule meets a DNA endpoint
> for ligation should also be super high -- resulting in roughly the same
> reaction time as a "typical" experiment.
> Am I right?  I've searched many places for the answer to this question,
> but could never find it.  Perhaps you or someone here can give some
> insight based on personal experiences and knowledge.

Doesn't sound quite right, it is the number of reaction that the ligase 
has to perform, is that not right?  Concentration is just one of the 
factors to consider when you talk about the rate of reaction.

Just something that you might consider - I read somewhere that using 
sonication (I think it's a sonication bath) can speed up the ligation 
reaction by making ligase detach faster from a ligated ends to another 
unligated ends.  So that a hour reaction can be done in 5 minutes or so. 
  Don't know if it will help in your case (would sonication affect the 
way your DNA anneal? I don't know), but perhaps you can look into that. 
  I don't have the paper, but have a search, I can probably find it 
somewhere if you can't find it.

>> 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
>> to me.
> Thanks for the tip.  I can certainly live with slightly less ligase at
> longer reaction times.  But again, the major concern here is, "a factor
> of hundreds of times".  If I had to use 2-3 times more ligase, even 10
> times more ligase, I'd be happy.  But hundreds of times more ligase?
> That's just too much, and will become a major problem.  If I conduct
> such an expensive experiment many times in the future, that would cost
> me thousands of dollars.
> I happen to like NEB and am very impressed with them.
> Andrew

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