Ligation failed (continued)

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Sat May 27 06:41:09 EST 2006

>> > Hi,
>> >
>> > I have transformed an insert/gene (2kb) into a cloning vector.
>> > Then cut the gene out to ensure it has sticky ends.
>> > Also, I have linearised an expression vector (11kb) with compatible
>> > sticky ends to the insert.
>> >
>> > Here is the ligation reaction:
>> >
>> > insert (12ng/ul)                     2ul
>> > linearised vector (6ng/ul)        4ul
>> > 10X lig. buffer                       1ul
>> > T4 ligase                              1ul
>> > Water                                  2ul
>> > TOTAL                                 10ul
>> >
>> > overnight incubation @ 4C in accordance to Roche T4 ligase information
>> > sheet.
>> > Total DNA in ligation reaction is 48ng.
>> > insert:vector ratio is 1:1 in accordance to Roche T4 ligase information
>> > sheet when insert to vector size are significantly different.
>> Actually, no, the insert to vector ratio is ~1:5...unless you are
>> cutting asymmetrically, then you don't have enough. You have to go by
>> molar ratio (or the equivalent of dividing mass by size), not mass ratio.
> That ratio doesn't look right to me. And please define molar ratio for 
> DNA?

molar ratio is the same whether you talk of DNA or whatever. It refers to 
number of molecules... you must know what a mol of a substance is, right?

the ratio is correct. 1:5.5 to be exact.

Your vector is 5.5x bigger than the insert, therefore if you have the same 
*mass* (24ng in your example) you have 5.5x more molecules of insert than of 

If vector and insert ar of the same size, then for a 1:1 ratio you use the 
same mass. If the vector is say, 3x bigger, then for a 1:1 ratio (in number of 
molecules, which is what matters here) you have to use a third of the amount 
of vector. Think about it.


This never made much sense to me.
I think he meant that if the vector is 3x bigger then you would use 3x more vector (OR 1/3 of insert) than insert in order to get the 1:1 molar ratio.

Sharp Tool

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