Ligation failed (continued)
customer at newsnet.com
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
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.
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