Ligation calculator
How many nanograms of insert to add for a given insert to vector molar ratio.
How many nanograms of insert to add for a given insert to vector molar ratio.
A ligation is set up by molar ratio, not by mass, because what matters is how many insert ends are available to each vector end. Molecules of a short fragment are lighter, so the same nanogram amount is many more molecules. Converting a ratio of molecules into the nanograms you can actually pipette takes one line:
The vector mass is whatever you choose to put in, usually 25 to 100 ng. The ratio is the number of insert molecules per vector molecule, written insert:vector.
"Try an example" loads a 5,000 bp vector at 50 ng, a 1,500 bp insert and the default 3:1 ratio. The calculator gives 45 ng of insert: 50 × (1500 ÷ 5000) × 3 = 45. In moles that is 0.0154 pmol of vector and0.0462 pmol of insert, which is the 3:1 ratio again. Add stock concentrations of 100 ng/µL for the vector and 50 ng/µL for the insert and the reaction becomes 0.5 µL of vector, 0.9 µL of insert, 2 µL of 10× buffer, 1 µL of ligase and 15.6 µL of water.
A base pair of double-stranded DNA averages 650 g/mol, counting both strands and the sodium counter-ions, so:
That is the same conversion thenucleic acid mass to moles converter uses, and it is why 1 µg of a 1 kb fragment is about 1.5 pmol. Picomole amounts in a ligation are tiny: a typical reaction holds 15 to 50 fmol of vector.
Ligase, phosphatase, polynucleotide kinase and adaptors all work on ends, not on whole molecules, so their protocols are written in picomoles of ends. A linear double-stranded fragment has exactly two, whatever its length; a circle has none until you cut it:
The figure sits under the result. In the worked example the cut 5,000 bp vector at 50 ngcarries 0.0308 pmol of ends and the 1,500 bp insert at 45 ng carries0.0923 pmol, so the tube holds 0.123 pmol of ends in total. In the units the older calculators use, 1 µg of a 5 kb linear fragment is 0.615 pmol of ends, and the same microgram of a 500 bp fragment is ten times that, 6.15 pmol, which is why short fragments eat so much more phosphatase or kinase than their mass suggests.
| Insert:vector | When |
|---|---|
| 1:1 | blunt ends, or a large insert |
| 2:1 | standard sticky-end cloning |
| 3:1 | standard sticky-end cloning |
| 5:1 | inserts under about 500 bp |
| 7:1 | small or stubborn inserts |
| 10:1 | very small inserts, adaptors and linkers |
Raising the ratio adds insert molecules, which helps when the insert is short, but too much insert gives tandem inserts, and too little gives empty vector. If self-ligation of the vector is the problem, treat the cut vector with a phosphatase instead of pushing the ratio past about 10:1. For a three-piece ligation, use this calculator once per insert and keep the vector mass the same for both.
Keep the total DNA volume below about 10 µL so that the buffer stays at 1×, and heat inactivate at 65 °C for 10 minutes before an electroporation, since salt carried over from the ligation causes arcing.
For Gibson assembly, NEBuilder or In-Fusion the pieces are joined by overlap rather than by compatible ends, the usual ratio is 2:1 and the total DNA is capped, so use theGibson assembly calculator. For a Golden Gate reaction the pieces are cut and joined in the same tube, and equimolar amounts are the norm. To check that your enzymes give compatible ends in the first place, use therestriction site finder.
Multiply the vector mass by the insert length divided by the vector length, then by 3. With 50 ng of a 5,000 bp vector and a 1,500 bp insert that is 50 x 1500 / 5000 x 3 = 45 ng of insert. The ratio is a ratio of molecules, so a short insert needs less mass than a long one to give the same number of ends.
This calculator uses insert to vector, the way protocols normally write it, so 3:1 means three insert molecules for every vector molecule. A few older protocols write the same thing as 1:3 vector to insert. If a number looks three or five times too small, that is usually the reason.
3:1 is the default for a sticky-end ligation and works for most inserts between about 0.5 and 3 kb. Go to 5:1 or higher for short inserts, where each molecule carries fewer ends per nanogram, and down to 1:1 for blunt ends or for an insert larger than the vector. If you see many empty vectors, dephosphorylate the vector rather than raising the ratio further.
It depends on the length. One base pair of double-stranded DNA averages 650 g/mol, so pmol = ng x 1000 / (bp x 650). 50 ng of a 5,000 bp plasmid is 0.0154 pmol, about 9.3 x 10^9 molecules. The same 50 ng of a 1,000 bp fragment is five times as many molecules.
Convert the mass to picomoles of molecules and double it, because every linear double-stranded fragment has two ends: pmol of ends = 2 x ng x 1000 / (bp x 650). One microgram of a 5,000 bp linear fragment is 0.308 pmol of molecules and 0.615 pmol of ends. A circular plasmid has no ends at all until it is cut, and a fragment cut once gives one linear molecule with two ends. The number is printed under the result and is what phosphatase, kinase and adaptor protocols are dosed against.
You can, but there is little point. T4 ligase is not the limiting factor in a 20 microlitre reaction, and the transformation is: most protocols use 25 to 100 ng of vector and transform 1 to 5 microlitres of the ligation. More DNA mostly means more background. Keep the total DNA volume under about half the reaction so the buffer stays at 1x.