45
ng
95
ng
4.75
ng/µL
0.9
x
45
ng
95
ng
4.75
ng/µL
0.9
x
The DNA Ligation Calculator determines the optimal amount of insert DNA needed for a ligation reaction based on the vector-to-insert molar ratio. Successful ligation requires the right stoichiometric balance between vector and insert molecules. Too little insert leads to empty vector background, while too much can result in multiple inserts or concatenation. This calculator takes the guesswork out of setting up ligation reactions.
The amount of insert DNA is calculated using the following formula:
Insert (ng) = (Insert size / Vector size) × Vector amount (ng) × Insert:Vector molar ratio
This formula works because:
Common molar ratios:
Inputs
Results
For a 5 kb vector (50 ng) with a 1.5 kb insert at 3:1 molar ratio, you need 45 ng of insert DNA for a total of 95 ng DNA in the reaction.
Inputs
Results
A small 500 bp insert requires 62.5 ng at a 5:1 ratio with 100 ng of a 4 kb vector.
For sticky-end (cohesive) ligations, a 3:1 insert-to-vector molar ratio is standard and works well in most cases. For blunt-end ligations, which are less efficient, try a 5:1 or even 10:1 ratio. If you are cloning a very large insert into a small vector, a 1:1 ratio may work better to reduce concatenation.
For a standard 10 µL T4 DNA ligase reaction, aim for 50–200 ng total DNA. Using too much DNA can actually inhibit ligation by crowding the reaction. A typical setup uses 50 ng of vector, which keeps total DNA in the optimal range for most insert:vector ratios.
Common reasons include: (1) incomplete restriction enzyme digestion leaving uncut vector, (2) insufficient dephosphorylation of the vector leading to self-ligation background, (3) degraded or inactive T4 DNA ligase, (4) incompatible buffer conditions, or (5) damaged DNA ends from prolonged gel extraction. Always include a vector-only control to assess background.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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