DNA Ligation Calculator
Enter vector mass, vector and insert lengths, and the target molar ratio to compute the insert DNA mass needed for a ligation reaction.
Input Data
Results
At a glance:Enter vector mass, vector and insert lengths, and the target molar ratio to compute the insert DNA mass needed for a ligation reaction.
Formula
ng insert = ng vector × (kb insert ÷ kb vector) × ratio.
$$m_{ins} = m_{vec} \times \frac{L_{ins}}{L_{vec}} \times R$$How to Use
- Enter the vector mass, vector and insert lengths (kb), and target molar ratio.
- The calculator returns the required insert mass (ng).
Insert mass for common ligation reactions (vector 50 ng)
| Vector length (kb) | Insert length (kb) | Molar ratio | Insert mass needed (ng) |
|---|---|---|---|
| 3 | 1 | 1 : 1 | 16.67 |
| 3 | 1 | 3 : 1 | 50.00 |
| 3 | 1 | 5 : 1 | 83.33 |
| 5 | 1 | 3 : 1 | 30.00 |
| 3 | 2 | 3 : 1 | 100.00 |
Shorter inserts and higher molar ratios change the needed insert mass; at the same ratio, a longer vector needs relatively less insert mass to reach equal moles.
Case Studies
Standard 3:1 sticky-end ligation
Vector 50 ng, length 3 kb; insert length 1 kb; use a 3:1 molar ratio.
ng insert = 50 × (1 ÷ 3) × 3 = 50 ng.
Mix 50 ng vector + 50 ng insert to reach 3:1, suitable for general sticky-end cloning.
1:1 ligation with long vector, short insert
Vector 100 ng, length 5 kb; insert length 1 kb; use a 1:1 molar ratio.
ng insert = 100 × (1 ÷ 5) × 1 = 20 ng.
Because the vector is long and the insert short, only 20 ng insert equals the vector in moles, avoiding excess insert DNA.
FAQ
Why use a molar ratio instead of a mass ratio for ligation?
Ligation success depends on the relative number of vector and insert molecules, not their weight. At the same mass, a longer DNA molecule is fewer in count. Mixing by equal mass would leave long fragments under-represented in molecules, so you must convert mass to molecule (mole) ratio via length.
What molar ratio should I use?
A common starting point is insert:vector 3:1. Compatible sticky ends usually need 1:1 to 3:1; blunt ends or low ligation efficiency may need 5:1 or higher to increase the chance the insert is captured. If you see many self-ligated empty vectors, adjust the ratio and add dephosphorylation.
How much vector should I add?
Typical reactions use about 20–100 ng vector. Too much vector raises self-ligation and background; too little lowers transformation efficiency. Fix a reasonable vector amount (e.g. 50 ng), then compute the matching insert mass from the target ratio and fragment lengths.
Is the computed mass a volume or a pure amount?
The result is the insert DNA mass (ng). In practice, measure the insert stock concentration (ng/μL) first, then convert to the volume to take (μL = needed ng ÷ concentration), and keep the total reaction volume and buffer/ligase proportions correct.
How do sticky-end and blunt-end ligations differ?
Sticky ends anneal via complementary overhangs, giving high efficiency and directional control; blunt ends have no overhang, low pairing probability, and no directionality, so blunt ligation often needs a higher insert:vector ratio, longer time, or more ligase to compensate.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.