Linear DNA ends calculator
Maintained by the Talindrew team · Last updated
Every linear double-stranded DNA molecule has two ends, so picomoles of ends is twice the picomoles of molecules: pmol ends = 2 × µg × 10⁶ / (N × 660), where N is the length in base pairs. One microgram of a 1,000 bp fragment has 3.03 pmol of ends.
Result
A linear double-stranded molecule has two ends, so pmol of ends = 2 × pmol of molecules. Circular DNA has zero free ends.
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What it looks like
Formula
pmol ends = 2 × µg × 10⁶ / (N bp × 660)
Worked example
Given: 1 µg of a 1,000 bp linear fragment
- 1.Molecules: 1 µg × 10⁶ / (1,000 × 660) = 1.52 pmol.
- 2.Ends: 2 × 1.52 = 3.03 pmol.
1 µg of 1,000 bp linear dsDNA = 3.03 pmol of ends
How the calculation flows
Units & constants
| Ends per linear molecule | 2 |
|---|---|
| Ends per circular molecule | 0 (until cut; one cut ⇒ 2 ends) |
| Average bp weight | 660 g/mol |
| Typical uses | End-labelling, phosphatase (CIP/SAP) units, adapter ligation stoichiometry |
| Price | Free |
When do you need pmol of ends rather than pmol of DNA?
Enzymes that act on termini are dosed per end, not per molecule.
- T4 polynucleotide kinase end-labelling: units and ATP are specified per pmol of 5′ ends.
- Dephosphorylation with CIP or SAP: vendor unit definitions are per pmol of ends.
- Adapter and linker ligations: the adapter:end molar ratio sets the outcome.
- Terminal transferase tailing: dosing is per pmol of 3′ ends.
How do restriction digests change the count?
Each cut adds two new ends. A circular plasmid cut once yields one linear molecule with 2 ends; cut n times it yields n fragments with 2n ends in total. A linear fragment cut once yields two fragments and 4 ends. If you digest before an end-dependent reaction, count ends on the digested species, not the starting molecule.
Frequently asked questions
No: a covalently closed circle has zero free ends, which is why undigested plasmid is invisible to end-labelling and phosphatase reactions. Linearising it with a single-cutter enzyme creates exactly two ends per molecule.
Because the only difference is the factor of two for the two termini of each linear duplex. The underlying mass-to-moles arithmetic, µg × 10⁶ / (N × 660), is identical.
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