Restriction Fragment Count Calculator
Work out how many fragments a restriction enzyme produces and their average size, plus how often a site is expected to occur.
How it works
Linear: fragments = sites + 1 · Circular: fragments = sites · Expected frequency = 1 site per 4^(recognition length) bp
A linear DNA molecule is broken into one more piece than the number of cuts, because the two free ends are already there; a circular plasmid gains no free end, so n cuts give exactly n fragments. The expected frequency of a recognition site assumes random, equal base composition: a 6-base site occurs on average once every 4⁶ = 4,096 bp, so real gels are compared against this baseline to judge whether a sequence is unusually rich or poor in a site.
Worked example
Five cuts on a 10,000 bp linear DNA give 5 + 1 = 6 fragments averaging 10,000 ÷ 6 ≈ 1,667 bp. A 6-base cutter is expected once per 4⁶ = 4,096 bp, so about 10,000 ÷ 4,096 ≈ 2.4 sites would be expected by chance.
Sources: Restriction site — Wikipedia
Frequently asked questions
Why does a plasmid give fewer fragments than a linear molecule?
A circle has no free ends, so the first cut only linearizes it. Each of n cuts then yields exactly n fragments, whereas a linear molecule of the same cut count yields n + 1.
How is the expected cut frequency calculated?
Assuming random sequence with four equally likely bases, a specific k-base site occurs with probability (1/4)^k per position, so on average once every 4^k bases — 256 bp for a 4-cutter, 4,096 bp for a 6-cutter.
Why does my real digest differ from the expectation?
Genomes are not random: GC content, methylation and repeated motifs all skew site frequency, so the expected value is a reference point, not a prediction of exact cut counts.
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This calculator is for educational purposes. Double-check important results before acting on them.