Understand the relationship
The reasoning behind the result
Occupancy uses the free concentration
Kd=RfreeLfree/B; Rtotal=Rfree+B; θ=B/Rtotal
At equilibrium, the one-to-one model balances complex formation and dissociation. Substituting site conservation into the equilibrium expression gives θ=Lfree/(Kd+Lfree). Kd therefore equals the free ligand concentration at half occupancy, regardless of how much ligand is bound to sites.
The site input counts equivalent binding sites. For a multivalent molecule, assuming that site count equals molecule count can be wrong; even a correct nominal site count does not justify independent-site behavior or reproduce avidity.
Finite totals require a mass balance
B²−(Rtotal+Ltotal+Kd)B+RtotalLtotal=0
When the input is total ligand, its free concentration is unknown because some ligand is in the complex. Combining ligand and site conservation gives a quadratic equation. Only the smaller physical root can represent a bound concentration no greater than either total.
The implementation uses a numerically stable form of that root and computes free quantities without subtracting nearly equal rounded totals. The ledger exposes the resulting ligand depletion; no generic depletion cutoff is used to decide whether a simpler approximation is acceptable.
A target requires free ligand plus the occupied sites
Lfree=Kd·θ/(1−θ); Ltotal=Lfree+θRtotal
An occupancy target first determines the required equilibrium free ligand. Filling the sites also consumes ligand, so the total must include the bound concentration. At 50% occupancy, total ligand is Kd+Rtotal/2, not simply Kd.
A zero target requires zero ligand. For positive Kd, exactly 100% occupancy is an asymptote with no finite concentration solution. This arithmetic is an equilibrium scenario, not an instruction to use a reagent concentration in an unvalidated experiment.
Follow the numbers
An 80% occupancy target with finite sites
- Use Kd=10 nM and total sites=100 nM, with θ=0.8.
- Required free ligand is 10×0.8/(1−0.8)=40 nM; bound ligand is 0.8×100=80 nM.
- Total ligand must be 40+80=120 nM. The remaining free sites are 20 nM, verifying Kd=20×40/80=10 nM.
Entering 40 nM as total ligand would solve a different experiment and would not produce 80% occupancy.
Quick guide
How to use this calculator
- Enter a positive Kd and the concentration of independent sites, which is not automatically antibody molecule concentration.
- Specify whether ligand measurements are free concentrations or total amounts added per volume.
- Enter named ligand scenarios, or a target fraction of occupied sites.
- Check the separate free, bound and total quantities; concentration-based fractions alone do not establish binding stoichiometry or assay suitability.
Calculation method
Calculation and interpretation
Keep free ligand, bound complex, unoccupied sites and total added ligand reconciled at equilibrium.
θ=Lfree/(Kd+Lfree); B=θRtotal; Ltotal=Lfree+B; Kd=(Rtotal−B)(Ltotal−B)/B.
Worked example
An 80% occupancy target with finite sites
Entering 40 nM as total ligand would solve a different experiment and would not produce 80% occupancy.
θ=Lfree/(Kd+Lfree); B=θRtotal; Ltotal=Lfree+B; Kd=(Rtotal−B)(Ltotal−B)/B.
Supported inputs
Precision and limits
A one-to-one experimental model
Assumes one class of independent equivalent sites, reversible one-to-one binding and stated assay conditions. It does not establish affinity in an organism, immune protection, antibody performance, clinical effect or a treatment dose.
Specific binding and free concentration
Multivalent avidity, cooperativity, nonspecific binding, multiple site classes, transport limitation and rebinding require other models. An antigen–antibody system can use this arithmetic only when the stated one-to-one independent-site approximation is justified.
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