The concentration placed in a binding experiment is a total concentration. The concentration in a dissociation-constant expression is the free concentration. They are close only when binding removes a negligible fraction of the available ligand. When the binding partner captures an appreciable amount, substituting total ligand for free ligand can distort an affinity estimate.
The repair is mass balance. Count the ligand that is free and the ligand held in complexes, then do the same for its binding partner. A simple one-to-one equilibrium model leads to a quadratic equation. That extra algebra represents molecules already present in the tube; it is not an optional mathematical refinement added after the experiment.
This guide develops one quantitative connection within The Gold Standard of Biophysics. The numerical systems are fictional, with concentrations expressed in nanomolar, or nM. They teach an equilibrium model, not a drug dose or cellular response. Follow the wider purpose through learning that transfers into life and the How X Works library.
eduKateSG · The Gold Standard · Biophysics
Choose the question you need to resolve
Choose a route from the contents of a tube to an equilibrium claim that conserves both binding partners.
Full chapter contents · Transfer checks · Biophysics subject guide
Chapter contents
Name the binding quantities · Chapters 1–3
Solve and compare · Chapters 4–6
Inspect the evidence · Chapters 7–9
Practise and continue · Chapters 10–12
CHAPTER 1 OF 12 · Name the binding quantities
1. Three concentrations can describe the same ligand
Imagine a tube containing a protein P and a ligand L. Some ligand remains free in solution; some forms the complex PL. If the tube is closed and our model contains no other ligand-bearing species, the total ligand concentration is the sum of free ligand and complex concentration.
Write Lₜ for total ligand, L𝒻 for free ligand and B for the concentration of complexes. Then Lₜ = L𝒻 + B. Likewise, if Pₜ is total active protein and P𝒻 is free protein, Pₜ = P𝒻 + B. One complex contains one of each partner in this model.
The subscript matters more than the letter chosen. A tube label reading “10 nM ligand” usually describes what was prepared, not an independent measurement that 10 nM remains free after binding. Ask which quantity was weighed, diluted, inferred or directly observed.
This bookkeeping also identifies the boundary of the explanation. Adsorption to a vessel, nonspecific binding, degradation, multiple sites and inactive protein can require more species or different concentration definitions. A two-species balance does not automatically describe every experiment that uses the word “binding”. Start with a model simple enough to inspect, then state what it omits.
CHAPTER 2 OF 12 · Name the binding quantities
2. Affinity and occupancy answer different questions
For reversible one-to-one binding at equilibrium, define the dissociation constant Kd = P𝒻L𝒻/B. In this concentration-based treatment, Kd has concentration units. The physical assumptions include a suitable dilute-solution approximation and specified solution conditions.
The fraction of active protein sites occupied is θ = B/Pₜ. Using the protein balance in the equilibrium expression gives θ = L𝒻/(Kd + L𝒻). At free ligand concentration equal to Kd, half the protein is occupied.
That statement uses free ligand. It does not say that half occupation occurs when the total ligand added equals Kd. Confusing these statements is the central error this article repairs. Affinity is a parameter of the chosen equilibrium model under stated conditions; occupancy is a state reached with particular available concentrations.
Hulme and Trevethick’s review of equilibrium ligand-binding assays discusses depletion and the interpretation of binding measurements. Its practical message for this calculation is that the ligand balance belongs inside model selection. Our numerical cases below make that relationship explicit without assuming a particular detector or biological effect.
CHAPTER 3 OF 12 · Name the binding quantities
3. Check whether the ligand pool is a reservoir
The familiar expression θ ≈ Lₜ/(Kd + Lₜ) treats total ligand as a useful approximation to free ligand. This can work when B is small relative to Lₜ. It fails when a substantial proportion of ligand has entered complexes.
Define the depletion fraction δ = B/Lₜ for positive Lₜ. Because L𝒻 = Lₜ(1 − δ), the fractional difference between total and free ligand is visible directly. A small ratio of protein to ligand can bound depletion: B cannot exceed Pₜ, so δ cannot exceed Pₜ/Lₜ. That upper bound is useful but does not estimate the actual bound amount by itself.
Do not decide adequacy from one ratio at the largest concentration in a titration. A series may have negligible depletion at its upper end and appreciable depletion near the transition where affinity information is being obtained. Check the model over the informative region.
There is also no single tolerance suitable for every purpose. A rough demonstration and a precise affinity comparison have different demands. If an approximation is used, state its expected effect relative to measurement uncertainty. A mass-balance calculation is often straightforward enough that a questionable approximation saves little work.
CHAPTER 4 OF 12 · Solve and compare
4. Derive the equation that conserves both partners
Substitute P𝒻 = Pₜ − B and L𝒻 = Lₜ − B into Kd = P𝒻L𝒻/B. Rearranging gives B² − (Pₜ + Lₜ + Kd)B + PₜLₜ = 0.
The physically acceptable root is B = [(Pₜ + Lₜ + Kd) − √((Pₜ + Lₜ + Kd)² − 4PₜLₜ)]/2. It must satisfy 0 ≤ B ≤ min(Pₜ,Lₜ). The larger root is rejected because it would consume more of a partner than is present.
For numerical work, an equivalent form can reduce cancellation when the bound amount is small: B = 2PₜLₜ/[Pₜ + Lₜ + Kd + √((Pₜ + Lₜ + Kd)² − 4PₜLₜ)]. A calculator result still needs the physical bounds checked.
Once B is known, calculate free concentrations from the two balances and substitute them back into the Kd expression. This gives three useful checks: neither free concentration is negative, neither total balance loses molecules, and the recovered dissociation constant matches the assumed one. An attractive fitted curve is less informative than an equation whose quantities survive these checks.
Set Pₜ = 10 nM, Lₜ = 10 nM and Kd = 1 nM. The quadratic gives B = (21 − √41)/2 ≈ 7.298 nM. Both free protein and free ligand are therefore about 2.702 nM.
The occupied fraction is 7.298/10 ≈ 0.730, or 73.0%. The equilibrium check is 2.702 × 2.702/7.298 ≈ 1 nM. The depletion fraction is also about 73.0%, because this example has equal protein and ligand totals.
If total ligand were substituted into the reservoir approximation, the predicted occupancy would be 10/(1 + 10) ≈ 90.9%. That estimate asks the protein to behave as though all 10 nM ligand were still free, despite much of it already being counted in complexes.
| Quantity | Mass-balance model | Total-as-free approximation |
|---|---|---|
| Total ligand | 10 nM | 10 nM |
| Free ligand used | 2.702 nM | 10 nM |
| Bound protein fraction | 73.0% | 90.9% |
| Conserves available ligand in this case? | Yes | No, if interpreted as an exact free concentration |
The error is conceptual before it is numerical. Adding a ligand to a tube does not guarantee that it remains in the free pool. Bound ligand cannot simultaneously serve as free ligand in the same mass balance.
CHAPTER 6 OF 12 · Solve and compare
6. Explain a shifting midpoint without changing affinity
At half occupation, B = Pₜ/2 and P𝒻 = Pₜ/2. The equilibrium expression then requires L𝒻 = Kd. The ligand balance gives Lₜ,half = Kd + Pₜ/2.
This is a precise result for the stated one-to-one model. With Kd = 1 nM and Pₜ = 10 nM, half occupation occurs at total ligand 6 nM. With the same Kd and Pₜ = 0.2 nM, it occurs at 1.1 nM. The midpoint on a total-concentration axis has moved even though the underlying affinity has not.
Suppose two laboratories report these different total-concentration midpoints and conclude that the protein preparations have different affinities. The first check should be their active protein concentrations and fitted equations. The observations might be consistent with the same Kd under different depletion conditions.
The midpoint result can also expose a model problem. If a proposed half-occupation point requires less than Pₜ/2 total ligand, a one-to-one closed system cannot supply enough ligand to occupy half the sites. Before fitting a more complicated explanation, check axis units, active concentration, site stoichiometry and whether the measured signal really represents half occupation.
CHAPTER 7 OF 12 · Inspect the evidence
7. The detector does not necessarily report the bound fraction
A fluorescence instrument measures light. A calorimeter measures heat. A surface sensor measures a response associated with material or optical change near a surface. Translating those observations into occupancy requires a measurement model as well as a binding model.
In a particularly simple fluorescence model, response is a linear mixture of free and bound signals, so θ = (R − Rfree)/(Rbound − Rfree). That conversion assumes appropriate baselines, a stable scale and no other concentration-dependent optical contribution. Quenching, aggregation or changing brightness can break it.
The existing guides to fluorescence anisotropy and microscale thermophoresis examine those method-specific signals. This article supplies the explicit molecular balance that a suitable fit may need.
A half-height response is therefore not automatically half occupation. Before using the midpoint equation, explain how the detector scale maps to bound fraction. A second instrument can provide useful independent evidence, but agreement between two numbers is strongest when both observation models are understood. Otherwise two similar fitting assumptions can reproduce the same mistake.
Our equations describe one class of independent sites with one ligand per site, a closed well-mixed volume, known active totals and equilibrium. Multiple binding sites, cooperativity, competing ligands or coupled conformational changes can require another model.
Equilibrium is especially important. Waiting the same number of minutes does not establish that every preparation has equilibrated. A time-dependent response can produce an apparent concentration curve whose midpoint depends on incubation history. Binding speed and equilibrium affinity are different properties.
The active concentration also deserves attention. If only part of a nominal protein preparation can bind, Pₜ in our balance is the concentration of active sites represented by the model. Replacing it with total protein mass converted using molecular weight may be unjustified. Some experiments can estimate an active fraction, but adding extra fitted parameters requires enough informative data.
Finally, a biochemical interaction in solution does not directly establish activity in a cell. Transport, localisation, competition and regulation can change the available pools and downstream consequences. The calculation is useful precisely because it limits the claim: these concentrations and this equilibrium model support this bound amount. It does not turn a binding constant into a clinical or biological outcome.
CHAPTER 9 OF 12 · Inspect the evidence
9. Read uncertainty through the mass balance
Imagine that nominal Pₜ is uncertain. In a strongly depleted regime, the position of the concentration transition depends on how many active sites must be filled. An error in that total can be absorbed into an apparently precise fitted Kd if the model treats Pₜ as exact.
Repeat measurements do not necessarily reveal this systematic error. Re-reading a single prepared titration mainly tests some measurement variation. Preparing independent concentration series, checking stocks and examining another valid experimental orientation address different parts of reproducibility.
Report the fitted equation, parameter definitions, concentration basis, temperature, buffer, incubation conditions, replicates and uncertainty. Preserve raw response as well as any normalised fraction. A smooth curve should not hide a baseline assumption or an inactive fraction imposed silently by software.
Where concentration uncertainty is important, assess its influence rather than displaying extra decimal places. For example, calculate predictions across a credible range of active protein totals and observe how the affinity estimate changes. That sensitivity analysis does not replace calibration, but it tells the reader whether the conclusion depends heavily on an uncertain input. The most useful next experiment may be a concentration check rather than another replicate of the same response.
CHAPTER 10 OF 12 · Practise and continue
10. Transfer checks with explained answers
Change the amount of protein. With Pₜ = 0.10 nM, Lₜ = 10 nM and Kd = 1 nM, bound ligand cannot exceed 0.10 nM. Depletion cannot exceed 1% of total ligand. The total-as-free approximation is therefore much more plausible here than in the 10 nM protein example, although the required accuracy still matters.
Change the midpoint. A one-to-one system has Kd = 2 nM and Pₜ = 8 nM. At half occupation, free ligand is 2 nM, bound ligand is 4 nM and total ligand is 6 nM. The midpoint on a total axis is not 2 nM.
Change the proposed result. A fit with Pₜ = 10 nM and Lₜ = 3 nM claims B = 5 nM. Reject it under the stated model: the tube contains only 3 nM ligand. No numerical fit statistic can repair the missing molecules.
Change the observation. A half-height fluorescence signal appears at 6 nM total ligand. Can you apply the midpoint equation immediately? No. Establish whether the normalised signal represents half occupancy and whether a one-site equilibrium model is appropriate. Transfer means recognising both the conserved structure and the new assumption requiring evidence.
Does depletion mean the ligand was destroyed? Here it means depletion of the free pool through binding. The ligand remains in the system as part of a complex. Destruction, adsorption or removal would need additional accounting.
Is Kd always the concentration producing half response? It is the free ligand concentration producing half occupation in the simple model used here. A detector midpoint, a total-ligand midpoint and a functional response midpoint can be different quantities.
Should I always fit a quadratic? Use the model required by the experiment. The quadratic is appropriate for the stated closed one-to-one equilibrium system. A negligible-depletion approximation may be adequate, while more complicated binding can require a different model.
Can tight binding make Kd difficult to estimate? Yes. When the transition is dominated by stoichiometric filling and concentration uncertainty, the data may constrain active concentration more strongly than a very small Kd. Report what the observations can identify rather than a falsely precise number.
Does lower Kd mean faster binding? Not by itself. Equilibrium and kinetic descriptions are related within particular mechanisms, but a Kd alone does not specify the individual association and dissociation time courses.
CHAPTER 12 OF 12 · Practise and continue
12. Connected reading and source boundaries
Build the physical and chemical account Return to Biophysics and Biochemistry. The biophysics mechanism guide connects models with biological observations.
Follow the method that produced the signal Use the guides to fluorescence anisotropy, microscale thermophoresis, isothermal titration calorimetry, surface plasmon resonance and bio-layer interferometry. Each has its own observation mechanism and experimental constraints.
Primary reference and the next check Hulme and Trevethick’s equilibrium-binding review is the methodological reference for depletion and assay interpretation. Links and source records were checked on 21 October 2026. The algebra, examples and physical-bound checks here follow the explicitly stated one-to-one model; they do not represent measured affinity for a real target.
For the next binding graph you read, identify which axis uses a total concentration and where the model obtains the corresponding free concentration. Reconstruct one point before interpreting the fitted parameter. Continue into The Gold Standard of Education or How X Works.