Why Estimate Binding Rates?
An equilibrium affinity summarizes how much ligand is bound after a system has equilibrated, but it does not by itself describe how quickly binding or unbinding occurs. For a reversible one-to-one interaction, the association rate constant \(k_{\mathrm{on}}\) and dissociation rate constant \(k_{\mathrm{off}}\) describe those processes separately. Their ratio determines the equilibrium dissociation constant \(K_d\). Estimating both rates therefore connects time-course evidence to the affinity scale introduced in the earlier material on dissociation constants.
The distinction matters when interpreting experimental binding curves. A slow change in measured binding could reflect slow association, slow dissociation, or both. A time course at a single ligand concentration generally identifies a combined rate, not both constants individually. Concentration variation or an independently measured dissociation phase is needed to separate them.
Association Time Courses and the Observed Rate
The basic model assumes a single homogeneous target population with one binding site per target and reversible one-to-one binding. The free ligand concentration \(L\) is constant during the association observation; ligand depletion is negligible or otherwise accounted for. The rate constants do not change over time, and the measured signal is proportional to bound fraction after any baseline has been removed. Under these assumptions, the fraction bound increases through association and decreases through dissociation:
The factor \(1-B\) is the unbound fraction available to bind. The association flux is therefore \(k_{\mathrm{on}}L(1-B)\), while the dissociation flux is \(k_{\mathrm{off}}B\). Rearranging gives a linear differential equation:
At equilibrium the derivative is zero, so the equilibrium bound fraction \(B_{\mathrm{eq}}\) is \(k_{\mathrm{on}}L/(k_{\mathrm{on}}L+k_{\mathrm{off}})\). Define \(k_{\mathrm{obs}}=k_{\mathrm{on}}L+k_{\mathrm{off}}\). Subtracting the equilibrium value from both sides of the differential equation yields \(d(B-B_{\mathrm{eq}})/dt=-k_{\mathrm{obs}}(B-B_{\mathrm{eq}})\). Separating variables and integrating from time zero to \(t\) gives the exponential time course:
Thus, at each fixed \(L\), an association curve can estimate \(k_{\mathrm{obs}}\), but \(k_{\mathrm{obs}}\) combines both rate constants. Across several free ligand concentrations, the model predicts a straight-line relationship: \(k_{\mathrm{obs}}\) has slope \(k_{\mathrm{on}}\) and intercept \(k_{\mathrm{off}}\). The intercept has units of time\(^{-1}\); the slope has units of concentration\(^{-1}\) time\(^{-1}\).
Worked Example: Estimating Both Rates Across Concentrations
Suppose hypothetical association experiments produce the following fitted observed rates. Assume free ligand concentrations are known, remain effectively constant, and each time course is adequately described by a single exponential.
| \(L\) (\(\mathrm{\mu M}\)) | \(k_{\mathrm{obs}}\) (\(\mathrm{min^{-1}}\)) |
|---|---|
| 0.25 | 0.125 |
| 0.50 | 0.150 |
| 1.00 | 0.200 |
| 2.00 | 0.300 |
The increase from \(L=0.25\) to \(L=0.50\ \mathrm{\mu M}\) is \(0.150-0.125=0.025\ \mathrm{min^{-1}}\) for a concentration increase of \(0.25\ \mathrm{\mu M}\). The slope is therefore \(0.025/0.25=0.100\ \mathrm{\mu M^{-1}min^{-1}}\). The intercept, calculated from the first point, is \(0.125-(0.100)(0.25)=0.100\ \mathrm{min^{-1}}\). The other points agree with this line.
Accordingly, \(k_{\mathrm{on}}=0.100\ \mathrm{\mu M^{-1}min^{-1}}\) and \(k_{\mathrm{off}}=0.100\ \mathrm{min^{-1}}\). Converting the association constant to molar units gives \(0.100\ \mathrm{\mu M^{-1}min^{-1}}\times10^6\ \mathrm{\mu M/M}=1.00\times10^5\ \mathrm{M^{-1}min^{-1}}\). The equilibrium dissociation constant is:
The units cancel to concentration, as required. At \(L=1.00\ \mathrm{\mu M}\), the model predicts \(B_{\mathrm{eq}}=1.00/(1.00+1.00)=0.500\). This agreement is an internal check on the rate estimates and their interpretation.
Estimating Rates From Dissociation or Combined Information
A dissociation phase can provide a direct estimate of \(k_{\mathrm{off}}\). If free ligand is rapidly removed so that rebinding is negligible, then \(L=0\) during the measurement and the model reduces to \(dB/dt=-k_{\mathrm{off}}B\). After baseline correction, the solution is \(B(t)=B(0)e^{-k_{\mathrm{off}}t}\). A plot of the logarithm of bound fraction against time has slope \(-k_{\mathrm{off}}\), provided the single-exponential assumptions hold. A slow washout, appreciable rebinding, or an unremoved signal baseline can bias this estimate.
Worked Example: Estimating \(k_{\mathrm{off}}\) From Washout
In a hypothetical washout experiment, baseline-corrected bound fractions are \(B(0)=0.800\) and \(B(20\ \mathrm{min})=0.439\). Assume ligand removal is rapid, rebinding is negligible, and the measured bound fraction follows a single exponential. Using the two measurements:
The ratio is \(0.439/0.800=0.54875\), so the estimated rate is approximately \(0.0300\ \mathrm{min^{-1}}\). The estimate describes the fractional dissociation rate under the assumed model, not a concentration-dependent association rate.
Suppose an independent association analysis estimates \(k_{\mathrm{on}}=0.100\ \mathrm{\mu M^{-1}min^{-1}}\). Then \(K_d=0.0300/0.100=0.300\ \mathrm{\mu M}\). At \(L=0.500\ \mathrm{\mu M}\), this implies \(B_{\mathrm{eq}}=0.500/(0.500+0.300)=0.625\). The concentration units and equilibrium prediction are consistent with the estimated rate constants.
It is also possible to recover both rates from a fitted \(k_{\mathrm{obs}}\) and equilibrium fraction at one known concentration, if both quantities are independently well estimated and the model applies. This is algebraically valid, but the estimates may be strongly dependent on the quality and duration of the time course.
Worked Example: Recovering Rates at One Concentration
Consider a hypothetical time course at \(L=0.500\ \mathrm{\mu M}\), with fitted \(k_{\mathrm{obs}}=0.160\ \mathrm{min^{-1}}\) and \(B_{\mathrm{eq}}=0.625\). Assume the free concentration is known and constant and that the observed curve meets the single-exponential model assumptions.
Because \(B_{\mathrm{eq}}=k_{\mathrm{on}}L/k_{\mathrm{obs}}\), the association contribution is \(k_{\mathrm{on}}L=(0.625)(0.160)=0.100\ \mathrm{min^{-1}}\). Therefore \(k_{\mathrm{on}}=0.100/0.500=0.200\ \mathrm{\mu M^{-1}min^{-1}}\). The remaining contribution is dissociation:
Consequently, \(K_d=0.0600/0.200=0.300\ \mathrm{\mu M}\). Substituting that result into the equilibrium equation gives \(0.500/(0.500+0.300)=0.625\), matching the fitted equilibrium fraction. The algebraic consistency check succeeds, although it cannot by itself establish that the underlying model is correct.
Assumptions, Diagnostics, and Alternatives
Rate estimates are meaningful only to the extent that the model's assumptions are credible. In addition to one-to-one reversible binding, the analysis assumes a homogeneous binding population, appropriate baseline handling, and no important delay between the free ligand concentration at the target and the concentration used in the calculation. If ligand depletion is substantial, nominal added concentration may differ from free concentration; the binding mass balance discussed earlier is then relevant. Ignoring depletion can distort the estimated concentration dependence.
Inspect the fitted time courses as well as the summary rates. Systematic residual patterns, a visibly multi-phase curve, concentration-dependent residual structure, or estimates that change markedly when a few time points are omitted can indicate model inadequacy. For the \(k_{\mathrm{obs}}\)-versus-\(L\) relationship, curvature or an intercept inconsistent with an independently measured dissociation rate suggests that the simple model may not capture the data. A negative estimated intercept or slope is also incompatible with the interpretation as positive rate constants; it is a warning about noise, concentration errors, or model mismatch, not evidence for a negative physical rate.
The principal alternatives answer different questions. Fitting equilibrium binding data estimates \(K_d\), but equilibrium data alone identify the ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\), not the two rates separately. Fitting association curves jointly across concentrations estimates the line relating \(k_{\mathrm{obs}}\) to \(L\). Adding a dissociation phase can strengthen identification of \(k_{\mathrm{off}}\). A global fit of raw time-course data can estimate shared rate constants directly and account for measurement errors, while separate curve fits followed by a line fit are easier to inspect but can propagate uncertainty less fully. Neither approach repairs a wrong binding model or inaccurate free-concentration measurements.
A common mistake is to treat \(k_{\mathrm{obs}}\) at one ligand concentration as \(k_{\mathrm{on}}\), or to call its reciprocal \(K_d\). This is dimensionally and mechanistically wrong: \(k_{\mathrm{obs}}=k_{\mathrm{on}}L+k_{\mathrm{off}}\) has units of time\(^{-1}\), whereas \(k_{\mathrm{on}}\) has units of concentration\(^{-1}\) time\(^{-1}\), and \(K_d\) has units of concentration. At a single concentration, a fitted \(k_{\mathrm{obs}}\) alone does not separate association from dissociation. Use multiple concentrations, independent dissociation information, or additional equilibrium information, and check the units before interpreting any estimate.
Finally, the rate constants are conditional on the experimental system and model. Multiple target states, conformational changes, nonspecific binding, transport limitations, or incomplete ligand removal can produce apparent rates that do not correspond to a single elementary binding step. A good numerical fit is not sufficient evidence for a one-to-one mechanism; mechanistic interpretation should be supported by the design and diagnostics.
Check Your Understanding
Try each question, then compare your reasoning with the solutions.
- If \(k_{\mathrm{on}}=0.040\ \mathrm{\mu M^{-1}min^{-1}}\) and \(k_{\mathrm{off}}=0.020\ \mathrm{min^{-1}}\), calculate \(K_d\), including units.
- At \(L=2.00\ \mathrm{\mu M}\), a model has \(k_{\mathrm{on}}=0.030\ \mathrm{\mu M^{-1}min^{-1}}\) and \(k_{\mathrm{off}}=0.010\ \mathrm{min^{-1}}\). Calculate \(k_{\mathrm{obs}}\) and its units.
- Why can equilibrium binding measurements alone not generally identify both \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\)?
- A dissociation experiment has \(B(0)=0.900\) and \(B(10\ \mathrm{min})=0.900e^{-0.40}\). Under the single-exponential washout assumptions, calculate \(k_{\mathrm{off}}\).
Solutions. (1) \(K_d=0.020/0.040=0.500\ \mathrm{\mu M}\); time units cancel and the result has concentration units. (2) \(k_{\mathrm{obs}}=(0.030)(2.00)+0.010=0.0700\ \mathrm{min^{-1}}\). (3) At equilibrium the occupancy depends on the ratio \(k_{\mathrm{off}}/k_{\mathrm{on}}\); multiplying both rates by the same positive factor leaves that ratio unchanged, so the equilibrium curve does not determine their absolute speeds. (4) The bound-fraction ratio is \(e^{-0.40}\), so \(k_{\mathrm{off}}=-\ln(e^{-0.40})/(10\ \mathrm{min})=0.0400\ \mathrm{min^{-1}}\).