Irreversible Binding and Time-Dependent Occupancy

Why Irreversible Binding Requires a Time Dimension

For reversible binding, occupancy can approach an equilibrium determined by free ligand concentration and affinity. Irreversible binding is different: a ligand first encounters its target, then converts the encounter complex into a bound state that does not dissociate on the timescale of interest. Consequently, the amount of target converted depends not only on concentration, but also on exposure duration. A concentration that produces modest conversion during a short exposure may produce much more during a longer one.

A useful mechanism has two steps. Free target \(T\) and ligand \(L\) form a reversible encounter complex \(TL\), which can then convert to an irreversibly bound target \(T^*\). The conversion may represent covalent bond formation or another process that makes dissociation negligible during the observation window.

Definition: In the two-step irreversible-binding model, \(T+L \rightleftharpoons TL \rightarrow T^*\). The association and dissociation rate constants are \(k_{\mathrm{on}}\) and \(k_{\mathrm{off}}\), and the irreversible conversion rate constant is \(k_{\mathrm{inact}}\). The irreversible target fraction is the fraction in state \(T^*\); the encounter complex \(TL\) is not itself irreversibly bound.

Assume one ligand binds one target site; free ligand concentration \(C\) is constant during exposure; target is not synthesized or degraded during the modeled interval; and the initial target-state fractions are known. Constant \(C\) is most plausible when ligand is in large excess over target or is maintained by the experimental design. If ligand depletion is material, \(C\) cannot simply be treated as fixed. The basic model also assumes that the measured signal distinguishes the states relevant to the analysis and that “irreversible” means dissociation is negligible over the observation period, not necessarily forever.

Mechanistic Equations and a Useful Approximation

Let \(U\), \(X\), and \(Y\) denote the fractions of total target that are respectively uncomplexed and unconverted, in the encounter complex \(TL\), and irreversibly converted to \(T^*\). Thus \(U+X+Y=1\). Under constant \(C\), define \(a=k_{\mathrm{on}}C\), \(b=k_{\mathrm{off}}\), and \(c=k_{\mathrm{inact}}\). Mass-action kinetics give:

\frac{dU}{dt}=-aU+bX,\qquad \frac{dX}{dt}=aU-(b+c)X,\qquad \frac{dY}{dt}=cX.

These equations show why irreversible occupancy is not generally a simple equilibrium function of \(C\): \(Y\) accumulates at a rate proportional to the amount currently in \(TL\). If \(C\) is removed, association stops, but pre-existing \(TL\) can still either dissociate or convert.

When the encounter-complex fraction relaxes rapidly relative to changes in the unconverted target pool, a quasi-steady-state approximation sets \(dX/dt\approx0\). Write \(R=U+X=1-Y\) for the fraction not yet irreversibly converted. Since \(U=R-X\), this condition gives \(X/R=k_{\mathrm{on}}C/(k_{\mathrm{on}}C+k_{\mathrm{off}}+k_{\mathrm{inact}})=C/(K_I+C)\), where \(K_I=(k_{\mathrm{off}}+k_{\mathrm{inact}})/k_{\mathrm{on}}\). Therefore:

\frac{dR}{dt}\approx-k_{\mathrm{inact}}\frac{C}{K_I+C}R =-k_{\mathrm{obs}}R,\qquad k_{\mathrm{obs}}=\frac{k_{\mathrm{inact}}C}{K_I+C}.

The \(k_{\mathrm{obs}}\) expression is commonly used for time-dependent inactivation. It is not the general exact solution to the three-state model: it relies on rapid equilibration of the encounter process relative to conversion and on constant free concentration. With \(R(0)=1\), its solution is:

R(t)=e^{-k_{\mathrm{obs}}t},\qquad Y(t)=1-e^{-k_{\mathrm{obs}}t}.

A short justification follows by separating variables: \(dR/R=-k_{\mathrm{obs}}dt\). Integrating from \(0\) to \(t\) gives \(\ln R(t)-\ln R(0)=-k_{\mathrm{obs}}t\). Substituting \(R(0)=1\) yields the expressions above. The irreversible fraction approaches one asymptotically in this idealized model, while the rate of accumulation slows as the remaining unconverted pool becomes smaller.

Worked Example: Time-Dependent Conversion at a Fixed Concentration

Suppose, hypothetically, that \(k_{\mathrm{inact}}=0.120\ \mathrm{min}^{-1}\), \(K_I=4.00\ \mathrm{nM}\), and free ligand is held at \(C=4.00\ \mathrm{nM}\) for 10 minutes. Assume the rapid-equilibration approximation applies and initially \(Y(0)=0\).

First calculate the apparent inactivation rate:

k_{\mathrm{obs}}= \frac{(0.120\ \mathrm{min}^{-1})(4.00\ \mathrm{nM})} {4.00\ \mathrm{nM}+4.00\ \mathrm{nM}} =0.0600\ \mathrm{min}^{-1}.

The predicted irreversible fraction at 10 minutes is:

Y(10)=1-e^{-(0.0600\ \mathrm{min}^{-1})(10\ \mathrm{min})} =1-e^{-0.600}=0.451.

Thus the model predicts that 45.1% of the target is irreversibly converted after 10 minutes at this concentration. This is a time-specific prediction, not an equilibrium occupancy at 4.00 nM. A longer exposure at the same concentration would generally produce more conversion under the stated assumptions.

Estimating the Binding-Rate Parameters

At a fixed concentration, a time course can estimate \(k_{\mathrm{obs}}\). If the assay measures the irreversible fraction \(Y(t)\), the approximation implies \(\ln[1-Y(t)]=-k_{\mathrm{obs}}t\) when the initial irreversible fraction is zero. A straight-line fit of log remaining fraction against time has slope \(-k_{\mathrm{obs}}\). With measurements at several concentrations, the resulting \(k_{\mathrm{obs}}\) values can be fit to the saturable relation \(k_{\mathrm{obs}}=k_{\mathrm{inact}}C/(K_I+C)\). The high-concentration plateau informs \(k_{\mathrm{inact}}\), while the concentration at half that plateau is \(K_I\).

Alternatively, fit all time-course observations jointly by nonlinear regression to \(Y(t,C)=1-\exp[-k_{\mathrm{inact}}Ct/(K_I+C)]\), provided the same approximation and initial condition apply. Fitting the full time courses avoids treating the estimated \(k_{\mathrm{obs}}\) values as error-free. The exact three-state equations should be fitted instead when encounter binding is not fast enough for the approximation, or when the experiment observes substantial transient \(TL\).

Worked Example: Estimating \(k_{\mathrm{inact}}\) and \(K_I\) From Time Courses

Consider hypothetical measurements at 10 minutes after constant exposures of 1.00, 4.00, and 16.0 nM. Suppose an assay that isolates irreversible target conversion gives \(Y=0.213\), \(0.451\), and \(0.617\), respectively. Assume initially unconverted target and use the rapid-equilibration model.

For each concentration, calculate \(k_{\mathrm{obs}}=-\ln(1-Y)/t\). At 1.00 nM, \(k_{\mathrm{obs}}=-\ln(0.787)/10=0.0240\ \mathrm{min}^{-1}\). At 4.00 nM, \(k_{\mathrm{obs}}=-\ln(0.549)/10=0.0600\ \mathrm{min}^{-1}\). At 16.0 nM, \(k_{\mathrm{obs}}=-\ln(0.383)/10=0.0960\ \mathrm{min}^{-1}\). The small rounding differences from the underlying values are expected from the reported fractions.

The largest observed rate, \(0.0960\ \mathrm{min}^{-1}\), is approaching the saturating rate. A joint nonlinear fit of the three rates to \(k_{\mathrm{obs}}=k_{\mathrm{inact}}C/(K_I+C)\) gives estimates close to \(k_{\mathrm{inact}}=0.120\ \mathrm{min}^{-1}\) and \(K_I=4.00\ \mathrm{nM}\). For a direct check, these parameters predict rates of \(0.0240\), \(0.0600\), and \(0.0960\ \mathrm{min}^{-1}\) at the three concentrations. At 4.00 nM, the prediction is half the plateau rate, consistent with \(K_I=4.00\ \mathrm{nM}\). In a real analysis, parameter uncertainty and assay error should be assessed rather than reporting these hypothetical fitted values as exact.

What Happens When Free Ligand Is Removed?

After washout, set free ligand concentration to zero, so no new \(TL\) forms. Any encounter complex already present has two possible fates: it converts at rate \(k_{\mathrm{inact}}\) or dissociates at rate \(k_{\mathrm{off}}\). If the post-washout initial encounter-complex fraction is \(X_0\), then for elapsed time \(\tau\):

X(\tau)=X_0e^{-(k_{\mathrm{off}}+k_{\mathrm{inact}})\tau},

The amounts converted and dissociated from that initial complex pool are respectively \(X_0\frac{k_{\mathrm{inact}}}{k_{\mathrm{off}}+k_{\mathrm{inact}}}[1-e^{-(k_{\mathrm{off}}+k_{\mathrm{inact}})\tau}]\) and \(X_0\frac{k_{\mathrm{off}}}{k_{\mathrm{off}}+k_{\mathrm{inact}}}[1-e^{-(k_{\mathrm{off}}+k_{\mathrm{inact}})\tau}]\). These are competing first-order exits from \(TL\); their sum plus the residual \(TL\) equals \(X_0\). Existing \(T^*\) remains in this model, since target turnover and reversal of the converted state were excluded.

Worked Example: Partitioning the Encounter Complex After Washout

Suppose, hypothetically, that immediately after free ligand removal, \(X_0=0.2996\) of total target is in \(TL\). Let \(k_{\mathrm{off}}=0.100\ \mathrm{min}^{-1}\), \(k_{\mathrm{inact}}=0.200\ \mathrm{min}^{-1}\), and consider 5.00 minutes after washout. The combined exit rate is \(0.300\ \mathrm{min}^{-1}\).

The residual encounter complex is \(0.2996e^{-(0.300)(5.00)}\approx0.0668\). The fraction of the initial complex pool that has exited is \(1-e^{-1.50}\approx0.7769\). Of that exited pool, two-thirds converts and one-third dissociates because the rate constants are in a 2:1 ratio. Therefore, conversion is \(0.2996[1-e^{-1.50}](0.200/0.300)\approx0.155\), and dissociation is \(0.2996[1-e^{-1.50}](0.100/0.300)\approx0.0776\).

The initial complex has been partitioned: approximately 0.0668 remains as \(TL\), 0.155 has converted to \(T^*\), and 0.0776 has dissociated. These individually rounded amounts sum to 0.2994; the unrounded amounts sum to 0.2996. The residual complex has neither converted nor dissociated; those outcomes apply only to the portions that have left \(TL\). In the absence of target turnover, the converted fraction persists after free ligand removal, whereas the encounter complex continues to resolve over time.

Assumptions, Diagnostics, and Alternatives

The model matters because a single occupancy measurement can confound concentration and exposure duration. A concentration-response curve measured at one fixed time may describe the experiment, but by itself it does not establish the time course or identify the mechanism. Time-course data at multiple concentrations help distinguish a saturable rate process from a reversible equilibrium response.

Check whether the model is behaving plausibly before interpreting fitted parameters. At each concentration, plot the measured remaining unconverted fraction on a log scale against time. Approximate linearity supports a constant \(k_{\mathrm{obs}}\) over that interval; systematic curvature can indicate an initial encounter-complex transient, changing free concentration, multiple target populations, or other misspecification. Plot residuals against time and concentration after fitting. Trends, rather than random scatter around zero, suggest the assumed concentration dependence or kinetics are inadequate. Also check that the fitted rates increase with concentration and approach a plausible plateau over the measured range.

If all concentrations lie well below \(K_I\), then \(k_{\mathrm{obs}}\approx(k_{\mathrm{inact}}/K_I)C\): the data may inform the ratio but may not separately determine \(k_{\mathrm{inact}}\) and \(K_I\). If all lie far above \(K_I\), they may estimate the plateau but provide little information about \(K_I\). A useful design spans concentrations below, near, and above the transition, and includes multiple times. Assay signals must also distinguish irreversible conversion from transient encounter binding; otherwise a measured “occupancy” may combine states the model treats separately.

The equilibrium reversible-binding model is a simpler alternative when binding and dissociation are reversible and the observation is near equilibrium; it does not represent cumulative conversion. The full reversible kinetic model is appropriate when occupancy rises and falls as free concentration changes without an irreversible state. The rapid-equilibration \(k_{\mathrm{obs}}\) model is simpler and often practical for covalent inactivation, but its approximation should be checked against time-course residuals or the exact three-state model. The exact model requires more kinetic information and may be harder to estimate from sparse data. Neither form automatically accommodates target turnover, heterogeneous target behavior, ligand depletion, or assay artifacts; these require additional structure or experimental controls.

Common Mistakes

Treating irreversible occupancy as an equilibrium value. The correction is to specify exposure duration and model the time-dependent accumulation of \(T^*\).
Calling every ligand-associated target irreversible. \(TL\) is a reversible encounter complex in this mechanism; only \(T^*\) is converted.
Assuming washout instantly removes all bound ligand. Washout prevents new association, but existing \(TL\) can still convert or dissociate.
Fitting the saturation equation without checking its conditions. Confirm that free ligand is approximately constant and encounter binding is fast relative to conversion, or fit the full kinetic equations.
Interpreting \(K_I\) automatically as an equilibrium dissociation constant. Here \(K_I=(k_{\mathrm{off}}+k_{\mathrm{inact}})/k_{\mathrm{on}}\) under the stated mechanism; it need not equal \(k_{\mathrm{off}}/k_{\mathrm{on}}\).

Key takeaway: Irreversible binding accumulates through a time-dependent conversion step. The commonly used \(k_{\mathrm{obs}}\) relation summarizes that process under rapid encounter equilibration, while washout leaves any pre-existing encounter complex to partition between conversion and dissociation.

Check Your Understanding

Use the model assumptions stated above. Solutions are included for checking your reasoning.

At \(C=2.00\ \mathrm{nM}\), \(k_{\mathrm{inact}}=0.100\ \mathrm{min}^{-1}\), and \(K_I=3.00\ \mathrm{nM}\), calculate \(k_{\mathrm{obs}}\) and \(Y\) after 6.00 minutes.
Why can measurements only at concentrations much lower than \(K_I\) fail to identify \(k_{\mathrm{inact}}\) and \(K_I\) separately?
After washout, an encounter-complex pool has \(X_0=0.200\), \(k_{\mathrm{off}}=0.050\ \mathrm{min}^{-1}\), and \(k_{\mathrm{inact}}=0.150\ \mathrm{min}^{-1}\). Calculate residual \(TL\), conversion, and dissociation after 4.00 minutes.
A fitted log remaining-fraction plot curves systematically over time. Give two possible explanations and a model-based diagnostic response.

Solutions. (1) \(k_{\mathrm{obs}}=(0.100)(2.00)/(3.00+2.00)=0.0400\ \mathrm{min}^{-1}\). Thus \(Y(6)=1-e^{-(0.0400)(6.00)}=1-e^{-0.240}=0.213\), or 21.3%. (2) At low concentration, \(k_{\mathrm{obs}}\approx(k_{\mathrm{inact}}/K_I)C\), so the data primarily identify the ratio \(k_{\mathrm{inact}}/K_I\); concentrations near and above the transition are needed to inform the plateau and curvature. (3) The combined rate is \(0.200\ \mathrm{min}^{-1}\), so residual \(TL=0.200e^{-0.800}=0.0899\). The exited amount is \(0.200(1-e^{-0.800})=0.1101\). The conversion share is \(0.150/0.200=0.75\), giving \(0.0826\); dissociation is \(0.25(0.1101)=0.0275\). These sum to \(0.200\), up to rounding. (4) Curvature may reflect a transient encounter-binding phase or changing free ligand concentration, among other causes. Inspect concentration and time residuals, verify exposure conditions, and compare the rapid-equilibration fit with the full three-state kinetic model.