Residence Time From Dissociation Kinetics

Two Timescales From One Dissociation Rate

A ligand’s dissociation rate describes how quickly bound ligand leaves its receptor under specified conditions. Two useful timescales follow from that rate: the mean receptor residence time, which is the average time a bound ligand remains bound before dissociating, and the dissociation half-life, the time by which half of an initially bound cohort has dissociated. They are related, but they are not the same summary.

Neither quantity is the duration of drug exposure. Exposure describes how long drug is present in a relevant compartment or at a relevant concentration. Residence time describes the duration of an individual bound episode under a dissociation model. Drug can be cleared while some ligand remains bound, or drug can remain available after particular ligand molecules have dissociated. The quantities answer different questions.

Definition: Under a constant, first-order dissociation model, \(k_{\mathrm{off}}\) is the dissociation rate constant, with units of inverse time. The mean receptor residence time is \(\tau=1/k_{\mathrm{off}}\). The dissociation half-life is \(t_{1/2}=\ln(2)/k_{\mathrm{off}}\). Both are conditional on a ligand being bound at the start of the dissociation period and on the model assumptions holding.

Model and Derivation

Consider a cohort of ligand–receptor complexes that are bound at time zero. Assume that each complex has a constant probability of dissociating per unit time, represented by the same \(k_{\mathrm{off}}\), and that dissociation events are independent. For this calculation, dissociation is treated as a one-way transition out of the bound state: after a ligand dissociates, it does not rebind during the observation period. The rate is constant in time, and the bound complexes are adequately described by a single kinetic population.

Let \(S(t)\) be the fraction of this initially bound cohort still bound at time \(t\). First-order dissociation means that the instantaneous loss is proportional to the amount still bound:

\frac{dS(t)}{dt}=-k_{\mathrm{off}}S(t), \qquad S(0)=1.

Separating variables and integrating gives \(\ln S(t)=-k_{\mathrm{off}}t\), so the bound fraction remaining is exponential:

S(t)=e^{-k_{\mathrm{off}}t}.

The dissociation-time probability density is the rate of loss from the surviving cohort, \(f(t)=k_{\mathrm{off}}e^{-k_{\mathrm{off}}t}\), for \(t\geq 0\). Its mean is obtained by integrating time against that density. Integration by parts gives the result directly:

\begin{aligned} E[T] &=\int_0^\infty t\,k_{\mathrm{off}}e^{-k_{\mathrm{off}}t}\,dt\\ &=\left[-t e^{-k_{\mathrm{off}}t}\right]_0^\infty +\int_0^\infty e^{-k_{\mathrm{off}}t}\,dt\\ &=0+\frac{1}{k_{\mathrm{off}}} =\tau. \end{aligned}

For the half-life, set \(S(t)\) equal to one-half and solve. The logarithm appears because the exponential model is being evaluated at the time when half the initial cohort remains:

\frac{1}{2}=e^{-k_{\mathrm{off}}t_{1/2}} \quad\Longrightarrow\quad t_{1/2}=\frac{\ln(2)}{k_{\mathrm{off}}}.

Thus \(t_{1/2}=\ln(2)\tau\), or approximately \(0.6931\tau\). The half-life is less than the mean in this model. The half-life is also the median dissociation time: half the complexes have dissociated by then, and half remain bound. It should not be reported as the mean.

Worked Example: Calculating Both Times From a Dissociation Rate

Suppose a hypothetical binding experiment gives \(k_{\mathrm{off}}=0.0250\ \mathrm{min}^{-1}\), and the single-rate, no-rebinding model is considered adequate. Calculate the mean residence time and the dissociation half-life.

The reciprocal rate has units of minutes, so the mean residence time is

\tau=\frac{1}{k_{\mathrm{off}}} =\frac{1}{0.0250\ \mathrm{min}^{-1}} =40.0\ \mathrm{min}.

The half-life uses the same rate and time unit:

t_{1/2} =\frac{\ln(2)}{0.0250\ \mathrm{min}^{-1}} =\frac{0.693147}{0.0250}\ \mathrm{min} =27.7\ \mathrm{min}.

These values describe different summaries of dissociation: the model’s mean bound duration is 40.0 minutes, while half of the initially bound cohort has dissociated by about 27.7 minutes. As a check, the model predicts \(S(40.0)=e^{-0.0250(40.0)}=e^{-1}=0.3679\), not one-half, at the mean time.

Units, Rounding, and Interpretation

A rate constant has units of inverse time. Its reciprocal therefore has units of time, and dividing \(\ln(2)\), a dimensionless number, by the rate gives the same time units. Keep units consistent throughout: a rate in \(\mathrm{h}^{-1}\) produces times in hours, while a rate in \(\mathrm{min}^{-1}\) produces times in minutes. Convert the rate or the final time when a different unit is needed.

Worked Example: Converting a Half-Life to a Mean Without Rounding Error

A hypothetical ligand has a dissociation half-life of 60.0 minutes under the stated first-order model. Find \(k_{\mathrm{off}}\) and the mean residence time. Retain the unrounded rate in the second calculation.

Rearranging the half-life equation gives

k_{\mathrm{off}}=\frac{\ln(2)}{t_{1/2}} =\frac{0.693147}{60.0\ \mathrm{min}} =0.0115525\ \mathrm{min}^{-1}.

The mean residence time is therefore

\tau=\frac{1}{k_{\mathrm{off}}} =\frac{60.0\ \mathrm{min}}{\ln(2)} =86.6\ \mathrm{min}.

The mean exceeds the half-life, as expected, and \(\tau/t_{1/2}=1/\ln(2)\approx1.443\). The displayed rate is rounded for reporting; using that rounded value as if it were exact can slightly change the reciprocal. Carrying the unrounded expression \(60.0/\ln(2)\) through the calculation gives 86.6 minutes.

A useful interpretation of the exponential model is that the fraction remaining depends on the product \(k_{\mathrm{off}}t\), not on either quantity alone. For example, after one mean residence time, \(S(\tau)=e^{-1}\approx0.3679\); after one half-life, \(S(t_{1/2})=0.5000\). An individual ligand does not remain bound for exactly the mean or exactly the half-life. These are population summaries of a distribution of dissociation times.

Residence Time Is Not Exposure Duration

In pharmacology, exposure commonly refers to the presence or concentration of drug in plasma, tissue, or another compartment over time. Residence time calculated from \(k_{\mathrm{off}}\) concerns the bound-state episode. It does not state how long free drug remains in the body, how long a dose maintains a chosen concentration, or how long a response persists. Those outcomes can depend on processes beyond dissociation, so they cannot be inferred from \(k_{\mathrm{off}}\) alone.

For instance, a treatment might maintain free drug exposure for 12 hours even if its mean bound-state residence time is much shorter. Conversely, a compound may have a long mean bound-state residence time even when free drug concentration declines sooner. The rate-based calculation does not by itself tell how many receptors are occupied at a later time in a system where free ligand is available, because dissociated ligand may rebind. It describes the fate of an initially bound cohort under the specified dissociation-only assumptions.

Worked Example: Comparing Residence Times With a Fixed Exposure Window

Two hypothetical ligands are studied under conditions supporting first-order dissociation without rebinding. Ligand A has \(k_{\mathrm{off}}=0.120\ \mathrm{h}^{-1}\), and ligand B has \(k_{\mathrm{off}}=0.0400\ \mathrm{h}^{-1}\). A separate hypothetical dosing profile maintains measurable free-drug exposure for 12 hours. Calculate the residence summaries and the fractions of each initially bound cohort remaining at 12 hours.

For ligand A:

\tau_A=\frac{1}{0.120\ \mathrm{h}^{-1}}=8.33\ \mathrm{h}, \qquad t_{1/2,A}=\frac{0.693147}{0.120\ \mathrm{h}^{-1}}=5.78\ \mathrm{h}.

At 12 hours, the fraction of the initially bound cohort still bound is

S_A(12)=e^{-(0.120\ \mathrm{h}^{-1})(12\ \mathrm{h})} =e^{-1.44}=0.2369.

For ligand B:

\tau_B=\frac{1}{0.0400\ \mathrm{h}^{-1}}=25.0\ \mathrm{h}, \qquad t_{1/2,B}=\frac{0.693147}{0.0400\ \mathrm{h}^{-1}}=17.3\ \mathrm{h}.

Its corresponding fraction at 12 hours is

S_B(12)=e^{-(0.0400\ \mathrm{h}^{-1})(12\ \mathrm{h})} =e^{-0.480}=0.6188.

Under this model, ligand B has the longer bound-state timescale. The 12-hour exposure window is a separate description of drug availability; it does not replace either calculated residence time, nor does it establish the fraction of receptors occupied in a system where rebinding can occur.

Assumptions, Diagnostics, and Alternatives

A \(k_{\mathrm{off}}\)-based mean and half-life are interpretable when a single, approximately constant first-order rate adequately describes dissociation over the time range of interest. The quantities also depend on the experimental conditions under which the rate was estimated. Temperature, buffer conditions, and the assay environment can affect a measured rate; a value should not be treated as a universal property independent of conditions.

A practical diagnostic is whether the observed dissociation curve is reasonably described by a single exponential over the relevant time range. Systematic curvature in a plot of the logarithm of the remaining bound fraction against time, or a clear multi-phase decline, suggests that one constant rate may not summarize the data adequately. In that case, reporting \(1/k_{\mathrm{off}}\) from a forced single-rate fit can conceal important heterogeneity. A suitable alternative may be to report empirical quantiles or to use a model that represents the observed phases, with its assumptions stated.

Rebinding is another important failure mode. If dissociated ligand can bind again during the observation period, the measured decline in bound signal can be slower than the underlying dissociation of individual binding episodes. Assays may reduce this problem by removing free ligand or otherwise preventing rebinding; whether that was achieved is an experimental question. If substantial rebinding remains, the apparent decay should not automatically be interpreted as the single-event \(k_{\mathrm{off}}\).

The mean and half-life are complementary rather than competing definitions. Use the mean when the question concerns the average duration of a bound episode under the model. Use the half-life when the question concerns the time for the bound cohort to fall to 50% of its initial amount. If a dissociation curve is not exponential, neither single-rate formula should be applied without qualification. Neither measure is a substitute for exposure duration or a complete prediction of pharmacological effect.

Common pitfall: “The dissociation half-life is 60 minutes, so the mean receptor residence time is also 60 minutes.” This confuses the time to lose half the cohort with the average duration of individual bound episodes. For first-order dissociation, \(t_{1/2}=\ln(2)/k_{\mathrm{off}}\), whereas \(\tau=1/k_{\mathrm{off}}\); therefore \(\tau=t_{1/2}/\ln(2)\), which is about 1.443 times the half-life.

Check Your Understanding

Work each problem using the single-rate, no-rebinding model unless the question says otherwise. Solutions are included below the questions.

A hypothetical ligand has \(k_{\mathrm{off}}=0.0800\ \mathrm{min}^{-1}\). Calculate its mean residence time and dissociation half-life.
A hypothetical dissociation half-life is 30.0 hours. Calculate \(k_{\mathrm{off}}\) in \(\mathrm{h}^{-1}\) and the mean residence time.
For \(k_{\mathrm{off}}=0.0500\ \mathrm{h}^{-1}\), what fraction of an initially bound cohort remains after 20.0 hours? State what this calculation does and does not describe.
A drug is present in plasma for 10 hours, while its measured \(k_{\mathrm{off}}\) implies a mean residence time of 25 hours. Is this necessarily contradictory? Explain.

Solutions. (1) \(\tau=1/0.0800=12.5\ \mathrm{min}\), and \(t_{1/2}=0.693147/0.0800=8.66\ \mathrm{min}\). (2) \(k_{\mathrm{off}}=0.693147/(30.0\ \mathrm{h})=0.0231\ \mathrm{h}^{-1}\); retaining the unrounded expression gives \(\tau=30.0/0.693147=43.3\ \mathrm{h}\). (3) \(S(20.0)=e^{-(0.0500)(20.0)}=e^{-1}=0.3679\), so about 36.8% of the initially bound cohort remains under the stated dissociation-only model. It does not determine occupancy when free ligand can rebind. (4) No. Plasma exposure duration and mean bound-state residence time describe distinct processes; the comparison alone does not establish an inconsistency or determine receptor occupancy or response.