Complex Internalization and the Fate of Bound Antibody

Why the Fate of a Bound Antibody Matters

Binding does not by itself remove antibody from the body. A drug molecule bound to its target may dissociate and return to the free-drug pool, or the complex may be internalized. After internalization, the antibody might be degraded, recycled, or partly recycled and partly degraded. These alternatives can produce different drug clearance even when binding affinity and target abundance are unchanged.

The canonical target-mediated drug disposition (TMDD) model and its mass-balance formulation provide the starting point. Here the focus is the fate of the complex after it forms: which terms transfer drug between free and bound states, and which terms represent irreversible loss? The distinction is essential because a complex can disappear from the measured extracellular pool without all of its antibody being eliminated.

Definition: Let \(D\) be free drug concentration, \(R\) free target concentration, and \(C\) drug–target complex concentration. Association occurs at rate \(k_{\mathrm{on}}DR\), dissociation at rate \(k_{\mathrm{off}}C\), and complex internalization at rate \(k_{\mathrm{int}}C\). The rate constants \(k_{\mathrm{on}}\), \(k_{\mathrm{off}}\), and \(k_{\mathrm{int}}\) have units of concentration\(^{-1}\)time\(^{-1}\), time\(^{-1}\), and time\(^{-1}\), respectively.

Concentrations are assumed to use a common volume basis, with drug and target in a one-to-one binding stoichiometry. The model treats association, dissociation, and internalization as well-mixed processes with mass-action rates. The input \(u(t)\) is drug input per volume per time; \(k_{\mathrm{el}}D\) is linear elimination of free drug. Target synthesis and free-target turnover are represented by \(k_{\mathrm{in,R}}\) and \(k_{\mathrm{out,R}}R\). These are deterministic compartmental balances, not statements that every molecule follows an identical path.

Mass Balance With Dissociation and Internalization

Suppose that internalized antibody is degraded and does not return to the extracellular drug pool. The free and complex drug equations are

\frac{dD}{dt} = u(t)-k_{\mathrm{el}}D-k_{\mathrm{on}}DR+k_{\mathrm{off}}C, \qquad \frac{dC}{dt} = k_{\mathrm{on}}DR-(k_{\mathrm{off}}+k_{\mathrm{int}})C.

The association term removes one drug molecule from the free pool and adds one to the complex pool. Dissociation does the reverse. Internalization removes complex from the modeled extracellular pools; under the stated assumption, that internalized drug is lost. Add the two equations to track total extracellular drug \(D_{\mathrm{tot}}=D+C\):

\frac{dD_{\mathrm{tot}}}{dt} = u(t)-k_{\mathrm{el}}D-k_{\mathrm{int}}C.

The binding and dissociation terms cancel exactly. Thus, dissociation changes the distribution between free and bound drug but is not itself a loss from \(D+C\). Internalization is a loss from this total only because this version assumes that internalized antibody is degraded rather than returned.

The corresponding target balance depends on target fate as well. If internalized target is also degraded, then

\frac{d(R+C)}{dt} = k_{\mathrm{in,R}}-k_{\mathrm{out,R}}R-k_{\mathrm{int}}C.

This target equation is not interchangeable with the drug equation: antibody and target can have different post-internalization fates. A model must specify each explicitly.

Worked Example: Competing Dissociation and Internalization

Consider a hypothetical cohort of \(5.00\) nM complexes at time zero. Assume no further association, with first-order dissociation \(k_{\mathrm{off}}=0.20\) h\(^{-1}\) and internalization \(k_{\mathrm{int}}=0.30\) h\(^{-1}\). During internalization, half of the antibody is recycled intact as free drug (\(q_D=0.50\)); the remainder is degraded. Find the amounts in each fate after \(2.00\) h.

For each complex, the first event has total rate \(k_{\mathrm{off}}+k_{\mathrm{int}}=0.50\) h\(^{-1}\). Under constant competing first-order rates, the probabilities of dissociation and internalization by time \(t\) are respectively \(k_{\mathrm{off}}/(k_{\mathrm{off}}+k_{\mathrm{int}})\) and \(k_{\mathrm{int}}/(k_{\mathrm{off}}+k_{\mathrm{int}})\), each multiplied by \(1-e^{-(k_{\mathrm{off}}+k_{\mathrm{int}})t}\). Therefore,

P(\mathrm{internalized\ by\ 2\,h}) = \frac{0.30}{0.50}(1-e^{-0.50(2.00)}) = 0.3793, \qquad P(\mathrm{dissociated\ by\ 2\,h}) = \frac{0.20}{0.50}(1-e^{-1}) = 0.2528.

The unrounded amounts are \(5.00(0.379272\ldots)=1.896\) nM internalized and \(5.00(0.252848\ldots)=1.264\) nM dissociated. The amount still bound is \(5.00e^{-1}=1.839\) nM. Of the internalized amount, \(q_D=0.50\) returns \(0.50(1.89636\ldots)=0.9482\) nM as free drug, and \(0.9482\) nM is degraded. The antibody balance is \(1.264+1.839+0.9482+0.9482=4.999\) nM, with the small discrepancy from \(5.00\) due to rounding. Dissociation returns drug to the free pool; recycling returns a separate fraction after internalization.

Partial Recycling and the Net Loss Term

Let \(q_D\) be the fraction of internalized antibody returned intact to the modeled free-drug pool. The remaining fraction, \(1-q_D\), is degraded or otherwise leaves the modeled drug system. Let \(q_R\) similarly denote the fraction of internalized target returned as free target. In this formulation, recycling returns drug and target as separate free species, not as an intact complex. The equations become

\begin{aligned} \frac{dD}{dt} &= u(t)-k_{\mathrm{el}}D-k_{\mathrm{on}}DR+k_{\mathrm{off}}C +q_Dk_{\mathrm{int}}C,\\ \frac{dR}{dt} &= k_{\mathrm{in,R}}-k_{\mathrm{out,R}}R-k_{\mathrm{on}}DR+k_{\mathrm{off}}C +q_Rk_{\mathrm{int}}C,\\ \frac{dC}{dt} &= k_{\mathrm{on}}DR-(k_{\mathrm{off}}+k_{\mathrm{int}})C. \end{aligned}

The full internalization flux is still \(k_{\mathrm{int}}C\): the whole complex leaves the extracellular complex pool. Only the fraction \(q_D\) returns to free drug. Adding the free- and complex-drug equations gives

\frac{d(D+C)}{dt} = u(t)-k_{\mathrm{el}}D-(1-q_D)k_{\mathrm{int}}C.

This derivation is a direct conservation check. Association and dissociation cancel, and recycling adds back the fraction \(q_D\) of the internalized flux. The net drug loss attributable to internalization is therefore \((1-q_D)k_{\mathrm{int}}C\), not \(k_{\mathrm{int}}C\), unless \(q_D=0\). If \(q_D=1\), internalization creates no net loss of antibody from \(D+C\) in this model, although free-drug elimination can still remove drug after recycling.

The analogous target balance is \(d(R+C)/dt=k_{\mathrm{in,R}}-k_{\mathrm{out,R}}R-(1-q_R)k_{\mathrm{int}}C\). The equations allow \(q_D\ne q_R\): recycling of antibody does not require recycling of target, or vice versa. This is a model assumption to be supported by the biology or data, not a consequence of binding stoichiometry.

Worked Example: Correcting Internalization Loss for Recycling

A hypothetical complex cohort begins at \(C(0)=4.00\) nM. Suppose there is no dissociation or reassociation during a \(2.00\)-h interval and \(k_{\mathrm{int}}=0.30\) h\(^{-1}\). The complex concentration declines as \(C(t)=4.00e^{-0.30t}\). If \(q_D=0.50\), calculate internalized drug and net drug loss.

The internalized amount is the time-integrated internalization flux:

\int_0^{2.00} k_{\mathrm{int}}C(t)\,dt = \int_0^{2.00}0.30(4.00)e^{-0.30t}\,dt = 4.00(1-e^{-0.60}) = 1.804754\ldots\ \mathrm{nM}.

The recycled amount is \(q_D\) times this value, \(0.50(1.804754\ldots)=0.902377\ldots\) nM. The net antibody loss is the remaining fraction:

(1-q_D)\int_0^{2.00}k_{\mathrm{int}}C(t)\,dt = 0.50(1.804754\ldots) = 0.9024\ \mathrm{nM}.

The final complex amount is \(4.00e^{-0.60}=2.195\) nM. The \(4.00\) nM starting amount is accounted for by \(2.195\) nM still bound, \(0.9024\) nM recycled, and \(0.9024\) nM lost, up to rounding. This calculation treats returned antibody as an intact drug molecule; it does not claim that it remains in the complex or avoids later elimination.

Worked Example: Checking the Instantaneous Drug Balance

At a hypothetical instant, let \(D=10.0\) nM, \(R=2.00\) nM, and \(C=4.00\) nM. Take \(k_{\mathrm{on}}=0.0400\) nM\(^{-1}\)h\(^{-1}\), \(k_{\mathrm{off}}=0.0750\) h\(^{-1}\), \(k_{\mathrm{int}}=0.100\) h\(^{-1}\), \(q_D=0.250\), \(k_{\mathrm{el}}=0.0500\) h\(^{-1}\), and \(u=0.200\) nM/h. Find the free-drug and total-drug rates of change.

The association flux is \(0.0400(10.0)(2.00)=0.800\) nM/h, dissociation is \(0.0750(4.00)=0.300\) nM/h, and internalization is \(0.100(4.00)=0.400\) nM/h. The free-drug equation gives

\frac{dD}{dt} = 0.200-0.500-0.800+0.300+0.250(0.400) = -0.700\ \mathrm{nM/h}.

The complex rate is \(dC/dt=0.800-0.300-0.400=0.100\) nM/h. Thus \(d(D+C)/dt=-0.700+0.100=-0.600\) nM/h. The aggregate balance gives the same result: \(u-k_{\mathrm{el}}D-(1-q_D)k_{\mathrm{int}}C=0.200-0.500-0.750(0.400)=-0.600\) nM/h. Agreement is an important check that recycling and internalization have not been counted twice.

Choosing and Checking a Complex-Fate Formulation

These formulations represent different biological assumptions, not merely different parameter values. With dissociation alone, the complex returns drug and target to their free pools; it creates no direct total-drug loss. With internalization followed by degradation, complex removal is a direct loss at rate \(k_{\mathrm{int}}C\). With partial recycling, only the non-recycled fraction is a net loss from total drug. A distinct possibility is intact-complex recycling: in that model, a fraction of internalized complex returns to the complex pool, so the recycling term belongs in the equation for \(C\), rather than as separate \(D\) and \(R\) terms. These mechanisms should not be conflated.

A practical comparison is:

Assumed complex fate	Drug consequence	Interpretation
Dissociation without internalization	No direct loss from \(D+C\)	Drug returns to the free pool when the complex dissociates.
Internalization and degradation	Loss at \(k_{\mathrm{int}}C\)	All internalized antibody is removed from the modeled drug system.
Internalization with fraction \(q_D\) recycled as free drug	Net loss at \((1-q_D)k_{\mathrm{int}}C\)	Recycled drug returns intact, while the remainder is lost.
Internalization with intact-complex recycling	Depends on the complex-return flux	Drug returns bound; the model must explicitly track that state.

When evaluating a model, inspect both state equations and the total-drug balance. A negative net internalization-loss term, or a total-drug balance that still subtracts all of \(k_{\mathrm{int}}C\) when \(q_D>0\), signals inconsistent accounting. If measured complex falls, that alone does not identify whether antibody was degraded, recycled as free drug, or returned bound. Measurements of drug, target, and complex over time may help distinguish mechanisms, but the model’s structural assumptions and available observations determine what can be inferred.

The main alternatives also have different strengths. A dissociation-only model is simpler and appropriate when internalization is negligible over the timescale of interest. The all-degraded model is a clear limiting case when internalized antibody is not returned. A fractional-recycling model represents partial recovery but adds a parameter that may be difficult to estimate separately from internalization. An intact-complex recycling model can represent another trafficking route but requires a distinct state transition. A more elaborate model is not automatically better: use added states only when the mechanism and data justify them.

The assumptions can fail if internalization is saturable, recycling fractions vary with concentration or time, the drug is cleaved rather than recycled intact, or the complex has multiple stoichiometries. A single constant \(q_D\) then may not describe the process. Likewise, this well-mixed model does not represent spatially distinct compartments or delays during intracellular trafficking. These limits concern the model structure; changing a fitted rate constant cannot repair a missing fate pathway.

Common Mistake and Diagnostic Checks

Wrong: “Every internalized complex removes one antibody from the body.” This is true only when \(q_D=0\). Correction: the modeled net loss is \((1-q_D)k_{\mathrm{int}}C\) when the returned fraction is intact drug.
Wrong: Add recycling to the free-drug equation but retain the full internalization loss in the total-drug balance. Correction: add the free- and complex-drug equations; the result must include the net loss fraction.
Wrong: Treat dissociation as drug elimination because complex concentration decreases. Correction: dissociation decreases \(C\) and increases \(D\) by the same amount, so it cancels from \(D+C\).
Wrong: Assume drug and target have the same recycling fraction. Correction: specify \(q_D\) and \(q_R\) separately unless the biological mechanism supports equality.
Diagnostic: Check units of every flux (concentration per time), verify that all recycling fractions lie between 0 and 1, and compare the summed state equations with an independently written total balance.

Key takeaway: Complex disappearance is not synonymous with antibody loss. Under fractional recycling, the drug mass-balance loss due to internalization is \((1-q_D)k_{\mathrm{int}}C\); the value of \(q_D\) and the form of the recycling pathway determine how bound antibody returns, if at all.

Check Your Understanding

Answer each question, then compare with the solutions below.

If a complex dissociates but neither drug nor target is otherwise eliminated, what happens to \(D+C\) at the moment of dissociation?
At \(C=3.00\) nM, \(k_{\mathrm{int}}=0.200\) h\(^{-1}\), and \(q_D=0.750\), what is the instantaneous net drug loss due to internalization?
With only dissociation and internalization as competing fates, \(k_{\mathrm{off}}=0.10\) h\(^{-1}\), \(k_{\mathrm{int}}=0.20\) h\(^{-1}\), and a long enough interval that every complex experiences one of the two events, what fraction internalizes?
Why is it incorrect to add \(q_Dk_{\mathrm{int}}C\) to the free-drug equation and then describe \(k_{\mathrm{int}}C\) as the net loss from total drug?

Solutions. (1) \(D+C\) is unchanged: dissociation contributes \(+k_{\mathrm{off}}C\) to \(dD/dt\) and \(-k_{\mathrm{off}}C\) to \(dC/dt\). (2) The internalization flux is \(0.200(3.00)=0.600\) nM/h, so net loss is \((1-0.750)(0.600)=0.150\) nM/h. (3) The eventual internalization fraction is \(k_{\mathrm{int}}/(k_{\mathrm{off}}+k_{\mathrm{int}})=0.20/(0.10+0.20)=0.667\), or about 66.7%. (4) The fraction \(q_D\) has been returned to the free-drug pool, so only \((1-q_D)k_{\mathrm{int}}C\) is a net loss from \(D+C\); subtracting the full flux would count recycled drug as lost.