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Bispecific Antibodies and Antibody-Drug Conjugates · Tutorial 475 of 799

Deriving ADC Internalization and Recycling Fluxes

Learn how internalization and endosomal sorting fluxes jointly determine surface-target loss and intracellular conjugated-ADC accumulation.

Advanced 11 min read

What You'll Learn

  • Write mass-action binding and ADC internalization fluxes with units.
  • Derive receptor recycling and degradation fluxes from an endosomal target compartment.
  • Relate internalization and sorting to the change in total surface target.
  • Calculate intracellular conjugated-ADC accumulation without assuming linker cleavage.
  • Distinguish a reduced effective-loss model from explicit endosomal sorting dynamics.
  • Identify assumptions and diagnostics for interpreting these fluxes.

From ADC Binding to Uptake and Sorting

ADC binding does not by itself remove target from the cell surface or deliver conjugated drug into the cell. Those changes depend on what happens after a surface target–ADC complex is internalized. A flux model keeps these events distinct: binding forms a surface complex, internalization transfers it into the cell, and endosomal sorting returns target to the surface or sends it toward degradation. Meanwhile, the conjugated ADC can accumulate inside the cell. The earlier material on ADC species states distinguishes conjugated ADC from released payload; here, intracellular conjugated ADC is tracked without modeling payload release.

We use a minimal model for one target and one ADC species. Let \(R\) be free target on the cell surface, \(C\) surface target–ADC complex, \(E\) internalized target awaiting a recycling-or-degradation fate, and \(A\) free extracellular ADC concentration. Let \(I\) be intracellular conjugated ADC amount. Surface and endosomal target amounts are measured in target molecules per cell; \(I\) is measured in ADC molecules per cell. The model assumes that internalized ADC remains intracellular after the target and ADC separate, while target sorting determines whether the target returns to the surface. This is a simplifying assumption, not a claim that all ADCs follow the same intracellular path.

Definition: A flux is an amount transferred or removed per unit time. In this model, binding flux is \(k_{\mathrm{on}}AR\), internalization flux is \(k_{\mathrm{int}}C\), recycling flux is \(k_{\mathrm{rec}}E\), and target degradation flux is \(k_{\mathrm{deg}}E\). The rate constants \(k_{\mathrm{int}}, k_{\mathrm{rec}}, k_{\mathrm{deg}}\) have units of inverse time; \(k_{\mathrm{on}}\) has units of concentration\(^{-1}\) time\(^{-1}\).

The surface complex forms by mass-action binding and is lost by dissociation or internalization. Free surface target is made at rate \(k_{\mathrm{syn}}\), turns over at rate \(k_{\mathrm{turn}}R\), binds ADC, gains target through dissociation, and receives recycled target from \(E\). The resulting equations are:

$$ \begin{aligned} \frac{dR}{dt} &= k_{\mathrm{syn}}-k_{\mathrm{turn}}R-k_{\mathrm{on}}AR+k_{\mathrm{off}}C+k_{\mathrm{rec}}E,\\ \frac{dC}{dt} &= k_{\mathrm{on}}AR-(k_{\mathrm{off}}+k_{\mathrm{int}})C,\\ \frac{dE}{dt} &= k_{\mathrm{int}}C-(k_{\mathrm{rec}}+k_{\mathrm{deg}})E,\\ \frac{dI}{dt} &= k_{\mathrm{int}}C-k_{\mathrm{clear}}I. \end{aligned} $$

Here \(k_{\mathrm{off}}\) is the complex dissociation rate, \(k_{\mathrm{syn}}\) is target synthesis, and \(k_{\mathrm{clear}}\) is a lumped rate of disappearance from the tracked intracellular conjugated-ADC pool. It is not a linker-cleavage rate. The ADC concentration \(A\) is treated as fixed over the time window of interest; if extracellular ADC changes appreciably, its concentration needs its own balance equation. All rate constants and amounts are nonnegative. The equations also assume a well-mixed cell population represented by one set of amounts and rate constants.

Deriving Surface-Target and Sorting Fluxes

The internalization flux \(k_{\mathrm{int}}C\) removes target from the surface complex and adds it to the endosomal target pool. The two competing exits from that pool are recycling and degradation. To see how they divide an internalized target cohort, suppose \(k_{\mathrm{rec}}+k_{\mathrm{deg}}>0\), and consider an amount \(E_0\) already in \(E\) with no further input. Its remaining amount after time \(t\) is \(E(t)=E_0e^{-(k_{\mathrm{rec}}+k_{\mathrm{deg}})t}\). Integrating each exit flux over time gives the eventual amount following that route:

$$ \begin{aligned} \int_0^\infty k_{\mathrm{rec}}E(t)\,dt &=E_0\frac{k_{\mathrm{rec}}}{k_{\mathrm{rec}}+k_{\mathrm{deg}}},\\ \int_0^\infty k_{\mathrm{deg}}E(t)\,dt &=E_0\frac{k_{\mathrm{deg}}}{k_{\mathrm{rec}}+k_{\mathrm{deg}}}. \end{aligned} $$

Thus, under these constant competing first-order rates, the fractions of endosomal target that recycle and degrade are \(k_{\mathrm{rec}}/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\) and \(k_{\mathrm{deg}}/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\), respectively. They sum to one. If both rates are zero, the target remains in \(E\) and neither eventual fraction is defined.

The total amount of surface target is \(S=R+C\), not just the free target \(R\). Adding the equations for \(R\) and \(C\) cancels binding and dissociation: these processes move target between free and bound surface states but do not change the total at the surface. The result is:

$$ \frac{dS}{dt} =k_{\mathrm{syn}}-k_{\mathrm{turn}}R-k_{\mathrm{int}}C+k_{\mathrm{rec}}E. $$

The internalization term removes target from the surface, while the recycling term restores it. Target degradation acts after entry into \(E\), so it does not appear directly in this surface equation. Adding the \(E\) equation as well shows the overall target balance:

$$ \frac{d(S+E)}{dt} =k_{\mathrm{syn}}-k_{\mathrm{turn}}R-k_{\mathrm{deg}}E. $$

Internalization and recycling cancel from the whole-cell target balance because they transfer target between compartments. Degradation is the endosomal route that removes target from the tracked surface-plus-endosomal amount. This distinction is useful: a large uptake flux can coexist with modest net surface-target loss when much of the internalized target recycles.

Worked Example: Calculating Instantaneous Fluxes

Consider a hypothetical cell with \(C=100\) surface complexes per cell and \(E=20\) internalized target molecules per cell. Let \(k_{\mathrm{int}}=0.20\ \mathrm{h}^{-1}\), \(k_{\mathrm{rec}}=0.30\ \mathrm{h}^{-1}\), and \(k_{\mathrm{deg}}=0.10\ \mathrm{h}^{-1}\). Let \(R=400\) targets per cell, \(k_{\mathrm{syn}}=10\) targets per cell per hour, and \(k_{\mathrm{turn}}=0.010\ \mathrm{h}^{-1}\). Calculate the uptake, recycling, degradation, and surface-target change rates.

The internalization flux is \(k_{\mathrm{int}}C=(0.20\ \mathrm{h}^{-1})(100)=20\) ADC complexes per cell per hour, and it transfers 20 target molecules per cell per hour into \(E\). Recycling is \(k_{\mathrm{rec}}E=(0.30)(20)=6\) target molecules per cell per hour. Degradation is \(k_{\mathrm{deg}}E=(0.10)(20)=2\) target molecules per cell per hour.

The total surface-target rate is synthesis minus basal turnover, plus recycling, minus internalization: \(dS/dt=10-(0.010)(400)-20+6=-8\) targets per cell per hour. Thus surface target is decreasing at 8 targets per cell per hour at this instant. The endosomal target rate is \(dE/dt=20-(6+2)=12\) targets per cell per hour. The result is consistent: total target \(S+E\) changes at \(-8+12=4\) targets per cell per hour, which also equals \(10-4-2=4\) from synthesis minus basal turnover and degradation.

Intracellular Conjugate Accumulation

The model’s intracellular conjugated-ADC input is the internalization flux, \(k_{\mathrm{int}}C\). The balance \(dI/dt=k_{\mathrm{int}}C-k_{\mathrm{clear}}I\) does not equate intracellular ADC with released payload. It only says that conjugated ADC enters the tracked pool with internalized complexes and leaves that pool at the specified lumped clearance rate. If each ADC carries a fixed drug-to-antibody ratio \(d\), the corresponding input of conjugated drug equivalents is \(d\,k_{\mathrm{int}}C\). If the ratio varies among ADC species, each species needs its own amount and ratio; a single average ratio is justified only when it represents the quantity being modeled.

Worked Example: Recycling Versus Degradation at Steady State

Suppose the internalization flux is held at 20 complexes per cell per hour, with \(k_{\mathrm{rec}}=0.30\ \mathrm{h}^{-1}\) and \(k_{\mathrm{deg}}=0.10\ \mathrm{h}^{-1}\). Find the steady endosomal target amount and route-specific fluxes. Assume the input remains constant long enough for \(E\) to reach steady state.

At steady state \(dE/dt=0\), so \(k_{\mathrm{int}}C=(k_{\mathrm{rec}}+k_{\mathrm{deg}})E\). Therefore \(E=20/(0.30+0.10)=50\) target molecules per cell. The recycling flux is \((0.30)(50)=15\) targets per cell per hour; the degradation flux is \((0.10)(50)=5\) targets per cell per hour.

Equivalently, the recycling fraction is \(0.30/0.40=0.75\) and the degradation fraction is \(0.10/0.40=0.25\). Of the 20 targets internalized per hour, 15 return to the surface and 5 are degraded. The internalization-related contribution to surface change is therefore \(-20+15=-5\) targets per cell per hour, equal in magnitude to the degradation flux at this steady state. If the ADC has fixed \(d=4\), its conjugated-drug input is \(4(20)=80\) drug equivalents per cell per hour.

Worked Example: Time-Dependent Intracellular Conjugate

Suppose a hypothetical cell has a constant uptake input of 20 ADC molecules per cell per hour, \(k_{\mathrm{clear}}=0.050\ \mathrm{h}^{-1}\), and \(I(0)=100\) ADC molecules per cell. Calculate \(I\) after 10 hours. Also calculate the endosomal target amount after 10 hours if the same 20-target-per-hour input enters \(E\), \(k_{\mathrm{rec}}=0.30\ \mathrm{h}^{-1}\), \(k_{\mathrm{deg}}=0.10\ \mathrm{h}^{-1}\), and \(E(0)=0\).

For constant uptake \(u\), the solution to \(dI/dt=u-k_{\mathrm{clear}}I\) is \(I(t)=I(0)e^{-k_{\mathrm{clear}}t}+(u/k_{\mathrm{clear}})(1-e^{-k_{\mathrm{clear}}t})\). Substitution gives \(I(10)=100e^{-0.5}+(20/0.050)(1-e^{-0.5})=60.65+157.39=218.0\) ADC molecules per cell. With \(d=4\), this corresponds to 872.2 intracellular conjugated drug equivalents per cell, not necessarily that many released payload molecules.

For endosomal target, the constant-input solution is \(E(t)=u/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\,[1-e^{-(k_{\mathrm{rec}}+k_{\mathrm{deg}})t}]\). Thus \(E(10)=20/0.40\,[1-e^{-4}]=49.08\) targets per cell. At that time, recycling is \((0.30)(49.08)=14.73\) targets per cell per hour and degradation is \((0.10)(49.08)=4.908\) targets per cell per hour. These are instantaneous fluxes at 10 hours; they are not the total amounts recycled or degraded over the entire 10-hour period.

Model Choice, Diagnostics, and Limitations

A reduced model can replace the explicit endosomal compartment with an effective loss from the surface complex. If endosomal sorting is much faster than changes in surface binding and uptake, \(E\) may be approximated as being near its quasi-steady value \(E\approx k_{\mathrm{int}}C/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\). The resulting recycling flux is approximately \(k_{\mathrm{int}}C\,k_{\mathrm{rec}}/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\), and the net target loss due to uptake and sorting is approximately \(k_{\mathrm{int}}C\,k_{\mathrm{deg}}/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\). This shortcut is useful when endosomal transients are unimportant. The explicit \(E\) model is preferable when the timing of recycling or degradation matters, or when data resolve endosomal accumulation.

The equations have interpretable failure modes. A predicted negative amount of \(R\), \(C\), \(E\), or \(I\) indicates a numerical or model-implementation problem, since nonnegative initial amounts and nonnegative flux rates should preserve nonnegative states. A fitted \(E\) that rises without bound under sustained uptake when \(k_{\mathrm{rec}}+k_{\mathrm{deg}}>0\) is inconsistent with this model’s finite steady state \(E=k_{\mathrm{int}}C/(k_{\mathrm{rec}}+k_{\mathrm{deg}})\), unless uptake itself is changing or another process is missing. A large mismatch between measured surface-target decline and the modeled balance can indicate omitted target synthesis, basal turnover, recycling, or changes in ADC concentration.

The model is intentionally limited. It uses first-order sorting, one target and ADC species, and a lumped intracellular-conjugate loss term. It does not represent multiple endosomal compartments, receptor-dependent changes in ADC trafficking, cell-to-cell heterogeneity, or the chemistry and location of linker cleavage. Those mechanisms require additional states and evidence to identify their rates. In particular, interpreting \(k_{\mathrm{clear}}\) as payload release would conflate distinct processes.

Common pitfall: “Internalization removes \(k_{\mathrm{int}}C\) targets from the surface, so that must be the net surface-target loss.” The first clause is correct for the instantaneous transfer out of surface states, but not for the net loss: recycled target returns through \(k_{\mathrm{rec}}E\). The complete surface balance includes both terms, \(-k_{\mathrm{int}}C+k_{\mathrm{rec}}E\), as well as synthesis and basal turnover.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. If \(C=60\) complexes per cell and \(k_{\mathrm{int}}=0.25\ \mathrm{h}^{-1}\), what is the internalization flux?
  2. For \(k_{\mathrm{rec}}=0.20\ \mathrm{h}^{-1}\) and \(k_{\mathrm{deg}}=0.30\ \mathrm{h}^{-1}\), what fractions of endosomal target recycle and degrade?
  3. With uptake 10 targets per cell per hour and \(k_{\mathrm{rec}}+k_{\mathrm{deg}}=0.50\ \mathrm{h}^{-1}\), what is the steady amount in \(E\)?
  4. Why is \(dI/dt=k_{\mathrm{int}}C-k_{\mathrm{clear}}I\) not, by itself, a payload-release model?

Solutions. (1) The flux is \((0.25)(60)=15\) complexes per cell per hour. (2) The recycling fraction is \(0.20/0.50=0.40\), and the degradation fraction is \(0.30/0.50=0.60\). (3) At steady state, \(E=10/0.50=20\) targets per cell. (4) \(I\) tracks intracellular conjugated ADC, and \(k_{\mathrm{clear}}\) is a lumped loss from that tracked pool; the equation does not define a released-payload state or a cleavage mechanism.

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