Modeling Payload Efflux and Acquired Resistance

Why Add Efflux to an ADC Model?

An ADC can deliver and release payload inside a cell, yet the resulting intracellular concentration depends on more than the delivery rate. Payload may leave the cell through transporters or other efflux processes before it reaches its intracellular targets. If cells acquire a higher efflux capacity during treatment, the same payload input can produce lower intracellular exposure and weaker killing. A model that represents delivery but omits efflux may therefore attribute a response change to the wrong mechanism.

The earlier material on ADC species states and intracellular payload release provides the context for the input to this model: payload becomes available inside a cell after ADC uptake and release. Here that process is summarized as an input rate. The focus is the subsequent intracellular balance, especially how an altered efflux rate changes payload exposure. The calculations do not establish that efflux is the cause of resistance in a real system; other changes can produce similar observations.

Definition: Let $C(t)$ be the intracellular concentration of free payload, in nM, and let $q(t)$ be its net intracellular input rate, in nM/h. Let $k_{\mathrm{eff}}$ be a first-order efflux rate constant and $k_{\mathrm{other}}$ the combined first-order rate constant for other intracellular losses, both in h$^{-1}$. The one-compartment balance is $$ \frac{dC}{dt}=q(t)-\left(k_{\mathrm{eff}}+k_{\mathrm{other}}\right)C. $$ The efflux flux is $k_{\mathrm{eff}}C$, in nM/h. “Acquired efflux resistance” means that efflux capacity increases over time or after selection, reducing payload concentration or exposure at a given input.

Constant Input and the Concentration Solution

The model assumes the intracellular payload pool is well mixed; the stated input reaches this pool; efflux and other losses are first order over the concentration range considered; and their rate constants are constant within the interval being analyzed. It also treats the cell population as a representative cell, with no explicit variation in uptake, release, or efflux between cells. These are modeling assumptions, not properties guaranteed by the ADC mechanism.

For constant input $q$, define $k=k_{\mathrm{eff}}+k_{\mathrm{other}}$, with $k>0$. Starting from $C(0)=C_0$, the balance has the solution below. It approaches the steady-state concentration $C_{\mathrm{ss}}=q/k$; the time constant $1/k$ determines how quickly it approaches that level.

C(t)=\frac{q}{k}+\left(C_0-\frac{q}{k}\right)e^{-kt}, \qquad C_{\mathrm{ss}}=\frac{q}{k}.

A short derivation follows by multiplying the differential equation $dC/dt=q-kC$ by the integrating factor $e^{kt}$. Then $d(e^{kt}C)/dt=qe^{kt}$. Integrating from 0 to $t$ gives $e^{kt}C(t)-C_0=(q/k)(e^{kt}-1)$; rearranging yields the stated solution. The result requires constant $q$ and $k$ over the interval. If efflux changes continuously, this constant-rate solution cannot be applied over the whole interval with a single $k$.

For an initially payload-free cell, the area under the intracellular concentration curve from 0 to duration $T$ is $$ \mathrm{AUC}_{0,T}=\int_0^T C(t)\,dt =\frac{q}{k}\left[T-\frac{1-e^{-kT}}{k}\right]. $$

At steady state, the input is balanced by total loss: $q=(k_{\mathrm{eff}}+k_{\mathrm{other}})C_{\mathrm{ss}}$. The efflux and other-loss fluxes are $k_{\mathrm{eff}}C_{\mathrm{ss}}$ and $k_{\mathrm{other}}C_{\mathrm{ss}}$, respectively. Increasing efflux therefore lowers steady-state concentration when input and other loss remain fixed. However, the observed concentration alone does not reveal which term changed.

Worked Example: Steady-State Payload and a Simplified Killing Calculation

Consider a hypothetical ADC-treated cell receiving a constant payload input of $q=10.0$ nM/h. Suppose $k_{\mathrm{other}}=0.0500$ h$^{-1}$. Before resistance, let $k_{\mathrm{eff}}=0.200$ h$^{-1}$; after resistance, let it increase to $0.700$ h$^{-1}$. Assume both rate constants remain constant in their respective conditions.

Before resistance, $k=0.200+0.0500=0.250$ h$^{-1}$, so $C_{\mathrm{ss}}=10.0/0.250=40.0$ nM. After resistance, $k=0.700+0.0500=0.750$ h$^{-1}$, so $C_{\mathrm{ss}}=10.0/0.750=13.333\ldots$ nM. The relative reduction, calculated with the unrounded concentration, is $(40.0-13.333\ldots)/40.0=0.6667$, or 66.7%.

The flux balance checks the arithmetic. Before resistance, efflux is $0.200(40.0)=8.00$ nM/h and other loss is $0.0500(40.0)=2.00$ nM/h, totaling the 10.0 nM/h input. After resistance, using $13.333\ldots$ nM rather than its rounded display, efflux is $0.700(13.333\ldots)=9.333\ldots$ nM/h and other loss is $0.0500(13.333\ldots)=0.6667$ nM/h. Their sum is 10.0 nM/h.

To illustrate a possible response consequence without treating it as a universal ADC response model, suppose cell killing follows a constant hazard $h(C)=h_{\max}C/(K_C+C)$, with hypothetical $h_{\max}=0.120$ h$^{-1}$ and $K_C=20.0$ nM. If concentration is held at steady state for 24.0 h, and there is no cell growth or other source of cell loss, the killed fraction is $1-e^{-h(C)T}$. At 40.0 nM, $h=0.120(40.0)/(20.0+40.0)=0.0800$ h$^{-1}$, giving $1-e^{-0.0800(24.0)}=0.853$, or 85.3%. At $13.333\ldots$ nM, $h=0.120(13.333\ldots)/(20.0+13.333\ldots)=0.0480$ h$^{-1}$, giving $1-e^{-0.0480(24.0)}=0.684$, or 68.4%. These are predictions under the stated illustrative hazard model, not measured or general ADC killing fractions.

Finite-Duration Exposure and Emerging Resistance

A steady-state comparison can exaggerate or understate the exposure difference during a finite treatment interval, especially when cells start with little payload. The transient solution accounts for the time needed to accumulate payload. If efflux changes during treatment, the balance can instead be solved over successive intervals, carrying the concentration at the end of one interval into the next. This is a useful approximation when resistance is represented as a switch; gradual induction requires a time-varying model.

Worked Example: Comparing Exposure Over a Finite Input Period

Use the hypothetical input $q=10.0$ nM/h and the two total loss rates from the previous example. Suppose each condition starts at $C(0)=0$ and receives constant input for $T=24.0$ h. For the pre-resistance condition, $k=0.250$ h$^{-1}$, so $$ \mathrm{AUC}_{0,24} =40.0\left[24.0-\frac{1-e^{-0.250(24.0)}}{0.250}\right] =40.0\left[24.0-\frac{1-e^{-6}}{0.250}\right] \approx 800.4\ \mathrm{nM\cdot h}. $$

For the resistant condition, $k=0.750$ h$^{-1}$, and $$ \mathrm{AUC}_{0,24} =\frac{10.0}{0.750}\left[24.0-\frac{1-e^{-0.750(24.0)}}{0.750}\right] \approx 302.2\ \mathrm{nM\cdot h}. $$

The resistant-to-pre-resistance AUC ratio is $302.2/800.4=0.3776$, so the finite-interval AUC is about 62.2% lower in this model. It is not reduced by the 66.7% steady-state concentration difference because the two concentration curves rise at different rates from zero. The calculation assumes the same input in both conditions throughout all 24 hours; altered payload release would violate that comparison.

Worked Example: Efflux Increases Partway Through Treatment

Suppose a hypothetical cell receives $q=10.0$ nM/h from $t=0$ to $t=24.0$ h, beginning at $C(0)=0$. Let $k=0.250$ h$^{-1}$ during the first 12.0 h, then let acquired resistance increase the total loss rate to $0.750$ h$^{-1}$. This piecewise-constant change is an approximation to an abrupt change in efflux capacity; the input and other loss are held fixed.

At 12.0 h, the first-interval solution gives $C(12)=40.0(1-e^{-0.250(12.0)})=40.0(1-e^{-3})=38.01$ nM. In the second interval, concentration starts at 38.01 nM and tends toward $10.0/0.750=13.333\ldots$ nM. Thus $C(24)=13.333\ldots+(38.0085-13.333\ldots)e^{-0.750(12.0)}\approx13.336$ nM.

The first-interval exposure is $40.0[12.0-(1-e^{-3})/0.250]=327.97$ nM·h. The second-interval exposure, using the concentration at 12 h as its initial value, is $$ \mathrm{AUC}_{12,24} =\frac{10.0}{0.750}(12.0) +\left(38.0085-\frac{10.0}{0.750}\right) \frac{1-e^{-0.750(12.0)}}{0.750} \approx 192.90\ \mathrm{nM\cdot h}. $$

The total is approximately $327.97+192.90=520.9$ nM·h. It lies between the 24-hour AUCs for constant pre-resistance efflux (800.4 nM·h) and constant resistant efflux (302.2 nM·h), as expected for this model: the cell spends half the interval under each condition, with the concentration carrying over between intervals.

Representing Acquired Efflux Resistance

There are several levels of model detail. A simple comparison assigns different fixed values of $k_{\mathrm{eff}}$ to sensitive and resistant conditions. This is appropriate for comparing measured conditions or representing a resistance state whose timescale is much longer than the modeled exposure. It does not explain how resistance develops. If transporter abundance is measured or its induction is of interest, let $E(t)$ denote transporter abundance and allow it to change over time. One possible model is $$ \frac{dE}{dt}=k_{\mathrm{syn}}(1+I)-k_{\mathrm{deg}}E, \qquad k_{\mathrm{eff}}(t)=k_0+\beta E(t), $$ where $I$ is a dimensionless induction signal, $k_{\mathrm{syn}}$ and $k_{\mathrm{deg}}$ are synthesis and degradation parameters, and $k_0$ and $\beta$ map transporter abundance to an efflux rate constant. The exact form must be supported by the biology and available data. It adds parameters and requires joint solution of the payload and transporter equations.

An alternative is a population model with sensitive and resistant subpopulations, each having its own efflux rate and abundance. That approach can represent selection of pre-existing resistant cells, whereas an induction model represents a change within cells. Bulk concentration or viability data may not distinguish these explanations. Measurements of transporter abundance, intracellular payload, or single-cell response can help, but each measurement has its own sampling and assay limitations.

Diagnostics, Alternatives, and Limitations

Check the model against both the direction and timescale of the data. Higher efflux should shorten the payload residence time and lower the concentration under the same input. In a constant-rate model, the approach to steady state has rate $k$; an estimated trajectory that is inconsistent with that timescale suggests that the one-compartment, constant-rate assumptions may be inadequate. Also check the flux balance at steady state and verify that concentration and rate units are consistent.

Confounding with delivery or release: Lower intracellular payload can result from lower uptake or slower payload release as well as greater efflux. A fit to intracellular concentration alone may not identify these mechanisms separately.
Concentration-dependent transport: If efflux saturates, a first-order term $k_{\mathrm{eff}}C$ may misrepresent the concentration dependence. Residual patterns across concentrations or direct transport measurements can reveal this failure.
Changing cell populations: A bulk-average response may reflect selection of a resistant subpopulation rather than induction of efflux in each cell. A single-cell or population model may be needed to distinguish them.
Limited sampling: Sparse early samples may miss the rise toward steady state, while sparse later samples may not establish whether transporter activity is changing. Sensitivity analysis and informative sampling times matter.
Response-model limitations: The illustrative hazard calculation assumes a specified concentration-response relationship and constant hazard. It does not account for delayed damage, cell-cycle dependence, regrowth, or heterogeneous sensitivity.

The main alternative to an efflux explanation is not a competing formula but a competing mechanism in the payload balance: altered input, altered release, metabolism, sequestration, or target sensitivity. Models should be compared using measurements that inform these processes where possible. Adding an efflux parameter can improve a numerical fit without proving an efflux mechanism.

Common pitfall: “The resistant cells have one-third the payload concentration, so efflux must have increased threefold.” This reverses the mass-balance logic. In the steady-state example, concentration falls from 40.0 to 13.333… nM because total loss increases from 0.250 to 0.750 h$^{-1}$ while input stays fixed; the efflux rate constant itself changes from 0.200 to 0.700 h$^{-1}$, not by a factor of three. Without independent information on input and other losses, the concentration ratio alone cannot identify the efflux change.

Check Your Understanding

Try each question, then compare your reasoning with the solutions below.

With $q=6.00$ nM/h, $k_{\mathrm{eff}}=0.300$ h$^{-1}$, and $k_{\mathrm{other}}=0.200$ h$^{-1}$, what is the steady-state concentration and each loss flux?
For the conditions in question 1, calculate $C(2.00)$ if $C(0)=0$.
What model change distinguishes gradual transporter induction from a fixed resistant-state model?
Give two mechanisms, other than increased efflux, that could lower measured intracellular payload under ADC treatment.

Solutions. (1) The total loss rate is $0.300+0.200=0.500$ h$^{-1}$, so $C_{\mathrm{ss}}=6.00/0.500=12.0$ nM. Efflux is $0.300(12.0)=3.60$ nM/h and other loss is $0.200(12.0)=2.40$ nM/h; together they equal the 6.00 nM/h input. (2) The transient solution gives $C(2.00)=12.0(1-e^{-0.500(2.00)})=12.0(1-e^{-1})=7.59$ nM. (3) Add a transporter-abundance state with synthesis and degradation dynamics, and link that state to $k_{\mathrm{eff}}(t)$, instead of assigning one fixed efflux rate to the whole interval. (4) Examples include reduced ADC uptake and slower intracellular payload release; increased metabolism or sequestration could also lower free payload concentration.