Why Separate Endosomal and Lysosomal Processing?
After an ADC is internalized, payload liberation need not occur immediately. The ADC may first reside in an endosomal state and then transfer to a lysosomal state, where processing liberates payload. Representing these stages separately makes the timing of liberation explicit: a slow endosomal transfer can delay lysosomal arrival, while slow lysosomal processing can delay liberation even after arrival.
The model below begins with intracellular ADC influx, not administered dose or plasma concentration. The earlier material on ADC species states and internalization provides the context for defining that influx. Linker stability outside the cell is a separate process; here, an ADC entering the model carries a fixed amount of payload until lysosomal processing liberates it.
\( \frac{dE}{dt}=u(t)-k_EE,\quad \frac{dL}{dt}=k_EE-k_LL,\quad \frac{dP}{dt}=qk_LL. \)
The model assumes well-mixed states, constant positive rate constants, and first-order transfer and processing. It also assumes that ADC is not lost through competing pathways after entry, that each ADC carries the same \(q\), and that liberated payload is counted cumulatively rather than removed from a free-payload pool. The standard starting condition is \(E(0)=L(0)=P(0)=0\). These assumptions make the model a transparent description of sequential delay, not a complete account of every intracellular fate.
Deriving the Time Course of Liberation
Consider first a single cohort of ADC molecules that enters the endosomal state at time zero. Its endosomal residence time is exponentially distributed with rate \(k_E\), and its lysosomal processing time is exponentially distributed with rate \(k_L\). The total time until processing is complete is the sum of these two residence times. For \(k_E\ne k_L\), the probability that a cohort molecule is still unprocessed at age \(a\) is
Thus, the fraction processed by age \(a\) is \(F(a)=1-S(a)\). One justification is to integrate over the possible time \(s\) spent in the endosomal state: completion by age \(a\) requires \(0\le s\le a\), with endosomal transfer density \(k_Ee^{-k_Es}\) and lysosomal completion probability \(1-e^{-k_L(a-s)}\). Therefore,
If the rates are equal, \(k_E=k_L=k\), the limiting expression is \(S(a)=e^{-ka}(1+ka)\), so \(F(a)=1-e^{-ka}(1+ka)\). This equal-rate form avoids division by zero and is useful when the two estimated rates are close.
Worked Example: Liberation From an Internalized Cohort
Suppose a hypothetical cohort of 100 ADC molecules per cell is internalized at time zero. Let \(k_E=0.8\ \mathrm{h}^{-1}\), \(k_L=0.4\ \mathrm{h}^{-1}\), and \(q=4\) payload molecules per ADC. How much payload has been liberated from this cohort after 3 hours?
The rates differ, so use the sequential survival expression. Substituting \(a=3\) hours gives
The processed fraction is \(F(3)=1-0.5117=0.4883\). The cohort therefore yields \(100(0.4883)=48.83\) processed ADC molecules per cell by 3 hours. Multiplying by \(q\), cumulative liberated payload is \(4(48.83)=195.3\) payload molecules per cell. The remainder, equivalent to 51.17 ADC molecules, is still unprocessed in one of the two modeled states.
For a time-varying influx, each input cohort contributes according to its age. The cumulative number of processed ADC molecules per cell by time \(t\) is \(\int_0^t u(s)F(t-s)\,ds\), and cumulative payload liberation is \(q\) times that amount. This convolution makes the delay explicit: recently internalized ADC contributes little to liberation when the processing steps take time.
Constant Input and the Effect of Rate Changes
For constant input \(u\) from time zero through time \(t\), the ages of the entered ADC molecules range from zero to \(t\). Integrating the cohort completion fraction gives the number processed. Equivalently, integrate the unprocessed fraction \(S(a)\) and subtract it from the total input \(ut\):
Here \(R(t)\) is processed ADC in ADC molecules per cell, and \(P(t)\) is cumulative liberated payload in payload molecules per cell. The expression follows because \(\int_0^t S(a)\,da=\{\frac{k_L}{k_E}(1-e^{-k_Et})-\frac{k_E}{k_L}(1-e^{-k_Lt})\}/(k_L-k_E)\). This formula applies when the input has been constant throughout the interval being evaluated; for an input that stops earlier, the cohort-convolution form remains applicable.
Worked Example: Comparing Endosomal Transfer Rates
Suppose a hypothetical cell receives a constant input of \(u=20\) ADC molecules per cell per hour for 4 hours, with \(q=4\). Compare two scenarios with the same lysosomal rate, \(k_L=1.0\ \mathrm{h}^{-1}\): scenario A has \(k_E=0.25\ \mathrm{h}^{-1}\), and scenario B has \(k_E=0.8\ \mathrm{h}^{-1}\). The model assumes no losses from either state.
For scenario A, using \(e^{-1}=0.3679\) and \(e^{-4}=0.01832\), the integrated unprocessed time per unit input is
Thus, \(R_A(4)=20[4-3.0441]=19.12\) processed ADC molecules per cell, and \(P_A(4)=4(19.12)=76.47\) payload molecules per cell.
For scenario B, \(e^{-3.2}=0.04076\), so
Therefore, \(R_B(4)=20[4-2.0685]=38.63\) processed ADC molecules per cell, and \(P_B(4)=4(38.63)=154.5\) payload molecules per cell. With these specified rates and input, the faster endosomal transfer produces about \(154.5/76.47=2.021\) times as much liberation by 4 hours. The result is specific to this time horizon and these rates; it is not a general claim that increasing \(k_E\) always doubles release.
The input supplies \(20(4)=80\) ADC molecules per cell in either scenario. At 4 hours, scenario A has \(80-19.12=60.88\) ADC molecules per cell still in the modeled states, while scenario B has \(80-38.63=41.37\). In this no-loss model, both scenarios eventually process all 80 ADC molecules, yielding \(4(80)=320\) payload molecules per cell. A finite input duration alone does not reduce eventual yield when both rates are positive and there are no competing losses. A finite observation time can, however, show less than that eventual amount.
What Happens When Input Stops?
Stopping input does not stop the processing of ADC already in \(E\) or \(L\). For example, if constant input ends at time \(T\), later liberation comes from the cohorts that entered before \(T\); it is not calculated by setting \(P(t)\) to its value at \(T\). In the no-loss model, the eventual processed amount equals the total ADC that entered. The time course differs with \(k_E\) and \(k_L\), even though the eventual yield does not.
Worked Example: Equal Rates and a Finite Input Window
A hypothetical cell receives \(10\) ADC molecules per cell per hour for 2 hours, then receives no further ADC. Let \(k_E=k_L=1.0\ \mathrm{h}^{-1}\) and \(q=3\). Calculate cumulative liberation at the moment input stops, and compare it with eventual liberation.
For equal rates, \(S(a)=e^{-a}(1+a)\). Its integral from 0 to 2 hours is
Hence the amount processed by 2 hours is \(R(2)=10[2-1.4587]=5.413\) ADC molecules per cell, and liberated payload is \(P(2)=3(5.413)=16.24\) payload molecules per cell. The input supplied \(10(2)=20\) ADC molecules per cell, so \(20-5.413=14.59\) ADC molecules per cell remain unprocessed at that time. With no competing losses, those molecules continue through the pathway after input stops. Eventual liberation is \(3(20)=60\) payload molecules per cell, not 16.24.
Diagnostics, Alternatives, and Limitations
The model provides direct checks on calculations and on fitted trajectories. With zero initial amounts, the ADC mass balance is
This balance follows by adding the three state equations after dividing the \(P\) equation by \(q\): transfer terms cancel, leaving the input. If a computed solution violates this identity, check units, signs, initial conditions, and numerical implementation. Also check that \(E\), \(L\), and \(P\) remain nonnegative, that \(P(t)\) does not decrease, and that liberation cannot exceed \(q\) times cumulative ADC input.
A poor fit can indicate that one or more assumptions are inadequate. For example, a persistent mismatch in the observed timing of lysosomal arrival may indicate that a single first-order endosomal state is too simple. A measured loss of ADC-derived material would violate the no-competing-loss assumption and alter eventual yield. Measurements must also distinguish ADC still carrying payload from liberated payload; otherwise the two modeled processes may not be separately identifiable from the data.
A one-step model can be useful when only the overall delay matters, but it cannot attribute that delay to endosomal transfer versus lysosomal processing. The two-step model preserves that distinction at the cost of additional parameters and data requirements. If observations support a broad distribution of transit times rather than exponential residence times, a more detailed transit representation may be considered; adding states without informative measurements, however, can make rate estimates difficult to interpret. This model also does not describe the concentration of free payload after liberation, its removal, or its effects on cell viability.
Check Your Understanding
Answer each question, then compare with the concise solutions below.
- In this model, what does \(P(t)\) represent, and why is it not necessarily the amount of free payload currently inside the cell?
- If \(k_E=k_L=k\), write the fraction of a cohort processed by age \(a\).
- A cell receives 12 ADC molecules per cell per hour for 3 hours, with \(q=2\). Under the no-loss assumptions, what is the eventual liberated payload?
- Which model quantity should be changed to represent a slower endosomal-to-lysosomal transfer, and what qualitative effect should occur at a fixed early observation time?
- Name one mass-balance or trajectory check that can reveal an implementation or model problem.
Solutions. (1) \(P(t)\) is cumulative payload liberated up to time \(t\). It does not subtract payload that may subsequently leave or be otherwise removed, because those processes are not represented. (2) \(F(a)=1-e^{-ka}(1+ka)\). (3) Total ADC input is \(12(3)=36\) ADC molecules per cell, so eventual liberation is \(2(36)=72\) payload molecules per cell, provided both rates are positive and there are no competing losses. (4) Decrease \(k_E\); at a fixed early time, slower transfer generally delays lysosomal processing and lowers cumulative liberation. (5) Check \(E+L+P/q=\int_0^t u(s)\,ds\), nonnegative states, or nondecreasing cumulative \(P\).