ADC Species States From Conjugated Drug to Released Payload

Why ADC States Need Explicit Accounting

An antibody-drug conjugate (ADC) does not remain one chemical species as it moves through a biological system. An intact conjugate can be outside a cell, become internalized, be converted into drug-containing catabolites, and ultimately yield released payload. A model that treats all of these as one “ADC concentration” cannot distinguish where the payload is or whether it remains attached to antibody-derived material.

A state-transition scheme makes those distinctions explicit. Its purpose here is to account for the movement of conjugate and payload among four states, not to specify detailed uptake mechanisms or intracellular processing pathways. Each transition is represented by a flux, and each state changes according to the fluxes entering and leaving it.

Definition: Let \(A_e\) be extracellular intact ADC concentration and \(A_i\) be intracellular intact ADC concentration, both measured in ADC-molecule equivalents per unit reference volume. Let \(M_i\) be intracellular payload-containing catabolite concentration, measured in payload equivalents per the same volume, and let \(P_i\) be intracellular released-payload concentration in the same payload-equivalent units. Let \(q\) be the fixed number of payload equivalents per intact ADC molecule in this simplified model.

The choice of units is consequential. \(A_e\) and \(A_i\) count ADC molecules, while \(M_i\) and \(P_i\) count payload equivalents. Multiplying an ADC-molecule concentration by \(q\) converts it to payload-equivalent concentration. All four concentrations must use the same reference volume, such as a specified tissue or cellular volume, for their amounts to be added in a total-payload balance.

States, Fluxes, and Assumptions

Consider the following aggregate scheme:

\text{input} \longrightarrow A_e \underset{k_u}{\longrightarrow} A_i \underset{k_c}{\longrightarrow} M_i \underset{k_r}{\longrightarrow} P_i

The arrows marked by rate constants indicate first-order transitions: their flux is the rate constant multiplied by the concentration in the source state. The input enters as an amount rate \(R(t)\). In addition to the displayed transfers, extracellular intact ADC can leave the modeled system at rate \(k_e A_e\); intracellular intact ADC can leave without forming the tracked catabolite pool at rate \(k_i A_i\); catabolites can be lost at rate \(k_m M_i\); and released payload can be lost at rate \(k_p P_i\). These loss terms represent removal from the states being modeled, not necessarily one particular biological process.

This is a deliberately lumped scheme. It assumes a well-mixed reference volume, nonnegative concentrations, constant \(q\), and nonnegative rate constants. It treats all intracellular catabolites that retain payload as one pool and all released payload as another. The first-order rates are assumed appropriate over the modeled conditions. The scheme does not distinguish multiple ADC drug-loading classes or resolve intermediate biological steps.

Model: With concentrations expressed in nM and time in hours, \(R(t)\) has units nM ADC molecules per hour, and each \(k\) has units h\(^{-1}\). The state equations are

\begin{aligned} \frac{dA_e}{dt} &= R(t)-(k_u+k_e)A_e,\\ \frac{dA_i}{dt} &= k_uA_e-(k_c+k_i)A_i,\\ \frac{dM_i}{dt} &= qk_cA_i-(k_r+k_m)M_i,\\ \frac{dP_i}{dt} &= k_rM_i-k_pP_i. \end{aligned}

The factor \(q\) appears in the formation of \(M_i\) because \(k_cA_i\) is a flux of ADC molecules, whereas \(M_i\) is measured in payload equivalents. The flux \(k_rM_i\), by contrast, already has payload-equivalent units. The same units check applies to every term in each equation: all terms on a given right-hand side must have units of that state’s concentration per time.

Worked Example: Calculating State Changes at One Time

Suppose a hypothetical model has \(q=3\) payload equivalents per ADC molecule, \(A_e=10\) nM, \(A_i=2\) nM, \(M_i=1.5\) nM payload equivalents, and \(P_i=0.4\) nM payload equivalents. Let \(R=0.40\) nM ADC molecules/h; \(k_u=0.10\), \(k_e=0.020\), \(k_c=0.20\), \(k_i=0.030\), \(k_r=0.50\), \(k_m=0.10\), and \(k_p=0.25\) h\(^{-1}\). These are hypothetical values for illustrating the accounting.

For extracellular conjugate, \(\frac{dA_e}{dt}=0.40-(0.10+0.020)(10)=0.40-1.20=-0.80\) nM ADC molecules/h.

For intracellular conjugate, \(\frac{dA_i}{dt}=0.10(10)-(0.20+0.030)(2)=1.00-0.46=0.54\) nM ADC molecules/h.

For payload-containing catabolites, \(\frac{dM_i}{dt}=3(0.20)(2)-(0.50+0.10)(1.5)=1.20-0.90=0.30\) nM payload equivalents/h.

For released payload, \(\frac{dP_i}{dt}=0.50(1.5)-0.25(0.4)=0.75-0.10=0.65\) nM payload equivalents/h. Thus, at this instant, extracellular ADC and all three intracellular states have the indicated local rates of change. A negative derivative for \(A_e\) means its concentration is decreasing at that instant; it does not imply a negative concentration.

Deriving the Payload Mass Balance

The total payload represented in the four states is not \(A_e+A_i+M_i+P_i\), because the first two states count ADC molecules and the last two count payload equivalents. The correctly scaled total is

$$ L=q(A_e+A_i)+M_i+P_i. $$

Differentiate this expression and substitute the state equations. The internal transfer from \(A_e\) to \(A_i\) cancels because it leaves one state and enters the other at the same ADC-molecule flux. The \(k_c\) conversion also cancels in payload units: it removes \(qk_cA_i\) payload equivalents from intact intracellular ADC and adds exactly \(qk_cA_i\) to the catabolite pool. Likewise, release from \(M_i\) enters \(P_i\) at equal payload-equivalent flux.

\begin{aligned} \frac{dL}{dt} &=q\left(\frac{dA_e}{dt}+\frac{dA_i}{dt}\right) +\frac{dM_i}{dt}+\frac{dP_i}{dt}\\ &=qR(t)-qk_eA_e-qk_iA_i-k_mM_i-k_pP_i. \end{aligned}

This identity is a mass-balance check: transitions within the modeled system redistribute payload but do not create or destroy it. Only payload entering through the input or leaving through a specified loss changes \(L\). If a modeled process actually converts payload into an untracked state, its outgoing flux must appear as a loss from \(L\), or that state must be added to the scheme.

An analogous accounting for intact antibody-bearing ADC states gives \(\frac{d(A_e+A_i)}{dt}=R(t)-k_eA_e-k_iA_i-k_cA_i\). Here, uptake is an internal transfer and cancels, while conversion at \(k_c\) removes material from the intact-ADC states. This balance does not track antibody fragments after that conversion.

Worked Example: Checking the Total Payload Rate

Use the hypothetical concentrations and rates from the preceding example. The total payload rate computed from the state derivatives is \(\frac{dL}{dt}=3(-0.80+0.54)+0.30+0.65=-0.78+0.95=0.17\) nM payload equivalents/h.

The same result follows directly from the external input and losses: \(3(0.40)-3(0.020)(10)-3(0.030)(2)-0.10(1.5)-0.25(0.4)\). The arithmetic is \(1.20-0.60-0.18-0.15-0.10=0.17\) nM payload equivalents/h. Agreement confirms that internalization, catabolite formation, and payload release have been treated as transfers rather than incorrectly counted as payload losses.

The positive net rate means total payload in the tracked system is increasing at this instant, even though \(A_e\) itself is decreasing. State-specific trends and total-payload trends need not have the same direction.

Steady State and a Second Accounting Check

If the input is constant and the model reaches a steady state, each derivative is zero. For positive removal rates, the equations can be solved in sequence: first \(A_e\), then \(A_i\), then \(M_i\), and finally \(P_i\). This sequence follows the direction of the state transitions.

Worked Example: Solving for Steady-State Concentrations

Let the hypothetical constant input be \(R=0.40\) nM ADC molecules/h, with the rates from the first example and \(q=3\). At steady state, \(A_e=R/(k_u+k_e)=0.40/0.12=3.333\) nM ADC molecules.

Then \(A_i=k_uA_e/(k_c+k_i)=0.10(3.333)/0.23=1.449\) nM ADC molecules.

For the catabolite pool, \(M_i=qk_cA_i/(k_r+k_m)=3(0.20)(1.449)/0.60=1.449\) nM payload equivalents. Finally, \(P_i=k_rM_i/k_p=0.50(1.449)/0.25=2.899\) nM payload equivalents.

The payload input is \(qR=1.20\) nM payload equivalents/h. At steady state, the accounted losses are \(qk_eA_e=0.200\), \(qk_iA_i=0.1304\), \(k_mM_i=0.1449\), and \(k_pP_i=0.7246\) nM payload equivalents/h. Their sum is \(1.200\) nM payload equivalents/h to rounding, as required. The steady state is therefore consistent with the total-payload balance.

Diagnostics, Alternatives, and Limits

A useful diagnostic is to calculate the payload-balance residual from model outputs: \(\frac{dL}{dt}-[qR-qk_eA_e-qk_iA_i-k_mM_i-k_pP_i]\). It should be zero apart from numerical rounding. A persistent residual usually points to an omitted or double-counted flux, inconsistent units, or a mismatch between the payload stoichiometry used in a transition and the state definitions. Numerical integration should also preserve nonnegative concentrations for nonnegative inputs and rates; substantial negative states indicate an unsuitable numerical step or an implementation error.

Several errors are especially easy to make:

Adding unlike quantities. \(A_e+A_i+M_i+P_i\) is not a total payload concentration. Convert ADC molecules to payload equivalents with \(q\).
Counting a transfer as a loss. Internalization, catabolite formation, and payload release move payload between modeled states. They cancel from the total-payload balance when both source and destination are included.
Using inconsistent volumes. Concentrations from different reference volumes cannot be added directly unless first converted to amounts or put on a common volume basis.
Interpreting a lumped state too literally. \(M_i\) aggregates payload-containing catabolites; it does not identify their chemical composition or distinguish the steps that produced them.

A lumped four-state scheme is a compact alternative to a more detailed species network. It is appropriate when the available data support aggregate states and the goal is to track overall payload disposition. A more explicit network can separate distinct conjugate or catabolite species, but it requires additional states, transition definitions, and information to constrain them. Conversely, measuring only total ADC-related material may not identify how much is extracellular intact conjugate, intracellular conjugate, catabolite-bound payload, or free payload.

The model’s assumptions also limit its interpretation. A fixed \(q\) does not represent a distribution of drug loading. First-order transition rates do not automatically describe saturation, changing cellular conditions, or spatial heterogeneity. The loss terms do not identify where material goes after leaving the tracked system. These are reasons to extend the state scheme only when the scientific question and data justify the added structure, rather than treating every possible molecular detail as known.

Common pitfall: “Internalization removes ADC from the extracellular compartment, so it must be a payload loss.” The correction is to distinguish a compartmental loss from a system-boundary loss. Internalization decreases \(A_e\) and increases \(A_i\) by the same ADC flux; it is not a loss from total modeled payload.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

If \(q=4\) and \(A_i=2\) nM ADC molecules, how many nM payload equivalents are represented in the intact intracellular ADC state?
Why does the flux \(k_cA_i\) require a factor of \(q\) in the equation for \(M_i\), but \(k_rM_i\) does not require that factor in the equation for \(P_i\)?
Suppose payload release increases while all other quantities are momentarily fixed. Which state derivative changes directly, and what happens to the total-payload balance if release is an internal transfer?
Name one reason why a persistent nonzero payload-balance residual can occur in an implemented model.

Solutions. (1) \(qA_i=4(2)=8\) nM payload equivalents. (2) \(k_cA_i\) is measured in ADC molecules per volume per time, so it must be multiplied by \(q\) to become payload equivalents; \(k_rM_i\) already has payload-equivalent units because \(M_i\) is defined in those units. (3) Increasing \(k_r\) directly increases \(\frac{dP_i}{dt}\) and increases the outgoing term in \(\frac{dM_i}{dt}\) by the same amount. The two contributions cancel in \(\frac{dL}{dt}\), so release alone does not change total tracked payload. (4) Examples include omitting a loss flux, counting an internal transfer as a loss, using inconsistent units, or applying different payload stoichiometry in the state equation and balance calculation.