Why Separate Cleavage by Compartment?
An ADC can encounter different chemical and biological conditions in plasma and tumor interstitial fluid. A single linker-cleavage rate therefore may not describe both compartments: cleavage in plasma can generate payload before the ADC reaches the tumor, while cleavage in tumor interstitial fluid can generate payload locally. To quantify these differences, track intact ADC and released free payload separately in each compartment. The earlier material on ADC internalization and recycling describes cellular fluxes; here, the focus is extracellular cleavage and its effect on intact-ADC and free-payload exposure.
The model below treats each compartment as well mixed and uses constant first-order rates. It represents free payload in plasma and tumor interstitial fluid, not payload inside cells. Amount AUCs are integrals of amount over time; concentration AUCs are obtained by dividing by the corresponding compartment volume when that volume is constant.
Linked Mass-Balance Model
For a simple model, ADC moves from plasma to tumor interstitial fluid with first-order rate constant \(k_{pt}\), with no return flux. Intact ADC is also removed by non-cleavage processes at rates \(k_{p,\mathrm{el}}\) and \(k_{t,\mathrm{el}}\). These may represent other intact-ADC loss processes; they are distinct from linker cleavage. After cleavage, free payload is removed locally at rates \(k_{F,p}\) and \(k_{F,t}\). Assume no free-payload transport between compartments in this base model.
The cleavage flux in a compartment is its cleavage rate constant multiplied by the intact-ADC amount there. Multiplying that flux by \(q\) gives payload generation in payload-equivalent amount per unit time. This formulation assumes a fixed effective payload loading \(q\); it does not describe a changing distribution of drug-to-antibody ratio.
The required assumptions are constant rate constants, well-mixed compartments, first-order cleavage and loss, and consistent amount units for ADC equivalents and payload equivalents. The equations also assume that cleavage-generated payload enters the modeled free-payload pool immediately. If initial free payload is present, it must be included as an initial condition rather than attributed to linker cleavage.
Integrated Exposures and a Useful Derivation
Consider a plasma bolus \(D\), so \(A_p(0)=D\), \(A_t(0)=0\). Define \(K_p=k_{p,\mathrm{cl}}+k_{p,\mathrm{el}}+k_{pt}\) and \(K_t=k_{t,\mathrm{cl}}+k_{t,\mathrm{el}}\). Integrating the plasma equation from zero to infinity, assuming \(A_p(t)\) eventually vanishes, gives:
Integrating the tumor equation gives \(A_t(\infty)-A_t(0)=k_{pt}\mathrm{AUC}_{A_p}-K_t\mathrm{AUC}_{A_t}\). With \(A_t(0)=A_t(\infty)=0\), this yields the tumor amount AUC. The same integration also shows why cleavage and non-cleavage loss must both appear in the denominator: both remove intact ADC from that compartment.
For payload, define \(G_j\) as the total payload generated by cleavage in compartment \(j\), where \(j\) is \(p\) or \(t\). Integrating the free-payload equation, assuming free payload eventually vanishes, gives a general result that includes initial payload:
Thus, the expression \(G_j/k_{F,j}\) alone is valid only when \(F_j(0)=0\). This distinction matters even when there is no ADC cleavage: initially present free payload can still contribute exposure as it is eliminated. For constant compartment volume \(V_j\), concentration AUC is \(\mathrm{AUC}_{F_j}/V_j\); the same conversion applies to intact ADC.
Worked Example: Baseline Intact-ADC and Payload Exposure
Suppose a hypothetical plasma bolus contains \(D=100\) nmol ADC equivalents with \(q=4\) payload equivalents per ADC equivalent. Let \(k_{p,\mathrm{cl}}=0.10\ \mathrm{h}^{-1}\), \(k_{p,\mathrm{el}}=0.05\ \mathrm{h}^{-1}\), and \(k_{pt}=0.05\ \mathrm{h}^{-1}\). In tumor interstitial fluid, let \(k_{t,\mathrm{cl}}=0.30\ \mathrm{h}^{-1}\) and \(k_{t,\mathrm{el}}=0.20\ \mathrm{h}^{-1}\). Assume both free-payload initial amounts are zero, with \(k_{F,p}=0.40\ \mathrm{h}^{-1}\) and \(k_{F,t}=0.20\ \mathrm{h}^{-1}\).
The total intact-ADC loss rates are \(K_p=0.10+0.05+0.05=0.20\ \mathrm{h}^{-1}\) and \(K_t=0.30+0.20=0.50\ \mathrm{h}^{-1}\). Therefore:
The payload generated in plasma is \(G_p=4\times0.10\ \mathrm{h}^{-1}\times500\ \mathrm{nmol\cdot h}=200\) nmol payload equivalents. In tumor interstitial fluid it is \(G_t=4\times0.30\ \mathrm{h}^{-1}\times50\ \mathrm{nmol\cdot h}=60\) nmol. Because the initial free-payload amounts are zero, the free-payload AUCs are:
These are amount AUCs, not concentration AUCs. For example, if the hypothetical tumor interstitial-fluid volume were \(0.50\) L and constant, the tumor free-payload concentration AUC would be \(300/0.50=600\ \mathrm{nmol\cdot h/L}\). The plasma and tumor amount AUCs should not be compared as if they were concentrations.
Worked Example: Increasing Tumor Cleavage
Keep the preceding example's plasma parameters and bolus, but increase the tumor cleavage rate to \(k_{t,\mathrm{cl}}=0.60\ \mathrm{h}^{-1}\), leaving \(k_{t,\mathrm{el}}=0.20\ \mathrm{h}^{-1}\). The plasma AUC remains \(500\ \mathrm{nmol\cdot h}\), so the amount of ADC delivered into tumor interstitial fluid over time, integrated as a flux, is \(k_{pt}\mathrm{AUC}_{A_p}=0.05\times500=25\) nmol ADC equivalents.
Now \(K_t=0.60+0.20=0.80\ \mathrm{h}^{-1}\). Thus the tumor intact-ADC AUC and tumor payload generation are:
With \(F_t(0)=0\) and \(k_{F,t}=0.20\ \mathrm{h}^{-1}\), \(\mathrm{AUC}_{F_t}=75/0.20=375\ \mathrm{nmol\cdot h}\). Compared with the baseline, increasing tumor cleavage lowers tumor intact-ADC AUC from 50 to 31.25 nmol·h while increasing tumor free-payload AUC from 300 to 375 nmol·h. Faster cleavage does not simply increase every exposure: it depletes the intact species as it increases local payload generation.
Worked Example: Increasing Plasma Cleavage
Return to the baseline parameters, but increase plasma cleavage from \(0.10\) to \(0.20\ \mathrm{h}^{-1}\). Keep \(k_{p,\mathrm{el}}=0.05\ \mathrm{h}^{-1}\), \(k_{pt}=0.05\ \mathrm{h}^{-1}\), and the tumor rates unchanged. Then \(K_p=0.30\ \mathrm{h}^{-1}\), so:
Plasma payload generation is \(G_p=4\times0.20\times333.3=266.7\) nmol, giving \(\mathrm{AUC}_{F_p}=266.7/0.40=666.7\ \mathrm{nmol\cdot h}\) for \(F_p(0)=0\). Tumor payload generation is \(G_t=4\times0.30\times33.33=40.00\) nmol, giving \(\mathrm{AUC}_{F_t}=40.00/0.20=200.0\ \mathrm{nmol\cdot h}\). Relative to baseline, more payload exposure occurs in plasma and less in tumor: the increased plasma cleavage competes with intact-ADC delivery to the tumor.
Initial Conditions, Diagnostics, and Alternatives
A frequent error is to calculate free-payload AUC as total generated payload divided by the payload elimination rate without checking the initial condition. For example, if \(F_j(0)=12\) nmol, \(G_j=20\) nmol, and \(k_{F,j}=0.40\ \mathrm{h}^{-1}\), then:
Using only \(G_j/k_{F,j}\) would give \(50\ \mathrm{nmol\cdot h}\), omitting the contribution from initial free payload. A related check is that when \(G_j=0\), the AUC must still be \(F_j(0)/k_{F,j}\), not zero, if the initial amount is positive.
The model gives useful diagnostic checks. Intact amounts should remain nonnegative; integrated tumor ADC delivery must equal \(k_{pt}\mathrm{AUC}_{A_p}\); and, with zero initial free payload, \(k_{F,j}\mathrm{AUC}_{F_j}\) must equal \(G_j\). If the calculated tumor payload generation exceeds \(q\) times the total ADC amount delivered to the tumor in this one-way model, check the rate definitions, units, and initial conditions. A fitted model that requires negative amounts, implausible rate constants, or systematic discrepancies between observed and predicted time courses may indicate that the assumptions or compartment structure are inadequate.
The one-way ADC transfer and no-free-payload-transport assumptions are deliberate simplifications, not universal properties. If ADC returns from tumor to plasma, add a return flux to both equations and include it in the corresponding compartment loss and gain terms. If free payload transports between compartments, its transfer terms must be added to the free-payload equations; then a local amount AUC is not generally its local generation divided by its local elimination rate. A numerical solution of the coupled equations, or an integrated mass balance across both compartments, is then needed.
An alternative is to use a single effective cleavage rate for the whole body. This may be adequate if the data cannot identify compartment-specific rates or the compartments behave similarly for the question at hand. It cannot represent a distinct plasma-versus-tumor cleavage pattern. A more detailed alternative can represent multiple ADC loading states or time-varying cleavage conditions, but those additions require data that can support the added parameters. The two-compartment first-order model is most informative when its rates and transfer terms can be estimated or constrained and when the question concerns exposure rather than intracellular payload processing.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A plasma dose is \(60\) nmol ADC equivalents. The plasma cleavage, non-cleavage elimination, and tumor-transfer rates are \(0.10\), \(0.10\), and \(0.10\ \mathrm{h}^{-1}\), respectively. What is the plasma intact-ADC amount AUC?
- Using the values in question 1, tumor cleavage is \(0.20\ \mathrm{h}^{-1}\) and tumor non-cleavage elimination is \(0.30\ \mathrm{h}^{-1}\). What is the tumor intact-ADC amount AUC?
- In a compartment, \(q=3\), \(k_{\mathrm{cl}}=0.25\ \mathrm{h}^{-1}\), and intact-ADC amount AUC is \(20\ \mathrm{nmol\cdot h}\). If free payload starts at \(5\) nmol and \(k_F=0.50\ \mathrm{h}^{-1}\), calculate payload generated and free-payload amount AUC.
- Why does faster plasma cleavage potentially reduce tumor payload generation, even though cleavage releases payload?
Solutions. (1) \(K_p=0.10+0.10+0.10=0.30\ \mathrm{h}^{-1}\), so \(\mathrm{AUC}_{A_p}=60/0.30=200\ \mathrm{nmol\cdot h}\). (2) Tumor input integrated over time is \(k_{pt}\mathrm{AUC}_{A_p}=0.10\times200=20\) nmol. Tumor total intact-ADC loss is \(0.20+0.30=0.50\ \mathrm{h}^{-1}\), so \(\mathrm{AUC}_{A_t}=20/0.50=40\ \mathrm{nmol\cdot h}\). (3) \(G=3\times0.25\times20=15\) nmol payload equivalents. Including the initial \(5\) nmol, \(\mathrm{AUC}_F=(5+15)/0.50=40\ \mathrm{nmol\cdot h}\). (4) Plasma cleavage removes intact ADC before it can transfer to the tumor. The tumor receives less intact ADC, which can reduce tumor cleavage flux and local payload generation, even as plasma payload generation rises.