Why antigen heterogeneity and turnover matter
Two tumor-cell populations can differ not only in how much antigen is present at baseline, but also in how quickly that antigen is replenished or removed. Consequently, a population with more antigen per cell need not have uptake proportional to that antigen level. Binding can occupy only a fraction of surface antigen, bound ADC can internalize, and target turnover can remove antigen—with or without its bound ADC—from the cell surface.
A useful calculation therefore keeps the subpopulations distinct and represents the surface states explicitly. The model below describes ADC molecules bound to surface antigen and the rate at which those complexes internalize. It does not equate surface binding with intracellular uptake, nor does it predict payload exposure or cell killing.
With \(k_{\mathrm{on}}\) expressed in concentration\(^{-1}\) time\(^{-1}\), and the other rate constants in time\(^{-1}\), the mass-action model is
Here \(u_j(t)\) is the internalized ADC rate in ADC molecules per cell per unit time. Association transfers antigen from \(R_j\) to \(C_j\); dissociation transfers it back. Internalization removes a surface complex and counts as ADC uptake. Turnover of bound antigen also removes a surface complex in this model, but is not counted as ADC internalization. That distinction is an assumption: it represents target loss by a route other than the modeled internalization flux.
Assumptions and steady-state calculation
The equations assume well-mixed free ADC at a constant concentration, one effective binding site per antigen molecule, and constant rate parameters within each subpopulation. Synthesis and turnover are represented by constant rates, and internalization is first order in surface complex. The calculation also treats the ADC concentration as unaffected by binding or uptake. These assumptions are most defensible over intervals where free ADC changes little and where subpopulations can be represented by reasonably stable parameter values.
For a steady-state calculation, set both time derivatives to zero. Define \(a=k_{\mathrm{on}}A\), the pseudo-first-order association rate, and \(b_j=k_{\mathrm{off}}+k_{\mathrm{int}}+k_{\mathrm{turn},C,j}\). The complex equation gives \(C_j=(a/b_j)R_j\). Substituting this into the unbound-antigen equation gives
The effective loss term involving \(a\) is not simply \(a\): dissociation returns antigen to the unbound state, while internalization and bound-target turnover remove it from the surface complex. The expression follows because \(1-k_{\mathrm{off}}/b_j=(k_{\mathrm{int}}+k_{\mathrm{turn},C,j})/b_j\). This derivation also shows why binding alone does not determine uptake: uptake depends on the fraction of complex that internalizes, as well as on how quickly antigen is supplied and lost.
Worked Example: Steady-State Uptake in Two Antigen Subpopulations
Consider a hypothetical tumor with \(1.00\times10^8\) cells: 30% are high-antigen cells and 70% are low-antigen cells. Without ADC, their surface antigen levels are 5,000 and 1,000 molecules per cell, respectively. Both have unbound-antigen turnover \(k_{\mathrm{turn},R}=0.20\) day\(^{-1}\), so their synthesis rates are \(q_{\mathrm{high}}=0.20(5{,}000)=1{,}000\) and \(q_{\mathrm{low}}=0.20(1{,}000)=200\) molecules per cell per day. Use \(A=2.00\) nM, \(k_{\mathrm{on}}=0.100\) nM\(^{-1}\)day\(^{-1}\), \(k_{\mathrm{off}}=0.300\) day\(^{-1}\), \(k_{\mathrm{int}}=0.500\) day\(^{-1}\), and \(k_{\mathrm{turn},C}=0.100\) day\(^{-1}\) for both groups.
First calculate \(a=k_{\mathrm{on}}A=0.100(2.00)=0.200\) day\(^{-1}\) and \(b=0.300+0.500+0.100=0.900\) day\(^{-1}\). The steady-state denominator for either group is \(0.200+0.200[(0.500+0.100)/0.900]=0.333333\) day\(^{-1}\).
For high-antigen cells, \(R_{\mathrm{high,ss}}=1{,}000/0.333333=3{,}000\) molecules per cell. Then \(C_{\mathrm{high,ss}}=(0.200/0.900)(3{,}000)=666.7\) molecules per cell, and \(u_{\mathrm{high,ss}}=0.500(666.7)=333.3\) ADC molecules per cell per day.
For low-antigen cells, \(R_{\mathrm{low,ss}}=200/0.333333=600\) molecules per cell. Thus \(C_{\mathrm{low,ss}}=(0.200/0.900)(600)=133.3\) molecules per cell, and \(u_{\mathrm{low,ss}}=0.500(133.3)=66.67\) ADC molecules per cell per day.
The population-average rate is the fraction-weighted mean: \(0.30(333.3)+0.70(66.67)=146.7\) ADC molecules per cell per day. For \(1.00\times10^8\) cells, this is \(1.467\times10^{10}\) ADC molecules per day. If those rates remain at steady state for 2.00 days, cumulative uptake is approximately \(2.933\times10^{10}\) ADC molecules. This total is conditional on constant free ADC and unchanged cell numbers and subpopulation fractions.
Uptake over time and aggregation
At a fixed ADC concentration, the differential equations describe transient changes as well as steady state. Immediately after ADC is introduced, antigen can begin binding before much complex has accumulated for internalization. Thus \(u_j(t)\) can initially be low even when the binding flux \(k_{\mathrm{on}}A R_j(t)\) is high. For a population with \(N\) cells and fixed subpopulation fractions \(f_j\), where \(\sum_j f_j=1\), the total uptake rate is \(N\sum_j f_j u_j(t)\). Cumulative uptake from time 0 to \(T\) is the time integral of that rate.
When subpopulations differ, calculate each \(u_j(t)\) using its own antigen synthesis, turnover, and initial conditions, then weight the rates. In general, substituting a mean antigen level into a single-cell model will not reproduce that weighted result: the steady-state equations are nonlinear functions of the group-specific rates and antigen supply. An average can hide a low-antigen fraction that contributes little uptake or a high-turnover fraction that replenishes surface target rapidly.
Worked Example: Early Uptake After ADC Exposure
Use the high-antigen parameters from the previous example. Before exposure, suppose the cells are at their no-ADC steady state: \(R(0)=5{,}000\) molecules per cell and \(C(0)=0\). At \(t=0\), set \(A=2.00\) nM. The initial derivatives are \(dR/dt=1{,}000-(0.200+0.200)(5{,}000)=-1{,}000\) molecules per cell per day and \(dC/dt=0.200(5{,}000)=1{,}000\) molecules per cell per day.
For a single forward-Euler step of \(\Delta t=0.100\) day, \(R(0.100)\approx5{,}000+(-1{,}000)(0.100)=4{,}900\) molecules per cell, and \(C(0.100)\approx0+1{,}000(0.100)=100\) molecules per cell. The internalization rate is 0 at the start and approximately \(0.500(100)=50.0\) ADC molecules per cell per day at the end of the step. A trapezoidal approximation to uptake over this step is \([(0+50.0)/2](0.100)=2.50\) ADC molecules per cell. For \(1.00\times10^8\) such cells, that is approximately \(2.50\times10^8\) ADC molecules.
This is a deliberately coarse numerical illustration, not an exact transient solution: forward Euler approximates the states using their initial derivatives, and the trapezoidal calculation approximates the changing uptake rate. Smaller time steps or a validated numerical solver are needed when transient accuracy matters.
Worked Example: How Turnover Changes Steady-State Uptake
Compare two hypothetical subpopulations that each have a no-ADC baseline of 5,000 surface-antigen molecules per cell, but different turnover rates. Keep \(a=0.200\) day\(^{-1}\), \(k_{\mathrm{off}}=0.300\) day\(^{-1}\), and \(k_{\mathrm{int}}=0.500\) day\(^{-1}\), and set bound-antigen turnover equal to unbound-antigen turnover. The synthesis rate is chosen as \(q=5{,}000k_{\mathrm{turn},R}\), preserving the stated no-ADC baseline in each case.
For the slow-turnover group, \(k_{\mathrm{turn},R}=k_{\mathrm{turn},C}=0.100\) day\(^{-1}\), so \(q=500\) molecules per cell per day and \(b=0.900\) day\(^{-1}\). The denominator is \(0.100+0.200(0.600/0.900)=0.233333\) day\(^{-1}\). Therefore \(R_{\mathrm{ss}}=500/0.233333=2{,}142.9\) molecules per cell, \(C_{\mathrm{ss}}=(0.200/0.900)(2{,}142.9)=476.2\) molecules per cell, and uptake is \(0.500(476.2)=238.1\) ADC molecules per cell per day.
For the faster-turnover group, both turnover rates are \(0.500\) day\(^{-1}\), so \(q=2{,}500\) molecules per cell per day and \(b=1.300\) day\(^{-1}\). The denominator is \(0.500+0.200(1.000/1.300)=0.653846\) day\(^{-1}\). Hence \(R_{\mathrm{ss}}=2{,}500/0.653846=3{,}823.5\) molecules per cell, \(C_{\mathrm{ss}}=(0.200/1.300)(3{,}823.5)=588.2\) molecules per cell, and uptake is \(0.500(588.2)=294.1\) ADC molecules per cell per day.
The faster-turnover group has lower complex per unbound antigen, but its greater synthesis replenishes the surface pool and yields higher steady-state uptake in this example. This comparison depends on holding the no-ADC antigen level fixed by changing synthesis along with turnover; changing turnover while holding synthesis fixed would answer a different question.
Diagnostics, alternatives, and limitations
Several checks help identify a model or calculation that is going wrong. First, the units must agree: \(k_{\mathrm{on}}A\), \(k_{\mathrm{off}}\), \(k_{\mathrm{int}}\), and turnover rates all have units of inverse time, while \(q_j\) has units of molecules per cell per time. Second, with nonnegative parameters and synthesis, steady-state \(R_j\), \(C_j\), and \(u_j\) should be nonnegative. Third, the surface equations should balance at steady state: the synthesis rate must equal the net loss from unbound antigen, including the net effect of binding and dissociation. Implausibly large uptake, negative states in a simulation, or substantial step-size dependence are reasons to check units, equations, initial conditions, and numerical resolution.
An alternative is to assume rapid binding equilibrium and calculate surface occupancy algebraically, then use a simpler uptake equation. That can be useful when binding and dissociation are much faster than internalization, turnover, and changes in ADC concentration. The explicit \(R_j,C_j\) model is preferable when those timescales may be comparable, because it preserves transient binding and dissociation rather than imposing instantaneous equilibrium. Neither representation is automatically appropriate: the choice should follow the timescales and data available.
A common mistake is to calculate uptake as \(k_{\mathrm{int}}\) times total antigen, or to multiply a single average antigen value by an occupancy estimated for the whole tumor. The correction is to calculate \(C_j\) and \(k_{\mathrm{int}}C_j\) within each subpopulation, then aggregate. Total antigen includes unbound antigen that is not being internalized as ADC complex, and heterogeneous groups can have different synthesis and turnover as well as different antigen levels.
The model also has important limits. A constant \(A\) ignores systemic and tumor-level ADC depletion and spatial concentration gradients. A fixed subpopulation fraction ignores changes in cell abundance. The first-order turnover terms do not represent variable synthesis, antigen regulation, or multiple trafficking routes. If bound-target turnover removes ADC without internalization, that loss must not be counted as uptake. Parameters and state definitions should be adjusted when measurements indicate a different mechanism; a calculated uptake rate is a model-based estimate, not a direct observation.
Check Your Understanding
Work each problem using the model and assumptions defined above. Solutions follow the questions.
- If \(k_{\mathrm{on}}A=0.12\) day\(^{-1}\), \(k_{\mathrm{off}}=0.20\) day\(^{-1}\), \(k_{\mathrm{int}}=0.30\) day\(^{-1}\), and \(k_{\mathrm{turn},C}=0.10\) day\(^{-1}\), what is \(C_{\mathrm{ss}}/R_{\mathrm{ss}}\)?
- At steady state, what is the uptake rate for a subpopulation with \(C_{\mathrm{ss}}=240\) molecules per cell and \(k_{\mathrm{int}}=0.25\) day\(^{-1}\)?
- A tumor has two equally sized subpopulations with uptake rates of 80 and 200 ADC molecules per cell per day. What is the mean rate per cell? Why should this result not be calculated from an averaged antigen level without further justification?
- For the high-antigen example, what is the initial binding flux immediately after ADC exposure, and why is the initial internalization rate different?
Solutions. (1) \(b=0.20+0.30+0.10=0.60\) day\(^{-1}\), so \(C_{\mathrm{ss}}/R_{\mathrm{ss}}=0.12/0.60=0.200\). (2) \(u_{\mathrm{ss}}=0.25(240)=60.0\) ADC molecules per cell per day. (3) The weighted mean is \(0.50(80)+0.50(200)=140\) ADC molecules per cell per day. The steady-state model is nonlinear in group-specific parameters, so a mean antigen level need not yield the mean group uptake. (4) The initial binding flux is \(k_{\mathrm{on}}A R(0)=0.200(5{,}000)=1{,}000\) complexes per cell per day. The initial internalization rate is zero because \(C(0)=0\), and \(u(0)=k_{\mathrm{int}}C(0)=0\).