Antigen-Density Effects on CAR T-Cell Killing

Why Antigen Density Matters

CAR T cells recognize accessible antigen at the tumor-cell surface. Two tumor cells may both express the target antigen yet present very different numbers of accessible molecules. If CAR activation rises with antigen density, those cells need not trigger the same cytolytic response. This distinction matters when comparing tumors, interpreting antigen-low residual disease, or predicting what a mixed population of antigen-high and antigen-low cells will experience.

The earlier material on CAR T-cell antigen recognition and CAR signaling describes recognition and downstream signaling. Here the focus is narrower: representing the dependence of activation and killing on antigen density. The model below is a tractable phenomenological approximation, not a claim that each biochemical step is a Hill function.

Definition: Let $A$ denote the density of accessible target antigen on a tumor cell, measured in molecules per cell or in a consistently calibrated equivalent unit. Let $r(A)$ be the relative antigen-dependent CAR activation, scaled from 0 to 1. A Hill response is $$ r(A)=\frac{A^h}{K_A^h+A^h}, $$ where $K_A>0$ is the antigen density giving half-maximal activation, and $h>0$ controls the steepness of the response. In a fixed assay or model condition, define the antigen-dependent killing hazard as $$ \lambda(A)=\lambda_{\max}r(A), $$ where $\lambda_{\max}$ is the maximum killing hazard under those conditions, with units of inverse time.

The parameters have distinct meanings. At $A=K_A$, the activation fraction is $r(A)=1/2$, regardless of $h$. A larger $h$ makes the response transition more sharply around $K_A$; it does not necessarily mean that the CAR has a larger maximum response. $K_A$ is a functional half-maximal density in this model, not automatically a receptor-binding dissociation constant. It can reflect multiple processes between antigen encounter and measured activation.

From Activation to Killing Over Time

A killing hazard is an instantaneous rate, not a fraction of targets killed. To obtain a killing probability, assume that antigen density remains constant during the observation interval, each target has a constant hazard $\lambda(A)$, and target fates are conditionally independent given the model conditions. Also assume that the CAR T-cell availability and function are effectively fixed over this interval. These assumptions are most plausible for a short, controlled assay; they are not guaranteed in a changing tumor environment.

If $S(t\mid A)$ is the probability that a target cell with antigen density $A$ remains alive at time $t$, a constant hazard means that over a short interval $dt$, the probability of death is approximately $\lambda(A)dt$ among targets still alive. Therefore,

\frac{dS(t\mid A)}{dt}=-\lambda(A)S(t\mid A), \qquad S(0\mid A)=1.

Separating variables and integrating gives $\ln S(t\mid A)=-\lambda(A)t$, using the initial condition $S(0\mid A)=1$. Thus survival and cumulative killing by time $\tau$ are

S(\tau\mid A)=e^{-\lambda_{\max}r(A)\tau}, \qquad P(\text{killed by }\tau\mid A)=1-e^{-\lambda_{\max}r(A)\tau}.

This conversion is important: a twofold increase in activation fraction does not generally produce a twofold increase in the fraction killed by a fixed time. The exponential relationship also means that predicted killing depends on observation duration, as well as antigen density and the killing-rate parameters.

Worked Example: Comparing Antigen-Low and Antigen-High Cells

Consider hypothetical short-assay parameters $K_A=1{,}000$ molecules per cell, $h=2$, and $\lambda_{\max}=0.80$ per day. Assume the accessible antigen densities are constant and the fixed assay conditions justify using the same $\lambda_{\max}$ for both cell types. Calculate predicted killing after one day for cells with $A=500$ and $A=2{,}000$ molecules per cell.

For $A=500$, the activation fraction is

r(500)=\frac{500^2}{1{,}000^2+500^2} =\frac{250{,}000}{1{,}250{,}000}=0.200.

The killing hazard is $0.80(0.200)=0.160$ per day. Therefore, the one-day killing probability is

1-e^{-0.160(1\ \mathrm{day})}=1-e^{-0.160}=0.1479.

The model predicts about 14.8% killing over one day for this antigen-low group.

For $A=2{,}000$, $r(2{,}000)=4{,}000{,}000/(1{,}000{,}000+4{,}000{,}000)=0.800$. The hazard is $0.80(0.800)=0.640$ per day, giving

1-e^{-0.640(1\ \mathrm{day})}=1-e^{-0.640}=0.4727.

Thus the predicted one-day killing is about 47.3% for the antigen-high group. The response differs because density changes the activation fraction, which changes the hazard; the final killing probabilities are not simply in the ratio $0.8:0.2$.

Response Steepness and Antigen Sensitivity

The Hill coefficient $h$ controls how quickly activation changes near $K_A$. When $h=1$, the response rises gradually. Larger values make the response more switch-like around $K_A$, while still varying continuously for positive $A$. The model predicts $r(0)=0$, $r(A)\to1$ as $A$ becomes large, and $r(K_A)=0.5$. These boundary properties are useful checks when implementing the equation.

Worked Example: How Steepness Changes a Low-Density Prediction

Use the same hypothetical $K_A=1{,}000$ molecules per cell and $A=300$ molecules per cell, but compare $h=2$ with $h=4$. Keep $\lambda_{\max}=0.80$ per day and calculate one-day killing.

With $h=2$,

r_2(300)=\frac{300^2}{1{,}000^2+300^2} =\frac{90{,}000}{1{,}090{,}000}=0.08257.

The hazard is $0.80(0.08257)=0.06606$ per day, so predicted one-day killing is $1-e^{-0.06606}=0.06393$, or about 6.39%.

With $h=4$,

r_4(300)=\frac{300^4}{1{,}000^4+300^4} =\frac{0.0081}{1+0.0081}=0.008035.

Here the powers are evaluated after expressing both densities relative to $1{,}000$, so the ratio is dimensionless. The hazard is $0.80(0.008035)=0.006428$ per day and one-day killing is $1-e^{-0.006428}=0.006407$, or about 0.641%. The two models agree at $A=K_A$, where both give $r=0.5$, but differ substantially below that density. This illustrates why low-antigen predictions can be sensitive to the assumed response steepness.

Heterogeneous Tumors Require Population Averaging

A tumor need not have one antigen density. Let $f(A)$ describe the fraction of target cells at each density, with the fractions over all densities summing to one. Under the same constant-hazard assumptions, the surviving fraction after time $\tau$ is the average of the density-specific survival probabilities:

S_{\mathrm{pop}}(\tau)=\int_0^\infty e^{-\lambda_{\max}r(A)\tau}f(A)\,dA, \qquad K_{\mathrm{pop}}(\tau)=1-S_{\mathrm{pop}}(\tau).

For discrete subpopulations with fractions $w_i$, where $\sum_i w_i=1$, the corresponding expression is $K_{\mathrm{pop}}(\tau)=\sum_i w_i[1-e^{-\lambda_{\max}r(A_i)\tau}]$. In general, this is not equal to the prediction obtained by putting the mean antigen density into the single-density formula. Both activation and survival are nonlinear functions of $A$, so replacing a distribution by its mean can discard biologically important information. The direction and size of the resulting discrepancy depend on the distribution and model parameters.

Worked Example: A Mixed Tumor Versus a Uniform Tumor

Use the hypothetical parameters $K_A=1{,}000$ molecules per cell, $h=2$, and $\lambda_{\max}=0.80$ per day for a one-day interval. Compare (a) a uniform population with $A=1{,}000$ on every cell and (b) a tumor with half its cells at $A=0$ and half at $A=2{,}000$. The mean density is $1{,}000$ molecules per cell in both cases.

For the uniform population, $r(1{,}000)=0.5$, so the hazard is $0.80(0.5)=0.40$ per day. Its predicted killing is

K_{\mathrm{uniform}}=1-e^{-0.40}=0.3297.

For the heterogeneous population, cells with $A=0$ have $r(0)=0$ and predicted killing of zero under this antigen-dependent model. Cells with $A=2{,}000$ have $r=0.80$ and predicted killing $1-e^{-0.640}=0.4727$. Weighting by the two equal subpopulation fractions gives

K_{\mathrm{heterogeneous}} =0.50(0)+0.50(0.4727)=0.2364.

Although both populations have the same mean density, the model predicts 32.97% killing for the uniform population and 23.64% for the heterogeneous one. This result follows from the specified distribution and parameters; it is not a rule that heterogeneity must always lower predicted killing. A different distribution, time interval, or response shape can change the comparison.

Assumptions, Diagnostics, and Model Choices

A density-response model is useful only when its assumptions are sufficiently compatible with the question and data. The Hill form compresses several biological processes into $K_A$ and $h$; the killing hazard then compresses signaling-to-cytolysis into $\lambda_{\max}$. This can be appropriate for comparing responses over a defined range, but it does not identify each underlying mechanism.

Check the density measurement. The model concerns accessible surface antigen, not simply transcript abundance or total protein. Verify that density units and calibration are comparable across samples.
Check the response range. Measurements should include enough low, intermediate, and high densities to distinguish $K_A$ from $h$. If all observations are far below or above $K_A$, those parameters may be poorly identified.
Check time-course behavior. If survival curves are not compatible with a roughly constant hazard over the interval, a single exponential survival model may be inadequate. Changing effector availability, antigen density, or cell state can produce time-varying hazards.
Check residual patterns and replicates. Systematic prediction errors concentrated in antigen-low cells may signal a misspecified response shape, a measurement problem, or mechanisms not represented by the model. Large replicate variation may require explicit variability rather than a single deterministic curve.
Check the assay context. $\lambda_{\max}$ is conditional on the CAR T-cell state, effector-to-target conditions, and other assay features. It should not be treated as a universal constant across different settings.

A hard-threshold alternative sets activation to zero below a chosen density and to a fixed level above it. This can approximate an abrupt functional boundary, but it creates a discontinuity and may misrepresent gradual responses. A linear alternative makes response proportional to density over its modeled range; without a cap it can exceed the intended maximum, and it may fit poorly near saturation. The Hill response is a continuous compromise with an explicit transition scale and steepness, but its flexibility does not establish that the underlying biology follows a Hill mechanism.

When density heterogeneity is substantial, a distribution-based calculation is preferable to a calculation based only on mean density. If only the mean is available, predictions for a heterogeneous tumor are underdetermined unless additional assumptions about the distribution are made. A richer mechanistic binding model can represent receptor and ligand interactions more explicitly, but it requires additional parameters and data; it is not automatically more reliable.

Common Pitfall and Limitations

Common pitfall: “The mean antigen density is $K_A$, so the whole tumor has half-maximal activation.” This is valid only for a uniform population at that density. For a heterogeneous tumor, the average response is calculated across the density distribution; it is generally not the response evaluated at the mean.

The model also predicts antigen-dependent killing only. It omits antigen-independent killing, spontaneous target death, changing CAR T-cell numbers, and changes in target or effector state. If such processes matter, they need to be represented and supported by data rather than implicitly attributed to antigen density. Likewise, this short-interval formulation does not describe effects of repeated stimulation or later changes in CAR T-cell function. Predicted killing is therefore conditional on the stated model and should not be interpreted as a clinical response probability without a broader, validated model.

Key takeaway: Antigen density can be linked to CAR activation with a saturating response and then to target killing through a hazard model. For heterogeneous tumors, calculate survival or killing within each density group and average across the population; a mean-density substitution can give a different answer.

Check Your Understanding

Answer each question using the assumptions and equations in this tutorial. Solutions follow each prompt.

For $A=K_A$, what is the activation fraction, and does that imply that half of the targets are killed by time $\tau$?
Solution: The activation fraction is $r(K_A)=1/2$. The killing probability is $1-e^{-\lambda_{\max}\tau/2}$, which is not generally 1/2; it depends on $\lambda_{\max}\tau$.
With $r(A)=0.25$, $\lambda_{\max}=0.60$ per day, and $\tau=2$ days, calculate the predicted killing probability.
Solution: The hazard is $0.60(0.25)=0.15$ per day. Thus $K=1-e^{-0.15(2)}=1-e^{-0.30}=0.2592$, or about 25.9%.
Why can two tumors with the same mean antigen density have different predicted killing?
Solution: The model’s survival is nonlinear in density. A uniform tumor uses survival at the mean, whereas a heterogeneous tumor’s survival is the weighted average of survival at each density. These quantities need not be equal.
What does increasing $h$ change, and what does it leave unchanged at $A=K_A$?
Solution: Increasing $h$ makes the transition around $K_A$ steeper. At $A=K_A$, the numerator and denominator are equal for any $h>0$, so $r=0.5$ remains unchanged.