CAR Signaling Strength, Costimulation, and Tonic Activation

From CAR Design to a Quantitative Signal

CAR design affects more than whether a T cell recognizes a target. The strength and persistence of intracellular signaling influence how much activation follows recognition, while signaling that occurs without target engagement can raise baseline activation. These effects create a design trade-off: stronger signaling may improve activation under a given stimulus, but it can also increase tonic activation or accelerate loss of functional competence during sustained drive. A quantitative model makes those competing effects explicit without assuming that any one costimulatory design is universally superior.

The model below is phenomenological. It represents the net effect of receptor-proximal signaling and costimulatory architecture with a small number of parameters. Those parameters should be estimated or constrained using measurements for the CAR construct and experimental context being modeled; they are not fixed biological constants.

Definition: Let $u(t)$ be a dimensionless effective antigen-engagement input, and let $\tau$ be a dimensionless input representing antigen-independent tonic signaling. Define the total input as $x(t)=u(t)+\tau$. The dimensionless signal is $S(t)=g x(t)$, where $g>0$ is an effective signal gain. For a threshold parameter $\theta>0$ and Hill coefficient $h>0$, define the fractional activation response as $$ q(t)=\frac{S(t)^h}{\theta^h+S(t)^h}. $$ Here $q$ is a modeled activation response between 0 and 1, not necessarily the probability that an individual cell is activated. The signal value $S=\theta$ gives $q=0.5$.

The Hill coefficient $h$ controls the steepness of the response around $\theta$. A larger $h$ produces a sharper transition, while a smaller $h$ produces a more gradual transition. The parameter $\theta$ is the half-maximal signal level, not necessarily a literal biological switch. Because $S$ is written in normalized signal units, the values of $g$ and $\theta$ must be interpreted in those same units.

Costimulatory design can affect the effective gain, threshold, and other parts of the model. For example, one construct might be represented by a larger $g$, a lower $\theta$, a different tonic input $\tau$, or a combination of these. These are possible model representations, not universal directions of effect: a particular costimulatory domain need not increase every parameter associated with activation, and parameters may need to change together to fit data.

Tonic Activation and the Design Trade-Off

In the absence of antigen engagement, $u=0$, so the tonic signal is $S_0=g\tau$. Its corresponding baseline activation response is $$ q_0=\frac{(g\tau)^h}{\theta^h+(g\tau)^h}. $$ A CAR with a high target-driven response can therefore also have a non-negligible baseline response if its tonic input or gain is large relative to its activation threshold. Tonic activation is not the same as antigen-driven activation: the model separates them through $\tau$ and $u$, even though both contribute to the same total input.

The half-maximal condition follows directly from the response equation. Setting $q=0.5$ gives $S^h/(\theta^h+S^h)=0.5$, hence $2S^h=\theta^h+S^h$, and therefore $S=\theta$ for nonnegative signal. More generally, for a desired response $q$ strictly between 0 and 1, the signal required by this model is

S=\theta\left(\frac{q}{1-q}\right)^{1/h}.

This relationship helps compare designs at a specified modeled activation level. It does not establish that the same $q$ corresponds to the same cytokine release, proliferation, or killing across constructs; those outcomes require additional observation models or data.

Worked Example: Comparing Target-Driven and Tonic Responses

Consider two hypothetical CAR designs with $h=2$. Design A has $g=1.0$, $\theta=0.60$, and $\tau=0.10$. Design B has $g=1.4$, $\theta=0.50$, and $\tau=0.12$. Compare their modeled baseline responses and their responses when the effective antigen-engagement input is $u=0.50$.

Design A at baseline. With $u=0$, $S_0=g\tau=(1.0)(0.10)=0.10$. Thus $$ q_0=\frac{0.10^2}{0.60^2+0.10^2} =\frac{0.0100}{0.3600+0.0100} =0.0270. $$

Design B at baseline. Here $S_0=(1.4)(0.12)=0.168$, so $$ q_0=\frac{0.168^2}{0.50^2+0.168^2} =\frac{0.028224}{0.2500+0.028224} =0.1014. $$

With antigen engagement. For Design A, $S=(1.0)(0.50+0.10)=0.60$, giving $q=0.60^2/(0.60^2+0.60^2)=0.500$. For Design B, $S=(1.4)(0.50+0.12)=0.868$, giving $$ q=\frac{0.868^2}{0.50^2+0.868^2} =\frac{0.753424}{0.2500+0.753424} =0.7509. $$

Under these hypothetical parameter values, Design B has the higher modeled response with antigen engagement, but it also has the higher baseline response. The calculation illustrates a trade-off within the model; it does not show that Design B is clinically better or that its tonic response would necessarily cause harm.

Representing Functional Decline Under Sustained Drive

An instantaneous activation response alone cannot represent a gradual change in functional capacity. To add a minimal dynamic state, let $F(t)$ denote normalized functional competence, with $F=1$ representing the reference level and $F=0$ representing no competence in this model. Let $A(t)$ be a dimensionless activation drive between 0 and 1; one possible assumption is $A(t)=q(t)$. A recovery process returns $F$ toward 1 at rate $k_{\mathrm{rec}}$, while activation-dependent loss reduces it at rate $k_d A(t)^m F$. Here $k_{\mathrm{rec}}$ and $k_d$ have units of inverse time, and $m>0$ controls how nonlinearly the loss rate depends on drive.

Model assumption: The functional state follows $$ \frac{dF}{dt}=k_{\mathrm{rec}}(1-F)-k_d A(t)^mF, $$ where $k_{\mathrm{rec}}\geq 0$, $k_d\geq 0$, and $A(t)\geq 0$. This is a lumped description of recovery and activation-associated functional loss, not a detailed model of the cellular processes that produce either effect.

For a constant activation drive $A$, define $L=k_{\mathrm{rec}}+k_dA^m$. The differential equation becomes $dF/dt=k_{\mathrm{rec}}-LF$. When $L>0$, equilibrium requires $0=k_{\mathrm{rec}}-LF_\infty$, so $F_\infty=k_{\mathrm{rec}}/L$. Subtracting the equilibrium from the state equation gives $d(F-F_\infty)/dt=-L(F-F_\infty)$. Separating variables and applying $F(0)$ yields the solution below.

F(t)=F_\infty+\bigl(F(0)-F_\infty\bigr)e^{-Lt}, \qquad F_\infty=\frac{k_{\mathrm{rec}}}{k_{\mathrm{rec}}+k_dA^m}, \qquad L>0.

The condition $L>0$ is essential. If $L=0$, nonnegativity of the rates and $A$ implies $k_{\mathrm{rec}}=0$ and $k_dA^m=0$. The differential equation is then $dF/dt=0$, so $F(t)=F(0)$ for all $t$; every constant value of $F$ is an equilibrium. Neither $k_{\mathrm{rec}}/L$ nor the exponential solution with a uniquely specified $F_\infty$ applies in this case.

When $L>0$, the equilibrium is unique. If $k_{\mathrm{rec}}>0$, it lies in $(0,1]$; it lies strictly below 1 when $k_dA^m>0$. If recovery is zero and the activation-dependent loss is positive, then $F_\infty=0$. For initial $F(0)$ in $[0,1]$, the model keeps $F(t)$ in $[0,1]$: at $F=0$, the derivative is $k_{\mathrm{rec}}\geq0$, and at $F=1$, it is $-k_dA^m\leq0$. These boundary checks support the interpretation of $F$ as a normalized state.

Worked Example: Calculating Functional Decline During Constant Drive

Suppose a hypothetical CAR T-cell population has constant activation drive $A=0.80$, $k_{\mathrm{rec}}=0.10\ \mathrm{day}^{-1}$, $k_d=0.25\ \mathrm{day}^{-1}$, $m=2$, and $F(0)=1.00$. Calculate the modeled equilibrium and functional competence after 10 days.

First, $A^m=0.80^2=0.640$. The total rate is $$ L=k_{\mathrm{rec}}+k_dA^m =0.10+0.25(0.640) =0.260\ \mathrm{day}^{-1}. $$ Since $L>0$, the equilibrium formula applies: $$ F_\infty=\frac{0.10}{0.260}=0.3846. $$

At 10 days, the solution is $$ F(10)=0.3846+(1.00-0.3846)e^{-(0.260)(10)} =0.3846+(0.6154)e^{-2.60} \approx 0.4303. $$

The model predicts a decline from 1.00 toward 0.3846, reaching approximately 0.4303 after 10 days under the stated constant-drive assumptions. The time unit is days because both rate constants are in inverse days. This result is a prediction of the specified state equation, not a direct measurement of any particular exhaustion marker or clinical response.

How to Use and Check the Model

A simple way to connect instantaneous activation and functional competence is to assume that a modeled output is proportional to their product, for example $Y(t)=Y_{\max}q(t)F(t)$. This is an additional observation assumption: it says that activation supplies the immediate drive and $F$ scales the available functional capacity. It should be checked against data rather than treated as a biological identity. A model can instead include separate links from $q$ and $F$ to proliferation, cytokine production, or killing.

For a constant drive, the trajectory of $F$ is a useful diagnostic. If $L>0$, it should approach $F_\infty$ exponentially with time scale $1/L$. A simulated trajectory that moves away from the equilibrium, leaves $[0,1]$ when initialized inside it, or changes when rates are expressed consistently in different time units signals an implementation or parameter-unit error. For time-varying drive, integrate the differential equation over time. In general, replacing $A(t)$ with its time average is not equivalent, because the loss term depends on $A(t)^mF(t)$, and $F$ itself changes over time.

Data should distinguish target-driven response from baseline signaling if both $g$ and $\tau$ are to be estimated. Measurements at multiple input levels can help identify the response threshold and steepness; baseline measurements are particularly informative about tonic input. Likewise, estimating $k_{\mathrm{rec}}$, $k_d$, and $m$ from a single endpoint may be difficult because different parameter combinations can produce similar values of $F$. Time-course data under more than one activation condition are more informative. These are identifiability considerations, not guarantees that a particular experiment will uniquely determine every parameter.

Approach	Useful when	Main limitation
Binary threshold response	A coarse on/off approximation is adequate.	It cannot represent graded responses or compare subthreshold activation.
Hill activation response	Graded, saturating activation is needed with a small number of parameters.	Its fitted threshold and steepness depend on the chosen input scale and data range.
Static activation model only	The question concerns an immediate response at a fixed condition.	It cannot represent recovery or time-dependent decline.
Activation plus dynamic functional state	The question concerns sustained drive and changing capacity.	It adds parameters and compresses complex biology into a single state.

The Hill model is often preferable to a binary threshold when the data show graded activation, but its extra flexibility can be poorly supported by sparse measurements. Similarly, the dynamic $F$ model is useful when time course matters, but a static model may be more defensible if only a single short-term response is available. Model complexity should match the question and the evidence.

Common Mistakes and Limitations

Assuming that stronger signaling is always better. A larger gain or lower threshold can raise target-driven response, but may also raise baseline response or activation-associated loss. Compare both baseline and stimulated behavior.
Treating tonic signaling as antigen engagement. In this model, tonic input is present when $u=0$. It is a separate source of signal, even though it enters the same response function.
Calling $q$ a measured clinical outcome. $q$ is a normalized activation response. A link to expansion, cytokines, killing, or patient benefit needs its own model and evidence.
Using an equilibrium formula when $L=0$. If both rates contributing to $L$ are zero, the state is constant and the equilibrium is not uniquely selected. Check $L$ before dividing by it.
Interpreting functional decline as a complete exhaustion model. The single state $F$ cannot distinguish cell subpopulations, memory and effector transitions, or different biological mechanisms of dysfunction.

The central limitation is that these equations are compact representations of complex signaling. Costimulatory effects can depend on expression, cell context, and stimulation history; a single effective gain or tonic input cannot encode all of those details. The model is most useful for clarifying assumptions, organizing measurements, and testing whether a proposed design change creates a plausible balance between stimulus response and baseline activation.

Check Your Understanding

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

For $h=2$, $\theta=0.40$, and $S=0.40$, what is $q$? What does the result mean in this model?
A design has $g=1.2$, $\tau=0.05$, $h=2$, and $\theta=0.50$. Calculate its tonic baseline response.
With constant $A=0.50$, $k_{\mathrm{rec}}=0.20\ \mathrm{day}^{-1}$, $k_d=0.40\ \mathrm{day}^{-1}$, and $m=2$, calculate $L$ and $F_\infty$.
If $k_{\mathrm{rec}}=0$ and $k_dA^m=0$, can the equilibrium be calculated as $k_{\mathrm{rec}}/L$? State the state trajectory.

Solutions. (1) $q=0.40^2/(0.40^2+0.40^2)=0.500$, the half-maximal modeled activation response. (2) $S_0=g\tau=(1.2)(0.05)=0.060$, so $q_0=0.060^2/(0.50^2+0.060^2)=0.00360/0.2536=0.0142$. (3) $A^m=0.25$, so $L=0.20+0.40(0.25)=0.300\ \mathrm{day}^{-1}$, and $F_\infty=0.20/0.300=0.667$. (4) No. Here $L=0$, and the equation gives $dF/dt=0$, so $F(t)=F(0)$ for all $t$; every constant value is an equilibrium.

CAR Signaling Strength, Costimulation, and Tonic Activation

What You'll Learn