Antigen Heterogeneity and T-Cell-Mediated Escape

Why Antigen Heterogeneity Changes the Response

A tumor can contain cells that express a target antigen and cells that do not. If tumor-reactive T cells recognize and kill antigen-positive cells more effectively, treatment can reduce the positive population while leaving antigen-negative cells relatively less affected. The tumor may then become enriched for antigen-negative cells even if their absolute number initially changes little. This is selection for antigen-loss escape: a change in population composition caused by differential fitness under treatment, not necessarily a treatment-induced change in an individual cell.

A useful first model separates the tumor into two populations. It treats antigen status as a state that determines susceptibility to T-cell killing, while keeping other processes as simple as the question allows. The distinction between population size and population frequency is central: a rising fraction of antigen-negative cells does not by itself mean that those cells are growing in absolute number.

Definition: Let \(N_+(t)\) and \(N_-(t)\) denote the abundances, in cells, of antigen-positive and antigen-negative tumor cells at time \(t\). The total tumor-cell abundance is \(N(t)=N_+(t)+N_-(t)\), and the antigen-negative fraction is \(f_-(t)=N_-(t)/N(t)\). Antigen-loss escape is an increase in the relative representation of antigen-negative cells during treatment because their net growth rate exceeds that of antigen-positive cells.

A Two-Population Model

Consider a treatment interval over which the effective T-cell exposure is specified as a function of time. Write \(q(t)\) for a dimensionless measure of effective T-cell-mediated killing, with \(0\leq q(t)\leq 1\). Its precise form depends on how exposure and target saturation are represented; the earlier material on effector-to-target ratios and saturable tumor killing provides one way to specify that exposure. Here the focus is the difference in susceptibility between the two tumor populations.

Let \(r_+\) and \(r_-\) be the untreated per-capita growth rates, in day\(^{-1}\), and let \(k_+\) and \(k_-\) be their killing-rate coefficients, also in day\(^{-1}\). A simple model is

\frac{dN_+}{dt}=\left[r_+-c(t)-k_+q(t)\right]N_+,\qquad \frac{dN_-}{dt}=\left[r_--c(t)-k_-q(t)\right]N_-.

The term \(c(t)\), in day\(^{-1}\), represents a shared per-capita growth constraint, such as a density-dependent limitation that acts equally on both populations. Antigen-positive cells are more susceptible when \(k_+>k_-\). The antigen-negative population need not be completely resistant: a positive \(k_-\) allows partial T-cell-mediated killing of cells lacking the modeled antigen, for example through other recognition mechanisms represented in aggregate.

This model assumes that each population is internally well mixed, that the rates adequately summarize the modeled time interval, and that antigen status does not change during that interval. It also assumes no population-specific effects hidden inside the shared term \(c(t)\). Those assumptions are modeling choices, not facts guaranteed by the data. If antigen loss or restoration occurs during treatment, additional transition terms are needed, as discussed below.

Interpretation: For population \(i\), the instantaneous net per-capita growth rate is \(g_i(t)=r_i-c(t)-k_iq(t)\). A negative \(g_i(t)\) means that population is declining at that time; a positive value means it is increasing. The difference \(g_-(t)-g_+(t)\), rather than either rate alone, determines the instantaneous direction of compositional change.

Deriving the Selection Rate

Define the odds of antigen-negative cells relative to antigen-positive cells as \(O(t)=N_-(t)/N_+(t)\), when both populations are present. Differentiating the logarithm of these odds gives

\frac{d}{dt}\log O(t) =\frac{1}{N_-}\frac{dN_-}{dt} -\frac{1}{N_+}\frac{dN_+}{dt} =g_-(t)-g_+(t).

Substituting the model rates cancels the shared constraint \(c(t)\):

\frac{d}{dt}\log O(t) =(r_--r_+)+(k_+-k_-)q(t).

This is the selection rate for the antigen-negative population on the log-odds scale. It can be positive even when both populations are declining: if antigen-positive cells decline faster, antigen-negative cells become a larger share of the remaining tumor. If growth rates are equal and \(k_+>k_-\), effective T-cell killing makes the selection rate positive whenever \(q(t)>0\).

Because \(f_-=O/(1+O)\), the fraction obeys

\frac{df_-}{dt}=f_-(1-f_-)\left[g_-(t)-g_+(t)\right].

When the selection rate \(s=g_--g_+\) is constant, integration yields \(O(t)=O(0)e^{st}\), and therefore

f_-(t)=\frac{f_-(0)e^{st}}{1-f_-(0)+f_-(0)e^{st}}.

For time-varying exposure, replace \(st\) by the cumulative selection \(\int_0^t [g_-(u)-g_+(u)]\,du\). This derivation also clarifies an important limitation: changing the frequency does not establish that antigen-negative cells were created during treatment. Pre-existing antigen-negative cells can become enriched through selection alone.

Worked Example: Predicting Enrichment During a Treatment Interval

Suppose a hypothetical tumor begins with \(9.00\times10^7\) antigen-positive cells and \(1.00\times10^7\) antigen-negative cells. Their untreated growth rates are both \(0.0300\) day\(^{-1}\). During a 14-day interval, assume constant effective T-cell exposure and killing rates of \(k_+=0.0800\) day\(^{-1}\) and \(k_-=0.0100\) day\(^{-1}\), with \(q=1\) and no shared density term. Calculate the populations and antigen-negative fraction at day 14.

The net rates are \(g_+=0.0300-0.0800=-0.0500\) day\(^{-1}\) and \(g_-=0.0300-0.0100=0.0200\) day\(^{-1}\). With constant net rates, \(N_i(t)=N_i(0)e^{g_it}\). Thus

N_+(14)=9.00\times10^7e^{-0.0500(14)} =9.00\times10^7e^{-0.700} \approx4.47\times10^7\ \text{cells},

N_-(14)=1.00\times10^7e^{0.0200(14)} =1.00\times10^7e^{0.280} \approx1.32\times10^7\ \text{cells}.

The selection rate is \(s=g_--g_+=0.0200-(-0.0500)=0.0700\) day\(^{-1}\). The day-14 antigen-negative fraction is

f_-(14)=\frac{1.32\times10^7}{4.47\times10^7+1.32\times10^7} \approx0.228.

The antigen-negative fraction therefore rises from \(0.100\) to about \(0.228\). In this hypothetical calculation, the negative population also increases in absolute abundance, while the positive population declines. The fraction change is consistent with the odds multiplier \(e^{0.0700(14)}\approx2.66\); the initial odds \(0.100/0.900=0.111\) become approximately \(0.296\).

Estimating Differential Susceptibility from Counts

If counts are available at the start and end of an interval, a constant-rate model gives an estimate of each population's net growth rate:

\widehat{g}_i=\frac{\log[N_i(t_1)/N_i(t_0)]}{t_1-t_0}.

These are net rates during the observed conditions. To attribute a difference specifically to T-cell killing, one must also account for differences in untreated growth and other treatment-associated effects. Counts alone identify net change; they do not automatically identify its mechanism.

Worked Example: Separating Net Change from Estimated Killing

In a hypothetical seven-day assay with a fixed T-cell exposure represented by \(q=1\), each population begins at \(2.00\times10^6\) cells. Antigen-positive cells end at \(1.00\times10^6\), while antigen-negative cells end at \(1.80\times10^6\). Separate untreated measurements suggest growth rates of \(r_+=0.0400\) day\(^{-1}\) and \(r_-=0.0350\) day\(^{-1}\). Estimate the net rates, the implied killing coefficients, and the selection rate, assuming no density term.

For the antigen-positive cells,

\widehat{g}_+=\frac{\log(1.00/2.00)}{7} =\frac{\log(0.500)}{7} \approx-0.0990\ \text{day}^{-1}.

For the antigen-negative cells,

\widehat{g}_-=\frac{\log(1.80/2.00)}{7} =\frac{\log(0.900)}{7} \approx-0.0151\ \text{day}^{-1}.

From \(g_i=r_i-k_iq\) and \(q=1\), the implied coefficients are \(k_+=r_+-\widehat{g}_+=0.0400-(-0.0990)=0.139\) day\(^{-1}\) and \(k_-=r_--\widehat{g}_-=0.0350-(-0.0151)=0.0501\) day\(^{-1}\). Their difference is about \(0.0890\) day\(^{-1}\). Independently, the observed selection rate is \(\widehat{g}_--\widehat{g}_+=-0.0151-(-0.0990)=0.0840\) day\(^{-1}\). These values differ because the untreated growth rates also differ: the model predicts \(s=(r_--r_+)+(k_+-k_-)q=-0.0050+0.0890=0.0840\) day\(^{-1}\).

Both populations declined in this assay, but the negative population declined more slowly. Its relative enrichment is not evidence that its absolute abundance increased.

How Quickly Can a Rare Escape Population Become Common?

For a constant positive selection rate \(s\), the time required for the negative fraction to reach a specified value can be calculated from the odds equation. If the target fraction is one half, its odds are one, so

t_{50}=\frac{\log[(1-f_-(0))/f_-(0)]}{s}.

Worked Example: Time to Reach Equal Population Fractions

Suppose a hypothetical tumor has an initial antigen-negative fraction of \(0.0100\) and a constant selection rate of \(0.120\) day\(^{-1}\). Under the two-population model, how long does it take for the negative fraction to reach \(0.500\)?

The initial odds are \(0.0100/0.990=0.0101\), so the log-odds change required to reach equal fractions is \(\log(1/0.0101)=\log(99)=4.595\). Therefore

t_{50}=\frac{\log(99)}{0.120} =\frac{4.595}{0.120} \approx38.3\ \text{days}.

This is a model-based time to equal fractions, not a prediction that the total tumor will reach a particular size or that clinical progression will occur at that time. The result depends on the selection rate remaining constant and on both populations persisting in the modeled conditions.

Assumptions, Diagnostics, and Model Extensions

The two-population model is most useful when antigen status can be classified reliably and when counts or fractions are measured over time. Before interpreting its selection rate, check whether the data support the model's key assumptions.

Check absolute counts as well as fractions. A rising antigen-negative fraction can result from negative-cell growth, positive-cell loss, or both. Report which populations increase or decline in absolute abundance when those data are available.
Check whether a constant rate is plausible. If log abundance versus time is strongly curved, a single exponential rate may not summarize the interval. Changing T-cell exposure, density dependence, or changing population susceptibility can produce such curvature.
Check antigen classification and sampling. Misclassification, assay detection limits, and sampling of only part of a heterogeneous lesion can distort estimated fractions. A small observed negative fraction may also be difficult to distinguish from zero.
Check for unequal growth unrelated to killing. If antigen-negative cells have an intrinsic growth advantage, then \(r_--r_+\) contributes to selection even without treatment. Untreated data or an appropriate comparison are needed to separate this contribution from differential killing.

If cells can lose or regain antigen, a model can include transition rates. Let \(\mu\) be the per-capita rate of transition from positive to negative and \(\nu\) the rate from negative to positive, both in day\(^{-1}\). The equations then include source and loss terms:

\frac{dN_+}{dt}=g_+N_+-\mu N_++\nu N_-, \qquad \frac{dN_-}{dt}=g_-N_-+\mu N_+-\nu N_-.

With these transitions, a negative population can increase because antigen-positive cells convert to the negative state, even when its net growth rate is not greater. The simple log-odds selection formula applies to the no-transition model; transition terms must be accounted for when fitting data in which antigen-state conversion is plausible.

Choosing a Model That Matches the Question

A two-population model is a transparent starting point, but it is not the only representation. The choice should follow the biological question and the resolution of the measurements.

Approach	Useful when	Main limitation
Two populations with distinct killing rates	Antigen status is measured as positive or negative, and the aim is to quantify differential selection.	Hides variation within each category and may not capture antigen-state transitions.
One aggregate tumor population with an average killing rate	Only total tumor burden is measured and composition is not central to the question.	Cannot predict antigen-negative enrichment or distinguish clone-specific responses.
Antigen-state transition model	Data or biological knowledge support loss or restoration of antigen during the interval.	Transition rates can be difficult to identify separately from differential growth and killing.

The aggregate approach may fit total burden while missing a clinically important change in composition. Conversely, adding transition rates without informative longitudinal measurements can make parameters difficult to distinguish. More detail is useful only when the data and question can support it.

Common Pitfall

Common mistake: “The antigen-negative fraction increased, so treatment caused antigen loss and the negative cells grew.”

Correction: A fraction increase establishes a compositional change, not its cause. Under the no-transition model, pre-existing negative cells become enriched whenever \(g_->g_+\), even if both populations decline. To claim that new antigen loss occurred, the analysis needs evidence or a model that separately represents antigen-state transitions.

Check Your Understanding

For each question, work through the calculation or interpretation before reading the solution.

A tumor has \(N_+=80\) million cells and \(N_-=20\) million cells. What is \(f_-\)?
Solution: \(f_-=20/(80+20)=0.200\), so 20.0% of the cells are antigen-negative.
Suppose \(g_+=-0.040\) day\(^{-1}\) and \(g_-=-0.010\) day\(^{-1}\). What is the selection rate, and what does its sign mean?
Solution: \(s=g_--g_+=-0.010-(-0.040)=0.030\) day\(^{-1}\). It is positive, so antigen-negative cells increase in relative representation, although both populations decline.
If \(r_+=r_-\), \(k_+=0.060\) day\(^{-1}\), \(k_-=0.020\) day\(^{-1}\), and \(q=0.5\), calculate \(s\).
Solution: \(s=(r_--r_+)+(k_+-k_-)q=0+(0.060-0.020)(0.5)=0.020\) day\(^{-1}\).
Can an antigen-negative fraction rise while the absolute number of antigen-negative cells falls? Explain.
Solution: Yes. If both populations decline but \(N_+\) declines faster than \(N_-\), the denominator \(N_++N_-\) can shrink faster than \(N_-\), increasing \(f_-\) even as \(N_-\) falls.
What additional process must be represented if positive cells can lose antigen during treatment?
Solution: Include a positive-to-negative transition, such as the term \(\mu N_+\) added to the negative-population equation and subtracted from the positive-population equation. Then negative-cell enrichment can reflect both selection and conversion.