Linear and Emax Cytotoxic Kill Functions

Why the Form of a Kill Function Matters

Cytotoxic therapy is represented in a tumor model by a drug-dependent rate of cell loss. The function chosen for that rate determines how predicted killing changes as concentration increases. A linear function continues to rise without bound; an Emax function rises toward a finite maximum. Those differences can be consequential when predictions extend beyond the concentrations that informed the model.

The earlier material on tumor growth inhibition through cytostasis distinguishes reduced growth from cell loss. Here the focus is specifically the concentration-dependent cytotoxic kill rate. The functions below describe a rate supplied at a given concentration; they do not specify how concentration changes over time or combine killing with tumor growth.

Definitions, Units, and Assumptions

Let \(C\) be the drug concentration supplied to the kill function, with \(C\geq 0\). Denote the instantaneous per-capita cytotoxic kill rate by \(K(C)\). A per-capita rate is the rate of loss relative to the viable tumor-cell burden, not a percentage of cells killed in a fixed interval.

Definition: The linear cytotoxic kill function is \(K_{\mathrm{lin}}(C)=k_{\mathrm{lin}}C\), where \(k_{\mathrm{lin}}\geq 0\) is the slope. The Emax cytotoxic kill function is \(K_{\mathrm{Emax}}(C)=E_{\max}C/(EC_{50}+C)\), where \(E_{\max}>0\) is the maximum kill rate and \(EC_{50}>0\) is the concentration at which the rate is half of \(E_{\max}\). Both functions have units of inverse time. If \(C\) is measured in mg/L and time in days, \(k_{\mathrm{lin}}\) has units L/(mg·day), while \(E_{\max}\) has units day\(^{-1}\) and \(EC_{50}\) has units mg/L.

The parameter constraints matter. Under these constraints, both functions are defined at \(C=0\) and give \(K(0)=0\). The Emax denominator is positive for every \(C\geq0\), including zero. If \(E_{\max}=0\) or \(EC_{50}=0\), the stated Emax parameterization would not satisfy the definition above.

These functions assume a direct, instantaneous relationship between the supplied concentration and the kill rate, with no baseline killing term. They are phenomenological descriptions: choosing one does not by itself establish a biological mechanism. The rate parameters should be interpreted in the time unit used by the tumor model, and concentrations should use the same unit as \(EC_{50}\).

How the Functions Behave

The linear function has constant slope \(k_{\mathrm{lin}}\). Doubling a positive concentration doubles its predicted rate, and the rate has no upper bound as \(C\) increases. If \(k_{\mathrm{lin}}=0\), it predicts zero drug-related killing at every concentration.

The Emax rate increases with concentration but approaches \(E_{\max}\) rather than exceeding it. At \(C=EC_{50}\), the numerator and denominator give:

K_{\mathrm{Emax}}(EC_{50}) =E_{\max}\frac{EC_{50}}{EC_{50}+EC_{50}} =\frac{E_{\max}}{2}.

The low-concentration behavior helps explain why these functions can appear similar over a restricted range. Dividing the Emax rate by \(C\), for \(C>0\), gives:

\frac{K_{\mathrm{Emax}}(C)}{C} =\frac{E_{\max}}{EC_{50}+C} \longrightarrow \frac{E_{\max}}{EC_{50}} \quad\text{as }C\longrightarrow 0^+.

Thus the Emax function has a low-concentration slope of \(E_{\max}/EC_{50}\). A linear function with slope \(k_{\mathrm{lin}}=E_{\max}/EC_{50}\) matches that initial behavior, but it does not match the saturating behavior at higher concentrations.

Worked Example: Comparing Rates Below and Above the Crossover

Suppose, hypothetically, that \(C\) is measured in mg/L and time in days, with \(E_{\max}=0.80\) day\(^{-1}\), \(EC_{50}=2.0\) mg/L, and \(k_{\mathrm{lin}}=0.12\) L/(mg·day). Compare the two rates at several concentrations.

At \(C=1.0\) mg/L:

K_{\mathrm{lin}}=0.12(1.0)=0.12\ \text{day}^{-1}, \qquad K_{\mathrm{Emax}}=0.80\frac{1.0}{2.0+1.0} =0.2667\ \text{day}^{-1}.

At \(C=2.0\) mg/L, the rates are \(0.12(2.0)=0.24\) day\(^{-1}\) and \(0.80(2.0)/(2.0+2.0)=0.40\) day\(^{-1}\), respectively. The Emax rate is larger at both concentrations.

The crossover calculation below gives \(C^*=0.80/0.12-2.0=4.667\) mg/L. At \(C=8.0\) mg/L, the linear rate is \(0.12(8.0)=0.96\) day\(^{-1}\), while the Emax rate is \(0.80(8.0)/(2.0+8.0)=0.64\) day\(^{-1}\). The ranking has reversed above the crossover.

When Do the Curves Intersect?

A positive-concentration intersection can be derived by setting the functions equal. First suppose \(k_{\mathrm{lin}}>0\). For \(C>0\), equating the rates and dividing by \(C\) gives:

k_{\mathrm{lin}}C =E_{\max}\frac{C}{EC_{50}+C} \quad\Longleftrightarrow\quad k_{\mathrm{lin}} =\frac{E_{\max}}{EC_{50}+C}.

Solving for \(C\) gives the only possible positive-concentration intersection:

C^*=\frac{E_{\max}}{k_{\mathrm{lin}}}-EC_{50}.

This is a positive concentration precisely when \(E_{\max}>k_{\mathrm{lin}}EC_{50}\). In that case, the Emax rate is larger below \(C^*\) and the linear rate is larger above \(C^*\). To see the ordering, compare their difference for \(C>0\):

K_{\mathrm{lin}}(C)-K_{\mathrm{Emax}}(C) =C\left(k_{\mathrm{lin}}-\frac{E_{\max}}{EC_{50}+C}\right).

The expression in parentheses is negative below \(C^*\), zero at \(C^*\), and positive above it. If \(k_{\mathrm{lin}}>0\) but \(E_{\max}\leq k_{\mathrm{lin}}EC_{50}\), there is no positive intersection: \(E_{\max}/(EC_{50}+C)<E_{\max}/EC_{50}\leq k_{\mathrm{lin}}\) for every \(C>0\), so the linear rate is greater. If \(k_{\mathrm{lin}}=0\), \(C^*\) is not defined by the formula; the linear rate is zero and the Emax rate is positive at every \(C>0\), so again there is no positive intersection.

Worked Example: Checking Whether an Intersection Is Positive

Consider hypothetical parameters \(E_{\max}=0.40\) day\(^{-1}\), \(EC_{50}=10\) mg/L, and \(k_{\mathrm{lin}}=0.05\) L/(mg·day). Here \(E_{\max}/k_{\mathrm{lin}}=8.0\) mg/L, so the formal solution is:

C^*=8.0-10=-2.0\ \text{mg/L}.

A negative concentration is outside the allowed domain. It does not represent a physically meaningful crossover. Equivalently, \(k_{\mathrm{lin}}EC_{50}=0.50\) day\(^{-1}\), which exceeds \(E_{\max}=0.40\) day\(^{-1}\). Therefore, at every positive concentration, the linear rate is greater than the Emax rate. For instance, at \(C=2.0\) mg/L the rates are \(0.10\) day\(^{-1}\) and \(0.40(2.0)/(10+2.0)=0.06667\) day\(^{-1}\), respectively.

Worked Example: Matching the Low-Concentration Slope

Suppose hypothetical Emax parameters are \(E_{\max}=0.60\) day\(^{-1}\) and \(EC_{50}=3.0\) mg/L. Its low-concentration slope is \(0.60/3.0=0.20\) L/(mg·day). Compare it with a linear function having \(k_{\mathrm{lin}}=0.20\) L/(mg·day).

At \(C=1.0\) mg/L, the linear rate is \(0.20\) day\(^{-1}\), while the Emax rate is \(0.60(1.0)/(3.0+1.0)=0.15\) day\(^{-1}\). At \(C=10\) mg/L, the linear rate is \(2.0\) day\(^{-1}\), whereas the Emax rate is \(0.60(10)/(3.0+10)=0.4615\) day\(^{-1}\).

The parameters give \(E_{\max}/k_{\mathrm{lin}}=3.0\) mg/L, so \(C^*=3.0-3.0=0\) mg/L. There is no positive-concentration intersection: for every \(C>0\), the linear rate exceeds the Emax rate. Matching the initial slope therefore does not make the two functions interchangeable across a broad range.

Diagnostics, Plausibility, and Model Choice

A kill-rate function should be checked over the full concentration range where it will be used, not only near the concentrations used to estimate its parameters. Useful checks include evaluating the rate at zero, at \(EC_{50}\) for an Emax model, at the minimum and maximum supported concentrations, and at any planned extrapolation points. Plotting both the rate and the implied difference between candidate functions can reveal a crossover or an unexpectedly steep rise.

A rate such as \(2.0\) day\(^{-1}\) is not automatically impossible: it is a rate, not a claim that more than 100% of the cells are killed in one day. Under a constant per-capita rate, the fraction lost over an interval depends on the interval length and the rate model, so a rate must not be reported as a percentage without specifying that interpretation. Nonetheless, a linear function that predicts very large rates at high concentrations may imply a biologically implausible timescale or exceed a plausible bound established for the system. That is a warning about the function's extrapolation, not a universal numerical cutoff.

The Emax function avoids unbounded growth in the predicted rate, but it is not automatically plausible. Its \(E_{\max}\) may be too large for the biological system, and the estimated \(EC_{50}\) and \(E_{\max}\) may be poorly determined if the available concentrations do not show enough curvature or approach saturation. Conversely, a linear function may be a reasonable local approximation over a narrow range even if it is unsuitable for distant extrapolation.

Question	Linear function	Emax function
Behavior as concentration approaches zero	Rate approaches zero with slope \(k_{\mathrm{lin}}\).	Rate approaches zero with slope \(E_{\max}/EC_{50}\).
Behavior at high concentration	Rate increases without bound if \(k_{\mathrm{lin}}>0\).	Rate approaches \(E_{\max}\) from below.
Useful when	The supported range is limited and proportionality is a plausible approximation.	A bounded maximum rate is biologically plausible and supported by the concentration range.
Main extrapolation concern	Predicted rates can become arbitrarily large.	The assumed plateau may be unsupported or incorrectly estimated.

The concentration range and biological context guide the choice. If observations cover only a narrow low-concentration range, both functions may fit similarly because the Emax function is approximately linear there. A good fit in that range alone does not demonstrate that the function's high-concentration behavior is correct. If data cover a wider range, examine whether the observed pattern supports proportional increase or a plateau, while checking parameter estimates, units, and predictions at the range boundaries. Neither function should be selected just because its form is familiar.

Common Mistakes

Calling the Emax parameter an observed rate at every concentration. \(E_{\max}\) is the limiting rate as concentration grows; at finite positive \(C\), the Emax rate is below it.
Using the crossover formula without checking its conditions. The formula \(C^*=E_{\max}/k_{\mathrm{lin}}-EC_{50}\) requires \(k_{\mathrm{lin}}>0\), and a positive intersection additionally requires \(E_{\max}>k_{\mathrm{lin}}EC_{50}\). When \(k_{\mathrm{lin}}=0\), compare the functions directly.
Treating a negative formal crossover as a physical concentration. A negative value signals that no positive-concentration intersection exists; it is not a prediction at an allowable concentration.
Reading a rate as a percentage killed per day. A rate has units of inverse time. Converting it to a fraction over an interval requires a specified time interval and rate interpretation.
Extrapolating a linear function indefinitely without checking plausibility. Its unbounded growth is a mathematical property, not evidence that killing continues to increase proportionally at arbitrarily high concentration.

Key takeaway: The linear function predicts proportional, unbounded increase; the Emax function predicts increasing but bounded killing. Their low-concentration slopes can match, so comparison requires attention to the supported concentration range, parameter units, and high-concentration behavior.

Check Your Understanding

Try each problem, then compare your work with the solutions below.

For \(E_{\max}=0.50\) day\(^{-1}\), \(EC_{50}=4.0\) mg/L, and \(C=4.0\) mg/L, calculate the Emax kill rate.
A linear function has \(k_{\mathrm{lin}}=0.08\) L/(mg·day). What rate does it predict at \(C=5.0\) mg/L?
For \(E_{\max}=0.90\) day\(^{-1}\), \(EC_{50}=2.0\) mg/L, and \(k_{\mathrm{lin}}=0.10\) L/(mg·day), find the crossover and state which function is larger below and above it.
For \(E_{\max}=0.40\) day\(^{-1}\), \(EC_{50}=10\) mg/L, and \(k_{\mathrm{lin}}=0.05\) L/(mg·day), explain whether a positive-concentration crossover exists.
A report describes a predicted kill rate of \(1.4\) day\(^{-1}\) as “140% of cells killed in one day.” Identify the error.

Solutions. (1) At \(C=EC_{50}\), the Emax rate is half the maximum: \(0.50/2=0.250\) day\(^{-1}\). (2) \(K_{\mathrm{lin}}=0.08(5.0)=0.400\) day\(^{-1}\). (3) \(C^*=0.90/0.10-2.0=7.0\) mg/L. The Emax rate is larger for \(0<C<7.0\) mg/L, the rates are equal at 7.0 mg/L, and the linear rate is larger above it. (4) The formal value is \(0.40/0.05-10= -2.0\) mg/L, so there is no positive intersection. Since \(E_{\max}<k_{\mathrm{lin}}EC_{50}\), the linear rate is larger at every \(C>0\). (5) A rate in day\(^{-1}\) is not a percentage of cells killed in a specified day. The statement confuses a continuous-time rate with a finite-interval fraction; that fraction requires a defined rate interpretation and interval.