From Drug Exposure to a Time-Varying Kill Rate
A tumor-cell kill function describes how strongly a drug can kill cells at a given exposure. In an actual treatment profile, exposure changes over time, so the corresponding kill rate changes too. The modeling task here is to take a specified concentration profile and map each concentration to an instantaneous, time-dependent rate. This link lets a pharmacometric model represent a changing drug effect without assuming that the drug concentration is constant.
The input is an exposure profile, such as concentration in plasma or at a specified effect site. The output is a kill rate with units of inverse time. This tutorial assumes the profile is already specified; it does not derive drug concentrations from a pharmacokinetic model. It also focuses on the exposure-to-effect link, not on the resulting tumor-size trajectory.
Two Exposure–Effect Relationships
The linear model assumes the drug-mediated kill rate increases in direct proportion to concentration:
Here \(\beta\) is the proportionality constant. If concentration is measured in mg/L and the rate in day\(^{-1}\), then \(\beta\) has units of L/(mg·day). In this form, \(k(0)=0\), and the model has no upper limit as concentration increases. It can be a useful approximation over a restricted range of concentrations, but its unbounded increase may be implausible when extrapolated far beyond that range.
The Emax model represents a saturating relationship:
Here \(k_{\max}\) is the maximum drug-mediated kill rate, and \(EC_{50}\) is the concentration at which the model predicts half of \(k_{\max}\). Both \(C\) and \(EC_{50}\) must use the same concentration units; \(k_{\max}\) has the units of a rate. At zero concentration the rate is zero, and at \(C=EC_{50}\), \(k(C)=k_{\max}/2\). As concentration increases without bound, \(k(C)\) approaches \(k_{\max}\).
These expressions give an instantaneous rate, not a percentage killed at each moment. Under a first-order killing assumption, the rate acts on the cells currently present: over a short interval \(dt\), the fraction surviving drug-mediated killing is approximately \(1-k(t)dt\). This interpretation assumes that the kill rate is a per-capita hazard and that the modeled rate is the drug-mediated component.
Assumptions and a Useful Derivation
Applying either relationship requires more than inserting numbers into a formula. The concentration must represent the exposure relevant to the effect model, the concentration units must match the parameter units, and the time units must be consistent with the rate. The instantaneous mapping also assumes no effect-site delay, turnover delay, or other memory of earlier concentrations. If such a delay matters, the same-time relationship may not describe the observed response adequately.
A useful summary of changing killing is the cumulative drug-mediated kill hazard over an interval. It is the integral of the instantaneous rate, not the rate evaluated at an average concentration:
To see why this integral is useful, suppose a cell has a constant per-capita kill rate \(k\) over a short interval of length \(dt\). Its probability of surviving drug-mediated killing over that interval is approximately \(1-kdt\). For many small intervals with potentially changing rates, multiplying those survival factors gives a product of terms \(1-k(t_i)dt\). Taking logarithms and using \(\log(1-z)\approx-z\) for small \(z\), the log survival approaches \(-\int k(C(t))dt\). Thus, under the first-order killing assumption, drug-only survival over the whole interval is \(\exp[-H(t_0,t_1)]\). This calculation does not include cell growth or other causes of cell loss.
For a piecewise-constant concentration profile, \(H\) is the sum of each rate multiplied by its interval duration. For a continuously varying or sparsely observed profile, the integral may require numerical approximation. A common approximation is the trapezoidal rule applied to the rate values, provided the interpolation and sampling assumptions are reasonable.
Worked Example: A Piecewise-Constant Profile With a Linear Kill Model
Consider a hypothetical concentration profile measured in mg/L: \(C=2\) from day 0 to day 2, \(C=5\) from day 2 to day 5, and \(C=1\) from day 5 to day 6. Suppose the linear model has \(\beta=0.080\) L/(mg·day). Calculate the kill rate in each interval and the cumulative drug-mediated kill hazard over six days.
Check the assumptions and units. Treat the stated concentrations as constant within each interval, and assume an instantaneous linear relationship. Multiplying mg/L by L/(mg·day) gives day\(^{-1}\), as required.
Calculate the interval rates. The rates are \(0.080(2)=0.160\) day\(^{-1}\), \(0.080(5)=0.400\) day\(^{-1}\), and \(0.080(1)=0.0800\) day\(^{-1}\), respectively.
The cumulative hazard is \(1.600\), a dimensionless quantity because rate multiplied by time has no units. Under the drug-only, first-order killing assumption, the surviving fraction is \(\exp(-1.600)=0.2019\), so the fraction killed by this modeled drug effect is \(1-0.2019=0.7981\), or about 79.8%. This is not a prediction of net tumor change because it excludes cell growth and any other modeled processes.
Nonlinearity Changes How Exposure Summaries Behave
With a linear relationship, the average rate over a fixed interval equals \(\beta\) times the average concentration over that interval. Under Emax, that generally does not hold: the rate must be calculated at each concentration and then averaged or integrated. High concentrations contribute less additional rate as the model approaches its maximum, while lower concentrations may lie on the steeper part of the curve.
For example, averaging concentrations first and then evaluating the Emax function gives \(k(\bar{C})\), which need not equal the time-average rate \(\overline{k(C)}\). A concentration summary such as a time average or area under the concentration curve may be useful for some analyses, but it does not, by itself, preserve the average effect under a nonlinear exposure–effect function.
Worked Example: Why Average Concentration Does Not Give Average Emax Effect
Suppose a hypothetical four-day profile has \(C=0\) mg/L for one day, \(C=2\) mg/L for one day, and \(C=8\) mg/L for two days. Let \(k_{\max}=0.60\) day\(^{-1}\) and \(EC_{50}=4\) mg/L. Calculate the time-average kill rate both from the profile and by incorrectly applying the Emax model to the average concentration.
Calculate each interval rate. At 0 mg/L, \(k=0\). At 2 mg/L, \(k=0.60(2)/(4+2)=0.200\) day\(^{-1}\). At 8 mg/L, \(k=0.60(8)/(4+8)=0.400\) day\(^{-1}\).
The rate-time area is \(0(1)+0.200(1)+0.400(2)=1.000\). Dividing by four days gives an average rate of \(1.000/4=0.250\) day\(^{-1}\).
The average concentration is \([0(1)+2(1)+8(2)]/4=18/4=4.50\) mg/L. Evaluating the Emax function at that mean gives \(0.60(4.50)/(4+4.50)=0.3176\) day\(^{-1}\), not \(0.250\) day\(^{-1}\). The latter is the profile's actual time-average rate under the stated model; \(k(\bar{C})\) is a different calculation.
Worked Example: Converting Concentration and Time Units Before Applying Emax
Suppose a hypothetical exposure is \(500\) nM for six hours. An Emax relationship is specified by \(k_{\max}=0.48\) day\(^{-1}\) and \(EC_{50}=0.25\) µM. Calculate the rate and cumulative hazard over the six-hour interval.
Convert the concentration. Since \(1\) µM equals \(1000\) nM, \(500\) nM equals \(0.500\) µM. The concentration and \(EC_{50}\) now use matching units.
Convert the duration. Six hours is \(6/24=0.250\) day. Treating the concentration as constant over that interval, the cumulative hazard is \(H=(0.320)(0.250)=0.0800\). The corresponding drug-only surviving fraction is \(\exp(-0.0800)=0.9231\), or about 92.3%; the modeled killed fraction is about 7.69%. The result is conditional on the specified concentration, the instantaneous Emax relationship, and the first-order killing assumption.
Diagnostics, Alternatives, and Limitations
Before trusting a calculated rate profile, check whether the model and exposure data support the calculation:
- Check dimensional consistency. A rate parameter must produce inverse-time units after applying the concentration function. In Emax, \(C\) and \(EC_{50}\) must share units.
- Check the exposure definition. Confirm whether the model uses plasma, free, or effect-site concentration. Substituting a different concentration measure changes the meaning of the fitted parameters.
- Inspect the concentration range. A linear model can imply implausibly large rates when extrapolated. An Emax curve may be poorly identified if the observed concentrations never approach saturation, or if they are all far above \(EC_{50}\).
- Check time resolution. Sparse concentration observations may miss peaks or troughs. Interpolation between them is an assumption, not additional data.
- Check the time course of response. If measured effects lag behind concentration changes, an instantaneous exposure–effect link may fail even when its concentration units and arithmetic are correct.
The linear and Emax functions offer different assumptions, not a universal ranking. A linear model is simpler and can be adequate over a narrow exposure range; Emax imposes a finite maximum effect but requires parameters for that saturation. Model choice should reflect the intended concentration range and be checked against response data, rather than selected solely because one curve looks familiar. More complex exposure–response structures may be needed when there is a delayed effect, a nonzero drug-independent component, or a changing sensitivity, but those features require additional assumptions and data.
A frequent mistake is to calculate the Emax rate at the mean concentration and call it the average rate. That is generally wrong because the Emax function is nonlinear: in general, \(k(\bar{C})\ne\overline{k(C)}\). Calculate the rate at each time or interval first, then average or integrate those rates over time. A second common error is to interpret \(k\) itself as the fraction killed. It is a rate; the fraction killed over a finite interval depends on the time-integrated rate under the stated first-order assumption.
Check Your Understanding
Try each problem, then compare your reasoning with the solutions below.
- A linear model has \(\beta=0.050\) L/(mg·day). What kill rate does it predict at \(C=4\) mg/L? State the units.
- An Emax model has \(k_{\max}=0.72\) day\(^{-1}\) and \(EC_{50}=3\) mg/L. What rate does it predict at \(C=3\) mg/L?
- For a nonlinear Emax model, explain why calculating the rate at average concentration may not give the time-average rate.
- A concentration is reported in nM, while \(EC_{50}\) is in µM. What must be done before using the Emax formula?
Solutions. (1) \(k=(0.050)(4)=0.200\) day\(^{-1}\); the concentration units cancel with those in \(\beta\). (2) Because \(C=EC_{50}\), the rate is half of \(k_{\max}\): \(0.72/2=0.360\) day\(^{-1}\). (3) Emax is nonlinear, so in general \(k(\bar{C})\ne\overline{k(C)}\); calculate rates over the profile and average those rates over time. (4) Convert one concentration quantity so \(C\) and \(EC_{50}\) are expressed in the same units, then substitute into the formula.