Why the Response Peak Needs Its Own Check
The time of maximum effect-site concentration is often an important landmark, but it is not automatically the time of maximum response. To predict a response peak, one must combine the effect-site trajectory \(C_e(t)\) with the specified concentration-response function. Whether their peaks coincide depends on the shape of that function over the concentrations actually reached.
The distinction matters even when the concentration trajectory is already known. For a conventional increasing Emax function, the response rises as effect-site concentration rises, so their peak times coincide. For inhibition expressed as an increasing amount of inhibition, the same is true; for a response scale that decreases with concentration, the concentration peak instead corresponds to a response minimum. A nonmonotone concentration-response function can produce a response peak before, after, or at more than one time distinct from the effect-site peak.
Assumptions and the Peak-Time Condition
The basic calculation assumes that the effect-site trajectory is specified over the time interval of interest and that the response at a given time is determined by the current effect-site concentration through a fixed function \(f\). Thus the model has no additional explicit time dependence, feedback from response to concentration, or separate delay between effect site and measured response. The concentration-response parameters are treated as fixed. A candidate peak should also be checked against the interval boundaries, because an endpoint may have a larger response than any interior stationary point.
At times where both functions are differentiable, the chain rule gives the key relationship:
An interior stationary point of response can therefore arise because the effect-site concentration is stationary, because the concentration-response function is stationary at the current concentration, or because both derivatives are zero. This is why locating the maximum of \(C_e(t)\) alone is not enough for a general response function.
Why Monotonicity Preserves a Peak
The result for a strictly increasing function follows directly from order preservation. Suppose \(C_e(t_*)\) is a global maximum, so \(C_e(t)\leq C_e(t_*)\) for every time \(t\) in the interval. Strict increase of \(f\) implies \(f(C_e(t))\leq f(C_e(t_*))\). Therefore \(R(t_*)\) is also a global maximum. The converse holds for a strictly decreasing function with maxima and minima reversed.
This argument concerns timing, not response magnitude. Even when the two peaks occur at the same time, the response at that time must still be calculated by substituting the peak concentration into the response function. In particular, the maximum-effect parameter in an Emax model is generally an asymptote, not the response actually attained at a finite concentration.
Worked Peak Calculations
Worked Example: Increasing Emax Response After a Bolus
Suppose a hypothetical effect-site trajectory, in mg/L with time \(t\) in hours, is \(C_e(t)=16.667(e^{-0.2t}-e^{-0.5t})\) for \(t\geq0\). The specified response is \(R(t)=10+80C_e(t)/(2+C_e(t))\), in response units. Find the time and value of the response peak.
First find the effect-site stationary point:
Thus \(t=-\ln(0.4)/0.3=3.0543\) h. The trajectory is zero at \(t=0\), tends to zero as \(t\) tends to infinity, and is positive between those limits; this interior stationary point is its maximum. The response function is strictly increasing for \(C_e\geq0\), since its derivative with respect to concentration is \(160/(2+C_e)^2>0\). Therefore the response peak occurs at the same time.
Using \(e^{-0.5t}=0.4e^{-0.2t}\) at the stationary point, \(C_{e,\max}=16.667(0.6)e^{-0.2(3.0543)}\approx5.4289\) mg/L. Substitution into the response function gives \(R_{\max}=10+80(5.4289)/(2+5.4289)\approx68.46\) response units. The parameter 80 response units is the Emax increment, approached only as concentration tends to infinity; it is not the achieved increment at this finite peak concentration.
Worked Example: Peak Inhibition Is Not the Maximum-Inhibition Parameter
For a hypothetical trajectory \(C_e(t)=5.2\exp[-(t-4)^2/8]\) mg/L, consider inhibition from baseline given by \(I(t)=60C_e(t)/(2+C_e(t))\), measured in percentage points. Find the time and magnitude of maximum inhibition.
Taking the derivative of the concentration gives \(\frac{dC_e}{dt}=C_e(t)(4-t)/4\). Since \(C_e(t)>0\), concentration increases for \(t<4\) and decreases for \(t>4\), so its maximum is \(5.2\) mg/L at \(t=4\) h. The inhibition function is strictly increasing in concentration because its derivative is \(120/(2+C_e)^2>0\). Its maximum over time is therefore at \(t=4\) h.
The achieved inhibition is \(I(4)=60(5.2)/(2+5.2)=312/7.2=43.33\) percentage points. The parameter 60 percentage points is the maximum-inhibition parameter: the model approaches it as concentration tends to infinity. It is not the inhibition at the finite peak concentration.
Worked Example: A Nonmonotone Response Has Two Peaks
Consider a hypothetical effect-site trajectory in mg/L: \(C_e(t)=t\) for \(0\leq t\leq4\) h, and \(C_e(t)=8-t\) for \(4<t\leq8\) h. Let the response be \(R(t)=100C_e(t)e^{-C_e(t)/2}\) response units. Locate the global response peaks.
As a function of concentration \(c\), the response is \(f(c)=100ce^{-c/2}\). Its derivative is \(f'(c)=100e^{-c/2}(1-c/2)\). The derivative is positive for \(c<2\), zero at \(c=2\), and negative for \(c>2\). Thus \(f\) reaches its maximum at \(c=2\) mg/L, not at the maximum concentration of 4 mg/L.
On the rising part of the trajectory \(C_e(t)=t\), the concentration reaches 2 mg/L at \(t=2\) h. On the falling part \(C_e(t)=8-t\), it reaches 2 mg/L again at \(t=6\) h. At both times, \(R=100(2)e^{-1}=200/e\approx73.58\) response units. At the effect-site peak, \(t=4\) h, the response is \(100(4)e^{-2}\approx54.13\) response units. The global response peaks are therefore at 2 h and 6 h, while the effect-site peak is at 4 h. Comparing the response at both concentration crossings and at the time of maximum concentration establishes that neither the concentration peak nor a single crossing can be assumed to describe the response peak.
A Practical Search and Its Diagnostics
For a specified trajectory and response function, a reliable peak search is a short sequence of checks. For an analytic trajectory, differentiate \(C_e(t)\), identify its stationary times and any boundary times, and evaluate the response there. For a nonmonotone \(f\), also solve \(f'(C_e(t))=0\) where those concentrations are attained. If the equations cannot be solved in closed form, use numerical root-finding or a sufficiently fine time grid to identify candidate regions, then refine candidate times and compare response values.
State the time interval, concentration units, response units, baseline, and concentration-response function. Decide whether the target is response, inhibition, or another defined quantity.
Use \(R'(t)=f'(C_e(t))C_e'(t)\). Include interior stationary times, points where the trajectory or response function changes differentiability, and interval endpoints.
Substitute the candidate concentrations into \(f\), check the resulting response values and units, and select the largest value for a global peak on the stated interval.
A numerical time grid is useful for seeing the broad shape, but it can miss a narrow peak or return only the closest grid time. A derivative-based search is more precise when the model is differentiable and its parameters are known. Both approaches should be checked by evaluating the response around a proposed maximum: values on either side should not be larger for a local maximum. For a global maximum, compare all candidate peaks and the boundaries as well.
Warning signs include a reported response peak at the concentration peak despite a nonmonotone response function, a result that changes substantially when the numerical time step is reduced, or an alleged global maximum that is lower than an endpoint value. Also check whether the response is defined as a change from baseline or as an absolute measurement. Adding a constant baseline does not change peak timing, but changing the direction or scale of the response can reverse whether the concentration peak is a response maximum or minimum.
Alternatives and Limits of the Calculation
When \(f\) is strictly monotone, the efficient approach is to locate the effect-site maximum or minimum and map that concentration through \(f\). There is no need to search the full response curve for additional concentration optima. When \(f\) is nonmonotone, the general stationary-point search is necessary; evaluating only the maximum effect-site concentration can miss multiple response peaks, as in the preceding example. Numerical evaluation on a time grid is a practical alternative when derivatives are unavailable, but it should be followed by refinement and boundary checks.
These calculations predict peaks conditional on the specified \(C_e(t)\) and \(f\); they do not establish that either model is adequate for observed data. Measurement noise, uncertain parameters, an incorrectly specified response scale, or a response that depends on past concentrations can make observed peak times differ from deterministic model predictions. If response has its own dynamics or depends explicitly on time, the static relation \(R(t)=f(C_e(t))\) is not sufficient, and the chain-rule result here cannot be used as though that relation still held.
Check Your Understanding
Answer each question using the stated response scale and time interval.
- If \(f\) is strictly increasing over the full concentration range, what is the relationship between the times of maximum \(C_e(t)\) and maximum \(R(t)\)?
- For an inhibitory response defined as the amount of inhibition, \(I=I_{\max}C_e/(IC_{50}+C_e)\), what must be true for the inhibition peak to coincide with the effect-site peak?
- A response function has a concentration optimum at \(c=3\) mg/L. The effect-site trajectory crosses 3 mg/L at 1 h and 5 h and reaches 6 mg/L at 3 h. Which times are candidates for response peaks, and what else must be checked?
- Why is it incorrect to report \(I_{\max}\) as the inhibition predicted at every finite concentration?
- A time-grid search reports its largest response at 2.0 h. Name one check that can improve confidence in the reported peak time.
Solutions. (1) The peak times coincide, because strict increase preserves the ordering of concentrations and responses. (2) \(I_{\max}>0\), \(IC_{50}>0\), and the stated inhibitory amount is strictly increasing in \(C_e\geq0\); under the static response model its maximum occurs when \(C_e\) is maximal. (3) Both 1 h and 5 h are candidates because the response function is stationary at 3 mg/L; evaluate the response at those times, at any other stationary or nondifferentiable points, and at the interval endpoints. (4) \(I_{\max}\) is the asymptotic inhibition approached as concentration tends to infinity; at finite concentration the predicted value is \(I_{\max}C_e/(IC_{50}+C_e)\), which is smaller when \(IC_{50}>0\). (5) Reduce the grid spacing and refine the time near 2.0 h, then compare the response on either side and against other candidate times and endpoints.