Why Decreasing Responses Need Careful Interpretation
When increasing drug concentration decreases a measured response, the curve’s lower asymptote is not automatically zero. An inhibitory Emax model describes both how much of the baseline response can be inhibited and how quickly that inhibition is approached as concentration increases. Keeping these quantities separate prevents a common error: treating the maximum inhibition as if it were the response that remains.
The worked calculations below use a model in which the maximum inhibition is expressed as a fraction of baseline response. This convention is stated explicitly because inhibitory Emax notation can vary. Concentration is denoted by \(C\), response by \(R(C)\), and the baseline response at zero concentration by \(R_0\).
At zero concentration the model gives \(R(0)=R_0\). As concentration increases without bound, \(C/(IC_{50}+C)\) approaches 1, so the limiting response is \(R_0(1-I_{\max})\). Thus, \(I_{\max}\) describes the inhibited fraction, while \(1-I_{\max}\) describes the fraction of baseline remaining at the high-concentration limit. These are different quantities.
Fitting the Curve From Decreasing Responses
A simple fit assumes that the baseline \(R_0\) is known and that observed responses are independent measurements around the model predictions. For additive errors with constant variance, nonlinear least squares estimates the parameters by minimizing the sum of squared response residuals:
The fitted values should respect the parameter constraints and be assessed against the observed concentration range. A low residual sum alone does not establish that the model is appropriate: the response pattern, residuals, and error scale also matter. Earlier material on nonlinear least squares, parameter constraints, and residual diagnostics provides the general fitting framework; here the emphasis is on the meaning and calculation of the inhibitory curve.
Worked Example: Estimating an Inhibitory Emax Curve
Suppose a hypothetical experiment has a known baseline response of 100 response units. The observed responses are 100, 80, 60, 40, and 28 units at concentrations of 0, 1, 3, 9, and 27 mg/L, respectively. Treat the baseline as fixed and use the decreasing observations to estimate \(I_{\max}\) and \(IC_{50}\).
Normalize each response by the baseline and define the fraction inhibited as \(q(C)=1-R(C)/R_0\). The model implies \(q(C)=I_{\max}C/(IC_{50}+C)\). At 3 mg/L, \(q=1-60/100=0.40\); at 9 mg/L, \(q=1-40/100=0.60\). Therefore:
Dividing the second equation by the first eliminates \(I_{\max}\):
Solving gives \(1.5(IC_{50}+9)=3(IC_{50}+3)\), so \(1.5IC_{50}=4.5\) and \(IC_{50}=3\) mg/L. Substituting into the first equation gives \(0.40=I_{\max}(3/6)\), hence \(I_{\max}=0.80\). The model therefore predicts \(R(C)=100[1-0.80C/(3+C)]\).
At 1 mg/L, the prediction is \(100[1-0.80(1/4)]=80\) units. At 27 mg/L, it is \(100[1-0.80(27/30)]=28\) units. These predictions agree with the listed observations, as do the predictions at 0, 3, and 9 mg/L. In noisy data, the estimates would generally be obtained by nonlinear fitting across all observations rather than by solving two selected points exactly.
What \(I_{\max}\) and \(IC_{50}\) Mean
The meaning of \(IC_{50}\) follows directly from the model: at \(C=IC_{50}\), the concentration term is \(1/2\), and the fraction of baseline inhibited is \(I_{\max}/2\). It does not mean that the response is reduced to half of baseline, unless \(I_{\max}=1\). Likewise, \(I_{\max}\) is not the response remaining after maximal inhibition.
A short derivation makes the distinction explicit. The model’s inhibited fraction is \(q(C)=1-R(C)/R_0\). Substitution of \(C=IC_{50}\) gives \(q(IC_{50})=I_{\max}(IC_{50})/(IC_{50}+IC_{50})=I_{\max}/2\). At the high-concentration limit, \(q(C)\) approaches \(I_{\max}\), while the response fraction \(R(C)/R_0\) approaches \(1-I_{\max}\).
Worked Example: Separating Inhibition From Residual Response
Consider a hypothetical curve with \(R_0=100\) units, \(I_{\max}=0.75\), and \(IC_{50}=3\) mg/L. Calculate the response at \(IC_{50}\), at 9 mg/L, and in the high-concentration limit.
At \(IC_{50}=3\) mg/L, half of the maximum inhibition is achieved. The inhibited fraction is \(0.75/2=0.375\), or 37.5% of baseline. The response is \(100(1-0.375)=62.5\) units, which is 62.5% of baseline—not 25%.
At 9 mg/L, \(C/(IC_{50}+C)=9/(3+9)=0.75\). The fraction of baseline inhibited is \(0.75(0.75)=0.5625\), or 56.25%. The response is \(100(1-0.5625)=43.75\) units, or 43.75% of baseline. In other words, 75% of the maximum inhibition has been achieved at 9 mg/L, but that does not mean 75% of baseline is inhibited.
As concentration tends to infinity, the maximum inhibited fraction is 0.75, leaving a residual response fraction of \(1-0.75=0.25\). The limiting response is therefore \(100(0.25)=25\) units. The 25% figure is the high-concentration residual response, not the response at \(IC_{50}\) or at 9 mg/L.
Additive and Proportional Residual Error
The concentration-response curve specifies the expected response; a residual-error model describes how observations vary around it. Under additive error, observations are modeled as the prediction plus an error with a constant standard deviation in response units. This is plausible when the absolute scatter is roughly similar across the response range. Under proportional error, the scatter scales with the prediction, so the coefficient of variation is approximately constant. A simple proportional-error fitting objective sums squared residuals divided by squared predictions (equivalently, squared relative residuals); a log-normal error model instead uses residuals on the log-response scale.
The choice should reflect measurement behavior, not which objective happens to look numerically smaller. Additive residuals have squared response units, whereas relative residuals are dimensionless, so their raw sums of squares are not directly comparable. For a formal comparison, the error distributions and their variances must be specified and assessed using an appropriate likelihood or other justified criterion.
Worked Example: Comparing Residual Scales
Suppose a candidate inhibitory curve predicts responses of 100, 80, 60, 40, and 28 units at five concentrations. Hypothetical observed responses are 101, 78, 63, 38, and 30 units. Calculate the residuals and compare their scale under additive and proportional error.
Define a residual as observed minus predicted. The residuals are \(1,-2,3,-2,2\) response units. Under an additive-error least-squares objective, the sum of squared residuals is \(1^2+(-2)^2+3^2+(-2)^2+2^2=22\) response-units squared.
For a proportional-error objective based on squared relative residuals, divide each residual by its prediction. The relative residuals are \(0.0100,-0.0250,0.0500,-0.0500,\) and \(0.07143\). Their squared sum is approximately \(0.01083\). This dimensionless value is not “better” merely because it is smaller than 22; it is on a different scale.
The observed absolute deviations range from 1 to 3 units, while their relative sizes are not constant: the deviation of 2 units at a prediction of 28 is about 7.14%, compared with 1% at a prediction of 100. This pattern may make proportional error worth examining, but five observations are too few to establish the error model. Residual plots and knowledge of the assay’s variability are also needed.
Diagnostics, Alternatives, and Limitations
A useful fit should show a decreasing response that is compatible with a saturating curve, without systematic residual patterns that the model leaves unexplained. Plot residuals against concentration and predicted response, and inspect both absolute and relative scales. A curved residual pattern can indicate that the assumed curve shape is inadequate; increasing spread with predicted response can be inconsistent with constant additive variance. Clustering or repeated measurements may also violate the simple independent-error assumption and require an analysis that accounts for that structure.
The main error-model alternatives are not interchangeable:
| Assumption | Expected residual pattern | Potential concern |
|---|---|---|
| Additive error | Roughly constant absolute spread in response units | Relative variability can become large when the response is small |
| Proportional error | Roughly constant relative spread across response levels | Can be unsuitable near zero or when responses may be nonpositive |
| Combined error | Both a baseline absolute component and a response-dependent component | Requires enough data to estimate the additional error parameters |
An alternative response model may be needed if the observed decline is approximately linear over the measured range rather than saturating, or if a simple Emax-shaped curve leaves systematic departures. A sigmoid Emax model allows a steeper or shallower transition through an additional shape parameter, but that added flexibility requires information across the transition region. It should not be introduced solely because one or two observations do not lie on an Emax curve.
Parameter estimates can be weakly determined when the data do not cover relevant parts of the curve. If all concentrations are far below \(IC_{50}\), the data may show an initial decline but provide little information about the plateau or maximum inhibition. If all are far above \(IC_{50}\), the high-concentration asymptote may be visible but the location of \(IC_{50}\) may be poorly resolved. If \(R_0\) is estimated rather than known, sparse low-concentration data can also make baseline and inhibition parameters difficult to distinguish. These are information limitations, not problems that can be repaired simply by reporting more decimal places.
Common Mistakes
- Calling \(I_{\max}\) the residual response. In this convention, \(I_{\max}\) is the maximum fraction inhibited; the high-concentration residual fraction is \(1-I_{\max}\).
- Interpreting \(IC_{50}\) as the concentration that halves baseline response. It is the concentration that achieves half of the maximum inhibition. The response at that concentration depends on \(I_{\max}\).
- Choosing an error model by comparing raw sums of squares. Additive and relative residual objectives use different scales. Examine the assumed variability and use a suitable likelihood-based comparison when warranted.
- Extrapolating an asymptote from concentrations that do not approach it. A fitted plateau may be highly uncertain even if the curve passes close to the observed points.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- If \(I_{\max}=0.60\), what fraction of baseline response remains in the high-concentration limit?
- For \(I_{\max}=0.80\) and \(IC_{50}=4\) mg/L, what fraction of baseline is inhibited at \(C=4\) mg/L?
- For \(R_0=80\) units, \(I_{\max}=0.50\), and \(IC_{50}=2\) mg/L, calculate the response at \(C=6\) mg/L.
- What residual pattern would make constant additive error questionable?
- Why may data collected only at concentrations far below \(IC_{50}\) be insufficient to estimate \(I_{\max}\) reliably?
Solutions. (1) The residual fraction is \(1-0.60=0.40\), or 40% of baseline. (2) At \(IC_{50}\), half the maximum inhibition is achieved, so the inhibited fraction is \(0.80/2=0.40\), or 40% of baseline. (3) The concentration term is \(6/(2+6)=0.75\); the inhibited fraction is \(0.50(0.75)=0.375\), so \(R(6)=80(1-0.375)=50\) units. (4) A clear increase or decrease in the absolute residual spread as predicted response changes is inconsistent with a constant additive standard deviation. (5) At concentrations much smaller than \(IC_{50}\), the curve mainly reflects its initial decline and may not reveal the high-concentration plateau, which determines \(I_{\max}\).