Why Curve Shape Needs Data Around the Transition
A sigmoid Emax model describes a response that approaches a maximum as concentration rises, with the steepness of the transition controlled by a shape parameter. Estimating that shape is different from merely locating the concentration that gives half the maximum effect: concentrations must cover enough of the transition to distinguish a gradual curve from a steep one.
This tutorial works through a hypothetical example in which the baseline and maximum drug effect have been established by calibration. That lets us focus on estimating the two shape parameters, \(EC_{50}\) and \(\gamma\). In an analysis where those calibration values are unavailable, they must generally be estimated along with the shape parameters, which can increase confounding and uncertainty.
The response fraction \(q=(E-E_0)/E_{\max}\) lies between zero and one for responses strictly between the baseline and the upper asymptote. Its logit has a useful relationship to concentration:
This identity explains why \(\gamma\) controls the slope on a log-concentration versus logit-response scale. It can help interpret a curve, but fitting a straight line to transformed noisy responses is not generally the same as minimizing squared errors on the original response scale. In particular, the transformation changes the error scale and is undefined at \(q=0\) or \(q=1\).
A Worked Fit With Calibrated Asymptotes
Assume an assay has established \(E_0=10\) response units and \(E_{\max}=80\) response units. The following hypothetical responses are generated from the sigmoid model with \(EC_{50}=4\ \mu\text{mol}/\text{L}\) and \(\gamma=2\). Values are displayed to four decimal places; calculations use the unrounded model values.
| Concentration \(C\) (\(\mu\text{mol}/\text{L}\)) | Response \(E(C)\) (response units) | Response fraction \(q\) |
|---|---|---|
| 1 | 14.7059 | 0.058824 |
| 2 | 26.0000 | 0.200000 |
| 3 | 38.8000 | 0.360000 |
| 4 | 50.0000 | 0.500000 |
| 5.3333 | 61.2000 | 0.640000 |
| 8 | 74.0000 | 0.800000 |
| 16 | 85.2941 | 0.941176 |
Worked Example: Estimating \(EC_{50}\) and \(\gamma\)
Use the calibrated values \(E_0=10\) and \(E_{\max}=80\) response units and the observations at \(C=2\) and \(C=8\ \mu\text{mol}/\text{L}\) to recover the shape parameters. The other observations provide a check that this relationship describes the transition across the measured range.
At \(C=2\), \(q=(26-10)/80=0.2\), so the logit is \(\log(0.2/0.8)=\log(0.25)=-1.3863\). At \(C=8\), \(q=(74-10)/80=0.8\), and the logit is \(\log(0.8/0.2)=\log(4)=1.3863\).
The slope between these logit values against log concentration is $$ \hat{\gamma}=\frac{1.3863-(-1.3863)}{\log(8)-\log(2)} =\frac{\log(4)-\log(0.25)}{\log(4)} =2.000. $$ The arithmetic uses the exact log-odds, \(\log(4)\) and \(-\log(4)\), rather than rounded intermediate values.
At \(C=4\), the observed response is 50, exactly halfway between 10 and 90 response units. Thus \(EC_{50}=4\ \mu\text{mol}/\text{L}\). Equivalently, the fitted logit line has intercept \(-\gamma\log(EC_{50})=-2\log(4)\), which gives \(\log(EC_{50})=\log(4)\). Therefore the estimates are \(EC_{50}=4\ \mu\text{mol}/\text{L}\) and \(\gamma=2.000\).
Substitution in the original model reproduces the listed values. For instance, at \(C=2\), the effect fraction is \(2^2/(4^2+2^2)=4/20=0.2\), giving \(E(2)=10+80(0.2)=26.0\) response units. All seven observations lie on the stated curve, so the response-scale sum of squared residuals is zero using unrounded values. This deliberately exact example demonstrates parameter recovery; it does not represent a realistic estimate of sampling uncertainty.
What Fixing \(\gamma=1\) Changes
A useful comparison is the ordinary Emax curve, obtained by setting \(\gamma=1\) while retaining the calibrated baseline and maximum effect. This is a restricted version of the sigmoid model: it cannot adjust steepness independently of \(EC_{50}\). For the example data, compare the observed responses with predictions from this restricted curve.
Worked Example: Comparing With the Emax Special Case
With \(\gamma=1\) and \(EC_{50}=4\ \mu\text{mol}/\text{L}\), the effect fraction is \(C/(4+C)\). The predicted response at \(C=1\), for example, is \(10+80(1/5)=26.0\) response units, compared with the observed 14.7059. The residual, defined here as observed minus predicted, is \(-11.2941\) response units.
A one-dimensional least-squares minimization over \(EC_{50}\) for the restricted curve gives \(EC_{50}=4\ \mu\text{mol}/\text{L}\) for these data. The paired concentrations are reciprocal around 4: each pair has product \(16\). At this value, the model fractions for each pair sum to one, as do the observed fractions. The paired residuals therefore have opposite signs, and their contributions to the derivative of the sum of squares with respect to \(\log(EC_{50})\) cancel. Numerical minimization confirms this stationary point is the minimum for this example.
The response-scale residuals at concentrations 1, 2, and 3 are approximately \(-11.2941\), \(-10.6667\), and \(-5.4857\) response units. At 4 the residual is zero; at 5.3333, 8, and 16 the residuals are the corresponding positive values. Thus, $$ \begin{aligned} SSE_{\gamma=1} &=2\left(11.2941^2+10.6667^2+5.4857^2\right)\\ &\approx 542.9\ \text{response-units}^2. \end{aligned} $$ The unrestricted curve has \(SSE=0\) for these exactly generated observations. This comparison shows that the curve shapes differ substantially across the transition. It does not, by itself, provide a significance test or a general model-selection rule: with real data, the residual error model and uncertainty must also be considered.
Derivation: Where the Data Inform Curve Shape
Let \(f=C^\gamma/(EC_{50}^\gamma+C^\gamma)\), so \(E=E_0+E_{\max}f\). Differentiating with respect to the log of \(EC_{50}\) and with respect to \(\gamma\) gives
These sensitivities explain a practical feature of fitting. At \(C=EC_{50}\), the response is maximally sensitive to the location parameter \(EC_{50}\), but \(\partial E/\partial\gamma=0\): a point exactly at the midpoint alone carries no local information about steepness. Far below or above the transition, \(f(1-f)\) is small, so changes in either shape parameter have little effect on predicted response. Measurements on both sides of the midpoint, within the transition, are especially useful for estimating \(\gamma\).
Worked Example: Diagnosing Sensitivity to \(\gamma\)
For the fitted curve, take \(C=2\ \mu\text{mol}/\text{L}\). Here \(f=0.2\), \(\gamma=2\), and \(\log(C/EC_{50})=\log(0.5)\). The local sensitivity to \(\gamma\) is $$ \frac{\partial E}{\partial\gamma} =80(0.2)(0.8)\log(0.5) \approx -8.872\ \text{response units per unit of }\gamma. $$ At \(C=8\ \mu\text{mol}/\text{L}\), \(f=0.8\) and \(\log(C/EC_{50})=\log(2)\), giving approximately \(+8.872\) response units per unit of \(\gamma\). The opposite signs show how observations on opposite sides of \(EC_{50}\) respond differently to a change in steepness. At \(C=4\), the sensitivity to \(\gamma\) is zero. A dataset concentrated only at the midpoint, or only at the nearly flat tails, would therefore be poor evidence about curve shape.
Assumptions, Diagnostics, and Limitations
The example relies on several assumptions. The measured concentration is the concentration relevant to the direct effect; the response follows a single increasing sigmoid curve; \(E_0\) and \(E_{\max}\) are known and applicable to these observations; and the response is measured on a scale where the chosen residual model is appropriate. In an ordinary unweighted nonlinear least-squares fit, the objective is the sum of squared response residuals. Its usual interpretation as a likelihood-based fit requires independent, approximately normal residuals with constant variance. If variability changes with response or concentration, an appropriate weighted or alternative error model is needed.
For noisy data, a numerical fit should minimize the stated objective over admissible values, typically \(EC_{50}>0\) and \(\gamma>0\). Check that the optimizer converged, that predictions and residuals are sensible across concentration, and that parameter estimates are not pinned to constraints. Residual patterns can reveal a poor curve or error model, but random-looking residuals alone do not prove that \(\gamma\) is well estimated.
A useful shape diagnostic is to examine the objective over a range of \(\gamma\) values, re-optimizing \(EC_{50}\) at each value. A broad, shallow profile indicates that many steepness values fit nearly as well; a narrow minimum indicates stronger information about shape. A fitted covariance matrix can also signal strong parameter correlation, but local approximations can be misleading for a nonlinear, weakly identified curve. The exact example above has zero residual error, so it cannot estimate residual variance or provide a meaningful uncertainty interval. Real data require an error model and an uncertainty assessment.
The main alternative here is the Emax model with \(\gamma=1\). It is simpler and may be adequate when the transition is gradual or data are too sparse to support an additional shape parameter. The sigmoid model is more flexible, but that flexibility does not guarantee reliable estimation: if measurements omit the steep transition, a wide range of \(\gamma\) and \(EC_{50}\) combinations may give similar predictions over the observed range. Nor should an estimated steep curve automatically be interpreted as evidence of a particular biological mechanism; it describes the observed concentration–response shape under the model assumptions.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For \(E_0=5\), \(E_{\max}=60\), and \(EC_{50}=3\ \mu\text{mol}/\text{L}\), what response does the model predict at \(C=3\), regardless of \(\gamma\)?
- Why does an observation exactly at \(C=EC_{50}\) provide no local sensitivity to \(\gamma\)?
- For the hypothetical example, what is the response fraction at \(C=2\), and what is its logit?
- What does a broad profile of the least-squares objective over \(\gamma\) indicate?
Solutions. (1) At \(EC_{50}\), the effect fraction is one half, so the predicted response is \(5+60/2=35\) response units. (2) The derivative with respect to \(\gamma\) contains \(\log(C/EC_{50})\), which equals zero at the midpoint. (3) The fraction is \((26-10)/80=0.2\), and its logit is \(\log(0.2/0.8)=\log(0.25)\approx-1.386\). (4) It indicates that a range of steepness values fit similarly after re-optimizing other parameters, so the data provide weak information about curve shape.