Deriving the Sigmoid Emax Model

Why Extend the Emax Model?

The ordinary Emax model describes a response that rises with concentration and approaches a maximum. Its fractional response is \(C/(EC_{50}+C)\). Some concentration-response patterns change more gradually around the midpoint; others change more sharply. The sigmoid Emax model extends the ordinary curve by adding an exponent, \(\gamma\), to control the shape of the fractional response.

Here \(C\) denotes the concentration driving the direct effect, expressed in a specified concentration unit. The model assumes that the measured response corresponds to that concentration without a relevant delay or indirect response mechanism. If a different concentration metric or a delayed effect is needed, the direct-effect equation alone does not resolve that modeling question.

Definition: The sigmoid Emax fractional response is \(f(C)=C^\gamma/(EC_{50}^\gamma+C^\gamma)\), for \(C\geq 0\), \(EC_{50}>0\), and \(\gamma>0\). Here \(f(C)\) is the fraction of the maximum baseline-relative response; \(EC_{50}\) is the concentration at which that fraction is one-half; and \(\gamma\) is a dimensionless shape parameter, often called the Hill coefficient.

For a stimulatory response, a common baseline-relative parameterization is \(E(C)=E_0+E_{\max}f(C)\), where \(E_0\) is the baseline response and \(E_{\max}\) is the maximum increase above baseline. Thus the model maximum is \(E_0+E_{\max}\), not \(E_{\max}\) alone. For an inhibitory response expressed as a decreasing measurement, one may write \(E(C)=E_0-I_{\max}f(C)\), where \(I_{\max}>0\) is the maximum decrease. In either case, \(f(C)\) describes the fraction of the modeled maximum change.

The exponent \(\gamma\) has no units because it acts on the concentration ratio \(C/EC_{50}\), which is dimensionless. \(EC_{50}\) has the same units as \(C\), such as mg/L. A fitted value of \(\gamma\) describes the steepness of this empirical curve; it does not, by itself, establish a particular molecular mechanism or number of binding sites.

Deriving the Fractional Response

A useful way to construct the curve is to relate the response fraction to its complement. The response fraction is \(f\), and the remaining fraction is \(1-f\). Assume that their ratio follows a power relationship with concentration relative to a reference concentration:

\frac{f(C)}{1-f(C)}=\left(\frac{C}{K}\right)^\gamma

Here \(K\) is a positive concentration scale and \(\gamma>0\). This power relationship is the modeling assumption: it says how the response odds change with concentration. To identify the scale in terms of the familiar half-maximal concentration, set \(C=K\). The right-hand side is then one, so \(f/(1-f)=1\), which implies \(f=1/2\). Therefore \(K\) is precisely the concentration at half of the modeled maximum change, and we name it \(EC_{50}\).

Solving the assumed relationship for \(f\) gives the sigmoid Emax fraction:

\begin{aligned} \frac{f}{1-f}&=\left(\frac{C}{EC_{50}}\right)^\gamma\\ f&=(1-f)\left(\frac{C}{EC_{50}}\right)^\gamma\\ f\left[1+\left(\frac{C}{EC_{50}}\right)^\gamma\right] &=\left(\frac{C}{EC_{50}}\right)^\gamma\\ f(C)&=\frac{(C/EC_{50})^\gamma}{1+(C/EC_{50})^\gamma} =\frac{C^\gamma}{EC_{50}^\gamma+C^\gamma}. \end{aligned}

This derivation makes clear what is and is not being asserted. The equation follows from the specified power relationship for response odds; that relationship is an assumption, not a consequence of saturation alone. For \(\gamma=1\), the fraction reduces to \(C/(EC_{50}+C)\), the ordinary Emax fraction. For positive concentration, the fraction increases with concentration: differentiating gives \(\gamma EC_{50}^{\gamma}C^{\gamma-1}/(EC_{50}^{\gamma}+C^\gamma)^2>0\). These statements require the stated positive-parameter conditions.

Worked Example: Deriving a Response and Interpreting Its Midpoint

Suppose a hypothetical stimulatory response has \(E_0=20\) response units, \(E_{\max}=80\) response units, \(EC_{50}=4\) mg/L, and \(\gamma=2\). Calculate the response at \(C=2\) mg/L and at \(C=EC_{50}\), and distinguish the model maximum from the baseline-relative increase.

At \(C=2\) mg/L, the fractional response is \(f(2)=2^2/(4^2+2^2)=4/(16+4)=0.2000\). The predicted response is therefore \(E(2)=20+80(0.2000)=36.00\) response units.

At \(C=EC_{50}=4\) mg/L, the fraction is \(f(4)=4^2/(4^2+4^2)=16/32=0.5000\). Thus \(E(4)=20+80(0.5000)=60.00\) response units. This is halfway between the baseline of 20 and the model maximum of \(20+80=100\) response units. Half of the absolute model maximum, 100, is 50 response units; 40 response units is half of the baseline-relative increase of 80, not half of the absolute response.

How Gamma Changes Predictions

When \(\gamma=1\), the sigmoid Emax model is exactly the ordinary Emax model. Values of \(\gamma\) different from one alter predictions away from the midpoint while preserving \(f(EC_{50})=0.5\). A value above one produces a more concentrated transition around the midpoint; a value below one produces a more gradual transition. These are descriptions of the curve’s shape, not mechanistic conclusions.

Worked Example: Comparing Three Curve Shapes

Consider hypothetical curves with \(EC_{50}=4\) mg/L. Compare \(\gamma=0.5\), \(1\), and \(2\) at \(C=2\) mg/L and \(C=8\) mg/L. These concentrations are respectively one-half and twice \(EC_{50}\).

At \(C=2\), the concentration ratio is \(C/EC_{50}=0.5\). For \(\gamma=0.5\), \(f=\sqrt{0.5}/(1+\sqrt{0.5})=0.4142\). For \(\gamma=1\), \(f=0.5/(1+0.5)=0.3333\). For \(\gamma=2\), \(f=0.5^2/(1+0.5^2)=0.2000\).

At \(C=8\), the ratio is 2. For \(\gamma=0.5\), \(f=\sqrt{2}/(1+\sqrt{2})=0.5858\). For \(\gamma=1\), \(f=2/(1+2)=0.6667\). For \(\gamma=2\), \(f=2^2/(1+2^2)=0.8000\). The predictions are summarized below.

Concentration	\(\gamma=0.5\)	\(\gamma=1\)	\(\gamma=2\)
2 mg/L	0.4142	0.3333	0.2000
4 mg/L	0.5000	0.5000	0.5000
8 mg/L	0.5858	0.6667	0.8000

For example, if the baseline-relative maximum increase were 60 response units, at 8 mg/L the predicted increases would be \(60(0.5858)=35.15\), \(60(0.6667)=40.00\), and \(60(0.8000)=48.00\) response units, respectively. The curves agree at \(EC_{50}\) but give distinct predictions on either side. In particular, the \(\gamma=1\) column is the non-sigmoid, ordinary Emax prediction.

Calculating a Concentration for a Target Response

The fractional equation can also be rearranged to find the concentration associated with a specified fraction \(f\), provided \(0<f<1\). Starting from the response-odds form gives \(C=EC_{50}[f/(1-f)]^{1/\gamma}\). For an observed response \(E\) in the stimulatory parameterization, first calculate \(f=(E-E_0)/E_{\max}\); this is a fraction of the baseline-relative increase, not a fraction of the absolute response.

Worked Example: Finding the Concentration for a Target Effect

A hypothetical stimulatory model has \(E_0=5\) response units, \(E_{\max}=40\) response units, \(EC_{50}=3\) mg/L, and \(\gamma=2\). Find the concentration predicted to produce 35 response units.

First calculate the fraction of the maximum increase: \(f=(35-5)/40=30/40=0.7500\). Then substitute into the rearranged equation: \(C=3[(0.7500)/(1-0.7500)]^{1/2}=3(3)^{1/2}=5.196\) mg/L.

Checking the result, \(C^2=27.00\) \((\text{mg/L})^2\) and \(EC_{50}^2=9.000\) \((\text{mg/L})^2\), so \(f=27/(9+27)=0.7500\). The predicted response is \(5+40(0.7500)=35.00\) response units. For comparison, the ordinary Emax model with the same \(EC_{50}\) would require \(C=3[0.75/0.25]=9.000\) mg/L for this fraction.

Assumptions, Diagnostics, and Alternatives

The sigmoid Emax curve is appropriate when a single direct concentration is being related to a bounded, monotone fractional change and the response can reasonably be represented by a baseline plus a maximum change. Concentration must be nonnegative, \(EC_{50}\) positive, and \(\gamma\) positive under the standard increasing stimulatory formulation. The concentration and \(EC_{50}\) must use the same units. For an inhibitory measured response, the sign of the response change is represented in the response equation, while \(f(C)\) remains a nonnegative fraction.

A practical diagnostic is to compare observed responses with the fitted curve across the concentration range, including the region around the transition. Persistent patterns in residuals—such as observations consistently above the curve at low concentrations and below it at intermediate concentrations—can indicate that the chosen shape does not represent the data well. A very sharp or gradual fitted transition should also be checked against the observed concentrations and responses rather than interpreted from the parameter alone. If the data do not cover the transition, predictions across it may be especially uncertain; a visually plausible fit over a narrow range does not verify the assumed shape.

Approach	Useful feature	Important limitation
Linear concentration-effect model	Simple description over a restricted range	Does not represent a saturating maximum over the full range
Ordinary Emax model, \(\gamma=1\)	Parsimonious saturating curve with a midpoint	Its transition shape is fixed
Sigmoid Emax model, \(\gamma\) estimated	Allows a more gradual or sharper transition	Adds a shape parameter and may be unsupported by sparse data

The more flexible curve is not automatically preferable. Use the ordinary Emax model when its fixed shape adequately represents the observations; the sigmoid form is useful when a different transition shape is supported by data and is scientifically relevant. Neither form by itself addresses delayed effects, hysteresis, exposure history, or other departures from a direct concentration-effect relationship.

Common mistake: Treating \(EC_{50}\) as the concentration that produces half of the absolute response, or treating \(E_{\max}\) as the absolute model maximum when a nonzero baseline is included. In the baseline-relative stimulatory model, \(EC_{50}\) produces half of the increase \(E_{\max}\), so \(E(EC_{50})=E_0+E_{\max}/2\). The absolute model maximum is \(E_0+E_{\max}\). Always identify whether a reported fraction refers to the baseline-relative change or to the absolute response.

Key takeaway: The sigmoid Emax fraction follows from an assumed power relationship between response odds and concentration. \(EC_{50}\) locates the half-maximal baseline-relative change, while \(\gamma\) modifies curve shape; setting \(\gamma=1\) recovers the ordinary Emax model.

Check Your Understanding

Answer each question, then check the solutions below.

For \(\gamma=1\), simplify the sigmoid fractional-response equation.
If \(E_0=12\) response units and \(E_{\max}=30\) response units, what response is predicted at \(EC_{50}\), and what is the absolute model maximum?
For \(C=EC_{50}\), calculate \(f(C)\) when \(\gamma=0.5\).
A target response corresponds to a baseline-relative fraction of 0.80. With \(EC_{50}=2\) mg/L and \(\gamma=2\), calculate the required concentration.

Solutions. (1) Setting \(\gamma=1\) gives \(f=C/(EC_{50}+C)\), the ordinary Emax fraction. (2) At \(EC_{50}\), the response is \(12+30(0.5)=27.00\) response units; the absolute model maximum is \(12+30=42.00\) response units. (3) For any positive \(\gamma\), \(f(EC_{50})=EC_{50}^{\gamma}/(EC_{50}^{\gamma}+EC_{50}^{\gamma})=0.5000\). (4) \(C=2[0.80/(1-0.80)]^{1/2}=2\sqrt{4}=4.000\) mg/L.