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Dose Justification Across Development Phases · Tutorial 225 of 596

Inverting Emax Models to Find Target Doses

Learn to invert standard and sigmoid Emax models to calculate target doses and determine when those estimates require extrapolation beyond observed doses.

Advanced 11 min read

What You'll Learn

  • Define a target as a fraction of the drug effect above baseline, rather than a fraction of the total response.
  • Derive the dose achieving a specified fraction of Emax in the standard Emax model.
  • Extend the inversion to a sigmoid Emax model with a Hill coefficient.
  • Calculate target doses from fitted parameters and verify them by substitution.
  • Check target doses against the observed dose range and recognize extrapolation.
  • Identify cases in which the target dose is undefined, unstable, or not meaningful.

From a Target Effect to a Dose

Dose selection often begins with a desired effect, not with a dose. For example, a development team may want a dose predicted to achieve 75% of the maximum drug-related effect. An Emax model makes this question algebraically tractable: once its parameters are fitted, the dose associated with a specified effect fraction can be calculated by rearranging the model.

The calculation is useful only if the target is defined on the correct scale and the fitted curve is suitable for the decision. In particular, a fraction of Emax refers to the drug effect above the baseline response, not generally to the total response. A calculated target dose also needs to be compared with the doses actually studied. A dose outside that range is an extrapolation, even if the formula returns a precise number.

Definition: In the standard increasing Emax model, the expected response at dose \(d\) is \(R(d)=E_0+E_{\max}d/(ED_{50}+d)\). Here \(E_0\) is the response at zero dose, \(E_{\max}>0\) is the maximum drug-related increase above \(E_0\), and \(ED_{50}>0\) is the dose producing half of that maximum increase. The dose \(d\) and \(ED_{50}\) have the same units.

This tutorial uses this standard, monotone model with nonnegative dose. The model assumes that the response follows the stated curve over the dose range being interpreted, that \(E_0\), \(E_{\max}\), and \(ED_{50}\) refer to the same endpoint and population, and that the fitted parameters are suitable inputs for the calculation. The model does not, by itself, establish that a particular response is clinically worthwhile or that the dose is acceptable for safety.

Deriving the Target-Dose Formula

Let \(f\) denote the desired fraction of the maximum drug effect, with \(0<f<1\). Under the standard model, the drug-related effect at dose \(d\) is \(R(d)-E_0\). Requiring this effect to equal \(fE_{\max}\) gives:

$$ \frac{R(d)-E_0}{E_{\max}} =\frac{d}{ED_{50}+d} =f. $$

Solving the final equality gives the dose directly:

$$ f(ED_{50}+d)=d \quad\Longrightarrow\quad fED_{50}=d(1-f) \quad\Longrightarrow\quad d_f=ED_{50}\frac{f}{1-f}. $$

Because \(ED_{50}>0\) and \(0<f<1\), this gives a positive finite dose. The expression increases with \(f\): a higher target fraction requires a higher dose. As \(f\) approaches 1, \(f/(1-f)\) grows without bound. Thus the model approaches, but does not reach, its asymptotic maximum effect at any finite dose.

If the target is specified as an absolute response \(R_T\), first convert it to a fraction of the maximum drug effect:

$$ f=\frac{R_T-E_0}{E_{\max}}, \qquad d_T=ED_{50}\frac{R_T-E_0}{E_0+E_{\max}-R_T}. $$

This inversion is defined for an increasing model only when \(E_0<R_T<E_0+E_{\max}\). A target at or below baseline does not specify a positive fraction of the drug effect; a target equal to or above the asymptotic maximum has no finite solution. These checks should be made before calculating a dose.

Worked Example: Finding a Dose for 75% of Emax

Suppose a fitted model for a hypothetical endpoint is \(R(d)=12+48d/(30+d)\), with response measured in endpoint units and dose in mg. The hypothetical study included doses from 10 to 120 mg. Find the dose predicted to achieve 75% of the maximum drug-related effect, and check it against the studied range.

Here \(E_0=12\) endpoint units, \(E_{\max}=48\) endpoint units, \(ED_{50}=30\) mg, and \(f=0.75\). The target is a fraction of the effect above baseline, so the formula is:

$$ d_{0.75}=ED_{50}\frac{f}{1-f} =30\text{ mg}\times\frac{0.75}{0.25} =30\text{ mg}\times3 =90\text{ mg}. $$

A substitution checks the result. At 90 mg, the modeled drug effect is \(48\times90/(30+90)=48\times0.75=36\) endpoint units. This is 75% of 48, and the total predicted response is \(12+36=48\) endpoint units. The target dose of 90 mg lies within the hypothetical studied range of 10–120 mg, so it is not an extrapolation beyond the range of tested doses. That comparison does not, by itself, establish that the dose is adequately supported or appropriate for clinical use.

When the Target Is an Absolute Response

The baseline matters whenever the target is stated in response units. Subtract \(E_0\) before dividing by \(E_{\max}\); otherwise, the calculation treats part of the baseline response as though it were drug effect. The resulting fraction must also be between zero and one for a finite positive target dose under this increasing model.

Worked Example: Converting an Absolute Target to a Dose

Consider a hypothetical fitted model with \(E_0=-2\) score units, \(E_{\max}=20\) score units, and \(ED_{50}=40\) mg. The desired total response is 13 score units. In a hypothetical dose-ranging study, doses ran from 25 to 100 mg. Calculate the target dose and assess the range.

First, the target drug effect above baseline is \(13-(-2)=15\) score units. The maximum drug effect is 20 score units, so:

$$ f=\frac{R_T-E_0}{E_{\max}} =\frac{13-(-2)}{20} =\frac{15}{20} =0.75. $$

The target is between the baseline and the asymptotic maximum, since \(-2<13<18\), where \(18=E_0+E_{\max}\). Its implied dose is:

$$ d_T=40\text{ mg}\times\frac{0.75}{1-0.75} =40\text{ mg}\times3 =120\text{ mg}. $$

At 120 mg, the modeled response is \(-2+20(120)/(40+120)=-2+15=13\) score units, as required. The calculated dose is 20 mg above the highest hypothetical studied dose of 100 mg. It is therefore an extrapolated target, not a dose directly supported by observations at or beyond 120 mg in this study.

Sigmoid Emax Models

Some dose-response analyses use a Hill coefficient, also called a shape parameter, to allow the curve to rise more or less steeply around \(ED_{50}\). For \(h>0\), the increasing sigmoid Emax model is:

$$ R(d)=E_0+E_{\max}\frac{d^h}{ED_{50}^{\,h}+d^h}. $$

In this model, \(ED_{50}\) still gives half the maximum drug effect: at \(d=ED_{50}\), the fraction \(d^h/(ED_{50}^{\,h}+d^h)\) equals \(1/2\). Setting that fraction equal to \(f\) and rearranging yields:

$$ f=\frac{d^h}{ED_{50}^{\,h}+d^h} \quad\Longrightarrow\quad \frac{f}{1-f}=\left(\frac{d}{ED_{50}}\right)^h \quad\Longrightarrow\quad d_f=ED_{50}\left(\frac{f}{1-f}\right)^{1/h}. $$

When \(h=1\), this reduces to the standard Emax formula. When \(h\) differs from 1, applying the standard formula without the exponent can give the wrong target dose. The parameterization must be checked: the formula above assumes the model uses \(d^h\) and \(ED_{50}^{\,h}\), not a differently defined scale or shape parameter.

Worked Example: Allowing for a Hill Coefficient

Suppose a hypothetical sigmoid Emax model has \(ED_{50}=50\) mg and \(h=2\), with an increasing drug effect. Find the dose predicted to achieve 80% of Emax. The hypothetical observed dose range was 10–150 mg.

Using the sigmoid-model formula with \(f=0.80\):

$$ d_{0.80} =50\text{ mg}\times \left(\frac{0.80}{0.20}\right)^{1/2} =50\text{ mg}\times\sqrt{4} =100\text{ mg}. $$

Checking the effect fraction at 100 mg gives \(100^2/(50^2+100^2)=10{,}000/12{,}500=0.80\). The target dose is within the hypothetical observed range of 10–150 mg. This calculation relies on the fitted model having the stated Hill parameterization and \(h=2\); a different estimated value of \(h\) would change the target dose.

Range Checks, Diagnostics, and Alternatives

A target-dose calculation is a transformation of fitted parameters, not a separate fit to data. Before treating its result as decision-relevant, check that the inputs and target are coherent:

  • Confirm that \(E_{\max}\), \(ED_{50}\), and the target refer to the same endpoint, population, dose definition, and model parameterization.
  • For the increasing standard model, confirm \(ED_{50}>0\), \(E_{\max}>0\), and \(0<f<1\). For an absolute target, verify \(E_0<R_T<E_0+E_{\max}\).
  • Compare the calculated dose with the minimum and maximum doses observed in the data used to fit the model. A target outside this range requires extrapolation.
  • Check model fit across the studied doses. Systematic departures between observed responses and fitted predictions, or a poorly supported plateau, can make the fitted parameters and their inverted target unreliable.
  • Check that the dose units match. The target dose inherits the units of \(ED_{50}\); a unit conversion changes the reported number, not the underlying dose.

The range check has a practical interpretation. A target within the observed range is an interpolation in dose, but it is not automatically precise or clinically validated. A target beyond the range is an extrapolation: its value depends more strongly on the assumed curve shape, especially on the estimated plateau and, for a sigmoid model, the Hill coefficient. In either case, the point estimate alone does not quantify uncertainty in the target dose.

Direct inspection of observed dose groups is a useful alternative when there are enough well-supported dose levels: it shows which studied doses approach the target without relying solely on a continuous curve. Interpolation between adjacent doses can be more transparent when the target lies between them, but it does not impose the Emax shape and may be unstable with sparse or noisy dose groups. The fitted Emax inversion uses the full parametric curve and can summarize information across doses, at the cost of relying on its model assumptions. Neither a group comparison nor an inverted curve removes the need to evaluate the clinical meaning of the target and the safety of the dose.

Common pitfall: A reader calculates a target fraction from \(R_T/E_{\max}\), then inverts the Emax equation, reasoning that the target is a fraction of the maximum response. This is wrong when the response has a nonzero baseline. The drug-effect fraction is \((R_T-E_0)/E_{\max}\); subtract the baseline first, then check that the fraction is strictly between zero and one.

Why the Inversion Matters

Inverting the Emax model translates a specified effect target into a dose on the scale used for development decisions. It makes the relation between a target fraction and \(ED_{50}\) explicit: under the standard model, 50% corresponds to \(ED_{50}\), 75% to \(3ED_{50}\), and 90% to \(9ED_{50}\). The rapid increase near the plateau is important: small increases in the desired fraction can imply large dose increases.

That mathematical relationship is not itself a dose recommendation. The target fraction must be justified for the endpoint, the fitted curve must be credible for the relevant doses, and the range check must be reported. An estimated dose within the observed range still needs to be considered alongside efficacy, safety, and the intended clinical use.

Check Your Understanding

Work each problem using the model stated, then compare with the solutions.

  1. A standard Emax model has \(ED_{50}=20\) mg. What dose gives 60% of Emax?
  2. A standard Emax model has \(E_0=5\) units and \(E_{\max}=30\) units. Find the dose for a target response of 20 units if \(ED_{50}=15\) mg.
  3. A model has \(E_0=4\), \(E_{\max}=10\), and \(ED_{50}=25\) mg. Can it produce a total response of 14 at a finite dose? Explain.
  4. A sigmoid Emax model has \(ED_{50}=40\) mg and \(h=2\). Calculate the dose for 75% of Emax and check the effect fraction by substitution.

Solutions. (1) \(d=20(0.60/0.40)=30\) mg. (2) The fraction is \(f=(20-5)/30=0.50\), so \(d=15(0.50/0.50)=15\) mg. (3) The asymptotic maximum total response is \(E_0+E_{\max}=14\). The model approaches 14 as dose increases but does not reach it at any finite dose, so there is no finite target dose. (4) \(d=40(0.75/0.25)^{1/2}=40\sqrt{3}=69.28\) mg. Substitution gives \(69.28^2/(40^2+69.28^2)\), approximately \(4{,}800/(1{,}600+4{,}800)=0.75\).