Assessment of non-linear combination effect terms for drug–drug interactions

Abstract

Drugs interact with their targets in different ways. A diversity of modeling approaches exists to describe the combination effects of two drugs. We investigate several combination effect terms (CET) regarding their underlying mechanism based on drug-receptor binding kinetics, empirical and statistical summation principles and indirect response models. A list with properties is provided and the interrelationship of the CETs is analyzed. A method is presented to calculate the optimal drug concentration pair to produce the half-maximal combination effect. This work provides a comprehensive overview of typically applied CETs and should shed light into the question as to which CET is appropriate for application in pharmacokinetic/pharmacodynamic models to describe a specific drug–drug interaction mechanism.

Appendices

Appendix 1: Drug receptor binding kinetics

General derivation

From the conservation of receptors and complexes Eq. (11) we obtain the receptor representation

$$R = \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) .$$

(51)

Rewriting the complexes Eqs. (8)–(10) with Eq. (51)

$$\begin{aligned} \frac{d}{dt}RC_{A} & = k_{onA} C_{A} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - k_{offA} RC_{A} - k_{onB} C_{B} RC_{A} \\ & \quad + k_{offB} RC_{AB} \\ \end{aligned}$$

(52)

$$\frac{d}{dt}RC_{B} = k_{onB} C_{B} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - k_{offB} RC_{B}$$

(53)

$$\frac{d}{dt}RC_{AB} = k_{onB} C_{B} RC_{A} - k_{offB} RC_{AB} .$$

(54)

Pseudo steady-state analysis of Eqs. (52)–(54) for the complexes and substituting Eq. (54) into Eq. (52) then leads to

$$\begin{aligned} 0 & = k_{onA} C_{A} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - k_{offA} RC_{A} \\ 0 & = k_{onB} C_{B} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - k_{offB} RC_{B} \\ 0 & = k_{onB} C_{B} RC_{A} - k_{offB} RC_{AB} . \\ \end{aligned}$$

With $K_{DA} = \frac{{k_{offA} }}{{k_{onA} }}$ and $K_{DB} = \frac{{k_{offB} }}{{k_{onB} }}$ we obtain

$$0 = C_{A} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - K_{DA} RC_{A}$$

(55)

$$0 = C_{B} \left( {R_{tot}^{0} - RC_{A} - RC_{B} - RC_{AB} } \right) - K_{DB} RC_{B}$$

(56)

$$0 = C_{B} RC_{A} - K_{DB} RC_{AB} \, .$$

(57)

Dividing Eq. (55) by $K_{DA}$ and Eqs. (56)–(57) by $K_{DB}$ we obtain the matrix notation Eq. (12)

$$\left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} & {\frac{{C_{B} }}{{K_{DB} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 0 & 1 \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {RC_{A} } \\ {RC_{B} } \\ {RC_{AB} } \\ \end{array} } \right) = R_{tot}^{0} \left( {\begin{array}{*{20}c} {\frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} \\ 0 \\ \end{array} } \right) .$$

We apply Cramer’s rule and obtain

$$M_{0} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} & {\frac{{C_{B} }}{{K_{DB} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 0 & 1 \\ \end{array} } \right) = 1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }} + \frac{{C_{A} C_{B} }}{{K_{DA} K_{DB} }} > 0$$

$$M_{1} = \det \left( {\begin{array}{*{20}c} {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} & {\frac{{C_{B} }}{{K_{DB} }}} \\ 0 & 0 & 1 \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }} \ge 0$$

$$M_{2} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}} & {\frac{{C_{B} }}{{K_{DB} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 0 & 1 \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }} \ge 0$$

$$M_{3} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} & {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} & {R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 0 & 0 \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }} \ge 0.$$

Note that $M_{0}$ is strictly positive and $M_{i}$, $i = 1,2,3$ are non-negative for $C_{A} ,C_{B} \ge 0$. The complexes then read

$$RC_{A} = \frac{{M_{1} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }} + \frac{{C_{A} C_{B} }}{{K_{DA} K_{DB} }}}}$$

(58)

$$RC_{B} = \frac{{M_{2} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }} + \frac{{C_{A} C_{B} }}{{K_{DA} K_{DB} }}}}$$

(59)

$$RC_{AB} = \frac{{M_{3} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }} + \frac{{C_{A} C_{B} }}{{K_{DA} K_{DB} }}}} .$$

(60)

Inserting Eqs. (58)–(60) in Eq. (13) then results in Eq. (14).

Appendix 2: Competitive CET

Derivation

Applying Cramer’s rule to Eq. (15) gives

$$M_{0} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} \\ \end{array} } \right) = 1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }} > 0$$

$$M_{1} = \det \left( {\begin{array}{*{20}c} {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ {R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}} & {1 + \frac{{C_{B} }}{{K_{DB} }}} \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }} \ge 0$$

$$M_{2} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} \\ {\frac{{C_{B} }}{{K_{DB} }}} & {R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}} \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }} \ge 0 .$$

The complexes then read

$$RC_{A} = \frac{{M_{1} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }}}}$$

(61)

$$RC_{B} = \frac{{M_{2} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{B} }}{{K_{DB} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{B} }}{{K_{DB} }}}} .$$

(62)

Inserting Eqs. (61)-(62) in Eq. (16) then results in Eq. (17).

Maximal effect

$$\begin{aligned} E^{Com} \left( {C_{A} ,C_{B} } \right) & \le \hbox{max} \left\{ {E_{maxA} ,E_{maxB} } \right\}\frac{{\left( {\frac{{C_{A} }}{{EC_{50,A} }}} \right)^{{\gamma_{A} }} + \left( {\frac{{C_{B} }}{{EC_{50,B} }}} \right)^{{\gamma_{B} }} }}{{1 + \left( {\frac{{C_{A} }}{{EC_{50,A} }}} \right)^{{\gamma_{A} }} + \left( {\frac{{C_{B} }}{{EC_{50,B} }}} \right)^{{\gamma_{B} }} }} \\ & < \hbox{max} \left\{ {E_{maxA} ,E_{maxB} } \right\} \, . \\ \end{aligned}$$

Diagonality

If drug $A$ equals drug $B$, we have

$$E^{Com} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right) = E_{maxA} \frac{{\frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }}}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }}}} = \frac{{E_{maxA} \frac{{C_{A} }}{{EC_{50A} }}}}{{1 + \frac{{C_{A} }}{{EC_{50A} }}}} = e\left( {C_{A} } \right) .$$

Agonistic-antagonist

For an antagonistic drug $B$ we have $e\left( {C_{B} } \right) = 0$ for $C_{B} \ge 0$ and therefore $E_{maxB} = 0$. Hence, we obtain with Eq. (18)

$$\begin{aligned} E^{Com} \left( {C_{A} ,C_{B} } \right) & = \frac{{E_{maxA} \left( {\frac{{C_{A} }}{{EC_{50,A} }}} \right)^{{\gamma_{A} }} +\, E_{maxB} \left( {\frac{{C_{B} }}{{EC_{50,B} }}} \right)^{{\gamma_{B} }} }}{{1 + \left( {\frac{{C_{A} }}{{EC_{50A} }}} \right)^{{\gamma_{A} }} + \left( {\frac{{C_{B} }}{{EC_{50B} }}} \right)^{{\gamma_{B} }} }} \\ & > \frac{{E_{maxA} \left( {\frac{{C_{A} }}{{EC_{50,A} }}} \right)^{{\gamma_{A} }} }}{{1 + \left( {\frac{{C_{A} }}{{EC_{50A} }}} \right)^{{\gamma_{A} }} + \left( {\frac{{C_{B} }}{{EC_{50B} }}} \right)^{{\gamma_{B} }} }} = E^{ComGad} \left( {C_{A} ,C_{B} } \right) . \\ \end{aligned}$$

Appendix 3: Uncompetitive CET

Derivation

The determinants for Cramer’s rule applied to Eq. (19) are

$$M_{0} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 1 \\ \end{array} } \right) = 1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }} > 0$$

$$M_{1} = \det \left( {\begin{array}{*{20}c} {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} & {\frac{{C_{A} }}{{K_{DA} }}} \\ 0 & 1 \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }} \ge 0$$

$$M_{2} = \det \left( {\begin{array}{*{20}c} {1 + \frac{{C_{A} }}{{K_{DA} }}} & {R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}} \\ { - \frac{{C_{B} }}{{K_{DB} }}} & 0 \\ \end{array} } \right) = R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }} \ge 0 .$$

The complexes then read

$$RC_{A} = \frac{{M_{1} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }}}}$$

(63)

$$RC_{B} = \frac{{M_{2} }}{{M_{0} }} = \frac{{R_{tot}^{0} \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }}}}{{1 + \frac{{C_{A} }}{{K_{DA} }} + \frac{{C_{A} }}{{K_{DA} }}\frac{{C_{B} }}{{K_{DB} }}}} .$$

(64)

Inserting Eqs. (63)–(64) in Eq. (20) then results in Eq. (21).

Appendix 4: Loewe CET

Derivation

Rearranging Eq. (27) with respect to $DC_{X}$ gives

$$DC_{X} = EC_{50X} \left( {\frac{E}{{E_{max} - E}}} \right)^{{\frac{1}{{\gamma_{X} }}}} .$$

(65)

Substituting Eq. (65) in Eq. (26) gives Eq. (28). Equation (28) can be written with $\gamma_{A} = \gamma_{B} = \gamma$ as

$$E^{{\frac{1}{\gamma }}} = \left( {E_{max} - E} \right)^{{\frac{1}{\gamma }}} \left( {\frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{B} }}{{EC_{50B} }}} \right)$$

resulting in

$$E = \left( {E_{max} - E} \right)\left( {\frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{B} }}{{EC_{50B} }}} \right)^{\gamma } .$$

(66)

Rearranging of Eq. (66) gives the Loewe CET Eq. (29).

Appendix 5: Bliss CET

Diagonality

We have

$$\begin{aligned} E^{Bliss} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right) & = \frac{{E_{max} \frac{{C_{A} }}{{2EC_{50A} }} + E_{max} \frac{{C_{A} }}{{2EC_{50A} }} + E_{max} \frac{{C_{A} }}{{4EC_{50A} }}\frac{{C_{A} }}{{EC_{50A} }}}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}}} \\ & = \frac{{E_{max} \left( {\frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}} \right)}}{{1 + \frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}}} = e\left( {\frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}} \right) . \\ \end{aligned}$$

Hence, the diagonality condition is satisfied with $\sigma = \frac{1}{4}$.

Appendix 6: Greco and Summation CET

Diagonality of the Greco CET

For $\gamma_{A} = \gamma_{B} = 1$ we obtain with Eq. (37)

$$\begin{aligned} E^{Greco} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right) & = \frac{{E_{max} \left( {\frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }} + \alpha \frac{{C_{A} }}{{4EC_{50A} }}\frac{{C_{A} }}{{EC_{50A} }}} \right)}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }} + \alpha \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}}} \\ & = e\left( {\frac{{C_{A} }}{{EC_{50A} }} + \alpha \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}} \right) \\ \end{aligned}$$

and the diagonality condition is fulfilled with $\sigma = \frac{\alpha }{4}$.

Diagonality of the summation CET

For equal drugs we can compute

$$E^{Sum} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right) = 2\frac{{E_{maxA} \frac{{C_{A} }}{{2EC_{50A} }}}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }}}} = 2e\left( {\frac{{C_{A} }}{2}} \right) > e\left( {C_{A} } \right) .$$

Thus, $E^{Sum} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right)$ cannot be written in the form assumed in the diagonality condition. To see the last inequality please note that with

$$f_{1} \left( x \right) = 2e\left( x \right) , \quad f_{2} \left( x \right) = e(2x)$$

we have

$$f_{1} \left( 0 \right) = f_{2} \left( 0 \right) ,$$

$$f_{1}^{{\prime }} \left( x \right) = 2e^{{\prime }} \left( x \right) ,\quad f_{2}^{{\prime }} \left( x \right) = 2e^{{\prime }} \left( {2x} \right) .$$

Since $e^{{\prime }} \left( x \right) > 0$ and $e'$ monotone decreasing, $f_{1} \left( x \right) > f_{2} (x)$, $x > 0$ follows and with $x = \frac{{C_{A} }}{2}$ the inequality.

Appendix 7: IDR CETs

Equivalent formulation for the inhibitory CET

To demonstrate the equivalence of Eq. (40) with Eqs. (41)–(42), we have to show

$$i\left( {C_{A} } \right) + i\left( {C_{B} } \right) - i\left( {C_{A} } \right)i\left( {C_{B} } \right) = I^{IDR} \left( {C_{A} ,C_{B} } \right) .$$

The denominator from Eq. (42) can be written as

$$De = \frac{{IC_{50A}^{{\gamma_{A} }} IC_{50B}^{{\gamma_{B} }} + C_{A}^{{\gamma_{A} }} IC_{50B}^{{\gamma_{B} }} + C_{B}^{{\gamma_{B} }} IC_{50A}^{{\gamma_{A} }} + C_{A}^{{\gamma_{A} }} C_{B}^{{\gamma_{B} }} }}{{IC_{50A}^{{\gamma_{A} }} IC_{50B}^{{\gamma_{B} }} }}$$

and the numerator reads

$$Nu = \frac{{I_{maxA} C_{A}^{{\gamma_{A} }} IC_{50B}^{{\gamma_{B} }} + I_{maxB} C_{B}^{{\gamma_{B} }} IC_{50A}^{{\gamma_{A} }} + \left( {I_{maxA} + I_{maxB} - I_{maxA} I_{maxB} } \right)C_{A}^{{\gamma_{A} }} C_{B}^{{\gamma_{B} }} }}{{IC_{50A}^{{\gamma_{A} }} IC_{50B}^{{\gamma_{B} }} }} .$$

Hence, we obtain

$$\begin{aligned} \frac{Nu}{De} & = \frac{{I_{maxA} C_{A}^{{\gamma_{A} }} \left( {IC_{50B}^{{\gamma_{B} }} + C_{B}^{{\gamma_{B} }} } \right) + I_{maxB} C_{B}^{{\gamma_{B} }} \left( {IC_{50A}^{{\gamma_{A} }} + C_{A}^{{\gamma_{A} }} } \right) - I_{maxA} I_{maxB} C_{A}^{{\gamma_{A} }} C_{B}^{{\gamma_{B} }} }}{{\left( {IC_{50A}^{{\gamma_{A} }} + C_{A}^{{\gamma_{A} }} } \right)\left( {IC_{50B}^{{\gamma_{B} }} + C_{B}^{{\gamma_{B} }} } \right)}} \\ & = \frac{{I_{maxA} C_{A}^{{\gamma_{A} }} }}{{\left( {IC_{50A}^{{\gamma_{A} }} + C_{A}^{{\gamma_{A} }} } \right)}} + \frac{{I_{maxB} C_{B}^{{\gamma_{B} }} }}{{\left( {IC_{50B}^{{\gamma_{B} }} + C_{B}^{{\gamma_{B} }} } \right)}} - \frac{{I_{maxA} C_{A}^{{\gamma_{A} }} }}{{\left( {IC_{50A}^{{\gamma_{A} }} + C_{A}^{{\gamma_{A} }} } \right)}}\frac{{I_{maxB} C_{B}^{{\gamma_{B} }} }}{{\left( {IC_{50B}^{{\gamma_{B} }} + C_{B}^{{\gamma_{B} }} } \right)}} . \\ \end{aligned}$$

Diagonality of the stimulation CET

If drug $A$ equals drug $B$, we have

$$\begin{aligned} E^{Stim} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right) & = \frac{{E_{maxA} \frac{{C_{A} }}{{2EC_{50A} }} + E_{maxA} \frac{{C_{A} }}{{2EC_{50A} }} + \left( {2E_{maxA} + E_{maxA}^{2} } \right)\frac{{C_{A} }}{{4EC_{50A} }}\frac{{C_{A} }}{{EC_{50A} }}}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A} }}}} \\ & = e\left( {\frac{{C_{A} }}{{EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A}^{2} }}} \right) + \frac{{\frac{1}{4}\left( {E_{maxA} + E_{maxA}^{2} } \right)\frac{{C_{A}^{2} }}{{EC_{50A}^{2} }}}}{{1 + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A} }}{{2EC_{50A} }} + \frac{{C_{A}^{2} }}{{4EC_{50A} }}}} . \\ \end{aligned}$$

Hence, $E^{Stim} \left( {\frac{{C_{A} }}{2},\frac{{C_{A} }}{2}} \right)$ cannot be written in form of a single drug effect term. With similar calculations we obtain the same conclusion for $I^{IDR}$.

Appendix 8: Relationships

To simplify the notation we set

$$x = \left( {\frac{{C_{A} }}{{EC_{50A} }}} \right)^{{\gamma_{A} }} \quad and \quad y = \left( {\frac{{C_{B} }}{{EC_{50B} }}} \right)^{{\gamma_{B} }} .$$

For $E_{maxA} ,E_{maxB} ,\gamma_{A} ,\gamma_{B} > 0$, Eq. (45) follows from

$$\frac{{E_{maxA} x}}{1 + x + y} + \frac{{E_{maxB} y}}{1 + x + y} < \frac{{E_{maxA} x}}{1 + x} + \frac{{E_{maxB} y}}{1 + y} < \frac{{E_{maxA} x}}{1 + x} + \frac{{E_{maxB} y}}{1 + y} + \frac{{E_{maxA} x}}{1 + x}\frac{{E_{maxB} y}}{1 + y} .$$

In the case of $E_{maxA} = E_{maxB} = E_{max} > 0$ and $\gamma_{A} ,\gamma_{B} > 0$ we have

$$E_{max} \frac{x + y}{1 + x + y} < E_{max} \frac{x + y + \eta xy}{1 + x + y + \eta xy}$$

(67)

$$< E_{max} \frac{x + y + 2\eta xy}{1 + x + y + \eta xy}$$

(68)

where $\eta > 0$ indicates existence of the multiplicative term $xy$. In Eqs. (46)–(47) the last inequality is obvious. Evaluation of Eq. (67) at $\eta = 1$ and Eq. (68) at $\eta = 1$ yields Eq. (46). In the case of $E_{maxA} = E_{maxB} = E_{max}$ and $\gamma_{A} = \gamma_{B} = 1$ we obtain Eq. (47) with evaluation of Eq. (67) at $\eta \in (0,1)$, and $\eta = 1$ in Eq. (68).

Appendix 9: Optimal concentration pair for half-maximal combination effect

In the following calculation we assume $\gamma_{A} = \gamma_{B} = \gamma = 1$ and set

$$x = \frac{{C_{A} }}{{EC_{50A} }} , \quad y = \frac{{C_{B} }}{{EC_{50B} }} .$$

(69)

Competitive CET

We assume $E_{maxB} = E_{maxA} = E_{max} > 0$. Using Eq. (18) by definition of the half maximal effect curve we have to solve

$$E^{Com} \left( {C_{A} ,C_{B} } \right) = \frac{{E_{max} x + E_{max} y}}{1 + x + y} = \frac{{E_{maxAB} }}{2} = \frac{{E_{max} }}{2}$$

which is equivalent to

$$2x + 2y = 1 + x + y .$$

This leads to

$$y = 1 - x$$

and the CET specific function for the half-maximal effect curve reads

$$\varphi_{Com} \left( x \right) = 1 - x .$$

(70)

The next step is to investigate the $CI$ values on the half-maximal effect curve which is due to Loewe Additivity Eq. (49) given by the objective function

$$h\left( x \right) = x + y = x + \varphi \left( x \right)$$

(71)

with its specific $\varphi$ under the constraints $x \ge 0, \varphi \left( x \right) \ge 0$. In case of the competitive CET we obtain

$$h_{Com} \left( x \right) = x + \left( {1 - x} \right) = 1$$

(72)

which gives the solution $CI^{Loewe} = 1$ and optimal concentration pairs are given by

$$\left( {\bar{x},\bar{y}} \right) = \left( {\frac{{\bar{C}_{A} }}{{EC_{50A} }},\frac{{\bar{C}_{B} }}{{EC_{50B} }}} \right) = \left( {\varepsilon ,1 - \varepsilon } \right), \quad 0 \le \varepsilon \le 1 \;arbitrary .$$

Greco CET

With Eq. (69) we obtain from Eq. (37) with $E = \frac{{E_{max} }}{2}$

$$1 = x + y + \alpha xy = x + \left( {1 + \alpha x} \right)y$$

resulting in

$$\varphi_{Greco} \left( x \right) = \frac{1 - x}{1 + \alpha x} , \quad - 1 \le \alpha \le 1 .$$

With Eq. (71) the objective function reads

$$h_{Greco} \left( x \right) = x + \varphi_{Greco} (x)$$

and we calculate

$$h_{Greco}^{'} \left( x \right) = 1 + \frac{{ - \left( {1 + \alpha x} \right) - \left( {1 - x} \right)\alpha }}{{\left( {1 + \alpha x} \right)^{2} }} = 1 + \frac{{ - \left( {1 + \alpha } \right)}}{{\left( {1 + \alpha x} \right)^{2} }} = 0$$

if and only if

$$\left( {1 + \alpha x} \right)^{2} = 1 + \alpha .$$

This leads to

$$\bar{x} = \frac{{\bar{C}_{A} }}{{EC_{50A} }} = \frac{{\sqrt {1 + \alpha } - 1}}{\alpha } , \quad 0 < \left| \alpha \right| \le 1 .$$

Using

$$\varphi_{Greco} \left( {\frac{{\sqrt {1 + \alpha } - 1}}{\alpha }} \right) = \frac{{\sqrt {1 + \alpha } - 1}}{\alpha } = \frac{{\bar{C}_{B} }}{{EC_{50B} }}$$

we obtain

$$CI^{Loewe} = 2\bar{x} = \frac{2}{\alpha }\left( {\sqrt {1 + \alpha } - 1} \right) , \quad 0 < \left| \alpha \right| \le 1 .$$

For $\alpha = 0$ we have $h_{Greco} \left( x \right) = x + \varphi_{Greco} \left( x \right) = 1$ which leads to $CI^{Loewe} = 1$.

Bliss CET

Using Eq. (33) we obtain

$$\frac{x + y + xy}{1 + x + y + xy} = \frac{1}{2}$$

which is equivalent to

$$x + y + xy = 1 .$$

(73)

This leads to

$$\varphi_{Bliss} (x) = \frac{1 - x}{1 + x} .$$

According to Eq. (50) and Eq. (73) the objective function reads

$$h_{Bliss} \left( x \right) = x + y + xy = 1 .$$

We obtain the solution

$$\left( {\bar{x},\bar{y}} \right) = \left( {\varepsilon ,\varphi_{Bliss} \left( \varepsilon \right)} \right) ,\quad 0 \le \varepsilon \le 1$$

with $CI^{Bliss} = 1$.

Summation CET

With Eq. (69) and $E_{maxA} = E_{maxB} = E_{max}$ we obtain from Eq. (38)

$$E_{max} \left( {\frac{x}{1 + x} + \frac{y}{1 + y}} \right) = \frac{{E_{maxAB} }}{2} = E_{max} .$$

Rearranging with respect to $y$

$$x + y + 2xy = 1 + x + y + xy$$

results in

$$y = \frac{1}{x} = \varphi_{Sum} \left( x \right) .$$

We set with Eq. (50)

$$h_{Sum} \left( x \right) = x + y + xy = x + \varphi_{Sum} \left( x \right) + x\varphi_{Sum} \left( x \right) = x + \frac{1}{x} + 1$$

and obtain

$$h_{Sum}^{'} \left( x \right) = 1 - \frac{1}{{x^{2} }} = 0$$

and therefore

$$\bar{x} = 1 .$$

Hence, we obtain

$$\varphi_{Sum} \left( 1 \right) = 1$$

and the optimal pair is

$$\left( {\bar{x},\bar{y}} \right) = \left( {1,1} \right)$$

with $CI^{Bliss} = x + y + 1 = 3$. Please note that a classification of the area of antagonistic, additive, or synergistic does not hold since the objective functions Eqs. (49), (50) defining the classifications have to be scaled appropriately, if $E_{maxAB} > \hbox{max} \left\{ {E_{maxA} ,E_{maxB} } \right\}$.