receptor kinetics laplace transform method pdf

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1 Deriving Ligand Binding Kinetics Equations Using The Laplace Transform 1 Introduction In pharmacological systems we often want to measure time-related drug parameters, such as the rates of activity, for example the rate of dissociation of a ligand from a receptor. Deriving equations that describe progress of activity over time typically starts with writing differential equations that describe the system. These differential equations describe the rate of change of time-dependent variables (“dy / dt”), for example the number of ligand-occupied receptors in a ligand dissociation experiment. The differential equations are then solved for the time-dependent variable by integration, yielding an analytic equation that can be used by curve-fitting programs to estimate the value of rate parameters. In all but the simplest pharmacological models, the integration process can be formidable and is often a frustrating barrier to the pharmacologist who wants to evaluate and explore time-dependent pharmacological processes. Fortunately, mathematical tools are available to solve differential equations, which require only a facility with simple algebra and the rules of the method. The Laplace transform is one such tool. It is used frequently in multi-compartment pharmacokinetic modeling (Popovic, 1999). In one of the simplest applications, the Laplace transform has been used to derive the equation defining drug levels on oral dosing. For a readily-understandable and detailed description of the method applied to pharmaceutical systems, see Mayersohn and Gibaldi, 1970. The goal here is to provide systematic instruction on the rules of the method by way of familiar pharmacological models, and to highlight the benefits of the approach. In the Laplace transform method, the differential equation is transformed into a mathematical framework that allows the time-dependent variables to be manipulated by simple algebra. The Laplace transform substitutes the time-derivative domain of the rate equation (the “dy / dt” term) with the complex domain of the Laplace operator, s. Once transformed into this domain, time- dependent variables can be handled using the same algebra pharmacologists employ to derive equilibrium-model equations. Once solved for the time-dependent variable of interest, a second transform is used to generate the analytic equation that can be used for curve fitting. The Laplace transform is used for solving first- or zero-order differential equations. It cannot be used for solving second-order differential equations, for example the kinetics of cooperative ligand binding in allosteric models. The method is exemplified here using four receptor-ligand binding models: Model 1: Ligand-receptor association Model 2: Dissociation of ligand from receptor Model 3: Unlabeled ligand pre-incubation and washout Model 4: Competition kinetics – labeled ligand association in the presence of unlabeled ligand 1 Sam Hoare, 2016, [email protected] or (US) 619-203-2886

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1

Deriving Ligand Binding Kinetics Equations Using The Laplace Transform1

Introduction

In pharmacological systems we often want to measure time-related drug parameters, such as the

rates of activity, for example the rate of dissociation of a ligand from a receptor. Deriving equations

that describe progress of activity over time typically starts with writing differential equations that

describe the system. These differential equations describe the rate of change of time-dependent

variables (“dy / dt”), for example the number of ligand-occupied receptors in a ligand dissociation

experiment. The differential equations are then solved for the time-dependent variable by

integration, yielding an analytic equation that can be used by curve-fitting programs to estimate the value of rate parameters. In all but the simplest pharmacological models, the integration process can

be formidable and is often a frustrating barrier to the pharmacologist who wants to evaluate and

explore time-dependent pharmacological processes.

Fortunately, mathematical tools are available to solve differential equations, which require only a

facility with simple algebra and the rules of the method. The Laplace transform is one such tool. It is

used frequently in multi-compartment pharmacokinetic modeling (Popovic, 1999). In one of the

simplest applications, the Laplace transform has been used to derive the equation defining drug

levels on oral dosing. For a readily-understandable and detailed description of the method applied to

pharmaceutical systems, see Mayersohn and Gibaldi, 1970. The goal here is to provide systematic

instruction on the rules of the method by way of familiar pharmacological models, and to highlight

the benefits of the approach.

In the Laplace transform method, the differential equation is transformed into a mathematical

framework that allows the time-dependent variables to be manipulated by simple algebra. The

Laplace transform substitutes the time-derivative domain of the rate equation (the “dy / dt” term)

with the complex domain of the Laplace operator, s. Once transformed into this domain, time-

dependent variables can be handled using the same algebra pharmacologists employ to derive

equilibrium-model equations. Once solved for the time-dependent variable of interest, a second

transform is used to generate the analytic equation that can be used for curve fitting.

The Laplace transform is used for solving first- or zero-order differential equations. It cannot be used

for solving second-order differential equations, for example the kinetics of cooperative ligand

binding in allosteric models.

The method is exemplified here using four receptor-ligand binding models:

Model 1: Ligand-receptor association

Model 2: Dissociation of ligand from receptor

Model 3: Unlabeled ligand pre-incubation and washout

Model 4: Competition kinetics – labeled ligand association in the presence of unlabeled ligand

1 Sam Hoare, 2016, [email protected] or (US) 619-203-2886

2

Model 1: Ligand-receptor association

R, receptor

L, labeled ligand; units, M

k1, labeled ligand association rate constant; units, M-1min-1

k2, labeled ligand dissociation rate constant; units, min-1

Minimal depletion of [𝐿] by [𝑅𝐿]

Step 1: Formularizing the model in a differential equation

𝑑[𝑅𝐿]

𝑑𝑡= [𝑅][𝐿]𝑘1 − [𝑅𝐿]𝑘2

Here the time dependent variables are [𝑅𝐿]and [𝑅]. Our goal is to obtain an analytic equation with one

time-dependent variable, that of the system component being measured, [RL]. The equation can be

reduced to a single time-dependent variable using the conservation of mass equation for the

receptor:

𝑁 = [𝑅] + [𝑅𝐿]

[𝑅] = [𝑁] − [𝑅𝐿]

Substituting into the differential equation gives,

𝑑[𝑅𝐿]

𝑑𝑡= 𝑁[𝐿]𝑘1 − [𝑅𝐿]([𝐿]𝑘1 + 𝑘2)

where N is the total concentration of receptors.

3

Step 2: Taking the Laplace transform

Six rules necessary for successful application of the Laplace transform method are given in Appendix 1. Three rules are followed in taking the Laplace transform of the differential equation of the ligand-

receptor association model:

1. Time-dependent variables are signified with an accent, as follows2:

[𝑅𝐿] → [𝑅𝐿]̅̅ ̅̅ ̅̅

2. The differential expression is substituted with the Laplace operator, s

𝑑[𝑅𝐿]

𝑑𝑡→ 𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅

3. Terms that are independent of time are divided by s

𝑁[𝐿]𝑘1 →𝑁[𝐿]𝑘1

𝑠

Applying these three rules we obtain the Laplace transform for [𝑅𝐿]:

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠− [𝑅𝐿]̅̅ ̅̅ ̅̅ ([𝐿]𝑘1 + 𝑘2)

The advantage of the Laplace transform over the differential equation is that the time-dependent

variable [𝑹𝑳]̅̅ ̅̅ ̅̅ is treated as a conventional algebraic term. Re-arrangement of the Laplace transform is

therefore simpler than re-arrangement of the differential equation. This property is particularly

useful when the expression for one time-dependent variable is substituted into the expression for a

second time-dependent variable. This benefit is evident in models that include binding of a

competitive inhibitor to the receptor, where the Laplace transform for unlabeled ligand is substituted

into that for the labeled ligand in order to reduce the equation to the time-dependent variable [𝑅𝐿]

(see Model 3 and Model 4).

Step 3. Solving the Laplace transform for [𝑅𝐿]

The Laplace transform is solved for [𝑅𝐿]̅̅ ̅̅ ̅̅ algebraically:

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠− [𝑅𝐿]̅̅ ̅̅ ̅̅ ([𝐿]𝑘1 + 𝑘2)

2 Alternatively, the accent character ^ is used.

4

[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠(𝑠 + [𝐿]𝑘1 + 𝑘2)

Step 4. Taking the inverse Laplace transform: Transformation to the time domain

In this final step the Laplace transform is transformed to the time domain, i.e. the domain in which t is a straightforward independent variable. This process is described as taking the inverse Laplace

transform. This step yields the analytic equation to which the experimental data can be fitted using

nonlinear regression programs that are commonly used by pharmacologists, for example Prism from

GraphPad; XLfit from IDBS; Solver (a plug-in for Microsoft Excel); and SigmaPlot. Finding the inverse

Laplace transform is usually straightforward; the re-arranged Laplace transform from Step 3 is

simply identified in a table of inverse Laplace transforms. These tables are available in textbooks, and

in teaching aids available on the internet. Appendix 2A is a table of inverse Laplace transforms

commonly-encountered in pharmacological models. The WolframAlpha computational knowledge

engine at http://www.wolframalpha.com/input/?i=inverse+Laplace+transform contains a large

library of Laplace transforms.

In these tables, the Laplace transform is written as a function of s and the inverse transform a function

of time, t. The Laplace transform for [𝑅𝐿] (above) takes the general form,

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)

where the constant is 𝑁[𝐿]𝑘1 and a is [𝐿]𝑘1 + 𝑘2.

This expression is then found in a table of inverse Laplace transforms – e.g. Appendix 2 or

http://www.wolframalpha.com/input/?i=inverse+Laplace+transform+c%2F(s*(s%2Ba)). In this

case, the inverse Laplace transform is,

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎(1 − 𝑒−𝑎𝑡)

(The numerator of the Laplace transform is the numerator of the inverse Laplace transform.

Consequently, in most Laplace transform tables the term constant is replaced by 1.)

Substituting with the terms of the model, we obtain the analytic equation for labeled ligand

association with the receptor:

[𝑅𝐿]𝑡 =𝑁[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2(1 − 𝑒−([𝐿]𝑘1+𝑘2)𝑡)

This equation is the same as that derived by integration. It is the familiar exponential equation for

ligand-receptor association.

5

Model 2: Labeled ligand dissociation

Variables defined in Model 1

In Model 1 above, the appearance of the variable of interest, [𝑅𝐿], is measured. The Laplace transform

approach to disappearance of the variable of interest is exemplified by the model of ligand dissociation

from receptor.

Step 1: Differential equation

−𝑑[𝑅𝐿]

𝑑𝑡= [𝑅𝐿]𝑘2

𝑑[𝑅𝐿]

𝑑𝑡= −[𝑅𝐿]𝑘2

Step 2: Laplace transform

Rule 4 is employed when taking the Laplace transform for disappearance of a time-dependent

variable: A constant is incorporated that defines the amount of the disappearing variable at the

initiation of the process, as follows:

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ = [𝑅𝐿]𝑡=0 − [𝑅𝐿]̅̅ ̅̅ ̅̅ 𝑘2

where [𝑅𝐿]𝑡=0 is the constant denoting the value of the time-dependent variable at the initiation of

the process being formularized, in this case the amount of ligand-occupied receptor at t = 0. Note this

constant is not divided by s.

Step 3. Solving the Laplace transform, [RL]

This equation can be solved for [𝑅𝐿]̅̅ ̅̅ ̅̅ , the time-dependent variable of interest:

6

[𝑅𝐿]̅̅ ̅̅ ̅̅ =[𝑅𝐿]𝑡=0

𝑠 + 𝑘2

Step 4. Inverse Laplace transform

The inverse Laplace transform is obtained from Laplace transform tables, e.g. Appendix 2 or

http://www.wolframalpha.com/input/?i=inverse+Laplace+transform+c%2F(s%2Ba)

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠 + 𝑎

𝑓(𝑡) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. 𝑒−𝑎𝑡

Substituting with the terms of the model, we obtain the analytic equation for labeled ligand

dissociation from receptor:

[𝑅𝐿]𝑡 = [𝑅𝐿]𝑡=0. 𝑒−𝑘2.𝑡

This is, of course, the familiar exponential ligand dissociation equation.

7

Model 3: Unlabeled ligand pre-incubation and washout

I, unlabeled ligand; units, M

k4, unlabeled ligand dissociation rate constant; units, min-1

Other variables given in Model 1

In the pre-incubation step, unlabeled ligand associates with the receptor to form the receptor-

unlabeled ligand complex RI. Free unlabeled ligand (that not bound to the receptor) is then washed

out. Subsequently, labeled ligand is presented to the receptor and their association measured

(Malany et al. 2009, Packeu et al. 2010, Uhlen at al. 2016).

In this model, the process of substituting one Laplace transform into another is exemplified. The

Laplace transform for dissociation of the receptor-unlabeled ligand complex is substituted into the

transform for labeled ligand association with the receptor, enabling the latter to be solved for the single time-dependent variable, [𝑅𝐿].

Step 1: Differential equations

The differential equation for unlabeled ligand dissociation is,

−𝑑[𝑅𝐼]

𝑑𝑡= [𝑅𝐼]𝑘4

The differential equation for labeled ligand association is, as given in Model 1:

𝑑[𝑅𝐿]

𝑑𝑡= [𝑅][𝐿]𝑘1 − [𝑅𝐿]𝑘2

We wish to reduce the expression for [𝑅𝐿] to a single time-dependent variable. Free receptor, [𝑅] can

be replaced by using the conservation of mass equation for the receptor:

𝑁 = [𝑅] + [𝑅𝐿] + [𝑅𝐼]

8

[𝑅] = 𝑁 − [𝑅𝐿] − [𝑅𝐼]

Substituting into the differential equation:

𝑑[𝑅𝐿]

𝑑𝑡= 𝑁[𝐿]𝑘1 − [𝑅𝐼][𝐿]𝑘1 − [𝑅𝐿]([𝐿]𝑘1 + 𝑘2)

This substitution introduces a second time-dependent variable, [𝑅𝐼]. It is not immediately obvious

how to integrate this differential equation to yield the analytic equation for [𝑅𝐿]. However, it is

straightforward to arrive at the analytical equation by using the Laplace transform method. The

transform for [𝑅𝐼] is substituted into that for [𝑅𝐿], yielding an expression in which [𝑅𝐿] is the only

time-dependent variable. This approach is described as follows:

Step 2: Laplace transforms

The Laplace transform for [𝑅𝐼] is, per Model 2,

𝑠[𝑅𝐼]̅̅ ̅̅ ̅ = [𝑅𝐼]𝑡=0 − [𝑅𝐼]̅̅ ̅̅ ̅𝑘4

Solving for [𝑅𝐼]̅̅ ̅̅ ̅,

[𝑅𝐼]̅̅ ̅̅ ̅ =[𝑅𝐼]𝑡=0

𝑠 + 𝑘4

The Laplace transform for [𝑅𝐿] is,

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠− [𝑅𝐼]̅̅ ̅̅ ̅[𝐿]𝑘1 − [𝑅𝐿]̅̅ ̅̅ ̅̅ ([𝐿]𝑘1 + 𝑘2)

Note that the [𝑅𝐼]̅̅ ̅̅ ̅ term is not divided by s because [𝑅𝐼]̅̅ ̅̅ ̅ is a time-dependent variable. The transform for [𝑅𝐼] is then substituted into the transform for [𝑅𝐿]:

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠−

[𝑅𝐼]𝑡=0[𝐿]𝑘1

𝑠 + 𝑘4− [𝑅𝐿]̅̅ ̅̅ ̅̅ ([𝐿]𝑘1 + 𝑘2)

This equation contains only [𝑅𝐿]̅̅ ̅̅ ̅̅ as a time-dependent variable.

Step 3. Solving the Laplace transform for [RL]

Solving for [𝑅𝐿]̅̅ ̅̅ ̅̅ :

9

[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠(𝑠 + [𝐿]𝑘1 + 𝑘2)−

[𝑅𝐼]𝑡=0[𝐿]𝑘1

(𝑠 + 𝑘4)(𝑠 + [𝐿]𝑘1 + 𝑘2)

Step 4. Inverse Laplace transform

This example demonstrates the fifth rule of the method. The sum of terms in a Laplace transform for

a single time-dependent variable is equal to the sum of terms in the inverse Laplace transform.

Generally,

𝐹(𝑠, �̅�)1 + 𝐹(𝑠, �̅�)2 = 𝑓(𝑡, 𝑋)1 + 𝑓(𝑡, 𝑋)2

In this case, we take the inverse Laplace transform for the first term and subtract the inverse

transform of the second.

First term: From Appendix 2 or as given at http://www.wolframalpha.com/input/?i= inverse

+Laplace+transform+c%2F(s*(s%2Ba))

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎(1 − 𝑒−𝑎𝑡)

Substituting with the model parameters,

𝑓(𝑡) =𝑁[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2(1 − 𝑒−([𝐿]𝑘1+𝑘2)𝑡)

Second term: From Appendix 2 or as given at http://www.wolframalpha.com/input/?i=inverse

++Laplace+transform+c%2F((s%2Ba)*(s%2Bb))

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡)

10

Here we apply the sixth rule of the method. Time-independent variables (in this case a and b) are

interchangeable when taking the inverse Laplace transform. In this model, the selection of which

terms should be a and which b in the inverse transformation makes no difference because these terms

are interchangeable. Specifically, it can be shown algebraically that,

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡) =

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎 − 𝑏(𝑒−𝑏.𝑡 − 𝑒−𝑎.𝑡)

This interchangeability is especially valuable when a and b are the positive and negative roots of

quadratic equations. This scenario arises in the derivation of the competition kinetics equation, as

shown in Model 4 below.

In this case, we can take the inverse Laplace transform as,

𝑓(𝑡) =[𝑅𝐼]𝑡=0[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2 − 𝑘4(𝑒−𝑘4.𝑡 − 𝑒−([𝐿]𝑘1+𝑘2).𝑡)

The analytic equation for [𝑅𝐿] is obtained by combining the two inverse Laplace transforms:

[𝑅𝐿]𝑡 =𝑁[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2(1 − 𝑒−([𝐿]𝑘1+𝑘2)𝑡) −

[𝑅𝐼]𝑡=0[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2 − 𝑘4(𝑒−𝑘4.𝑡 − 𝑒−([𝐿]𝑘1+𝑘2).𝑡)

Rearranging gives the equation in Malany et al. 2009:

[𝑅𝐿]𝑡 =𝑁[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2(1 − 𝑒−([𝐿]𝑘1+𝑘2)𝑡) +

[𝑅𝐼]𝑡=0[𝐿]𝑘1

[𝐿]𝑘1 + 𝑘2 − 𝑘4(𝑒−([𝐿]𝑘1+𝑘2).𝑡 − 𝑒−𝑘4.𝑡)

This equation has been validated by showing that data simulated using the equation matches that

from numerical solution of the differential equations (Packeu et al. 2010). Experimentally it has been

shown that fitted values of k4 match those from alternative, well-validated approaches (Malany et al.

2009, Packeu et al. 2010, Uhlen at al. 2016).

11

Model 4: Competition kinetics

k3, association rate constant of unlabeled ligand; units, M-1min-1

Minimal depletion of [𝐼] by [𝑅𝐼].

Other variables given in Model 1 and Model 3.

The competition kinetics equation (Motulsky and Mahan, 1984) is widely used to measure the

association rate constant and dissociation rate constant of unlabeled ligands. This equation was

derived using Laplace transforms. Here the derivation is presented in detail. It exemplifies the rules

and benefits of the method that are given in the three models above.

There are three time-dependent variables in the competition kinetics model – [𝑅], [𝑅𝐿], and [𝑅𝐼]. The

goal of the derivation is an analytic equation in terms of one time-dependent variable, [RL]. As for the

pre-incubation kinetics model above, [𝑅] is replaced using the conservation of mass equation for the

receptor, and [𝑅𝐼] is replaced by substituting the Laplace transform for [𝑅𝐼] into that for [𝑅𝐿].

Step 1: Differential equations

The differential equations for [𝑅𝐿] and [𝑅𝐼] are, respectively,

𝑑[𝑅𝐿]

𝑑𝑡= [𝑅][𝐿]𝑘1 − [𝑅𝐿]𝑘2

𝑑[𝑅𝐼]

𝑑𝑡= [𝑅][𝐼]𝑘3 − [𝑅𝐼]𝑘4

Replacing [𝑅] using the conservation of mass equation,

𝑁 = [𝑅] + [𝑅𝐿] + [𝑅𝐼]

[𝑅] = 𝑁 − [𝑅𝐿] − [𝑅𝐼]

12

𝑑[𝑅𝐿]

𝑑𝑡= 𝑁[𝐿]𝑘1 − [𝑅𝐼][𝐿]𝑘1 − [𝑅𝐿]([𝐿]𝑘1 + 𝑘2)

𝑑[𝑅𝐼]

𝑑𝑡= 𝑁[𝐼]𝑘3 − [𝑅𝐿][𝐼]𝑘3 − [𝑅𝐼]([𝐼]𝑘3 + 𝑘4)

Step 2: Laplace transforms

For the purpose of clarity, the following new variables are used, per Motulsky and Mahan, 1984:

𝐾𝐴 = [𝐿]𝑘1 + 𝑘2

𝐾𝐵 = [𝐼]𝑘3 + 𝑘4

The Laplace transforms for [𝑅𝐿] and [𝑅𝐼] are, respectively,

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠− [𝑅𝐼]̅̅ ̅̅ ̅[𝐿]𝑘1 − [𝑅𝐿]̅̅ ̅̅ ̅̅ 𝐾𝐴

𝑠[𝑅𝐼]̅̅ ̅̅ ̅ =𝑁[𝐼]𝑘3

𝑠− [𝑅𝐿]̅̅ ̅̅ ̅̅ [𝐼]𝑘3 − [𝑅𝐼]̅̅ ̅̅ ̅𝐾𝐵

Solving the [𝑅𝐼] transform for [𝑅𝐼]̅̅ ̅̅ ̅ so that it can be substituted into the transform for [𝑅𝐿],

[𝑅𝐼]̅̅ ̅̅ ̅ =𝑁[𝐼]𝑘3

𝑠(𝑠 + 𝐾𝐵)−

[𝑅𝐿]̅̅ ̅̅ ̅̅ [𝐼]𝑘3

𝑠 + 𝐾𝐵

Substituting into the transform for [𝑅𝐿],

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

𝑠−

𝑁[𝐿][𝐼]𝑘1𝑘3

𝑠(𝑠 + 𝐾𝐵)+

[𝑅𝐿]̅̅ ̅̅ ̅̅ [𝐿][𝐼]𝑘1𝑘3

𝑠 + 𝐾𝐵− [𝑅𝐿]̅̅ ̅̅ ̅̅ 𝐾𝐴

This equation contains only [𝑅𝐿]̅̅ ̅̅ ̅̅ as a time-dependent variable.

Step 3. Solving the Laplace transform for [RL]

Solving for [𝑅𝐿]̅̅ ̅̅ ̅̅ involves a factorization procedure. Before this, we arrive at an intermediate step:

13

[𝑅𝐿]̅̅ ̅̅ ̅̅ {(𝑠 + 𝐾𝐴)(𝑠 + 𝐾𝐵) − [𝐿][𝐼]𝑘1𝑘3} =𝑁[𝐿]𝑘1(𝑠 + 𝐾𝐵) − 𝑁[𝐿][𝐼]𝑘1𝑘3

𝑠

Factorization step

The [𝑅𝐿]̅̅ ̅̅ ̅̅ multiplier on the left-hand side is, as written, not readily amenable to transformation to an

inverse Laplace transform, not least because expressions of this form are not found in transform

tables. Motulsky and Mahan used a factorization procedure to enable straightforward

transformation, as follows:

Expanding the [𝑅𝐿]̅̅ ̅̅ ̅̅ multiplier,

(𝑠 + 𝐾𝐴)(𝑠 + 𝐾𝐵) − [𝐿][𝐼]𝑘1𝑘3 = 𝑠2 + 𝑠(𝐾𝐴 + 𝐾𝐵) + 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3

This expression can be factorized by introducing two compound variables, 𝐾𝐹 + 𝐾𝑆 and 𝐾𝐹𝐾𝑆, defined

as follows:

𝐾𝐹 + 𝐾𝑆 = 𝐾𝐴 + 𝐾𝐵

𝐾𝐹𝐾𝑆 = 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3

Substituting into the expanded [𝑅𝐿]̅̅ ̅̅ ̅̅ multiplier,

𝑠2 + 𝑠(𝐾𝐴 + 𝐾𝐵) + 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3 = 𝑠2 + 𝑠(𝐾𝐹 + 𝐾𝑆) + 𝐾𝐹𝐾𝑆

This expression can be readily factorized:

𝑠2 + 𝑠(𝐾𝐹 + 𝐾𝑆) + 𝐾𝐹𝐾𝑆 = (𝑠 + 𝐾𝑆)(𝑠 + 𝐾𝐹)

Resolving KF and KS in terms of the model parameters

In order to obtain an analytical equation that can yield estimates of the model parameters, KF and KS

need to be resolved in terms of the model parameters. KF can be resolved as follows,

Solving 𝐾𝐹 + 𝐾𝑆 = 𝐾𝐴 + 𝐾𝐵 for KS,

𝐾𝑆 = 𝐾𝐴 + 𝐾𝐵 − 𝐾𝐹

14

Substituting into 𝐾𝐹𝐾𝑆 = 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3,

𝐾𝐹(𝐾𝐴 + 𝐾𝐵 − 𝐾𝐹) = 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3

Solving for 𝐾𝐹 gives a quadratic equation:

0 = 𝐾𝐹2 − 𝐾𝐹(𝐾𝐴 + 𝐾𝐵) + 𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3

KF can be found as the root of the quadratic equation:

𝐾𝐹 =𝐾𝐴 + 𝐾𝐵 ± √(𝐾𝐴 + 𝐾𝐵)2 − 4(𝐾𝐴𝐾𝐵 − [𝐿][𝐼]𝑘1𝑘3)

2

which simplifies to,

𝐾𝐹 = 0.5 {𝐾𝐴 + 𝐾𝐵 ± √(𝐾𝐴 − 𝐾𝐵)2 + 4[𝐿][𝐼]𝑘1𝑘3}

The same procedure resolves KS into:

𝐾𝑆 = 0.5 {𝐾𝐴 + 𝐾𝐵 ± √(𝐾𝐴 − 𝐾𝐵)2 + 4[𝐿][𝐼]𝑘1𝑘3}

KF and KS differ only in whether the square root term is added or subtracted. Selection of which term

has the root added and which has the root subtracted makes no difference because KF and KS are

interchangeable when the inverse Laplace transform is taken (as shown below). In the original

derivation, the square root term is added in KF and subtracted in KS.

Final form of Laplace transform for [RL]

Now the Laplace transform for [𝑅𝐿] can be formularized in an equation readily-amenable for taking

the inverse Laplace transform. As given above, the intermediate equation of the Laplace transform

for [𝑅𝐿] is,

[𝑅𝐿]̅̅ ̅̅ ̅̅ {(𝑠 + 𝐾𝐴)(𝑠 + 𝐾𝐵) − [𝐿][𝐼]𝑘1𝑘3} =𝑁[𝐿]𝑘1(𝑠 + 𝐾𝐵) − 𝑁[𝐿][𝐼]𝑘1𝑘3

𝑠

Substituting the factorization expression for the [𝑅𝐿]̅̅ ̅̅ ̅̅ multiplier,

[𝑅𝐿]̅̅ ̅̅ ̅̅ (𝑠 + 𝐾𝑆)(𝑠 + 𝐾𝐹) =𝑁[𝐿]𝑘1(𝑠 + 𝐾𝐵) − 𝑁[𝐿][𝐼]𝑘1𝑘3

𝑠

15

Solving for [𝑅𝐿]̅̅ ̅̅ ̅̅ ,

[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1(𝑠 + 𝐾𝐵) − 𝑁[𝐿][𝐼]𝑘1𝑘3

𝑠(𝑠 + 𝐾𝑆)(𝑠 + 𝐾𝐹)

Rearranging yields the final version of the Laplace transform:

[𝑅𝐿]̅̅ ̅̅ ̅̅ =𝑁[𝐿]𝑘1

(𝑠 + 𝐾𝑆)(𝑠 + 𝐾𝐹)+

𝑁[𝐿]𝑘1𝑘4

𝑠(𝑠 + 𝐾𝑆)(𝑠 + 𝐾𝐹)

Step 4. Inverse Laplace transform

We now take the inverse Laplace transform.

First term: From Appendix 2 or as given at http://www.wolframalpha.com/input/?i=inverse

++Laplace+transform+c%2F((s%2Ba)*(s%2Bb))

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡)

Selection of whether a is 𝐾𝐹 or 𝐾𝑆, and respectively b is 𝐾𝑆 or 𝐾𝐹, does not affect the inverse

transformation, because,

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡) =

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎 − 𝑏(𝑒−𝑏.𝑡 − 𝑒−𝑎.𝑡)

Substituting with model parameters,

𝑓(𝑡) =𝑁[𝐿]𝑘1

𝐾𝐹 − 𝐾𝑆

(𝑒−𝐾𝑆 .𝑡 − 𝑒−𝐾𝐹.𝑡)

Second term: From Appendix or as given at http://www.wolframalpha.com/input/?i=inverse

+Laplace+transform+c%2F(s*(s%2Ba)*(s%2Bb))

16

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎𝑏[1 −

𝑏

𝑏 − 𝑎𝑒−𝑎𝑡 +

𝑎

𝑏 − 𝑎𝑒−𝑏𝑡]

Selection of whether a is 𝐾𝐹 or 𝐾𝑆, and respectively b is 𝐾𝑆 or 𝐾𝐹, does not affect the inverse

transformation, because,

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎𝑏[1 −

𝑏

𝑏 − 𝑎𝑒−𝑎𝑡 +

𝑎

𝑏 − 𝑎𝑒−𝑏𝑡] =

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎𝑏[1 −

𝑎

𝑎 − 𝑏𝑒−𝑏𝑡 +

𝑏

𝑎 − 𝑏𝑒−𝑎𝑡]

Substituting with model parameters,

𝑓(𝑡) =𝑁[𝐿]𝑘1𝑘4

𝐾𝐹𝐾𝑆[1 −

𝐾𝐹

𝐾𝐹 − 𝐾𝑆𝑒−𝐾𝑆𝑡 +

𝐾𝑆

𝐾𝐹 − 𝐾𝑆𝑒−𝐾𝐹𝑡]

We obtain an analytical equation for [𝑅𝐿] by adding the two inverse Laplace transforms:

[𝑅𝐿]𝑡 =𝑁[𝐿]𝑘1

𝐾𝐹 − 𝐾𝑆

(𝑒−𝐾𝑆.𝑡 − 𝑒−𝐾𝐹.𝑡) +𝑁[𝐿]𝑘1𝑘4

𝐾𝐹𝐾𝑆[1 −

𝐾𝐹

𝐾𝐹 − 𝐾𝑆𝑒−𝐾𝑆𝑡 +

𝐾𝑆

𝐾𝐹 − 𝐾𝑆𝑒−𝐾𝐹𝑡]

This equation can be rearranged to yield the competition kinetics equation in Motulsky and Mahan,

1984:

[𝑅𝐿]𝑡 =𝑁[𝐿]𝑘1

𝐾𝐹 − 𝐾𝑆[𝑘4(𝐾𝐹 − 𝐾𝑆)

𝐾𝐹𝐾𝑆+

𝑘4 − 𝐾𝐹

𝐾𝐹𝑒−𝐾𝐹𝑡 −

𝑘4 − 𝐾𝑆

𝐾𝑆𝑒−𝐾𝑆𝑡]

where,

𝐾𝐹 = 0.5 {𝐾𝐴 + 𝐾𝐵 + √(𝐾𝐴 − 𝐾𝐵)2 + 4[𝐿][𝐼]𝑘1𝑘3}

𝐾𝑆 = 0.5 {𝐾𝐴 + 𝐾𝐵 − √(𝐾𝐴 − 𝐾𝐵)2 + 4[𝐿][𝐼]𝑘1𝑘3}

𝐾𝐴 = [𝐿]𝑘1 + 𝑘2

𝐾𝐵 = [𝐼]𝑘3 + 𝑘4

17

Appendix 1: Rules for Laplace transform method

1. Time-dependent variables are signified with an accent3. For labeled ligand association (Model 1):

[𝑅𝐿] → [𝑅𝐿]̅̅ ̅̅ ̅̅

2. The differential expression is substituted with the Laplace operator, s. From Model 1:

𝑑[𝑅𝐿]

𝑑𝑡→ 𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅

3. Terms that are independent of time are divided by s. From Model 1:

𝑁[𝐿]𝑘1 →𝑁[𝐿]𝑘1

𝑠

4. In taking the Laplace transform of a disappearing time-dependent variable, a constant defining the

amount of the disappearing variable at the initiation of the process is incorporated. This constant is

not divided by s. For labeled ligand dissociation (Model 2):

𝑠[𝑅𝐿]̅̅ ̅̅ ̅̅ = [𝑅𝐿]𝑡=0 − [𝑅𝐿]̅̅ ̅̅ ̅̅ . 𝑘2

5. The sum of terms in a Laplace transform for a single time-dependent variable is equal to the sum

of terms in the inverse Laplace transform. From Model 3, unlabeled ligand pre-incubation and

washout:

𝐹(𝑠) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)−

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑓(𝑡) =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎(1 − 𝑒−𝑎𝑡) −

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡)

6. Time-independent variables (for example, a and b in Model 3) are interchangeable when taking

the inverse Laplace transform.

3 Alternatively, the accent character ^ is used.

18

Appendix 2: Inverse Laplace transforms commonly encountered in

pharmacological models

Laplace transform F (s) Inverse Laplace transform f (t)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠 + 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. 𝑒−𝑎𝑡

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎(1 − 𝑒−𝑎𝑡)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎(𝑒−𝑎.𝑡 − 𝑒−𝑏.𝑡)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎𝑏[1 −

𝑏

𝑏 − 𝑎𝑒−𝑎𝑡 +

𝑎

𝑏 − 𝑎𝑒−𝑏𝑡]

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)(𝑠 + 𝑐) 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [

𝑒−𝑎𝑡

(𝑏 − 𝑎)(𝑐 − 𝑎)+

𝑒−𝑏𝑡

(𝑐 − 𝑏)(𝑎 − 𝑏)+

𝑒−𝑐𝑡

(𝑎 − 𝑐)(𝑏 − 𝑐)]

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠(𝑠 + 𝑎)(𝑠 + 𝑏)(𝑠 + 𝑐) 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [

1

𝑎𝑏𝑐−

𝑒−𝑎𝑡

𝑎(𝑏 − 𝑎)(𝑐 − 𝑎)−

𝑒−𝑏𝑡

𝑏(𝑐 − 𝑏)(𝑎 − 𝑏)−

𝑒−𝑐𝑡

𝑐(𝑎 − 𝑐)(𝑏 − 𝑐)]

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠2(𝑠 + 𝑎)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎2(𝑎𝑡 − 1 + 𝑒−𝑎𝑡)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑠2(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎𝑏2[𝑏𝑡 − 1 −

𝑏

𝑎−

𝑎2

𝑎(𝑏 − 𝑎)𝑒−𝑏𝑡 +

𝑏2

𝑎(𝑏 − 𝑎)𝑒−𝑎𝑡]

𝑠. 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑎 − 𝑏(𝑎𝑒−𝑎.𝑡 − 𝑏𝑒−𝑏.𝑡)

(𝑠 + 𝛼)𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

(𝑠 + 𝑎)(𝑠 + 𝑏)

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑏 − 𝑎((𝛼 − 𝑎)𝑒−𝑎.𝑡 − (𝛼 − 𝑏)𝑒−𝑏.𝑡)

19

Sources

Charles Sullivan, Dartmouth College,

http://www.dartmouth.edu/~sullivan/22files/New%20Laplace%20Transform%20Table.pdf

WolframAlpha computational knowledge engine,

http://www.wolframalpha.com/input/?i=inverse+Laplace+transform

20

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