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:1 Sam Hoare, 2016, sam.hoare@pharmechanics.com or (US) 619-203-2886

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Model 1: Ligand-receptor associationModel 2: Dissociation of ligand from receptorModel 3: Unlabeled ligand pre-incubation and washoutModel 4: Competition kinetics – labeled ligand association in the presence of unlabeled ligand Model 1: Ligand-r eceptor association

R, receptorL, labeled ligand; units, Mk1, labeled ligand association rate constant; units, M-1min-1k2, labeled ligand dissociation rate constant; units, min-1Minimal 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,

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d [RL ]dt

=N [ L ]k 1−[RL] ([L]k1+k2 )

where N is the total concentration of receptors.

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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, sd [RL ]dt

→ s[RL ]

3. Terms that are independent of time are divided by sN [L]k1→

N [L]k 1s

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

N [L]k1s

−[RL] ([L]k 1+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:2 Alternatively, the accent character ^ is used.

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s[RL ]=N [L]k1s

−[RL] ([L]k 1+k2 )

[RL ]=N [L] k1

s ( s+[L]k 1+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)= constants ( 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 )= constanta

(1−e−at )

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(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.

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Model 2: Labeled ligand dissociation

Variables defined in Model 1In 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]k 2

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:

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[RL ]=¿¿

Step 4. Inverse Laplace transformThe 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)= constants+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.

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Model 3: Unlabeled ligand pre-incubation and washout

I, unlabeled ligand; units, Mk4, unlabeled ligand dissociation rate constant; units, min-1Other variables given in Model 1In 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 equationsThe 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

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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 ]

[R ]=N−[RL ]−[RI ]

Substituting into the differential equation:d [RL ]dt

=N [ L ]k 1−[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 ]=¿

Solving for [RI ],[RI ]=

[RI ]t=0s+k4

The Laplace transform for [RL ] is,s[RL ]=

N [L ] k1s

−[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 ]:10

s[RL ]=N [L ] k1s

−[RI ]t=0 [ L ] k1s+k4

−[RL] ([L]k 1+k2 )

This equation contains only [RL ] as a time-dependent variable.Step 3. Solving the Laplace transform for [RL] Solving for [RL ]:

[RL ]=N [ L ]k1

s ( s+[L]k 1+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 , X )1+F ( s , X )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)= constants ( s+a )

f ( t )= constanta

(1−e−at )

Substituting with the model parameters,

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f ( t )=N [ L ]k1

[L]k1+k 2(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 )= constantb−a

(e−a .t−e−b . t )

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, constantb−a

(e−a .t−e−b .t )= constanta−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 ]k 1[L]k1+k 2−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]k 1+k2 ) t)−[RI ]t=0 [L ]k1[L]k1+k2−k4

(e−k4 .t−e−([L] k1+k2 ) .t )

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Rearranging gives the equation in Malany et al. 2009:[RL ]t=

N [ L ] k1[L ]k1+k 2

(1−e−( [ L ]k1+k2) t )+ [RI ]t=0 [L ] k1[L] k1+k2−k 4

(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).

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Model 4: Competition kinetics

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

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[R ]=N−[RL ]−[RI ]

d [RL ]dt

=N [ L ]k 1−[RI ] [ L ]k 1−[RL ]( [L ]k1+k2 )

d [RI ]dt

=N [ I ]k3−[RL ] [ I ]k 3−[RI ] ( [ I ] k3+k 4 )

Step 2: Laplace transforms For the purpose of clarity, the following new variables are used, per Motulsky and Mahan, 1984:

K A=[ L ]k1+k2

K B=[ I ]k3+k4

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

N [L ]k1s

−[RI ] [L ] k1−[RL ]K A

s [RI ]=N [ I ] k3s

−[RL] [ I ]k3−[RI ]K B

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 ]k3s+K B

Substituting into the transform for [RL ],s [RL ]= N

[L ]k1s

−N [ L ] [ I ] k1k3s ( s+KB )

+[RL] [ L ] [ I ]k 1k3

s+KB−[RL ]K A

This equation contains only [RL ] as a time-dependent variable.15

Step 3. Solving the Laplace transform for [RL] Solving for [RL ] involves a factorization procedure. Before this, we arrive at an intermediate step:

[RL ] {( s+K A ) (s+K B )−[ L ] [ I ]k 1k3 }=N [L ] k1 ( s+KB )−N [L ] [ I ] k1k3

s

Factorization stepThe [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+K A ) (s+KB )−[ L ] [ I ]k1 k3=s2+s (K A+KB )+K A KB−[ L ] [ I ]k1k3

This expression can be factorized by introducing two compound variables, K F+K S and K F KS, defined as follows:K F+K S=K A+KBK F KS=K AKB−[ L ] [ I ]k1 k3

Substituting into the expanded [RL ] multiplier,s2+s (K A+K B )+K AK B−[ L ] [ I ] k1k 3=s2+s (KF+KS )+K FK S

This expression can be readily factorized:s2+s (KF+KS )+KF K S=( s+KS ) ( s+K F )

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Resolving KF and KS in terms of the model parametersIn 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 K F+K S=K A+KB for KS,

K S=K A+K B−KF

Substituting into K F KS=K A KB−[ L ] [ I ]k1 k3,K F (K A+K B−K F )=K A K B−[ L ] [ I ]k 1k3

Solving for K F gives a quadratic equation:0=KF

2−K F (K A+K B )+K A K B− [L ] [ I ] k1k 3

KF can be found as the root of the quadratic equation:K F=

K A+KB±√(K A+KB )2−4 (K AKB−[ L ] [ I ]k1 k3 )2

which simplifies to,K F=0.5 {K A+KB±√(K A−K B )2+4 [L ] [ I ] k1 k3}

The same procedure resolves KS into:K S=0.5 {K A+K B±√ (K A−KB )2+4 [ L ] [ I ]k 1k3}

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.

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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+K A ) (s+K B )−[ L ] [ I ]k 1k3 }=N [L ] k1 ( s+KB )−N [L ] [ I ] k1k3

s

Substituting the factorization expression for the [RL ] multiplier,[RL ] (s+KS ) ( s+K F )=

N [ L ] k1 ( s+K B )−N [L ] [ I ] k1 k3s

Solving for [RL ],[RL ]=

N [ L ]k1 ( s+K B )−N [ L ] [ I ]k1 k3s ( s+KS ) ( s+K F )

Rearranging yields the final version of the Laplace transform:[RL ]=

N [L ]k1( s+K S ) (s+K F )

+N [ L ] k1 k4

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 )= constantb−a

(e−a .t−e−b . t )

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Selection of whether a is K F or KS, and respectively b is KS or K F, does not affect the inverse transformation, because,constantb−a

(e−a .t−e−b .t )= constanta−b

(e−b .t−e−a . t )

Substituting with model parameters,f ( t )=

N [ L ] k1K F−K S

(e−K S .t−e−K F .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))F ( s)= constant

s ( s+a ) ( s+b )

f (t )= constantab [1− b

b−ae−at+ a

b−ae−bt ]

Selection of whether a is K F or KS, and respectively b is KS or K F, does not affect the inverse transformation, because,constantab [1− b

b−ae−at+ a

b−ae−bt ]= constantab [1− a

a−be−bt+ b

a−be−at ]

Substituting with model parameters,f ( t )=

N [L ]k1 k4K FKS [1− KF

K F−K Se−KS t+

KSK F−K S

e−K F t]We obtain an analytical equation for [RL ] by adding the two inverse Laplace transforms:

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[RL ]t=N [ L ]k1KF−KS

(e−KS . t−e−K F .t )+N [L ]k1 k4K F KS [1− K F

K F−K Se−KS t+

KSK F−KS

e−K F t]

This equation can be rearranged to yield the competition kinetics equation in Motulsky and Mahan, 1984:[RL ]t=

N [ L ]k1KF−KS [ k4 (K F−KS )

K FK S+k4−KFK F

e−K F t−k 4−K S

KSe−KS t]

where,K F=0.5 {K A+KB+√ (K A−KB )2+4 [ L ] [ I ] k1k 3}KS=0.5 {K A+K B−√ (K A−KB )2+4 [ L ] [ I ] k1k 3}

K A=[ L ]k1+k2K B=[ I ]k3+k4

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Appendix 1: Rules for Laplace transform method1. 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]k 1s

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] .k 2

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 )= constanta

(1−e−at )− constantb−a

(e−a .t−e−b . t )

3 Alternatively, the accent character ^ is used.21

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

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Appendix 2: Inverse Laplace transforms commonly encountered in pharmacological modelsLaplace transform F (s) Inverse Laplace transform f (t)

constants constant

constants+a constant . e−at

constants(s+a)

constanta

(1−e−at )

constant(s+a )(s+b)

constantb−a

(e−a .t−e−b .t )

constants ( s+a )(s+b)

constantab [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 [ 1abc− e−at

a (b−a ) (c−a )− e−bt

b ( c−b ) (a−b )− e−ct

c (a−c ) (b−c ) ]constants2(s+a)

constanta2

(at−1+e−at )

constants2(s+a)(s+b)

constantab2 [bt−1− ba− a2

a (b−a )e−bt+ b2

a (b−a )e−at ]

s . constant(s+a )(s+b)

constanta−b

(ae−a .t−be−b .t )

(s+α )constant( s+a )(s+b)

constantb−a ( (α−a ) e−a . t−(α−b ) e−b .t )

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SourcesCharles Sullivan, Dartmouth College, http://www.dartmouth.edu/~sullivan/22files/New%20Laplace%20Transform%20Table.pdfWolframAlpha computational knowledge engine, http://www.wolframalpha.com/input/?i=inverse+Laplace+transform

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ReferencesMalany, S, Hernandez, LM, Smith, WF, Crowe, PD and Hoare, SRJ (2009) “Analytical method for simultaneously measuring ex vivo drug receptor occupancy and dissociation rate: application to (R)-dimethindene occupancy of central histamine H1 receptors.” J. Recept. Signal, Transduct. Res. 29: 84-93Mayersohn, M and Gibaldi, M (1970) "Mathematical Methods in Pharmacokinetics. I. Use of the Laplace Transform in Solving Differential Rate Equations." Amer. J. Pharm. Ed. 34: 608-614Motulsky, HJ and Mahan, LC (1984) “The kinetics of competitive radioligand binding predicted by the law of mass action.” Mol. Pharm. 25: 1-9Packeu, A, Wennerberg, A, Balendran, A and Vauquelin, G (2010) “Estimation of the dissociation rate of unlabelled ligand–receptor complexes by a ‘two-step’ competition binding approach.” Brit. J. Pharmacol. 161: 1311-1328Popovic, J (1999) “Derivation of Laplace transform for the general disposition deconvolution equation in drug metabolism kinetics.” Exp. Toxicol. Pathol. 51: 409-411Uhlen, S, Schioth, HB and Jahsen, JA (2016) “A new, simple and robust radioligand binding method used to determine kinetic off-rate constants for unlabeled ligands. Application at α2A- and α2C-adrenoceptors.” Eur. J. Pharmacol. 788: 113-21

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