1) The document describes using the Laplace transform method to derive equations for receptor-ligand binding kinetics models. 2) The Laplace transform allows time-dependent differential equations to be solved using simple algebra by transforming them into another mathematical domain. 3) Four binding models are used as examples: ligand association, dissociation, unlabeled ligand pre-incubation and washout, and competition kinetics. Through taking the Laplace transform, substituting terms, and taking the inverse transform, analytical solutions are obtained for each model.

1. 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-fittingprograms to estimate the
value of rate parameters. In all but the simplest pharmacologicalmodels, 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 Laplacetransform is used forsolving first-or zero-orderdifferential equations. Itcannot 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, sam.hoare@pharmechanics.com or (US) 619-203-2886

2. 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 toobtain 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. 3
Step 2: Taking the Laplace transform
Six rules necessary forsuccessful application of the Laplace transform method are given in Appendix
1. Three rules are followedin 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 forone 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
competitiveinhibitor tothe receptor,where theLaplace transform forunlabeled 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. 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 whicht
is a straightforward independent variable. This process is described as taking the inverse Laplace
transform. This step yields the analytic equation to whichthe experimental data can be fitted using
nonlinear regression programs that are commonly used by pharmacologists, forexample Prism from
GraphPad; XLfitfromIDBS;Solver (a plug-in forMicrosoftExcel); 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 inverseLaplace transforms. These tables are availablein 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.
Inthese tables, theLaplace transformis written as afunctionof sand the inverse transforma 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. 5
Model 2: Labeled ligand dissociation
Variables defined in Model 1
In Model1 above, the appearance of the variable of interest, [𝑅𝐿],is measured. The Laplace transform
approach todisappearance of the variable of interestis exemplified by the model of liganddissociation
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 receptorat 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. 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. 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
Wewish 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. 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. 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 fifthrule 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, 𝑋̅)1 + 𝐹( 𝑠, 𝑋̅)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 withthe 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. 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 nodifference 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
fromnumerical solutionof the differentialequations (Packeuet al. 2010). Experimentally it has been
shown that fitted values of k4 match those fromalternative, well-validatedapproaches (Malany et al.
2009, Packeu et al. 2010, Uhlen at al. 2016).

11. 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 competitionkinetics model – [ 𝑅], [𝑅𝐿], and [𝑅𝐼]. The
goal of the derivation is an analytic equation in terms of one time-dependent variable, [RL]. As forthe
pre-incubation kinetics model above, [ 𝑅] is replaced using the conservation of mass equation forthe
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 conservationof mass equation,
𝑁 = [ 𝑅] + [ 𝑅𝐿] + [𝑅𝐼]
[ 𝑅] = 𝑁 − [ 𝑅𝐿] − [𝑅𝐼]

12. 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 followingnew variables are used, per Motulsky and Mahan, 1984:
𝐾𝐴 = [ 𝐿] 𝑘1 + 𝑘2
𝐾 𝐵 = [ 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]̅̅̅̅̅ 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 + 𝐾 𝐵)
−
[𝑅𝐿]̅̅̅̅̅̅[ 𝐼] 𝑘3
𝑠 + 𝐾 𝐵
Substituting into the transform for[ 𝑅𝐿],
𝑠[ 𝑅𝐿]̅̅̅̅̅̅ =
𝑁[ 𝐿] 𝑘1
𝑠
−
𝑁[ 𝐿][ 𝐼] 𝑘1 𝑘3
𝑠( 𝑠 + 𝐾 𝐵)
+
[𝑅𝐿]̅̅̅̅̅̅[ 𝐿][ 𝐼] 𝑘1 𝑘3
𝑠 + 𝐾 𝐵
− [𝑅𝐿]̅̅̅̅̅̅ 𝐾𝐴
This equation contains only [𝑅𝐿]̅̅̅̅̅̅ as a time-dependent variable.
Step 3. Solving the Laplace transform for [RL]
Solving for [ 𝑅𝐿]̅̅̅̅̅̅ involvesa factorizationprocedure. Before this, we arrive at an intermediate step:

13. 13
[ 𝑅𝐿]̅̅̅̅̅̅{( 𝑠 + 𝐾𝐴)( 𝑠 + 𝐾 𝐵) − [ 𝐿][ 𝐼] 𝑘1 𝑘3} =
𝑁[ 𝐿] 𝑘1( 𝑠 + 𝐾 𝐵) − 𝑁[ 𝐿][ 𝐼] 𝑘1 𝑘3
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 + 𝐾 𝐵) − [ 𝐿][ 𝐼] 𝑘1 𝑘3 = 𝑠2 + 𝑠( 𝐾𝐴 + 𝐾 𝐵) + 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3
This expression canbe factorizedby introducingtwo compoundvariables, 𝐾𝐹 + 𝐾𝑆 and 𝐾𝐹 𝐾𝑆,defined
as follows:
𝐾𝐹 + 𝐾𝑆 = 𝐾𝐴 + 𝐾 𝐵
𝐾𝐹 𝐾𝑆 = 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3
Substituting into the expanded [ 𝑅𝐿]̅̅̅̅̅̅ multiplier,
𝑠2 + 𝑠( 𝐾𝐴 + 𝐾 𝐵) + 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3 = 𝑠2 + 𝑠( 𝐾𝐹 + 𝐾𝑆)+ 𝐾𝐹 𝐾𝑆
This expression can be readily factorized:
𝑠2 + 𝑠( 𝐾𝐹 + 𝐾𝑆)+ 𝐾𝐹 𝐾𝑆 = ( 𝑠 + 𝐾𝑆)( 𝑠 + 𝐾𝐹)
Resolving KF and KS in terms of the model parameters
In order to obtain an analyticalequation 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 + 𝐾 𝐵 forKS,
𝐾𝑆 = 𝐾𝐴 + 𝐾 𝐵 − 𝐾𝐹

14. 14
Substituting into 𝐾𝐹 𝐾𝑆 = 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3,
𝐾𝐹( 𝐾𝐴 + 𝐾 𝐵 − 𝐾𝐹) = 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3
Solving for 𝐾𝐹 gives a quadratic equation:
0 = 𝐾𝐹
2
− 𝐾𝐹( 𝐾𝐴 + 𝐾 𝐵) + 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3
KF can be found as the root of the quadratic equation:
𝐾𝐹 =
𝐾𝐴 + 𝐾 𝐵 ± √( 𝐾𝐴 + 𝐾 𝐵)2 − 4( 𝐾𝐴 𝐾 𝐵 − [ 𝐿][ 𝐼] 𝑘1 𝑘3)
2
whichsimplifies to,
𝐾𝐹 = 0.5{𝐾𝐴 + 𝐾 𝐵 ± √( 𝐾𝐴 − 𝐾 𝐵)2 + 4[ 𝐿][ 𝐼] 𝑘1 𝑘3}
The same procedure resolves KS into:
𝐾𝑆 = 0.5 {𝐾𝐴 + 𝐾 𝐵 ± √( 𝐾𝐴 − 𝐾 𝐵)2 + 4[ 𝐿][ 𝐼] 𝑘1 𝑘3}
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 + 𝐾 𝐵) − [ 𝐿][ 𝐼] 𝑘1 𝑘3} =
𝑁[ 𝐿] 𝑘1( 𝑠 + 𝐾 𝐵) − 𝑁[ 𝐿][ 𝐼] 𝑘1 𝑘3
𝑠
Substituting the factorizationexpression forthe [ 𝑅𝐿]̅̅̅̅̅̅ multiplier,
[ 𝑅𝐿]̅̅̅̅̅̅( 𝑠 + 𝐾𝑆)( 𝑠 + 𝐾𝐹) =
𝑁[ 𝐿] 𝑘1( 𝑠 + 𝐾 𝐵) − 𝑁[ 𝐿][ 𝐼] 𝑘1 𝑘3
𝑠

15. 15
Solving for [ 𝑅𝐿]̅̅̅̅̅̅,
[ 𝑅𝐿]̅̅̅̅̅̅ =
𝑁[ 𝐿] 𝑘1( 𝑠 + 𝐾 𝐵) − 𝑁[ 𝐿][ 𝐼] 𝑘1 𝑘3
𝑠( 𝑠 + 𝐾𝑆)( 𝑠 + 𝐾𝐹)
Rearranging yields the final version of the Laplace transform:
[ 𝑅𝐿]̅̅̅̅̅̅ =
𝑁[ 𝐿] 𝑘1
( 𝑠 + 𝐾𝑆)( 𝑠 + 𝐾𝐹)
+
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) =
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 withmodel parameters,
𝑓( 𝑡) =
𝑁[ 𝐿] 𝑘1
𝐾𝐹 − 𝐾𝑆
( 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))

16. 16
𝐹( 𝑠) =
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑠( 𝑠 + 𝑎)( 𝑠 + 𝑏)
𝑓( 𝑡) =
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑎𝑏
[1 −
𝑏
𝑏 − 𝑎
𝑒−𝑎𝑡 +
𝑎
𝑏 − 𝑎
𝑒−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 − 𝑎
𝑒−𝑎𝑡 +
𝑎
𝑏 − 𝑎
𝑒−𝑏𝑡] =
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑎𝑏
[1 −
𝑎
𝑎 − 𝑏
𝑒−𝑏𝑡 +
𝑏
𝑎 − 𝑏
𝑒−𝑎𝑡]
Substituting withmodel parameters,
𝑓( 𝑡) =
𝑁[ 𝐿] 𝑘1 𝑘4
𝐾𝐹 𝐾𝑆
[1 −
𝐾𝐹
𝐾𝐹 − 𝐾𝑆
𝑒−𝐾 𝑆 𝑡 +
𝐾𝑆
𝐾𝐹 − 𝐾𝑆
𝑒−𝐾 𝐹 𝑡]
We obtain an analytical equation for [ 𝑅𝐿] by adding the two inverse Laplace transforms:
[ 𝑅𝐿]𝑡 =
𝑁[ 𝐿] 𝑘1
𝐾𝐹 − 𝐾𝑆
( 𝑒−𝐾 𝑆 .𝑡 − 𝑒−𝐾 𝐹.𝑡) +
𝑁[ 𝐿] 𝑘1 𝑘4
𝐾𝐹 𝐾𝑆
[1−
𝐾𝐹
𝐾𝐹 − 𝐾𝑆
𝑒−𝐾 𝑆 𝑡 +
𝐾𝑆
𝐾𝐹 − 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] 𝑘1
𝐾𝐹 − 𝐾𝑆
[
𝑘4( 𝐾𝐹 − 𝐾𝑆)
𝐾𝐹 𝐾𝑆
+
𝑘4 − 𝐾𝐹
𝐾𝐹
𝑒−𝐾 𝐹 𝑡 −
𝑘4 − 𝐾𝑆
𝐾𝑆
𝑒−𝐾 𝑆 𝑡]
where,
𝐾𝐹 = 0.5{𝐾𝐴 + 𝐾 𝐵 + √( 𝐾𝐴 − 𝐾 𝐵)2 + 4[ 𝐿][ 𝐼] 𝑘1 𝑘3}
𝐾𝑆 = 0.5 {𝐾𝐴 + 𝐾 𝐵 − √( 𝐾𝐴 − 𝐾 𝐵)2 + 4[ 𝐿][ 𝐼] 𝑘1 𝑘3}
𝐾𝐴 = [ 𝐿] 𝑘1 + 𝑘2
𝐾 𝐵 = [ 𝐼] 𝑘3 + 𝑘4

17. 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 fora 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 (forexample, a and b in Model 3) are interchangeable when taking
the inverse Laplace transform.
3 Alternatively, the accent character ^ is used.

18. 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 − 𝑎
𝑒−𝑎𝑡 +
𝑎
𝑏 − 𝑎
𝑒−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. 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. 20
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