Exercise 13
Deconvolution analysis vs.
modeling to document the
process of drug absorption
1
Objectives
• To implement a WNL user model for a
monocompartmental model with two processes
of absorption either with an algebraic equation
or with a set of differential equations.
• To compare two concurrent models:
monocompartmental with a single site of
absorption vs. a monocompartmental model with
two sites of absorption using the AIC criterion
• To obtain the input function of a drug using
deconvolution to reveal information about it
process of absorption.
2
Overview
• Double peaks in the plasma concentration–time
profile following oral administration have been
reported for several compounds.
• To describe a complicate drug invasion in
plasma like a double peak, there are two
possible approaches:
• (i) curve fitting using a modified customary
compartmental model
• (ii) numerical deconvolution to identify the
input rate of the drug in the central compartment.
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The data
• The goal of this experiment was to
compare two formulations (A and B) of a
new drug product administered by IM
route at a dose of 100 µg/kg.
• Twelve (12) animals were investigated
following a cross over design and raw data
are given in the corresponding Excel
sheet.
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Visual inspection of data
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Fitting using conventional model
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Apparently a good fitting
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Inspection of residuals:
not randomly scattered
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Model with two sites of absorption
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The system can be very easily described by
a set of differential equations :
DZ(1) = -Ka1*Z(1)
DZ(2) = -Ka2*Z(2)
DZ(3) = Ka1*Z(1)+Ka2*Z(2)-K10*Z(3)
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Initial conditions
• The next step to describe the model is to qualify
the so-called initial condition of the system to tell
the numerical solver how to proceed at time 0
• here we know that a fraction of the dose (noted
FR) is in the first site of absorption and the rest
of the dose (noted 1-FR) in the other site thus:
• Z(1) = FR*Dose
• Z(2) = (1-FR)*Dose
• Z(3) = 0
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Parameters to estimate
• The number of parameters to estimate is
of 4 namely 'FR', 'Ka1', 'Ka2', and 'K10'.
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The model as a set of differential equation
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MODEL
remark ******************************************************
remark Developer:
PL Toutain
remark Model Date:
12-12-2009
remark Model Version:
1.0
remark ******************************************************
remark
remark - define model-specific commands
COMMANDS
NCONSTANT 1
NFUNCTIONS 3
NDERIVATIVES 3
NPARAMETERS 4
PNAMES 'FR', 'Ka1', 'Ka2', 'K10'
END
remark - define temporary variables
TEMPORARY
Dose=CON(1)
END
remark - define differential equations starting values
START
Z(1) = FR*Dose
Z(2) = (1-FR)*Dose
Z(3) = 0
END
remark - define differential equations
DIFFERENTIAL
DZ(1) = -Ka1*Z(1)
DZ(2) = -Ka2*Z(2)
DZ(3) = Ka1*Z(1)+Ka2*Z(2)-K10*Z(3)
END
remark - define algebraic functions
FUNCTION 1
F= Z(1)
END
FUNCTION 2
F= Z(2)
END
FUNCTION 3
F= Z(3)
END
remark - define any secondary parameters
remark - end of model
EOM
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The command block
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MODEL
remark ******************************************************
remark Developer:
PL Toutain
remark Model Date: 12-12-2009
remark Model Version:
1.0
remark ******************************************************
remark
remark - define model-specific commands
COMMANDS
NCONSTANT 1
NFUNCTIONS 3
NDERIVATIVES 3
NPARAMETERS 4
PNAMES 'FR', 'Ka1', 'Ka2', 'K10'
END
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The temporary block
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TEMPORARY
Dose=CON(1)
END
remark - define differential equations
starting values
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The block for initial condition
• remark - define differential equations
starting values
• START
• Z(1) = FR*Dose
• Z(2) = (1-FR)*Dose
• Z(3) = 0
• END
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The differential block
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DIFFERENTIAL
DZ(1) = -Ka1*Z(1)
DZ(2) = -Ka2*Z(2)
DZ(3) = Ka1*Z(1)+Ka2*Z(2)-K10*Z(3)
END
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The function block
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FUNCTION 1
F= Z(1)
END
FUNCTION 2
F= Z(2)
END
FUNCTION 3
F= Z(3)
END
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The analytical expression of the
model
• This model using differential equations is
easy to write but it is preferable, whenever
possible, to use the corresponding
algebraic equations; here using the
Laplace transform,
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Equation describing our model
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The integrated model in WNL
• This model does not exist in the WNL library and
we will implement it.
• The following set of statements corresponds to
the 2 sites model written with an algebraic
equation.
• The structure of the model is the same as for the
preceding one; I just add t=X as a temporary
variable to tell WNL that the independent
variable (always X) can be written with a t in my
equation
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The integrated model in WNL
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MODEL
remark ******************************************************
remark Developer:
Pl toutain
remark Model Date:
03-22-2011
remark Model Version: 1.0
remark ******************************************************
remark
remark - define model-specific commands
COMMANDS
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NFUNCTIONS 1
NCON 1
NPARAMETERS 5
PNAMES 'ka1', 'ka2', 'k10', 'V', 'F1'
END
remark - define temporary variables
TEMPORARY
Dose=CON(1)
remark: t is the independent variable
t=X
END
remark - define algebraic functions
FUNCTION 1
F= (Dose/V)*(((ka1*F1*exp(-ka1*t)/(k10-ka1)) + (ka2*(1-F1)*exp(-ka2*t)/(k10-ka2)) + (((ka1*F1*(ka2k10))+(ka2*(1-F1)*(ka1-k10)))*exp(-k10*t)/((ka1-k10)*(ka2-k10)))))
END
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remark - define any secondary parameters
remark - end of model
EOM
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Compare results obtained with
the two models
• Compare results obtained with differential
equation vs. integrated model
• Compare fitting obtain for a given animal
using classical monocompartmental model
vs integrated model vs. differential model
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A deconvolution question for a simple
linear and time invariant system
Q: What is the instantaneous
rate of water flow associated to
this shower
The bucket as
an integral operator
Response: the high of water in the bath
(superposition principle holds)
Disposition function (a plug) giving a first
order emptying; K10=size of the cork
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Convolution for a simple linear and time
invariant system
Bolus imput (Dirac)
time

The bucket as
an integral operator
0
e
 k 10time
Disposition function
(the plug as an exponential
operator)
}
Observed Response:
the height of water in the bath
Output rate is function of the size
of the plug and of the height of
water in the bucket
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An other example of deconvolution problem
• To describe the release rate of an hormone as
LH from the hypophysis (LAH)
LH
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An other example of deconvolution problem
• To describe the release rate of an hormone as
LH from the hypophysis (LAH)
• What I can observe :
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LH release rate :
The physiological system
Hypotalamus
Hypophysis
Release rate
(amount / time)
Plasma concentration
LH
LH (ng/mL)
K10 (elimination rate) first order 28
Physiological system
Analogous system
infusion
LH
LH
Kinput
LH (ng/mL)
response: R
Kout (first order)
dLH
dt = Kin - Kout * R
K0 (input)
LH (ng/mL)
Concentration
blood
K10 (first order)
dc
dt = K0 - K10 * C
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The question
What you observe
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HYPOPHYSIS
Input signal
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The basic idea
to know how the body transform a dose into a
plasma concentration profile
Known IV Dose
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HYPO
Vc
Input signal
K10
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The question again but we know
the system
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HYPOPHYSIS
Vc
Input signal
K10
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The LH unit disposition function
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The LH secretion rate
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The convolution integral
• The convolution integral is the main
mathematical tool for analyzing system in
series and its reprentation by Laplace
transform
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The convolution integral
t
Re sponse (t )   g (t  u )i(u )du
0
Where:
• Response is most often for us plasma concentration at time t;
•g(t) is the response to a unit impulse input i.e. the unit disposition function
as EXP(-k10*t) ;
•i(u)is the input function
This convolution integral is a linear operator
This convolution integral is not convenient to use and it is
practical to make a Laplace transformation of it
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Unit amount of drug is injected as
an impulse at time 0
Time of injection of the
rectangular impulse unit is
at time 0
Elementary response
G(u)
i(u)du=1
G(u)=exp(-k10(t))
t
Re sponse (t )   g (t  u )i(u )du
0
Time
Time t
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The same stimulus (injected amount)
i(u)du applied at time u to u+du.
Time of injection of the
rectangular impulse unit is
at time u
Elementary response
G(u)
i(u)du=1
G(t-u)=exp(-k10(t-u))
t
Re sponse(t )   g (t  u )i(u )du
0
Time
u
Time t
U
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Stimulus of different sizes are injected
at different times (u1, u2, u3…)
Q: what will be observed?
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The system is assumed to be linear and
stationary.
• Stationarity means that if the unit amount is
injected at time u then the corresponding
response function , denoted g(t-u) is the same
as before, only shifted by the time u
• Linearity means that if the sum of the individual
of two or more stimuli is applied as a single
stimulus, then the response will be the sum of
the individual response (principle of
superposition); similarly if the stimulus is
multiplied by a constant a, the response
(concentration) is also multiplied by a
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Stimulus of different sizes are injected at different
times (u1, u2, u3…)
t
Re sponse (t )   g (t  u )i(u )du
0
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The convolution integral
• Consider now different stimulus (injected
amount) i1(u1)du ; i2(u2)du ….applied at
time u1, u2 etc.
• In accord with linearity the response
associated to each impulse will be i(u)du
times the unit impulse response g(t).
• But in accord with stationarity each
elementary response is shifted by the
time u1,u2….
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The convolution integral
t
Re sponse (t )   g (t  u )i(u )du
0
This convolution integral is not convenient to use and it is
practical to make a Laplace transformation of it
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The Laplace transform of the
convolution integral

L[ g (t )]  g ( s)   g (t )e dt
 st
0
What does the Laplace transformation is to replace the time
domain of a rate expression by the complex domain of the
Laplace operator s
The Laplace transfoem enble complex rate expression to be
manipulated easily by conventional algebraic technic one the the
time variabe has been replaced by the Laplace operator s
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Convolution integral after a Laplace
transformation
Re sponse( s)  g ( s) * i ( s)
The complicated convolution integral is now change into a
simple product of two functions ;
thus if i(t) and and response(t) were known, then analytically
it is much simple to obtain g(t) in the Laplace domain than
with the convolution integral
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Deconvolution is the inverse
problem of convolution
Re sponse ( s)
i( s) 
g ( s)
Deconvolution consists to compute the input from
measuremnt of the response and unit impulse response
For example to compute drug release rate from an
observed plasma concentration profile and knowing the
model of drug disposition from an IV study
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A deconvolution question for a simple
linear and time invariant system
Q: What is the instantaneous
rate of water flow [i(s)]
associated to this shower
The bucket as
an integral operator
Re sponse ( s)
i( s) 
g ( s)
Response: the high of water in
the bath
(superposition principle holds)
Disposition function (a plug describe by
g(s) ) giving a first order emptying;
K10=size of the cork
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Deconvolution: numerical aspect
• The practical handling of experimental
data is complicated because
deconvolution is very senstive to minor
error (but perfect with simulated data)
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Deconvolution analysis
• Open WNL with our data sheet.
• Select Deconvolution from the Tools
menu:
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The Deconvolution dialog
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Enter the dosing units in the Dose
units field (here µg).
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• In the Settings fields, select the number of
exponential terms (N) in the unit impulse
response here 1 because it is a monoexponential model)
• The IV bolus (unit impulse response) after
a 100 µg/kg bolus dose was described by
the following equation:
C (t )  800  Exp(0.1 time)
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unit impulse response
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Data analysis for animal 21
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Deconvolution results:
smoothed curve
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The instantaneous input rate
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The Cumulative Input plot
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