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Exercise 13

Deconvolution analysis vs. modeling to document the

process of drug absorption

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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.

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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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2

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 • 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• 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

• 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

• 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

• 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• 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 • 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*(ka2-k10))+

(ka2*(1-F1)*(ka1-k10)))*exp(-k10*t)/((ka1-k10)*(ka2-k10)))))• END• 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

Response: the high of water in the bath(superposition principle holds)

The bucket as an integral operator

Disposition function (a plug) giving a first order emptying; K10=size of the cork

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time

0

Convolution for a simple linear and time invariant system

Bolus imput (Dirac)

Observed Response: the height of water in the bath

The bucket as an integral operator

Disposition function (the plug as an exponential

operator) Output rate is function of the size

of the plug and of the height of water in the bucket

}timeke 10

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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

LH

LH (ng/mL)

Release rate (amount / time)

Hypophysis

Hypotalamus

Plasma concentration

K10 (elimination rate) first order

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Physiological system Analogous system

dLHdt = Kin - Kout * R dc

dt = K0 - K10 * C

K10 (first order)Kout (first order)

LH

LH (ng/mL)response: R

Kinput

LH

infusion

LH (ng/mL)Concentration

K0 (input)

blood

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The question

HYPOPHYSIS

?

Input signal

What you observe

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The basic ideato know how the body transform a dose into a

plasma concentration profile

HYPO

?

Input signal

Known IV Dose

K10

Vc

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The question again but we know the system

HYPOPHYSIS

?

Input signal

K10

Vc

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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

duuiutgtsponse0

)()()(Re

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 operatorThis 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

Time of injection of the rectangular impulse unit is at time 0

Elementary response

i(u)du=1

G(u)

Time t

G(u)=exp(-k10(t))

t

duuiutgtsponse0

)()()(Re

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The same stimulus (injected amount) i(u)du applied at time u to u+du.

u

Time

Time of injection of the rectangular impulse unit is at time u

U

Elementary response

i(u)du=1

G(u)

Time t

G(t-u)=exp(-k10(t-u))

t

duuiutgtsponse0

)()()(Re

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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

duuiutgtsponse0

)()()(Re

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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

duuiutgtsponse0

)()()(Re

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

0

)()()]([ dtetgsgtgL st

What does the Laplace transformation is to replace the time domain of a rate expression by the complex domain of the Laplace operator sThe 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 sisgssponse 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)(

sg

ssponsesi

Deconvolution consists to compute the input from measuremnt of the response and unit impulse responseFor 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

Response: the high of water in the bath(superposition principle holds)

The bucket as an integral operator

Disposition function (a plug describe by g(s) ) giving a first order emptying; K10=size of the cork

)(

)(Re)(

sg

ssponsesi

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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 mono-exponential model)

• The IV bolus (unit impulse response) after a 100 µg/kg bolus dose was described by the following equation:

)1.0(800)( timeExptC

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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