1. 1 Exercise 13 Deconvolution analysis vs. modeling to document the process of drug absorption

2. 2 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.

3. 3 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.

4. 4 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.

5. 5 Visual inspection of data

6. 6 Fitting using conventional model

7. 7 Apparently a good fitting

8. 8 Inspection of residuals: not randomly scattered

9. 9 Model with two sites of absorption

10. 10 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) 1 2 3

11. 11 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

12. 12 Parameters to estimate The number of parameters to estimate is of 4 namely 'FR', 'Ka1', 'Ka2', and 'K10'.

13. 13 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

14. 14 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

15. 15 The temporary block TEMPORARY Dose=CON(1) END remark - define differential equations starting values

16. 16 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

17. 17 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

18. 18 The function block FUNCTION 1 F= Z(1) END FUNCTION 2 F= Z(2) END FUNCTION 3 F= Z(3) END

19. 19 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,

20. 20 Equation describing our model

21. 21 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

22. 22 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

23. 23 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

24. 24 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

25. 25 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 }

26. 26 An other example of deconvolution problem To describe the release rate of an hormone as LH from the hypophysis (LAH) LH

27. 27 An other example of deconvolution problem To describe the release rate of an hormone as LH from the hypophysis (LAH) What I can observe :

28. 28 LH release rate : The physiological system LH LH (ng/mL) Release rate (amount / time) Hypophysis Hypotalamus Plasma concentration K10 (elimination rate) first order

29. 29 Physiological system Analogous system dLH dt = Kin - Kout * R dc dt = K 0 - K 10 * C K 10 (first order) K out (first order) LH LH (ng/mL) response: R K input LH infusion LH (ng/mL) Concentration K 0 (input) blood

30. 30 The question HYPOPHYSIS ? Input signal What you observe

31. 31 The basic idea to know how the body transform a dose into a plasma concentration profile HYPO ? Input signal Known IV Dose K10 Vc

32. 32 The question again but we know the system HYPOPHYSIS ? Input signal K10 Vc

33. 33 The LH unit disposition function

34. 34 The LH secretion rate

35. 35 The convolution integral The convolution integral is the main mathematical tool for analyzing system in series and its reprentation by Laplace transform

36. 36 The convolution integral 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

37. 37 The convolution integral Denote time by u (we will see why later) Consider the influx (input of drug) as a sequence of rectangular impulses Each impulse has a height i(u) and small ‘infinetisimal) duration du lasting from time u to u+du; amount injected is i(u)du If a unit amount of drug is injected as an impulse at time 0, then the the unit impulse response (in term of plasma concentartion) of the system is g(u)

38. 38 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))

39. 39 The convolution integral: linearity principle Consider now different stimulus (injected amount) i (u)du ; ….applied at time 0 is X times higher than the unit impulsse stimulus In accord with linearity principle, the response will be i(u)du times the unit impulse response g(t). If you double the input, you double the response

40. 40 The convolution integral Consider now that the same stimulus (injected amount) i(u)du is not applied at time 0 but at time u to u+du. the response will be also the unit impulse response g(t). But the response is shifted by the time u (d denotes only that the response is only the response to very small amount of drug i(u)du –Response(t)=g(t-u)i(u)du

41. 41 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))

42. 42 Stimulus of different sizes are injected at different times (u1, u2, u3…) Q: what will be observed?

43. 43 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

44. 44 Stimulus of different sizes are injected at different times (u1, u2, u3…)

45. 45 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….

46. 46 The convolution integral This convolution integral is not convenient to use and it is practical to make a Laplace transformation of it

47. 47 The Laplace transform of the convolution integral 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

48. 48 Convolution integral after a Laplace transformation 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

49. 49 Deconvolution is the inverse problem of convolution 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

50. 50 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

51. 51 Deconvolution: numerical aspect The practical handling of experimental data is complicated because deconvolution is very senstive to minor error (but perfect with simulated data)

52. 52 Deconvolution analysis Open WNL with our data sheet. Select Deconvolution from the Tools menu:

53. 53 The Deconvolution dialog

54. 54 Enter the dosing units in the Dose units field (here µg).

55. 55 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:

56. 56 unit impulse response

57. 57 Data analysis for animal 21

58. 58 Deconvolution results: smoothed curve

59. 59 The instantaneous input rate

60. 60 The Cumulative Input plot