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Edit Comment Close Premium member Presentation Transcript ONE COMPARTMENT MODEL (EXTRAVASCULAR ADMINISTRATION): ONE COMPARTMENT MODEL (EXTRAVASCULAR ADMINISTRATION) NEHA SOOD M.PHARMACY SEM-2 M100400031 1PHARMACOKINETICS OF DRUG ABSORPTION: PHARMACOKINETICS OF DRUG ABSORPTION Extravascular : In extravascular administration, a drug undergoes the process of absorption before it reaches systemic circulation.The absorption of a drug from the gastrointestinal tract or any other extravascular site is dependent on the following: Physiochemical properties of the drug Physiochemical properties of the dosage form Anatomy and physiology of the absorption site 2Slide 3: The rate of change of the amount of the drug ( dX / dt ) in the body following extravascular administration is equal to the difference between the rate of absorption ( dXa / dt ) and the rate of elimination ( dXe / dt ): dX = dXa - dXe dt dt dt The curve can be divided into various parts based on relative magnitudes of the rate of absorption and the rate of elimination at different time points. 3Slide 4: Absorption ph ase : The whole dose is available for absorption at zero time.The amount of drug at the absorption site declines as a function of time.Therefore , rate of drug absorption is greater than the rate of drug elimination. dXa > dXe dt dt 4Slide 5: Plateau phase: Absorption of a drug into the systemic circulation gradually increases the plasma level of the drug.This increases the rate of elimination of a drug,dXe/ dt.The rate of absorption of a drug decreases with time and the rate of elimination of drug increases with time.The time at which rate of absorption becomes equal to rate of elimination is called as time of peak drug concentration( t max ) in plasma and concentration is peak concentration( C max ). dxa = dXe dt dt 5Slide 6: Post Absorption Phase: The rate of elimination of the drug is faster than the rate of absorption. dXa < dXe dt dt Elimination Phase: When the drug at the absorption site becomes depleted, the rate of drug absorption approaches zero.Therefore , the rate of change in the amount of the drug in the body during the elimination phase is described by the first order process. dXe = -K e X dt 6 X= Amount of drug in the body at any time K= Overall elimination rate constantOral administration of unchanged drug in blood/plasma: Oral administration of unchanged drug in blood/plasma Ka K XO X= Vd.C Where, XO- is an oral dose of the drug administered X-is the amount of drug in the body Ka and K-are the first order rate constants of drug absorption and elimination For a drug that enters the body by an apparent first order absorption first order process and is eliminated by first order process and distributes in the body according to one compartment model. 7Slide 8: Rate of drug change in the body=Rate of drug absorption- Rate of drug elimination dX = dXa - dXe dt dt dt dX = Ka Xa - K e X dt Where, Xa is the amount of the drug available at the absorption site. Let us assume that ‘F’ is the fraction of the oral dose (XO) that will be absorbed. 8Slide 9: At time zero, Fxo amount is available for absorption . Therefore, Xa 0 = F XO Xa 0 = amount of dru g available for absorption at t=0 The rate of disappearance of the drug from the g.i.t is given by, dXa = -Ka Xa dt The time course of the amount of the drug in the body is given by X = Ka F X0 (e -Kt - e -Kat ) (Ka- K e ) 9Slide 10: Dividing the above equation by the volume of distribution of drug ( Vd ), X = C = Ka FX0 (e -Kt - e -Kat ) Vd Vd (Ka-K e ) Above equation describes the time course of the drug concentration in plasma following the oral route of administration. At larger values of t,the above equation will reduce to, C = Ka F X0 e -Kt Vd (Ka-K e ) 10Slide 11: Above equation explains the elimination phase, i.e. the time when absorption no longer occurs. The equation can be written in common logarithms as: log C = log [ Ka F X0 ] – K t/2.303 Vd (Ka- K e ) 11Determination of Absorption rate constant from oral data: Determination of Absorption rate constant from oral data Method of Residuals: The technique is also known as feathering, peeling and stripping. It is commonly used in pharmacokinetics to resolve a multiexponential curve into its individual components. The time course of drug concentration in plasma is given by equation: The value of Ka is obtained by using the following procedure: Plot the drug concentration versus time with the concentration values on the logarithmic axis. 12 X = C = Ka FX0 (e -Kt - e -Kat ) Vd Vd (Ka-K e )Slide 13: b)Calculate the slope and from the slope calculate ‘k’ and elimination half-life. c)The intercept is equal to Ka F X0/ Vd (Ka-K e ) d)The slope of the residual line is equal to –Ka/2.303. Wagner - Nelson Method: In this method, the absorption process is assumed to be the first order kinetics.This method requires the assumption that the body behaves as a single homogeneous compartment and drug elimination obeys the first order kinetics. 13Slide 14: Derivation: The mass balance equation can be written as the amount administered equals the amount absorbed (A) plus the amount unabsorbed,( Aun ). X0= A + Aun The amount absorbed (A) to any time t, is equal to the sum of the amount of the drug in body (X) and the amount of the drug eliminated from the body to any time, t( Xe ). A= X + Xe Taking the derivative with respect to time gives, dA = dX + dXe dt dt dt 14Slide 15: X= Vd C, hence dX = Vd dC dt dt dXe / dt = KX, but X= Vd C Therefore, dXe = K Vd C dt 15Slide 16: dA = Vd dC + K Vd C dt dt dA = Vd dC + K Vd C dt This method tells the absorption kinetics and also the release of drugs from dosage forms. 16Determination of cmax and tmax : Determination of c max and t max The time needed to reach the maximum concentration C max is called peak time, t max . The time is dependent on the rate constants for absorption (Ka) and elimination(K e ).At maximum concentration, the rate of drug absorption is equal to the rate of drug elimination,so rate of concentration change, dC / dt is equal to zero. The rate of concentration change can be obtained by differentiating equation, C= Ka F X0 (e -Kt - e -Kat ) Vd (Ka-K e ) 17Slide 18: Differentiating with respect to time, dC = - K Ka F X0 e -Kt + Ka 2 F X0 e -Kat dt Vd (Ka-k) Vd (Ka-K) When plasma concenteration reaches C max at time t max , dC / dt =0,therefore K Ka F X0 e - Kt max = Ka 2 FX0 e -Ka t max Vd (Ka-k) Vd (Ka-K) On simplifiction , K e - Kt max = Ka e - Kat max or Ka = e - Kt max K e - Kat max 18Slide 19: In common logarithms: Log Ka = - Kt max + Kat max K 2.303 2.303 Therefore, t max = 2.303 log Ka (Ka-K e ) K e Substituting t p for t: C max = Ka F X0 (e - Kt max – e - Kat max ) Vd (Ka- K e ) 19Volume of distribution(vd): Volume of distribution( vd ) Extrapolation of the semilog plot of the plasma drug concentration versus time gives an intercept, I = Ka FX0 Vd (Ka-K e ) Or Vd = Ka FX0 I (Ka-K e ) The calculation of the apparent volume of distribution of a drug from the plasma drug concentration- time data obtained following the first order absorption can also be done from: Vd = F X0 K e [ AUC] 0 20Lag time: Lag time The time delay prior to the commencement of the first order drug absorption is known as lag time ( t l ).The lag time, t l , represents the beginning of drug absorption.This can be represented by the following equation, C = Ka F X0 (e -K(t- t l ) – e -Ka(t- t l ) Vd (Ka-K e ) Flip-flop of Ka and K : The estimation of the rate constants for absorption and elimination by method of residuals is based on the assumption that Ka>> K.The terminal phase of the plasma level time curve represents elimination and the residual line represents absorption.If K>> Ka,then the terminal phase represents 21Slide 22: Absorption and residual phase represents elimination.This phenomenon is called as the flip-flop of the absorption and elimination rate constants. Most of the drugs observed to have flip-flop characteristics are drugs with a fast elimination.The larger the elimination rate constant of a drug, the greater is the chance for the flip-flop of Ka and K. Special case where Ka = K: A situation may occur where the absorption rate constant and elimination rate constant of a drug are equal. For such a situation an equation that describes the time course of drug concentration in plasma is, 22Slide 23: In concentration terms, C = K’ F X0 t . e - K’t Vd In logarithmic form, log C = log [ K’ F X0 t ] – K’t Vd 2.303 A plot of log C versus time will be a curve from time zero to infinity, because ‘t’ is present in intercept value which is a variable. 23Oral administration of drug-unchanged drug in urine: Oral administration of drug-unchanged drug in urine Excretion Rate Method: The amount of the unchanged drug excreted in urine is only dependent on the amount of unchanged drug in the body.The rate of urinary excretion of the unchanged drug is given by, dXu = Ke X ………(1) dt but, X= Ka FX0 (e -Kt – e -Kat )…….(2) (Ka- K e ) Substituting (2) in (1), dXu = Ke Ka F X0 (e -Kt - e -Kat ) dt (Ka- K e ) 24Slide 25: The rate of urinary drug excretion, dXu / dt cannot be determined directly, so Xu /t is used,which is the average rate of urinary drug excretion at the mid point of urine collection period (t’). Therefore, ∆ Xu = Ke Ka F X0 (e -Kt’ – e -Kat’ ) ∆ t (Ka- K e ) In logarithmic form, log ∆ Xu = log [ Ke Ka FX0 ] – Kt ’ ∆ t (Ka- K e ) 2.303 25Slide 26: Sigma-Minus Method: Cumulative amount of the unchanged drug excreted in urine upto any time can be obtained by the following equation: Xu t = Ke FX0 + KeKa F X0 ( e -Kat - e -Kt ) K (Ka-K) Ka K At t= ∞, equation reduces to, Xu ∞ = Ke F X0 K Where, Xu ∞ = is the total amount of the unchanged drug excreted in urine at infinite time. 26ADVANTAGES OF COMPARTMENT MODELING: ADVANTAGES OF COMPARTMENT MODELING It is a simple and flexible approch . Visual representation of various rate processes involved in drug disposition. It shows how many rate constants are necessary to describe these processes. It is useful in predicting drug concentration-time profile in both normal physiologic and in pathologic conditions. It is important in the development of dosage regimens. 27DISADVANTAGES OF COMPARTMENT MODEL: DISADVANTAGES OF COMPARTMENT MODEL No relationship with the physiologic functions or the anatomic structure of the species. Extensive efforts are required in the development of an exact model The model may vary within a study population. Compartment model may change with route of administration Difficulties generally arise when using model to interpret the differences between results from human and animal experiments. 28references: references “ Biopharmaceutics And Pharmacokinetics” by D.M Brahmankar , Sunil B.Jaiswal , 1 st edition, pp-244-252 “ Biopharmaceutics And Pharmacokinetics” by PL Madan,1 st edition,pp-234-248 29Slide 30: THANK YOU 30 You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
One compartment model(extravascular) nea24 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 810 Category: Science & Tech.. License: All Rights Reserved Like it (1) Dislike it (0) Added: May 29, 2011 This Presentation is Public Favorites: 1 Presentation Description No description available. Comments Posting comment... By: 2060pk (8 month(s) ago) i need this ppt...so plz make me allow to download this ppt.....plz...thanks Saving..... Post Reply Close Saving..... Edit Comment Close By: nea24 (9 month(s) ago) Snd m u'r id rajan Saving..... Post Reply Close Saving..... Edit Comment Close By: rajan290189 (9 month(s) ago) plz make me allow to download this ppt file.. thnx Saving..... Post Reply Close Saving..... Edit Comment Close Premium member Presentation Transcript ONE COMPARTMENT MODEL (EXTRAVASCULAR ADMINISTRATION): ONE COMPARTMENT MODEL (EXTRAVASCULAR ADMINISTRATION) NEHA SOOD M.PHARMACY SEM-2 M100400031 1PHARMACOKINETICS OF DRUG ABSORPTION: PHARMACOKINETICS OF DRUG ABSORPTION Extravascular : In extravascular administration, a drug undergoes the process of absorption before it reaches systemic circulation.The absorption of a drug from the gastrointestinal tract or any other extravascular site is dependent on the following: Physiochemical properties of the drug Physiochemical properties of the dosage form Anatomy and physiology of the absorption site 2Slide 3: The rate of change of the amount of the drug ( dX / dt ) in the body following extravascular administration is equal to the difference between the rate of absorption ( dXa / dt ) and the rate of elimination ( dXe / dt ): dX = dXa - dXe dt dt dt The curve can be divided into various parts based on relative magnitudes of the rate of absorption and the rate of elimination at different time points. 3Slide 4: Absorption ph ase : The whole dose is available for absorption at zero time.The amount of drug at the absorption site declines as a function of time.Therefore , rate of drug absorption is greater than the rate of drug elimination. dXa > dXe dt dt 4Slide 5: Plateau phase: Absorption of a drug into the systemic circulation gradually increases the plasma level of the drug.This increases the rate of elimination of a drug,dXe/ dt.The rate of absorption of a drug decreases with time and the rate of elimination of drug increases with time.The time at which rate of absorption becomes equal to rate of elimination is called as time of peak drug concentration( t max ) in plasma and concentration is peak concentration( C max ). dxa = dXe dt dt 5Slide 6: Post Absorption Phase: The rate of elimination of the drug is faster than the rate of absorption. dXa < dXe dt dt Elimination Phase: When the drug at the absorption site becomes depleted, the rate of drug absorption approaches zero.Therefore , the rate of change in the amount of the drug in the body during the elimination phase is described by the first order process. dXe = -K e X dt 6 X= Amount of drug in the body at any time K= Overall elimination rate constantOral administration of unchanged drug in blood/plasma: Oral administration of unchanged drug in blood/plasma Ka K XO X= Vd.C Where, XO- is an oral dose of the drug administered X-is the amount of drug in the body Ka and K-are the first order rate constants of drug absorption and elimination For a drug that enters the body by an apparent first order absorption first order process and is eliminated by first order process and distributes in the body according to one compartment model. 7Slide 8: Rate of drug change in the body=Rate of drug absorption- Rate of drug elimination dX = dXa - dXe dt dt dt dX = Ka Xa - K e X dt Where, Xa is the amount of the drug available at the absorption site. Let us assume that ‘F’ is the fraction of the oral dose (XO) that will be absorbed. 8Slide 9: At time zero, Fxo amount is available for absorption . Therefore, Xa 0 = F XO Xa 0 = amount of dru g available for absorption at t=0 The rate of disappearance of the drug from the g.i.t is given by, dXa = -Ka Xa dt The time course of the amount of the drug in the body is given by X = Ka F X0 (e -Kt - e -Kat ) (Ka- K e ) 9Slide 10: Dividing the above equation by the volume of distribution of drug ( Vd ), X = C = Ka FX0 (e -Kt - e -Kat ) Vd Vd (Ka-K e ) Above equation describes the time course of the drug concentration in plasma following the oral route of administration. At larger values of t,the above equation will reduce to, C = Ka F X0 e -Kt Vd (Ka-K e ) 10Slide 11: Above equation explains the elimination phase, i.e. the time when absorption no longer occurs. The equation can be written in common logarithms as: log C = log [ Ka F X0 ] – K t/2.303 Vd (Ka- K e ) 11Determination of Absorption rate constant from oral data: Determination of Absorption rate constant from oral data Method of Residuals: The technique is also known as feathering, peeling and stripping. It is commonly used in pharmacokinetics to resolve a multiexponential curve into its individual components. The time course of drug concentration in plasma is given by equation: The value of Ka is obtained by using the following procedure: Plot the drug concentration versus time with the concentration values on the logarithmic axis. 12 X = C = Ka FX0 (e -Kt - e -Kat ) Vd Vd (Ka-K e )Slide 13: b)Calculate the slope and from the slope calculate ‘k’ and elimination half-life. c)The intercept is equal to Ka F X0/ Vd (Ka-K e ) d)The slope of the residual line is equal to –Ka/2.303. Wagner - Nelson Method: In this method, the absorption process is assumed to be the first order kinetics.This method requires the assumption that the body behaves as a single homogeneous compartment and drug elimination obeys the first order kinetics. 13Slide 14: Derivation: The mass balance equation can be written as the amount administered equals the amount absorbed (A) plus the amount unabsorbed,( Aun ). X0= A + Aun The amount absorbed (A) to any time t, is equal to the sum of the amount of the drug in body (X) and the amount of the drug eliminated from the body to any time, t( Xe ). A= X + Xe Taking the derivative with respect to time gives, dA = dX + dXe dt dt dt 14Slide 15: X= Vd C, hence dX = Vd dC dt dt dXe / dt = KX, but X= Vd C Therefore, dXe = K Vd C dt 15Slide 16: dA = Vd dC + K Vd C dt dt dA = Vd dC + K Vd C dt This method tells the absorption kinetics and also the release of drugs from dosage forms. 16Determination of cmax and tmax : Determination of c max and t max The time needed to reach the maximum concentration C max is called peak time, t max . The time is dependent on the rate constants for absorption (Ka) and elimination(K e ).At maximum concentration, the rate of drug absorption is equal to the rate of drug elimination,so rate of concentration change, dC / dt is equal to zero. The rate of concentration change can be obtained by differentiating equation, C= Ka F X0 (e -Kt - e -Kat ) Vd (Ka-K e ) 17Slide 18: Differentiating with respect to time, dC = - K Ka F X0 e -Kt + Ka 2 F X0 e -Kat dt Vd (Ka-k) Vd (Ka-K) When plasma concenteration reaches C max at time t max , dC / dt =0,therefore K Ka F X0 e - Kt max = Ka 2 FX0 e -Ka t max Vd (Ka-k) Vd (Ka-K) On simplifiction , K e - Kt max = Ka e - Kat max or Ka = e - Kt max K e - Kat max 18Slide 19: In common logarithms: Log Ka = - Kt max + Kat max K 2.303 2.303 Therefore, t max = 2.303 log Ka (Ka-K e ) K e Substituting t p for t: C max = Ka F X0 (e - Kt max – e - Kat max ) Vd (Ka- K e ) 19Volume of distribution(vd): Volume of distribution( vd ) Extrapolation of the semilog plot of the plasma drug concentration versus time gives an intercept, I = Ka FX0 Vd (Ka-K e ) Or Vd = Ka FX0 I (Ka-K e ) The calculation of the apparent volume of distribution of a drug from the plasma drug concentration- time data obtained following the first order absorption can also be done from: Vd = F X0 K e [ AUC] 0 20Lag time: Lag time The time delay prior to the commencement of the first order drug absorption is known as lag time ( t l ).The lag time, t l , represents the beginning of drug absorption.This can be represented by the following equation, C = Ka F X0 (e -K(t- t l ) – e -Ka(t- t l ) Vd (Ka-K e ) Flip-flop of Ka and K : The estimation of the rate constants for absorption and elimination by method of residuals is based on the assumption that Ka>> K.The terminal phase of the plasma level time curve represents elimination and the residual line represents absorption.If K>> Ka,then the terminal phase represents 21Slide 22: Absorption and residual phase represents elimination.This phenomenon is called as the flip-flop of the absorption and elimination rate constants. Most of the drugs observed to have flip-flop characteristics are drugs with a fast elimination.The larger the elimination rate constant of a drug, the greater is the chance for the flip-flop of Ka and K. Special case where Ka = K: A situation may occur where the absorption rate constant and elimination rate constant of a drug are equal. For such a situation an equation that describes the time course of drug concentration in plasma is, 22Slide 23: In concentration terms, C = K’ F X0 t . e - K’t Vd In logarithmic form, log C = log [ K’ F X0 t ] – K’t Vd 2.303 A plot of log C versus time will be a curve from time zero to infinity, because ‘t’ is present in intercept value which is a variable. 23Oral administration of drug-unchanged drug in urine: Oral administration of drug-unchanged drug in urine Excretion Rate Method: The amount of the unchanged drug excreted in urine is only dependent on the amount of unchanged drug in the body.The rate of urinary excretion of the unchanged drug is given by, dXu = Ke X ………(1) dt but, X= Ka FX0 (e -Kt – e -Kat )…….(2) (Ka- K e ) Substituting (2) in (1), dXu = Ke Ka F X0 (e -Kt - e -Kat ) dt (Ka- K e ) 24Slide 25: The rate of urinary drug excretion, dXu / dt cannot be determined directly, so Xu /t is used,which is the average rate of urinary drug excretion at the mid point of urine collection period (t’). Therefore, ∆ Xu = Ke Ka F X0 (e -Kt’ – e -Kat’ ) ∆ t (Ka- K e ) In logarithmic form, log ∆ Xu = log [ Ke Ka FX0 ] – Kt ’ ∆ t (Ka- K e ) 2.303 25Slide 26: Sigma-Minus Method: Cumulative amount of the unchanged drug excreted in urine upto any time can be obtained by the following equation: Xu t = Ke FX0 + KeKa F X0 ( e -Kat - e -Kt ) K (Ka-K) Ka K At t= ∞, equation reduces to, Xu ∞ = Ke F X0 K Where, Xu ∞ = is the total amount of the unchanged drug excreted in urine at infinite time. 26ADVANTAGES OF COMPARTMENT MODELING: ADVANTAGES OF COMPARTMENT MODELING It is a simple and flexible approch . Visual representation of various rate processes involved in drug disposition. It shows how many rate constants are necessary to describe these processes. It is useful in predicting drug concentration-time profile in both normal physiologic and in pathologic conditions. It is important in the development of dosage regimens. 27DISADVANTAGES OF COMPARTMENT MODEL: DISADVANTAGES OF COMPARTMENT MODEL No relationship with the physiologic functions or the anatomic structure of the species. Extensive efforts are required in the development of an exact model The model may vary within a study population. Compartment model may change with route of administration Difficulties generally arise when using model to interpret the differences between results from human and animal experiments. 28references: references “ Biopharmaceutics And Pharmacokinetics” by D.M Brahmankar , Sunil B.Jaiswal , 1 st edition, pp-244-252 “ Biopharmaceutics And Pharmacokinetics” by PL Madan,1 st edition,pp-234-248 29Slide 30: THANK YOU 30