logging in or signing up one compartment open model drumadevi 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: Embed: Flash iPad Copy Does not support media & animations WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 1927 Category: Science & Tech.. License: All Rights Reserved Like it (4) Dislike it (0) Added: September 05, 2011 This Presentation is Public Favorites: 5 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Slide 1: BY Dr. S.K. Umadevi M.Pharm.,Ph.D ; Professor Rao’s college of Pharmacy, Nellore Andhra pradesh One compartment open modelIntroduction: Introduction In one compartment open model the body is considered as a single well- mixed compartment with out barriers to the movement of drugs. The drug administered is assuming to be following first-order disposition process. The term open indicates that the input and output are unidirectional and the drug can be eliminated from the body.Cont…: Cont… Applies to the drugs that distribute rapidly through out the body. Absorption and elimination processes are dynamic. The drug concentration reference compartment is Plasma. Any change in plasma drug concentration reflects a proportional change in drug concentration throughout the body.Representation of one-compartment open model (oral): Representation of one-compartment open model (oral)Types of one-compartment open models: Types of one-compartment open models One-compartment open models can be defined based on the rate of input. OCOM i.v. bolus administration. OCOM i.v. infusion. OCOM e.v. administration, Zero-order absorption. OCOM e.v. administration, First-order absorption.OCOM i.v. bolus administration: OCOM i.v. bolus administration Drug that distributes rapidly in the body is given in the form of a rapid intravenous injection (i.e. i.v. bolus or slug). It takes about 1 to 3 minutes for complete circulation and therefore the rate of absorption is neglected in calculations.Representation of one-compartment open model- i.v bolus: Representation of one-compartment open model- i.v bolus dX/dt = Rate in – Rate out (1) Since rate in or absorption is absent, equation (1) becomes dX/dt = – Rate out (2)Cont…: Cont… If the rate out or elimination follows first-order kinetics, then: dX /dt = -K E X (3) where K E = first-order elimination rate constant, and X = amount of drug in the body at any time t remaining to be eliminated. Negative sign indicates that the drug is being lost from the body.Estimation of Pharmacokinetic Parameters – i.v Bolus Administration : Estimation of Pharmacokinetic Parameters – i.v Bolus Administration For a drug that follows one-compartment kinetics and administered as rapid i.v . injection, the decline in plasma drug concentration is only due elimination phase. Elimination phase can be characterized by 3 parameters :- Elimination rate constant Elimination half-life Apparent volume of Distribution ClearanceElimination rate constant : Elimination rate constant Integration of equation 3 yields In X = In X 0 - K E t (4) where, X 0 = amount of drug at time t = zero Equation (4) in the exponential form as: X = X 0 e -K E t (5) The equation shows that disposition of a drug is mono exponential .Cont..: Cont.. Taking log on both sides logX = logX 0 – K E t /2.303 (6) Since, X= V d C (7) V d = proportionality constant popularly known as the apparent volume of distribution logC = logC 0 – K E t /2.303 (8) where, C 0 = plasma drug concentration immediately after i.v. injectionOCOM- I.V BOLUS- Conc. Vs Time PLOT: OCOM- I.V BOLUS- Conc. Vs Time PLOT K E = K e + K m + K b + K 1 + …..Elimination half-life : Elimination half-life It is also called as biological half life. It is defined as the time taken for the amount of drug in the body as well as plasma concentration to decline by one-half or 50% its initial value t 1/2 = 0.693/K E (9)Cont…: Cont… half-life is a secondary parameter that depends upon the primary parameters clearance and apparent volume of distribution, according to following equation: t 1/2 = 0.693V d /Cl T (10)Apparent volume of Distrubution : Apparent volume of Distrubution V d is a measure of the extent of distribution of drug and is expressed in litres. V d = Amount of drug in the body /Plasma drug concentration = X/C (11) = X 0 /C 0 = i.v. bolus dose/C 0Clearance : Clearance Clearance is the most important parameter in clinical drug applications and is useful in evaluating the mechanism by which a drug is eliminated by the whole organism or by a particular organ. Clearance=Rate of elimination/Plasma drug concentration = dX/dt (12) CCont…: Cont… Cl T = Cl R + Cl H + Cl others Cl T = Rate of elimination by kidney/C Cl H = Rate of elimination by liver/C Cl others = Rate of elimination by other organs/C From eqn (3), dX / dt = K E X Cl T = K E X/C = K E V d = 0.693V d /t 1/2 (13)OCOM i.v. infusion: OCOM i.v. infusion Advantages of zero-order infusion of drugs include: Ease of control of rate of infusion to fit individual patient needs. Prevents fluctuating maxima and minima (peak and valley) plasma level. This is desired especially when the drug has a narrow therapeutic index. Other drugs, electrolytes and nutrients can be conveniently administered simultaneously by the same infusion line in critically ill patients.OCOM – i.v INFUSION- MODEL: OCOM – i.v INFUSION- MODEL At any time during infusion, the rate of change in the amount of drug in the body, dX/dt is the difference between the zero-order rate of drug infusion R 0 and first-order rate of elimination, -K E X: dX/dt = R 0 -K E X (14)Cont…: Cont… On integration, X = R 0 (1-e -K E t ) K E (15) since X= V d C, C= R 0 (1-e -K E t ) K E V d = R 0 (1-e -K E t ) Cl T (16)Cont…: Cont…Cont…: Cont… At steady-state, the rate of change in amount of drug in the body is zero, hence, the equation (14) becomes: Zero = R 0 – K e X ss K e X ss = R 0 (15) Transforming to concentration terms and rearranging the equationCont…: Cont… C ss = R 0 / K E V d = R 0 / C l T (16) = Infusion rate/Clearance C = C ss ( 1-e -K E t ) (17) Rearrangement: C ss – C/ C ss = e - K E t (18) Taking log on both sides: log [ C ss – C/ C ss ] = - K E t /2.303 (19)SEMILOG PLOT OF i.v -INFUSION: SEMILOG PLOT OF i.v -INFUSIONEstimation of Pharmacokinetic Parameters – IV infusion: Estimation of Pharmacokinetic Parameters – IV infusion The first-order elimination rate constant and elimination half-life can be computed from a semilog plot of post-infusion concentration-time data.OCOM e.v. administration: OCOM e.v. administration When a drug is administered by extravascular route (e.g. oral, i.m., rectal, etc.), absorption is a prerequisite for its therapeutic activity. The rate of absorption may be described mathematically as a zero-order or first-order process.Differences between zero-order and first-order kinetics are illustrated here..: Differences between zero-order and first-order kinetics are illustrated here..Cont…: Cont… After e.v. administration, the rate of change in the amount of drug in the body dX/dt is the difference between the rate of input (absorption) dX ev /dt and rate of output (elimination) dX E /dt. dX/dt = Rate of absorption —Rate of elimination dX/dt = dX ev dt – dX E /dt (20)PLASMA CONC. TIME PROFILE- e.v: PLASMA CONC. TIME PROFILE- e.v For a drug that follows one-compartment kinetics, the plasma concentration-time profile is characterized by absorption phase, post-absorption phase and elimination phaseCont…: Cont… During the absorption phase , the rate of absorption is greater than the rate of elimination dX ev /dt > dX E /dt (21) At peak plasma concentration , the rate of absorption equals the rate of elimination and the change in amount of drug in the body is zero. dX ev /dt = dX E /dt (22)Cont…: Cont… During the post-absorption phase , there is some drug at the extravascular site still remaining to be absorbed and the rate of elimination at this stage is greater than the absorption rate. dX ev /dt < dX E /dt (23) After completion of drug absorption, its rate becomes zero and the plasma level time curve is characterized only by the elimination phaseZero-order Absorption model - e.v administration: Zero-order Absorption model - e.v administration This model is similar to that for constant rate infusion. The rate of drug absorption, as in the case of several controlled drug delivery systems, is constant and continues until the amount of drug at the absorption site is depleted. All equations that explain the plasma concentration-time profile for constant rate i.v. infusion are also applicable to this model.First-order Absorption model - e.v administration: First-order Absorption model - e.v administration For a drug that enters the body by a first-order absorption process, gets distributed in the body according to one-compartment kinetics and is eliminated by a first-order process, the model can be depicted as follows On differentiating the equation (20) dX/dt = K a X a – K E X (24) where F = fraction of drug absorbed systemically after e.v. administrationCont…: Cont… On integrating.. X = K a FX 0 [e -K E t -e -Kat ] (25) (K a –K E ) Transforming into concentration terms, the equation becomes C = K a FX 0 [e -K E t -e -Kat ] (26) V d (K a –K E ) where F = fraction of drug absorbed systemically after e.v. administrationAssessment of Pharmacokinetic Parameters – e.v. administration: Assessment of Pharmacokinetic Parameters – e.v. administration At peak plasma concentration, the rate of absorption equals rate of elimination i.e. K a X a = K E X and the rate of change in plasma drug concentration dC/dt = zero. This rate can be obtained by differentiating equation (26). dC = K a FX 0 [-K E e -K E t +K a e -Kat ] = Zero dt V d (K a –K E ) (27)Cont…: Cont… On simplification: K E e -K E t = K a e -Kat (28) Converting to log form: Log K E - K E t/2.303 = log K a - K a t/2.303 (29) Rearranging the above eqn.. t max = 2.303log(K a /K E ) K a –K E (30)Cont…: Cont… The above equation shows that as K a becomes larger than K E , t max becomes smaller since (K a - K E ) increases much faster than log K a /K E . C max can be obtained by substituting equation (30) in equation (26 ) C max = FX 0 e –K E t max V d (31)Cont…: Cont… It has been shown that at C max , when K a = K E , t max = 1/K E . Hence, the above equation further reduces to: C max = FX 0 e –1 = 0.37FX 0 (32) V d V dCont…: Cont… Elimination Rate Constant : This parameter can be computed from the elimination phase of the plasma level time profile. For most drugs administered e.v., absorption rate is significantly greater than the elimination rate i.e. K a t » K E t. Hence, one can say that e -Kat approaches zero much faster than does e -K E t At such a stage, when absorption is complete, the change in plasma concentration is dependent only on elimination rate and equation (26) reduces to C = K a FX 0 e -K E t (33) V d (K a –K E )Cont…: Cont… Absorption Rate Constant : It can be calculated by the 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. For a drug that follows one-compartment kinetics and administered e.v., the concentration of drug in plasma is expressed by a biexponential equation (26) C = K a FX 0 [e -K E t -e -Kat ] (26) V d (K a –K E ) K a FX 0 / V d (K a –K E ) = A, a hybrid constant, then: C = Ae -K E t - A e -Kat (34)Cont…: Cont… During the elimination phase, when absorption is almost over, K a »K E and the value of second exponential e -Kat approaches zero whereas the first exponential e -K E t retains some finite value. At this time, the equation (27) reduces to: C = Ae -K E t (35) Log on both sides: log C = log A – K E t/2.303 (36)Curve fitting method- OCOM-e.v : Curve fitting method- OCOM-e.vCont…: Cont… (C-C) = C r = Ae -Kat (37) Log form: log C r = log A – K a t/2.303 (38) A plot of log C r versus t yields a straight line with slope -K a /2.303 and y-intercept log A. Absorption half-life can then be computed from K a using the relation 0.693/K a . Thus, the method of residuals enables resolution of the biexponential plasma level-time curve into its two exponential components. The technique works best when the difference between K a and K E is large (K a /K E > 3).Cont…: Cont… In some instances, the K E obtained after i.v. bolus of the .same drug is very large, much larger than the K a obtained by the method of residuals (e.g. isoprenaline) and if K E /K a > 3, the terminal slope estimates K E and not K a whereas the slope of residual line gives K E and not K a . This is called as flip-flop phenomenon since the slopes of the two lines have exchanged their meanings. Ideally, the extrapolated and the residual lines intersect each other on y-axis i.e. at time t = zero and there is no lag in absorption. However, if such an intersection occurs at a time greater than zero, it indicates time lag. It is defined as the time difference between drug administration and start of absorption. It is denoted by symbol t 0 and represents the beginning of absorption process.Slide 45: Your’s Umadevi You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.