SATURATION AND FIRST ORDER KINETICS PRESENTED BY: POKIYA ANKIT M.PHARM PHARMACOLOGY SEM-II 1

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In some cases, the rate process of drug’s ADME are dependent upon carrier or enzymes that are substrate-specific, have definite capacities & susceptible to saturation at high drug concentration. Such pharmacokinetics are said to be DOSE DEPENDENT, MIXED ORDER, NONLINEAR, & CAPACITY LIMITED KINETICS. A number of drugs demonstrate saturation or capacity- limitedmetabolism in humans. Ex. glycine conjugation of salicylate . 2

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Drugs that demonstrate saturation kinetics usually show the following characteristics. Elimination of drug does not follow simple first-order kinetics—that is, elimination kinetics are nonlinear. The elimination half-life changes as dose is increased. Usually, the elimination half-life increases with increased dose due to saturation of an enzyme system. However, the elimination half-life might decrease due to "self"-induction of liver biotransformation enzymes, as is observed for carbamazepine . The area under the curve (AUC) is not proportional to the amount of bioavailable drug. The saturation of capacity-limited processes may be affected by other drugs that require the same enzyme or carrier-mediated system ( i.e , competition effects). The composition and/or ratio of the metabolites of a drug may be affected by a change in the dose. 3

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conti … DETECTION OF NONLINEARITY:- Two simple tests for detection of nonlinearity Determination of steady state plasma conc. at different doses. Determination of some of the important pharmacokinetic parameters such as fraction bioavailable, elimination half life, or total systemic clearance at different doses of the drug. CAUSES OF NONLINEARITY: Drug absorption Drug distribution Drug metabolism Drug excretion 4

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conti … DRUG ABSORPTION : Sources of nonlinearity: When absorption is solubility or dissolution rate limited. Ex:- Griseofulvin . Intestinal metabolism. Ex:- Salicylamide , propranolol Drugs with low solubility in GI but relatively high dose Ex:- Chorothiazide , griseofulvin , danazol Saturable gastric or GI decomposition Ex:- Penicillin G, omeprazole , saquinavir Saturable transport in gut wall Ex:- Riboflavin, gebapentin , L-dopa, baclofen , ceftibuten When absorption involves carrier-mediated transport systems. Ex:- Riboflavin, Ascorbic acid, Cyanocobalamin . 5

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conti … When presystemic gut wall or hepatic metabolism attains saturation Ex: Proprenolol, Hydralazine & Verapamil . Other causes include change in gastric emptying, GI blood flow & other physiological factors. 6

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conti … DRUG DISTRIBUTION: Sources of nonlinearity in drug distribution: Saturation of binding sites on plasma proteins. Ex:- Phenylbutazone & naproxen, lidocaine , salicylic acid, ceftriaxone , diazoxide , phenytoin , warfarin . Saturation of tissue binding sites. Ex:- thiopental & Fentanyl ., disopyramide Cellular uptake Ex:- Methicillin (rabbit) Tissue binding Ex:- Imiprimine (rat) CSF transport Ex:- Benzylpenicillins Saturable transport into or out of tissues Ex:- Methotrexate Clearance is also altered depending upon extraction ratio of drug. 7

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conti … DRUG METABOLISM : Causes of nonlinearity in metabolism are: Capacity limited metabolism due to enzyme & cofactor saturation. Ex:- Phenytoin, alcohol, theophylline. Enzyme induction. Ex:- Carbamazepine, where a decrease in peak plasma concentration has been observed on repetitive administration. Saturable metabolism Ex:- Phenytoin , salicyclic acid, theophylline , valproic acid. Cofactor or enzyme limitation Ex:- Acetaminophen, Altered hepatic blood flow Ex:- Propranolol , verapamil Metabolite inhibition Diazepam. Other causes, saturation of binding sites & pathological situation such as hepatotoxicity . 8

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conti … DRUG EXCRETION: Two active saturable processes are: Active tubular secretion. Ex:- Penicillin G. Active tubular reabsorption . Ex:- Water soluble vitamins & glucose. Biliary secretion Ex:- Iodipamide , sulfobromophthalein sodium. Enterohepatic recycling Ex:- Cimetidine , isotretinoin . Other causes include forced diuresis, change in pH, nephrotoxicity & saturation of binding sites. Kinetic of capacity-limited or saturable processes is best described by Michaelis-menten equation. 9

MICHAELIS MENTEN EQUATION::

MICHAELIS MENTEN EQUATION: 10 Where, dCp / dt is rate of decline of drug concentration with time. C p is the concentration of drug in the plasma V max is the maximum elimination rate K M is the Michaelis constant

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conti … Three situation can be considered depending upon values of K M & C When K M = C :- under this condition, the equation reduces to i.e. rate process is equal to one half its maximum rate. 11 a PLOT of michaelis-menten equation( elimination rate versus concentration).initially ,the rate increases linearly with concentration & then reaches maximum beyond which it proceeds at a constant rate .

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12 A plot of michaelis menten equation ( elimination rate dc/ dt vs. C ). Initially the rate increase linerly (first order) with concentration become mixed order at higher concentration. And then reaches maximum ( Vmax ) beyond which it proceeds at constant rate( zero order).

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conti … When K M >> C :- C p in the denominator is negligible , The above equation is identical to the one that describes first order elimination of a drug where Vmax /K M =K’. 13

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conti … When C p >> K M saturation of the enzymes occurs and the value for K M is negligible. This is identical to one that describes a zero order process i.e. the rate process occurs at a constant rate Vmax & is independent of drug concentration e.g. metabolism of ethanol. 14

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conti … Estimation of Km & Vmax : This parameters can be assessed from plasma concentration-time data collected after i.v. bolus administration of a drug with nonlinear elimination characteristics. Integration of Michaelis Menten equation given below, 15

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conti … Integration followed by conversion to log base 10 yields : log Cp= log C 0 p + ( C 0 p -C) / 2.303 K M – Vmax / 2.303 K M A semi log plot of Cp vs. t yields a curve with a terminal linear portion having Slope of – Vmax /2.303 K M & when back extrapolated to time zero gives y-intercept log Co extrapolated. The equation describes this line is : At low plasma conc., both above equations are identical. 16

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conti … Equating both the 2 & simplifying, we get : K M thus can be obtained from above equation. Vmax can be obtained by substituting the value of K M in the slope value. 17

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conti … An alternating approach of estimating Vmax & K M is determining rate of change of plasma drug conc. at different times & using the reciprocal of Michaelis Menten Equation : where,V = dCp / dt , C= plasma drug conc. At midpoint of the sampling interval. This is known as double reciprocal plot or Lineweaver-Burke plot of 1/ dCp / dt vs. 1/C which yields a straight line with slope= K M /Vmax & y-intercept = 1/Vmax. 18

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conti … A disadvantage of Lineweaver-Burke plot is that the points are clustered . More reliable plots in which points are uniformly scattered are Hanes-Woolf plot (equation 1) & Woolf- Augustinsson - Hofstee plot ( equation 2) …..equation 1 ……equation 2 19

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conti … K M & Vmax from steady state concentration: W hen a drug is administered as a constant rate i.v. infusion or in a multiple dose regimen, the steady state concentration Css is given in terms of dosing rate DR as: DR= C ss Cl T where, DR=R 0 when drug is administered as zero order i.v. infusion At steady state, dosing rate equals rate of decline(elimination b’coz of metabolism) in plasma drug conc.& if decline is due to a single capacity limited process, then; R = dose/day or dosing rate(DR) 20

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conti … Graphical computation of K M & Vmax: Lineweaver-Burke plot / Klotz plot: A plot of 1/R vs. 1/Css yields a straight line with slope K M / Vmax & y-intercept 1/Vmax. 21

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conti … 22 LINEWEAVER-BURKE/KLOTZ PLOT

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conti … Direct linear plot: Here, a pair of Css i.e. Css1 & Css2 obtained with two different dosing rates DR1 & DR2 is plotted. The points Css1 &DR1 are joined to form a line and a second line is obtained similarly by joining Css2 & DR2. The points where these two lines intersect each other is extrapolated on DR axis to obtain Vmax & on x-axis to get K M . 23

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conti … The third graphical method: A plot of R vs. R/ Css yields a straight line with slope –K M & y-intercept Vmax. 24

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conti … There are certain limitation of K M & Vmax estimated by assuming one-compartment system & a single capacity-limited process. More complex equations will results & the computed K M & Vmax will usually be larger when : The drug is eliminated by more than one capacity limited process. The drug exhibits parallel capacity limited and first-order elimination process. The drug follows multi compartment kinetics. 25

First order kinetics:

First order kinetics 26

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First-Order Reactions If the amount of drug A is decreasing at a rate that is proportional to the amount of drug A remaining, then the rate of disappearance of drug A is expressed as ……….equation 1 where k is the first-order rate constant and is expressed in units of time – 1 From the equation it is clear that a first order process is the one whose rate is directly proportional to the concentration of drug undergoing reaction. i.e. greater the concentration faster the reaction. 27

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It is because of such proportionality between rate of reaction of drug that first order process is said to be followed linear kinetics. ln A = - kt + ln A O ……….equation 2 Integration of Equation yields the following expression: ……….equation 3 Since equation 3 has only one exponent the first order process is also called as monoexponantial rate process. Thus, first order process is characterized by logarithmic or exponantial kinetics. i.e. a constant fraction of the drug undergoes reaction per unit time. 28

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Because ln = 2.3 log, Equation becomes ……….equation 4 When drug decomposition involves a solution, starting with initial concentration C 0 , it is often convenient to express the rate of change in drug decomposition, dC / dt , in terms of drug concentration, C , rather than amount because drug concentration is assayed. 29

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Hence, ……….equation 5 ……….equation 6 Above equation may be expressed as ……….equation 7 Because ln = 2.3 log, Equation becomes ……….equation 8 30

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According to Equation 4 ,A graph of log A versus t will yield a straight line , the y intercept will be log A 0 , and the slope of the line will be – k /2.3 . Similarly, a graph of log C versus t will yield a straight line according to Equation 8. The y intercept will be log C 0 , and the slope of the line will be – k /2.3. For convenience, C versus t may be plotted on semilog paper without the need to convert C to log C . An example is shown in . 31

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This graph demonstrates the constancy of the t 1/2 in a first-order reaction. 32

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Half-Life Half-life ( t 1/2 ) expresses the period of time required for the amount or concentration of a drug to decrease by one-half. First-Order Half-Life The t 1/2 for a first-order reaction may be found by means of the following equation: It is apparent from this equation that, for a first-order reaction, t 1/2 is a constant. No matter what the initial amount or concentration of drug is, the time required for the amount to decrease by one-half is a constant. 33

references:

references “SHARGEL” L. , “WU-PONG” S. , “YU ANDREW” B. C. , NONLINEAR PHARMACOKINETICS, APPLIED BIOPHARMACEUTICS & PHARMACOKINETICS , FIFTH EDITION, THE MCGRAW-HILL COMPANIES, 219-248. “BRAHMANKAR” D.M. , “JAISWAL” S. B. , NONLINEAR PHARMACOKINETICS, BIOPHARMACEUTICS AND PHARMACOKINETICS-A TREATIES, SECOND EDITION-2009, VALLABH PRAKASHAN , 316-317. “GIBALDI” M., “PERRIER” D., NONLINEAR PHARMACOKINETICS, PHARMACOKINETICS , SECOND EDITION, REVISED & EXPANDED, INFORMA HEALTHCARE, 271-314. 34

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