# Determination of Absorption-Rate Constant

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Determination of Absorption-Rate Constant
For more presentations and information visit Determination of Absorption-Rate Constant By: Arooj Khalid Alvi

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Drug Absorption “Absorption is defined as the appearance of drug in the body fluids.”

Kinetics of oral absorption of drug
For more presentations and information visit Kinetics of oral absorption of drug ka ke dDB/dt = dDG/dt – dDE/dt Graphically: A plasma-level time curve showing drug absorption and elimination can be made. Absorption phase: Rate of drug absorption is greater than the rate of drug elimination dDG/dt > dDE/dt Peak-drug concentration: Rate of drug absorption equals rate of drug elimination dDG/dt = dDE/dt Post-absorption phase: Rate of drug elimination at this time is faster than the rate of absorption dDG/dt < dDE/dt Elimination phase: Drug absorption appears zero. The plasma level time curve only represents the rate of drug elimination dDB/dt =-kDB DGi DB VD DE Post-absorption phase Absorption phase Elimination phase

ABSORPTION MODELS Zero order absorption models
For more presentations and information visit ABSORPTION MODELS Zero order absorption models Zero-order drug absorption from the dosing site into the plasma occurs when a zero-order controlled release delivery system is used. Mathematically: The rate of drug input is k0 and the rate of drug elimination is k. Therefore net change per unit time in the body is: dDB/dt = -kDB

First-order absorption model
For more presentations and information visit Contd… First-order absorption model Normally absorption process in the body is assumed to follow first-order kinetics. This model applies to the drugs those are in solution form or rapidly dissolving dosage forms.

Differential Equations:
For more presentations and information visit Differential Equations: The differential equations describing the rates of change of the three components DG, DB and DE are: the rate of disappearance of drug from the gut: dDG/dt = -kaDGF (where F is the fraction absorbed the rate of drug eliminated: dDE/dt = +kelDB  the rate of drug change in the body is the rate of drug in – the rate of drug out of the body:  dDB/dt =F kaDG - kelDB dDB/dt =FkaDG-kelV.Cp (DB=V.Cp) The first term, ka •DG, represents absorption and the second term, kel.V.Cp, represents elimination. Using Laplace transforms, our equation becomes:

Absorption rate constant
For more presentations and information visit Absorption rate constant DEFINITION: “It may be described as a value describing how much drug is absorbed per unit of time”.  DETERMINATION: Absorption rate constant can be determined by the “Method of residuals” by plotting the oral absorption data. 1. In One-Compartment Model by plotting amount of drug absorbed versus time by plotting the amount of drug unabsorbed versus time 2. In Two-Compartment Model.

1. In One-Compartment Model
For more presentations and information visit 1. In One-Compartment Model Amount of drug absorbed vs. time: Using the equation We can get the y-intercept. As with the passage of time, drug absorption is virtually complete so exp(-kat)=0 As = B So Cp=Be-kt This equation representing the first-order drug elimination will yield a linear plot on semi log paper. This slope is equal to –k/2.303.The value of ka can be obtained using the method of residuals, or feathering technique

Method of residuals: Steps involved in method of residuals:
For more presentations and information visit Steps involved in method of residuals: The value of ka can be obtained by the following procedure: 1.Plot the drug-concentration versus time, with time on x-axis and conc. on y-axis. 2.Obtain the slope of the terminal phase by extrapolation 3.Take any points on the upper part of the line(x1,x2,and x3) and drop vertically to obtain corresponding points on the curve(x’1,x’2,x’3) 4.Read the conc. values at x1 and x’1 and so on and plot the difference at the corresponding time point’s t1,t2,t3.a straight line will be obtained with a slope of –ka/2.303.

1. In One-Compartment Model (Cont…)
For more presentations and information visit 1. In One-Compartment Model (Cont…) Lag-time: In few individuals, the absorption of drug after a single oral-dose does not start immediately the time delay prior to the commencement of first-order absorption is known as lag-time. Flip-Flop Of Ka And Kel: Using the method of residuals to estimate ka and kel, the terminal phase of the absorption curve is usually represented by kel whereas steeper slope is represented by ka .In few cases kel obtained from oral absorption does not agree with that obtained after I/v bolus. Apparently the ka and kel obtained by this method has been interchanged. This phenomenon is termed as “flip-flop” of ka and kel. Drugs having flip-flop characteristics: are generally the one that have fast elimination.(ke>ka).so for drugs with large kel, the chance of flip-flop is much greater.

1. In One-Compartment Model (Cont…)
For more presentations and information visit 1. In One-Compartment Model (Cont…) B. Fraction of drug un-absorbed vs. time: (Wagner-Nelson Method) From Blood-Data: Let Ab = DB+DU (amount of drug absorbed) And Ab∞=amount of drug absorbed at t∞ Ab= CpVD+ kVD[AUC]t 0 Ab∞=0+ kVD[AUC]∞ 0 as at t=∞,Cp∞=0 Ab/Ab∞= Cp+k[AUC]t 0/ k[AUC]∞ 0 The drug remaining in the GIT at any time t is: DGI=D0e-kat and DGI/ D0= e-kat log DGI/ D0=-kat/2.303 (taking log) as DGI/ D0is actually the fraction of drug unabsorbed that is (1-Ab/Ab∞), so the plot of this fraction unabsorbed against t yields the slope-ka/2.303.

1. In One-Compartment Model (Cont…)
For more presentations and information visit 1. In One-Compartment Model (Cont…) Steps in determination of ka: Plot the fraction of drug unabsorbed vs. time. find k from the terminal part of the slope when the slope is –k/2.303. Find [AUC]0t by plotting Cp vs. t. Find k[AUC]0t by multiplying each [AUC]0t with k. Find all [AUC]∞t by adding up all [AUC]0t from t=0 to t=∞ Determine (1- Ab/Ab∞) Plot this (1- Ab/Ab∞) vs. time on semi log graph paper with (1- Ab/Ab∞) on y-axis. If this graph gives a linear regression. Then the rate of drug absorption DGI/ Dt is a first-order process. 1-(Ab/Ab∞)

1. In One-Compartment Model (Cont…)
For more presentations and information visit 1. In One-Compartment Model (Cont…) Advantages The absorption and elimination processes can be quite similar and still accurate determinations of ka can be made. The absorption process doesn't have to be first order. This method can be used to investigate the absorption process. This type of method has been used to investigate data obtained after IM administration and it was found that two absorption steps maybe appropriate. Possibly a fast step from drug in solution and a slower step from drug precipitated at the injection site. Disadvantages The major disadvantage of this method is that you need to know the elimination rate constant, from data collected following intravenous administration.

1. In One-Compartment Model (Cont…)
For more presentations and information visit 1. In One-Compartment Model (Cont…) From the Urinary Data Absorption rate constant can also be determined from urinary-excretion data using the plot of amount of drug un-absorbed vs. time. And the fraction of drug absorbed is Abt/ Ab∞=(dDU/dt)+k(DU)t /kDU∞ And a plot of the fraction of drug unabsorbed (1- Abt/Ab∞), vs. t gives –k/2.3 as a slope from which the absorption rate constant can be determined.

2. In Two-Compartmental Model (Loo-Riegelman Method)
For more presentations and information visit 2. In Two-Compartmental Model (Loo-Riegelman Method) Plotting the percent of drug unabsorbed vs. time can be used for two-compartment model as well. This method requires that the drug is given i/v and orally. Total amount of drug in the body : Ab=Dp+Dt+Du Ab/Ab∞=( Cp +Dt/Vp+ k[AUC]0t)/ k[AUC]0∞ Limitation: For calculation of ka by this method, the drug must be given i/v to allow evaluation of the distribution and the elimination rate constants .For the drugs that cannot be given i/v, ka cannot be calculated by Loo-Reiegelman Method. So for those drugs, Wagner-Nelson method may be used that assumes ka in one-compartment model.

Effects of Absorption-Rate Constant
For more presentations and information visit Effects of Absorption-Rate Constant Higher the absorption rate constant, faster the absorption of the drug. As the peak concentration is a function of both the rate of absorption and the rate of elimination. If the elimination rate is the same, as the absorption rate increases, the peak concentration increases. And the product with highest peak conc, will exhibit highest intensity of effects. so If the value for ka and kel are reversed ,tmax is same but Cmax and AUC are different. If kel is constant and only ka is increased, then tmax becomes shorter and Cmax is increased and AUC remains same. And if ka is kept constant but ke is increased, then tmax ,Cmax and AUC decreases.

Significance of Absorption-Rate Constant
For more presentations and information visit Significance of Absorption-Rate Constant For many immediate-release dosage forms, the absorption process is first-order due to the physical nature of drug diffusion. The calculation of ka is useful in designing a multiple-dosage regimen. Knowledge of ka and k allows for the prediction of peak and trough plasma drug concentrations following multiple dosing. In bio-equivalent studies, time of peak concentrations can be very useful in comparing respective rates of absorption of a drug from chemically equivalent drug products.

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References Applied Biopharmaceutics and Pharmacokinetics by Leon Shargel.Chapter no.7,page no.161. Pharmacology and Experimental Therapeutics Department Glossary at Boston University School of Medicine. Journal of Pharmacokinetics and Pharmacodynamics, Volume 8, Number 2 / April, 1980, pages Mayersohn, M and Gibaldi, M Mathematical Methods in Pharmacokinetics. I Use of the Laplace Transform for Solving Differential Rate Equations, Amer. J. Pharm. Education, 34, Mosby's Medical Dictionary, 8th edition. © 2009, Elsevier.