One-Compartment Open Model: Intravenous Bolus PDF

One-Compartment Open Model:Intravenous Bolus Prof. Dr. Fatima A. Tawfiq 1 ❖ The one-compartment open model assumes that the body can be described as a single, uniform compartment (ie, one compartment), and that drugs can enter and le...

One-Compartment Open Model:Intravenous Bolus Prof. Dr. Fatima A. Tawfiq 1 ❖ The one-compartment open model assumes that the body can be described as a single, uniform compartment (ie, one compartment), and that drugs can enter and leave the body (ie, open model). ❖ The simplest drug administration is when the entire drug is given in a rapid IV injection, also known as an IV bolus. Thus, the one-compartment open model with IV bolus administration is the simplest pharmacokinetic model. ❖ It assumes that the drug is administered instantly into the body, it is instantaneously and rapidly distributed throughout the body, and drug elimination occurs immediately upon entering the body. 2 ❖ The rate and extent of distribution to the tissue organs depends on several processes and properties: - The blood flow to that organ. - The drug may have different abilities to cross from the vasculature to the organ depending on the molecular weight of the drug. - Tissues also have different affinity for the drug, depending on lipophilicity and drug binding. - Finally, large organs may have a large capacity for drugs to distribute to. 3  The volume of this single compartment is termed the apparent volume of distribution, VD. The apparent volume of distribution is a proportion between the dose and the concentration of the drug in plasma, Cp0, at that time immediately after being injected. Pharmacokinetic model for a drug administered by rapid intravenous injection. DB = drug in body; VD = apparent volume of distribution; k = elimination rate constant. k IV D,V B D 4  Most drugs are eliminated from the body by liver metabolism and/or renal excretion. All of the processes of drug elimination can be described by the elimination rate constant, k. The elimination rate constant is the proportion between the rate of drug elimination and the amount of drug in the body. Because the amount of drug in the body changes over time, the rate of drug elimination changes, but the elimination rate constant remains constant for first-order elimination. 5 ELIMINATION RATE CONSTANT The rate of elimination for most drugs from a tissue or from the body is a first-order process, in which the rate of elimination at any point in time is dependent on the amount or concentration of drug present at that instance. The elimination rate constant, k, is a first-order elimination rate constant with units of time–1 (eg, h–1 or 1/h). Total removal or elimination of the injected drug from this compartment is affected by metabolism and excretion. The elimination rate constant represents the sum of each of these processes: 𝑘 = 𝑘 𝑚 + 𝑘𝑒 where km = first-order rate process of metabolism and ke = first-order rate process of excretion. 6  𝑑𝐷𝐵 = −𝑘𝐷𝐵 𝑑𝑡 This expression shows that the rate of elimination of drug in the body is a first-order process, depending on the overall elimination rate constant, k, and the amount of drug in the body, DB, remaining at any given time, t. Integration of Equation gives the following: −kt logDB = + logD0B 2..3 Where DB = the drug in the body at time t and DB0 is the amount of drug in the body at t = 0. When log DB is plotted against t for this equation, a straight line is obtained. In practice, instead of transforming values of DB to their corresponding logarithms, each value of DB is placed at logarithmic intervals on semilog paper. 7 100 D0 B 10 Slope = – k 2.3 1 0 1 2 3 4 5 Time 8  APPARENT VOLUME OF DISTRIBUTION The volume of distribution represents a volume that must be considered in estimating the amount of drug in the body from the concentration of drug found in the sampling compartment. Because the value of the volume of distribution does not have a true physiologic meaning in terms of an anatomic space, the term apparent volume of distribution is used. The VD relates the concentration of drug in plasma (Cp) and the amount of drug in the body (DB) DB = V Cp 9 a similar expression based on drug concentration in plasma is obtained for the first-order decline of drug plasma level −𝑘𝑡 𝑙𝑜𝑔𝐶𝑝 = + 𝑙𝑜𝑔𝐶𝑝0 2.3 where Cp = concentration of drug in plasma at time t and Cp0 = concentration of drug in plasma at t = 0 10 Example: Exactly 1 g of a drug is dissolved in an unknown volume of water. Upon assay, the concentration of this solution is 1 mg/mL. What is the original volume of this solution? remembering that 1 g = 1000 mg: Therefore, the original volume was 1000 mL or 1 L. 11 Calculation of Volume of Distribution In a one-compartment model (IV administration), the VD is calculated with the following equation: 𝐷 0 𝐷𝑜𝑠𝑒 𝐵 𝑉𝐷 = 0 = 0 𝐶𝑝 𝐶𝑃 When Cp0 is determined by extrapolation, Cp0 represents the instantaneous drug concentration after drug equilibration in the body at t = 0. The dose of drug given by IV bolus (rapid IV injection) represents the amount of drug in the body, DB0, at t = 0. Because both DB0 and Cp0 are known at t = 0, then the apparent volume of distribution, VD, may be calculated. 12 100 Cp0 10 1 0 1 2 3 4 5 Time 13  Significance of the Apparent Volume of Distribution Most drugs have an apparent volume of distribution smaller than, or equal to, the body mass. If a drug is highly bound to plasma proteins or the molecule is too large to leave the vascular compartment, then Cp0 will be higher, resulting in a smaller apparent VD. For example, the apparent volume of distribution of warfarin is small, approximately 0.14 L/kg, much less than the total body mass. This is because warfarin is highly bound to plasma proteins, making it hard to leave the vascular compartment. 14 ❖ For some drugs, the volume of distribution may be several times the body mass. ❖ In this case, a very small Cp0 may occur in the body due to concentration of the drug in peripheral tissues and organs, resulting in a large VD. ❖ Drugs with a large apparent VD are more concentrated in extravascular tissues and less concentrated intravascularly. ❖ For example, the apparent volume of distribution of digoxin is very high, 7.0 L/kg, much greater than the body mass. This is because digoxin binds extensively to tissues, especially muscle tissues. Consequently, binding of a drug to peripheral tissues or to plasma proteins will significantly affect the VD 15 ✓ The apparent VD is a volume term that can be expressed as a simple volume or in terms of percent of body weight. ✓ If VD is a very large number—that is, >100% of body weight—then it may be assumed that the drug is concentrated in certain tissue compartments. In the digoxin example above, 7.0 L/kg is estimated to be 700% of body weight. Thus, the apparent VD is a useful parameter in considering the relative amounts of drug in the vascular and in the extravascular tissues. ✓ For example, if the VD is 5% of body weight, this is approximately the volume of plasma, and it would be assumed that this drug occupies the vascular compartment with very little distributing to tissues outside the vascular compartment. 16 ❖ For each drug, the apparent VD is a constant. In certain pathologic cases, the apparent VD for the drug may be altered if the distribution of the drug is changed. ❖ For example, in edematous conditions, the total body water and total extracellular water increases; this is reflected in a larger apparent VD value for a drug that is highly water soluble. ❖ Similarly, changes in total body weight and lean body mass (which normally occur with age, less lean mass, and more fat) may also affect the apparent VD. 17 CLEARANCE Clearance is a measure of drug elimination from the body without identifying the mechanism or process. Drug Clearance in the One-Compartment Model: Drug clearance refers to the volume of plasma fluid that is cleared of drug per unit time. Clearance may also be considered the fraction of drug removed per unit time. Drug elimination expressed as amount per unit time or as volume per unit time or as fraction eliminated per unit Time 𝐶𝑙 = 𝑘 × 𝑉𝐷 18 ClT may be defined as the rate of drug elimination divided by the plasma drug concentration. Thus, clearance is expressed in terms of the volume of plasma containing drug that is eliminated per unit time. 𝑒𝑙𝑖𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐶𝑙 𝑇 = 𝑝𝑙𝑎𝑠𝑚𝑎 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛(𝐶𝑝 ) 𝐶𝑙𝑇 = 𝐾𝑉 𝐷 = Dose / AUC ClT, is the product of VD and k, both of which are constant. As the plasma drug concentration decreases during elimination, the rate of drug elimination, dDE/dt, will decrease accordingly, but clearance will remain constant. Clearance will be constant as long as the rate of drug elimination is a first-order process. 19  Clearance from Drug-Eliminating Tissues Clearance may be applied to any organ that is involved in drug elimination from the body. As long as first-order elimination processes are involved, clearance represents the sum of the clearances for both renal and nonrenal clearance ClT = ClR +ClNR where ClR is renal clearance or drug clearance through the kidney, and ClNR is nonrenal clearance through other organs. ClNR is assumed to be due primarily to hepatic clearance (ClH) in the absence of other significant drug clearances, such as elimination through the lung or the bile ClT = ClR +ClH 20  CALCULATION OF k FROM URINARY EXCRETION DATA The elimination rate constant k may be calculated from urinary excretion data. In this calculation the excretion rate of the drug is assumed to be first order. The term ke is the renal excretion rate constant, and Du is the amount of drug excreted in the urine. 𝑑𝐷𝑢 −𝑘𝑡 𝑙𝑜𝑔 = + 𝑙𝑜𝑔𝑘 𝑒 𝐷𝐵𝑂𝑂 𝑑𝑡 2 3 A straight line is obtained from this equation by plotting log dDu/dt versus time on regular paper. The slope of this curve is equal to – k/2.3 and the y intercept is equal to 𝑘𝑒 D B0. For rapid intravenous administration, DB0 is equal to the dose D0. Therefore, if DB0 is known, the renal excretion rate constant (ke) can be obtained. Because both ke and k can be determined by this method. 21 Log k D0 e B –k Slope = 2.3 Time 22