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HO pharmacokinetics 9-9-24.pdf

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M. Valentovic, Ph.D. 1 PHARMACOKINETICS GOALS and OBJECTIVES At the end of this section you should be able to: 1. Define all pharmacokinetics parameters list...

M. Valentovic, Ph.D. 1 PHARMACOKINETICS GOALS and OBJECTIVES At the end of this section you should be able to: 1. Define all pharmacokinetics parameters listed in bold type. A student should be able to list how each parameter is calculated. These parameters include but are not limited to: bioavailability half-life clearance volume of distribution elimination rate constant steady state concentration 2. Define a one-compartment and two-compartment model and be able to graphically recognize these models. 3. Define zero and first order kinetics and be able to graphically recognize these kinetic parameters. 4. Describe the relationship between clearance, half-life, volume of distribution and plasma drug concentration. 5. Predict the change in steady state drug concentration if a change occurs in another pharmacokinetics parameter. 6. List and define all formulas presented that is boxed or in bold type. 7. Calculate any pharmacokinetics parameter when given appropriate data. I. INTRODUCTION A. Pharmacokinetics is the quantitative expression of the relationship between absorption, distribution and elimination of drugs as a function of time. B. Relationship between drug concentration and effect 1. Basic principle of pharmacokinetics is that the effect of a drug is related to the blood concentration of the drug. 2. Changes in blood concentration will modify the effect of the drug M. Valentovic, Ph.D. 2 MODELS OF DRUG DISPOSITION C. ONE COMPARTMENT MODEL 1. A drug rapidly equilibrates between plasma and tissue relative to its elimination rate. The body acts as a single compartment. Dose Central Compartment Elimination 2. The semi-log plot of plasma concentration over time is a straight line when the drug follows first order kinetics 3. Changes in plasma concentration reflect a proportional change in tissue concentration. -the plasma and tissue concentrations may be different -the drug does not inherently distribute to all tissues D. TWO COMPARTMENT MODEL 4. A drug takes a longer time period to establish equilibrium between plasma and tissue -drug has two compartments for distribution: a central and a peripheral compartment - drugs can move to and from the peripheral compartment back to the central compartment -elimination occurs from the central compartment DOSE Central compartment Peripheral compartment Elimination M. Valentovic, Ph.D. 3 4. The semi-log plot results in two phases when the drug follows first order kinetics a. Initial distribution phase (more rapid decline in plasma drug concentration over time) b. Elimination phase Distribution Log Drug concentration phase Elimination phase Time (hr) ELIMINATION KINETICS E. FIRST ORDER 1. Definition, the rate of decrease in drug concentration is directly proportional to the amount of drug in the body 2. A constant fraction of the drug is eliminated 3. Most drugs follow first order kinetics DC/dt proportional to ke{C} C = (dose/Vd) e-kt 4. The elimination rate (ke) is a constant - fraction of drug removed at any given time is constant and independent of dose - constant fraction eliminated 5. Graphic examination of first order kinetics The semi-log plot for one compartment first order kinetics will be a straight linear line M. Valentovic, Ph.D. 4 F. ZERO ORDER Definition, a constant amount of a drug is eliminated independent of dose; the enzymes involved in elimination are saturated, PHARMACOKINETIC PARAMETERS G. BIOAVAILABILITY 1. Fraction of unchanged drug reaching the systemic circulation after drug administration 2. Drugs administered by the intravenous route have a bioavailability of 1 all other routes of administration have a bioavailability of < 1. 3. Bioavailability is denoted in most pharmacokinetics formulas by the letter “F” 4. Determination of bioavailability -Administer drug by the IV route and measure plasma concentrations over time -Administer same dose by another route (usually oral) and measure plasma concentrations as a function of time -The area under the curve (AUC) vs. time is an index of the extent of drug that enters the bloodstream and thus the body M. Valentovic, Ph.D. 5 AUC oral F= AUC iv 5. For oral administration, bioavailability is an indication of the extent of absorption and/or first pass effect 6. Importance of bioavailability -comparison of different products with the same active ingredients -method to assess influence of food or drugs on bioavailability H. Half-Life of Elimination (t1/2) 1. The half-life is the time needed for a 50% decrease in plasma drug concentration 2. Half-life is denoted as t1/2 3. Half-life is expressed in units of time (usually hr) 4. Half-life can be determined graphically or using a formula - half-life is calculated using the following formula for a one compartment model when the elimination rate constant (ke ) is known t1/2 = Ln 2 = 0.693 ke ke 5. Half-life can be determined using a semi-log plot (first order kinetics) 6. Half-life can be used to measure how long it takes for a drug to be totally eliminated from the body Ct = C0 e(-kxt) Ln Ct = Ln C0 - kt M. Valentovic, Ph.D. 6 Log Ct = Log C0 –kt/2.303 Ct plasma concentration at a given time and t is time C0 plasma concentration at time 0 7. Half life can be used to determine the fraction of drug present in the body after a dose Number of half- Percent Percent removed from lives remaining in body body 1 50 50 2 25 75 3 12.5 87.5 4 6.25 93.75 5 3.13 96.87 6 1.56 98.44 7 0.78 99.9 Four half lives rule A drug is considered essentially eliminated from the body after 4 half- lives. Following 4 half lives, 94% of the drug is eliminated and only 6% remains in the body The fraction of drug eliminated from the body can be calculated as Fraction eliminated = 1 - 1/(2n) n is the number of half lives The fraction remaining in the body can be calculated as: Fraction remaining = 1/(2n) M. Valentovic, Ph.D. 7 I. VOLUME OF DISTRIBUTION (Vd) 1. Definition, The apparent volume that a drug would be distributed within the body once an equilibrium is established with plasma 2. Volume of distribution is denoted Vd 3. Volume of distribution is expressed usually in Liters or Liters/kg 4. The volume of distribution is an apparent volume. Large volumes of distribution are an indication of a drug that is highly lipophilic or has low protein binding 5. Certain agents can have a volume of distribution greater than total body Water Volumes of some body compartments in a 70 kg person Plasma volume 3 L Blood volume 5.5 L Total body water 42 L Extracellular fluid volume 12 L a) Warfarin has a Vd of 9 L (99% plasma protein binding) b) Morphine Vd is 245 L; Thiopental Vd 140 L -distributes to fat and brain very rapidly 6. The formula to determine Vd Dose(iv) Vd = C0 The Vd can be graphically determined using the plasma concentration at time 0 40 C0 Drug conc 10 ug/ml 1 1 2 4 6 8 Time (hr) M. Valentovic, Ph.D. 8 PRACTICE PROBLEM 1. What is the C0 if 400 mg of a drug is administered IV to a 70 kg male if the Vd is 40 L? J. Clearance 1. Clearance is the volume of plasma from which a drug is totally removed per unit of time. Clearance is usually expressed as L/hr or ml/min -drug can be removed by metabolism and or excretion -routes of excretion (urine, bile, expired air, sweat, etc.) 2. Total body clearance is the sum of all clearance values for a drug and are additive CL total = CL renal + CLhepatic + CLother 3. Formula for clearance CL = Ke x Vd Substituting the 0.693/t1/2 for ke generates the following formula CL = 0.693/t1/2 x Vd and rearranged to solve for half-life t1/2 = (0.693 x Vd) CL 4. The half-life of a drug is dependent on clearance and Vd 5. Cannot use t1/2 alone to predict Vd or CL -changes in half-life can be due to changes in Vd or Clearance -half-life will increase if clearance decreases, provided Vd remains the same DOSING REGIMENS A. Assuming a single drug administration, plasma concentration is directly proportional to the dose administered A = C x Vd A= amt of drug C= plasma concentration Vd= volume of distribution B. Concentration at steady state (Css) 1. When a drug following first order kinetics is given, the blood concentration will have a peak and then it will decrease over time. If additional drug is given, the amount absorbed from the second dose will be added to the amount remaining from the first dose in the body. M. Valentovic, Ph.D. 9 2. If a drug is given at regular intervals, the amount of drug in the body will rise and plasma concentration will increase until the rate of administration equals the rate of elimination. When the rate of administration equals the rate of elimination, then the drug attains steady state. Css is the plasma plasma concentration maintained at Steady State. Css is attained after 4 half-lives. Css (rate of administration = rate of Drug Concentration elimination) TIME C. Maintenance Dose 1. Often drugs are given multiple times at regular intervals for 7, 10, 14 days or longer. 2. Drugs, when given at multiple, regular intervals will have a peak and a trough value. The trough is considered the lowest concentration immediately prior to the next dose. -the fluctuations between the peak and trough are dependent on the dosing interval -the longer the dosing interval the greater the variation in peak and trough values EXAMPLE Tetracycline is given for an upper respiratory infection as 1,500 mg per capsule 2 times/day for 7 days. The half-life is 12 h. How much tetracycline remains in the body when the 2nd dose is given assuming F=1? How long will it take to get to Css? How long will it take to get to Css if a 250 mg capsule was given 4 times/day for 7 days? M. Valentovic, Ph.D. 10 3. The following formula expressed the relationship between CL and Css if the maintenance dose (MD) is known. Css can be mathematically determined by Css = MD MD,maintenance dose MD = Dose x F CL τ is the dosing interval Dosing interval (τ) 4. Css for constant IV infusion When a drug is given by a constant IV infusion, the infusion rate is substituted for the maintenance dose Css = Infusion rate CL D. Loading Dose 1. So far we have discussed that for a drug that follows first order kinetics, it takes 4 half-lives to get to a steady state concentration. However, if someone has a very serious condition, it may not be desirable to wait 4 half-lives to get to the steady concentration, as they need an immediate effect. For example, if someone has an asthmatic attack and needs immediate treatment with theophylline to prevent another attack. Theophylline has a half-life of 9 hr and it would take 36 h to get to steady state, that would be 1 ½ days without adequate drug therapy to prevent an asthmatic attack. M. Valentovic, Ph.D. 11 Another example is digoxin, a drug used for congestive heart failure. An individual with a failing heart needs immediate care to enhance the function of the heart. However, digoxin has a half-life of 40 hr. How many days would it take to get to steady state? 2. A single large dose called a loading dose can be given to produce an immediate rise in plasma drug concentration. 3. Calculation of the loading dose Loading dose = Css x Vd F Rearranging the formula to solve for Css Css = Loading dose x F Vd Although this may appear to be a new formula it is a rearrangement of a previous formula for the Co. C0 = Dose x F Vd M. Valentovic, Ph.D. 12 II. Summary A. Time to Css for first order kinetics 1. Independent of dose 2. Steady state occurs after 4 half-lives B. Css 1. Css is directly proportional to dose (Css = MD/CL) provided all other parameters stay the same 2. Css is directly proportional to half-life (Css = MD/CL= MD/(kexVd)= MD or Css = MD x t1/2.693/ t1/2 xVd.693x Vd 3. Css is inversely related to Clearance (CL) 4. Css inversely related to Vd C. Fluctuations in peak and trough values 1. Variation in peak and trough values dependent on dosing interval 2. Longer the dosing interval (τ) the greater the variation in peak and trough values 3. Constant iv infusion eliminates peak and trough values D. Loading Dose 1. Loading dose provided IV to quickly attain Css (Css = LD/Vd) 2. Css requires a maintenance dose in order to maintain Css E. First and Zero Order kinetics 1. Zero order kinetics a) drug is removed at a constant amount b) as dosing is increased drug will accumulate 2. First order kinetics a) Drug is eliminated as a constant fraction b) The rate of elimination is proportional to the amount of drug present M. Valentovic, Ph.D. 13 PROBLEM SET 1 1. What is the ke for a drug with a half-life of 3 hours? 2. How long will it take for Drug Y, with a half-life of 5 hours, to be removed from the body following an iv injection of 200 mg? Will the time to remove Drug Y from the body be different if 100 mg was injected as a single dose iv? 3. A 69 year old woman is switched to once a day dosing with Amlodipine (FYI, Norvasc) to control her hypertension. Previous treatment with a diuretic was not successful due to a compliance problem. She is started on 5 mg oral, once a day. Amlodipine has a half-life of 1.5 days. How long will it take to achieve steady state concentration in this individual? 4. The maintenance dose for a drug is 120 mg as a tablet/twice per day. Bioavailability (F=1) and the clearance is 100 ml/min. What is the steady state concentration (mg/L)? 5. A concentration versus time plot on a linear scale depicts a straight line. The half-life is 4 hours for a single administration of 100 mg. Can you calculate the steady state concentration for this drug? 6. Phenytoin is an anticonvulsant with a half-life of 20 hr and a volume of distribution of 45 L. What would be the loading dose for phenytoin, in Mg, to maintain a steady state level of 10 µg/ml? 7. A cystic fibrosis patient is given 80 mg every 8 hr iv of the antibiotic tobramycin. The drug is eliminated totally by renal clearance which equals the glomerular filtration rate of 120 ml/min. What is the steady state concentration (µg/ml) in this patient? What would the state state concentration be if glomerular filtration suddenly was reduced to 60 ml/min? 8. An individual needs an iv infusion of a drug to maintain a steady state concentration of 3 mg/L. The drug has a clearance of 60 L/hr. Calculate the infusion rate in mg/hr needed to maintain the steady state concentration. M. Valentovic, Ph.D. 14 Problem Set 2 1. In a 70 kg patient, drug X has a half-life of 7 hours and a volume of distribution of 1 liter/kg. What is the plasma clearance of this drug in L/ hr? 2. Drug Z has the following kinetic parameters F = 1.0 t½ = 1 day Vd = 10 liters What will be the approximate steady-state plasma concentration, in mg/L, if 100 mg is given daily. 3. A drug has a half-life of 2 days, a volume of distribution equal to 35 L, and is effective at a steady state plasma concentration of 10 mcg/ml. What should the iv loading dose be, in mg, to rapidly achieve steady state plasma levels? Use the following information to answer questions 4-7. An iv bolus dose of 100 mg is administered to an individual. Blood is drawn at various times to determine drug concentration and the values are plotted below. M. Valentovic, Ph.D. 15 4. What is the half-life of the drug? 5. Does the drug follow zero or first order kinetics? Is this a one or two compartment model? 6. What is the elimination rate constant? 7. What is the clearance of the drug in L/hr? M. Valentovic, Ph.D. 16 ANSWERS & EXPLANATIONS FOR PROBLEM SET #1 1. What is the ke for a drug with a half-life of 3 hours? Ke = 0.693/t½ Ke = (0.693/3) = 0.231 hrs-1 2. How long will it take for Drug Y, with a half-life of 5 hours, to be removed from the body following an iv injection of 200 mg? Will the time to remove Drug Y from the body be different if 100 mg was injected as a single dose iv? Applying the 4 half-lives rule, in 4 half-lives or 20 hr 94% of the drug will be eliminated from the body. It will take 20 hr whether 100 mg or 200 mg of drug is injected provided that the drug follows first order kinetics. 3. A 69 year old woman is switched to once a day dosing with Amlodipine (FYI, Norvasc) to control her hypertension. Previous treatment with a diuretic was not successful due to a compliance problem. She is started on 5 mg oral, once a day. Amlodipine has a half-life of 1.5 days. How long will it take to achieve steady state concentration in this individual? It will take 4 half-lives or 1.5 days x 4 = 6 days to attain steady state concentration 4. The maintenance dose for a drug is 120 mg as a tablet/twice per day. Bioavailability is (F=1) and the clearance is 100 ml/min. What is the steady state concentration (mg/L)? The formula to use is: MD The dose is 120 mg/12 hr or 10 mg/hr or 240 mg/day Css = The clearance is 100 ml/min or CL 100 ml/min x 60 min/hr = 6000 ml/hr or 6 L/hr 10 mg/hr x (1) Css = 6 L/hr = 1.67 mg/l NOTE: Check your units to be sure that they are the same (ml/min changed to L/hr) 5. A concentration versus time plot on a linear scale depicts a straight line. The half-life is 4 hours for a single administration of 100 mg. Can you calculate the steady state concentration for this drug? A drug with a linear scale straight line follows zero order kinetics. Cannot determine steady state concentration M. Valentovic, Ph.D. 17 6. Phenytoin is an anticonvulsant with a half-life of 20 hr and a volume of distribution of 45 L. What would be the loading dose for phenytoin (expressed in mg) maintain a steady state level of 10 µg/ml? Loading dose Css = Vd NOTE: Css of 10 µg/ml is the same as 10 mg/L Loading dose = 10 mg/L x 45 L = 450 mg 7. A cystic fibrosis patient is given 80 mg every 8 hr iv of the antibiotic tobramycin. The drug is eliminated totally by renal clearance which equals the glomerular filtration rate of 120 ml/min. What is the steady state concentration (µg/ml) in this patient? MD Css = CL The Maintenance Dose is 80 mg every 8 hr or 10 mg/hr with F=1 since it is iv Css = 10 mg/hr x 1 120 ml/min x 60 min/hr = 0.00139 mg/ml but the Css is requested in µg/ml which is 1.39 ug/ml If the clearance is decreased from 120 ml/min to 60 ml/min what would happen to Css if the dosing rate stays the same? If the clearance is decreased by 50% then the Css will double as CL is inversely related to Css. 8. An individual needs an iv infusion of a drug to maintain a steady state concentration of 3 mg/L. The drug has a clearance of 60 L/hr. Calculate the infusion rate in mg/hr needed to maintain the steady state concentration. Infusion rate Css = CL rearranging the formula Infusion rate = Css x CL Infusion rate = 3 mg/L x 60 L/h = 180 mg/h M. Valentovic, Ph.D. 18 ANSWERS AND EXPLANATIONS FOR PROBLEM SET #2 1. In a 70 kg patient, drug X has a half-life of 7 hours and a volume of distribution of 1 liter/kg. What is the plasma clearance of this drug in L/ hr? Wt = 70 kg t1/2 = 7 hr Vd = 1 L/kg = 70 Kg * 1 L/kg = 70 L CL = ? L/hr CL = ke x Vd Vd = 70 L ke =.693/t1/2 =.693/(7 hr) =.1/hr CL =.1/hr x 70 L = 7 L/hr 2. Drug Z has the following kinetic parameters F = 1.0 t½ = 1 day Vd = 10 liters What will be the approximate steady-state plasma concentration, in mg/L, if 100 mg is given daily. MD = 100 mg per day Css = ? mg/L Css = MD/CL CL = ke x Vd =.693/(1 day) x 10 L = 6.93 L/day Css = 100 mg/day / (6.93 L/day) = 14.4 mg/L 3. A drug has a half-life of 2 days, a volume of distribution equal to 35 L, and is effective at a steady state plasma concentration of 10 mcg/ml. What should the iv loading dose be, in mg, to rapidly achieve steady state plasma levels? t1/2 = 2 days Vd = 35 L = 35,000 ml Css = 10 mcg/ml = 10 mg/L Loading dose iv = ? mg F = 1 since dose will be administered iv Loading dose = (Vd x Css)/F = 35 L x 10 mg/L = 350 mg M. Valentovic, Ph.D. 19 Use the following information to answer questions 4-7. An iv bolus dose of 100 mg is administered to an individual. Blood is drawn at various times to determine drug concentration and the values are plotted below. 4. What is the half-life of the drug? Using the graph, it is 2 hr M. Valentovic, Ph.D. 20 5. Does the drug follow zero or first order kinetics? Is this a one or two compartment model? First order kinetics (straight line on semi-log plot) One compartment model 6. What is the elimination rate constant? The half-life is 2 hrs ke = 0.693/2 hr = 0.35 hr-1 7. What is the clearance of the drug in L/hr? CL = ke x Vd CL = 0.35 hr-1 x Vd The Vd can be determined using the C0 from the graph Using the graph the C0 is 500 µg/ml = 500 mg/L Vd = Dose/C0 Vd = 100 mg/500 mg/L = 0.2 L CL = 0.35 hr-1 x 0.2 L = 0.07 L/hr

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