Introduction to Clinical Pharmacokinetics PDF

Introduction to Clinical Pharmacokinetics Baba Sulemana Mohammed Department of Pharmacology & Toxicology School of Pharmacy and Pharmaceutical Sciences University for Development Studies 01/12/2024 BSM...

Introduction to Clinical Pharmacokinetics Baba Sulemana Mohammed Department of Pharmacology & Toxicology School of Pharmacy and Pharmaceutical Sciences University for Development Studies 01/12/2024 BSM 1  Outline Clinical PK: What is it? Review of Concepts Bioavailability Volume of Distribution Clearance, Half-Life and K Cases Discussion/Questions  What is clinical pharmacokinetics ? Application of the principles of pharmacokinetics in relation to pharmacodynamics to the individual patient. A means of prescribing the right dose Optimising therapeutic effect Minimizing adverse effects Considers the physiological body compartments in the mathematical modelling of the interaction of the drug and the body  Pharmacokinetics (PK) & pharmacodynamics (PD) PK - What the body does to the drug? Absorption; distribution, metabolism, excretion (ADME) PD - What the drug does to the body? Drug concentration at the site of action or in the plasma is related to the magnitude of effect ADME - Summary  Pharmacokinetics (PK) and pharmacodynamics (PD) Plasma Site Concen- of Dose Effects tration Action PK PD Absorption & Ionization Non-ionised drug More lipid soluble drug Diffuse across cell membranes more easily Effect of pH on Absorption Acid medium of the stomach Bicarbonates (NaHCO3)  Weakly acidic drugs absorbed faster in stomach Aspirin Barbiturates. Warfarin. Valproic Acid. Phenytoin. Tolbutamide. Phenylbutazone Ibuprofen Paracetamol Frusemide  Weakly basic drugs absorbed faster in intestine Quinine Quinidine, Chlorpromazine, Propranolol, Pentazocin, Antipyrine, Diazepam Lignocaine/Xylocaine  Basic Parameters The basic concepts and parameters have been dealt with already under General Principles (SPS 301) and will be reviewed here. In pharmacokinetics the body is represented as a single or multiple compartments in to which the drug is distributed. Some of the parameters are therefore a little abstract as we know the body is much more complicated !  Concentration-Time Curve X Blood Concentration (mg/L) Time (hr) Blood Concentra 12 tion (mg/L) 0 0 1 7 10 10 2 10 8 3 5 7 4 2.5 6 5 1.25 5 6 0.6 4 7 0.2 2.5 2 8 0 1.25 0.6 0 0 0.2 0 0 1 2 3 4 5 6 7 8 9  The Trapezoidal Rule Area = I/2( h1 +h2) x 12 b Blood Concentration in 10 Total Area = A1+A2+A3…. An 8 plsama 6 4 A1 2 An 0 0 1 2 3 4 5 6 7 8 9 Time (hr) The Trapezoidal Rule Time Blood Concentration Time Interval (mg/L) (hr) Average concentration Area (mghr/L) (mg/L) 0 0 0 0 0.00 1 7 1 3.5 3.50 2 10 1 8.5 8.50 3 5 1 7.5 7.50 4 2.5 1 3.75 3.75 5 1.25 1 1.875 1.88 6 0.6 1 0.925 0.93 7 0.2 1 0.4 0.40 8 0 1 0.1 0.10 26.55 Determination of AUC Time (hr) Blood Concentration Time Interval (mg/L) (hr) Average concentration Area (mghr/L) (mg/L) 0 0 0 0 0.00 1 7 1 3.5 3.50 2 10 1 8.5 8.50 3 5 1 7.5 7.50 4 2.5 1 3.75 3.75 5 1.25 1 1.875 1.88 Blood Concentration (mg/L) 6 0.6 1 0.925 0.93 7 0.2 1 0.4 0.40 8 0 1 0.1 0.10 26.55 Uses of the concentration time curve Extent of Absorption (AUC) Bioavailability (AUCx/AUCIV X 100) Rate of Absorption (Slope of Absorption phase) Evaluating Exposure (AUC) Clearance ( Dose/AUC) Tmax & Cmax  Bioavailability High Bioavailability considered to be above 80% In case of oral, oral dose almost equal to IV dose (paracetamol 1g vs 1g) Low Bioavailability below 40% In case of oral, oral dose considerably higher than IV dose (Propranolol, 40-80 mg vs, 1 mg)  Distribution The movement of drug from the blood to and from other compartments of the body Determined by: partitioning across various membranes binding to tissue components (large Vd) binding to blood components (RBC, plasma protein) (small Vd)  Volume of Distribution Apparent volume of distribution is the theoretical volume that would have to be available for drug to disperse in, if the concentration everywhere in the body were the same as that in the plasma or serum (where drug concentration sampling generally occurs).  Volume of Distribution  Volume of Distribution An abstract concept Gives information on HOW the drug is distributed in the body Used to calculate a loading dose  Loading Dose Dose = Cp(Target) x VD  Question What is the loading dose required for drug A if; Target concentration is 10 mg/L VD is 0.75 L/kg Patients weight is 75 kg Answer is on the next slide  Answer: Loading Dose of Drug A Dose = Target Concentration x VD VD = 0.75 L/kg x 75 kg = 56.25 L Target Conc. = 10 mg/L Dose = 10 mg/L x 56.25 L = 565 mg This would probably be rounded to 560 or even 500 mg.  Clearance Ability of organs of elimination (e.g. kidney, liver to “clear” drug from the bloodstream Volume of blood in a defined region of the body that is completely cleared of a drug per unit time. Clearance is a more useful concept in reality than t1/2 or kel since it takes into account blood flow rate Clearance varies with body weight Also varies with degree of protein binding Units are in L/hr or L/hr/kg Pharmacokinetic term used in determination of maintenance doses  Clearance Rate of elimination = kel D, Remembering that Vd= D/C And therefore D= C Vd Rate of elimination = kel C Vd For first order Rate of elimination is proportional to Plasma Concentration and the constant of proportionality represents CL Therefore, Rate of elimination for whole body = CLT C Combining the two, CLT C = kel C Vd and simplifying gives: CLT = kel Vd......................(1) For a drug eliminated with first-order kinetics, clearance is a constant, i.e., the ratio of rate of elimination to plasma concentration is the same regardless of plasma concentration. The elimination rate will be rapid at first and slow as the concentration decreases. Steady-State Steady-state occurs after a drug has been given for approximately five elimination half-lives. At steady-state the rate of drug administration equals the rate of elimination and plasma concentration - time curves found after each dose should be approximately superimposable.  Accumulation to Steady State 100 mg given every half-life 194 … 200 187.5 175 150 100 97 … 100 87.5 94 75 50 C Cpav t Four half lives to reach steady state  What is Steady State (SS) ? Why is it important ? Rate in = Rate Out Reached in 4 – 5 half-lives (linear kinetics) Important when interpreting drug concentrations in TDM or assessing clinical response  Maintenance Dose Calculation Maintenance Dose = CL x CpSSav CpSSav is the target average steady state drug concentration The units of CL are in L/hr or L/hr/kg Maintenance dose will be in mg/hr so for total daily dose will need multiplying by 24  Question What maintenance dose is required for drug A if; Target average SS concentration is 10 mg/L CL of drug A is 0.015 L/kg/hr Patient weighs 75 kg Answer on next slide.  Answer Maintenance Dose = CL x CpSSav CL = 0.015 L/hr/kg x 75 = 1.125 L/hr Dose = 1.125 L/hr x 10 mg/L = 11.25 mg/hr So will need 11.25 x 24 mg per day = 270 mg  Half-Life and k Half-life is the time taken for the drug concentration to fall to half its original value The elimination rate constant (k) is the fraction of drug in the body which is removed per unit time.  Commonly seen Elimination Kinetics Zero Order First Order Kinetics Kinetics Rate = k C Rate = k C = Co e-kt C = Co - kt C vs. t graph is NOT C vs. t graph is linear, decaying LINEAR exponentially. k is elimination LogC vs. t graph is rate constant linear. log10C = log10C0 – k.t/2.303 lnC = lnCo – kt Conc. Vs. time plots C = Co - kt; lnC = lnCo – kt log10C = log10C0 – k.t/2.303  Comparison Zero Order First Order Elimination Elimination [drug] decreases [drug] decreases linearly with time exponentially w/ Rate of elimination time is constant Rate of elimination Rate of elimination is proportional to is independent of [drug] [drug] Plot of log [drug] or No true t 1/2 ln[drug] vs. time are linear t 1/2 is constant regardless of [drug]  Half-Life (i.e for first order elimination kinetics) C = Coe - kt C/Co = 0.50 for half of the original amount (ie at the half-life) 0.50 = e – k t 1/2 ln 0.50 = -k t ½ -0.693 = -k t ½ t 1/2 = 0.693 / k