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PK-Fourth ou t year yea One Compartment Model with First order Absorption Single extravascular dose The plasma concentration–time profile of a large number of drugs can be described by ao one-compartment e co pa t e t model ode with t first-order st o de absorption and elimination. Consider the concentration versus time profile in the following slide following a single oral dose. ١ Extravascular Administration Scheme of the Model ٢ The Model Assuming first-order absorption and firstorder elimination, the rate of change of amount of drug (X) in the body is described by dX = KaXa − KX dt Cp = ٣ [ KaFXo e − Kt − e − Kat Vd ( Ka − K ) ] Pharmacokinetic Parameters Back projection or stripping technique (Method of Residuals) • K, K Ka • Half-lives • AUC • Clearance • Volume • Tmax and Cmax Method of Residuals ٤ ٥ tp Cp max = = 2 . 303 Ka log ( Ka − K ) K [ KaFXo e − Kt max − e − Kat max Vd ( Ka − K ) FXo − Kt max C max = Vd e AUC = KaFXo 1 ⎤ ⎡ 1 − Vd ( Ka − K ) ⎢⎣ K Ka ⎥⎦ AUC = FXo KVd Cl = FXo AUC FXo Vd = K . AUC ٦ ] ٧ Normal or Flip-Flop?? Pharmacokinetic parameters of oral absorption Tmax, Cmax and AUC ¾ These three parameter are important for evaluation of an extravascular dosage forms and evaluation of bioequivalence of drugs. d ¾ Tmax : indicator of how fast a drug is absorbed, if the rate of absorption is fast, Tmax is short ¾ Cmax represents the highest plasma concentration that can be achieved with a single dose. ¾ AUC: a parameter that covers the entire period of sampling and it represents the extent of absorption and how complete is the absorption ¾ AUC and Cmax are dependent on the dose, as the dose increases the AUC and Cmax increases ٨ Pharmacokinetic parameters of oral absorption ¾ Effect of ka and k on Cmax, tmax, and AUC ¾ Changes in ka and k may affect t max, C max, and AUC ¾ If the values for ka and k are reversed, then the same t max is obtained but the C max and AUC are different obtained, different. ¾ If the elimination rate constant is kept at 0.1 /hr and the ka changes from 0.2 to 0.6 hr– 1 (absorption rate increases), then the tmax becomes shorter (from 6.93 to 3.58 hr), the C max increases (from 5.00 to 6.99 µg/mL), but the AUC remains constant (100 µg hr/mL). ¾ In contrast, contrast when the absorption rate constant is kept at 0.3 0 3 hr hr– 1 and k changes from 0.1 to 0.5 hr– 1 (elimination rate increases), then the tmax decreases (from 5.49 to 2.55 hr), the C max decreases (from 5.77 to 2.79 µg/mL), and the AUC decreases (from 100 to 20 µg hr/mL). ٩ Elimination rate constant is kept constant and the ka is increased; tmax becomes shorter, the C max increases but the AUC remains constant. ١٠ ¾when the absorption rate constant is kept constant, and k increases, then elimination rate increases ¾The tmax decreases ,the C max decreases and th AUC decreases. the d Effect of K on Tmax and Cmax F=1, Dose=500, Ka=1.0 hr-1 40 35 30 25 20 15 10 5 0 0 1 2 3 4 5 6 A K=0.1 ١١ K=0.4 K=0.6 7 8 9 10 Effect of K on Tmax and Cmax F=1, Dose=500, Ka=1.0 hr-1 100.0 10.0 1.0 0.1 0 1 2 3 4 5 6 A K=0.1 ١٢ K=0.4 K=0.6 7 8 9 10 ١٣ Bioavailability ¾ Bioavailability and area under plasma conc. time curve are important factors in absorption kinetics ¾ The completeness of absorption is of primary importance in therapeutic situations. ¾ The bioavailability, F, is proportional to the total area under plasma-time curve (AUC) irrespective to its shape. Total amount eliminated =clearance* AUC But the Total amount eliminated is the amount absorbed, F*dose ١٤ F*dose =clearance* AUC Calculation of F, bioavailability FXooral AUCoral = KVd AUCIV = since KVd = Cl Xoiv KVd Cl IV = Cloral = FXooral AUCoral Xo iv AUC IV FXooral Xoiv = AUCoral AUCIV F= Xo iv AUC oral If the two doses were equal * Xo oral AUC IV AUC oral F= ١٥ AUC IV Pharmacokinetics 475 Dr. M. Al-Ghazawi Case 1 A 500-mg dose of the sulfonamide sulfamethoxazole is administered as an oral tablet to a human subject. Eighty percent of the drug is absorbed, and the balance is excreted unchanged in feces. The drug distributes into an apparently homogeneous body volume of 12 L, and has an absorption half-life of 15 min and overall elimination half-life of 12 h. 1) Calculate the following: (i) AUC0→∞,(ii) tmax and (iii) C max. 2) Recalculate the values in Problem 1 if all parameter values remained unchanged, but the elimination half-life was increased to 18 h. ١٦ Case 2 • A patient received a single dose of 500 mg erythromycin in the form of a tablet that is known to have 80% bioavailability. Calculate • the time to reach the maximum concentration, • the maximum concentration, • AUC and Clearance after this single dose If K is 0.2 hr-1, Ka is 1.3 hr-1, and Vd is 40 liters. ١٧ ١٨ ١٩ ٢٠ Determination of elimination rate constant from urinary drug excretion data: Approach I: The urinary excretion rate: Urinary drug excretion data may also be used for calculation of the first-order elimination rate constant. The rate of drug excretion after a single oral dose of dXu dt drug is given by = KrX (1) Considering the amount of the drug in the body after oral administration is given by the equation: X = KaFXo (Ka − K) [e − Kt − e − Kat ] and substituting for X in equation 1 as given by equation (2) dXu dt = KrKaFXo (Ka − K) [e − Kt − e − Kat (2) ] (3) This equation suggests that if Ka >> K, the term e-Kat will approach zero after enough time, while e-Kt still assumes a finite value, then equation 3 reduces to dXu dt ٢١ = KrKaFXo (Ka − K) [e − Kt ] (4) dXu dt log dXu dt = KrKaFXo − K) (Ka = log KrKaFXo (Ka − K) [e − Kt ] − Kt 2.303 (4) (5) Because the rate of urinary drug excretion, dxu/dt, cannot be determined directly for any given time point, an average rate of urinary drug excretion (∆X/∆t) is obtained and this value is plotted against the midpoint of the collection period for each urine sample. log Δ Xu KrKaFXo = log Δt (Ka − K) − Kt * 2.303 (5) This means that plots of log(∆Xu/ ∆t) versus time will yield a curvilinear profile with the terminal phase representing the elimination rate constant (slope = K/2.303) and the residual line representing the absorption rate constant (slope = - Ka/2.303). By contrast, drugs exhibiting flip-flop kinetics, the terminal segment of the plot d drawn b between llog(∆Xu/ (∆X / ∆t) ∆ ) versus time i will ill represent the h absorption b i rate constant (slope = -Ka/2.303) and the residual represent the elimination rate constant (slope = -K/2.303). ٢٢ log( ΔXu ) residual Δt = log KrKaFXo (Ka − K) − Ka 2.303 Approach II: Drug remaining to be excreted (Sigma Minus Method): This represent an alternative approach to the rate approach. It makes use of the amount of drug remaining to be excreted. dXu dt = Xu = KrKaFXo ((Ka − K)) KrKaFXo K [e − Kt − e − Kat ⎡ 1 e − Kt Ke − K a t ⎢ + − (K − Ka) Ka(K − Ka) ⎢⎣ Ka Since the amount of the drug to be excreted at infinity may be expressed as: Xu ∞ = KrFXo K [ Xu∞ (Xu − Xu)= Kae−Kt − Ke−Kat Ka− K ∞ (Xu ٢٣ ∞ ] [ ] Xu ∞ K a − Kt − Xu) = e Ka − K ] (3) ⎤ ⎥ ⎥⎦ (Xu ∞ − Xu) = [ Xu ∞ K a e − Kt − Ke −K at Ka − K (Xu ∞ − Xu) = ] [ ] Xu ∞ K a − Kt e Ka − K * Potting the logarithm of the amount remaining to be eliminated ((Xu∞ - Xu ) versus time will yield a curvilinear plot with terminal segment representing elimination rate constant (slope = -K/2.303) * The residual line representing the absorption rate constant (slope = -Ka/2.303) and vice versa in flip-flop. y collection of sufficient number of urine samples p during g the * Practically, absorption phase to enable a pharmacokinetics analysis of this phase is generally not possible unless the absorption half-life is long. A serious shortcoming for this approach is manifested by the need to collect the urine for a long period of time (seven t1/2), and also that samples cannot be skipped. ٢٤ Case 1 The following information is available for ampicillin: 90% is excreted unchanged and a 250 mg IV bolus dose yields an AUC of 11 mic/mL*hr. The following blood level profile has been reported for two brands of ampicillin which were given as 500 mg oral capsules. ٢٥ ٢٦ Case 2 In a controlled, crossover bioavailability study of oral dosage forms, formulation A, containing 250 mg of a drug, was tested against i fformulation l i B B, containing i i 1 0 mg off 150 the same drug. The drug is extensively metabolized in the body to one major metabolite. Urinary recovery of the metabolite was 210 mg g from formulation A and 130 from formulation B. Calculate the ratio of systemic availability from formulation A to availability from formulation B. Case 3 After 400 mg of a drug administered orally, 50% is absorbed into the circulation. The drug has a first- order absorption rate constant of 2.0 h-1 and d is i eliminated li i t d ffrom th the b body d iin equall proportions by metabolism and by excretion in unchanged form in urine. The overall biological half-life is 3.46 h. If a sigma-minus plot was constructed from the urinary excretion data, what would be values of (i) the rate constant from the slope of the terminal and residual lines, and (ii)the intercept of each of the two lines on the ordinate ( t = 0 )? ٢٧ Dose 400 Bioavailable Not abs 200 200 In urine 100 Metabolite 100 Case 4 An oral dose of 100 mg of drug X in a solution form was administered to a 70 Kg subject. Assuming g the administered dose is completely absorbed, and knowing that absorption half-life =0.2hr, the elimination half-life =1.5hr, the y-intercept=3.9mg/ml in the conc.time profile. Calculate the amount of drug in body at Tmax And if 76% of the drug is excreted unchanged find Kr & renal clearance ٢٨ Case 5 Following oral administration of a dose in solution form, an elimination half-life of 4hrs & an absorption half-life of 0.5 hrs were obtained by plotting the urinary y excretion rate against g tmid for urine collection. The intercept of the elimination phase extension was 38.5 mg/hr. Assuming complete absorption of the drug, 1- what was the administered dose if 30% of the available dose is excreted unchanged in urine? 2- What is the renal clearance iff Vd =22L 3- Find Tmax ,Cmax, and 4- If we double the dose what is the new Tmax 6- A patient receives a drug dose of 400 mg orally, the drug is 100% bioavailable, 25% is metabolized and 75% is excreted as unchanged drug by the kidneys. If K=0.2/hr, Ka=1.5/hr, AUC=105.8 mg.hr/ml. Find Vd, total clearance, Cmax ,Xu∞. 7- Using the sigma-minus method of constructing urine data after the administration of 400 mg oral dose we found that K=0.2/hr and Ka=1.5/hr, and that 75% of the dose is excreted unchanged in urine urine. 1-Calculate the renal clearance, (Vd = 18.9L and F=1) and find Tmax ,Cmax 2-If we changed the dose to 600 mg what is the new Tmax ,Cmax ٢٩