Pharmacokinetic Dosing Calculations PDF

PHARMACOKINETIC DOSAGE REGIMEN CALCULATIONS MATTHEW AIDOO Department of Phar macolog y and Toxicolog y School of Phar macy and Pharmaceutical Sciences University for Development Studies, Tamale  INTRODUCTION ❑The inter and intra individual variations to the effects of...

PHARMACOKINETIC DOSAGE REGIMEN CALCULATIONS MATTHEW AIDOO Department of Phar macolog y and Toxicolog y School of Phar macy and Pharmaceutical Sciences University for Development Studies, Tamale  INTRODUCTION ❑The inter and intra individual variations to the effects of drugs can be avoided by tailoring a dose or dosage to a given patient through the use of clinical pharmacokinetics Generally the initial dosage of the drug is estimated using average population pharmacokinetics parameters obtained from literature (e.g. drug package leaflet) The dosage is then modified according to the patient’s diagnosis, demographics and any other known factor that might affect the patient’s response to the dosage. ❑After initiation of the dose or dosage, the patient is then monitored for therapeutic response by clinical and physical assessment. After assessment of the patient, an adjustment of the dosage regimen may be needed. A measurement of plasma drug concentration may be used to obtain the patient’s individual PK parameters which can be used to modify the dosage regimen. Modification of the dosage regimen is determined by pharmacokinetic calculations either manually or by software programs.  DOSAGE REGIMEN ❑Factors affecting the dosage regimen of a drug: ❖Patient related factors Individual patient’s tolerance of the drug Genetic predisposition Concurrent administration of other drugs Patient’s age, body weight, gender Liver and kidney function of the patient  DOSAGE REGIMEN ❑Factors affecting the dosage regimen of a drug: ❖Pharmacokinetic factors – Rate and extent of Absorption, Distribution Metabolism and Excretion ❖Pharmaceutical factors Type of dosage form Route of administration ADME  ADME ❑Upon administration, a drug moves from the site of administration and gets absorbed into the systemic circulation where it will then gets distributed throughout the body. The process of distribution involves the movement of a drug between the central [intravascular (blood/plasma) & highly perfused organs (liver, kidneys, etc.)] and peripheral [extravascular (intracellular & extracellular)] compartments of the body. Within each compartment of the body, a drug exists in equilibrium between a protein-bound and free form. ADME  VOLUME OF DISTRIBUTION ❑SINGLE COMPARTMENT MODEL These drugs appear to remain in the central compartment and not distribute to peripheral compartments. Vc = Dose administered (mg) / Co (mg/L) Because the drug is said to distribute instantaneously, the initial plasma concentration of drug at time = 0 (Co) is difficult to measure and is therefore estimated via extrapolation to time = 0 on a plasma concentration vs. time curve.  VOLUME OF DISTRIBUTION ❑MULTIPLE COMPARTMENT MODEL These drugs move from the central compartment into peripheral compartments before elimination. Distribution phase: following administration plasma drug concentration will initially decline while the total amount of drug in the body remains the same. Terminal elimination phase: following the distribution phase, the drug will be eliminated from the central compartment (by the kidneys/liver) causing changes in the amounts of the drug in plasma (central compartment).  VOLUME OF DISTRIBUTION ❖Basic drugs: have strong interactions with negatively charged phospholipid membranes. ↑ Vd ❖Acidic drugs: have a higher affinity to bind to albumin and remain in the plasma. ↓ Vd ❖Lipophilic drugs: have a higher lipid membrane permeability and thus, more likely to distribute from plasma and interact with peripheral tissue. ↑ Vd ❖Hydrophilic drugs: less likely to pass through lipid bilayers and therefore more likely to remain in the plasma. ↓ Vd  HALF-LIFE & VD ❑Half-life is dependent on the elimination rate constant (k), which is related to Vd & clearance (CL). T1/2 = 0.693/k CL = Vd x k, k = CL/Vd T1/2 = 0.693 x Vd/ CL T1/2 directly proportional to Vd and inverse to CL Only the drug located in the central compartment can be eliminated from the body because the process of elimination is primarily carried out by the liver and kidneys.  HALF-LIFE & VD ❑Drugs with a high Vd = less fraction of drug in central compartment compared to peripheral compartment. The fraction of drug in the plasma will be eliminated, causing a shift of equilibrium of the drug in the peripheral compartment into the central compartment. This shift will cause the plasma concentration to remain at a steady-state concentration. Thus, at a constant rate of CL, a drug with high Vd will have a longer elimination T1/2 than drug with lower Vd.  LOADING DOSE ❑Loading dose (mg) = [Cp x Vd]/ F Cp = desired plasma con. of drug, F = bioavailability of drug After administration of a loading dose, additional maintenance doses can be administered to maintain the desired plasma concentration of the drug. Loading dose is dependent on the drug's Vd Loading doses are usually indicated where a drug needs to reach steady-state rapidly. Loading dose usually does not need to be modified based on patient characteristics  MAINTENANCE DOSE ❑Maintenance dose rate = [Cp x CL x τ ]/ F Cp = desired plasma con. of drug, CL = clearance rate τ = DI MD is dependent on plasma drug clearance (CL) MD are required for most drugs to maintain the steady-state plasma concentration MD need to be adapted depending on various characteristics of the patient. Because MD are dependent on drug clearance which is a variable dictated by each individual patient.  STEADY STATE CONCENTRATION ❑Steady-state concentration (Css) occurs when the amount of a drug being given/absorbed is the same amount that’s being cleared from the body when the drug is given continuously or repeatedly. Steady-state concentration is the time during which the concentration of the drug in the body stays constant. For most drugs, the time to reach steady state is approximately 4–5 half-lives if the drug is given at regular dosing intervals—no matter the number of doses, or the dose size.  STEADY STATE CONCENTRATION After Dose 1: There are 0.5 doses left at the end of the dosing interval. This means we’re at 50% steady state. After Dose 2: There are 1.5 doses in the body, then half is eliminated to leave 0.75 doses (75% steady state). After Dose 3: There are 1.75 doses in the body, then half is eliminated to leave 0.875 doses (88% steady state). After Dose 4: There are 1.875 doses in the body, then half is eliminated to leave 0.9375 doses (94% steady state). After Dose 5: There are 1.9375 doses in the body, then half is eliminated to leave 0.96875 (97% steady state).  STEADY STATE CONCENTRATION ❑At 97% we’re considered to be at approximate steady state, where the rate of input equals the rate of elimination at one dose per dosing interval. ❖If we continue dosing a drug at the same dosing interval, at steady state (Css), the amount we dose will be eliminated during each dosing interval. ❖As a result, drug concentrations in the body remain constant (steady).  STEADY STATE CONCENTRATION ❑At Css with a dosing interval = half-life ❖The plasma con fluctuates two-fold over the dosing interval ❖The amount of drug in the body shortly after each dose is equivalent to 2x the maintenance dose ❖The Css averaged over the dosing interval is the same as the Css for a continuous infusion at the same dose rate.  STEADY STATE CONCENTRATION ❑Factors that affect Css: drug clearance (CL), dosing interval (τ), and dose. Css = [(dose/τ) x F]/CL Dosing rate k0=(dose/τ) = (Css x CL)/F Dose = (Css x CL x τ)/F Dosing interval (τ) = (F x dose)/ (Css x CL)  MULTIPLE DOSE AND STEADY STATE EQUATIONS ❑In most cases, medications are administered to patients in multiple doses, and drug serum concentrations for therapeutic drug monitoring are obtained when steady state is achieved. For these reasons, multiple dose equations that reflect steady- state conditions are usually more useful in clinical settings than single dose equations. Fortunately, it is simple to convert steady-state single dose (one compartment model) equations to their multiple dose (one compartment model) equations.  CONCENTRATION AT TIME EQUATIONS USING ONE COMPARTMENT MODEL (STEADY SATE)  STEADY STATE CONCENTRATION ❑INTRAVENOUS BOLUS C = (D/Vd)[e–kt/(1 – e–kτ)] ❑CONTINOUS IV INFUSION Css = k0/(kVd) = k0/CL C is drug serum concentration at time = t, Vd = volume of distribution, k = elimination rate constant, D = dose, τ = dosing interval, Css =steady-state con k0 = infusion rate, CL = clearance (Vd x k)  STEADY STATE CONCENTRATION ❑INTERMITENT IV INFUSION C = [k0/(kVd)][(1 – e–kt′)/(1 – e–kτ)] ❑EXTRAVASCULAR ADMINISTRATION C = (FD/Vd)[e–kt/(1 – e–kτ)] C is drug serum concentration at time = t, Vd = volume of distribution, k = elimination rate constant, k0 = infusion rate, t′ = infusion time, D = dose, τ = dosing interval, F = bioavailability  INTRAVENOUS BOLUS ❑INTRAVENOUS BOLUS C = (D/Vd)[e–kt/(1 – e–kτ)] A patient with tonic-clonic seizures is given phenobarbital 100mg IV daily until steady-state occurs. PK parameters in the patient for phenobarbital are k = 0.116/day, Vd =75 L. a. Find the steady-state con of phenobarbital 6 hours after the last dose? b. Find the steady-state con of phenobarbital 23 hours after the last dose?  INTRAVENOUS BOLUS ❑D = 100mg τ = 1day, t = 6 hrs = [(6 h/(24h/d) = 0.25 day C = (D/Vd)[e–kt/(1 – e–kτ)] k = 0.116/day, Vd =75 L (D/Vd) = (100mg/75 L) = 1.33mg/L [e–kt/(1– e–kτ)] = [e–(0.116/d)(0.25 d)/(1– e–(0.116/d x (1 d)] [e–kt/(1– e–kτ)] = 0.97/0.11 = 8.82 C = 1.33 mg/L x 8.82 = 11.73 mg/L The steady state of phenobarbital 6 hour after the last dose in the patient is 11.73 mg/L  INTRAVENOUS BOLUS ❑D = 100mg τ = 1day, t = 23 hrs = [(23 h/(24h/d) = 0.96 day C = (D/Vd)[e–kt/(1 – e–kτ)] k = 0.116/day, Vd =75 L (D/Vd) = (100mg/75 L) = 1.33mg/L [e–kt/(1– e–kτ)] = [e–(0.116/d)(0.96 d)/(1– e–(0.116/d x (1 d)] [e–kt/(1– e–kτ)] = 0.89/0.11 = 8.09 C = 1.33 mg/L x 8.09 = 10.76 mg/L The steady state of phenobarbital 23 hour after the last dose in the patient is 10.8 mg/L  INTERMITENT INFUSION ❑C = [k0/(kVd)][(1 – e–kt′)/(1 – e–kτ)] A patient with pneumonia is administered tobramycin 140mg every 8 hours until steady-state is achieved. PK parameters for tobramycin in the patient: Vd = 16 L, k =0.3/h. Find the steady-state con. immediately after a 1 hour infusion?  INTERMITENT INFUSION ❑Vd = 16 L, k =0.3/h, τ = 8 hrs t′ = 1 hr amount = 140mg K0 = dose/t′ = 140mg/h C = [k0/(kVd)][(1 – e–kt′)/(1 – e–kτ)] [k0/(kVd)] = [(140 mg/h)/(0.30/h x 16 L)] = 29.17 mg/L [(1 – e(–0.30/h x 1 h)) = 0.26 (1 – e(–0.30/h x 8 h))] = 0.91 [(1 – e–kt′)/(1 – e–kτ)] = 0.26/0.91 = 0.29 C = 29.17 mg/L x 0.29 = 8.46 mg/L  EXTRAVSACULAR ADMIN ❑A patient with an arrhythmia is administered 250 mg of quinidine orally (as 300 mg quinidine sulfate tablets) every six hours until steady-state occurs. PK constants for quinidine in the patient: Vd = 180 L, k = 0.0693/h, F = 0.7. Find the steady-state concentration at 12 hours? C = (FD/Vd)[e–kt/(1 – e–kτ)]  EXTRAVSACULAR ADMIN ❑Dose = 250 mg, Vd = 180 L, k =0.0693/h, τ = 6 hrs t = 12 hr, F= 0.7 C = (FD/Vd)[e–kt/(1 – e–kτ)] (FD/Vd) = [(0.7 x 250 mg)/180 L] = 0.97 mg/L [e–kt] = e – (0.0693/h x 12 h) = 0.44 [(1 – e–kτ)] = (1 – e(–0.0693/h x 6 h))]= 0.34 [e–kt/(1 – e–kτ)] = 0.44/0.34 = 1.29 C= 0.97 mg/L x 1.29 = 1.26 mg/L  ANY ROUTE OF ADMINISTRATION ❑Average steady state concentration (Css) Css = [(dose/τ) x F]/CL Dosing rate k0=(dose/τ) = (Css x CL)/F Dose = (Css x CL x τ)/F Dosing interval (τ) = (F x dose)/ (Css x CL) This equation works for any single or multiple compartment model, and thus it is deemed a model independent equation.  ANY ROUTE OF ADMINISTRATION ❑The Css computed by this equation is the concentration that would have occurred if a dose, adjusted for bioavailability, was given as a continuous IVF. Css = k0 x F/CL, k0= (dose/τ), F = 1, Css = k0/CL For example, 600 mg of theophylline tablets given orally every 12 hours (F = 0.8). Css = (600mg/12) x 0.8 = 40mg/h This would be equivalent to give to 40mg/h by continuous IVF.  ANY ROUTE OF ADMINISTRATION The average steady-state concentration equation is very useful when: 1. The half-life of the drug is long compared to the dosage interval or 2. If sustained dosage form is used.  ANY ROUTE OF ADMINISTRATION ❑Long half-life compared to dosage interval A patient is administered 250μg of digoxin tablets (t1/2= 1.5– 2.0 days) daily for heart failure until steady state. The PK constants for digoxin in the patient CL = 2880 L/h, F = 0.7 What is the steady-state concentration?  ANY ROUTE OF ADMINISTRATION ❑Long half-life compared to dosage interval Dose = 250μg, t1/2= 1.5–2.0 days, τ = daily, CL = 2880 L/h = 2880/24 = 120 L/day, F = 0.7 What is the steady-state concentration? Css = [(dose/τ) x F]/CL Css = [(250μg/1d) x 0.7]/120 L/d Css = 175μg/d/120L/d Css = 1.5 μg/L  ANY ROUTE OF ADMINISTRATION ❑Sustained-release dosage form A patient is given 1500 mg of procainamide sustained- release tablets every 12 hours until steady state for the treatment of an arrhythmia. The PK parameters of procainamide in the patient : F =0.85, CL = 720 L/d. What is the steady-state concentration?  ANY ROUTE OF ADMINISTRATION ❑Sustained-release dosage form Dose = 1500 mg, τ = 12 hours, F = 0.85 CL = 720 L/d = 30L/h. What is the steady-state concentration? Css = [(dose/τ) x F]/CL Css = [(1500 mg/12 h) x 0.85]/(30 L/h) Css = [106.25 mg/h]/30 L/h Css = 3.5 mg/L k, t1/2,Vd, CL, STEADY STATE EQUATIONS USING ONE COMPARTMENT MODEL  STEADY STATE EQUATIONS ❑IV BOLUS k = – (ln C1 – ln C2)/(t1 – t2) t1/2 = 0.693/k, Vd = D/(C0 – Cpredose), C0= C/e–kt CL = Vd x k C1 = drug serum concentration at time = t1, C2 = drug serum concentration at time = t2, C0 = drug concentration at time = 0 Cpredose = concentration before the dose  STEADY STATE EQUATIONS ❑CONTINOUS IV INFUSION CL = k0/Css ❑INTERMITENT CONTINOUS IV INFUSION k = – (ln C1 – ln C2)/(t1 – t2) t1/2 = 0.693/k Vd = [k0(1 – e–kt′)]/{k[Cmax – (Cpredosee–kt′)]} CL = Vd x k Cmax = maximum concentration  STEADY STAE EQUATIONS ❑EXTRAVASCULAR ROUTE k = – (ln C1 – ln C2)/(t1 – t2) t1/2 = 0.693/k Vd/F = D/(C0 – Cpredose) = Vd = [D/(C0 – Cpredose)] x F CL/F = Vd x k CL = Vd x k x F ❑AVERAGE STEADY STATE OF ANY ROUTE CL/F = (D/τ)/Css, CL = [(D/τ) x F]/Css  IV BOLUS DOSING ❑A patient receiving theophylline 300 mg IV every 6 hours has a pre-dose concentration of 2.5 mg/L and post-dose concentrations of 9.2 mg/L one hour and 4.5 mg/L five hours after the dose is given. 1. What is the patient’s elimination rate constant? 2. What is the patient’s volume of distribution? k = – (ln C1 – ln C2)/(t1 – t2) t1/2 = 0.693/k, CL = Vd x k Vd = D/(C0 – Cpredose), C0= C/e–kt  IV BOLUS DOSING ❑Elimination rate constant = k ❖C1 = 9.2 mg/dL C2 = 4.5 mg/dL ❖k = – (ln C1 – ln C2)/(t1 – t2) (ln C1 – ln C2) = ln 9.2 mg/L) – (ln 4.5 mg/L) (ln C1 – ln C2) = 2.2 mg/L – 1.5 mg/L = 0.7 mg/L (t1 – t2) = 1h – 5h = - 4h k = – [(0.7)/-4 h ] k = 0.175/h  IV BOLUS DOSING ❑Volume of distribution = Vd Vd = D/(C0 – Cpredose) C0= C/e–kt if C =9.2mg/L t = 1h, C = 2.5mg/L t = 5 h C0 = 9.2mg/L/e–0.18/h x 1h C0 = 9.2 mg/L/0.84 = 10.95 mg/L = 11.0 mg/L Cpredose = 2.5mg/L Vd = 300mg/(11.0mg/L – 2.5mg/L) Vd = 35.3 L  INTERMITENT IV INFUSION ❑A patient is prescribed gentamicin 100 mg infused over 1 hour every 12 hours. A predose at steady-state concentration (Cpredose) is drawn and is 2.5 mg/L. After the 1hour infusion, a steady-state maximum con. (Cmax) is obtained = 7.9 mg/L. What is the patient’s elimination rate constant k? What is the volume of distribution? k = – (ln C1 – ln C2)/(t1 – t2) Vd = [k0(1 – e–kt′)]/{k[Cmax – (Cpredosee–kt′)]}  INTERMITENT IV INFUSION ❑A patient is prescribed gentamicin 100 mg infused over 1 hour every 12 hours. A predose at steady-state concentration (Cpredose) is drawn and is 2.5 mg/L. After the 1hour infusion, a steady-state maximum con. (Cmax) is obtained = 7.9 mg/L. Since the patient is at steady state, it can be assumed that all predose steady-state concentrations are equal. Because of this the concentration at 12 hours (i.e. Cpredose steady-state concentration during the dosing interval 12 hours) can also be considered equal to 2.5 mg/L and used to compute the elimination rate constant (k) of gentamicin for the patient.  INTERMITENT IV INFUSION ❑k = – (lnC1 – lnC2)/(t1 – t2) k = – [(ln 7.9 mg/L) – (ln 2.5 mg/L)]/(1 h – 12 h) k = 0.105/h Vd = [k0(1 – e–kt′)]/{k[Cmax – (Cpredosee–kt′)]} Vd = [100mg/1 h (1 – e-(0.105/h x 1h))] = 9.97mg/h Vd = 0.105/h [ 7.9mg/L – (2.5mg/L x e-(0.105/h x 1h))] = 0.59mg/L/h Vd = [9.97mg/h]/[0.59mg/L/h] = 16.89 L  EXTRAVASCULAR ROUTE ❑A patient is given procainamide capsules 750 mg (F = 0.8) every 6 hour. The ff con. are obtained before and after the second dose. Cpredose = 1.1 mg/L, con. 2 hours and 6 hour post dose equal 4.6 mg/L and 2.9 mg/L. What is the patient’s elimination rate constant, & volume of distribution? k = – (ln C1 – ln C2)/(t1 – t2) t1/2 = 0.693/k, C0= C/e–kt Vd/F = D/(C0 – Cpredose)  EXTRAVASCULAR ROUTE ❑Elimination rate constant k = – (ln C1 – ln C2)/(t1 – t2) k = – [(ln 4.6 mg/L) – (ln 2.9 mg/L)]/(2 h – 6 h) k = – [(0.46)/(-4)] k = 0.115/h  EXTRAVASCULAR ROUTE ❑Volume of distribution C0 = C/e–kt C0 = (2.9 mg/L)/e(–0.115/h)(6 h) = 5.8mg/L C0 = (4.6 mg/L)/e(–0.115/h)(2 h) = 5.8mg/L Vd/F = D/[C0 – Cpredose] Vd/F = (750 mg)/(5.8 mg/L –1.1mg/L) Vd/F (hybrid Vd) = 160 L, F = 0.8 = 160 x 0.8 Vd = 128 L INDIVIDUALIZED DOSAGE REGIMEN USING ONE COMPARTMENT MODEL  IV BOLUS DOSING ❑STEP – 1: Estimating target average steady state concentration (Cssave): The following equation defines Css∞ based on the minimum (Cssmin) and maximum (Cssmax) steady state concentrations equal to MEC and MTC, respectively. Cssave = Cssmax – Cssmin/ ln (Cssmax/Cssmin)  IV BOLUS DOSING ❑STEP – 2: maximum allowable dosing interval (τmax) The rate of decline in the plasma concentration from Cssmax to Cssmin is governed by the drug elimination half life (t1/2) or elimination rate constant (k). Therefore, we can estimate how long it would take for the plasma con. to decline from a maximum to a minimum Cssmin = Cssmax e-kτmax : The equation can be rearranged, τmax = ln (Cssmax/Cssmin )/k  IV BOLUS DOSING ❑The (τ max) calculated is the longest interval that may be selected for the patient But, the drug administration at every (τmax) hour is not practical Hence, a τmax should be selected from one of the following more practical values: 4, 6, 8, 12, or 24 hourly Obviously, we will choose a longer τ if possible (for patient and staff convenience). However, a selected practical τ cannot be more than calculated τmax if the desired outcome is to keep the plasma con between Cssmin and Cssmax  IV BOLUS DOSING ❑STEP – 3: Estimation of the maintenance dose (MD): MD = Cssmax Vd x (1 – e–kτ) ❑STEP – 4: Estimation of the loading dose (LD): LD = Cssmax x Vd  IV BOLUS DOSING ❑An example of this approach is a patient that needs to be treated for complex seizures with IV phenobarbital. An initial dosage regimen is designed using the population PK parameters (k = 0.139/d, Vd = 50 L) to achieve (Cssmax) = 30 mg/L and (Cssmin) =25 mg/L What is the target average Css? What is the practical dosing interval? What is the maintenance dose? τmax = ln (Cssmax/Cssmin )/k, MD = Cssmax Vd x (1 – e–kτ)  IV BOLUS DOSING The target Cssave can be estimated Cssave= Cssmax – Cssmin/ln (Cssmax/Cssmin) Cssave = (30mg/L – 25mg/L)/In(30mgL/25mg/L) Cssave = (5mg/L)/(0.18) = 27.8 ≈ 28 mg/L  IV BOLUS DOSING ❑Estimation of the maximum allowable dosing interval τmax = (ln Cssmax – ln Cssmin)/k : k = 0.139/d τmax = [ln (30mg/L) – ln (25mg/L)]/0.139/d τmax = 0.18/0.139/d τmax = 1.3 d Round to a practical dosage interval of 1 day  IV BOLUS DOSING ❑Determining the maintenance dose (MD) MD = Cssmax V(1 – e–kτ) V = 50 L MD = (30 mg/L x 50 L) (1 – e(–0.139/d)(1 d)) MD = (1500mg) x (0.13) = 195mg MD, round to practical dose of 200mg The patient would be prescribed intravenous phenobarbital 200 mg daily.  CONSTINOUS IV INFUSION ❑This is the simplest case, as one deals with the infusion rate constant (k0) only [no need to estimate τ] ❖STEP – 1: Estimation of infusion rate constant (k0): k0 can be calculated based on the desired steady state concentration (Css) and the drug clearance (CL) k0 = Css x CL, CL = Vd x k k0 = Css x Vd x k The desired Css is normally a concentration within the MEC and MTC values  CONSTINOUS IV INFUSION ❑STEP – 2: Estimation of loading dose(LD): LD can be calculated based on the Css and Vd of the drug LD = Css x Vd Administration of LD should produce a concentration of Css which is maintained by simultaneous start of the infusion at a rate of infusion (k0)  CONSTINOUS IV INFUSION ❑A patient with a ventricular arrhythmia after a MI needing treatment with lidocaine at a Css of 3.0 mg/L (population PK used: Vd = 50 L, CL = 1.0 L/min). 1. What loading dose is required to achieved the Css? 2. What rate of infusion is required to maintenance the Css? LD = Css x Vd k0 = Css x CL  CONSTINOUS IV INFUSION ❑Loading dose LD = Css x Vd LD = 3mg/L x 50 L = 150 mg Rate of infusion k0 = Css x CL k0 = (3 mg/L) x (1.0 L/min) = 3 mg/min The patient would be prescribed lidocaine 150 mg IV followed by a 3 mg/min continuous infusion.  INTERMITTENT IV INFUSION ❑Few drugs (aminoglycosides and some other antibiotics such as vancomycin) have to be usually administered via multiple shorts (30-60 min) IV infusions at regular intervals Generally, for antibiotics, the desired Cmax is a value several fold above the MIC of the drug for the responsible organism However, Cmin is usually significantly lower than the MIC. For these antibiotics, it is desired to design a dosage regimen to have a Cssmin value above the MIC of the drug and a Cssmax value at or below the MTC.  INTERMITTENT IV INFUSION ❑The approach for selection of a dosage interval is, therefore, slightly different from that used earlier for the design of dosage regimens to produce concentrations within a therapeutic range (MEC and MTC). Dosing interval (τ) = [ln Cssmax – In Cssmin)/k] + t′ Maintenance dose k0 = Cssmax x k x Vd[(1−e-kτ)/ (1−e-k t′)] Loading dose = k0/(1−e-kτ) t′ = infusion time τ = dosing interval  INTERMITTENT IV INFUSION ❑A patient receiving tobramycin for the treatment of intra- abdominal sepsis. Using PK (Vd = 20 L, k = 0.087/h) previously measured in the patient using serum concentrations. 1. Compute a tobramycin dose (infused over 1 hour) that would provide (Cssmax) and (Cssmin) stead-state con of 6 mg/L and 1mg/L, respectively. Dosing interval (τ) = [ln Cssmax – In Cssmin)/k] + t′ Maintenance dose k0 = Cssmax x k x Vd[(1−e-kτ)/ (1−e-k t′)] Loading dose = k0/(1−e-kτ)  INTERMITTENT IV INFUSION ❑Dosing interval τ = [(ln Cssmax – ln Cssmin)/ k] + t′ τ = [(ln 6 mg/L – ln 1 mg/L)/0.087/h] + 1 h τ = 20.57/h + 1h = 21.57 hr: round to practical dosage interval of 24 hourly. τ = 24 hourly  INTERMITTENT IV INFUSION ❑Maintenance dose k0 = Cssmax k x Vd [(1 – e–kτ)/(1 – e–kt′)] Cssmax k x Vd = [(6 mg/L)(0.087/h)(20 L)] = 10.44 mg/h [(1 – e–kτ)/(1 – e–kt′)] = [(1 – e(–0.087/h)(24 h))/(1 – e(–0.087/h)(1 h))] = 10.8 k0 = 10.44 mg/h x 10.8 k0 = 112.8 mg, round to nearest dose of 100mg The patient would be prescribed tobramycin 100mg infused over a period of 1 hour every 24 hours  EXTRAVASCULAR ADMIN. ❑The estimation of dose and dosing rate after extravascular dosing (e.g., oral administration) is more complicated than that after IV bolus doses because the rate and extent bioavailability (F) of extravascular availability would also be important factors in addition to other kinetic parameters One extreme case for extravascular dosing is when the absorption is so fast that it can be assumed as instantaneous for practical purposes This case would be similar to IV bolus administration with F=1  EXTRAVASCULAR ADMIN. ❑Because of the complexity of calculations involving absorption rate constant, in practice, the absorption of most immediate release formulations is assumed to be instantaneous Therefore, the equations used for IV bolus dosing can also be used for design of extravascular dosage regimens with reasonable accuracy Dosing interval (τ) = [ln (Cssmax/Cssmin)/k] + Tmax Maintenance dose = [(Cssmaxx Vd)/F][(1−e−kτ)/e−kTmax] Loading dose = (Cssmax x Vd)/F  EXTRAVASCULAR ADMIN. ❑A patient with simple partial seizures that needs to receive valproic acid capsules (population PK: Vd = 12 L, k = 0.05/h, Tmax = 3 h, F = 1.0) and maintain steady-state (Cssmax) and (Cssmin) concentrations of 80 mg/L and 50mg/L resp. 1. Find the dosing interval 2. Find the maintenance dose Dosing interval (τ) = [ln (Cssmax/Cssmin)/k] + Tmax Maintenance dose = [(Cssmaxx Vd)/F][(1−e−kτ)/e−kTmax]  EXTRAVASCULAR ADMIN. ❑Dosing interval τ = [ln (Cssmax/ln Cssmin)/k] + Tmax τ = [(ln (80 mg/L/50 mg/L)/0.05/h] + 3 h = 12.4 h τ = 12.4 h, round to practical dosage interval of 12 h τ = 12 hours  EXTRAVASCULAR ADMIN. ❑Maintenance dose MD = [(Cssmaxx Vd)/F][(1−e−kτ)/e−kTmax] MD = [(80 mg/L x 12 L)/1.0)] = 960mg MD = [(1 – e(–0.05/h)(12 h))/e(–0.05/h)(3 h)] = 0.523 MD = 960mg x 0.51 = 502 mg, Round to practical dose of 500 mg The patient will be prescribed valproic acid orally very 12 hours  ANY ROUTE OF ADMINISTRATION ❑Dosing rate = dose/dosing interval (τ) Dosing rate (dose/τ) = (CL x Css)/F Dosing interval (τ) = (dose x F)/(CL x Css) Dose = [(CL x Css)/F] x τ Find dose, if Css = 5mg/L, CL = 2L/h, τ = 12 h F= 50% Dose = [(5mg/L x 2 L/h)/0.5] x 12 h Dose = 20 mg/h x 12 h = 240mg Dose = 240 mg  ANY ROUTE OF ADMINISTRATION ❑Maintenance dose = dosing rate x dosing interval Maintenance dose = [(Css x CL)/F] x τ CL = Vd x k, k = 0.693/half-life(T1/2) CL = (Vd x 0.693)/half-life(T1/2) Find MD, if Css = 5mg/L, CL = 2L/h, τ = 12h F= 50% MD = (5mg/L x 2L/h)/0.5 = 20mg/h MD = 20mg/h x 12h = 240mg MD = 240mg every 12 hours  ANY ROUTE OF ADMINISTRATION ❑Loading dose LD = [Css x Vd]/ F Vd = (LD x F)/ Css Vd = CL/k, k = 0.693/half-life(T1/2) Vd = CL x half-life(T1/2)/ 0.693 Vd = Dose/Plasma concentration ❖Find LD when Vd= 50 L, Css = 5mg/L and F =50% LD = (5mg/L x 50L)/0.5 LD = 250mg/0.5 = 500mg THANK YOU