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W.L. Backes CSRB 4D3 Principles of Pharmacology Drug Metabolism A. Characteristics 1. 2. 3. B. chemical change in a drug forms more water soluble metabolites usually terminates drug action Types of non-synthetic (phase 1) reactions 1. oxidation reactions – Mostly by cytochrome P450 NADPH RH O2 NADPH cyt. P450 reductase Fe3+ FAD FMN P450 NADP+ a. types of microsomal oxidations 1) aromatic hydroxylation OH CH2 CH 3 2) 3) CH 2 aliphatic hydroxylation epoxidation CH2OH H 2O ROH Principles of Pharmacology Backes Page 2 O 4) N-, O-, and S-dealkylations HO CH 3O + HCHO O O NCH 3 NCH 3 HO HO Morphine Codeine b. a. b. 5) N-hydroxylation (not P450) 6) N-oxidation (not P450) 7) oxidative deamination 8) sulfoxide formation 9) desulfuration broad selectivity of cytochrome P-450 1) multiple forms 2) mechanistic broad selectivity classification based on sequence similarity 1) 40% families 2) 59% subfamilies regulation of P450 expression 1) enzyme induction a. transcriptional activation b. also stabilization of mRNA, protein, and increased protein synthesis 2) polymorphisms a. are frequently silent b. expression polymorphisms c. polymorphisms in structural gene Principles of Pharmacology Backes Page 3 c. d. f. g. h. CYP1A 1) inducible by polycyclic hydrocarbons – large compounds that look like cholesterol 2) found in cigarette smoke 3) 1A1 – inducible 4) 1A2 constitutively expressed - little expression in extrahepatic tissues 5) CYP1A1 polymorphisms a) Ah receptor polymorphism b) polymorphisms in structural gene 6) CYP1A2 polymorphisms a) detected by caffeine metabolism b) CYP1A2 is expressed only in liver, but appears to be associated with cancer at distal sites CYP2A 1) hormonally regulated forms 2) 2A1 - increased in female and young male rats 3) 2A2 - not found in female rats 4) 2A6 in humans 5) CYP2A6 polymorphism a) significant ethnic variability b) variant 2A6 alleles c) coumarin 7-hydroxylation CYP2B - induced by phenobarbital 1) 2B1 inducible in liver but constitutive in lung 2) 2B2 constitutive and inducible 3) 2B6 human P450 isozyme CYP2C – constitutive hormonally regulated isozymes 1) gender specific expression in rodents 2) regulated by growth hormone levels 3) CYP2C19 polymorphisms a) there are 7 different 2C isozymes in human b) mephenytoin metabolism is marker activity (1) 3 – 5 % of Caucasians (2) 20 – 25% of Asians c) CYP2C19 metabolizes many drugs of pharmacologic significance (1) omeprazole, imipramine, propranolol, hexobarbital, diazepam, etc. (2) taxol CYP2D6 – human P450 isozyme – exhibits polymorphisms 1) debrisoquine polymorphism in 4-hydroxylation autosomal recessive trait related to increased cancer risk – extensive metabolizers were more susceptible to cancer P450 2D appears to have an abnormal substrate binding site Principles of Pharmacology Backes Page 4 2. 2) dextromethorphan 3) codeine i. CYP2E1 1) ethanol-inducible isozyme 2) production of reactive oxygen species - toxicity 3) polymorphisms 4) role in cancer 5) role in alcohol-related liver dysfunction j. CYP3A 1) hormonally regulated P450 isozymes 2) CYP3A4 major isozyme found in humans 3) no discernable polymorphisms 4) 10% of patients show poor response to immunosuppressant therapy with the 3A4 substrate cyclosporine k. non-microsomal oxidations 1) alcohol dehydrogenase 2) aldehyde dehydrogenase 3) xanthine oxidase 4) tyrosine hydroxylase 5) monoamine oxidase reduction a. location of reductases 1) microsomes 2) cytosol 3) anaerobic microorganisms in the gut b. 3. types of reduction reactions 1) azoreduction 2) nitroreduction 3) ketoreduction hydrolysis a. esterase 1) succinylcholine apnea a) due to an esterase deficiency b) enzyme appears to have an altered affinity for esters c) affects about 1 in 3000 b. amidase c. epoxide hydrolase Principles of Pharmacology Backes Page 5 B. synthetic (phase II) reactions 1. conjugation reactions OSO3H OH O-glucuronide a. b. d. glucuronidation 1) found in endoplasmic reticulum 2) reacts with hydroxyl, amine or carboxylic acid functions sulfate conjugation 1) found in the cytosol 2) reacts with alcohols, phenols, and aromatic amines 3) saturable acetylation Principles of Pharmacology Backes Page 6 CH3 C O O N NH2 C NH NH Acetyl CoA N C NH O 1) 2) 3) d. e. found in the cytosol reacts with primary amines Acetylation polymorphism a) differences not due to differences in quantity of enzyme b) in U.S. mostly slow acetylators (0.72) c) Japanese have a higher percentage of rapid acetylators, whereas peoples of Middle Eastern descent, Scandanavians and Finns have more slow acetylators d) slow acetylators more likely to exhibit toxic effects e) specific acetylation disorders (1) Isoniazid-induced neurotoxicity – slow acetylators (2) Isoniazid-induced hepatitis – rapid acetylators (3) Drug-induced lupus erythematosus – slow acetylators (4) Arylamine induced bladder cancer – higher incidence in slow acetylators (5) colorectal cancer – higher incidence in rapid acetylators particularly when associated with rapid CYP1A2 phenotype methylation 1) found in both the cytosol and the endoplasmic reticulum 2) S-adenosyl methionine is the methyl donor 3) reacts with hydroxyl, sulfhydryl, and amine functions glutathione conjugation 1) superfamily of genes found on 11 chromosomes 2) found in the cytosol and mitochondria (kappa) of liver and numerous other tissues 3) exists as a functional dimer 4) mu, pi, sigma (tyrosine) and theta (serine) 5) drug binds to glutathione (glu - - cys-gly) 6) can ultimately form mercapturic acids 7) reacts with: a) aromatic hydrocarbons (epoxides) b) arylamines c) organic halides a) phenols 8) Principles of Pharmacology Backes Page 7 b) some have ligand-binding capabilities (ligandins) alpha class GSTs are inducible by ingestion of high levels of brussels spouts H S C H2 O H OH H C C H2 H N H2 C COOH NH C CH2 CH2 CHNH2 COOH 9) f. polymorphisms a) four classes – GSTM1 to GSTM4 b) GSTM1 is polymorphic (*0, *a, and *b) 1) null gene, increased risk of cancer 2) essentially absent in 20 – 50% of all individuals c) GSTP1 (four allelic variants) d) ethnic component (1) higher percentage of null gene in Asians than Caucasians (2) lower percentage in blacks e) links between GSTM1 null phenotype and pituitary adenomas, head and neck cancer, malignant melanoma, colorectal cancer and bladder carcinoma f) association of lack of GSTM1 with lung cancer in heavy smokers g) appears to be linked to a combined GSTM1 null and CYP1A1 Msp I. glycine conjugation 1) found in both the cytosol and mitochondria 2) reacts with aliphatic acids and aromatic carboxylic acids 3) can be saturated 4) developmentally induced Factors that Affect Drug Metabolism I. Age A. B. Developmental changes Changes with senescence Principles of Pharmacology Backes Page 8 II. Induction and Inhibition by Foreign Compounds A. B. III. Dietary factors A. B. C. D. IV. Administration of drugs and chemicals can induce P450 and other drug metabolizing enzymes 1. phenobarbital and other drugs 2. polycyclic hydrocarbons and carcinogenesis 3. ethanol Inhibitors of drug metabolism 1. ethanol 2. cimetidine charcoal broiled beef vitamin deficiencies brussel sprouts grapefruit Species differences A. B. quantitative differences qualitative differences Principles of Pharmacology Backes Page 9 Clinical Pharmacokinetics I. Compartmental Modeling A. One compartment model Total Body Volume A B. absorption elimination A A Two compartment model Central compartment A absorption elimination A distribution A Peripheral compartment A Principles of Pharmacology Backes Page 10 [drug] in plasma g/100 ml Intravenous Bolus (One compartment model) 30 20 50 5 10 2 4 6 8 10 12 0.5 0 14 4 Time (h.) 8 12 Time (h.) 1. Elimination rate constant 2. Half-life 3. Relationship between elimination rate constant and half-life t1/2 = 0.693/kel 4. Zero order kinetics zero order 30 20 [drug] in plasma g/100 ml 0 0 [drug] in plasma g/100 ml [drug] in plasma g/100 ml C. zero order 50 5 1st order 0.5 0 4 8 12 Time (h.) 10 first order 0 0 2 4 6 8 Time (h.) 10 12 14 Principles of Pharmacology Backes Page 11 Single Extravascular Administration 30 100 [drug] in plasma g/100 ml [drug] in plasma g/100 ml C. 20 10 1 0 10 10 20 30 40 50 Time (h.) 0 0 10 20 30 40 50 Time (h.) 1. 2. 3. Elimination rate constant is calculated the same way as for an i.v. bolus Absorption rate constant Effect of absorption rate [drug] in plasma g/100 ml Effect of absorption rate constant ka=1.39 ka=0.693 20 ka=0.277 ka=0.069 10 0 0 10 20 30 40 50 Ti me (h.) Intravenous Bolus (Two compartment model) 75 [drug] in plasma g/100 ml 100 [drug] in plasma g/100 ml D. 50 1 0 25 0 0 10 4 8 12 16 20 24 Time (h.) 2 4 6 8 Time (h.) 10 12 14 Principles of Pharmacology Backes Page 12 1. 2. E. calculation and significance of the disposition rate constant distribution rate constant Single Extravascular Dose (Two compartment model) 50 40 [drug] in plasma g/Liter [drug] in plasma g/100 ml 100 30 20 10 1 0 10 20 30 40 Time (h.) 10 0 0 4 8 12 16 20 24 Time (h.) Volume of Distribution A. B. Apparent volume into which a drug distributes in the body How it is calculated 1. Inject a known quantity of drug as an i.v. bolus 2. Take blood samples at various times after injection 3. Plot data on a semilogarithmic plot and extrapolate back to t = 0 to calculate the concentration prior to elimination 4. Take values and plug into the following formula: VD = __amount of drug___ plasma concentration C. III. Factors affecting volume of distribution Clearance A. Total clearance 1. ClT = kel * VD 2. ClT = X0/Area under the curve = X0/Cp [drug] in plasma g/100 ml II. 50 Co p 5 0.5 0 4 8 Time (h.) 12 Principles of Pharmacology Backes Page 13 B. C. Bioavailability A. B. Intravenous Infusion A. Conditions 1. constant rate of drug administration (zero order) 2. proportional elimination of drug (first order) [drug] in plasma g/100 ml V. area under the curve time to peak drug concentration 30 20 10 t1/2 = 4 h 0 0 2 4 6 8 10 12 14 16 18 20 22 24 Time (h.) B. C. D. E. F. Css = infusion rate/ClT Adjustments to dosage – the Css is directly proportional to the infusion rate (dose) Time required to achieve Css Relationship between i.v. infusion and i.v. bolus Effect of alteration in infusion rate [drug] in plasma g/100 ml IV. 3. ClT = Clliver + Clkidney + Clother Hepatic clearance 1. Extraction ratio 2. Relationship between Cl Renal clearance 30 infusion 20 10 bolus 0 0 2 4 6 8 10 12 14 16 18 20 22 24 Time (h.) Principles of Pharmacology Backes Page 14 VI. Repeated intravenous injections A. Relationship between frequency of administration and Css 10 [drug] in plasma [drug] in plasma 20 5 0 0 50 100 150 10 0 0 200 Time (h) 10 20 30 40 Time (h) Left panel – A drug with a t1/2 = 5 h is administered after virtually all of the drug is eliminated. Right panel – A drug with a t1/2 = 5 is administered about once every half-life. Note the greater accumulation of drug and decreased fluctuations. B. Relationship between dose and Css C. Relationship between t1/2 and Css [drug] in plasma 75 50 t1/2 = 50 h 25 t1/2 = 5 h 0 0 10 20 30 40 50 60 70 Time (h) The above graph depicts two drugs that are administered with the same frequency, but have different half-lives. The drug with the shorter half-life reaches its steady state concentration more rapidly. The drug with the longer half-life exhibits a much greater accumulation in the blood and shows smaller fluctuations (on a percentage basis). Principles of Pharmacology Backes Page 15 D. Loading dose 20 [drug] in plasma administered an initial loading dose 10 no loading dose 0 0 10 20 30 Time (h) VII. Repeated extravascular administration 1 5 [drug]inblod 1 0 5 0 0 5 1 0 1 5 2 0 2 5 3 0 3 5 T i m e ( h r s ) Principles of Pharmacology Backes Page 16 Principles of Pharmacology Pharmacokinetics Problem Set January 2009 1. A patient was administered an analgesic drug by a single 12 mg intravenous injection. Blood samples were then taken at various times and the following results were obtained. [drug] in plasma g/Liter 100 10 1 0.1 0 4 8 12 Time (h.) What type of pharmacokinetic model is followed by this data? Calculate the t1/2, the elimination rate constant and the apparent volume of distribution for the drug. What is the total body clearance for this drug? 2. You have decided to administer the drug described in question #1 by intravenous infusion at a rate of 3.7 mg/min. How long will it take before steady state is attained? What is the Css after intravenous infusion? The drug was administered to the patient for 10 days, after which administration was discontinued. What would the drug concentration be 9 hrs. after administration was stopped? You want to give the drug orally in four separate doses, and you know that 20% of the drug will be inactivated by first pass metabolism. What dose of this drug would you prescribe if you wanted to achieve a steady state plasma concentration of 0.29 mg/L. A new drug used in the treatment of asthma was administered by i.v. infusion at a rate of 1 g/min. A graph of the infusion is shown. 20 [drug] (g/liter) 3. 15 10 5 0 0 10 20 30 40 50 60 Time (hours) 70 80 90 100 Principles of Pharmacology Backes Page 17 a. What is the half-life of this drug? b. Also calculate the elimination rate constant, the volume of distribution, and total body clearance for this drug. c. The drug is eliminated 20% by renal and 80% by hepatic mechanisms. If there is a 50% hepatic failure, then what is the new steady state concentration? d. If the drug was administered by oral administration how many hours would be required to attain steady state? e. If the drug (no hepatic failure) when taken orally has only a 40% bioavailability when compared to the intravenous route, then what oral dose would be required to attain a steady state concentration of 2 g/ml (the drug is to be administered every 12 h). f. Provide a possible explanation for the low bioavailability of this drug. 4. Blood ethanol data were obtained from a patient on admission and at various times afterward. The following graph of the data was obtained. [Ethanol] (mg/100 ml) 400 350 300 250 200 150 100 50 0 0 1 2 3 4 5 6 7 8 Time (hours) What is the pharmacokinetic model? Calculate the rate of disappearance of the drug. Answers 1. One compartment Model t1/2 = 3 hrs kel = 0.231/ hr Vd = 375 liters ClT = 86.6 Liter/h or 1444 ml/min 2. Css = 2.56 µg/ml 12 – 15 hrs to attain steady state C9 hrs = 0.32 µg/ml Principles of Pharmacology Backes Page 18 Dose = 31.4 mg/hr = 753.6 mg/day (or 4 x 188 mg) 3. t1/2 = 10 hrs (from graph) kel = 0.0693/hr ClT = 50 ml/min Vd = 43.3 liters ClR = 10 ml/min ClH = 40 ml/min with 50% hepatic failure, ClH is decreased to 20 ml/min therefore ClTmodified = 30 ml/min Css = 33µg/liter Recalculate time to achieve steady state: t1/2 = 16.7 hrs time to steady state = 66 - 83 hrs Calculation of oral dose 4. 180 mg every 12 hrs zero order decay 20 mg (100 ml)-1(hr)-1 for zero order reactions v = kel, therefore kel = 20 mg/100ml/hr.