bài tập IV bolus tiếng việt

1 Acyclovir is an antiviral drug used in the treatment of herpes simplex, varicella zoster, and in suppressive therapy In this study, three male cynomolgus monkeys were each given a 10 intravenous dose The monkeys weighed an average of 3.35 each Blood samples were collected and the following data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (giờ) Nồng độ huyết C (µg/mL) LnC 0.167 26.0 3.258 0.3 23.0 3.135 0.5 19.0 2.944 0.75 16.0 2.773 1.0 12.0 2.485 1.5 7.0 1.946 2.0 5.0  Phương trình LnC theo thời gian: y = -0.9258x + 3.4152 1.609 Hằng số tốc độ thải trừ: K = 0.9258 h-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.9258 = 0.75 h LnC0 = 3.4152 → C0 = e3.4152 = 30.4 µg/mL AUC = C0/K = 30.4/0.9258 = 32.84 àg/mLìh Th tớch phõn bố: Vd = D/C0 = 3.35×10/30.4 = 1.1 L Độ thải: Cl = K×Vd= 0.9258×1.1 = 1.02 L/h k= 0.93hr1 tẵ= 0.75hr C0 = 30.4ug /mL AUC= 32.75ug/mL ì hr Vd = 1.1 L Cl= 1.02L /hr A study by Xu, Pai, and Melethil establishes the pharmacokinetics of Aluminum in Rats In this study, four rats with an average weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg) Blood samples were taken at various intervals and the following data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (giờ) Nồng độ huyết C (ng/mL) LnC 0.4 19000 9.852 0.6 18000 9.798 1.4 15000 9.615 1.6 14500 9.582 2.3 12500 9.433 3.0 10500 9.259 4.0 8500 9.047 5.0 6500 8.779 6.0 5000 8.517 8.0 3250 8.086 10.0 2000 7.601 12.0 1250  Phương trình LnC theo thời gian: y = -0.235x + 9.9555 7.131 Hằng số tốc độ thải trừ: K = 0.235 h-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.235 = 2.95 h LnC0 = 9.9555 → C0 = e9.9555 = 21067.8 ng/mL Diện tích đường cong: AUC = C 0/K = 21067.8/0.235 = 89646.8 ng/mL×h Thể tích phân bố: Vd = D/C0 = 0.375×106/21067.8= 17.8 mL Độ thải: Cl = K×Vd= 0.235×17.8 = 4.183 mL/h k= 0.234hr–1 t½= hr C0= 21000ng /mL AUC= 89285ng /mL × hr Vd = 17.86mL Cl= 4.18mL /hr Amgen (r-Epo) is a form of recombinant erythropoetin Erythropoetin is a hormone that is produced in the kidneys and used in the production of red blood cells The kidneys of patients who have end-stage renal failure cannot produce erythropoetin; therefore, r-Epo is being investigated for use in these patients in order to treat the anemia that results from the lack of erythropoetin In a study by Salmonson et al, six healthy volunteers were used to demonstrate that both IV and subcutaneous administration of erythropoetin have similar effects in the treatment of anemia due to chronic renal failure The six volunteers were each given a 50 U/kg intravenous dose of Amgen The average weight of the six volunteers was 79 kg Blood samples were drawn at various times and the data obtained is summarized below: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (giờ) Nồng độ huyết C (mU/mL) LnC 700 6.551 600 6.397 400 5.991 300 5.704 12 24 150 5.010 40 3.689  Phương trình LnC theo thời gian: y = -0.1332x + 6.8004 Hằng số tốc độ thải trừ: K = 0.1332 h-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.1332 = 5.2 h LnC0 = 6.8004→ C0 = e6.8004 = 898.2 mU/mL Diện tích đường cong: AUC = C0/K = 898.2/0.1332 = 6.743 mU/mL×h Thể tích phân bố: Vd = D/C0 = 79×50/898.2 = 4.398 L Độ thải: Cl = K×Vd= 0.1332×4.398 = 0.586 L/h k= 0.134hr–1 tẵ= 5.2hr C0= 900mU /mL AUC= 6945mU/mL ì hr Vd = 4.44L Cl= 0.6L/hr 4.A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats Samples of blood were taken at various intervals throughout the length of the study and the following data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (phút) Nồng độ huyết C (pg/mL) LnC 380 5.94 10 280 5.63 20 170 5.13 30 130 4.87 40 100 4.61 50 70 4.25 60 50  Phương trình LnC theo thời gian: y = -0.034x + 5.9574 Hằng số tốc độ thải trừ: K = 0.034 min-1 3.91 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.034 = 20.38 LnC0 = 5.9574→ C0 = e5.9574 = 386.6 pg/mL Diện tích đường cong: AUC = C0/K = 386.6/0.034= 11370 pg/mL×min Thể tích phân bố: Vd = D/C0 = 45×103/386.6 = 116.4 mL Độ thải: Cl = K×Vd= 0.034×116.4 = mL/min k= 0.0345min1 tẵ= 20.09min C0= 386.6pg ÔmL AUC= 11206.4pg /mL × Vd = 116.4mL Cl= 4.02mL ¤min Aztreonam is a monolactam structure which is active against aerobic, gramnegative bacilli The pharmacokinetic parameters of Aztreonam were established in a study presented in by Cuzzolim et al in which Aztreonam (100 mg/ kg) was administered intravenously to 30 premature infants over minutes every 12 hours The group of neonates had an average weight of 1639.6g The following set of data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (phút) Nồng độ huyết C (µg/mL) LnC 40.5 3.7 34.99 3.555 29.99 3.4 23.88 3.173 22.2 3.1 19.44 2.967 16.55 2.806 14.99  Phương trình LnC theo thời gian: y = -0.1437x + 3.8233 2.707 Hằng số tốc độ thải trừ: K = 0.1437 min-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.1437= 4.8225 LnC0 = 3.8233 → C0 = e3.8233 = 45.755 µg/mL AUC = C0/K = 45.755/0.1437= 318.406 àg/mLìmin Th tớch phõn b: Vd = D/C0 = 100×1.6396/45.755 = 3.435 L Độ thải: Cl = KìVd= 0.1437ì3.435= 0.5L/min k= 0.144min1 tẵ= 4.81 C0= 45.75ug ¤mL AUC= 317.7ug/mL × Vd = 3.58L Cl= 0.516L /min Bovine placental lactogen (bPL) is a hormone similar to growth hormone and prolactin It binds to both prolactin and growth hormone receptors in the rabbit and stimulates lactogenesis in the rabbit In a study by Byatt, et al., four cows (2 pregnant and nonpregnant) were given IV bolus injections of mg and the following data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (phút) Nồng độ huyết C (µg/L) LnC 3.8 117 4.762 6.8 72 4.277 12 43 3.76 16.0 27 3.296 20.0 18  Phương trình LnC theo thời gian: y = -0.113x + 5.123 2.89 Hằng số tốc độ thải trừ: K = 0.133 min-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.133 = 6.13 LnC0 = 5.123 → C0 = e5.123 = 167.8 µg/L Diện tích đường cong: AUC = C 0/K = 21067.8/0.235 = 89646.8 àg/Lìmin Th tớch phõn b: Vd = D/C0 = 4×103/167.8= 23.84L Độ thải: Cl = KìVd= 0.133ì23.84= 2.69 L/min k= 0.113min1 tẵ= 6.13min C0= 167.8ug /L AUC= 1484.9ug/ L × Vd = 23.84L Cl= 2.69L /min This study examines the pharmacokinetics of caffeine in the rabbit In this study type I New Zealand White rabbits were given an mg intravenous dose of caffeine Blood samples were taken and the following data was obtained: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Thời gian (phút) Nồng độ huyết C (µg/mL) LnC 12 3.75 1.32 40 2.8 1.03 65 2.12 0.75 90 1.55 0.438 125 1.23 0.207 173 0.72 -0.328 243 0.37 -0.994  Phương trình LnC theo thời gian: y = -0.01x + 1.4122 Hằng số tốc độ thải trừ: K = 0.01 h-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.01 = 69.3 LnC0 = 1.4122→ C0 = e1.4122 = 4.105 µg/mL Diện tích đường cong: AUC = C0/K = 4.105/0.01 = 410.5àg/mLìmin Th tớch phõn b: Vd = D/C0 = ×103 /4.105=1.95(L) Độ thải: Cl = K×Vd= 0.01×1.95 = 0.0195L/min k= 0.00997min–1 t½= 69.51 C0= 4.105ug /mL AUC= 411.7ug /mL × Vd = 1.95L Cl= 19.44mL/min Ceftazidime is a third generation cephalosporin which is administered parenterally In this study, eight patients with chronic renal failure were each given g of ceftazidime intravenously Both blood samples were taken the data obtained from the study is summarized in the following table From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Nồng độ huyết C (mg/L) 50 45 38 24 21 36 14 48 11 60 72  Phương trình LnC theo thời gian: y = -0.0324x + 3.8622 Thời gian (giờ) LnC 3.91 3.8 3.64 3.04 2.64 2.4 2.079 1.386 Hằng số tốc độ thải trừ: K = 0.0324h-1 Thời gian bán thải: t1/2 = 0.693/K = 0.693/0.0324 = 21.39 h LnC0 = 3.8622→ C0 = e3.8622 = 47.57 mg/L Diện tích đường cong: AUC = C0/K = 47.57/0.0324= 1468.21mg/L×h Thể tích phân bố: Vd = D/C0 = 103/47.57 = 21.02 L Độ thải: Cl = KìVd= 0.0324ì21.02 = 0.68 L/h k= 0.0324hr1 tẵ= 21.39hr C0= 47.57mg/L AUC= 1468.2mg/L × hr Vd = 21.02L Cl= 0.681 L/hr Ciprofloxacin is a fluoroquinolone antibiotic which is used in the treatment of infections of the urinary tract, lower respiratory tract, skin, bone, and joint In this study, twelve healthy, male volunteers were each given 300 mg intravenous doses of Ciprofloxacin Blood and urine samples were collected at various times throughout the day and the following data was collected: From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Dose X = 300mg Nồng độ huyết C (mg/L) 1.20 0.85 0.70 0.50 0.35 10 0.25  Phương trình LnC theo thời gian: y = -0,1875x + 0,454  lnCp = lnCpo – Kt = 0,454 – 0,1875t K = 0,1875 hr-1 t1/2 = = = 3,697 hr Cpo = e0,454 = 1,5746 (mg/L) AUC = = = 8,398 (mg.hr/L) Thời gian (giờ) LnC 0.182 -0.163 -0.357 -0.693 -1.05 -1.386 Vd = = = 190,52 L Cl = K.Vd = 0,1875 190,52 = 35,72 (L/hr) k= 0.1875hr–1 t½= 3.7hr C0= 1.57mg/L AUC= 8.395mg/L× hr Vd = 190.6L Cl= 35.74L/hr 10 Diprophylline is used as a bronchodilator A study by Nadai et al was designed to determine whether or not coadministration of Diprophylline with Probenecid affected the pharmacokinetic parameters of Diprophylline In this study, male rats (average weight: 300 g) were given 60 mg/kg of Diprophylline intravenously and a mg/kg loading dose of Probenecid followed by a continuous infusion of 0.217 mg/min/kg of Probenecid The following set of data was obtained for Diprophylline (DPP) From the data presented in the Preceding table, determine the following: Find the elimination rate constant, Find the half life, Find C0 Find the Area Under the Curve, Find the volume of distribution, Find the clearance, Solution Dose X = 60.0,3 = 18 mg Nồng độ huyết C (µg/mL) 16 40.00 31 27.00 60 13.00 91 6.50 122 3.50  Phương trình LnC theo thời gian: y = -0,023x + 4,01  lnCp = lnCpo – Kt = 4,01 – 0,023t K = 0,023 min-1 t1/2 = = = 30,077 Cpo = e4,01 = 55,15 (µg/mL) AUC = = = 2398 (µg.min/mL) = 2,398 (mg.min/mL) Vd = = 103 = 326,38 mL Cl = K.Vd = 0,023 326,38 = 7,507 (mL/min) Thời gian (phút) LnC 3.689 3.296 2.565 1.872 1.253 k= 0.023min–1 t½= 30.13min C0= 55.13ug/mL AUC= 2396.96ug/mL × Vd = 326.5mL Cl= 7.5mL /min 11 (Q3 page 78) The following cumulative amounts of drug in the urine (Xu) were obtained after an intravenous bolus injection of 500 mg of the drug (X0), which is eliminated exclusively by urinary excretion Plot the data in as many ways as possible and, by means of your plots determine the following a The elimination half life (t1/2) b The elimination rate constant (K) c The cumulative amount of drug eliminated (Xu) in the urine at h following the administration of a 500 mg dose Time (h) 2.0 4.0 6.0 8.0 10.0 12.0 Infinity Xu (mg) 190.0 325.0 385.0 433.0 460.0 474.0 500.0 Solution Dose X = 500 mg Time (h) Xu (mg) ARE (mg) 2.0 190.0 310 4.0 325.0 175 6.0 385.0 115 8.0 433.0 67 10.0 460.0 40 12.0 474.0 Infinity 500.0  Phương trình LnX theo thời gian: y = -0,248x + 6,202  ln(ARE) = lnXo – Kt = 6,202 – 0,248t K = 0,248 hr-1 t1/2 = = = 2,795 hr Xu = X0.(1-e-Kt) = 500.(1-e-0,248.7) = 411,888 mg ln(ARE) 5.737 5.164 4.745 4.205 3.689 1.792 12 (Q5 page 78) Israel et al (1978) reported urinary excretion data for cinoxacin following intravenous bolus administration of 250 mg of drug to nine healthy male volunteers Plot the data in suitable manners and, by means of your plot, determine the following a The elimination rate constant (K) b A comparison of the elimination rate constant obtained by both methods c A comparison of the elimination rate constant obtained by these methods with that obtained in question d The excretion (Ku) and metabolite rate (Km) constants Time interval (h) Mass cinoxacin recovered 0–2 2–4 4–6 6–8 8–24 in urine (mg ("SD)) 88.0 ± 34 25.0 ± 13 10.0 ± 3.0 ± 0.4 ± 0.5 Solution Dose X = 250 mg Xu’ (mg) = ΔXu taverage (h) (mg) 0–2 88.0 ± 34 2–4 25.0 ± 13 4–6 10.0 ± 6–8 3.0 ± 8–24 0.4 ± 0.5 16  Phương trình Ln() theo thời gian: y = -0,552x + 4,291  ln() = ln(Ku.X0) – Kt = 4,291 – 0,552t K = 0,552 hr-1 Ku.X0 = Ku.250 = e4,291 = 73,039 Ku = = 0,292 hr-1 Km = K – Ku = 0,552 – 0,292 = 0,26 hr-1 Δt (h) ln() 3.784 2.526 1.609 1.253 ... pharmacokinetics of Aluminum in Rats In this study, four rats with an average weight of 375g, were given an IV bolus dose of aluminum (1 mg/kg) Blood samples were taken at various intervals and the following... 6945mU/mL × hr Vd = 4.44L Cl= 0.6L/hr 4.A study by Brier and Harding a dose of 45 ng was given by IV bolus to rats Samples of blood were taken at various intervals throughout the length of the... lactogenesis in the rabbit In a study by Byatt, et al., four cows (2 pregnant and nonpregnant) were given IV bolus injections of mg and the following data was obtained: From the data presented in the Preceding