applied pharmacokinetics i, Exercises of Pharmacology

Download applied pharmacokinetics i and more Exercises Pharmacology in PDF only on Docsity! 1 COLLEGE OF MEDICINE AND HEALTH SCIENCES SCHOOL OF MEDICINE AND PHARMACY DEPARTMENT OF PHARMACY HUYE CAMPUS CLASS: PHARMACY IV ACADEMIC YEAR 2018-2019 NO REG.NO NAMES Others 1. 215007907 MINANI Theobald 2. 216130735 MUGISHA Carmelle 3. 216074142 NIYIRORA Clementine 4. 215012997 HAGABIMANA Pierre 5. 216163242 MURIISA Moise 6. 216074991 NIYONKURU Emmanuel 7. 216351537 NDEREYIMANA Irene Fischer EXERCISES/ASSIGNMENT MODULE : APPLIED PHARMACOKINETICS AND TOXICOLOGY COMPONENT : APPLIED PHARMACOKINETICS AND BIOPHARMACY LECTURER : PROF KADIMA NTOKAMUNDA GROUP 5 PHARM IV 2 THERAPEUTIC DRUG MONITORING SECTION I. ANTIEPILEPTIC DRUGS I.CARBAMAZEPINE Given  Elimination: active anticonvulsant metabolite, 10,11-epoxide.  Therapeutic window (Cp): 4 – 12 μg/ml ,(F)=0.8, ( S)=1, (Vd)=1.4 L/kg (Cl)=0.064 L/h/kg, (T1/2) =15 h en steady state , (T1/2)=30-35 h en dose unique., (𝛼) =0.2 – 0.3, Peak absorption =6h Demand: 1. Da and Dm=?? 2. Dm to adjust from 4mcg/ml to 6mcg/ml Solution and formula 1. 𝐴𝑡 𝑡ℎ𝑒 𝑏𝑒𝑔𝑔𝑖𝑛𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 , 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑛𝑒𝑒𝑑 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒  𝐴𝑡 𝑏𝑒𝑔𝑔𝑖𝑛𝑖𝑛𝑔 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 , 𝐷𝐴𝐼𝐿𝑌 𝐷𝑂𝑆𝐸 = 𝐷𝑎(𝐿𝑜𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒)  𝐷𝑎 = 𝑉𝑑∗𝑐𝑝 𝑆∗𝐹  𝑡ℎ𝑒 𝑑𝑜𝑠𝑒 𝑤𝑎𝑠 𝑔𝑖𝑣𝑒𝑛 𝑑𝑎𝑖𝑙𝑦 𝑖𝑚𝑝𝑙𝑦 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑑𝑜𝑠𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 Ʈ = 24ℎ  𝑇ℎ𝑢𝑠 , 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛𝑠 (𝑉𝑑) = 60𝑘𝑔 ∗ 1.4𝑙 𝑘𝑔 = 84𝑙  A) 𝐴𝑡 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑤𝑒𝑒𝑘 , 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑤𝑖𝑙𝑙 𝑛𝑒𝑒𝑑 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛𝑡𝑎𝑖𝑛𝑒𝑛𝑐𝑒 𝑑𝑜𝑠𝑒  𝐶𝑙 = 0.064𝑙/ℎ/𝑘𝑔 ∗ 60𝑘𝑔 = 3.84𝑙/ℎ  𝐾𝑒 = 0.693 𝑡½ = 0.693 15ℎ = 0.0462/ℎ  𝑑𝑟𝑢𝑔 𝑤𝑎𝑠 𝑎𝑑𝑚𝑖𝑛𝑖𝑠𝑡𝑒𝑟𝑒𝑑 𝑑𝑎𝑖𝑙𝑦 , 𝑖𝑚𝑝𝑙𝑦 𝑡ℎ𝑎𝑡 Ʈ = 24ℎ B. 𝑴𝒂𝒊𝒏𝒕𝒂𝒊𝒏𝒆𝒏𝒄𝒆 𝒅𝒐𝒔𝒆 ( 𝑫𝒎)  𝐶𝐸 = 𝑄𝑜 𝑘𝑒𝑉𝑑Ʈ 𝑻𝒉𝒆 𝒍𝒐𝒂𝒅𝒊𝒏𝒈 𝒅𝒐𝒔𝒆 𝑫𝒂 = 𝟖𝟒𝒍 ∗ 𝟔𝒎𝒈/𝒍 𝟏 ∗ 𝟎. 𝟖 = 𝟔𝟑𝟎𝒎𝒈 𝒄𝒂𝒓𝒃𝒂𝒎𝒂𝒛𝒆𝒑𝒊𝒏𝒆 𝑫𝒎 = 𝑸𝒐 = 𝑪𝒍 ∗ Ʈ ∗ 𝑪𝑬 = 𝟑. 𝟖𝟒𝒍 𝒉 ∗ 𝟐𝟒𝒉 ∗ 𝟔𝒎𝒈 𝒍 = 𝟓𝟓𝟐. 𝟗𝟔𝒎 𝒐𝒇 𝒄𝒂𝒓𝒃𝒂𝒎𝒂𝒛𝒆𝒑𝒊𝒏𝒆 5  𝑇ℎ𝑢𝑠 𝐶𝐸 = 𝑄𝑜 𝑉𝑑∗𝑘𝑒∗Ʈ = 𝑄𝑜 𝐶𝑙∗Ʈ ↔ 𝐶𝑙 = 𝑄𝑜 𝐶𝐸∗Ʈ  𝐴𝑡 𝑐𝑝 = 35𝑚𝑔/𝑙 , 𝐶𝑙 = 250𝑚𝑔 35𝑚𝑔/𝑙∗12ℎ = 0.595𝑙/ℎ  𝑡ℎ𝑢𝑠 𝑡ℎ𝑒 𝑑𝑜𝑠𝑒 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎𝑑𝑗𝑢𝑠𝑡𝑒𝑑 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡ℎ𝑖𝑠 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 III.PHENOBARBITAL Given Pharmacokinetics - Inducer of metabolism ; - (Cp):10 – 30 mg/L , (F)=1, ( S)=0.9 (sodium salt), (Vd)=0.6 à 0.7 L/kg - (Cl) adult=4 ml/h/kg, (Cl) child >1 year =2 times adult dose - (T1/2) adult= 5 days, (𝛼)=0.5, Cl (hemodialysis)=3 L/h Demand Da, Dm, time , D replacement Solution and formula 𝟓. 𝑇ℎ𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝑓𝑜𝑟 𝑀𝑟 𝑋 𝑎𝑔𝑒𝑑 39 𝑤𝑖𝑡ℎ 70𝑘𝑔  𝐿𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝐷𝑎 = 𝑉𝑑∗𝐶𝑝 𝑆∗𝐹 ↔ 𝑉𝑑 =? ? 𝐶𝑝 =? ? , 𝑆 = 0.9 𝑎𝑛𝑑 𝐹 = 1 o 𝐼𝑑𝑒𝑎𝑙 𝑉𝑑 𝑖𝑠 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 0.6 𝑡𝑜 𝑜. 7𝑙 𝑘𝑔 𝑡ℎ𝑢𝑠 𝑙𝑒𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑉𝑑 o 𝑇ℎ𝑒 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑟𝑒𝑎𝑙 𝑉𝑑 = 0.6𝑙 𝑘𝑔 ∗ 70𝑘𝑔 = 42𝑙 o 𝑇ℎ𝑒 𝐶𝑝 = 30𝑚𝑔 𝑙 − 10𝑚𝑔 𝑙 = 20𝑚𝑔/𝑙  𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝑎𝑡 𝑡ℎ𝑒 𝑚𝑎𝑥 𝑐𝑙𝑒𝑎𝑟 = 49𝑙∗20𝑚𝑔/𝑙 0.9∗1 = 1088.88𝑚𝑔 𝟔. 𝑀𝑎𝑖𝑛𝑡𝑎𝑖𝑛𝑒𝑛𝑐𝑒 𝑑𝑜𝑠𝑒 𝐷𝑚 = 𝐷𝑎(1 − 𝑒−𝑘𝑒Ʈ) ↔ 𝐷𝑎 = 933.3𝑚𝑔 𝑎𝑛𝑑 𝑘𝑒 =? , Ʈ =?  𝑄𝑜 = 𝐶𝐸 ∗ 𝐶𝑙 ∗ Ʈ ⇒ 44.5𝑚𝑔 𝑙 ∗ 0.595𝑙 𝑙 ∗ 12ℎ = 𝟑𝟏𝟕. 𝟕𝟓𝐦𝐠 𝐨𝐟 𝐞𝐭𝐡𝐨𝐬𝐮𝐱𝐢𝐦𝐢𝐝𝐞 𝑳𝒐𝒂𝒅𝒊𝒏𝒈 𝒅𝒐𝒔𝒆 𝑫𝒂 𝒂𝒕 𝐦𝐢𝐧 𝑪𝒍𝒆𝒂𝒓𝒏𝒄𝒆 = 𝟒𝟐𝒍 ∗ 𝟐𝟎𝒎𝒈 𝒍 𝟎. 𝟗 ∗ 𝟏 = 𝟗𝟑𝟑. 𝟑𝒎𝒈 6 o Ʈ𝑚𝑎𝑥 = Ʈ = 𝑙𝑛 𝐶𝑚𝑎𝑥−𝑙𝑛 𝐶𝑚𝑖𝑛 𝐾𝑒 ↔ 𝑘𝑒 = 0.693 𝑡½ ⇒ Ʈ = (𝑙𝑛𝐶𝑚𝑎𝑥−𝑙𝑛𝐶𝑚𝑖𝑛)∗𝑡½ 0.693 o Ʈ = (𝑙𝑛30−𝑙𝑛10)∗(5∗24) 0.693 = 190ℎ ↔ 7.92𝑑𝑎𝑦 o 𝑘𝑒 (𝑓𝑜𝑟 𝑎𝑑𝑢𝑙𝑡) = 0.693 𝑡½ = 0.693 (5∗24) = 0.005775/ℎ  𝐼𝑓 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝑖𝑠 933.3𝑚𝑔 𝑎𝑡 𝑉𝑑 𝑜𝑓 42𝑙  𝑀𝑎𝑖𝑛𝑡𝑎𝑖𝑛𝑒𝑛𝑐𝑒 𝑑𝑜𝑠𝑒 𝐷𝑚 𝑤𝑖𝑙𝑙 𝑏𝑒 = 𝐷𝑎(1 − 𝑒−𝑘𝑒Ʈ)  ⇒ 𝐷𝑚 = 933.3𝑚𝑔(1 − 𝑒−0.005775∗190) = 933.3𝑚𝑔(1 − 0.33378) 𝟕 𝑇ℎ𝑒 𝑡𝑖𝑚𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒 𝑡𝑜 𝑎𝑐ℎ𝑖𝑒𝑣𝑒 𝑡ℎ𝑒 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑜𝑓 20𝑚𝑔/𝑙 𝑖𝑓 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑑𝑚𝑖𝑛𝑖𝑠𝑡𝑒𝑟𝑒𝑑  𝐶𝑝 = 𝑑𝑜𝑠𝑒 𝑉𝑑 ( 1 1−𝑒−𝑘𝑒Ʈ ) 𝑒𝑘𝑒𝑡  20 = 621.78 42 ( 𝑒−0.005775𝑡 1−𝑒−0.005775∗190 ) ⇒ 20 = 14.785 ( 𝑒−0.005775𝑡 0.6662 )  ⇒ 𝑒−0.005775𝑡 = ( 20 14.785 ) ∗ 0.6662 ⇒ 𝑒−0.005775𝑡 = 0.90118  −0.005775𝑡 = 𝑙𝑛0.90118 𝟖 8 o 𝑇ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 , 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑎𝑔𝑒𝑑 25𝑦𝑒𝑎𝑟𝑠 𝑎𝑛𝑑 𝑤𝑒𝑖𝑔ℎ 70𝑘𝑔 o 𝑡𝑎𝑘𝑖𝑛𝑔 𝐷𝑜𝑠𝑒 = 60𝑚𝑔 𝑝ℎ𝑒𝑛𝑜𝑏𝑎𝑟𝑏𝑖𝑡𝑎𝑙 , 𝑡𝑤𝑖𝑐𝑒 𝑑𝑎𝑦 ↔ Ʈ = 12ℎ o 𝐶𝑝 = 20𝑚𝑔/𝑙 𝑎𝑡 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝟑𝒎𝒐𝒏𝒕𝒉 𝒍𝒂𝒕𝒆  ⇒ 𝑫𝒎 = 𝟔𝟐𝟏. 𝟕𝟖𝒎𝒈 𝒐𝒇 𝑷𝒉𝒆𝒏𝒐𝒃𝒂𝒓𝒃𝒊𝒕𝒂𝒍/ 𝟕. 𝟗𝟐𝒅𝒂𝒚  ⇒ 𝒅𝒂𝒊𝒍𝒚 𝑫𝒎 = 𝟔𝟐𝟏.𝟕𝟖𝒎𝒈 𝟕.𝟗𝟐𝒅𝒂𝒚 = 𝟕𝟖. 𝟓𝒎𝒈/𝒅𝒂𝒚 𝒕 = − 𝒍𝒏𝟎. 𝟗𝟎𝟏𝟏𝟖 𝟎. 𝟎𝟎𝟓𝟕𝟕𝟓 ⇒ 𝒕 = 𝟏𝟖𝒉 , 𝒏𝒆𝒆𝒅 𝒕𝒐 𝒓𝒆𝒂𝒄𝒉 𝒔𝒕𝒆𝒂𝒅𝒚 𝒊𝒔 𝑫𝒂 𝒊𝒔 𝒂𝒃𝒔𝒆𝒏𝒕 7  𝑻𝒉𝒆 𝒑𝒂𝒕𝒊𝒆𝒏𝒕 𝒊𝒔 𝒑𝒖𝒕 𝒖𝒏𝒅𝒆𝒓 𝑯𝒆𝒎𝒐𝒅𝒊𝒂𝒍𝒚𝒔𝒊𝒔 𝒆𝒗𝒆𝒓𝒚 𝟔𝒉, 𝟑𝒙 𝒘𝒆𝒆𝒌 𝑨𝒏𝒔𝒘𝒆𝒓 𝟖𝒂, 𝒅𝒐𝒆𝒔 𝑫 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒏𝒆𝒄𝒆𝒔𝒔𝒂𝒓𝒚? ?  1. 𝑡ℎ𝑒 𝑑𝑟𝑢𝑔 ℎ𝑎𝑙𝑓 𝑙𝑖𝑓𝑒 𝑖𝑠 5𝑑𝑎𝑦 ↔ 120ℎ > 2ℎ, 𝑐𝑜𝑛𝑑𝑖𝑜𝑛 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑  𝑑𝑟𝑢𝑔 𝑐𝑙𝑒𝑎𝑟𝑛𝑐𝑒 = 4𝑚𝑙/ℎ/𝑘𝑔 ∗ 70𝑘𝑔 = 280𝑚𝑙/ℎ = 4.666𝑚𝑙/𝑚𝑖𝑛 +𝑑𝑖𝑎𝑙𝑦𝑧𝑎𝑏𝑙𝑒 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 3𝑙 ℎ = 3000𝑚𝑙 60𝑚𝑖𝑛 = 50𝑚𝑙/𝑚𝑖𝑛  2. 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝐶𝑙 = 50𝑚𝑙/𝑚𝑖𝑛 + 4.666𝑚𝑙/𝑚𝑖𝑛 = 54.666𝑚𝑙/𝑚𝑖𝑛 < 500𝑚𝑙/𝑚𝑖𝑛 𝑖. 𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝐷 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑  𝑇ℎ𝑒 𝑑𝑖𝑎𝑙𝑖𝑠𝑎𝑏𝑙𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 (𝑉𝑑′) = 𝑉𝑑 𝛼 = 42𝑙 0.5 = 84𝑙  3. 𝑉𝑑′ < 250𝑙 𝑡ℎ𝑢𝑠 3𝑟𝑑 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑢𝑙𝑙𝑓𝑖𝑙𝑙𝑒𝑑  𝐴𝑛𝑠𝑤𝑒𝑟 8𝑏, 𝐷 𝑟𝑒𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑉𝑑 ∗ 𝐶𝑝 ( 1 − 𝑒 −𝐶𝑙 𝑡𝑜𝑡𝑎𝑙∗𝑇𝑑 𝑉𝑑 )  𝑇𝑜𝑡𝑎𝑙 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 𝐶𝑙 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 + 𝐶𝑙 ℎ𝑒𝑚𝑜𝑑𝑖𝑎𝑙𝑦𝑠𝑖𝑠 = 0.28𝑙/ℎ + 3𝑙/ℎ = 3.28𝑙/ℎ  𝑇𝑑: 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑑𝑖𝑎𝑙𝑦𝑠𝑖 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 6ℎ  𝑇ℎ𝑢𝑠 𝐶𝑙𝑒𝑎𝑟𝑛𝑐𝑒 𝑡𝑜𝑡𝑎𝑙∗𝑇𝑑 𝑉𝑑 = 3.28𝑙 ℎ ∗6ℎ 42𝑙 = 0.46857 SECTION II. ANTIARRYTHMIC DRUGS IV.LIDOCAINE Pharmacotherapeutics : - Antiarrythmic indicated in severe ventricular arrhythmias. - Local Anesthetic. Pharmacokinetics - Important first pass effect. - The myocardium is located in the central compartment (two compartment model) - (CE)=1 –5 mg/L , (F)=0.3 , S)=1 - patient condition Normal CHF Cirrhosis  𝑫 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 = 𝟒𝟐𝒍 ∗ 𝟐𝟎𝒎𝒈 𝒍 (𝟏 − 𝒆−𝟎.𝟒𝟔𝟖𝟓𝟕) ⇒ 𝑫𝒓𝒆𝒑 = 𝟖𝟒𝟎𝒎𝒈(𝟏 − 𝟎. 𝟔𝟐𝟓𝟖𝟗) ⇒ 𝑫 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 = 𝟑𝟏𝟒. 𝟐𝟒𝒎𝒈 𝒄𝒐𝒏𝒄𝒍𝒖𝒔𝒊𝒐𝒏 𝒊𝒔 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒅𝒐𝒔𝒆 𝒊𝒔 𝒏𝒆𝒆𝒅𝒆𝒅(𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅)𝒇𝒐𝒓 𝒕𝒉𝒆 𝒑𝒂𝒕𝒊𝒆𝒏𝒕 10  𝐻𝑎𝑠 𝑚𝑦𝑜𝑐𝑎𝑟𝑑𝑖𝑢𝑚 𝑖𝑛𝑓𝑎𝑟𝑐𝑡𝑖𝑜𝑛 𝑎𝑛𝑑 𝑟𝑒𝑛𝑎𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑤𝑖𝑡ℎ 𝑆𝐶𝑟𝑠𝑠 = 1.3𝑚𝑔/𝑑𝑙 𝑎𝑛𝑑 𝐶𝑟𝑒𝑎𝑡𝑖𝑛𝑖𝑛𝑒 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝐶𝑙𝐶𝑟 = 50𝑚𝑙/𝑚𝑖𝑚  𝐼𝑉 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 𝑑𝑜𝑠𝑒 𝑡𝑜 𝑎𝑐ℎ𝑖𝑒𝑣𝑒 𝐶𝑝 𝑜𝑓 8𝑚𝑔/𝑙 =? ? ?  𝐹𝑜𝑟 𝑖𝑛𝑗𝑒𝑐𝑡𝑖𝑜𝑛(𝐼𝑉) , 𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝐹 = 0.85 , 𝑡ℎ𝑢𝑠 𝐹 𝑤𝑖𝑙𝑙 𝑏𝑒 = 1 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑎𝑙𝑡 𝑓𝑎𝑐𝑡𝑜𝑟 𝑤𝑖𝑙𝑙 𝑟𝑒𝑚𝑎𝑖𝑛 𝑎𝑠 𝑢𝑠𝑢𝑎𝑙 𝑆 = 0.87 , 𝑐𝑝 = 8𝑚𝑔 𝑙 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑉𝑑 = 2𝑙 𝑘𝑔 ∗ 70𝑘𝑔 = 140𝑙 𝟏𝟐  𝐶𝐸 = 𝑄𝑜 𝑉𝑑∗𝐾𝑒∗Ʈ ⇒ 𝑄𝑜 = 𝐶𝐸 ∗ 𝐾𝑒 ∗ 𝑉𝑑 ∗ Ʈ = 𝐶𝐸 ∗ 𝐶𝑙(𝑡𝑜𝑡𝑎𝑙) ∗ Ʈ  𝐹𝑜𝑟 𝐼𝑉, 𝐹 = 1 𝑎𝑛𝑑 𝑃𝑟𝑜𝑐𝑎𝑖𝑛𝑎𝑚𝑖𝑑𝑒 𝐻𝐶𝑙 𝑤𝑖𝑡ℎ 𝑆 = 0.87 , 𝑄𝑜 = 𝑆𝐹𝐷𝑚  𝐷𝑚 = 𝐶𝐸∗𝐶𝑙 𝑇𝑜𝑡𝑎𝑙∗Ʈ 𝑆∗𝐹 𝐶𝑙 𝑡𝑜𝑡𝑎𝑙 =? ? 𝑎𝑛𝑑 Ʈ = 3ℎ 𝑜𝑟 4ℎ  𝐶𝑙 𝑡𝑜𝑡𝑎𝑙 = 𝑅𝑒𝑛𝑎𝑙 𝐶𝑙 + 𝐿𝑖𝑣𝑒𝑟 𝐶𝑙 + 𝑂𝑡ℎ𝑒𝑟 𝐶𝑙  𝑅𝑒𝑛𝑎𝑙 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 3 ∗ 𝑐𝑟𝑒𝑎𝑡𝑖𝑛𝑖𝑛𝑒 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑜𝑟  . 𝐶𝑙𝐶𝑟𝑠𝑠 = 50𝑚𝑙/𝑚𝑖𝑛  𝑃𝑎𝑡𝑖𝑒𝑛𝑡 𝑟𝑒𝑛𝑎𝑙 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 3 ∗ 50𝑚𝑙/𝑚𝑖𝑛 = 150𝑚𝑙/𝑚𝑖𝑛 = 9𝑙/ℎ  𝐻𝑒𝑝𝑎𝑡𝑖𝑐 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒(𝑎𝑐𝑒𝑡𝑦𝑙𝑎𝑡𝑖𝑜𝑛) = 0.13𝑙/ℎ /𝑘𝑔 ∗ 70𝑘𝑔 = 9.1𝑙/ℎ  𝑂𝑡ℎ𝑒𝑟 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = 0.1𝑙/ℎ/𝑘𝑔 ∗ 70𝑘𝑔 = 7𝑙/ℎ  𝑪𝒍 𝒕𝒐𝒕𝒂𝒍 = 𝟗𝒍/𝒉 + 𝟗. 𝟏𝒍/𝒉 + 𝟕𝒍/𝒉 = 𝟐𝟓𝒍/𝒉 𝑫𝒂(𝑳𝒐𝒂𝒅𝒊𝒏𝒈 𝒅𝒐𝒔𝒆 𝑰𝑽) = 𝑽𝒅∗𝑪𝒑 𝑺∗𝑭 = 𝟏𝟒𝟎𝒍∗ 𝟖𝒎𝒈 𝒍 𝟎.𝟖𝟕∗𝟏 = 𝟏𝟐𝟖𝟕. 𝟑𝒎𝒈 ↔ IV(F=1) 𝑷𝒓𝒐𝒄𝒂𝒊𝒏𝒂𝒎𝒊𝒅𝒆 𝒐𝒓𝒂𝒍 (𝑫𝒂) = 𝟏𝟒𝟎𝒍 ∗ 𝟖𝒎𝒈/𝒍 𝟎. 𝟖𝟕 ∗ 𝟎. 𝟖𝟓 = 𝟏𝟓𝟏𝟒. 𝟓𝒎𝒈 𝑫𝒎 (𝑰𝑽 𝒂𝒏𝒅 Ʈ = 𝟑) = 𝑪𝑬 ∗ 𝑪𝒍 𝑻𝒐𝒕𝒂𝒍 ∗ Ʈ 𝑺 ∗ 𝑭 = 𝟔𝒎𝒈 𝒍 ∗ 𝟐𝟓𝒍 𝒉 ∗ 𝟑𝒉 𝟎. 𝟖𝟕 ∗ 𝟏 = 𝟓𝟏𝟕. 𝟐𝟒𝒎𝒈 𝑫𝒎 (𝒐𝒓𝒂𝒍 𝒂𝒏𝒅 Ʈ = 𝟑) = 𝑪𝑬 ∗ 𝑪𝒍 𝑻𝒐𝒕𝒂𝒍 ∗ Ʈ 𝑺 ∗ 𝑭 = 𝟔𝒎𝒈 𝒍 ∗ 𝟐𝟓𝒍 𝒉 ∗ 𝟑𝒉 𝟎. 𝟖𝟕 ∗ 𝟎. 𝟖𝟓 = 𝟔𝟎𝟖. 𝟓𝟐𝒎𝒈 11 VI.QUINIDINE Pharmacotherapeutics : - Antiarrythmic drug for atrial fibrillation and other forms of arrhythmias. - Usual doses (200 mg to 300 mg per os in 3 - 4 doses/day. Pharmacokinetics - Renal Elimination; - (Cp)=1 –4 mg/L (specific method) ; (F)=0.73 (sulfate), 0.70 (gluconate) - ( S)=0.82 (sulfate), 0.62 (gluconate) - patient conditions Normal CHF Cirrhosis Vd 2.8 L/kg 1.8 L/kg 3.8 L/kg - Clearance (Cl) 4.7ml/min/kg 3.5ml/min/kg 3.5ml/min/kg - (T1/2) 7 h 7 h 7 h Problems 13.A male patient 70 kg aged 50 years had heart attack with auricular fibrillation. He is treated with 300 mg quinidine every 6 hours. Calculate Cpmax and Cpmin reached after 3 doses are given. 11. Discuss if it is relevant to take a blood sample during the first 24 hours to measure the Cpss. 𝟏𝟑 𝑸𝑼𝑰𝑵𝑰𝑫𝑰𝑵𝑬 𝑺𝑼𝑳𝑭𝑨𝑻𝑬  𝐶𝑝 𝑚𝑎𝑥 = 𝑆𝐹𝐷 𝑉𝑑 ( 1−𝑒−𝑛𝑘𝑒Ʈ 1−𝑒−𝑘𝑒Ʈ ) 𝑡ℎ𝑢𝑠 𝑛 = 3, 𝑄𝑜 = 300𝑚𝑔, Ʈ = 6ℎ, 𝑉𝑑 = ? 𝑎𝑛𝑑 𝐾𝑒 =? ? . 𝑆 = 0.83 𝑎𝑛𝑑 𝐹 = 0.73  𝑉𝑑 (𝐶𝐻𝐹) = 1.8𝑙 𝑘𝑔 ∗ 70𝑘𝑔 = 126𝑙 , 𝐶𝑙 = 3.5𝑚𝑙/𝑚𝑖𝑛/𝑘𝑔 ∗ 70𝑘𝑔 = 245𝑚𝑙/𝑚𝑖𝑛 ⇒ 𝐶𝑙 = 14.7𝑙/ℎ 𝑡ℎ𝑒𝑛 ℎ𝑎𝑙𝑓 𝑙𝑖𝑣𝑒 ( 𝑡½) = 7ℎ  𝐾𝑒 = 𝐶𝑙 𝑉𝑑 𝑎𝑛𝑑 𝑎𝑙𝑠𝑜 𝐾𝑒 = 0.693 𝑡½ 𝑫𝒎 (𝑰𝑽 𝒂𝒏𝒅 Ʈ = 𝟒) = 𝑪𝑬 ∗ 𝑪𝒍 𝑻𝒐𝒕𝒂𝒍 ∗ Ʈ 𝑺 ∗ 𝑭 = 𝟔𝒎𝒈 𝒍 ∗ 𝟐𝟓𝒍 𝒉 ∗ 𝟒𝒉 𝟎. 𝟖𝟕 ∗ 𝟏 = 𝟔𝟖𝟗. 𝟔𝟓𝒎𝒈 𝑫𝒎 (𝒐𝒓𝒂𝒍 𝒂𝒏𝒅 Ʈ = 𝟒) = 𝑪𝑬 ∗ 𝑪𝒍 𝑻𝒐𝒕𝒂𝒍 ∗ Ʈ 𝑺 ∗ 𝑭 = 𝟔𝒎𝒈 𝒍 ∗ 𝟐𝟓𝒍 𝒉 ∗ 𝟒𝒉 𝟎. 𝟖𝟕 ∗ 𝟎. 𝟖𝟕 = 𝟖𝟏𝟏. 𝟑𝟒𝒎𝒈 12  𝑂𝑢𝑟 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑑𝑟𝑢𝑔 𝑃𝑙𝑎𝑠𝑚𝑎 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑎𝑓𝑓𝑒𝑐𝑡𝑒𝑑 𝑏𝑦 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑛𝑜𝑡 ℎ𝑎𝑙𝑓𝑙𝑖𝑓𝑒 𝑡ℎ𝑢𝑠 𝑘𝑒 = 14.7𝑙/ℎ 126𝑙 = 0.1166/ℎ Ke=0.693/7h= 0.099h-1 𝐶𝑝 𝑚𝑎𝑥 = 300𝑚𝑔 ∗ 0.73 ∗ 0.82 126𝑙 ( 1 − 𝑒−3∗0.1166∗6 1 − 𝑒−0.1166∗6 ) = 300𝑚𝑔 126𝑙 ( 1 − 0.1226 1 − 0.4967 ) 𝐶𝑝 𝑚𝑎𝑥 = 181.77 126𝑙 ( 0.8774 0.5033 ) = 2.515𝑚𝑔/𝑙  𝐶𝑝𝑚𝑖𝑛 = 𝐶𝑝𝑚𝑎𝑥𝑒−𝑘𝑒Ʈ = 2.515𝑒−0.1166∗6 = 1.25𝑚𝑔/𝑙 𝑸𝑼𝑰𝑵𝑰𝑫𝑰𝑵𝑬 𝑮𝑳𝑼𝑪𝑶𝑵𝑨𝑻𝑬  𝐶𝑝 𝑚𝑎𝑥 = 𝑆𝐹𝐷 𝑉𝑑 ( 1−𝑒−𝑛𝑘𝑒Ʈ 1−𝑒−𝑘𝑒Ʈ ) 𝑡ℎ𝑢𝑠 𝑛 = 3, 𝑄𝑜 = 300𝑚𝑔, Ʈ = 6ℎ, 𝑉𝑑 = 126𝑙 𝑎𝑛𝑑 𝐾𝑒 = 0.1166/ℎ. 𝑆 = 0.62 𝑎𝑛𝑑 𝐹 = 0.70 𝐶𝑙 = 14.7𝑙/ℎ 𝐶𝑝 𝑚𝑎𝑥 = 300𝑚𝑔 ∗ 0.70 ∗ 0.62 126𝑙 ( 1 − 𝑒−3∗0.1166∗6 1 − 𝑒−0.1166∗6 ) = 300𝑚𝑔 126𝑙 ( 1 − 0.1226 1 − 0.4967 ) 𝐶𝑝 𝑚𝑎𝑥 = 130.2 126𝑙 ( 0.8774 0.5033 ) = 1.8𝑚𝑔/𝑙 𝐶𝑝𝑚𝑖𝑛 = 𝐶𝑝𝑚𝑎𝑥𝑒−𝑘𝑒Ʈ = 1.8 𝑒−0.1166∗6 = 0.894𝑚𝑔/𝑙 𝑄𝑢𝑖𝑛𝑖𝑑𝑖𝑛𝑒 𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑄𝑞𝑢𝑖𝑛𝑖𝑑𝑖𝑛𝑒 𝐺𝑙𝑢𝑐𝑜𝑛𝑎𝑡𝑒 𝐶𝑝𝑚𝑎𝑥 2.515𝑚𝑔/𝑙 1.8𝑚𝑔/𝑙 𝐶𝑝𝑚𝑖𝑛 1.25𝑚𝑔/𝑙 0.894𝑚𝑔/𝑙 𝑪𝑶𝑴𝑴𝑬𝑵𝑻: 𝑄𝑢𝑖𝑛𝑖𝑑𝑖𝑛𝑒 𝐺𝑙𝑢𝑐𝑜𝑛𝑎𝑡𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑚𝑒𝑑𝑖𝑐𝑖𝑛𝑒 𝑡𝑜 𝑡ℎ𝑖𝑠 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 𝑓𝑜𝑟 𝑤𝑒𝑙𝑙 𝑚𝑜𝑛𝑖𝑡𝑜𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡ℎ𝑒𝑟𝑎𝑝𝑒𝑢𝑡𝑖𝑐 𝐶𝑝 𝑓𝑜𝑟 𝑡ℎ𝑖𝑠 𝑑𝑜𝑠𝑒 , 𝑄. 𝑆𝑈𝐿𝐹𝐴𝑇𝐸 𝐼𝑆 𝑇𝐻𝐸 𝐵𝑒𝑠𝑡 𝑜𝑟 𝑐ℎ𝑜𝑜𝑠𝑒 𝑡ℎ𝑒 𝑔𝑙𝑢𝑐𝑜𝑛𝑎𝑡𝑒 𝑤𝑖𝑡ℎ 𝑑𝑜𝑠𝑒 𝑎𝑑𝑗𝑢𝑠𝑡𝑖𝑛𝑔 15 - (T1/2) in non renal failure =2- 3 hrs - (T1/2) in renal failure =30 - 60 hrs Problems 17. Mss BETY 30 years old and 70 kg carrying infections received a first dose of gentamicin 100 mg in infusion during 30 min. Her Scrss is 0.9 mg/dL. Calculate the concentration Cp achieved one hour after starting infusion. 18. Considering the infusion as rapid intravenous injection, re-calculate the concentration achieved one hour after injection taking CL-6.06 L/h, Vd=17.5 L and ke=0.346 h-1. 19. Calculate Cpssmax and Cpssmin if the infusion is given every 8 hours. Answers 𝟏𝟕  𝐶𝑝 (𝑡 − 𝑇) = 𝐾 𝑉𝑑∗𝑘𝑒 (1 − 𝑒−𝑘𝑒Ʈ)𝑒−𝑘𝑒(𝑡−𝑇)  𝑅𝑎𝑡𝑒 𝑜𝑓 𝑖𝑛𝑓𝑢𝑠𝑖𝑛 (𝐾) = 𝐷𝑂𝑆𝐸 𝐼𝑛𝑓𝑢𝑠𝑖𝑜𝑛 𝑡𝑖𝑚𝑒 = 100𝑚𝑔 0.5ℎ = 200𝑚𝑔/ℎ  𝑉𝑑 = 7.3𝑙 𝑘𝑔 ∗ 70𝑘𝑔 = 511𝑙  𝑡 − 𝑇 = 1ℎ − 0.5ℎ = 0.5ℎ  𝑘𝑒 = 0.693 𝑡½ = 0.693 2.5ℎ = 0.2772/ℎ or  𝐾𝑒 = 𝐶𝑙 𝑉𝑑 ⇒ 𝑐𝑙 = ( 𝑤𝑒𝑖𝑔ℎ(140−𝑎𝑔𝑒) 72𝑆𝑟𝐶𝑟𝑠𝑠 ) ∗ 0.85 = ( 70𝑘𝑔(140−30) 72∗0.9 ) ∗ 0.85  ⇒ 𝐶𝑙 = 101𝑚𝑙/ 𝑚𝑖𝑛 = 6.06𝑙/ℎ  𝐾𝑒 = 6.06𝑙/ℎ 511𝑙 = 0.011859/ℎ , 𝑓𝑜𝑟 𝑜𝑢𝑟 𝑝𝑎𝑡𝑖𝑒𝑛𝑡 , 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑛𝑎𝑙 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 ↔ 𝑑𝑟𝑢𝑔 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑖𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑟𝑒𝑎𝑡𝑖𝑛𝑖𝑛𝑒 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒. 16  𝐶𝑝 (𝑡 − 𝑇) = 200𝑚𝑔/ℎ 6.06𝑙/ℎ (1 − 𝑒−0.011859∗0.5)𝑒−0.011859∗0.5  𝐶𝑝 (𝑡 − 𝑇) = 200𝑚𝑔/ℎ 6.06𝑙/ℎ (0.00591) ∗ 0.00591 = 0.1949𝑚𝑔/𝑙 𝟏𝟖  𝐶𝑝 (𝑚𝑎𝑥) = 𝑄𝑜 𝑉𝑑 ( 1−𝑒−𝑛𝑘𝑒Ʈ 1−𝑒−𝑘𝑒Ʈ ) = 100𝑚𝑔 17.5𝑙 ( 1−𝑒−2∗0.346∗0.5 1−𝑒−0.346∗0.5 ) = 10.55𝑚𝑔/𝑙  𝐶𝑝 (𝑡) = 𝐶𝑝(𝑚𝑎𝑥)𝑒−𝑘𝑒𝑡  𝐶𝑝 (𝑡) = 100𝑚𝑔 17.5𝑙 ( 1−𝑒−2∗0.346∗0.5 1−𝑒−0.346∗0.5 ) 𝑒−0.346∗1  𝐶𝑝 (𝑡) = 100𝑚𝑔 17.5𝑙 ( 0.2924 0.1588 ) ∗ 0.7075  ⇒ 𝐶𝑝 (𝑡) = 10.52 ∗ 0.7075 = 7.44𝑚𝑔/𝑙 𝟏𝟗  𝐶𝑝 𝑚𝑎𝑥 𝑠𝑠 = 𝑄𝑜 𝑉𝑑 ( 1 1−𝑒−𝑘𝑒Ʈ )  𝐶𝑝 (𝑡) = 100𝑚𝑔 17.5𝑙 ( 1 1−𝑒−0.346∗8 )= 100𝑚𝑔 17.5𝑙 ( 1 1−𝑒−0.346∗8 )  𝐶𝑝 𝑚𝑎𝑥 𝑠𝑠 = 100𝑚𝑔 17.5𝑙 ( 1 0.9372 ) = 6.09𝑚𝑔/𝑙  𝐶𝑝 𝑚𝑖𝑛 𝑠𝑠 = 𝑄𝑜 𝑉𝑑 ( 1 1−𝑒−𝑘𝑒Ʈ ) 𝑒𝑘𝑒Ʈ = 𝑐𝑝𝑚𝑎𝑥𝑠𝑠𝑒𝑘𝑒Ʈ  𝐶𝑝 𝑚𝑖𝑛 𝑠𝑠 = 𝑐𝑝𝑚𝑎𝑥𝑠𝑠𝑒𝑘𝑒Ʈ = 6.097𝑚𝑔/𝑙𝑒−0.346∗8  𝐶𝑝𝑚𝑖𝑛𝑠𝑠 = 0.3828𝑚𝑔/𝑚𝑔