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Pharmacy 732 Winter 2003. Instructors William HaytonPharmaceutics James CoylePPAD Cari BrackettPPAD Kristin LugoPPAD Juhyun KimPharmaceutics.

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Presentation on theme: "Pharmacy 732 Winter 2003. Instructors William HaytonPharmaceutics James CoylePPAD Cari BrackettPPAD Kristin LugoPPAD Juhyun KimPharmaceutics."— Presentation transcript:

1 Pharmacy 732 Winter 2003

2 Instructors William HaytonPharmaceutics James CoylePPAD Cari BrackettPPAD Kristin LugoPPAD Juhyun KimPharmaceutics

3 Contact Information E-mail name.n@osu.edu Phone 292- Office Parks Hall HaytonWebCT-1288232 Coyle.1@osu.edu-7103A208 Brackett.2@osu.edu-5718A216 Lugo KNLgator@aol.com-5335425 Kim.1452 @osu.edu 247-6469243

4 Exams 3 Exams, 100 points each –Exam 1: Tues. Jan. 28, –Exam 2: Tues. Feb. 18, –Exam 3: Tue. March 18, 9:30 – 11:18 Quizzes, – seven total, 15 points each, drop the lowest score - 90 points Workshop participation - 10 points Total:---------------------------------- 400 points

5 Grading Total Points  88.0A78.0B- 86.0A-75.0C+ 83.0B+71.0C 80.0B67.0C-

6 Remediation and Make-Up Exams 1 & 2 will be offered again during the 10 th week class. Remediation –Open to all students; manditory for students scoring <67. –Scoring: 0.2 x first score + 0.8 x second score (0.2)(65) + (0.8)(85) = 81 Make-Up –If exam was missed for a qualified reason, full score will count; if not, remediation scoring will be used with the first score being zero.

7 Post-Course Remediation Course average between 60.0 and 66.9 –Eligible to take a comprehensive exam over Pharmacy 732 during workshop of the first week of Pharmacy 733.  67.0: C- for the course <67.0: E for the course Course average <60 –No post-course remediation allowed and E grade stands.

8 Workshops Monday Afternoon, 1:30 – 3:18 –Rm. 550 –Rm. 257 Wednesday Afternoon 1:30 – 3:18 –Rm. 550 Parks Hall –Rm. 257 Parks Hall

9 Course Materials Clinical Pharmacokinetics, Rowland and Tozer, third edition Calculator Semilog graph paper Course Pack

10 Classroom Conduct Quiet is appreciated Cell phones turned off Pagers in quiet mode Refrain from exiting and re-entering

11 Academic Misconduct: Exams No caps No use of memory in calculators No PDAs, laptops, etc. No exchanging information with others No cell phones or pagers in “on” mode No crib notes, cheat sheets, etc.

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13 WebCT Course Materials –Slides from class –Narrated slides from class –Reference materials; e.g., PK parameters –Exam Keys old and current –Quiz Keys old and current Bulletin board Email Quiz and exam scores; i.e., grade book Self tests Student home page - optional

14 WebCT Logon (1) Open a web browser and go to: http://class.osu.eduhttp://class.osu.edu (2) Log on: Enter your “username” and “password.” Username = lastname.123 (all lowercase, No CAPITAL LETTERS!) Example: If your E-mail address is smith.303@osu.edu = Username is “smith.303”smith.303@osu.edu Password = same one you use to check your OSU e-mail, to register online for classes, and to enable Internet access in OIT student computer labs.

15 Dosing Regimen Design Infusion regimen

16 Why is a dosing regimen necessary? To replace the drug that the body eliminates. How can drug be replaced? Continuously or intermittently.

17 Continuous Input Learning Objectives –input rate to achieve a desired plasma concentration. –kinetics of accumulation; i.e., how long to steady state. –loading dose. –determination of CL, V, K E and t 1/2.

18 Examples of continuous input i.v. drip i.v. infusion transdermal patch sustained release oral dosage forms Ocusert Norplant

19 Kinetics of continuous input A = amount of drug in body C p = plasma concentration K o = input rate (amt/time) K E = elimination rate constant, (CL/V) v CL KoKo dA/dt = rate in - rate out dA/dt = K o - K E A A = V x C p and K E V = CL dA/dt = K o - CLC p dA/dt = 0 at a plateau C p Rate In = Rate Out K o = CL C p,ss C p,ss = K o /CL

20 Example: Diazepam C p,t profile in young and old: Ln C p Time young old What’s different? V ss CLt 1/2 f up [L/kg][L/h/kg][h][%] 0.880.017444.52.49 1.390.015671.52.76 Herman and Wilkinson. Br. J. Clin. Pharmacol. 42:147,1996. #2919 DR adjustment?

21 Principles When infused at the same rate (one compartment model assumed): –all drugs with the same half life will have the same steady-state amount of drug in the body. –all drugs with the same clearance will have the same steady-state plasma concentration.

22 Accumulation Kinetics

23 Time to steady state 3.3 t 1/2 is time to 90% of C p,ss t = nt 1/2 where n = no. of half lives that have passed

24 Example This drug has a V = 45 L and a CL = 12 L/h. What infusion rate is needed to achieve a C p,ss of 25 mg/L? K o = CL x C p,ss = 12 L/h x 25 mg/L = 300 mg/h How long will it take to get to 90% of steady state? t 1/2 = ln 2 V/CL = (0.693)(45)/(12) = 2.6 h t 90% = 3.3 t 1/2 = (3.3)(2.6) = 8.6 h How much drug is in the body at steady state? A ss = C p,ss V = 45 L x 25 mg/L = 1,125 mg

25 Summary C p,ss A ss Time to SS  Ko Ko   CL   V 

26 Diagram A ss C ss KoKo CL t 90% V V P, V E, V R R E/I, f ur, f up Q H, f up, CL int,u GFR, etc.

27 Review With a constant rate of input, K o Rate In = _______ Rate Out Rate In = C ss x ___ CL Rate In = A ss x ___ KEKE All drugs with same CL will have same ____ All drugs with same t 1/2 will have same ____ C ss A ss

28 Post-Infusion C p Profile

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30 Changing to a new C p,ss  CL by 25%  K o by 25%  V by 25% CpCp Time

31 Bolus and Infusion A “loading dose” may be used to start at steady state immediately. Loading Dose = A ss = K o /K E = C ss V

32 Rowland and Tozer, Figure 6-5. p. 74

33 Time to Steady State 3.3 t 1/2 is time to 90% of C p,ss. When C p,0 is 0, this is within ±10% of C p,ss. When C p,0 is  0, the time to ±10% of C p,ss differs from the t 90%. What is the appropriate endpoint for calculation of the time to steady state?

34 Calculation of time to ±10% C p,ss Plasma concentration at any time after bolus + infusion: Given C p,0 and C ss, the time to reach any C p can be calculated from:

35 Example C p,0 = 500 mg/L and C ss = 100 mg/L, how long does it take to reach 110 mg/L? (i.e., 110% of C ss ) (1/2) n = (100 – 110)/(100 – 500) = -10/-400 = 0.025 (n)[ln (0.5)] = ln (0.025) n = ln (0.025) / ln (0.5) = 5.32 half lives What would be the C p after 3.3 half lives? 140 mg/L

36 Assessment of PK parameters 1. CL = K o / C p,ss Time Log (C p,ss - C p ) -2.3 slope = K E 2. Get K E from the slope of a semilog plot of (C p,ss – C p ) vs. t. 3. V = CL / K E


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