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Dosing Regimen Design Infusion regimen.

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Presentation on theme: "Dosing Regimen Design Infusion regimen."— Presentation transcript:

1 Dosing Regimen Design Infusion regimen

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

3 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, KE and t1/2.

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

5 Kinetics of continuous input
A = amount of drug in body Cp = plasma concentration Ko = input rate (amt/time) KE = elimination rate constant, (CL/V) dA/dt = rate in - rate out dA/dt = Ko - KEA A = V x Cp and KEV = CL dA/dt = Ko - CL•Cp v CL Ko dA/dt = 0 at a plateau Cp Rate In = Rate Out Ko = CL •Cp,ss Cp,ss = Ko/CL

6 Example: Diazepam Cp,t profile in young and old: DR adjustment? Ln Cp
Time young old What’s different? Vss CL t1/2 fup [L/kg] [L/h/kg] [h] [%] 0.88 0.0174 44.5 2.49 1.39 0.0156 71.5 2.76 Herman and Wilkinson. Br. J. Clin. Pharmacol. 42:147, #2919

7 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.

8 Accumulation Kinetics

9 Time to steady state 3.3 t1/2 is time to 90% of Cp,ss
t = nt1/2 where n = no. of half lives that have passed

10 Example This drug has a V = 45 L and a CL = 12 L/h. What infusion rate is needed to achieve a Cp,ss of 25 mg/L? Ko = CL x Cp,ss = 12 L/h x 25 mg/L = 300 mg/h How long will it take to get to 90% of steady state? t1/2 = ln 2 V/CL = (0.693)(45)/(12) = 2.6 h t90% = 3.3 t1/2 = (3.3)(2.6) = 8.6 h How much drug is in the body at steady state? Ass = Cp,ssV = 45 L x 25 mg/L = 1,125 mg

11 Summary Cp,ss Ass Time to SS  Ko  CL  V

12 Diagram VP, VE, VR RE/I, fur, fup Ass Css Ko CL t90% V
QH, fup, CLint,u GFR, etc.

13 Review With a constant rate of input, Ko Rate Out Rate In = _______ CL
Rate In = Css x ___ Rate In = Ass x ___ KE All drugs with same CL will have same ____ Css All drugs with same t1/2 will have same ____ Ass

14 Post-Infusion Cp Profile

15 Post-Infusion Cp Profile

16 Changing to a new Cp,ss CL by 25% Ko by 25% V by 25% Cp Time

17 Bolus and Infusion A “loading dose” may be used to start at steady state immediately. Loading Dose = Ass = Ko/KE = CssV

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

19 Time to Steady State 3.3 t1/2 is time to 90% of Cp,ss.
When Cp,0 is 0, this is within ±10% of Cp,ss. When Cp,0 is  0, the time to ±10% of Cp,ss differs from the t90%. What is the appropriate endpoint for calculation of the time to steady state?

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

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

22 Assessment of PK parameters
1. CL = Ko / Cp,ss Time Log (Cp,ss - Cp) -2.3 slope = KE 2. Get KE from the slope of a semilog plot of (Cp,ss – Cp) vs. t. 3. V = CL / KE


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