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Dosing Regimen Design
Infusion regimen
Why is a dosing regimen necessary?
• To replace the drug that the body eliminates.
How can drug be replaced?
• Continuously or intermittently.
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.
Examples of continuous input
• i.v. drip
• i.v. infusion
• transdermal patch
• sustained release oral dosage forms
• Ocusert
• Norplant
Kinetics of continuous input
A = amount of drug in body
Cp = plasma concentration
Ko = input rate (amt/time)
KE = elimination rate constant, (CL/V)
v
CL
Ko
dA/dt = rate in - rate out
dA/dt = Ko - KEA
A = V x Cp and KEV = CL
dA/dt = Ko - CL•Cp
dA/dt = 0 at a plateau Cp
Rate In = Rate OutKo = CL •Cp,ss
Cp,ss = Ko/CL
Example: Diazepam Cp,t profile in young and old:
Ln C
p
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. 42:147,1996. #2919
DR adjustment?
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.
Accumulation Kinetics
tKss
tK
E
o EE eAeK
KA 11inf
tKssp
tKop
EE eCeCL
KC 11 ,inf,
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
n
ntttK ee E
21
2/12/12ln
nssptK
sspp CeCC E 2111 ,,inf,
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
Summary
Cp,ss Ass Time to SS
Ko
CL
V
Diagram
Ass
Css
Ko CL
t90%
V
VP, VE, VR
RE/I, fur, fup
QH, fup, CLint,u
GFR, etc.
Review
With a constant rate of input, Ko
Rate In = _______ Rate Out
Rate In = Css x ___ CL
Rate In = Ass x ___ KE
All drugs with same CL will have same ____
All drugs with same t1/2 will have same ____
Css
Ass
Post-Infusion Cp Profile
0.000
0.005
0.010
0.015
0.020
0.025
0.030
0.035
0 20 40 60 80 100 120
Time
Infusion
Post-Infusion Cp Profile
0.00001
0.00010
0.00100
0.01000
0.10000
0 20 40 60 80 100 120
Time
Infusion
Changing to a new Cp,ss
CL by 25%
Ko by 25%
V by 25%
Cp
Time
Bolus and Infusion
A “loading dose” may be used to start at steady state immediately.
Loading Dose = Ass = Ko/KE = CssV
Rowland and Tozer, Figure 6-5. p. 74
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?
Calculation of time to ±10% Cp,ss
tKss
tKpp
EE eCeCC 10,
Plasma concentration at any time after bolus + infusion:
Given Cp,0 and Css, the time to reach any Cp can be calculated from:
ntK
pss
pss EeCC
CC
2
1
0,
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)
ntK
pss
pss EeCC
CC
2
1
0,
(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 Cp after 3.3 half lives? 140 mg/L
Assessment of PK parameters
1. CL = Ko / Cp,ss
TimeL
og (
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 tKsspp
EeCC 1,inf,