Dosing Regimen Design

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Transcript Dosing Regimen Design

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
dA/dt = rate in - rate out
A = amount of drug in body
Cp = plasma concentration
dA/dt = Ko - KEA
Ko = input rate (amt/time)
A = V x Cp and KEV = CL
KE = elimination rate
constant, (CL/V)
dA/dt = Ko - CL•Cp
Ko
dA/dt = 0 at a plateau Cp
Rate In = Rate Out
Ko
= CL •Cp,ss
Cp,ss
= Ko/CL
v
CL
Example: Diazepam
Cp,t profile in young and old:
Ln Cp
young
What’s different?
old
Time
DR adjustment?
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
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
Ko
KEt
KEt
1  e   Ass 1  e 
Ainf 
KE
C p ,inf
Ko
KEt
KEt
1  e   C p,ss 1  e 

CL
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
e
KEt
e
  ln 2 t1/ 2 nt1/ 2
C p ,inf  C p , ss 1  e
1

 
2
KEt
n
  C 1  1 2 
n
p , ss
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
VP, VE, VR
RE/I, fur, fup
V
Ass
Ko
t90%
CL
Css
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 ____
Css
All drugs with same t1/2 will have same ____
Ass
Post-Infusion Cp Profile
Infusion
0.035
0.030
0.025
0.020
0.015
0.010
0.005
0.000
0
20
40
60
Time
80
100
120
Post-Infusion Cp Profile
Infusion
0.10000
0.01000
0.00100
0.00010
0.00001
0
20
40
60
Time
80
100
120
Changing to a new Cp,ss
CL by
25%
Cp
V by
25%
Ko by
25%
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
Plasma concentration at any time after bolus + infusion:
C p  C p ,0 e
KEt
 Css 1  e
KEt

Given Cp,0 and Css, the time to reach any Cp can be
calculated from:
C
C
ss
ss
 Cp 
 C p ,0 
e
KEt
1

 
2
n
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)
C
C
ss
ss
 Cp 
 C p ,0 
e
KEt
1

 
2
n
(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
Log (Cp,ss - Cp)
2. Get KE from the
slope of a semilog
plot of (Cp,ss – Cp)
vs. t.
-2.3 slope = KE
Time
3. V = CL / KE
C p ,inf  C p ,ss 1  e  K E t 