Dosing Regimen Design Multiple Dosing: Intermittent or multiple dose regimen.
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Transcript of Dosing Regimen Design Multiple Dosing: Intermittent or multiple dose regimen.
Dosing Regimen Design
Multiple Dosing:Intermittent or multiple dose
regimen
100 mg q. t1/2 via i.v. bolus
Time [t1/2]: 1 2 3 4 5 6
Am
ou
nt
in B
od
y
[mg
]
50
100
150
200
Principle: 1 dose lost per At steady state, Rate In = Rate Out
F•Dose = Ass,max - Ass,min
Ass,min = Ass,maxe-KE
F•Dose = Ass,max (1 - e-KE)
Ass,max = F•Dose / (1 - e-KE)
Ass,min = Ass,max - F•Dose
AN,max & AN,min
E
E
ENK
ssK
NK
N eADoseFee
A
1
11
max,max,
N = 1 2 3 4 5 6 7
EKNN eAA max,min,
Not AN,min = AN,max - F•Dose Why?
Css,max & Css,min
EKss eVDoseF
C
1max,
V
DoseFCeCC ss
Kssss
E
max,max,min,
Average amount of drug in the body at steady state
At steady state, Rate In = Rate Out
FDose/ = KE
Ass,av
2/1
,
44.1 tDoseF
KDoseF
AE
avss
1/KE = t1/2/ln 2 = 1.44 t1/2
Average steady-state plasma concentration
avsCCLDoseF
,
CLDoseF
C avss,
AUC
Equal Areas
AUCAUC 0
0
,
AUCC avss
Css,av
Concept applies to all routes of administration and it is independent of absorption rate.
Dosing rate from AUC0-
Given AUC after a single dose, D, the maintenance dose, DM , is:
DAUC
CD avssM
0
,
Where D produced the AUC0-, Css,av is the desired steady-state plasma concentration,
and is the desired dosing interval.
Example: 50 mg p.o. dose
Cp
mg/L
1 2 4 8Time [h]
2 3 2.4 1.8 1012
201
0
AUC
5.2122
322
1
AUC
4.5242
4.234
2
AUC
4.8482
8.14.28
4
AUC
63.08.18
Elast KCAUC
AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L
Example, continued
Calculate a dosing regimen for this drug that would provide an average steady-state plasma concentration of 15 mg/L.
mgLhmghLmg
DAUC
CD avssM 50
/3.236/15
0
,
DM = 193 mg
Dosing regimen: 200 mg q. 6 h.
Example, continued - 2
There is a problem with this approach ??
Peak and trough concentrations are unknown.
Css,max
Css,min
Another example - digitoxin
t1/2 = 6 days; usual DR is 0.1 mg/day
Assuming rapid and complete absorption of digitoxin,What would be the average steady-state body level?
Maximum and minimum plateau values?
Is there accumulation of digitoxin?
Ass,av = 1.44 F Dose t1/2/= (1.44)(1)(0.1mg/d)(6d)/(1d)
= 0.864 mg
Ass,max = 0.1/(1-e-(0.116)(1))
= 0.909 mg
Ass,min = 0.909 – 0.1
= 0.809 mgHow long to reach steady state?
Rate of Accumulation, AI, FI
The rate of accumulation depends on the half-life of the drug: 3.3 x t1/2 gives 90% of the steady-state level.
Accumulation Index (AI): Ass,max/F DM = (1 – e-KE)-1
Fluctuation Index (FI): Ass,max/Ass,min = e+KE
AI
FI
KE
KE
When = t1/2
AI = FI = 2
Absorption Rate influence on Rate of Accumulation
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
0 20 40 60 80 100 120
Time
ka=0.5.pwo
ka=0.01.pwo
v
CL
ka
KE = 0.1
Ea
NKa
Ea
NkE
ssss
N
Kk
ek
Kk
eKf
A
A Ea
1max,
max,
Ea
NKa
Ea
NkE
ssss
N
Kk
ek
Kk
eKf
A
A Ea
1max,
max,
When ka >> KE, control is by drug t1/2:
When ka << KE, control is by absorption t1/2:
ENKss ef 1
aNkss ef 1
Loading Dose (LD)
Whether a LD is needed depends upon:
•Accumulation Index
•Therapeutic Index
•Drug t1/2
•Patient Need
= Ass,max F
Dosing Regimen Design
OBJECTIVE: Maintain Cp within the therapeutic window.
Cp
Time
Dosing Regimen DesignAPPROACH: Calculate max and DM,max.
Cp
Time
Cu
Clmax
l
u
E
l
u
Kul
C
Ct
K
CC
eCC E
ln44.1ln
2/1max
max
DM,max and Dosing Rate
From the principle that one dose is lost over a dosing interval at steady state:
DM,max = (V/F)(Cu - Cl)
The Dosing Rate (DR) is DM,max max
lu
luE
Elu
luM
CC
CCFVK
KCC
CCFVDDR
ln/lnmax
max,
Cp
Time
KEV = CL
(Cu - Cl)/ln(Cu/Cl) = Css,av = logarithmic average of
Cu and Cl.
The log average is the concentration at the midpoint of the dosing interval; it’s less than the arithmetic average.DR = (CL/F)Css,av
lu
luE
Elu
luM
CC
CCFVK
KCC
CCFVDDR
ln/lnmax
max,
Average Concentration Approach
1. Choose the average to maintain:Css,av = (Cu - Cl)/ln (Cu/Cl)
2. Choose :
max ;usually 4, 6, 8, 12, 24 h
3. Calculate DR:DR = (CL/F)Css,av
4. Calculate DM:
DM = DR•
Example
V = 35L KE = 0.143 h-1 t1/2 = 4.85 hCu = 10 mg/L
CL = 5L/h
F = 0.80 Cl = 3 mg/L
Css,av = (10 – 3)/ln (10/3) = 5.8 mg/Lmax = (1.44)(4.85)[ln (10/3)] = 8.41 h Choose < max: 8
hDR = (5 L/h)(5.8 mg/L)/(0.8) = 36.25 mg/hDM = (36.25 mg/h)(8 h) = 290 mg 300 mgDosing Regimen: 300 mg q 8 h
Peak Concentration Approach
1. Choose the peak concentration to maintain.
2. Choose :
max ;usually 4, 6, 8, 12, 24 h
3. Calculate DM:
DM = (V•Cpeak/F)(1 - e-KE)
from: EKss eVDoseF
C
1max,
Example
V = 35L KE = 0.143 h-1 t1/2 = 4.85 hCu = 10 mg/L
CL = 5L/h
F = 0.80 Cl = 3 mg/L
Cpeak = 8 mg/L
max = (1.44)(4.85)[ln (10/3)] = 8.41 h Choose < max: 6
h set to 6 h so that Css,min > Cl
DM = [(35 L)(8 mg/L)/0.8](1 – e-(0.143)(6)) = 202 mg
Dosing Regimen: 200 mg q 6 h
Check EKss eVDoseF
C
1max,
VDoseF
CC ssss
max,min,
Css,max = [(0.8)(200)/35]/(1 – e-(0.143)(6)) = 7.93 mg/l
Css,min = 7.93 - (0.8)(200)/35 = 3.4 mg/L
Rationale for controlled release dosage forms
Compliance vs. fluctuation: when the dosing interval is less than 8 h, compliance drops. For short half-life drugs, either must be small
(2, 3, 4, 6 h), or the fluctuation must be quite large, when conventional dosage forms are used.
Use of controlled release permits long while maintaining low fluctuation.
Not generally of value for drugs with long half lives (> 12 h). Due to extra expense, they should not be recommended.
Assessment of PK parameters
CL:
CL/F = (DM/)/Css,av and Css,av = AUCss,/
Relative F:
B
M
A
M
Aavss
Bavss
A
B
D
D
C
C
F
F
,,
,,
CLR:
CLR = (Ae,ss/ x Css,av) where Ae,ss is the amount of drug excreted in the urine over one .