Transcript 06_Multiple dosing Extravascular Administration

Multiple dosing:
Extravascular
Administration
Dr Mohammad Issa Saleh
Multiple dosing: Extravascular
Administration
Multiple dosing: Extravascular
Administration

Concentration after n doses:
KaFXo  (1  e  nK )e  Kt (1  e  nKa )e  Kat 
Cn 



 K
 Ka
Vd ( Ka  K )  1  e
1 e


Concentration at steady state
  Kt
 Kat 
e
KaFXo
e
Css 


 K
 Ka 
Vd ( Ka  K ) 1  e
1 e

n: number of doses, T: dosing interval
Multiple dosing: Extravascular
Administration

Time to get to maximum conc after n doses:
 Ka(1  e  nKa ) 1  e  K 
2.303
(tp)n 
log 
*
 nK
 Ka 
Ka  K
) 1 e
 K (1  e


Time to get to maximum conc at steady state:
 K
 Ka(1  e ) 
2.303
(tp) ss 
log 
 Ka 
Ka  K
)
 K (1  e
Accumulation ratio

Accumulation can be determined by
comparing the minimum plasma
concentrations of drug at steady state and
following the first dose:
min
CSS
R  min
C N 1
 e  K

FXo
e  Ka


 K
Vd ( Ka  K )  (1  e
) (1  e  Ka ) 

FXo
e  K  e  Ka
Vd ( Ka  K )
1
R
(1  e  K )(1  e  Ka )


Accumulation ratio

If the dose was administered in the
elimination phase (no significant
absorption occurs, accumulation ratio can
be simplified into:
1
R
 K
1 e
Time to reach steady state
conc.

Fraction of steady state conc obtained after the
nth dose is given by:
K e  nKa
Ka e  nK
fss  1 

Ka  K
Ka  K

At very large values of Ka (i.e. Ka/K≥10), time to
get to steady state can be represented as:
n
 1.44 t 0.5 ln( 1  fss)
The AUC during a dosing interval at steady
state is equal to the total AUC following a
single dose (For linear PK)
Average steady state conc

The ‘‘average’’ plasma concentration of a
drug at steady state for extravascularly
administered dose can be calculated from:

C
average
SS

C
0
SS

.dt
FXo 1.44t0.5 FXo


*
KVd
Vd

Example 1

QJ is a 67-year-old, 80-kg male being treated for chronic
obstructive pulmonary disease. Sustained-release oral
theophylline is being added to his drug regimen.
Assuming F = 1.0, V = 40 L, and t1/2 = 5 hours, compute
an oral theophylline dose to be administered every 12
hours that would achieve a Css = 8 mg/L using the
average steady-state concentration equation
Xo  C
ss
Vd
8 mg/L * 40 L *12 hr

 533.3 mg
1.44t0.5 F
1.44 * 5 hr *1
av
Example 2

A patient is taking 1000 mg sulfamethoxazole tablet
every 12 hours for the treatment of urinary tract infection.
Sulfamethoxazole is rapidly and completely absorbed
after oral administration. What is the amount of
sulfamethoxazole eliminated during one dosing interval
at steady state ?
Solution:
amount eliminated during one dosing interval at steady
state = amount administered during one dosing interval
at steady state = maintenance dose= 1000 mg
Example 3

During repeated administration of 60 mg
indomethacin tablet every 12 hours (F=1) for the
treatment of rheumatoid arthritis the AUC during
one dosing interval was 8 mg-hr/L. What is the
average indomethacin steady state plasma
concentration during this dosing regimen (60 mg
every 12 hours) ?
C 
ss
av
AUC0 (at steady state)

8 mg.hr/L

 0.67 mg/L
12 hr
Example 4

JB is a 78-year-old, 100-kg male being treated
with digoxin for heart failure. While receiving
digoxin tablets 125 μg daily, an average steadystate digoxin concentration equal to 0.6 μg/L is
obtained. Assuming F = 0.7,
 Compute
digoxin clearance for the patient using the
average steady-state concentration equation.
 Compute a new digoxin tablet dose for the patient
that will achieve Css = 1.2 μg/L..
Example 4
1.Calculate clearance (Cl):
At steady state: input rate = output rate
 dosing rate = elimination rate
FX 0
average
 C SS
* Cl

FX 0
0.7 *125 g
Cl 

 6.1 L/hr
average
 * CSS
24 hr * 0.6 g/L
Example 4
2- Calculate a new dose patient that will
achieve Css = 1.2 μg/L:
0.6 µg/L  125 µg
1.2 µg /L  ?? µg
1.2
Dose 
*125  250 g
0.6