Transcript One open compartement model

One open compartement
model
Problems
Basic Equations
0  kt
p
Cp  C e
Cp 
D0
VD e
 CL / VD t
Basic Equations
C p 2  C p1e
t1/ 2
 k t 2 t1 
0.693t 2  t1 

ln C p1  ln C p 2
Intravenous bolus administration
• A single iv dose of an antibiotic was given
to a 50 kg woman at a dose level of 20
mg/kg. Urine and blood samples were
removed periodically and assayed for
parent drug. The following data were
obtained:
Time (hr)
Cp (mg/mL)
Du (mg)
0.25
4.2
160
0.50
3.5
140
1.0
2.5
200
2.0
1.25
250
4.0
0.31
188
6.0
0.08
46
Penentuan parameter farmakokinetik
berdasarkan data Cp
• Data diolah menurut grafik log Cp vs t
• Diperoleh persamaan garis:
y = 0.6956-0.2994 x (r=-1.0000)
• Jika : log Cp= log Co – k/2.303 t
maka: - k/2.303 = -0.2994 dan k =
0.6895/jam, t1/2 = 0.693/k = 1.0051 jam;
Co= antilog 0.6956 = 4.9614 mg/mL
Penentuan parameter farmakokinetik
berdasarkan data Du
Time(hr)
Du(mg)
Du/∆t
mg/hr
t* (hr)
0.25
160
160/0.25
640
0.125
0.5
140
140/0.25
560
0.375
1.0
200
200/0.5
400
0.750
2.0
250
250/1
250
1.50
4.0
188
188/2
94
3.0
6.0
46
46/2
23
5.0
• t* =nilai tengah waktu pengambilan sampel
• ∆t=interval waktu pengambilan sampel
• Pers garis:
dDu
 kt
log

 log ke DBo
dt
2.303
• Grafik antara log Du/∆t versus t* memberikan garis lurus
dengan slope = -k/2.303
• Persamaan garis log Du/∆t vs t:
y = 2.8437 – 0.2952 x (r=-0.9997)
-k/2.303 = -0.2952, jadi k = 0.6798/jam dan t1/2 = 1.0194
jam.
Du/∆t pada t* = 0 adalah = antilog 2.8437
Du/ ∆t = 697.7502 mg/jam=log keDoB
Dengan sigma-minus method (metoda berd
jumlah obat yang akan dieksresikan)
• Data diolah menurut pers:
 kt

log D  Du 
 log Du
2.303


u

Homeworks
Shargel & Yu (5th ed), Learning Questions
page 69 – 70 no: 1, 2, 3, 4 and 6
Exercise
A single iv bolus injection containing 500 mg
cefamandole nafate is given to an adult
female patientb(63 years, 55 kg) for a
septicemic infection. The apparent Vd is 0.1
L/kg and t1/2 is 0.75 hour. Calculate:
a. C0
b. The amount of drug in the body 4 hrs
after the dose given
c. The time for the drug to decline to 0.5
mcg/mL, the MIC for streptococci
Intra-venous infusion
1. An antibiotic has a volume of distribution
of 10 L and a k of 0.2 hr-1. A steady state
plasma conc of 10 mg/mL is desired.The
infusion rate needed to maintain this
conc can be determined as follows.
R = Css . Vd . K
R = (10 mg/mL)(10.000mL)(0.2 hr-1)
R = 20 mg/hr
What is the rate of infusion (mL/min) if the
drug concentration = 25 mg/mL
2. A patient receiving IV theophylline (or
aminophylline) which has a half-life of 6 hours
(K=0.116/hr) and Vd of 30 L (Cl = 3.48 L/hr). The
steady-state plasma conc for a continuous IV
infusion of 50 mg/hr :
CSS
K0

VK
50mg/hr
CSS 
 14.4mg/L
1
30L  0.116hr
If we wish to increase the Css to 18
mg/L, we would use the same
equation to determine K0:
K0
18mg/L 
1
30L  0.116hr
1
K 0  18mg/L  30L  0.116hr  62.6mg/hr


or with equation:
K 0new  
CSSdesired
CSSmeasured 
 K 0original
18mg/L
K 0new  
 50mg/hr  62.5mg/hr
14.4mg/L