Transcript 05_Multiple dosing IV bolus

Multiple dosing: intravenous bolus
administration
Dr Mohammad Issa Saleh
Multiple dosing calculations using
Superposition
Let:
Dose 1  Conc. 1
and:
Dose 2  Conc. 2
then the response system behaves according to
the superposition principle if:
Dose 1 +Dose 2  Conc. 1 + Conc. 2
and in that case the response system is a linear
response system
2
Multiple dosing calculations using
Superposition

A patient is to be given 100 mg of a
drug intravenously. Assuming that K
= 0.10 hr-1 and a V = 15 L,
estimate the following:
1.
The half life
t1/ 2
ln( 2) ln( 2)


 6.93 hr
K
0.1 hr
3
Multiple dosing calculations using
Superposition
2.
The concentration 2 hrs after the dose
D  K (t  2 )
C (t  2)  e
 5.46 mg/L
V
3.
The concentration 10 hrs after the
dose
D  K (t 10)
C (t  10)  e
 2.45 mg/L
V
4
Multiple dosing calculations using
Superposition
4.
The concentration 18 hrs after the
dose
D  K (t 18)
C (t  18)  e
 1.10 mg/L
V
5
7
6
Conc. (mg/L)
5
4
3
2
1
0
0
8
16
Time (hr)
24
32
6
Multiple dosing calculations using
Superposition
5.
Assuming that 100 mg of the drug is
administered every 8 hrs, estimate
the concentration 2 hrs after the third
dose using the values calculated in
parts 2-4. What property of the linear
systems did you use to answer this
question?
7
12
10
Conc. (mg/L)
8
Conc. After the first dose
C1 (t )
6
4
2
0
0
8
16
24
32
Time (hr)
8
12
10
Conc. (mg/L)
8
Conc. After the second dose
6
C2 (t )
4
2
0
0
8
16
24
32
Time (hr)
9
12
10
Conc. (mg/L)
8
Conc. After the third dose
C3 (t )
6
4
2
0
0
8
16
24
32
Time (hr)
10
12
10
Total Conc.
Cn3 (t ' )
Conc. (mg/L)
8
6
4
2
0
0
8
16
24
32
Time (hr)
11
Cn3 (t '  2)  C1 (t  18)  C2 (t  10)  C3 (t  2)
12
t = 2 hrs after third dose
= 10 hrs after second dose
= 18 hrs after first dose
10
Conc. (mg/L)
8
6
4
2
0
0
8
16
24
32
Time (hr)
12
Multiple dosing calculations using
Superposition
5.
Assuming that 100 mg of the drug is
administered every 8 hrs, estimate the
concentration 2 hrs after the third dose
using the values calculated in parts 2-4.
What property of the linear systems did
you use to answer this question?
Cn 3 (t '  2)  C1 (t  18)  C2 (t  10)  C3 (t  2)
Cn 3 (t '  2)  1.10  2.45  5.46
Cn 3 (t '  2)  9.01 mg/L
13
Multiple dosing calculations using
Superposition


The principle of superposition assumes
that early doses of drug do not affect the
pharmacokinetics of subsequent doses.
Therefore, the blood levels after the
second, third, or nth dose will overlay or
superimpose the blood level attained after
the (n – 1)th dose
Multiple administration every 4 hrs
Dose
Number
Time
(hr)
Dose 1
1
0
0
0
1
21.0
21.0
3
19.8
19.8
4
16.9
0
16.9
5
14.3
21.0
35.3
7
10.1
19.8
29.9
8
8.50
16.9
0
25.4
9
7.15
14.3
21.0
42.5
11
5.06
10.1
19.8
35.0
12
4.25
8.50
16.9
0
29.7
13
3.58
7.15
14.3
21.0
46.0
15
2.53
5.06
10.1
19.8
37.5
2
3
4
Dose 2
Dose 3
Dose 4
Total
Multiple IV bolus administration

Concentration after n doses:
 D  K ( t ) 
Cn  r  e

V

where r:
 nKT
1 e
r
1  e  KT
n: number of doses, T: dosing interval
Multiple IV bolus administration

Concentration at steady state:
 D  K (t ) 
C ss  R e

V

where R is the accumulation ratio:
1
R
1  e  KT
T: dosing interval
Multiple IV bolus administration: useful
equations

Maximum concentration after n doses:
max
CN

D
 r 
V 
Maximum concentration at steady state:
max
C SS
D
 R 
V 
Multiple IV bolus administration: useful
equations

Minimum concentration after n doses:
min
CN

 D KT 
 r  e 
V

Minimum concentration at steady state:
min
C SS
 D KT 
 R  e 
V

Conc time profile:
The AUC during a dosing interval at
steady state is equal to the total AUC
following a single dose (For linear PK)
Multiple IV bolus administration: useful
equations

Average concentration at steady state:

C SSaverage 
C
0
SS
.dt

τ

0
0
As explained in the previous slide,  CSS .dt   C(single dose).dt

X0
AUC for a single dose is:  C(single dose).dt 
KVd
0
average
C SS
X0

KVd
Predicting average Css using single
dose data
Time to reach steady state conc.

The time required to reach to a certain
fraction of the steady-state level is given
by:
 1.44 t 0.5 ln( 1  fss)
n

Time required to achieve steady-state
depends on the half-life and is
independent of the rate of dosing and the
clearance

To get to 95% of the steady-state: 5 halflives are needed

To get to 99% of the steady-state: 7 halflives are needed
Different doses regimen have the same
average steady state conc: The same dosing
rate (Dose/ T)
Multiple IV bolus dosing compared to
IV infusion
Multiple IV
bolus
IV infusion
Multiple IV bolus dosing compared to
IV infusion

For IV infusion:
average
C SS

K0
Dosing rate


KVd
clearance
For multiple IV bolus (dosing rate =
dose/ dosing interval):
average
C SS

X0
dosing rate


KVd
clearance
The steady-concentration depends
on the rate of dosing and the
clearance
Example 1

1.
2.
3.
4.
To a patient 250 mg penicillin with t½ of
1 h and Vd of 25 L is administered every
6 h intravenously
Estimate Cmax, Cmin and Cav at steady
state
Has the objective of maintaining
concentration above minimum inhibitory
concentration (4 mg/L) been achieved in
this therapy? Elaborate!
How long did it take to reach 95% of Css?
Is the idea of giving a bolus dose to
achieve Css in a shorter time feasible
with regard to this drug?
Example 1
K
0.693 0.693

 0.693 hr 1
t 0.5
1
R
1
1

 1.016
 KT
 0.693*6
1 e
1 e
D
 250 
C SS  R    1.016
  10.16 mg/L
V
 25 
 D  KT 
 250 0.693*6 
min
C SS  R   e   1.016
e
  0.16 mg/L
V

 25

X0
250
average
C


 2.41 mg/L
KVdτ 0.693 * 25 * 6
max
SS
Example 1


Drug concentration cannot be maintained above
the MIC if it is being administered every 6 h (6 x
t½). Because almost 98% of the dose is out of
the body at the time of the next administration.
However, conventionally penicillins are given
q.i.d. and it is known that they are effective.
Therefore, there is no need for keeping the
concentration above MIC during the entire
therapy.
nτ  1.44 t0.5 ln(1  fss)
nτ  1.44 *1 * ln(1  0.95)  4.3 hr
 4.3 hrs are needed to get to 95% of Css (i.e.
Css was obtained as a result of the first dose)
Example 1

The steady-state is achieved very
rapidly (after the first dose). Since
there is no need for accumulation,
there is little justification for giving
a loading dose.
Example 2

A patient is receiving 1000 mg of
sulfamethoxazole iv every 12 hours
for the treatment of severe gramnegative infection. At steady state
the maximum and minimum serum
sulfamethoxazole concentrations
were 81.5 mg/L and 40 mg/L,
respectively. Estimate the values of
K and VD
Example 2
max
min
)  ln( CSS
)
ln( C1 )  ln( C2 ) ln( CSS
K

t 2  t1
T
ln(81.5)  ln(40)
K
 0.059 hr 1
12
1
1
R

 1.97
 KT
 0.059*12
1 e
1 e
D
C max

R
 
SS
V
 D
 V  R  max
C
 SS

  1.97  1000   24.2 L

 81.5 
