IV-Bolus-and-Infusion.ppt

IV-Bolus-and-Infusion

1. 1
Clinical pharmacokinetic
equations and calculations
Mohammad Issa Saleh

2. 2
One vs. two compartments

3. 3
One vs. two compartments
Two compartments
IV bolus
One compartment
IV bolus

4. 4
One vs. two compartments
In many cases, the drug distributes
from the blood into the tissues
quickly, and a pseudoequilibrium of
drug movement between blood and
tissues is established rapidly. When
this occurs, a one-compartment
model can be used to describe the
serum concentrations of a drug.

5. 5
One vs. two compartments
In some clinical situations, it is
possible to use a one-compartment
model to compute doses for a drug
even if drug distribution takes time
to complete. In this case, drug
serum concentrations are not
obtained in a patient until after the
distribution phase is over.

6. 6
One-compartment model equations
IV bolus
IV infusion
Extravascular

7. 7
Intravenous Bolus Equation
When a drug is given as an
intravenous bolus and the drug
distributes from the blood into the
tissues quickly, the serum
concentrations often decline in a
straight line when plotted on
semilogarithmic axes (Figure next
slide).

8. 8
Intravenous Bolus Equation
In this case, a one-compartment
model intravenous bolus equation
can be used:tK
e
Vd
D
C



9. 9
Intravenous Bolus Equation tK
e
Vd
D
C



10. 10
Intravenous Bolus
Most drugs given intravenously cannot be
given as an actual intravenous bolus
because of side effects related to rapid
injection.
A short infusion of 5–30 minutes can
avoid these types of adverse effects
If the intravenous infusion time is very
short compared to the half-life of the drug
so that a large amount of drug is not
eliminated during the infusion time,
intravenous bolus equations can still be
used.

11. 11
Very short infusion time compared to
the half-life
For example, a patient is given a
theophylline loading dose of 400 mg
intravenously over 20 minutes.
Because the patient received
theophylline during previous
hospitalizations, it is known that the
volume of distribution is 30 L, the
elimination rate constant equals
0.116 h−1, and the half-life (t1/2) is
6 hours (t1/2 = 0.693/ke =
0.693/0.115 h−1 = 6 h).

12. 12
Very short infusion time compared to
the half-life
To compute the expected
theophylline concentration 4 hours
after the dose was given, a one-
compartment model intravenous
bolus equation can be used:mg/L 8.4e
L 30
mg 400
e
Vd
D
C
40.115tK



13. 13
Slow infusion and/or distribution
If drug distribution is not rapid, it is
still possible to use a one
compartment model intravenous
bolus equation if the duration of the
distribution phase and infusion time
is small compared to the half-life of
the drug and only a small amount of
drug is eliminated during the
infusion and distribution phases

14. 14
Slow infusion and/or distribution
For instance, vancomycin must be infused
slowly over 1 hour in order to avoid
hypotension and red flushing around the
head and neck areas. Additionally,
vancomycin distributes slowly to tissues
with a 1/2–1 hour distribution phase.
Because the half-life of vancomycin in
patients with normal renal function is
approximately 8 hours, a one
compartment model intravenous bolus
equation can be used to compute
concentrations in the postinfusion,
postdistribution phase without a large
amount of error.

15. 15
Slow infusion and/or distribution
As an example of this approach (previous slide), a
patient is given an intravenous dose of
vancomycin 1000 mg. Since the patient has
received this drug before, it is known that the
volume of distribution equals 50 L, the elimination
rate constant is 0.077 h−1, and the half-life
equals 9 h (t1/2 = 0.693/ke = 0.693/0.077 h−1
= 9 h).
To calculate the expected vancomycin
concentration 12 hours after the dose was given,
a one compartment model intravenous bolus
equation can be used: mg/L 7.9e
L 50
mg 1000
e
Vd
D
C
120.077tK



16. 16
Estimating individual Pharmacokinetic
parameters: IV bolus
1.Plotting
2.Simple fitting (will not be
discussed)
3.Calculation

17. 17
Estimating individual PK parameters:
plotting
For example, a patient was given an
intravenous loading dose of phenobarbital
600 mg over a period of about an hour.
One day and four days after the dose was
administered phenobarbital serum
concentrations were 12.6 mg/L and 7.5
mg/L, respectively.
By plotting the serum concentration/time
data on semilogarithmic axes, the time it
takes for serum concentrations to
decrease by one-half can be determined
and is equal to 4 days.

18. 18
Estimating individual PK parameters:
plotting

19. 19
Estimating individual PK parameters:
plotting
The elimination rate constant can be
computed using the following
relationship:
The extrapolated concentration at
time = 0 (C
0= 15 mg/L in this
case) can be used to calculate the
volume of distribution:1-
2/1
day 173.0
4
0.693
t
0.693
Ke L 40
15
600D
V
0

C

20. 20
Estimating individual PK parameters:
Calculation
Alternatively, these parameters could be
obtained by calculation without plotting
the concentrations.
The elimination rate constant can be
computed using the following equation:
where t1 and C1 are the first
time/concentration pair and t2 and C2 are
the second time/concentration pair day 0.173
4-1
ln(7.5)-ln(12.6)
-
t-t
)ln(C-)ln(C
-Ke
1-
21
21

















21. 21
Estimating individual PK parameters:
Calculation
The elimination rate constant can be
converted into the half-life using the
following equation: days 4
173.0
0.6930.693
t
2/1 
Ke

22. 22
Estimating individual PK parameters:
Calculation
The serum concentration at time = zero
(C0) can be computed using a variation of
the intravenous bolus equation:
where t and C are a time/concentration pair
The volume of distribution (V) :tKe
e
C
C
.0 
 mg/L 15.0
e
12.6
C
0.173.10 
 L 40
15
600D
V
0

C

23. 23
Continuous intravenous infusion
(one-compartment model)
Dr Mohammad Issa

24. 24
IV infusion0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Time
Concentration
During infusion Post infusion

25. 25
IV infusion: during infusion0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Time
Concentration
where K
0is the infusion
rate, K is the elimination
rate constant, and Vd is
the volume of distribution)1(e
Kto
K
K
X

 )1(e
Kto
p
KVd
K
C



26. 26
Steady state0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Time
Concentration
≈ steady state
concentration (C
ss)KVd
K
C
o
ss

27. 27
Steady state
At steady state the input rate (infusion
rate) is equal to the elimination rate.
This characteristic of steady state is valid
for all drugs regardless to the
pharmacokinetic behavior or the route of
administration.
At least 5 t
1/2are needed to get to 95% of
Css
At least 7 t
1/2are needed to get to 99% of
Css
5-7 t
1/2are needed to get to Css

28. 28
IV infusion + Loading IV bolus
To achieve a target steady state
conc (C
ss) the following equations
can be used:
For the infusion rate:
For the loading dose:ssCClK 
0 ssCVdLD 

29. 29
Concentration
ConcentrationConcentration
Concentration
Half-lives
Half-lives
Half-lives
Half-lives
Case A
Infusion alone
(K0= Css∙Cl)
Case B
Infusion (K0= Css∙Cl)
loading bolus (LD= Css∙Vd)
Case D
Infusion (K0= Css∙Cl)
loading bolus (LD< Css∙Vd)
Case C
Infusion (K0= Css∙Cl)
loading bolus (LD > Css∙Vd)
Scenarios with different LD

30. 30
Post infusion phase0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Time
Concentration
During infusion Post infusion
C
end
(Concentration
at the end of
the infusion)e
Kt
endpostinf
CC

 )1(e
KTo
end
KVd
K
C

 t is the post infusion time
T is the infusion duration

31. 31
Post infusion phase data
Half-life and elimination rate
constant calculation
Volume of distribution estimation

32. 32
Elimination rate constant calculation
using post infusion data
K can be estimated using post
infusion data by:
Plotting log(Conc) vs. time
From the slope estimate K:303.2
k
Slope

33. 33
Volume of distribution calculation using
post infusion data
If you reached steady state conc
(C* = C
SS):
where k is estimated as described in
the previous slidess
ss
CK
K
Vd
VdK
K
C




00

34. 34
Volume of distribution calculation using
post infusion data
If you did not reached steady state
(C* = C
SS(1-e
-kT
)):)1(
*
)1(*
00 kTkT
e
Ck
k
Vde
Vdk
k
C







35. 35
Example 1
Following a two-hour infusion of 100
mg/hr plasma was collected and
analysed for drug concentration.
Calculate kel and V.
Time relative
to infusion
cessation
(hr)
1371016 22
Cp (mg/L) 129853.91.7

36. 36Post infusion data
y = -0.0378x + 1.1144
R
2
= 0.9664
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Time (hr)
Log(Conc) mg/L
Time is the time after stopping the infusion

37. 37
Example 1
From the slope, K is estimated to
be:
From the intercept, C* is estimated
to be:1/hr 0.0870.03782.303Slope2.303k  mg/L 1310C*
1.1144interceptlog(C*)
1.1144



38. 38
Example 1
Since we did not get to steady
state:L 14.1)e(1
(13)(0.087)
100
Vd
)e(1
*Ck
k
Vd
20.087*
kT0









39. 39
Example 2
Estimate the volume of distribution
(22 L), elimination rate constant
(0.28 hr-1), half-life (2.5 hr), and
clearance (6.2 L/hr) from the data in
the following table obtained on
infusing a drug at the rate of 50
mg/hr for 16 hours.
Time
(hr)
0 2 4 610121516182024
Conc
(mg/L)
03.485.476.67.67.88 84.62.620.85

40. 40
Example 2

41. 41
Example 2
1.Calculating clearance:
It appears from the data that the
infusion has reached steady state:
(CP(t=15) = CP(t=16) = C
SS)L/hr 25.6
mg/L 8
mg/hr 50
00

SS
SS
C
K
Cl
Cl
K
C

42. 42
Example 2
2.Calculating elimination rate constant
and half life:
From the post infusion data, K and t1/2 can
be estimated. The concentration in the
post infusion phase is described
according to:
where t1 is the time after stopping the
infusion. Plotting log(Cp) vs. t1 results
in the following:1
303.2
)log()log(
1
t
K
CCeCC
SSP
tK
SSP 


43. 43
Example 2y = -0.1218x + 0.9047
R
2
= 1
-0.2
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5 6 7 8 9
Post infusion time (hr)
log(Conc) (mg/L)

44. 44
Example 2
K=-slope*2.303=0.28 hr-1
Half life = 0.693/K=0.693/0.28=
2.475 hr
3.Calculating volume of distribution:L 3.22
hr 28.0
L/hr 25.6
1-

K
Cl
V
D

45. 45
Example 3
For prolonged surgical procedures,
succinylcholine is given by IV
infusion for sustained muscle
relaxation. A typical initial dose is 20
mg followed by continuous infusion
of 4 mg/min. the infusion must be
individualized because of variation in
the kinetics of metabolism of
suucinylcholine. Estimate the
elimination half-lives of
succinylcholine in patients requiring
0.4 mg/min and 4 mg/min,
respectively, to maintain 20 mg in the
body. (35 and 3.5 min)

46. 46
Example 3
For the patient requiring 0.4 mg/min:
For the patient requiring 4 mg/min:0
2/1
2/1
0
693.0
693.0 K
A
t
t
K
K
K
A
SS
o
ss

 min 65.34
4.0
)693.0)(20(693.0
0
2/1 


K
A
t
SS min 465.3
4
)693.0)(20(693.0
0
2/1 


K
A
t
SS

47. 47
Example 4
A drug is administered as a short
term infusion. The average
pharmacokinetic parameters for
this
drug are:
K = 0.40 hr-1
Vd = 28 L
This drug follows a one-
compartment body model.

48. 48
Example 4
1) A 300 mg dose of this drug is given
as a short-term infusion over 30
minutes. What is the infusion rate?
What will be the plasma
concentration at the end of the
infusion?
2) How long will it take for the plasma
concentration to fall to 5.0 mg/L?
3) If another infusion is started 5.5
hours after the first infusion was
stopped, what will the plasma
concentration be just before the
second infusion?

49. 49
Example 4
1) The infusion rate (K0) =
Dose/duration = 300 mg/0.5 hr =
600 mg/hr.
Plasma concentration at the end of
the infusion:
Infusion phase: 
mg/L 71.9)1(
L) 28)(hr 4.0(
mg/hr 600
hr) 5.0(
1C
)5.0)(4.0(
1-
0
P






etC
e
VK
K
P
tK
D

50. 50
Example 4
2) Post infusion phase:
The concentration will fall to 5.0 mg/L
1.66 hr after the infusion was
stopped.hr 1.66
0.4
ln(5)ln(9.71)
K
)ln(C-infusion)) of end (at theln(C
t
tK-infusion)) of end (at theln(C)ln(C
einfusion) of end (at theCC
PP
2
2PP
tk
PP
2







51. 51
Example 4
3) Post infusion phase (conc 5.5 hrs
after stopping the infusion):mg/L 08.19.71)e(hr) 5.5(tC
einfusion) of end (at theCC
-.4)(5.5)(-0
p
tk
PP
2




52. 52