Transcript 03_Intravenous infusion and Laplace

Laplace transformation
1
Laplace transformation
• The Laplace transform is a mathematical
technique used for solving linear
differential equations (apparent zero order
and first order) and hence is applicable to
the solution of many equations used for
pharmacokinetic analysis.
2
Laplace transformation
procedure
1. Write the differential equation
2. Take the Laplace transform of each differential
equation using a few transforms (using table in
the next slide)
3. Use some algebra to solve for the Laplace of the
system component of interest
4. Finally the 'anti'-Laplace for the component is
determined from tables
3
Important Laplace transformation
(used in step 2)
Expression
Transform
dX/dt
sX  X 0
K (constant)
K
s
X (variable)
X
K∙X (K is constant)
KX
where s is the laplace operator,
X is the laplace integral
, and X0 is the amount at time zero
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Anti-laplce table (used in step 4)
5
Anti-laplce table continued (used in
step 4)
6
Laplace transformation:
example
• The differential equation that describes the
change in blood concentration of drug X is:
dX
 kX
dt
Derive the equation that describe the
amount of drug X??
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Laplace transformation:
example
1. Write the differential equation:
dX
 kX
dt
2. Take the Laplace transform of each
differential equation:
sX  X 0   kX
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Laplace transformation:
example
3. Use some algebra to solve for the
Laplace of the system component of
interest
X0
X
ks
4. Finally the 'anti'-Laplace for the
component is determined from tables
X (t )  X 0 e
 kt
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Laplace transformation:
example
• The derived equation represent the
equation for IV bolus one compartment
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Continuous intravenous infusion
(one-compartment model)
Dr Mohammad Issa
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Theory of intravenous infusion
•
The drug is administered at a selected
or calculated constant rate (K0) (i.e.
dX/dt), the units of this input rate will
be those of mass per unit time
(e.g.mg/hr).
•
The constant rate can be calculated
from the concentration of drug solution
and the flow rate of this solution, For
example, the concentration of drug
solution is 1% (w/v) and this solution is
being infused at the constant rate of
10mL/hr (solution flow rate). So 10mL
of solution will contain 0.1 g (100 mg)
drug.
•
The infusion rate (K0) equals to the
solution flow rate multiplied by the
concentration
•
In this example, the infusion rate will
be 10mL/hr multiplied by 100 mg/10
mL, or 100mg/hr. The elimination12of
drug from the body follows a first
IV infusion
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During infusion
Post infusion
30
Concentration
25
20
15
10
5
0
0
5
10
15
20
Time
25
30
35
40
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IV infusion: during infusion
35
30
Ko
 Kt
X
(1  e )
K
Concentration
25
20
Ko
 Kt
Cp 
(1  e )
KVd
15
10
5
0
0
5
10
15
20
Time
where K0 is the infusion
rate, K is the elimination
rate constant, and Vd is
25
30
35
40
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the
volume
of distribution
Steady state
35
30
Concentration
25
≈ steady state
concentration (Css)
20
15
Ko
Css 
KVd
10
5
0
0
5
10
15
20
Time
25
30
35
40
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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.
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Fraction achieved of steady state
concentration (Fss)
Ko
 Kt
Cp 
(1  e )
KVd
since Css  Ko
KVd
be represented as:
, previous equation can
 Kt
Cp  Css (1  e )
Cp
 Kt
 Fss 
 1 e
C ss
t 
 Fss  1  e
ln(2 )
t1 / 2
t
 1  t0.5
 1  
2
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Fraction achieved of steady state
concentration (Fss)
t
 1  t0.5
Fss  1   
2
or
 Kt
Fss  1  e
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Time needed to achieve steady
state
t   1.44 t0.5  ln( 1  FSS )
time needed to get to a certain fraction of
steady state depends on the half life of the
drug (not the infusion rate)
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Example
• What is the minimum number of half lives
needed to achieve at least 95% of steady
state?
t   1.44  t0.5  ln( 1  FSS )
  1.44  t0.5  ln( 1  0.95)  4.3  t0.5  5 t 0.5
• At least 5 half lives (not 4) are needed to
get to 95% of steady state
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Example
• A drug with an elimination half life of 10
hrs. Assuming that it follows a one
compartment pharmacokinetics, fill the
following table:
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Example
Time
Fss
10
30
50
70
90
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Example
Time
Number of elapsed
half-lives
Fss
10
1
0.5
30
3
0.875
50
5
0.969
70
7
0.992
90
9
0.998
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IV infusion + Loading IV bolus
• During constant rate IV administration, the drug
accumulates until steady state is achieved after five to
seven half-lives
• This can constitute a problem when immediate drug effect
is required and immediate achievement of therapeutic drug
concentrations is necessary such as in emergency
situations
• In this ease, administration of a loading dose will be
necessary. The loading dose is an IV holus dose
administered at the time of starting the IV infusion to
achieve faster approach to steady state. So administration
of an IV loading dose and starting the constant rate IV
infusion simultaneously can rapidly produce therapeutic
drug concentration. The loading dose is chosen to produce
Plasma concentration similar or close to the desired plasma
concentration that will be achieved by the IV infusion at 24
steady state
IV infusion + Loading IV bolus
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IV infusion + Loading IV bolus
• To achieve a target steady state conc (Css)
the following equations can be used:
– For the infusion rate:
K 0  Cl  Css
– For the loading dose:
LD  Vd  Css
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IV infusion + Loading IV bolus
• The conc. resulting from both the bolus
and the infusion can be described as:
Ctotal =Cinfusion + Cbolus
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IV infusion + Loading IV bolus:
Example
• Derive the equation that describe plasma
concentration of a drug with one
compartment PK resulting from the
administration of an IV infusion (K0=
Css∙Cl ) and a loading bolus (LD= Css∙Vd)
that was given at the start of the infusion
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IV infusion + Loading IV bolus:
Example
Ctotal =Cinfusion + Cbolus
• Cinfusion:
Cinfusion  Css (1  e
 K t
)
• Cbolus:
X 0  Kt Css  Vd  Kt
 K t
C bolus 
e

e
 Css  e
Vd
Vd
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IV infusion + Loading IV bolus:
Example
Ctotal =Cinfusion + Cbolus
C total  Css (1  e
 Kt
)  Csse
 C total  Css  Csse
 K t
 K t
 Csse
 K t
 C total  Css
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Scenarios with different LD
Infusion alone
(K0= Css∙Cl)
Concentration
Concentration
Case A
Case C
Infusion (K0= Css∙Cl)
loading bolus (LD > Css∙Vd)
Half-lives
Half-lives
Concentration
Concentration
Half-lives
Case B
Infusion (K0= Css∙Cl)
loading bolus (LD= Css∙Vd)
Case D
Infusion (K0= Css∙Cl)
loading bolus (LD< Css∙Vd)
Half-lives
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Changing Infusion Rates
Concentration
Increasing the infusion rate
results in a new steady state
conc. 5-7 half-lives are
needed to get to the new
steady state conc
Half-lives
5-7 half-lives are
needed to get to
steady state
Decreasing the infusion rate
results in a new steady state conc.
5-7 half-lives are needed to get to
the new steady state conc
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Changing Infusion Rates
• The rate of infusion of a drug is sometimes
changed during therapy because of
excessive toxicity or an inadequate
therapeutic response. If the object of the
change is to produce a new plateau, then
the time to go from one plateau to
another—whether higher or lower—
depends solely on the half-life of the drug.
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Post infusion phase
35
During infusion
Post infusion
30
Concentration
25
20
 Kt
Cp  C *  e
C*
(Concentration
at the end of
the infusion)
15
10
5
0
0
5
10
15
20
Time
25
30
35
40
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Post infusion phase data
• Half-life and elimination rate constant
calculation
• Volume of distribution estimation
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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:
k
Slope  
2.303
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Volume of distribution calculation
using post infusion data
• If you reached steady state conc (C* =
CSS):
K0
K0
Css 
 Vd 
K  Vd
K  Css
• where k is estimated as described in the
previous slide
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Volume of distribution calculation
using post infusion data
• If you did not reached steady state (C* =
CSS(1-e-kT)):
k0
k0
 kT
C* 
(1  e )  Vd 
(1  e  kT )
k  Vd
k C *
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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
1
cessation (hr)
Cp (mg/L)
12
3
7 10
16
22
9
8
3.9
1.7
5
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Post infusion data
1.2
Log(Conc) mg/L
1
y = -0.0378x + 1.1144
R2 = 0.9664
0.8
0.6
0.4
0.2
0
0
5
10
15
20
Time (hr)
Time is the time after stopping the infusion
25
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Example 1
• From the slope, K is estimated to be:
k  2.303  Slope  2.303  0.0378  0.087 1/hr
• From the intercept, C* is estimated to be:
log(C*)  intercept  1.1144
C*  10
1.1144
 13 mg/L
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Example 1
• Since we did not get to steady state:
k0
 kT
(1  e )
Vd 
k C*
100
 0.087*2
Vd 
(1  e
)  14.1 L
(0.087)  (13)
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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
Conc
(mg/L)
0
2
4
3.48 5.47
6
10
12
15
16
18
6.6
7.6
7.8
8
8
4.6
20
24
2.62 0.85
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Example 2
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Example 2
1. Calculating clearance:
It appears from the data that the infusion
has reached steady state:
(CP(t=15) = CP(t=16) = CSS)
C SS
K0
K 0 50 mg/hr

 Cl 

 6.25 L/hr
Cl
C SS
8 mg/L
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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:
C P  CSS  e
 K t1
K
 log( C P )  log( CSS ) 
t1
2.303
where t1 is the time after stopping the infusion.
Plotting log(Cp) vs. t1 results in the following:
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Example 2
1
log(Conc) (mg/L)
0.8
y = -0.1218x + 0.9047
R2 = 1
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
-0.2
Post infusion time (hr)
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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:
Cl 6.25 L/hr
VD 

 22.3 L
-1
K
0.28 hr
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Example 3
• A drug that displays one compartment
characteristics was administered as an IV
bolus of 250 mg followed immediately by a
constant infusion of 10 mg/hr for the
duration of a study. Estimate the values of
the volume of distribution (25 L), elimination
rate constant (0.1hr-1), half-life (7), and
clearance (2.5 L/hr) from the data in the
following table
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Example 3
Time(hr)
0
5
20
45
50
Conc(mg/L)
10
7.6
4.8
4.0
4.0
50
Example 3
• The equation that describes drug
concentration is:
C P  Drug from IV bolus  Drug from IV infusion

X 0  K t
K0
 K t

e

1 e
VD
K  VD

a- Calculating volume of distribution:
At time zero,
X0
X0
250 mg
C P (t  0) 
 VD 

 25 L
VD
C P (t  0) 10 mg/L
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Example 3
b- Calculating elimination rate constant and
half life:
Since the last two concentrations (at time 45
and 50 hrs) are equal, it is assumed that a
steady state situation has been achieved.
C SS
K0
K0
10 mg/hr

K 

 0.1 hr -1
K  VD
C SS  VD 4 mg/L  25 L
Half life = 0.693/K=0.693/0.1= 6.93 hr
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Example 3
c- Calculating clearance:
Cl  K  VD  0.1 25  2.5 L/hr
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Example 4
• 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)
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Example 4
t1 / 2
ASS  0.693
Ko
Ass 
 K0 
 t1 / 2 
K
0.693
K0
For the patient requiring 0.4 mg/min:
t1 / 2
ASS  0.693 (20)(0.693)


 34.65 min
K0
0.4
For the patient requiring 4 mg/min:
t1 / 2
ASS  0.693 (20)(0.693)


 3.465 min
K0
4
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Example 5
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.
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Example 5
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?
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Example 5
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:

K0
CP 
1  e  K t
K  VD

600 mg/hr
 ( 0.4 )( 0.5 )
C P (t  0.5 hr) 
(
1

e
)  9.71 mg/L
-1
(0.4 hr )( 28 L)
58
Example 5
2) Post infusion phase:
C P  C P (at the end of infusion)  e  kt 2
 ln(C P )  ln(C P (at the end of infusion)) - K  t 2
ln(C P (at the end of infusion)) - ln(C P ) ln(9.71)  ln(5)
 t2 

 1.66 hr
K
0.4
The concentration will fall to 5.0 mg/L 1.66
hr after the infusion was stopped.
59
Example 5
3) Post infusion phase (conc 5.5 hrs after
stopping the infusion):
C P  C P (at the end of infusion)  e  kt 2
C p (t  5.5 hr)  (9.71)e (-0-.4)(5.5)  1.08 mg/L
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