Download Clinical pharmacokinetic equations and calculations

Clinical pharmacokinetic
equations and calculations
Mohammad Issa Saleh
1
One vs. two compartments
2
One vs. two compartments
Two compartments
IV bolus
One compartment
IV bolus
3
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.
4
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.
5
One-compartment model equations



IV bolus
IV infusion
Extravascular
6
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).
7
Intravenous Bolus Equation

In this case, a one-compartment
model intravenous bolus equation
can be used:
D  K t
C
e
Vd
8
Intravenous Bolus Equation
D  K t
C
e
Vd
9
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.
10
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).
11
Very short infusion time compared to
the half-life

To compute the expected
theophylline concentration 4 hours
after the dose was given, a onecompartment model intravenous
bolus equation can be used:
D  Kt 400 mg 0.1154
C
e

e
 8.4 mg/L
Vd
30 L
12
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
13
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.
14
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:
D  Kt 1000 mg 0.07712
C
e

e
 7.9 mg/L
Vd
50 L
15
Estimating individual Pharmacokinetic
parameters: IV bolus
1.
2.
3.
Plotting
Simple fitting (will not be
discussed)
Calculation
16
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.
17
Estimating individual PK parameters:
plotting
18
Estimating individual PK parameters:
plotting


The elimination rate constant can be
computed using the following
relationship:
0.693 0.693
Ke 

 0.173 day -1
t1/ 2
4
The extrapolated concentration at
time = 0 (C0 = 15 mg/L in this
case) can be used to calculate the
volume of distribution:
D 600
V

 40 L
19
C0 15
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:
 ln(C 1 ) - ln(C 2 )   ln(12.6) - ln(7.5)
  -
Ke  -
t1 - t 2
1- 4

 

-1

0.173
day


where t1 and C1 are the first
time/concentration pair and t2 and C2 are
the second time/concentration pair
20
Estimating individual PK parameters:
Calculation

The elimination rate constant can be
converted into the half-life using the
following equation:
t1/ 2
0.693 0.693


 4 days
Ke
0.173
21
Estimating individual PK parameters:
Calculation

The serum concentration at time = zero
(C0) can be computed using a variation of
the intravenous bolus equation:
C0 
C
e  Ke.t
where t and C are a time/concentration pair
12.6
C0  0.173.1  15.0 mg/L
e
The volume of distribution (V) :
D 600
V

 40 L
C0 15
22
Continuous intravenous infusion
(one-compartment model)
Dr Mohammad Issa
23
IV infusion
35
During infusion
Post infusion
30
Concentration
25
20
15
10
5
0
0
5
10
15
20
Time
25
30
35
40
24
IV infusion: during infusion
35
30
Ko
 Kt
X
(1  e )
K
Concentration
25
20
Ko
 Kt
Cp 
(1  e )
KVd
15
10
where K0 is the infusion
rate, K is the elimination
rate constant, and Vd is
the volume of distribution
5
0
0
5
10
15
20
Time
25
30
35
40
25
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
26
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 t1/2 are needed to get to 95% of
Css
At least 7 t1/2 are needed to get to 99% of
Css
5-7 t1/2 are needed to get to Css
27
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
28
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
29
Post infusion phase
35
During infusion
Post infusion
30
 Kt
C postinf  Cend  e
Concentration
25
20
Cend
(Concentration
at the end of
the infusion)
15
10
Cend
Ko
 KT

(1  e )
KVd
t is the post infusion time
T is the infusion duration
5
0
0
5
10
15
20
Time
25
30
35
40
30
Post infusion phase data


Half-life and elimination rate
constant calculation
Volume of distribution estimation
31
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
32
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
33
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 *
34
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)
Cp (mg/L)
1
3
7 10
16
22
12
9
8
3.9
1.7
5
35
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
36
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*  101.1144  13 mg/L
37
Example 1

Since we did not get to steady
state:
k0
(1  e  kT )
Vd 
k C*
100
Vd 
(1  e 0.087*2 )  14.1 L
(0.087)  (13)
38
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
39
Example 2
40
Example 2
1.
C SS
Calculating clearance:
It appears from the data that the
infusion has reached steady state:
(CP(t=15) = CP(t=16) = CSS)
K0
K 0 50 mg/hr

 Cl 

 6.25 L/hr
Cl
C SS
8 mg/L
41
Example 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:
2.
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:
42
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)
43
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
44
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)
45
Example 3
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
46
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 onecompartment body model.
47
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?
48
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:

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)
49
Example 4
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.
50
Example 4
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
51
52