Transcript PK Simulation

Index of Simulations
• 1-Compartment, IV bolus
• 1-Compartment, IV infusion: SteadyState
• 1-Compartment, IV infusion: NonSteady-State
• 1-Compartment, IV infusion: No
Elimination Phase
1-Compartment, IV Bolus
• The following data were obtained
after 100 mg of Drug X was
administered to a healthy volunteer.
Blood was collected starting at onehour post-dose for a total of 12
hours. Calculate Cl, V and t1/2
Data (X0 = 100 mg)
Time (h)
[Drug X] (ug/L)
1
1.5
2
2.5
314.7
280.0
246.5
219.5
3
4
6
194.1
152.8
94.1
8
10
12
59.0
35.6
22.1
Step 1) Graph on Semi-log Paper
[Drug X] (ug/L)
1000
100
10
0
Log scale
Linear scale
5
10
Time (h)
15
Step 2) Draw best fit line
[Drug X] (ug/L)
1000
100
10
0
5
10
Time (h)
15
Step 3) Find V
• Use the relationship:
X0
C0 
V
• Rearranged for V:
X0
V
C0
– We know X0 (100 mg) and C0 we can get from
the graph
Step 3) Find V
[Drug
[DrugX]X](ug/L)
(ug/L)
1000
1000
C0
C0 = 400 ug/L
400
100
300
200
100
10 0
0
1
2
5
3
Time (h)
Time (h)
4
10
5
15
Step 3) Find V
• We know X0 (100 mg) and C0 we can
get from the graph (C0 = 400 ug/L)
(please watch units)
X 0 100000 ug
V

 250 L
C0
400 ug/L
• Now we have our volume (250 L)
Step 4) Find t1/2 (and k)
• Half-life (t1/2) can be obtained directly
from the graph by reading how long
it takes for the concentration to be
reduced by 50%
Step 4) Find t1/2
[Drug X] (ug/L)
1000
C0 = 400 ug/L
200
100
Half
ofwith
400Cis0 which
200.
At
Start
the
intesection,
Draw
line
from
draw
equals
aa line
400 down
ug/L to
200
across
until
it
the
X-axis
Your
t1/2 is(time).
~3
hours
intersects
your the
Read the value
best
fit line
line intersects
the
axis…this is t1/2
10
0
t1/2 = 3
5
10
Time (h)
15
Step 5) Find Cl
• Clearance (Cl) can be calculated from
k (Step 4) and V (Step 4) and using
the following equation:
Cl  k  V  0.2311/h250 L  57.8 L/h
Summary
• Cl = 57.8 L/h
• V = 250 L
• t1/2 = 3 h
Return to Table of Contents
Onto Steady-State Infusion
1-Compartment, IV Infusion:
Steady-State
• The following data were obtained
after 100 mg of Drug X was infused
over 15 hours to a healthy volunteer.
Blood was collected starting at onehour post-dose for a total of 24
hours. Calculate Cl, V and t1/2
Data (X0 = 100 mg)
Time (h)
[Drug X] (ug/L)
0
1
3
6
0
23.7
57.3
85.0
10
15
18
100.5
109.8
54.1
20
24
32.6
12.5
Step 1) Graph on Semi-log Paper
[Drug X] (ug/L)
1000
100
10
1
0
Log scale
Linear scale
10
20
Time (h)
30
Step 2) Find t1/2 (and k)
• Half-life (t1/2) can be obtained directly
from the graph by reading how long
it takes for the concentration to be
reduced by 50%.
– For infusions, you must use the
terminal portion where concentrations
are falling!!
• First however, draw a best fit line
through the terminal portion
Step 2) Find t1/2
[Drug X] (ug/L)
1000
100
10
1
0
10
20
Time (h)
30
Step 2) Find t1/2
[Drug X] (ug/L)
1000
C ~ 110 ug/L
100
55
10
At the intesection,
Your
t1/2
isCthe
timeto
Start
with
which
155.
Half
is
drawof
a110
line
down
you
just
read
you
C(time).
at 15
h
Draw
a line
from
the know
X-axis
0 55
infusion
=minus
110 ug/L
across
until
it the
Read
the
value
time
(18 h –your
15 h
=
intersects
line intersects
the
3
hours)
best
fit line
axis…
10
20
t1/2 = 18 - 15
Time (h)
30
Step 2) Find t1/2 (and k)
• Half-life (t1/2) from the graph is 3
hours. We can find k by the
following equation:
0.693 0.693
k

 0.231 1/h
t1
3h
2
Step 3) Find Cl
• Clearance (Cl) can be calculated from
the steady-state concentration (Css)
and the infusion rate (k0) using the
equation:
CSS
• Rearranged to:
k0

Cl
k0
Cl 
C ss
Step 3) Find Cl
• We know the dose (100 mg) and
infusion time (T=15 h), therefore
infusion rate is:
X 0 100 mg
k0 

 6.67 mg/h
T
15 h
Step 3) Find Cl
• We can obtain Css from the graph by
looking to see when concentrations stop
changing.
• How do we know for sure this is steadystate? Remember steady-state is 3-5 halflives.
– Half-life from Step 2 = 3h
– 3 x 5 (or 3 or 4) = 15 h
– Infusion was stop at 15 hours therefore we are
at steady-state and this approach is valid
Step 3) Find Cl
[Drug X] (ug/L)
1000
Css= 110 ug/L
100
10
1
0
10
20
Time (h)
30
Step 3) Find Cl
• We have k0 (6.67 mg/h), we have CSS
(110 ug/L), now we can calculate Cl
k 0 6670 ug/h
Cl 
 60.6 L/h
C ss 110 ug/L
Step 4) Find V
• Volume (V) can be calculated from k
(Step 3) and Cl (Step 1) and using the
following equation:
Cl  k  V
• Rearrange and solve for V
Cl 60.6 L/h
V

 262 L
k 0.231 1/h
Summary
• Cl = 60.6 L/h
• V = 262 L
• t1/2 = 3 h
Return to Table of Contents
Onto Non-Steady-State
Infusion
1-Compartment, IV Infusion:
Non-Steady-State
• The following data were obtained
after 100 mg of Drug X was infused
over 6 hours to a healthy volunteer.
Blood was collected starting at onehour post-dose for a total of 24
hours. Calculate Cl, V and t1/2
Data (X0 = 100 mg)
Time (h)
[Drug X] (ug/L)
0
1
3
6
0
61.6
159.1
260.0
10
15
18
136.2
61.7
38.2
20
24
25.2
14.4
Step 1) Graph on Semi-log Paper
[Drug X] (ug/L)
1000
100
10
1
0
Log scale
Linear scale
10
20
Time (h)
30
Step 2) Find t1/2 (and k)
• Half-life (t1/2) can be obtained directly
from the graph by reading how long
it takes for the concentration to be
reduced by 50%.
– For infusions, you must use the
terminal portion where concentrations
are falling!!
• First however, draw a best fit line
through the terminal portion
Step 2) Find t1/2
[Drug X] (ug/L)
1000
100
10
1
0
10
20
Time (h)
30
Step 2) Find t1/2
[Drug X] (ug/L)
1000
C ~ 260 ug/L
130
100
10
At the intesection,
Start
with
which
1130.
Half
is
Your
drawof
ta1/2260
line
isCthe
down
time
to
you
C(time).
at 0
6h=
Draw
a line
from
you
the know.
X-axis
just
read
260
ug/L
130
across
untilthe
it
minus
Read
the
infusion
value
intersects
time
line intersects
(10 h your
– 6 hthe
=4
best
fit line
hours)
axis…
t1/2 = 10 - 6
10
Time (h)
20
30
Step 2) Find t1/2 (and k)
• Half-life (t1/2) from the graph is 3
hours. We can find k by the
following equation:
0.693 0.693
k

 0.173 1/h
t1
4h
2
Step 3) Find Cl
• Clearance (Cl) can be calculated twoways. Please select a method to
calculate clearance
• AUC Method – More exact but more
calculations
• Equation Method – Quicker but less
exact
Clearance Via AUC
• To calculate clearance via the AUC,
you must first calculate the AUC via
the trapezoidal rule
Trapezoidal Rule
[Drug X] (ug/L)
300
C2
For
this
method,
The
area
of the
we
break the
trapezoid
(or
curve
into a
this case
individual
triangle)
is the
trapezoids
as
average height
shown
here…
(C +C
)/2
250
200
150
100
1
2
multiplied by the
base (t2-t1)
50
C1
0
0
10
t1
20
t2
Time (h)
30
Step 2) Setup the table
Time [Drug X]
A
B
A*B
(C2+C1
)/2
(t2-t1)
AUC
0
0
---
---
---
1
61.6
30.8
1
30.8
3
159.1
110.4
2
220.8
6
260.0
209.6
3
628.8
10
136.2
198.1
4
792.4
15
61.7
99.0
5
495.0
18
38.2
50.0
3
150.0 AUC tail 
20
25.2
31.7
2
63.4
24
14.4
39.6
4
158.4
---
---
83.1
Tail
SUM 2622.7
CLast
k
Step 3: Calculate Cl
• Since we now have AUC, using the
dose (100 mg), and the equation:
X0
AUC 
Cl
• Solve for Cl:
X0
100000 ug
Cl 

 38.1 L/h
AUC 2622.7 ug.h/L
Select another Cl calculation
Go to Volume calculation
Clearance via Infusion
Equation
• We can use the equation that describes an
infusion and solve for Cl.
– During Infusion (t = time during infusion)
k0
( k  t)
C
 (1  e
)
Cl
– Solving for Cl
k0
Cl 
 (1  e ( k t) )
C
Clearance via Equation
• Now plug in the values we know
(infusion rate, C, t, k)
k0
( k  t)
Cl 
 (1  e
)
C
100000 ug
( 0.173 1/h3 h)
6h

 (1  e
)
159.1 ug/L
 42.4 L/h
Select another Cl calculation
Go to Volume calculation
Step 3) Find Cl
• We know the dose (100 mg) and
infusion time (T=15 h), therefore
infusion rate is:
X 0 100 mg
k0 

 6.67 mg/h
T
15 h
Step 3) Find Cl
• We can obtain Css from the graph by
looking to see when concentrations stop
changing.
• How do we know for sure this is steadystate? Remember steady-state is 3-5 halflives.
– Half-life from Step 2 = 3h
– 3 x 5 (or 3 or 4) = 15 h
– Infusion was stop at 15 hours therefore we are
at steady-state and this approach is valid
Step 3) Find Cl
[Drug X] (ug/L)
1000
Css= 110 ug/L
100
10
1
0
10
20
Time (h)
30
Step 3) Find Cl
• We have k0 (6.67 mg/h), we have CSS
(110 ug/L), now we can calculate Cl
k 0 6670 ug/h
Cl 
 60.6 L/h
C ss 110 ug/L
Step 4) Find V
• Volume (V) can be calculated from k
(Step 2) and Cl (Step 3) and using the
following equation:
Cl  k  V
• Rearrange and solve for V
Cl 38.1 L/h
V

 220 L
k 0.173 1/h
1-Compartment, IV infusion:
No elimination phase
• The following data were obtained
after 100 mg of Drug X was infused
over 15 hours to a healthy volunteer.
Blood was collected starting at onehour post-dose for a total of 15
hours. Calculate Cl, V and t1/2
Data (X0 = 100 mg)
Time (h)
[Drug X] (ug/L)
0
1
3
6
0
23.7
57.3
85.0
10
12
14
100.5
104.9
106.8
15
108.9
Step 1) Graph on Semi-log Paper
[Drug X] (ug/L)
1000
100
10
0
Log scale
Linear scale
5
10
Time (h)
15
Step 2) Find t1/2 (and k)
• Since we do not have an elimination
phase, we must find another way to
estimate half-life. We will use the
approach to steady-state method.
ln(CSS  Ct )  kt
• So first we need to estimate CSS
Step 2) Find CSS
We can estimate
CSs either
Just
read Ctaking
SS from
theintercept
average of
of the
the
the
last few
Y-axis
concentrations or
use a best fit line
[Drug X] (ug/L)
1000
106.5
ug/L
100
10
0
5
10
Time (h)
15
Step 2) Find CSS
• Our CSS = 106.5 ug/L
• We can find K by plotting
– CSS-Ct versus time on semi-log
paper…but first lets find CSs-Ct
Data (X0 = 100 mg)
Time (h) [Drug X] (ug/L)
(CSs-Ct)/CSS
0
1
3
6
0
23.7
57.3
85.0
(106.5-0)/106.5 = 1
(106.5-23.7)/106.5 = 0.78
(106.5-57.3)/106.5 = 0.46
(106.5-85.0)/106.5 = 0.20
10
12
14
100.5
104.9
106.8
(106.5-100.5)/106.5 = 0.06
~0
~0
15
108.9
~0
Step 2) Find t1/2
~1
(Css-C)/Css
1
At the intersection,
draw
at line
down
to
Yourof
Half
1 is the
0.5. time
1/2
Start
with
(Cssthe
X-axis
(time).
you
just
read
(2.5
Draw a line from
0)/Css
which
you
Read
the
value
the
hours)
0.5 across until it
know
1
line
intersects
intersects yourthe
axis…
best fit line
0.5
t1/2 = 2.5
0.1
0.01
0
5
10
Time (h)
15
Step 2) Find t1/2 (and k)
• Half-life (t1/2) from the graph is 2.5
hours. We can find k by the
following equation:
0.693 0.693
k

 0.277 1/h
t1
2.5 h
2
Step 3) Find Cl
• Clearance (Cl) can be calculated from
the steady-state concentration (Css)
and the infusion rate (k0) using the
equation:
CSS
• Rearranged to:
k0

Cl
k0
Cl 
C ss
Step 3) Find Cl
• We know the dose (100 mg) and
infusion time (T=15 h), therefore
infusion rate is:
X 0 100 mg
k0 

 6.67 mg/h
T
15 h
Step 3) Find Cl
• We can obtain Css from the graph as
we just did and had a value of 106.5
ug/L
Step 3) Find Cl
• We have k0 (6.67 mg/h), we have CSS
(110 ug/L), now we can calculate Cl
k 0 6670 ug/h
Cl 
 62.6 L/h
C ss 106.5 ug/L
Step 4) Find V
• Volume (V) can be calculated from k
(Step 3) and Cl (Step 1) and using the
following equation:
Cl  k  V
• Rearrange and solve for V
Cl 62.7 L/h
V

 226 L
k 0.277 1/h
Summary
• Cl = 62.6 L/h
• V = 226 L
• t1/2 = 2.5 h
Return to Table of Contents
Lesson Done!