Transcript C 2

Module 2
#3
Pharmacokinetics
2004-2005
• how to make sense of the previous lesson
or
• why do I have to learn this stuff?
2004-2005
drug dosing calculations
Concepts:
target plasma concentration
therapeutic window
15
Cp (free)
approximates [D]
plasma conc. (mg/L)
After distribution:
12
9
Cptox
6
Cpeff
3
0
0
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10
time (hr)
20
30
• Therapeutic window: Difference between the
minimum effective concentrations (MEC) required
for a desired response and one that produces an
adverse effect.
• For some drugs it is small (only two- to three-fold
difference)
• E.g. digoxin, theophylline, lidocaine,
aminoglycosides, cyclosporine, anticonvulsants.
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the loading dose
If we know the target plasma concentration, and
the Vd, we can calculate the i.v.loading dose:
L = Cp.Vd
For the oral loading dose we have to take the
fraction bioavailable into account (0-1)
L = Cp.Vd/F
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Loading Dose (DL) = a dose of drug sufficient to
produce a plasma concentration of drug that would
fall within the therapeutic window after only one or
very few doses over a very short interval. It is larger
than the dose rate needed to maintain the
concentration within the window and would produce
toxic concentrations if given in repeated doses.
Information needed to calculate a DL
• Volume of distribution, usually in L or L/Kg - from literature For example: Goodman & Gilman’ Pharmacology,
2001, Table A-II-1
• Desired concentration - from literature
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Loading dose
• DL = target Cp Vd/ F
e.g. lidocaine t1/2 = 1-2 h
Post MI arrythmias – life threatening – can’t wait 4-8
h – DL is used
• Another example: Digoxin for heart failure
• If only maintenanace dose is given it takes 10 days
t1/2 = 61 h
DL = 1.5 ng/ml x 580 / 0.7 = 1243 micg ~ 1 mg
Can be given iv divided – 0.5 mg, aft 6-8 h 0.25 mg,
aft 6h 0.125 mg and then 0.125 mg (to avoid
overdigitalization and toxicity)
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Volume of distribution (Vd)
Drug with low Vd
Drug with high Vd
high tissue
binding
So now if we want an equation,
W
C = -------V
or
W
=
C
V
i.e
D
=
Cp
Vd
where C = concentration (plasma), W = weight of drug
or dose of drug, and V = volume of solution or vol. of
distribution
Reference books say the Vd of theophylline
is 0.5 L per kg of body weight.
What is the IV loading dose to give a serum
theophylline concentration of 12 mg/L in
a 62 kg man?
W
= C V
or
LD = Cp Vd
LD = (12 mg/L) (0.5 L/kg) (62 kg)
= 372 mg
we usually use mcg/mL (or ug/mL), which is the same thing
as mg/L
Actually, theophylline for injection is available only as Aminophylline
(theophylline+ethylenediamine), which contains
80% theophylline. In this problem, we need
to give a larger weight of Aminophylline to
get the same weight of theophylline.
372 mg Theo
0.80
= 465 mg Amino
This gives us a loading dose
Half-life ( t 1/2 )
time interval after which the concentration
is half that at the beginning of the time
interval
has real-life meaning only in first-order
kinetics!
Cp (mg/L)
iv bolus
half life (t1/2)
100
slope=-kel (k)
10
2
4
6
8
10
time (hr)
the time taken for the Cp to fall by half
mathematically after i.v. bolus: Cpt = Cp0e-kt
where k is the rate constant (in hrs-1) so:
ln Cpt = ln Cp0 -kt
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when Cpt/ Cp0 = 1/2 = t1/2 = ln2/k = 0.693/k
most drugs follow first order kinetics:
rate of change of drug in the body
= -kel.Amount of
Drug in the body
.
where kel (k) is the rate constant of elimination
a few drugs (e.g. ethanol) follow 0 order kinetics:
rate of change = k
a few drugs (e.g. phenytoin) follow 1st order kinetics:
at low conc. and 0 order at high conc.
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Rate processes (mainly elimination)
zero-order
constant amount of drug eliminated per
unit of time
elimination rate constant has units of weight/time:
Ke = 50 mg/h, for example
typical drug: ethanol
Rate processes (mainly elimination)
first-order
constant fraction or percent of drug eliminated per
unit of time
elimination rate constant has units of weight/time:
Ke = 0.25/h or 0.25 h-1 , for example.
typical drug: theophylline, most other drugs
first-order elimination following
single IV bolus dose
log
Cp
Cp
Time
Time
Special properties of first-order kinetics
Cp is 90% of the way to new steady-state
level after change of dose in 3 t1/2 ,
96% in 5 t1/2
this is true regardless of
initial concentration
final concentration
dose
but only for first-order kinetics !
First-order is also known as linear
kinetics:
there is a linear relationship between
dose and steady-state concentration
Cpss
Dose
A patient’s peak serum phenobarbital level is
12.0 ug/mL at steady state on a dose of 60 mg/day.
If we want a level of 20 ug/mL, what new dose
should we order if we know that phenobarbital
has linear kinetics?
12.0 ug/mL
20 ug/mL
X = 100 mg/day
=
60 mg/day
X
If we change the dose to 90 mg/day, when can
we get another plasma level to check our calculations?
In adults, the half-life averages 100 hours. In
3 half-lives (300 h, or 12.5 days) we will be 90%
of the way from 12.2 ug/mL to the new steadystate level (we hope it’s 20 ug/mL). This is close
enough to get a level. Reschedule the patient
to come back in 2 weeks for this level.
change
dose
level
not OK
Cp
Time
level OK
clearance
defined as the volume of blood from which drug is
irreversibly removed per unit time.
ml/min
can calculate: UV/P = excretion rate/plasma concentration
at steady state: rate of excretion = rate in (the dose, M)
M = rate of excretion = Cldrug. Cp
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Clearance
 Clearance is the VOLUME of plasma that
can be freed of drug per unit time, i.e., gives
estimate of function of organs of elimination
and rate of removal of drug from the body
•
Rate of Elimination (mg/hr)
CL = -------------------------------- = vol/time
Concentration (mg/L)
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Clearance calculations
From renal physiology:
CL = (Ux * V) / Px
L/h = ( mg/L * L/h) / mg/L
Where:
Ux = urine concentration of "x" (mg/L)
V = urine flow rate (L/h)
Px = Systemic venous plasma concn. of "x"
(mg/L)
CL = renal clearance (L/hr)
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Clearance
• Whole body clearance is simply the sum of
all organ clearances
CL(body) = CL(renal) + CL(hepatic) +
CL(pulmonary) + CL(etc)
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relation between Cl, k and Vd
• the rate of drug removal (drug out) depends on the
amount of drug in the body and the rate constant of
excretion (kel, k)
•drug out = k*Amount of drug in the body (A)
• A = Cp*Vd
• drug out = k*Cp*Vd
• but drug out = Cl*Cp
• Cl*Cp = k*Cp*Vd
• therefore: Cl = k*Vd
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plasma conc. (mg/L)
calculating clearance
t1/2 = 0.693 Vd
Cl
100
Cldrug =
80
Dose
AUC
60
40
20
0
0
10
20
30
time (hr)
Cldrug = k.Vd = (0.693/t1/2 ).Vd
Cltotal = Cl renal + Clnon-renal
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t1/2 = 0.693
k
or
k = 0.693
t1/2
Relationship between CL(body) and Ke
oWhole body clearance and the elimination rate constant are
related, but emphasize different aspects of the same processes.
oElimination Rate constant (Ke) is the FRACTION of drug in
the body that is eliminated each unit of time, e.g., "fraction per
hour".
oCalculate Ke from CL & Vd: If one knows the Vd of a drug
and the CL (body), one can calculate the FRACTION of the drug
in the body that is removed per unit time. This is the same fraction
as the fraction of "plasma equivalent" that is completely cleared
of drug per unit time. This fraction is the Ke.
Ke = CL(body) / Vd (fraction/hour)
CL(body) = Ke x Vd (volume/hour)
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the maintenance dose
•Maintenance Dose (DM) = The dose needed to
maintain the concentration within the therapeutic
window when given repeatedly at a constant
interval
M = Cldrug.Cp
for oral dosing
M = Cldrug.Cp/F
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Why clearance and Vd?
• clearance and volume of distribution are
independent variables.
• they determine the half-life (Cl = k*Vd)
t1/2 = 0.693
k
or
k = 0.693
t1/2
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t1/2 = 0.693 Vd
Cl
plasma conc. (mg/L)
bioavailability
absolute bioavailability = F
150
iv
= AUCoral / AUCi.v.
100
= a fraction between 0 and 1
50
po
0
0
5
10
15
time (hrs)
Some drugs with low "F" in humans (F < .5) - Alprenolol ,
5-fluorouracil, Lidocaine, Morphine, Nifedipine,
Propranolol, Salicylamide
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oral dosing
the time taken
to reach
steady state
depends only
on t1/2
(= 4-5 x t1/2)
M (average Cp) = Cl.Cp/F
for drugs with short t1/2, must dose
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frequently
Css = F.dose
CL.T
Time to steady-state or elimination is
independent of dosage
Time required to reach steady state
Depends on elimination rate
Requires 5 elimination half-lives to reach 97% of steady state
Requires 5 elimination half-lives to eliminate 97% of drug
number of half-lives
0
100
500
1
50
250
2
3
25
12.5
125
62.5
4
6.25
31.25
5
3.125
15.625
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Drug
remaining
summary
• drug dosing should aim for the target plasma
concentration.
• the volume of distribution is useful in calculating the
loading dose
• the clearance is useful in calculating the maintenance
dose
• the time to reach steady state depends only on the
half life
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how can a drug have >100%
bioavailability? (F>1.0)
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About the linear model....
It is based on the exponential function
y = ex
where e is the base of natural logarithms ( ln )
on your calculator, e x is
ex
or
inv
ln x
C1 = C2 e kt
C1/C2 =
2/1 =
e kt
e kt
ekt = 2
kt = 0.693
t = 0.693 / k
log
Cp
( C1, t1 )
t
Time
C2 = C1 e - kt
( C2, t2 )
A patient’s acetaminophen level at 2:00 PM is
86.2 ug/mL and at 6:00 the same day is 27.8
ug/mL. If we know that ke of acetaminophen
in this patient is 0.283 h -1 , what was the level
at 10:00 AM?
log
Cp
( C1, t1 )
( C2, t2 )
t
Time
C1 = C2 e kt
C1 = 86.2 e ( 0.283 ) ( 4 )
C1 = 267.3 ug/mL
How did we know that ke of acetaminophen
in this patient is 0.283 h -1 ?
Either of the two previous equations can also be
stated as
ln
ke
=
C1
C2
t2 - t1
If the patient’s acetaminophen level at 2:00 PM (1400)
is 86.2 ug/mL and at 6:00 (1800) the same day
is 27.8 ug/mL,
ln
ke
=
C2
t2 - t1
ln
ke
C1
86.2 ug/mL
27.8 ug/mL
( 1800 - 1400 ) h
ke
=
0.283 h - 1
What is the half-life of acetaminophen in this
patient?
One last equation:
t 1/2 =
So
0.693
t 1/2 =
0.283 h - 1
t 1/2 =
2.45 h
0.693
ke
What good is all this?
Well, we started with the bottom line
and worked backward so that the importance
of each equation might be clear.
The next slide is the problem
as it would occur in real life. Try working
through it toward the bottom line, starting
with information existing at the start of the
case.
A patient is admitted following a suicide attempt in
which he took an unknown quantity of acetaminophen.
After appropriate gastric lavage and supportive therapy, he is being considered for therapy with the specific antidote (N-acetylcysteine). This antidote is recommended if the serum acetaminophen concentration
4 h after the ingestion is > 200 ug/mL. The overdose
is reported to have occurred at 10 AM (1000); at
2:00 PM (1400) his serum acetaminophen level was
86.2 ug/mL and at 6:00 (1800) the same day it is
27.8 ug/mL. Should he receive N-acetylcysteine? Is
this patient’s half-life consistent with the population
average (2.5 h) or is there something unusual about
his metabolism? Work it out.
A patient who is taking 400 mg of carbamazepine
every 12 hours has a trough serum carbamazepine
level of 4.2 ug/mL. Her seizure control is
inadequate; if we want to achieve a serum drug
level of 8.0 ug/mL, what should the new dose be?
We know the drug usually has linear kinetics.
Hint: we usually use trough levels with this drug.
With linear kinetics, the trough levels are proportional to the dose just like peak levels are.
761 mg every 12 h; you would probably do this
increase in two steps a week apart.
A patient weighs 88 pounds. He needs a
loading dose of phenytoin for status epilepticus.
If the average Vd is 0.65 L/kg, what is the correct
loading dose? (Although this drug has nonlinear
kinetics, this fact only influences accumulation
during repeated dosing; it has no effect on loading
doses.) The therapeutic range of this drug is
10-20 ug/mL, and we would like to achieve a level
of 12 ug/mL.
312 mg - Did you forget to convert pounds
to kilograms?
A 76-kg patient has received an intravenous bolus
of 400 mg of a drug with the following population
averages: Vd 2.3 L/kg, t 1/2 12 h, minimum effective
concentration 1 ug/mL. How long will it be before
another dose is needed? Assume linear kinetics.
(Hint: first calculate ke , then calculate the drug
concentration after the bolus dose. Then use the
C2 equation and try various values for the elapsed
time to see the largest value that keeps C2 above
1.0 ug/mL. [If you like math, you may want to solve
this equation for t ; first take the logarithm of both
sides and then solve.]
I get about 14 hours.
First-order: steady state following
repeated IV bolus dosing
all peaks same
Cp
all troughs same
Time