EQUATIONS AND CALCULATIONS ONECOMPARTMENT MODEL EQUATIONS FOR LINEAR PHARMACOKINETICS Intravenous Bolus Equation t time after the intravenous bolus was given t 0 at the time the dose was administered ID: 933829
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Slide1
CLINICAL PHARMACOKINETIC
EQUATIONS AND CALCULATIONS
Slide2ONE-COMPARTMENT MODEL EQUATIONS FOR
LINEAR PHARMACOKINETICS
Slide3Intravenous
Bolus
Equation
t= time
after the intravenous bolus was given (t = 0 at the time the dose was administered), C is the concentration at time = t, V is the volume of distribution, ke is the elimination rate constant.
C = (
D/V)e
−
ket
Slide41) A patient is given a theophylline loading
dose of 400 mg
intravenously over
20 minutes
, volume of distribution is 30 L, the elimination rate constant equals 0.115 h-1. compute the expected theophylline concentration 4 hours after the dose was given.t1/2 = 0.693/ke = 0.693/0.115 h-1 = 6 hC = (D/V)e−ket = (400 mg/30 L)e −(0.115 h)(4 h)
=
8.4
mg/L
Slide5Distribution phase & Elimination phase
Slide62) A patient
is given an intravenous dose of vancomycin 1000 mg. the volume of distribution equals 50 L, the elimination rate constant is 0.077
h
−
1, calculate the expected vancomycin concentration 12 hours after the dose was given.t1/2 = 0.693/ke = 0.693/0.077 h−1 = 9 hC = (D/V)e −ket
=
(1000 mg / 50 L
)
e
−(0.077 h−1)(12 h)
=
7.9
mg/L
Slide7Pharmacokinetic
parameters for patients can also be computed for use in the equations.
The elimination
rate constant, half-life and volume of distribution can be calculated
Slide83) 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
t1/2
is determined by measuring the timeneeded for serum concentrations to decline by 1/2 2. ke = 0.693/t1/2 = 0.693/4d = 0.173d−13. The concentration/time line can be extrapolated to the concentration axis to derive the concentration at time zero (C0 = 15 mg/L) volume of distribution
V = D/C0
=
600 mg
/ (
15 mg/L
)
= 40
L
Slide9k
e
= −(ln C
1 − ln C2)/(t1 − t2)t1 and C1 are the first time/concentration pair t2 and C2 are the second time/concentration pairke=−[ln(12.6 mg/L) − ln(7.5 mg/L)]/(1 d − 4 d) = 0.173 d−1t1/2 = 0.693/ke = 0.693/0.173
d
−1
= 4
d
C
0
= C/e
−
ket
= (12.6 mg / L) / e−(0.173
d−1)(1 d) = 15.0 mg/LV = D/C0 = 600 mg / (15 mg/L) = 40 L
Alternatively
Slide10Continuous and Intermittent Intravenous Infusion Equations
Slide11If infusion is running
C = (k
0
/Cl)(1-e
-ket) = [k0/(keV)](1 – e-ket) k0 is the drug infusion rate (amount per unit time,
mg/h
or
μg
/min),
Cl
=
k
e
V
, this substitution was made in the second version of the equation),
ke
is the
elimination rate
constant
,
t is
the time that the infusion has been running.
Css
= k0 / Cl = k0 / (keV)
C = (k
0
/Cl)(1-e
-ket
)
Slide12If the infusion is
stopped
C
postinfusion
= Cende−ketpostinfusiont post infusion is the post infusion time (t post infusion = 0 at end of infusion and increases from that point).
Slide134) A
patient is administered
60
mg/h of
theophylline, V = 40 L and ke = 0.139 h−1. calculate The serum concentration of theophylline in this patient after receiving the drug for 8 hours and at steady state.C = [k0/(keV)](1 − e−ket
)
= [(60 mg/h)/(0.139
h
−
1
⋅ 40 L)](1 −
e
−(0.139 h−1)(8 h
)
)
= 7.2
mg/L
Css
= k
0
/(
ke
V) = (60 mg/h)/(0.139 h−1 ⋅ 40 L) = 10.8 mg/L
Slide14Continue…
If the
infusion only ran for 8 hours,
C
postinfusion = Cend e −ke tpostinfusion
=
(7.2 mg/L
) e
−(0.139 h−1)(6 h
)
=
3.1
mg/L
If the
infusion ran
until steady state was achieved
,
Cpostinfusion
=
C
end
e
−ke tpostinfusion = (10.8 mg/L) e−(0.139 h−1)(6 h) = 4.7 mg/L
compute the theophylline serum concentration 6 hours after the infusion stopped
Slide155) A
patient was given a single 120-mg dose of tobramycin as a 60-minute infusion, and concentrations at the end of infusion (6.2 mg/L) and 4 hours after the infusion ended (1.6 mg/L) were obtained.
1-
t
1/2 can be determined by measuring the time it takes for serum concentrations to decline by one-half = 2hr2- ke = 0.693/t1/2 = 0.693/2 h = 0.347 h−1
Slide16Or without a plot
1- k
e
= −(ln C1− ln C2)/(t1 − t2) = −[ln (6.2 mg/L) − ln (1.6 mg/L)] / (1 h − 5 h) =0.339 h
−1
2- t
1/2
= 0.693/
k
e
= 0.693/0.339
h
−
1
= 2
h
Slide17The
volume of distribution (V)
Can
be computed using the following
equations: At steady state and when we know C0 V= Dose / C0 or V= Cl/K in I.V infusion or before steady state[k0(1-e
-ket’
)] / k
e
[
Cmax
-(
Cpredose
e
-
ket
’
)]
V= Dose / C
0
or V= Cl/K
[k0(1-e
-ket’)] / ke[Cmax-(
Cpredose
e
-
ket
’
)]
Slide18Extravascular Equation
The absorption rate
constant describes
how quickly drug is absorbed with a large number indicating fast absorption
and a small number indicating slow absorption
Slide19Extravascular Equation
C = {(
FkaD
) / [V(
ka − ke)]}(e−ket − e−kat) F is the bioavailability fraction, ka
is the absorption rate constant,
D
is the dose
Slide20example
500 mg
of
oral procainamide -----
Da half-life equal to 4 hours ----- t1/2an elimination rate constant of 0.173 h−1 ---- kevolume of distribution of 175 L -----
V
absorption
rate constant equal to 2
h
−
1
------
k
a
oral
bioavailability fraction
of 0.85 ----- F
C = {(
Fk
a
D
) / [V(
ka − ke)]}(e
−
ket
− e
−
kat
)
Slide21After the end of absorption phase the C can be calculated by equation of I.V bolus
C
= [(FD)/
V]e −ket C is the concentration at any post absorption, post distribution time
Slide22The hybrid volume
of distribution/bioavailability (V/F
) parameter
Since
volume of distribution relate the dose given with the obtained concentration and since in extravascular route not all the dose enter the blood stream so we use (V/F) to indicate the value of volume of distribution V/F = D/C0…………….or V = D/ [C0 − Cpredose], if not first dose C0 = C/e −
ket
K
e
= − (ln C
1
− ln C
2
) / (t
1
− t
2
)
Slide23After
graphing the serum
concentration/time data----
k
e = 0.693 / t1/2 = 0.693/14 h----At time = 0 ----- hybrid volume of distribution/bioavailability V/F = D/C0 = 750 mg/70 L = 10.7 L
t
1/2
=14 hours
(C
0
= 70 mg/L)
k
e
= 0.0495
h
−
1
6)
An
oral dose of valproic acid 750 mg as capsules. 6 and 24 hours after the dose, the valproic acid serum concentrations are 51.9 mg/L and 21.3 mg/L, respectively.
Slide24Slide25ke
= −(ln C
1
− ln C
2)/(t1 − t2)t1/2 = 0.693/keC0 = C/e −ketV = D/C0
Alternatively
Slide26Multiple-Dose and Steady-State Equations
change
a single dose equation to the multiple
dose version
, multiply each exponential term in the equation by the multiple dosing factor: n is the number of doses administered, ki is the rate constant, τ
is the
dosage interval
.
(1 −
e
−
nk
i
τ
)/(1 − e
−
k
i
τ
)
Slide27At steady state
number
of doses (n) is large,
the
exponential term in the numerator of the multiple dosing factor (−nkiτ) becomes a large negative number-------exponent approaches zero. Therefore, the steady-state version of the multiple dosing factorbecomes the following:1/(1 − e
−
k
i
τ
)
Slide28Multiple-Dose and Steady-State Equations
C = (
D/V)e
−
ketC = (D/V)[e−ket/ (1 − e −keτ)]
Slide29To be
continued….