PK ( WOLDI.pdf11111111111111111122222233s

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About This Presentation

Pk


Slide Content

Part II
Pharmacokinetics
1/18/20251
ELIAS S. (BA. ACC. , B.Pharm.)
Elias s. ( BA ACC. , BPHARM )

Session objectives
At the end of this session , you are able to
➢Define pharmacokinetics (pk)
➢Discuss the areas & applications of pk
➢Explain kinetic homogeneity
➢Compare zero order and 1st order kinetics
➢Define different pk models
➢Analyse one & two compartment open models
➢Explain physiologic model
➢Differentiate compartmental models and physiologic models
➢Clinical pharmacokinetics
1/18/20252
Elias s. ( BA ACC. , BPHARM )

1. Introduction
1.1 Definitions
Pharmacokinetics: derived form two Greek words:
Pharmakon →drug
Kinesis → motion or change of rate
The study and characterization of the time course of drug
[ADME]
Explores the quantitative relationship between Absorption,
Distribution, Metabolism, and Excretion.
1/18/20253
Elias s. ( BA ACC. , BPHARM )

Schematic Presentation of ADME
1/18/20254
Elias s. ( BA ACC. , BPHARM )

Cont…
Clinical Pharmacokinetics:
The application of p’kprinciples to the safe and effective
therapeutic management of drugs in an individual
patient
Goals
➢enhancing efficacy
➢decreasing toxicity
▪Important?
70 to 80% of ADRs are dose related
▪Scope
Drug dosing in special
population
TDM
1/18/20255
Elias s. ( BA ACC. , BPHARM )

Cont…
Basic pharmacokinetics
➢Determination of optimum dose & dosing regimens in
drug/formulation development
➢General population is considered
Population pharmacokinetics
➢Reasoning of differences in therapeutic response of
drugs in various population groups
➢Age, gender, genetic, ethnic, etc
Toxicokinetics
➢Drug safety evaluation
➢Validating dose-related toxicities
1/18/20256
Elias s. ( BA ACC. , BPHARM )

1.2 Kinetic Homogeneity
Drug effect is often related to its conc. at the
site of action/receptors
Receptor sites
widely distributed
inaccessible
Direct measurement of [drug] at the site of
action is not practical
[drug] in dynamic equilibrium b/n plasma
and site of action
Blood is the fluid most often sampled for
[drug] determination
1/18/20257
Elias s. ( BA ACC. , BPHARM )

Cont…
Predictable r/p b/n [drug]in
plasma and at the receptor site
Changes in the plasma [drug]
reflect changes in [drug]in other
tissues
As [drug]in plasma increases,
the [drug]in most tissues will
increase proportionally
Major assumption in clinical p’k
Not true for some drugs
1/18/20258
Elias s. ( BA ACC. , BPHARM )

1.3 Rate & order of kinetics
Rate
The speed with which the reaction/process occurs
Drug A → Drug B
Rate = -dA/dt = +dB/dt
Order
How conc. or amount of the drug influences the rate of a
process
Zero order, first order, ….
1/18/20259
Elias s. ( BA ACC. , BPHARM )

Cont…
❖Zero-order kinetics
Rate is independent of the remaining conc/amount
dA/dt = -K
o
K
o
✓The zero-order rate constant
✓Expressed in units of mass/time (eg, mg/min)
Integration
A = -K
ot + A
o
✓A
0 &
A are the amount of drug at t = 0 & time t
▪Some kinetic processes in the body follow zero order
1/18/202510
Elias s. ( BA ACC. , BPHARM )

Cont…
Graph of A versus t yields a
straight line
The y intercept is equal to A
0,
and the slope of the line is equal
to -k
0
Zero-Order Half-Life (t
1/2)
t
1/2 = 0.5A
o/K
o
1/18/202511
Elias s. ( BA ACC. , BPHARM )

Cont…
In terms of drug concentration, which can be
measured directly, the above equation can be expressed
as:
C
0 is the drug concentration at time 0, C is the drug
concentration at time t, and k
0 is the zero-order
decomposition constant.
1/18/202512
Elias s. ( BA ACC. , BPHARM )

Cont..
Example: A pharmacist weighs exactly 10 g of a drug and dissolves
it in 100 mL of water. The solution is kept at room temperature,
and samples are removed periodically and assayed for the drug. The
pharmacist obtains the following data:
Drug Concentration (mg/mL)Time (hr)
100 0
95 2
90 4
85 6
80 8
75 10
70 12
Q:
Order of
reaction?
Reaction
constant?
1/18/202513
Elias s. ( BA ACC. , BPHARM )

Cont…
If C
0 = concentration of 100mg/mL at t=0 And C =
concentration of 90 mg/mL at t=4hr
Then: 90 =- k
04+100
K
0 = 2.5 mg/mL hr
Careful examination of the data will also show that the
concentration of drug declines 5 mg/mL for each 2-hour
interval.
Therefore, the zero-order rate constant may be obtained by dividing 5
mg/mL by 2 hours:
What was the half life of the drug?
1/18/202514
Elias s. ( BA ACC. , BPHARM )

❖First-order kinetics
Amount of drug A is decreasing at a rate that is
proportional to the amount of drug A remaining
dA/dt = -KA; dC/dt = -kC
K : first-order rate constant; units of time
–1
(eg, hr
–1
)
Integration:
lnA = -Kt + lnA
o; log A = - (Kt/2.3) + logA
o
lnC = lnC
o – kt; log C = log C
o - (Kt/2.3)
Most kinetic processes in the body follow 1
st
- order
1/18/202515
Elias s. ( BA ACC. , BPHARM )

Cont…
A graph of log A versus t
✓straight line
✓y-intercept = log A
0
✓slope= –k/2.3
First order t
1/2
t
1/2 = 0.693/k
Constant, independent of A
O
1/18/202516
Elias s. ( BA ACC. , BPHARM )

Examples
1. A solution of a drug was freshly prepared at a concentration of
300 mg/mL. After 30 days at 25°C, the drug concentration in
the solution was 75 mg/mL.
a)Assuming first-order kinetics, when will the drug decline to
one-half of the original concentration?
b)Assuming zero-order kinetics, when will the drug decline to
one-half of the original concentration?
1/18/202517
Elias s. ( BA ACC. , BPHARM )

Cont…
2. Assume first-order kinetics, how many half-lives
(t
1/2) would it take for 99.9% of any initial
concentration of a drug to be decomposed?
1/18/202518
Elias s. ( BA ACC. , BPHARM )

Comparisons
First Order kinetics
✓[drug] decreases
exponentially w/ time
✓Rate is proportional to
[drug]
✓Plot of log [drug] or ln[drug]
vs. time are linear
✓t
1/2 is constant regardless of
[drug]
Zero Order kinetics
✓[drug] decreases linearly
with time
✓Rate is constant
✓Rate is independent of
[drug]
✓Plot of [drug] vs. time is
linear
✓t
1/2 is not constant,
depends on [drug]
1/18/202519
Elias s. ( BA ACC. , BPHARM )

2. PHARMACOKINETIC
MODELS
1/18/202520
Elias s. ( BA ACC. , BPHARM )

Introduction
The handling of a drug by the body can be very complex
Several processes (ADME) work to alter [drug] in tissues and
fluids
ADME often happen simultaneously
Simplifications of body processes are necessary to
predict a drug's behavior in the body
One way to make these simplifications is to apply
mathematical principles to the various processes.
Pharmacokinetic models are used
1/18/202521
Elias s. ( BA ACC. , BPHARM )

Cont…
P’K models
•Simulation of the rate processes of the drug in the body
•Mimic closely the physiologic processes in the body
✓But seldom consider all the rate processes in the body
Hypothesis using mathematical terms describing the
quantitative r/p b/n drug concentration & time so as to
estimate p’k parameters of the drug
✓Simplified mathematical expressions
1/18/202522
Elias s. ( BA ACC. , BPHARM )

Cont…
Used to develop equations which describe p’k processes
Predict [drug] in the body as function of time
Express quantitative relationship between various p’k
parameters (calculate the parameters)
Two types of pharmacokinetic models
Compartmental models
Physiologic models /perfusion models
1/18/202523
Elias s. ( BA ACC. , BPHARM )

I. Compartmental Models
Compartment
An entity which can be described by a definite volume &
concentration
Simulation of group of tissues/organs which have similar
perfusion and drug affinity
It is not a real anatomic or physiologic region of the body
Body parts are grouped as highly perfused, poorly perfused and
negligible perfused to form one or more compartments
1/18/202524
Elias s. ( BA ACC. , BPHARM )

Cont…
A. Highly perfused tissue group
There is high blood flow to these tissues
Lung, brain, Heart, liver, kidney, endocrine glands
Drug concentration rapidly equilibrates with blood
B. Poorly perfused tissue group
Poor blood flow to the tissues.
Muscle, skin, fatty tissue, bone marrow
C. Negligible perfusion tissue group
Negligible blood supply to the tissues
Bone, nail, hair, ligaments, tendons, cartilages, teeth
1/18/202525
Elias s. ( BA ACC. , BPHARM )

Assumptions and Simplifications of compartmental model:
1.Body is series of compartments joined reversibly together
➢Drug moves to- and from- each compartment
2.Drug is administered to the central compartment (CC)
3.Elimination is from the CC
4.Distribution of drug in a compartment is rapid and uniform
5.Each drug molecule in a compartment has equal probability of leaving
the compartment at any time
6.All rate processes follow first order kinetics
Linear kinetics
1/18/202526
Elias s. ( BA ACC. , BPHARM )

Cont…
According to Compartmental Models
Amount of drug in the body is the sum of the amount of
drug in each compartment
Two types of compartmental models
Catenary model
Mammilary model
1/18/202527
Elias s. ( BA ACC. , BPHARM )

Cont…
1.Catenary Model
Compartments are joined together like compartments of a
train.
The drug cannot go to 3
rd
compartment without going to the
2
nd
compartment.
It doesn’t apply to the way most organs are connected to the
plasma
It is rarely used (no universal acceptance)
1/18/202528
Elias s. ( BA ACC. , BPHARM )

Cont…
2. Mammilary Model
It is the most commonly used p’k model
Each peripheral compartment (PC) is reversibly joined to the CC
➢Like the connection between plasma and other body organs
CC: Plasma and highly perfused organs
1/18/202529
Elias s. ( BA ACC. , BPHARM )

Cont…
Each PC is independently & reversibly connected to the CC
Drug is introduced to the CC and then goes to the PC
No, one or more PCs
three types of compartment models in pharmacokinetics:
▪One-compartment model
▪Two-compartment model
▪Multicompartmentmodel
▪Open→ the drug can enter and leave the body
1/18/202530
Elias s. ( BA ACC. , BPHARM )

i. One-compartment Open Model
Simplest and mostly used
The body is considered as a single
well-mixed container
The drug distributes (equilibrates)
instantly b/n blood and other body
tissues/fluids
Equilibrium (steady state) is reached
rapidly
1/18/202531
Elias s. ( BA ACC. , BPHARM )

ii. Two compartment open model
The body is viewed as two
compartments
The central compartments
➢consists of the plasma and tissues
that take up the drug so rapidly that
distribution can be considered to be
instantaneous.
The peripheral compartments
➢consists of tissues that up take the
drug at slower rate than tissues in the
central compartment
1/18/202532
Elias s. ( BA ACC. , BPHARM )

iii. Three compartment open model
An extension of the two compartment model, where a sizable
amount of the drug distributes to certain very poorly perfused
tissues, such as fat and bone, at an extremely slow rate.
The three compartment model has three group of tissues:
▪Central compartment tissues
▪Peripheral tissues
▪Deep tissue compartment
1/18/202533
Elias s. ( BA ACC. , BPHARM )

II. Physiologic models
Physiologic pharmacokinetic models, also known as blood flow
or perfusion models
➢ pharmacokinetic models based on known anatomic and
physiologic data.
 These models kinetically describe the data with the
consideration that the blood flow is responsible for distributing
a drug to various parts of the body.
 Uptake of a drug into organs is determined by binding of the
drug in these tissues.
In contrast to the tissue volume of distribution, the actual
tissue volume is used.
1/18/202534
Elias s. ( BA ACC. , BPHARM )

Cont…
Because there are many tissue organs in the body, each tissue
volume must be estimated and its drug concentration
described.
The model would potentially predict a realistic tissue
drug concentration, which a compartment model
fails to do.
 Unfortunately, much of the information that is required for
adequately describing a physiologic model is experimentally
difficult to obtain.
 In spite of the limitation, a physiologic pharmacokinetic
model does provide a much better insight of how physiologic
factors may change drug distribution from one animal species
to another.
1/18/202535
Elias s. ( BA ACC. , BPHARM )

Comparison
Compartmental models
Simple and flexible
Limited amount of data to
determine p’k parameters
Empirical/ lacking physiologic
relevance
Don’t consider
pathophysiologic changes
Plama conc. vs time
P’k parameters
Physiologic models
Complicated
Large amount of data
required
Real physiology and
anatomy
Can evaluate effect of
pathophysiologic changes
Drug distribution
Uptake & clearance by
organs
1/18/202536
Elias s. ( BA ACC. , BPHARM )

i. One-compartment Open Model
One-compartment Open Model used to estimate PK
parameters from:
➢Intravenous bolus administration
➢Continuous infusion
➢Extravascular administration
➢Multiple dosage regimens
1/18/202537
Elias s. ( BA ACC. , BPHARM )

❑One-compartment Model-
Intravenous Bolus Administration
Drug is injected all at once into a box, or compartment,
 The drug distributes instantaneously and
homogenously throughout the compartment.
Drug elimination also occurs from the compartment in
first order fashion.
1/18/202538
Elias s. ( BA ACC. , BPHARM )

Cont…
Consider a single IV bolus injection of drug X. As
time proceeds, the amount of drug in the body is:
dCp/dt = rate in (availability) – rate out
(elimination)
Since rate in or absorption is absent, equation
becomes

dCp/dt = - rate out
If rate out or elimination follows first order kinetic
1/18/202539
Elias s. ( BA ACC. , BPHARM )

Cont…pel
p
C .k
dt
dC
−=
Integrating this formula gives tk0
t
el
eCpCp

=
❑Where,
➢ C
pt is the concentration at
any time t
➢C
p
0
is the concentration at
time 0
➢ k is the elimination rate
constant
1/18/202540
Elias s. ( BA ACC. , BPHARM )

Cont…
Example: Ifweknowthattheplasmadrug
concentrationjustafteragentamicindoseis8mg/Land
thepatient'seliminationrateconstantis0.25hr
-1
,
calculate theconcentration8hourslater?
solution:
C=C
0e
-Kt
, C
0=8mg/L,K=0.25hr
-1
,andt=8hours

1/18/202541
C
at 8 hr= 8 mg/L X e
-0.25 hr-1(8 hr)
= 8 mg/L (0.718) = 6.23mg/L
Elias s. ( BA ACC. , BPHARM )

Cont…
NB: Theterme
-Kt
indicatesthefractionoftheinitial
doseofdrugthatremainsinthebodyattimet;
➢0.718(or71.8%)remainsinthebody8hoursafter
theinitialdoseinthisexample.
Conversely,theterm1-e
-Kt
[(1-0.718)= 28.2%] would
indicatethepercentorfractionexcretedaftertime(t).
1/18/202542
Elias s. ( BA ACC. , BPHARM )

Cont…
The above equation describes the single exponential
decline in drug concentration as a function of time.
This fall in plasma concentration is called mono-
exponential decay.
If we know kel and Cp
0
we could calculate Cp at any
time after a single IV bolus dose.
However, it still isn't very convenient for estimating a
value of kel from concentration versus time data.
1/18/202543
Elias s. ( BA ACC. , BPHARM )

Cont…
For estimation purposes it is preferable to use a straight
line equation.
A straight line equation can be achieved by taking the
natural logarithm of both side of Equationtk Cpln Cpln
el
0
t −= 2.303
tk
Cp logCp log
el0
t
−=
1/18/202544
Elias s. ( BA ACC. , BPHARM )

Cont…
Hence, Plotting ln(Cp) versus t should give a straight
line with a slope of - k
el and an intercept of ln(Cp
0
) (y =
mX + b with b = intercept and m = slope)
1/18/202545
Elias s. ( BA ACC. , BPHARM )

Con…
C
0canbedeterminedfromadirectmeasurementor
estimatedbyback-extrapolation
K is the slope
1/18/202546
Elias s. ( BA ACC. , BPHARM )

❑ Elimination rate constant
Elimination rate constant (k) represents the fraction of
drug removed per unit of time and has units of
reciprocal time (e.g., minute
-1
, hour
-1
, and day
-1
). ( )
21
21
lnln
tt
CC
k
el

−−
=
•where t
1 and C
1 are the first
time/concentration pair and
t
2 and C
2 are the second
time/concentration
1/18/202547
Elias s. ( BA ACC. , BPHARM )

❑ Half life
It isthetimerequired fortheconcentrationofdrugin
theplasmatodecreasebyhalf.
Consider:lnC=lnC
0-Kt
Bydefinition,atonehalf-life,theconcentration(C)at
thetime(t)ishalfofwhatitwasinitially (C
0).
Sowecansaythatatt=t1/2,C=1/2C
0.
1/18/202548
Elias s. ( BA ACC. , BPHARM )

Cont…elelel
oo
el
oo
1/2
k
0.693
k
ln2
k
/Cln2C
k
1/2CCln
t ==== 21el
00
tkCpln Cp
2
1ln −=
Example: Ifadoseofgentamicinisadministeredanda
peakplasmaconcentrationis6mg/Laftertheinfusionis
completedandis1.5mg/L4hourslater,calculate half
life of the drug?
1/18/202549
Elias s. ( BA ACC. , BPHARM )

Cont…
Solution: Firsttheeliminationrateconstant(K)is
calculatedasshownpreviously then Half-lifecanbe
calculatedfrom k.hr
k
hr
hr
Lmg
tt
CC
k
elo
o
el
2
693.0
then t348.0
4
39.1
04
/6ln5.1lnlnln
1/2
1
1
===−=


−=


−=
1/18/202550
Elias s. ( BA ACC. , BPHARM )

❑ Amount of Drug in the Body, D
B
Can not be determined directly
Less significance in clinical practice
Blood sample is removed at periodic intervals and analyzed
for [drug]
V
D relates [drug] in plasma (C
p) and the amount of drug in
the body (D
B)
1/18/202551
Elias s. ( BA ACC. , BPHARM )

❑Apparent volume of distribution V
D
The volume of distribution represents a volume that
must be considered in estimating the amount of drug in
the body from the concentration of drug found in the
sampling compartment.
Apparent: not real physiologic volume
Measure of extent of distribution
Relates C
P & D
B
1/18/202552
Elias s. ( BA ACC. , BPHARM )

Cont…
V
D for a given drug is constant
(unchanged)
V
D = D
B
0
/C
p
0
=

dose/C
p
0
C
p
0
: instantaneous [drug]
➢[drug] at t = 0
➢determined by extrapolation
Exercise : V
D
Exactly 1 g of a drug is dissolved in an unknown volume of
water. Upon assay, the concentration of this solution is 1
mg/mL. What is the original volume of this solution?
1/18/202553
Elias s. ( BA ACC. , BPHARM )

Cont…
Solution:
The original volume of the solution may be obtained
by the following proportion, as 1 g = 1000 mg:
1000mg/x mL = 1mg/mL
X = 1000mL
Therefore, the original volume was 1000 mL or 1 L.
1/18/202554
Elias s. ( BA ACC. , BPHARM )

Significance of the Apparent Volume of Distribution
The apparent volume of distribution is not a true
physiologic volume.
V
D may be:
Smaller than, or equal to, the body mass: for most
drugs
several times the body mass: for some drugs
➢For a given dose, a very small C
p
0
may occur in the
body due to high concentration of the drug in
peripheral tissues and organs.
For this dose, the small C
p
0
will result in a large V
D.
1/18/202555
Elias s. ( BA ACC. , BPHARM )

Cont…
If a drug is highly bound to plasma proteins or remains in the
vascular region, then C
p
0
will be higher, resulting in a smaller
apparent V
D.
Distribution coefficient, Δ’
V
D divided by body weight
Δ’ = V
D/BW …….ml/g, L/kg
Yields V
D for a particular patient (when multiplied by wt)
Used for calculation of individual doses and dosing intervals
►except in cases of obesity, edema, …..
1/18/202556
Elias s. ( BA ACC. , BPHARM )

❑ Drug Clearance, Cl
Total body clearance is an important pharmacokinetic
parameter that is often defined as the volume of
blood or plasma completely cleared of the drug
per time.
Fixed volume of fluid (containing the drug) cleared of drug
per unit of time
Unit ….. volume/time (eg, ml/min, L/hr)
1/18/202557
Elias s. ( BA ACC. , BPHARM )

Cont…
Describes drug elimination from the body without
identifying the mechanism of the process
Example, penicillin (Cl = 15 ml/min & V
D = 12L)
➢15 ml of the 12L will be cleared of drug per min
Sum of the clearances for each drug-eliminating organ
1/18/202558
Elias s. ( BA ACC. , BPHARM )

1/18/202559
Elias s. ( BA ACC. , BPHARM )

Cl vs elimination rate
Cl = elimination rate/C
P = (dD
E/dt)/C
P
dD
E/dt = rate of elimination
D
E = amount of drug eliminated
Elimination rate = Cl C
P
Cl is a constant
Elimination rate is not constant
1/18/202560
Elias s. ( BA ACC. , BPHARM )

Cl, k & V
D
dD
E/dt = KD
B = K C
p V
D
 k = 1
st
order elimination rate constant
D
B = amount of drug in the body
Substituting for elimination rate, Cl C
P
Cl = K C
pV
D/C
p = KV
D
1/18/202561
Elias s. ( BA ACC. , BPHARM )

Cl, V
D & t
1/2
t
1/2 can be estimated from Cl and V
D
Cl = K V
D and K = 0.693/t
1/2
Therefore, by substitution
Cl = V
D 0.693/t
1/2
t
1/2= 0.693V
D/Cl
1/18/202562
Elias s. ( BA ACC. , BPHARM )

Exercise-1:
A new drug was given in a single intravenous dose
of 200 mg to an 80-kg adult male patient.
After 6 hours, the plasma concentration of the drug
was 1.5 mg/100 mL of plasma.
i.Assuming that the apparent V
D is 10% of body
weight, compute the total amount of drug in the
body fluids after 6 hours.
ii.What is the half-life of this drug?
1/18/202563
Elias s. ( BA ACC. , BPHARM )

Exercise-2: Homework
A drug has an elimination half-life of 8 hours and
follows first-order elimination kinetics. If a single 600-
mg dose is given to an adult female patient (62 kg) by
rapid IV injection, what percent of the dose is
eliminated (lost) in 24 hours assuming the apparent V
D
is 400 mL/kg? What is the expected plasma drug
concentration (C
p) at 24 hours post dose?
1/18/202564
Elias s. ( BA ACC. , BPHARM )

❑ One compartment model-
Multiple IV bolus dosing
Outlines
•MDR-introduction
•C
Max, C
min,C
P
•Steady state

•Accumulation factor
•D
L & D
m
1/18/202565
Elias s. ( BA ACC. , BPHARM )

MDR-Introduction
Most drugs are given as MDR
Single-dose → C
p rises above and then falls below the MEC
➢Decline in therapeutic effect
MDR → prolonged therapeutic activity
In designing a MDR, the target C
p must be related to the
therapeutic response
►C
P must be maintained within the TW
p’k parameters are first obtained from single-dose studies
1/18/202566
Elias s. ( BA ACC. , BPHARM )

Cont…
Two main parameters that can be adjusted in MDR
1.The dose (D
o)
2.Dosing interval (t)
Frequency of administration
Superimposition is assumed
►Early doses don’t affect the p’k of subsequent doses
►C
P after the 2
nd
, 3
rd
or n
th
dose will overlay/superimpose
the C
P attained after the (n–1)
th
dose
►Allows prediction of C
P after multiple doses from a dose
1/18/202567
Elias s. ( BA ACC. , BPHARM )

Predicted Plasma Drug Concentrations for Multiple-Dose Regimen Using the Superposition Principle
Dose NumberTime (hr) Plasma Drug Concentration (g/mL)
Dose 1 Dose 2 Dose 3 Dose 4 Dose 5 Dose 6 Total
1 0 0 0
1 21.0 21.0
2 22.3 22.3
3 19.8 19.8
2 4 16.9 0 16.9
5 14.3 21.0 35.3
6 12.0 22.3 34.3
7 10.1 19.8 29.9
3 8 8.50 16.9 0 25.4
9 7.15 14.3 21.0 42.5
10 6.01 12.0 22.3 40.3
11 5.06 10.1 19.8 35.0
4 12 4.25 8.50 16.9 0 29.7
13 3.58 7.15 14.3 21.0 46.0
14 3.01 6.01 12.0 22.3 43.3
15 2.53 5.06 10.1 19.8 37.5
5 16 2.13 4.25 8.50 16.9 0 31.8
17 1.79 3.58 7.15 14.3 21.0 47.8
18 1.51 3.01 6.01 12.0 22.3 44.8
19 1.27 2.53 5.06 10.1 19.8 38.8
6 20 1.07 2.13 4.25 8.50 16.9 0 32.9
21 0.90 1.79 3.58 7.15 14.3 21.0 48.7
22 0.75 1.51 3.01 6.01 12.0 22.3 45.6
23 0.63 1.27 2.53 5.06 10.1 19.8 39.4
24 0.53 1.07 2.13 4.25 8.50 16.9 33.4
1/18/202568
Elias s. ( BA ACC. , BPHARM )

Cont…
Situations of superimposition principle does not
apply.
The pharmacokinetics of the drug change after multiple dosing
due to various factors, including:
➢Changing pathophysiology in the patient,
➢Saturation of a drug carrier system,
➢Enzyme induction, and
➢Enzyme inhibition.
1/18/202569
Elias s. ( BA ACC. , BPHARM )

C
max, C
min, C
p
Constant dosing interval, T
1
st
dose
C
t= C
0
e
-Kt
C
P at the end of the 1
st
T
C
p = C
min1= C
0
e
-KT
Fraction remaining after T
C
min1/C
0
=

e
-KT
F ig 2517
1/18/202570
Elias s. ( BA ACC. , BPHARM )

Cont…
n
th
dose-at the end of T
C
max, n = C
0
[(1-e
-nkT
)/(1-e
-kT
)]
= D/V
D[(1-e
-nkT
)/(1-e
-kT
)]
C
min, n = C
0
[(1-e
-nkT
)/(1-e
-kT
)]e
-kT
= D/V
D[(1-e
-nkT
)/(1-e
-kT
)] e
-kT
C
P t hours after the n
th
dose
C
P = D
o/V
D[1-e
-nkT
/1-e
-kT
]e
-kt
Concentration contributed by the missing dose
is
C
p= D/V
De
-Ktmiss
F ig 2517
1/18/202571
Elias s. ( BA ACC. , BPHARM )

Exercise, Cp: Missed dose
A cephalosporin (k = 0.2 hr


1
, V
D = 10 L) was administered by IV
multiple dosing; 100 mg was injected every 6 hours for 6 doses.
1.What was the plasma drug concentration 4 hours after the 6th dose?
2.What was the plasma drug concentration 4 hours after the 6th dose if:
(a)the 5th dose was omitted?
(b)the 6th dose was omitted?
(c)the 4th dose was omitted?
1/18/202572
Elias s. ( BA ACC. , BPHARM )

Cont…
Solutions:
C
P = D
o/V
D[1-e
-nkT
/1-e
-kT
]e
-kt
Substitute k = 0.2 hr


1
, V
D = 10 L, D = 100 mg, n =
6, t = 4 hr, and T= 6
 Cp = 6.425 mg/L
If no dose was omitted, then 4 hours after the 6th
injection, C
p would be 6.425 mg/L.
1/18/202573
Elias s. ( BA ACC. , BPHARM )

Cont…
(a)Missing the 5th dose, its contribution must
be subtracted off, t
miss = 6 + 4 = 10 hours
(the time elapsed since missing the dose):
C
p= D/V
De
-Ktmiss
= 100/10e
-0.2X10
Cp = 1.353mg/L
Drug concentration correcting for the
missing dose = 6.425 – 1.353 = 5.072 mg/L.
1/18/202574
Elias s. ( BA ACC. , BPHARM )

Cont…
(b) If the 6th dose is missing, t
miss = 4 hours:
C
p= D/V
De
-Ktmiss
= 100/10e
-0.2X4
C
p = 4.493mg/L
Drug concentration correcting for the missing
dose = 6.425 – 4.493 = 1.932 mg/L
1/18/202575
Elias s. ( BA ACC. , BPHARM )

Cont…
(c) If the 4th dose is missing, t
miss = 12 + 4 = 16
hours:
C
p= D/V
De
-Ktmiss
= 100/10e
-0.2X16
C
p = 0.408mg/L
The drug concentration corrected for the missing
dose = 6.425 – 0.408 = 6.017 mg/L.
1/18/202576
Elias s. ( BA ACC. , BPHARM )

Steady state/plateau
Fixed dose & T
Successive doses result in
accumulation
Rate of elimination increases
After some doses:
Rate of elimination = rate
in
Called steady state/plateau
F ig 2537
1/18/202577
Elias s. ( BA ACC. , BPHARM )

Cont…
At steady state
Amount eliminated
over a T = dose
Constant C
max & C
min

C
max & C
min must be
within the TW
F ig 2537
1/18/202578
Elias s. ( BA ACC. , BPHARM )

Cont…
Peak or max. conc. at steady state
C
max,ss = (D/V
D)(1/(1-e
-kT
))
Minimum or trough conc. at steady state
C
min,ss = (D/V
D)(1/(1-e
-kT
) e
-kT
C
max,ss & C
min,ss must be with in the TW
 4 to 5 t
1/2 to reach steady state
1/18/202579
Elias s. ( BA ACC. , BPHARM )

Exercise, steady state
The patient received 1000 mg of an antibiotic every
6 hours by repetitive IV injection. The drug has V
D
of 20 L and elimination half-life of 3 hours.
Q: Calculate:
(a)the plasma drug concentration C
p at 3 hours after
the second dose,
(b)The steady-state plasma drug concentration C

p at
3 hours after the last dose,
(c)C
max,ss
(d)C
min,ss,
1/18/202580
Elias s. ( BA ACC. , BPHARM )

Cont…
a.The C
p at 3 hours after the second dose. n = 2,
t = 3 hours, and making other appropriate
substitutions.
K = 0.693/t
1/2 = 0.693/3 = 0.231hr
-1
C
P = D
o/V
D[1-e
-nkT
/1-e
-kT
]e
-kt
C
P = 1000/20([1-e
-2(0.231)(6)
/1-e
-(0.231)6
]e
-0.231(3)

= 31.3mg/L
1/18/202581
Elias s. ( BA ACC. , BPHARM )

Cont…
b. The C
p at 3 hours after the last dose. Because steady
state is reached:
C

p, ss

= (D/V
D)(1/(1-e
-kT
)) e
-kt
C

p, ss

= (1000/20)(1/(1-e
-0.231(6)
)) e
-0.231(3)

= 33.3 mg/L
c. The C
max,ss is calculated from
C
max,ss = (D/V
D)(1/(1-e
-kT
))
= (1000/20)(1/(1-e
-0.231(6)
)
= 66.7mg/L
1/18/202582
Elias s. ( BA ACC. , BPHARM )

Cont…
d. The C
min,ss may be estimated as the drug concentration
after the dosage interval , or just before the next dose.
C
min,ss = (D/V
D)(1/(1-e
-kT
) e
-kT
= C
max,ss e
-kT
= 66.7 e
-kT
= 16.7mg/L

1/18/202583
Elias s. ( BA ACC. , BPHARM )

Loading and Maintenance Doses
Loading dose/priming
the required C
p is quickly achieved
drugs with long t
1/2
D
L = D
M/(1-e
-kT
)
✓D
L→ loading dose
✓D
m→ maintenance dose
▪D
M depends on the C
max,ss or MTC
▪D
M = C
max,ss V
D = MTC.
V
D
1/18/202584
Elias s. ( BA ACC. , BPHARM )

Exercise: D
L
What is the loading dose for an antibiotic (k = 0.23
hr


1
) with a maintenance dose of 200 mg every 3
hours?
Dose the patient with 200 mg every 3 hours.
D
L = D
M/(1-e
-kT
)
= 200/(1-e
-0.23(3)
)
= 400mg
1/18/202585
Elias s. ( BA ACC. , BPHARM )

❑ One compartment model- IV Infusion
Outlines
Introduction
C
p Vs t
Steady state conc.
Time to reach steady state
Cp after infusion period
 D
L and infusion
1/18/202586
Elias s. ( BA ACC. , BPHARM )

IV infusion-introduction
1/18/202587
Given slowly at a constant or
zero-order rate
Maintains an effective constant
plasma [drug]
For drugs with a narrow TW
Allows precise control of plasma
[drug] to fit the individual needs
of the patient
[Drug] is constant after a
plateau or steady-state
Elias s. ( BA ACC. , BPHARM )

Cont…
1/18/202588
As no drug was present in the body at zero time, drug
level rises from zero drug concentration and gradually
becomes constant when a plateau or steady-state drug
concentration is reached
Elias s. ( BA ACC. , BPHARM )

C
P - t relationship
1/18/202589
At steady state:
Rate at which the drug is leaving the body is equal
to the rate entering the body
Infusion rate = Elimination rate
Zero-order input and first-order output
No net change in the Cp
dC
p/dt = 0
Elias s. ( BA ACC. , BPHARM )

C
P - t relationship…
1/18/202590
Change in the amount of drug in the body at any time
Rate of input minus the rate of output
dD
B/dt = R-kD
B
R = infusion rate → Dose/duration of infusion
D
B = amount of drug in the body
K = 1
st
order elimination rate constant
C
p = (R/kV
D)(1-e
-kt
)
Elias s. ( BA ACC. , BPHARM )

Steady state Concentration (C
SS)
1/18/202591
no net change in the amount of
drug in the body
C
p = (R/kV
D)(1-e
-kt
)
Theoretical C
ss is at t = ∞
At t = ∞, e
–kt
approaches zero
C
ss = R/kV
D = R/Cl
C
SS depends on R, k & V
D
Elias s. ( BA ACC. , BPHARM )

Exercise: R
1/18/202592
1. An antibiotic has a volume of distribution of 10 L
and a k of 0.2 hr


1
.
A steady-state plasma conc. of 10 µg/mL is desired.
Determine the infusion rate needed to maintain this
concentration.
R = C
ssKV
D
R = (10µg/mL)(10)(1000mL)(0.2hr
-1
)
= 20 mg/hr
Elias s. ( BA ACC. , BPHARM )

Cont…
1/18/202593
2. Assume the patient has a uremic condition and the
elimination rate constant has decreased to 0.1 hr
–1
.
To maintain the steady-state concentration of 10
µg/mL , we must determine a new rate of infusion as
follows.
R = (10 µg/mL)(10)(1000mL)(0.1hr
-1
)
= 10 mg/hr
When the elimination rate constant decreases, the
infusion rate must decrease proportionally to maintain
the same C
SS.
However, because the elimination rate constant is smaller (ie,
the elimination t
1/2 is longer), the time to reach C
SS will be
longer.
Elias s. ( BA ACC. , BPHARM )

Time to Reach C
SS
1/18/202594
Drug elimination is exponential (1
st
order)
 C
P becomes asymptotic to the theoretical C
SS
Mathematically, the time to reach true steady-state
drug concentration, C
SS, would take an infinite time
The time for the C
P to reach more than 95% of the C
SS
is often taken as time for C
SS
About 4 to 5 t
1/2
= Time to reach desired therapeutic conc.

Elias s. ( BA ACC. , BPHARM )

t
1/2 to reach C
SS
1/18/202595
Example: no of t
1/2 to reach 99% C
SS
C
ss = R/kV
D
C
p = (R/kV
D)(1-e
-kt
)
0.99 C
ss = C
ss (1-e
-kt
)
0.01 = e
-kt
t
99%SS = ln(0.01)/-k
= 4.61/k
= (4.61/0.693) t
1/2
= 6.65t
1/2
Not dependent on R
Depends on K or t
1/2
Elias s. ( BA ACC. , BPHARM )

Cont…
1/18/202596
Percent of C
SS
reached Number of t
1/2
90 3.32
95 4.32
99 6.65
Number of t
1/2
to reach a fraction of C
SS

Elias s. ( BA ACC. , BPHARM )

Exercise: Cl
1/18/202597
A patient was given an antibiotic (t
1/2 = 6 hr) by constant
IV infusion at a rate of 2 mg/hr. At the end of 2 days, the
serum drug concentration was 10 mg/L.
 Q: Calculate the total body clearance Cl
T for this antibiotic.
The serum sample was taken after 2 days or 48 hours of
infusion, which time represents 8 x t
1/2, therefore, this
serum drug concentration approximates the C
SS.
Cl = R/ C
SS = (2mg/hr)/(10mg/L) = 200mL/hr
Elias s. ( BA ACC. , BPHARM )

Cp After Infusion Period
1/18/202598
If infusion stops at steady
state or before steady, C
P
declines according to first-
order kinetics
Stopped after steady state
C
t = C
sse
-kt
t → time after the
infusion
Similar to IV bolus, C
o= C
ss
Elias s. ( BA ACC. , BPHARM )

Loading Dose (D
L) Plus IV Infusion
1/18/202599
Initial bolus dose
➢used to obtain desired conc.
as rapidly as possible
[drug] in the body after an
IV bolus dose
C
1 = (D
L/V
D)e
-kt
[drug] by infusion at the rate
R C
2 = (R/V
DK) (1-e
-kt
)
Elias s. ( BA ACC. , BPHARM )

D
L Plus IV Infusion…
1/18/2025100
If D
L and infusion is started at the same time,
➢total [drug], C
p at t hours
C
1 + C
2
C
p =(D
L/V
D) e
-kt
+ (R/V
D K) (1-e
-kt
)
D
L is usually equal to D
SS
D
L = C
ss V
D = R/k
Elias s. ( BA ACC. , BPHARM )

Exercise: D
L
1/18/2025101
1. A physician wants to administer an anesthetic agent at a rate of
2 mg/h by IV infusion. The elimination rate constant is 0.1 h-1
and the volume of distribution (one compartment) is 10 L.
A.How much is the drug plasma concentration at the steady
state?
B.What loading dose should be recommended to reach steady
state immediately?
Elias s. ( BA ACC. , BPHARM )

Cont…
1/18/2025102
Solution
A.C
ss = R/kV
D = 2000mcg/h/(10,000mL *0.1h-1)
C
ss = 2mcg/mL
A.To reach Css instantly,
D
L = C
ss V
D = R/k =(2mg/h)/( 0.1h-1)
D
L =20 mg
Elias s. ( BA ACC. , BPHARM )

C
pt after D
L and infusion
1/18/2025103
2. What is the concentration of a drug at 6 hours after infusion
administration at 2 mg/h, with an initial loading dose of 10 mg
(the drug has a t1/2 of 3 hours and a volume of distribution of 10
L)?
Solution:
K = 0.693/3hr
C
p =(D
L/V
D) e
-kt
+ (R/V
D K) (1-e
-kt
)
= (10/10) e
-(0.693/3)(6)
+(2/10(0.693/3) (1-e
-(0.693/3)(6)
)
= 0.90mg/mL
Elias s. ( BA ACC. , BPHARM )

❑ One compartment model-
Extravascular Dose
1/18/2025104
Outline
Introduction
Amount of Drug in the Body
Plasma level–time curve
Plasma concentration, C
p
Time for maximum plasma conc., t
max
Determination of k and k
a
Elias s. ( BA ACC. , BPHARM )

Introduction
1/18/2025105
Drug absorption from the site of
administration
Complicated by variables at the
absorption site
Drug degradation
inter- and intra-patient
differences in the rate and extent
of absorption
Determination of P’K is not as
easy as IV route
Elias s. ( BA ACC. , BPHARM )

Amount of Drug in the Body
1/18/2025106
Depend on absorption and elimination rates
Regardless of whether absorption is zero-order or
first-order
➢dD
B/dt is dependent on the relative rates of drug
absorption and elimination
The net rate of drug accumulation in the body at any
time is equal to the rate of absorption less the rate
of elimination
Elias s. ( BA ACC. , BPHARM )

Amount of drug in the body…
1/18/2025107
dD
B/dt = dD
GI/dt –dD
E/dt
D
GI= amount of drug in GIT, in solution, at time t
D
E= amount of drug eliminated at time t
D
B = amount of drug in the body at time t
Elias s. ( BA ACC. , BPHARM )

❖ Plasma level–time curve
1/18/2025108
Four different regions
A. Absorption phase
Rate of drug absorption is greater
than the rate of drug elimination
dD
GI/dt >> dD
E/dt
During the absorption phase,
elimination occurs even though
absorption predominates
Elimination occurs whenever
drug is present in the plasma
Elias s. ( BA ACC. , BPHARM )

Plasma level–time curve …
1/18/2025109
B. Peak drug conc. (C
max)
The rate of absorption equals
the rate of elimination
dD
GI/dt
= dD
E/dt
No net change in the amount
of drug in the body
Elias s. ( BA ACC. , BPHARM )

Plasma level–time curve …
1/18/2025110
C. Post absorption phase
Immediately after the time of
C
max
Some drug may still be at the
absorption site
But the rate of drug
elimination at this time is
faster than the rate of
absorption
dD
GI/dt
<< dD
E/dt
Elias s. ( BA ACC. , BPHARM )

Plasma level–time curve …
1/18/2025111
D. Elimination phase
The rate of drug absorption
approaches zero
dD
GI/dt = 0
Only the elimination of drug
from the body
Usually 1
st
-order process
dD
B/dt = dD
E/dt = -KD
B
k = 1
st
-order elimination
rate constant
Elias s. ( BA ACC. , BPHARM )

❖ Plasma concentration, C
p
1/18/2025112
A.zero-order absorption: the drug is absorbed by
A saturable process, or
Controlled-release delivery system is used
The rate of input (absorption) = k
0
The rate of first-order elimination at any time = -kD
B
Therefore, net change per unit time in the body
dD
B/dt = K
O – KD
B

C
p = k
o/(kV
D)(1-e
-kt
)
Elias s. ( BA ACC. , BPHARM )

Plasma concentration, C
p…
1/18/2025113
B. 1
st
-Order absorption model
Absorption is usually assumed to be a first-order
process
Rapidly dissolving (immediate release) oral DFs,
suppositories as well as SC & IM injections
The rate of disappearance of drug from the GIT
dD
GI/dt = - K
aD
GI; D
GI = FD
0e
-kat
K
a = 1
st
-order absorption rate constant
dD
GI/dt = FK
aD
oe
-kat
Elias s. ( BA ACC. , BPHARM )

Plasma concentration, C
p…
1/18/2025114
dD
B/d
t =FK
aD
oe
-kat
-kD
B
C
P = FK
aD
o(Ke
-kt
-Kae
-kat
)
V
D(K
a-k)
Elias s. ( BA ACC. , BPHARM )

❖ t
max
1/18/2025115
Time for maximum plasma conc.
At t
max, net rate of conc. change is equal to zero

Elias s. ( BA ACC. , BPHARM )

t
max…
1/18/2025116
dC
P/dt = Fk
aD
o(-ke
-kt
+k
ae
-kat
) = 0
V
D(k
a-k)
k
ae
-kat
-ke
-kt
= 0
ke
-kt
= k
ae
-kat
lnk-kt = lnk
a-k
at
lnk-lnk
a = kt-k
at
ln(k
a/k) = t(k
a-k)
t
max = ln(k
a/k)/(k
a-k)
t
max = 2.3log(k
a/k)/K
a-k
Independent of D
o
Depend on k
a and k
Elias s. ( BA ACC. , BPHARM )

Exercise: t
max, C
max and t
1/2
1/18/2025117
A single oral dose (100 mg) of an antibiotic was given to an
adult male patient (43 years, 72 kg). From the literature, the
pharmacokinetics of this drug fit a one-compartment open
model. The equation that best fits the pharmacokinetics of
the drug is
From the equation above, calculate
a)t
max,
b)C
max, and
c)t
1/2 for the drug in this patient.
Assume C
p is in mg/mL and the first-order rate constants are in
hours


1
.
C
P = 45(e
-0.17t
-e
-1.5t
)
Elias s. ( BA ACC. , BPHARM )

Non-Linear Pharmacokinetics
1/18/2025118
Outlines
Introduction
Causes of nonlinear behavior
Michaelis-Menten Kinetics
Determination of V
max & K
m
Calculation of dose
Time for steady state
Elias s. ( BA ACC. , BPHARM )

Introduction
1/18/2025119
Linear p’k
First order elimination
Cl remains constant with any dose size
C
p and AUC increase proportionally/linearly with dose
Superimposition is assumed in multiple dosing
C
ss = FD/(Cl.T) = F(dose rate)/Cl
Hence C
ss is directly proportional to the dose rate (D/T)
C or AUC vs dose plot is linear
Elias s. ( BA ACC. , BPHARM )

Introduction…
1/18/2025120
Non-linear p’k
C
P change more or less than expected
No proportional increase in AUC and C
P with dose
P’k of the drug changes with the dose given
Zero-order, or dose-dependent p’k at high dose or after multiple
doses
Capacity-limited behavior
Superimposition cannot be assumed in multiple dosing
C
ss cannot be predicted from p’k of a single dose
No predictable relationship between C
ss and dose rate
Elias s. ( BA ACC. , BPHARM )

Causes of Nonlinear p’k Behavior
1/18/2025121
Capacity limited behavior could occur at any phase of p’k
process
✓Absorption, distribution, metabolism, excretion.
Absorption
✓Poor solubility/ and slow dissolution/release
✓Saturation of carriers
✓Saturation of pre-systemic metabolism
Changes F and hence Cl and C
ss
Distribution
✓Saturation of plasma protein binding
✓Saturation of tissue binding
✓Saturation of transport systems into tissue compartments
Leads to increase in F
U and hence increases Cl
Elias s. ( BA ACC. , BPHARM )

Causes of Nonlinear p’k Behavior …
1/18/2025122
Metabolism
✓Saturation of metabolism
✓Product inhibition
✓Autoinduction
Change the clearance and hence C
ss
Excretion
✓Saturation of carriers for active secretion and
reabsorption
Change the clearance and hence C
ss
nonlinear pharmacokinetics usually refers to the
processes of drug elimination
Elias s. ( BA ACC. , BPHARM )

1/18/2025123
Causes Drug
GI Absorption
Saturable transport in gut wall Riboflavin, gebapentin, L-dopa, baclofen,
Intestinal metabolism Salicylamide, propranolol
Drugs with low solubility in GI but
relatively high dose
Chorothiazide, griseofulvin
Saturable gastric or GI decompositionPenicillin G, omeprazole, saquinavir
Distribution
Saturable plasma protein binding Phenylbutazone, lidocaine, salicylic acid,
ceftriaxone, diazoxide, phenytoin, warfarin,
disopyramide
Cellular uptake Methicillin (rabbit)
Tissue binding Imiprimine (rat)
CSF transport Benzylpenicillins
Saturable transport into/ out of tissuesMethotrexate
Examples of Drugs Showing Nonlinear Kinetics
Elias s. ( BA ACC. , BPHARM )

1/18/2025124
Renal Elimination
Active secretion Mezlocillin, para-aminohippuric acid
Tubular reabsorption Riboflavin, ascorbic acid, cephapirin
Change in urine pH Salicylic acid, dextroamphetamine
Metabolism
Saturable metabolism Phenytoin, salicyclic acid, theophylline, valproic acid
Cofactor or enzyme limitationAcetaminophen, alcohol
Enzyme induction Carbamazepine
Altered hepatic blood flowPropranolol, verapamil
Metabolite inhibition Diazepam
Biliary Excretion
Biliary secretion Iodipamide, sulfobromophthalein sodium
Enterohepatic recycling Cimetidine, isotretinoin
Examples of Drugs Showing Nonlinear Kinetics…
Elias s. ( BA ACC. , BPHARM )

Causes of nonlinear p’k behavior…
1/18/2025125
Many drugs exhibit mixed-order pharmacokinetics
First-order p’k at low drug concentrations
Zero-order pharmacokinetics at high concentrations
It is important to know the conc. at which a drug
"order" switches from first to zero
For most drugs the switch of ‘order’ don’t occur at
therapeutic conc.
Eg. Theophylline does not switch until concentrations
reach the toxic range
Elias s. ( BA ACC. , BPHARM )

Michaelis-Menten Kinetics
1/18/2025126
Describe the kinetics of saturable systems
enzymes, active transporters
It allows prediction of C
Ss of drugs with
saturable elimination
dC/dt
= Vmax C
p
K
m + C
p
V
max→ the theoretical maximum rate of
the elimination (amount per unit of time )
C
p→ total plasma [drug]
Elias s. ( BA ACC. , BPHARM )

Michaelis-Menten Kinetics….
1/18/2025127
K
m → Michaelis constant
C
P when the rate of elimination is half the maximum
rate (V
max)
C
P above which saturation of drug metabolism is likely
C
p = K
m →
C
p<<K
m→ V
max = 1
st
order elimination rate constant
C
P >> K
m
elimination of drug becomes a zero-order process
 the rate of drug elimination is approximated by V
max
 elimination is zero-order
Elias s. ( BA ACC. , BPHARM )

Determination
of V
max &k
m
1/18/2025128
V
max &k
m are important in calculation of doses
Drugs with saturable elimination at C
p achieved
with therapeutic doses
 At steady state (after multiple drug doses)
the rate of drug loss from the body (milligrams
removed per day) is equal to dose rate
Michaelis-Menten equation
dC/dt indicates the rate of drug loss from the body
Elias s. ( BA ACC. , BPHARM )

Determination
of V
max &k
m…
1/18/2025129
Therefore, at steady state:
dC/dt = DR = V
max C/[K
m + C]
DR = dose rate = daily dose
C = steady state conc. for a given dose
Rearranging
DR= -K
m(DR/C) + V
max
Plot DR vs DR/C
Slope = -K
m
Y-intercept =V
max
Steady state conc at various dose rates is determined
Elias s. ( BA ACC. , BPHARM )

Dose and steady state conc.
1/18/2025130
C = K
m(dose)/(V
max- dose)
~The desired average C
p at steady state
Dose = V
maxC/(V
max + C)
➢ Daily dose required to get the desired steady state conc.
Elias s. ( BA ACC. , BPHARM )

1/18/2025131
Thank
you!
Elias s. ( BA ACC. , BPHARM )
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