1. Dr/Omnia Sarhan 1
PHARMACOKINETICS
Lecture 2

2. Introduction
2
2- First order kinetics (linear kinetics)
The rate of reaction is directly proportional to the concentration. i.e., the change depends on the
concentration.
ln A – ln Ao = - k t
ln (A/Ao) = - k t
A /Ao = e-Kt………….. (1)
Where e is the natural logarithm, the equation has one exponent and called monoexponential rate
process.
Since ln = 2.303 log, equation 1 can be written as;
Log A = log Ao – k t / 2.303 ............ (2)
A plot of log A versus time, a straight line is obtained whose
slope = k / 2.303
intercept = log Ao
Dr/Omnia Sarhan

3. Introduction
3
T½ for first order reaction
t ½ =0.693/ k
Dr/Omnia Sarhan

4. Graphically on
semilog paper
Co
1
2
1
2 log
log
t
t
y
y
slope



4
Dr/Omnia Sarhan

5. v
Conc.
Time
100
1
90
2
14
3
9
4
4.2
5
0.08
6
5
Dr/Omnia Sarhan

6. %
Remaining
Time
C0
Slope = -k/2.303
= (log y2 – log y1) / (x2 – x1)
No zero on y axis
6
Dr/Omnia Sarhan

7. k
t
693
.
0
2
/
1 
Half life t1/2 is constant
& doesn’t depend on
concentration.
2
log
303
.
2
2
/
1
k
t 
303
.
2
log
log 0
kt
C
Ct 

303
.
2
log
2
/
1
log 2
/
1
0
0
kt
C
C 

0
0
2
/
1
2
1
log
303
.
2
C
C
k
t 
0
0
2
/
1
2
/
1
log
log
303
.
2
C
C
kt


0
0
2
/
1
2
/
1
log
303
.
2 C
C
kt

7
Dr/Omnia Sarhan

8. 100%A
10
min
50%
A
25%
A
12.5%
A
10
min
10
min
Half life t1/2 is constant
& doesn’t depend on
concentration.
8
Dr/Omnia Sarhan

9. ln ln o
c c kt
  
 
 
ln ln5 0.0005 120
ln 1.549 ln
4.71 /
C
C shift
C mg ml
  


90%
0.105 0.105
210
0.0005
t
k
  
Don’t forget to
check the units of
the problem
An ophthalmic solution dispensed at 5 mg/ml conc. exhibits a first order
reaction, with a rate constant of 0.0005 day-1. How much drug remaining
after 120 days? & how long will it take for the drug to degrade to 90% of its
initial concentration?
Answer:
A first order reaction
days
Example 1
9
Dr/Omnia Sarhan

10. The original concentration of one sample was 50 mg/ml. When assayed
20 months later the concentration was found to be 40 mg/ml, assuming
that the decomposition is first order, What is the half-life of this
product?
Ans.:
Co= 50mg/mL Ct= 40mg/mL t= 20 mon.
log Ct= log Co- Kt/2.303
log 40= log 50- 20K/2.303
K= 0.0111mon-1
T1/2= 0.693/K= 0.693/ 0.0111= 62.43 mon.
Example 2
10
Dr/Omnia Sarhan

11. 1/ 2
0.693 0.693
0.1215
5.7
k
t
  
At 25 °C, the half-life period for the decomposition of N205 is 5.7
hours and is independent of the initial pressure of N205,
calculate:
a- The time for the completion of the reaction.
b- The specific rate constant for the reaction.
Answer:
•t1/2 is independent on the initial conc. So, it is 1st order, hence the
reaction never goes to an end and the life time (time for completion
of the reaction) is infinity.
hr-1
Example 3
11
Dr/Omnia Sarhan

12. Intercept
Slope
(x-
axis)
(y-axis)
Half-life
eqn.
Integrated
rate eqn.
Orde
r
Co
-Ko
t
C
t1/2 = Co
2k
Co – C = kt
0
Log Co
-K
2.303
t
LogC
t1/2 = 0.693
k
1
303
.
2
log
log 0
kt
C
Ct 

Dr/Omnia Sarhan

13. Introduction
13
Pharmacokinetic Model
Is a mathematical model devised to simulate the rate process of drug absorption, distribution and
elimination. These mathematical models make possible the development of equations to describe drug
concentration in the body as a function of time.
There are three different approaches to pharmacokinetic analysis of experimental data;
*Compartment modeling
*Non-compartment modeling
*Physiological modeling
1- Compartment modeling;
Are hypothetical structures that are used to describe the fate of a drug in a biological system
following its administration.
Dr/Omnia Sarhan

14. Introduction
14
Compartment
*Is a tissue or a group of tissues that have similar blood flow or drug affinity.
*A compartment is not real physiological or anatomic region.
*The rat constants are used to describe drug movement in and out of the compartments.
*The nature and behavior of the drug determines the number of the compartments, e.g some drugs
go to 2 places so we have two compartments
*The model is open system because drug can be eliminated from the system
a- One compartment open model, IV injection and with first order absorption
Following drug administration, the body is depicted as a kinetically homogeneous unit. This
assumes that the drug achieves instantaneous distribution throughout the body and that the drug
equilibrates instantaneously between tissues. Thus the drug concentration–time profile shows a
monophasic response. It is important to note that this does not imply that the drug concentration in
plasma (Cp) is equal to the drug concentration in the tissues.
Dr/Omnia Sarhan

15. Introduction
15
k
1
However, changes in the plasma concentration
quantitatively reflect changes in the tissues.
k
1
ka
Dr/Omnia Sarhan

16. Introduction
16
b- Two-compartment model IV injection and with first order absorption;
The two-compartment model resolves the body into a central compartment and a peripheral
compartment. It is assumed that the central compartment comprises tissues that are highly perfused such
as heart, lungs, kidneys, liver and brain. The peripheral compartment comprises less well-perfused
tissues such as muscle, fat and skin.
A two-compartment model assumes that, following drug administration into the central
compartment, the drug distributes between that compartment and the peripheral compartment. However,
the drug does not achieve instantaneous distribution, i.e. equilibration, between the two compartments.
k
12
1 2
k
k
12
1 2
k
k
21
ka
Dr/Omnia Sarhan

17. Introduction
17
The log drug concentration–time plot shows a biphasic response and can be
used to distinguish whether a drug shows a one- or two-compartment model.
The curve shows a profile in which initially there is a rapid decline in the
drug concentration owing to elimination from the central compartment and
distribution to the peripheral compartment. Hence during this rapid initial phase
the drug concentration will decline rapidly from the central compartment, rise to
a maximum in the peripheral compartment, and then decline.
After a time interval (t), a distribution equilibrium is achieved between the
central and peripheral compartments, and elimination of the drug is assumed to
occur from the central compartment. As with the one compartment model, all the
rate processes are described by first-order reactions.
Dr/Omnia Sarhan

18. Introduction
18
Dr/Omnia Sarhan

19. Introduction
19
c- Multicompartment model
In this model the drug distributes into
more than one compartment and the
concentration–time profile shows more than
one exponential (polyphasic response). Each
exponential on the concentration–time profile
describes a compartment. For example,
gentamicin can be described by a three-
compartment model following a single IV
dose.
Dr/Omnia Sarhan

20. Problems
Dr/Omnia Sarhan 20
*The linear relationship that describes the change in the drug amount (A) in mg, as a
function of time (t) in h, can be described by the following equation: A = 120 mg − 2
(mg/h) t.
a. Calculate the amount of the drug remaining after 10 h.
b. What is the time required for the drug amount to decrease to zero?

21. Time (months) Concentration remaining
(mg/ml)
0
4
8
12
16
20
24
50
44.1
36.4
30.2
25.3
18.1
12.6
Problem 1
Degradation of a suspension of triamcinolone acetonide, 50 mg/mL, was followed over a
year at 40oC and the following results were obtained:
a. Plot the data and prove that the degradation follows zero order.
b. Determine the constant of degradation.
21
Dr/Omnia Sarhan

22. Time (months) Conc.
0 100
4 81.4
8 68.8
12 56.4
16 44.6
20 36.2
24 31.2
Problem 2
The following are the concentrations of sulphadiazine remaining at various
times during the thermal decomposition at 40 C
Using the above data determine:
1.the order of the decomposition reaction.
2.The rate constant of the reaction.
3.The shelf life of the product.
Dr/Omnia Sarhan

23. Introduction
23
VII- Types of pharmacokinetics
When drugs are given on a constant basics such as a continuous
intravenous infusion or an oral administration gives every fixed
time intervals, serum drug concentration increases until the rate of
drug administration equal the rate of drug metabolism and
excretion. At that point serum drug concentration become constant,
the state which called a steady state. In this case, the steady state
achieved and presented as a solid line or dashed oscillated line,
respectively.
1- Linear: If a plot of steady state concentration versus dose yields
a straight line, the drug is said to follow linear pharmacokinetics. In
this situation, steady state serum concentration increase or decrease
proportionally with dose. Most of drugs follow linear
pharmacokinetics.
Dr/Omnia Sarhan

24. Introduction
24
2- Non-linear: Some drugs show a steady state
disproportionate with the drug dose. This state called
nonlinear pharmacokinetics, which is two types:
a- Saturable or Michaelis-Menten pharmacokinetics, a state
in which a steady state concentration increase more than
expected after a dosage increase, this because process
removing drug from the body have become saturated.
b- Saturable protein binding or autoinduction: when
steady state concentration increase less than expected after a
dosage increase, which is due to these drugs increase their own
rat of metabolism.
Dr/Omnia Sarhan

25. Introduction
25
IIX- Pharmacokinetic parameters
1- Plasma concentration of unbound versus bound drugs
**Once absorbed into the circulation, a portion of the total drug plasma concentration
(CT) is bound to plasma proteins (usually albumin), and a portion remains unbound, or free.
It is the unbound drug (CU) that is available for further transport to its site of action in the
body. The fraction of unbound drug compared with bound drug (CB) is primarily a function
of the affinity of the drug molecules for binding to the plasma proteins and the concentration
of the latter. Some drug molecules may be more than 90% bound to plasma proteins,
whereas others may be bound only slightly. Any change in the degree of binding of a given
drug substance can alter its distribution and elimination and thus its clinical effects.
**The fraction of unbound drug in the plasma compared with the total plasma drug
concentration (bound and unbound), is termed alpha (or ).
 = CU / CU + CB = CU / CT
Dr/Omnia Sarhan

26. Introduction
26
2- Elimination Rate Constant, kel
**The elimination rate constant (kel) is the first order rate constant describing drug elimination from the
body. This is an overall elimination rate constant describing removal of the drug by all elimination processes
including excretion and metabolism. Metabolites are different chemical entities and have their own elimination
rate constant.
**The elimination rate constant is the proportionality constant relating the rate of change drug
concentration and concentration OR the rate of elimination of the drug and the amount of drug remaining to be
eliminated.
Defining Equations:
dCp / dt = - Kel  Cp or dX / dt = - Kel  X0
where X amount of drug, X0 dose and Kel first-order elimination rate constant. The units for kel are time-1, for
example hr-1, min-1, or day-1.
Dr/Omnia Sarhan

27. Introduction
27
Determining Values of kel:
a- From the integrated of the following equation:
Ln Cp = ln Cp
0 –Kel t
Plotting ln(Cp) values versus t, time should result in a
straight line with a slope equal to -kel, thus kel can be
calculated as:
Kel = ln Cp
1 – ln Cp
2 / t2 – t1
Note: It is important to distinguish between the
elimination rate and the elimination rate constant. The
rate (tangent or slope, dCp/dt) changes as the
concentration changes, however, for a linear model the
rate constant (kel) is constant, it does not change.
Dr/Omnia Sarhan

28. Introduction
28
b-Also, elimination rate constant can be calculated from
the following equation:
Kel = ln (Cp
1 / ln Cp
2 )/ t2 – t1
With more than two Cp - time data points it is
possible to plot the data on semi-log graph paper and
draw a 'best-fit' line through the data points. The best
answer for kel can be calculated by taking points at either
end of the 'best-fit' line. Always try to use the extremes
of the line for the best accuracy.
c- From the following equation:
Kel = 0.693 / t1/2
Dr/Omnia Sarhan

29. Introduction
29
Example use of semi-log graph paper: Plot the data, draw a
line "through the data", and calculate the slope of the line.
From the figure:
kel = [ln(22.8) - ln(1.0)]/(12.8 - 0) = [3.127 - 0.000]/12.8 =
3.127/12.8 = 0.244 hr-1
Orkel = ln(22.8/1.0)/(12.8 - 0) = ln(22.8)/12.8 = 3.127/12.8 =
0.244 hr-1
1 2 4 8 12
Time (hr)
20 15 6.8 3.2 1.3
Cp (mg/L)
t2
t1
Cp
2
Cp
1
12.8
0
1.0
22.8
Dr/Omnia Sarhan

30. Introduction
30
3- Volume of distribution, Vd
*The volume of distribution (Vd) has no direct physiological meaning; it is not a ‘real’
volume and is usually referred to as the apparent volume of distribution.
*It is defined as that volume of plasma in which the total amount of drug in the body
would be required to be dissolved in order to reflect the drug concentration attained in
plasma.
*It is important to realize that the concentration of the drug (Cp) in plasma is not
necessarily the same in the liver, kidneys or other tissues.
Defining equation:
The changes in the drug concentration in plasma (Cp) are proportional to changes in the
amount of drug (X) in the tissues.
C (tissues)  Cp (plasma) i.e. X (tissues)  Cp (plasma)
Dr/Omnia Sarhan

31. Introduction
31
Then
X (tissues)  Vd  Cp (plasma)
Where:
Vd is the constant of proportionality and is referred to as the volume of distribution,
which thus relates the total amount of drug in the body at any time to the corresponding
plasma concentration. Thus
Vd = X / Cp
and Vd can be used to convert drug amount X to concentration (Cp
t ).
Since X = X0 e –kel t
Then
X / Vd = X0 e –kel t / Vd
Thus
Cp
t = Cp
0 e –kel t
Dr/Omnia Sarhan

32. Introduction
32
This formula describes a monoexponential decay, where Cp
t plasma concentration at
any time t. The curve can be converted to a linear form using natural logarithms (ln):
ln Cp
t = ln Cp
0 –kel t
Where the slope is -kel , and the y intercept is ln Cp
0
Since Vd = X / Cp
0
then at zero concentration (Cp
0 ), the amount administered is the dose, D, so that
Cp
0 = D / Vd
*If the drug has a large Vd that does not equate to a real volume, e.g. total plasma
volume, this suggests that the drug is highly distributed in tissues.
*If the Vd is similar to the total plasma volume this will suggest that the total amount of
drug is poorly distributed and is mainly in the plasma.

33. Introduction
33
Units:
The units for the apparent volume of distribution are volume units. Most commonly Vd
is expressed in liters, L. On occasion the value for the apparent volume of distribution will
be normalized for the weight of the subject and expressed as a percentage or more usually
in liters/kilogram, L/Kg.
Determining Values of Vd:
The usual method of calculating the apparent volume of distribution of the one
compartment model is to extrapolate concentration versus time data back to the y-axis
origin. for an example. This gives an estimate of Cp
0. When the IV bolus dose is know the
apparent volume of distribution can be calculated from.
Vd = X / Cp
0
Dr/Omnia Sarhan

34. Introduction
34
4- Half-life (t1/2)
*It is the time required to reduce the plasma concentration to one half its initial
value. Units for this parameter are units of time
* Many drug substances follow first-order kinetics in their elimination from the
body, meaning that the rate of drug elimination per unit of time is proportional to the
amount present at that time, but the fraction of a drug in the body eliminated over a
given time remains constant.
*The elimination half-life is independent of the amount of drug in the body, and the
amount of drug eliminated is less in each succeeding half-life.
*After five elimination half-lives, it may be expected that virtually all of a drug
(97%) originally present will have been eliminated.
Dr/Omnia Sarhan

35. Introduction
35
Defining equation:
ln Cpt = ln Cp
0 - kelt
Let Cp
0 decay to Cp
0 / 2 and solve for t = t1/2
ln Cp
0 / 2= ln Cp
0 - kel t1/2
Hence
kel t1/2 = ln Cp
0 - ln Cp
0 / 2
t1/2 = ln 2 / kel
t1/2 = 0.693 / kel
*This parameter is very useful for estimating how long it will take for levels to be reduced by half the
original concentration.
Dr/Omnia Sarhan

36. Introduction
36
*It can be used to estimate for how long a drug should be stopped
if a patient has toxic drug levels, assuming the drug shows linear one-
compartment pharmacokinetics.
*The equation of t1/2 can be used as an approximate method of
calculating kel. If we look at a plot of Cp versus time on semi-log graph
paper.
The steps to take are:
Draw a line through the points (this tends to average the data)
Pick any Cp and t1 on the line
Determine Cp /2 and t2 using the line
Calculate t1/2 as (t2 - t1)
And finally kel = 0.693/t1/2
Dr/Omnia Sarhan

37. Introduction
37
5- Total Body Clearance, CL
**Total body clearance or clearance is an important pharmacokinetic parameter that describes
how quickly a drug is eliminated from the body.
**It is often defined as the volume of blood or plasma completely cleared of the drug per
time. However, it may be easier to view it as the proportionality constant relating the rate of
elimination and drug concentration. The units are volume per time such as ml/min, L/hr. The rate
of elimination of a drug can be described by this equation:
dX/dt = - CL  Cp
CL = - dX/dt  Cp Or CL = Dose / AUC
Clearance can be calculated from this equation by measuring the amount of drug eliminated
during some time interval and the drug concentration at the midpoint of this collection interval.
Dr/Omnia Sarhan

38. Introduction
38
*Also CL can be calculated from the following equation ( in case of IV bolus injection)
CL = Kel  Vd
CL = 0.693  Vd / t 1/2
*The clearance of a drug can be used to understand the processes involved in drug
elimination, distribution and metabolism.
*Relating clearance to a patient renal or hepatic function can be used in the determination of
suitable drug dosage regimens.
*If a drug has a CL of 2L/h, this tells you that 2 liters of the Vd is cleared of drug per hour.
*If the Cp is 10 mg/L, then 20 mg of drug is cleared per hour.
Dr/Omnia Sarhan

39. Problems
39
1- A drug has an elimination half life of 4 hours. After the administration of a 250 mg
dose, plasma concentration at zero time point was found to be 5.65 g / mL. What is
the volume of distribution of the drug? (44.2 L)
2- The plasma concentration of a drug immediately following a 50 mg intravenous bolus
dose of the drug was found to be 0.84 g / mL. What is the apparent volume of
distribution of the drug? (59.5 L)
3- If the apparent volume of distribution of a drug is 30 L, and the plasma concentration
extraploted to zero time point (CP
0) is 0.15 g / mL. What is the dose of the drug? (4.5
mg)
4- The alpha () value for a drug in the blood is 0.90, equating to 0.55 ng/mL. What is the
concentration of total drug in the blood?
Dr/Omnia Sarhan

40. Problems
40
5- If a 6-mg dose of a drug is administered intravenously and produces a blood
concentration of 0.4 g/mL, calculate its apparent volume of distribution.
6- If at equilibrium, two thirds of the amount of a drug substance in the blood is bound to
protein, what would be the alpha () value of the drug?
7- A patient received an intravenous dose of 10 mg of a drug. A blood sample was drawn,
and it contained 40 g/100 mL. Calculate the apparent volume of distribution for the
drug.
8- The volume of distribution for a drug was found to be 10 liters with a blood level
concentration of 2 g/mL. Calculate the total amount of drug present in the patient.
9- If the half-life of a drug is 4 hours, approximately what percentage of the drug
administered would remain in the body 15 hours after administration?
Dr/Omnia Sarhan

41. Problems
41
10- The volume of distribution for chlordiazepoxide has been determined to be 34 liters. Calculate the expected drug
plasma concentration of the drug, in micrograms per deciliter, immediately after an intravenous dose of 5 mg.
11- In normal subjects, blood makes up about 7% of the body weight.
(a) Calculate the approximate blood volume, in liters, for a man weighing 70 kg.
(b) If the drug ZANTAC reached peak blood levels of about 500 ng/mL 2 to 3 hours after an oral dose,
calculate the total amount of the drug, in milligrams, in the blood of the patient described in (a) when
peak blood levels are achieved
12- If a drug is poorly distributed to tissues, its apparent volume of distribution is probably:
A. large.
B. small.
Dr/Omnia Sarhan

42. Problems
42
12- Plasma refers only to the fluid portion of blood, including soluble proteins but not formed elements.
A. True B. False
13- The units for clearance are:
A. concentration/half-life. B. dose/volume.
C. half-life/dose. D. volume/time.
14- Total body clearance is the sum of clearance by the kidneys, liver, and other routes of elimination.
A. True B. False
15- To determine drug clearance, we must first determine whether a drug best fits a one- or two-compartment
model.
A. True B. False
Dr/Omnia Sarhan

43. Problems
43
16- With a drug that follows first-order elimination, the amount of drug eliminated per
unit time:
A. remains constant while the fraction of drug eliminated decreases.
B. decreases while the fraction of drug eliminated remains constant.
Dr/Omnia Sarhan

44. Dr/Omnia Sarhan 44