1

Pharos University Faculty of Pharmacy and Drug Manufacturing, Department of Pharmaceutics

Yousry M. El-Sayed, Ph.D. Professor of Pharmaceutics Spring Semester 2015

Basic Pharmacokinetics

PHR 417

Lectures 1 - 3

2

Basic Pharmacokinetics

PHR 417

Drug Kinetics Following an Intravenous Bolus Dose

Objectives: At the end of this sequence you will be able to:

1. Explain the assumption of linear one-compartment model.

2. Describe the meaning of an elimination rate constant, and a first order process.

3. Describe the following terms:

a. Apparent volume of distribution. b. Clearance. c. Half-life. d. Elimination rate constant.

4. Determine the plasma concentrations, the amount of the drug in the body, and the apparent volume of distribution when any two of these values are known.

5. Define; in words and equations; plasma clearance, blood clearance.

6. Explain the statements:

a. The elimination rate constant, or the fraction rate of elimination, is dependent on the value of the clearance of a drug and on its apparent volume of distribution.

3

b. The clearance and the apparent volume of distribution are independent of the elimination rate constant or the elimination half-life.

7. Estimate the values of the elimination half-life, elimination rate

constant, apparent volume of distribution and total clearance from plasma, serum, or blood concentrations following an intravenous bolus dose.

8. Derive the interrelationship between elimination half-life, elimination

rate constant, apparent volume of distribution and total clearance.

9. Derive and understand the relationship between the administered dose, total body clearance and area under the plasma concentration-time curve.

10. Calculate plasma levels at different times, given appropriate data.

Reference:

Chapter 3: Applied Biopharmaceutics and Pharmacokinetics. Edited by:

Leon Shargel, Andrew B.C. Yu ; 5th edition. Publisher: Mc Graw-Hill,

2005

4

Definition Pharmacokinetics is the study of drug and/or metabolite kinetics in the body. It deals with a mathematical description of the rates of drug movement into, within and exit from the body. It also includes the study of drug metabolism or biotransformation rates. The body is a very complex system and a drug undergoes many steps as it is being absorbed, distributed through the body, and metabolized or excreted (ADME).

Drug Disposition

The drug also interacts with receptors and causes therapeutic and/or toxic responses. Although the details of drug kinetics are complicated it is fortunate that we can often approximate drug kinetic processes using "simple" mathematical models.

5

One Compartment IV Bolus

Assumptions We will start the course with a one compartment - linear model. Also, we will first consider drug kinetics after a rapid intravenous injection, an IV bolus injection. According to this model we will consider the body to behave as a single well-mixed container. To use this model mathematically we need to make a number of assumptions:

1. One compartment The drug in the blood is in rapid equilibrium with drug in the extravascular tissues. The drug concentration may not be equal in each tissue or fluid however they are proportional to the concentration of drug in the blood at all times. 2. Rapid Mixing The drug is mixed instantaneously in blood or plasma. The actual time taken for mixing is usually very short, within few minutes, and in comparison with normal sampling times it is insignificant. We usually don't sample fast enough to see drug mixing in the blood. 3. Linear Model

I V

Dose

K

Drug Elimination

Ab

Body

6

For most drugs the elimination follows first order kinetics. First order kinetics means that the rate of change of drug concentration by any process is directly proportional to the drug concentration remaining to undertake that process. Remember first order kinetics is an assumption of a linear model not a one-compartment model. If we have a linear model, when we double the dose, the concentration will double at each time point.

Figure of a body before and after a rapid IV bolus injection

The body behaves as a single container or compartment

In order to simplify the mathematics it is often possible to assume that when a drug is given by rapid intravenous injection, the drug is rapidly mixed in the body. The Figure above represents the uniformly mixed drug very shortly after. Linear Model - First Order Kinetics First-order kinetics To illustrate first order kinetics we might consider what would happen if we were to give a drug by IV bolus injection, collect blood samples at various times and measure the plasma concentrations of the drug. We might see a steady decrease in concentration as the drug is eliminated, as shown in Figure below.

7

1 3 5 7 9 11 13

0

20

40

60

80

100

TIME (Hours)

Con

cent

rati

on in

the

Bod

y (

mg/

L)

Plot of the Concentration versus Time

This can be described mathematically with regard to the amount of drug in the body as: Rate of change of the amount of drug in the body:

Change in Ab Change in time Proportionality Constant Called Elimination Rare Constant

Rate of Change in the amount, Ab, with time

Where k is proportionality constant we call a rate constant and Ab is the amount remaining to be eliminated. Note the use of the symbol’d’ to represent a very small increment in Ab or t. Thus dAb/dt represents the rate of change of the amount of drug in the body. Unit of K: The elimination rate constant (K) is also called the fractional rate of

elimination:

Ab . K - =

dt

dAb

1-Time Time

1 Amount . K

Time

Amount Ab . K - =

dt

dAb

8

(Constant) (First-Order Kinetics)

As the amount of drug in the body decreases so the rate of change

decreases. “K” remains constant.

First-Order Elimination

Time after Drug Administration

(hours)

Amount of Drug in Body (mg)

Amount of Drug Eliminated Over Preceding Hour

(mg)

Fraction of Drug Eliminated Over Preceding Hour

0 1000 - -

1 880 120 0.12

2 774 106 0.12

3 681 93 0.12

4 599 82 0.12

5 527 72 0.12

6 464 63 0.12

7 408 56 0.12

With the first-order elimination process, although the amount of drug eliminated may change with the amount of drug in the body, the fraction of a drug in the body eliminated over a given time remains constant. In practical terms, the fraction or percentage of drug being removed is the same with either high or low drug concentrations. For example, if 1000 mg of a drug is administered and the drug follows first-order elimination, we might observe the patterns in Table above.

K =dt

dAb

Ab

9

The actual amount of drug eliminated is different for each fixed time period, depending on the initial amount in the body, but the fraction removed is the same, so this elimination is first order. Because the elimination of this drug (like most drugs) occurs by a first-order process, the amount of drug eliminated decreases as the concentration in plasma decreases. The actual fraction of drug eliminated over any given time (in this case 12%) depends on the drug itself and the individual patient’s capacity to eliminate the drug.

This is the differential equation for a first-order process and K is the “First-

order rate constant”

Differential calculus is involved with the study of rates of processes. The calculus part comes in when we look at these processes in detail, that is, during small time intervals. Rearranging the above equation:

Upon integration from t = 0 to t = t

Ab K - =dt

dAb

dt K - =Ab

dAb

t K - Ab Ln-Abt Ln

t K - Ab LnAbt Ln

t K - Ab Log303.2Abt Log2.303

2.303

t K - Ab LogAbt Log

Linear First-Order Kinetics

10

Where Ln is the natural Logarithm

The antilog:

Where Ab0 is the amount of the drug in the body at zero time i.e. time

immediately after injection (DoseIV).

Linear Scale

1 3 5 7 9 11 130

20

40

60

80

100

TIME (Hours)

Am

ou

nt i

n t

he B

od

y (

Ab

) (

mg)

Semi-Logarithmic Scal

1 2 3 4 5 6 7 8 9 10 11 12 13 141

2

5

10

20

50

100

TIME (Hours)

Am

ou

nt

in t

he B

od

y (

Ab

) (m

g)

t K - =)Ab

Abt( Ln

tK-e =Ab

Abt

tK-e IV

Dose=Abt

tK-eAb =Abt 2.303

t K - AbLogLogAbt

2.303

K -Slope

tK-e Ab =Abt

11

Relationship between Amount and Concentration

V is called Apparent Volume of Distribution (V)

Volume of Distribution

The apparent volume into which a drug distributes in the body at equilibrium is called the (apparent) volume of distribution. The Apparent volume of distribution is the factor that relates the plasma concentration to the amount of drug in the body. Volume of distribution is useful in estimating plasma concentration when a known amount of drug is in the body or, conversely, in estimating the dose required to achieve a desired plasma concentration.

Cp Ab

Cp V Ab

ionConcentrat Plasma

Body in the Drug ofAmount

Cp

Ab V

12

Immediately after the intravenous dose is administered the amount of drug

in the body is the IV dose. Thus:

And

Therefore:

The one compartment model assumption is that there is a rapid equilibration in drug concentrations throughout the body; however, this does not mean that the concentration is the same throughout the body.

Drug distribution means the reversible transfer of drug from one location to another within the body. Once a drug has entered the vascular system it becomes distributed throughout the various tissues and body fluids in a pattern that reflects the physiochemical nature of the drug and the ease with which it penetrates different membranes. The ONE COMPARTMENT model assumes rapid distribution but it does not preclude extensive distribution into various tissues.

Cp

Dose V

IV

V

Dose Cp

IV

tK-e V

Dose =Cp

IV

tK-e Cp =Cp

13

The volume of distribution is an important parameter for determining proper drug dosing regimens. Often referred to as the apparent volume of distribution, it does not have an exact physiologic significance, but it can indicate the extent of drug distribution and aid in determination of dosage required to achieve a desired plasma concentration. Loading Dose = V X Cpdesired

Where V is the volume of distribution, Cp is the desired plasma level

Linear Scale

1 3 5 7 9 11 130

20

40

60

80

100

TIME (Hours)

Dru

g C

on

cen

trati

on

in

Pla

sma (

mg/L

)

Semi-Logarithmic Scale

1 3 5 7 9 11 131

2

5

10

20

50

100

200

TIME (Hours)

Dru

g C

on

cen

trati

on

in

Pla

sma (

mg/L

)

tK-e Cp0 =Cp

t K- LnCp= LnCp

K- Slope

“monoexponential decline”

14

Linear plot of ln(Cp) versus time

NOTICE, there are no UNITS for ln(Cp) in the above Figure. There are units of hour for time (X axis) so the slope has units of time-1 e.g. min-1, hr-1.

Now we can measure the elimination rate constant (k) by determining Cp

versus time and plotting Ln(Cp) versus time.

t K- LnCp=LnCp

2.303

tK - LogCp=LogCp

2.303

K - =Slope

K- Slope

15

Semi-Logarithmic Scale

0 2 4 6 8 10 122

5

10

20

50

100

200

TIME (Hours)

Dru

g C

on

cen

trati

on

in

Pla

sma (

mg/L

)

Determining the Value of K

From the integrated equation:

Plotting Ln(Cp) values versus t, time should result in a straight line with a slope equal to -k, thus k can be calculated as: Note that Cp1 and Cp2 changed for K to be positive

Thus with two value for Cp and time data a value for k can be determined. 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. This plot is shown on the Figure above. Taking points at either end of the ‘best-fit’ line can calculate the best answer for k. Using semi-log graph paper the scale on the y-axis is proportional to the log of the number, not the number itself. Note that you can change the sign by calculating the numerator as Ln (Cp1) – Ln (Cp2). Thus k can be calculated using the above equation.

CpIntercept

t1- t2

Cp(1) Ln- Cp(2) Ln Slope

K- Slope

t K- Cp Ln=Cp Ln

t1- t2

Cp(2) Ln- Cp(1) Ln K

Cp1

Cp2

t1 t2

t K- Cp Ln=Cp Ln

16

Note: It is important to distinguish between the elimination rate and the elimination rate constant. The rate of elimination (dCp/dt) changes as the concentration changes, however, for a linear model the rate constant (k) is constant, it does not change. Determining the Value of the Apparent Volume of Distribution (V)

We can use the Equation above to calculate the plasma concentration at any time when we know k and Cp˚. However, usually we don't know Cp˚ ahead of time, but we do know the dose. To calculate Cp˚ we need to know the volume that the drug is distributed into. That is, the apparent volume

of the mixing container, the body. This apparent volume of distribution is not a physiological volume. It is never lower than blood or plasma volume but for some drugs it can be much larger than body volume. It is a mathematical factor relating the amount of drug in the body and the concentration of drug in the measured compartment, usually plasma.

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. See below Figure for an example. This gives an

estimate of Cp˚. When the IV bolus dose is known the apparent volume of

distribution can be calculated from the Equation above.

tK-e Cp =Cp

V

Dose Cp

IV

tK-e V

Dose =Cp

IV

17

CP

0

= 40 µg/ml

Cp

Dose V

IV

V

Dose Cp

IV

Cp

Dose V

IV

18

Elimination Half-life (t1/2)

The half-life is the time taken for the plasma concentration to fall to half its original value. Units for this parameter are units of time such as hour, minute, or day. Thus if Cp is the concentration at time one (t1), Cp/2 is the concentration at time one half-life later (t2):

These Equations can be used to calculate K and t1/2

Semi-log Plot of Cp versus Time illustrating t1/2 Calculation

tK-e Cp =Cp

1/2tK-

e Cp =2

Cp

1/2

t K- Cp Ln=2

Cp Ln

1/2t K - =(0.5) Ln

1/2

K t- =0.693 -

K

0.693 =t

1/21/2t

0.693 =K

19

How long it takes to eliminate 90% of the drug from the body? Or how

long it takes for the concentration to fall 90% of the original value?

Useful values to know:

Time Remaining fraction of drug in the body

Fraction lost

1/3 t1/2 0.8 0.2

1/2 t1/2 0.7 0.3

1 t1/2 0.5 0.5

2 t1/2 0.25 0.75

3.32 t1/2 0.1 0.90

5 t1/2 3.125 % 96.875 %

6.64 t1/2 0.01 0.99

10 t1/2 0.001 0.999

Thus over 95 % is lost or eliminated after 5 half-lives. Typically, with pharmacokinetic processes, this is considered the completion of the process [Although in theory it takes an infinite time].

tK-e Cp =Cp

tK-e Cp =0.1Cp

t K- =0.1 Ln

tt1/2

0.693 - =2.303-

t3.32 =t 1/2

Note that 90% of the drug eliminated and 10% remaining

20

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 rate of elimination of a drug can be described by the following Equation:

In above Equation the elimination rate, dAb/dt, is related to the concentration of drug remaining. The proportionality constant for this relationship is Clearance. The symbol for clearance is CL and the units are volume per time such as ml/min, L/hr. Units of Clearance:

Cp CL =dt

dAb

Cp dt

dAb

L/hror ml/min

Time

Volume

Volume

Amount CL =

Time

Amount

Cp CL =dt

dAb

21

Dependence of Drug Elimination on Clearance and Distribution

Rate of Drug Elimination =

Note that V is constant

Note that CL is constant

I V

Dose

K

Drug Elimination

Ab

Body

Cp CL =

dt

dAb

Cp CL

Cp CL =

dt

dcp V

Cp V = Ab

Cp V

CL =

dt

dcp

Cp K =

dt

dcp

22

Clearance and Elimination Half-life

The previous equation was purposely arranged in the above manner to

stress that, the half-life and elimination rate constant reflects rather than

control the volume of distribution and drug clearance.

One can independently alter the volume of distribution or clearance and

hence change the t1/2 but not vice versa.

In some instance, the volume of distribution and clearance can change to

essentially the same extent, in which case the t1/2 remains unchanged.

K

0.693 = t1/2

V K = CL

1/2t

V 0.693 = CL

CL

V 0.693 = t1/2

V

CL = K

V K = CL

23

Example to demonstrate the dependence of the t1/2 on CL and V:

Although the clearance of Quinacrine is very large = 1500 ml/min, it takes

16 days to eliminate 50% of the drug from the body.

Whereas Gentamicin clearance = 120 ml/min, it takes only 1.5 hours to

eliminate 50% of the drug from the body.

The major difference is in Volume of distribution (V):

L50,000 = V Quinacrine

L/hr90 = ml/min 1500 = CL Quinacrine

days 16 = hours 385 = 90

50,000 * 0.693 = t1/2

L16 = V Gentamicin

L/hr7.2 = ml/min 120 = CL Gentamicin

hours 1.54 = 7.2

16 * 0.693 = t1/2

24

Area under the plasma concentration time curve (AUC) The area under the plasma (serum, or blood) concentration versus time curve (AUC) has a number of important uses in toxicology, biopharmaceutics and pharmacokinetics.

Toxicology: AUC can be used as a measure of drug exposure. It is

derived from drug concentration and time so it gives a measure how much - how long a drug stays in a body.

Biopharmaceutics: The AUC measured after administration of a drug

product is an important parameter in the comparison of drug products. Studies can be performed whereby different drug products may be given to a panel of subject on separate occasions. These bioequivalence or bioavailability studies can be analyzed by comparing AUC values (This will be discussed in detail in kinetics of oral administration).

Pharmacokinetics: Drug AUC values can be used to determine other pharmacokinetic parameters, such as clearance or bioavailability, F.

The area under the plasma concentration versus time curve (AUC) has units of concentration times time. For example, mg.hr/L or mg.hr.L-1. The area under the curve is calculated by the Trapezoidal rule.

25

Calculation of AUC using the Trapezoidal Rule

Linear Plot of Cp versus Time showing AUC and AUC segment

If we have a smooth line for concentration versus time or an equation for Cp versus time from a pharmacokinetic model we could slice the area into

vertical segments. Each segment would be very thin, t and in extreme dt, in width (much smaller than the segment in Figure above). The total AUC is calculated by adding these segments together (each Trapezoid). In calculus this would be the integral. Each very narrow segment has an area = Cp X dt. Thus the total area (AUC) is given by Equation:

26

The area under the plasma concentration time curve (AUC) is very useful for calculating the relative efficiency of different drug products. It can used to calculate the total body clearance (CL) and the apparent volume of distribution.

Clearance, Area under the curve and Volume of distribution

So far, clearance has been estimated from the half-life or the elimination

rate constant and volume of distribution:

Clearance can be better estimated in another way: By rearranging the following equation:

Rate of Drug Elimination =

V K = CL

1/2t

V 0.693 = CL

Cp CL

Cp CL =

dt

dAb

dtCp CL = dAb

dtCpt

0t

AUC

dtCp AUC )32()3t2t(

Total AUC0

27

Upon integration from t = 0 t = ∞

Amount lost over all time period (t=0 ∞) is the dose intravenously

administered.

Calculation of drug clearance in this way is independent of the

shape of the plasma concentrations-time profile.

Cpdt

o

i.v. Dose

CL

0AUC i.v. Dose CL

∞

0AUC Cpdt

o

dtCp

o

CL dAb

o

dtCp

o

CL Ab

28

Important terms to remember

Apparent volume of distribution (V):

The apparent volume into which a drug distributes in the body at equilibrium is called the (apparent) volume of distribution. The Apparent volume of distribution is the factor that relates the plasma concentration to the amount of drug in the body. Volume of distribution is useful in estimating plasma concentration when a known amount of drug is in the body or, conversely, in estimating the dose required to achieve a desired plasma concentration. Clearance (CL):

The factor that relates the amount of the drug eliminated per time (elimination rate) to plasma concentrations. Or: the volume of plasma the body has to clear of the drug to account for the total amount removed per time. Drug clearance is the factor that relates the plasma concentration to rate of drug elimination.

Elimination Rate Constant (K):

AUC0 Dose = CL IV

KCp

= AUC0

V Cp Dose IV

V K = CL

29

The elimination rate constant is defined as the fraction of the total amount

of drug in the body that is eliminated per unit time.

Elimination Half-life (t1/2):

The elimination half-life is the time taken for the concentration and amount

of drug in the body to fall by one-half.

How to obtain the pharmacokinetic parameters from the plasma

concentrations-time curves?

1. In the table below are the plasma concentration data following an I.V. bolus dose of a 1 gm of Drug X to a 70-Kg patient:

Time (hr) 1.00 2.00 3.00 4.00 6.00 8.00 10.00 12.00

Concentration (µg/ml) 33.6 28.3 23.8 20.0 14.2 10.0 7.1 5.0

Prepare linear and semi-logarithmic plots of the plasma

concentrations versus times

a. Calculate the volume of distribution.

b. From the slope of the line estimate the elimination half-life and the elimination rate constant.

c. Calculate the total clearance.

d. Calculate the total area under the plasma concentration-time curve by the trapezoidal rule.

30

e. Calculate the plasma concentration and the amount of the drug in the body 24 hours following administration of the drug.

Semi-Logarithmic Plot of the Plasma Concentrations-Time Curve

following I.V. Bolus of 1 gm of Drug X to 70-kg Patient

CP

0

= 40 µg/ml

31

A. Calculation of the volume of distribution:

The apparent volume of distribution (V) is obtained by dividing the

amount of the drug in the body by the plasma concentration. This

calculation requires that distribution equilibrium was achieved

between the drug in tissues and that in plasma. The amount of drug

in the body is known immediately after an intravenous bolus, it is the

dose administered. However, distribution equilibrium has not yet

been achieved. An estimate is needed of the plasma concentration

that would have resulted when the entire drug spontaneously

distributed into its final volume of distribution. To do this, use is

made of the linear decline during the elimination phase seen in the

semi-logarithmic plot (Figure above).

The decline in the plasma concentrations during the elimination

phase can be described by the linear equation:

Cp0 is an extrapolated value and is an estimate of the concentration

which when multiplied by the volume term, V, accounts for the dose

administered,

t K- Cp Ln= Cp Ln

tk-e Cp= Cp

1 = -e as 0 =at t Cp= Cp 0

V Cp= Dose

CpDose = V

32

B. Estimation of the elimination rate constant and elimination half-life

from the slope of the line:

tk-e Cp= Cp

t1)- (t2

Cp2) Ln- Cp1 (Ln =K

1-hr 0.173 =

)1-(12

)5 Ln- 6.33 (Ln =K

hrs 4.0 =

K

693.0 = t1/2

0) at t ted(extrapola Cp = Cp

Cpiv Dose = V

L25 =mg/L 40

mg 1000 = V

tK -CpLn = CpLn

33

C. Total Clearance:

D. Total AUC0→∞ by the Trapezoidal Rule:

Another method

AUC0→∞ determination from plasma concentration equations:

E. Calculate the plasma concentration and the amount of the drug in the

body 24 hours following administration of the drug:

L/hr4.33 25L 1-hr 0.173 CL

V K CL

k

CpAUCAUC nt

t

t n

n

10

μg/ml.hr 232.20 0.173

5 203.3 0 tAUC

∞→0AUC CL i.v. Dose

∞→0AUC V K V Cp

μg/ml.hr ∞→0 231.2 1-hr 0.173

g/ml 40

K

0Cp AUC

tk-e Cp= Cp

μg/ml 0.63 = 24 0.173-e 40 = Cp

V Cp = Ab

34

2. Drug: Ampicillin Clearance: 17 L/hr

Apparent volume of distribution: 25 L

Minimum plasma concentration needed to inhibit Haemophilus

Influenza approximately 0.5 µg/ml.

Questions:

A. After a 500 mg i.v. bolus dose of ampicillin, will the plasma concentration be above the inhibitory concentration of Haemophilus influenza 6 hours later?

mg 15.8 L25 g/ml 0.63 = Ab

1-hr 0.68 =

25L

L/hr17 =

V

CL = K

μg/ml 20 = mg/L 20 =

25L

mg 500 =

V

Dose = Cp

tk-e Cp= Cp

μg/ml 0.34 = 6 0.68-e 20 = Cp

35

B. At what time will the concentration drop below 0.5 µg/ml?

C. What dose should be given to achieve a concentration of 0.5 µg/ml 6 hours after the dose was given?

tk-e Cp= Cp * t0.68-e 20 = 0.5

* t0.68 - = 20

0.5 ln

hours 5.4= *t

* t0.68 - = 3.69-

tk-e Cp= Cp

6 0.68-e Cp= 0.5

0.0169 Cp= 0.5

mg/L 29.6 = μg/ml 29.6 = Cp

mg 740≈ 25L mg/L 29.6 =V Cp= Dose