מצגת שניה 18.3 ושיעור שני.pdf

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Pharmacokinetics 2:

IV Bolus Injection inOne CompartmentModel

Book Chapter 3

1

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Lecture Outline

Model description Apparent volume of distribution Elimination half-life Elimination rate constant Plotting plasma levels Monitoring urinal excretion

- amount remaining to be excretedmethod- rate of excretion method

2

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The Scheme of IV Bolus

1 compartment

3

This is the

simplestform of the model:

Distribution isimmediate

No metabolism 100% excretedunchanged inurine

.

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The General Equations of the Model

4

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Mass vs. Concentration

5

And therefore

Is the same as

And also

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Mass vs. Concentration Mass Graphs

6

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Mass vs. Concentration Concentration

Graphs

7

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Mass vs. Concentration

8

What is the difference between the two methods inpractice?

.

.

Which one is more practical and important? -

- .

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Application of the Model

9

Now that we have introduced the theoretical model,we will learn to apply it to find the following:

Apparent volume of distribution ( V or V d ) The elimination half-life ( t ) The elimination rate constant ( K or K el ) The systemic clearance ( Cl )

s

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Apparent Volume of Distribution

105-6

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12

If we know the amount (=mass; = g, mg or g) of Iodine weput in the baker and measure the concentration of Iodinein the water (analogous to plasma or serum), we cancalculate the volume of the beaker irrespectively of itsactual volume.

According to the calculation, beaker B will have largervolume

?

. , , .

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Apparent Volume of Distribution - Calculation

13

To calculate the volume of distribution, we need toknow the drug dose X 0 (mass) and the initial drugconcentration in the plasma (C

p )

0(mass/volume)

Then, (C p )0 = X 0 /V d V d = X 0 / (C p )0 Because actual (C p )0 cannot be measured, and wecant measure Xt either, we use the extrapolation of the C p vs. t graph

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Actual Factors That Determine the V d

14

The volume of distribution is determined by the physico-chemical properties of the drug

The most important factor is the lipophilicity of the drug: themore lipophilic, the larger the V d

Also important is the membrane permeability of the drug (oftenrelated to lipophilicity)

Many drugs are highly bound to plasma proteins, or are toopolar to penetrate from blood to tissues and therefore have verylow Vd these include many acidic drugs like salycilates,sulfonamides and penicillins

Theoretically, V d can vary between 3.5 and 500L. Actual V d svary between 7-10L to 200L

.

.

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The Nature of V d

15

Vd is a parameter, which means that for one drug andone individual it is a constant

It is thus independent of dose and time in all cases

It is similar, but not identical between persons It may vary with age and gender you can use apooled population average in some cases

Vd for a drug in populations is often expressed asV/body mass in order to eliminate weight differencesbetween individual

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The Nature of V d

16

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Elimination Half life

175-6

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Elimination Half Life - Definition

18

The elimination half life t is the time when (C p)0 (and X0) is reduced by half

t , like Vd is a parameter it varies between drugs

and individuals, but is the same with time and dose .

[] .

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Calculation of t

19

So, when t= t

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Calculation of t

20

Notice that we can calculate t with equation only if we know the elimination rate constant

We can calculate both graphically from the semi-logconcentration vs. time graph

In such a graph, you can use any two concentrationswhere one is half of the other instead of (C p )0 and0.5 (C p )0

2 .

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Calculation of t

21

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Examples of t

22

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The Elimination Rate Constant

23

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24

The elimination rate constant K or K el can becalculated using several related methods

1. From the equation:

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25

2. Directly from the slope of the graph:-K = (slope) x 2.303, so

Where Y represents C p

and K = - (slope) x 2.3033. From t , because we know that:t = 0.693 /K K = 0.693/ t

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26

3. From t , because we know that:t = 0.693 /K K = 0.693/ t

The unit of K is reciprocal time: time -1 in minutes orhours

K represents the total elimination (irreversible loss) of drug from the body

Therefore, it is composed of the rate constant forelimination of unchanged drug via the urine (K u) andmetabolism (K m), so that:

K = K u + K m , or for two metabolites: K = K u + K m1 + K m2

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Plotting Plasma levels

27

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Example of plotting C p vs. time

28

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From the semi-log graph we first determine (C p )0 bydrawing the line from our first sample point at 1h to t =0

(C p )0 = 63 g/mL

Then, we determine t graphically using convenientconcentrations: 20, 10 and 5 g/mL t = 1.3h

We can now calculate V d :and X0 a= 600mg and (C p )0 = 63 g/mLso

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Now we can calculatethe overall eliminationrate constant from t:

K = 0.693/ t = 0.693/1.3h= 0.533h -1

or, from the slope of thegraph:

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How we will calculate in our exercise

32

The use of log-paper and graphical methods for findingslopes and other values are outdated

Therefore, we will use Microsoft Excel and a different

method of calculating K , t When we get a set of values of plasma levels vs. timepoints, we first convert the values to log, then put themin a graph of log(C

p) against time

Because we expect a straight line, we can use linearregression to find the best trend line ( )

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Time (h) C p(g/dL) log(Cp)

0.5 211 2.32

1.0 189 2.28

1.5 163 2.21

2.0 141 2.152.5 111 2.05

3.0 97 1.99

4.0 78 1.89

5.0 62 1.797.0 39 1.59

10.0 18 1.26

15.0 6 0.78

24.0 0.7 -0.15

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y = -0.1048x + 2.3384R = 0.9983

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

0 5 10 15 20 25 30

L o g

[ X ] ( g

/ d L

)

time (h)

Plasma Levels of Drug X

Now we can use the trend line equation to determineour pharmacokinetic parameters

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The trend line equation is

where -0.1048 represents the slope, and 2.338 the

y-axis intercept Now

And

xC y p 10483384.2)log(

1241.0303.21048.0303.2303.2

h slope K K

slope

h

h K

t 88.2

241.0

693.0693.01

21

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If we also know the dose ( X 0), we can now calculatethe apparent volume of distribution V from X 0 and ( C p )0

We know that the Y-axis intercept represents (C p )0 , but

we have to remember that our line-equation is log( C p ) and therefore, we have to convert our intercept back to(C p )0

Given the dose was 100mg, then the apparent volumeof distribution is:

dL g Y antiC ercept p /0.21810)log()(3384.2

int0

L

Lmg

mg

dL g

mg

C

X V

V

X C

p

d

d

p 9.45

/18.2

100

/0.218

100

)(

)(0

000

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Performing an IV Bolus Experiment

37

When performing a real experiment:1. We administer a known dose of drug in a suitable vehicle(concentration x volume = dose) 2. Collect blood samples at least up to t x 4.32, more often at

the beginning 4.32 3. Clean and analyze the samples with a suitable analytical

method (most commonly bioassay with HPLC ) anddetermine C p at each time-point

4. Plot the data in a semilogarithmic plot

5. Calculate our PK parameters: V d , K el , t and (C p )0

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Performing an IV Bolus Experiment

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Why using t x 4.32 ?

4.32 5%

Because t x 4.32 = t 5%

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Monitoring the Urine After IV Bolus

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Why Should We Do It?

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Monitoring the urine for unchanged drug andmetabolites is a way to investigate drug metabolism

It is an easy and non-invasive method obtain PK data

It has the obvious disadvantages that we can not samplevery often, and we can not determine the exact sampletime

The data obtained reflects average excretion of aninterval

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Cumulative Excretion in Urine Graph

41

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Cumulative Excretion in Urine

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First of all, we can determine if there is a drugmetabolism: if (X u ) X 0, there is metabolism,and K Ku

, . If the metabolite is excreted in urine (which is almost

always the case), then the mass balance dictates:where: (X) t is the amount of drug in the body at time t,(Xu) is the amount excreted in the urine, (X m) is theamount of metabolite in the body, and (X mu ) is themetabolite excreted in the urine

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Cumulative Excretion in Urine

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How Urinary Excretion Data are used

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There are two methods we can use to calculateparameters from urinary excretion data:1. The method based on the Amount Remaining to be

Excreted, called the ARE, or the Sigma Minus method 2. The Rate of Excretion Method

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Amount Remaining To BeExcreted (ARE) Method

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The ARE method

465-4

In this method, we use the fact that the amount of drugremaining in the body is equal to the dose minus theexcreted amount to make an elimination plot and

calculate the elimination constant. So all the drug isexcreted in urine:

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The ARE method

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We can describe the amount remaining in the body by

where (X u ) - (X u )t is the amount remaining in the body and (X u ) = X 0 the given dose Now we can calculate the excretion rate constant from the graphof the remaining amount vs. time

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The ARE method

48

5:00

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The ARE method Limitations

495-4

The ARE method is difficult to apply because:1. Urine must be collected until there is almost no drug

in urine because we need to know that (X u) X0

We need about 7t - this can be very long time7

, !!! .

2. We need to collect all the urine in all samples3. There is a cumulative build up of errorThe ARE method is thus not very precise and very timeconsuming method especially for drugs with long t

, ARE

,

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ARE When Drug Metabolism Occurs

50

When there is drug metabolism, then (X u) X0 andKu K

Using Laplace transform to integrate, weget

This equation allows for predicting the amount excretedat time t the if elimination rate constant K and excretionrate K

u are known

_XK

KU

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When t = , the exponential term progresses to 0

and the equation becomes

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If we substitute (K uX0)/K with (Xu) in

we get:

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Now, we transform

into logarithmic equation

where the term on the left represents the amountremained to be excreted

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From the equation, we can draw the graph, calculate Kand extrapolate to find (X u) even if we dont know if there is metabolism

Notice thesimilarity to C p calculations

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Rate of Excretion (ROE) Method

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ROE Method: No Metabolism

56

We have already seen thatand that

if Ku= K

Now, we substitute the X from the differential equation,we get

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However, in practice we measure and average rate overthe period of urine collection.

Therefore, we express the equation as:

where we take the average time (t 2+t 1 )/2 to mach theaverage time to the average rate

.

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The graphs show the rate of urinary excretion vs. the average timeover which the excretion was measured in linear scale (a) andsemilog scale (b). Notice that the intercept is not X 0, but KuX0

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An example:

Notice that you calculate the rate over the time interval,

but you plot it against the average time measured from t 0 From the graphs of d Xu/ dt, we can calculate t , X 0 and K

K

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ROE Method: With Metabolism

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The fact that X 0 (Xu) has no consequences for thecalculation because we extrapolate to X 0 and not to (X u)

Naturally, K u K

We can use and if the drug is also

monitored in the blood

Notice that we have to know K and X 0

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ROE Method: With Metabolism

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The logarithmic form

can be used todetermine K u at the intercept of thegraph

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ROE Method

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This method tends to overestimate the interceptbecause of the way the time average is calculated

The error is proportional to the sample interval

Comparison of K and K u can tell us if X is metabolized The table shows the

overestimation of the

intercept K u(X0) in %as function of theinterval betweensamples rel. to t

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AREvs. ROE Methods: an Example

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A bolus injection of 80mg of drug U is administered The data and calculation are given in the table below forboth methods

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AREvs. REM Methods: an Example

64 ARE ROE

-----------------------------------=====================-

ROE

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AREvs. REM Methods: an Example

Both methods are useful In the ARE method, we have to collect samples for verylong time in order to avoid error in the estimation of X

X In the ROE method, we have to take samples frequently inorder to avoid overestimation of the intercept K u(X0) andtherefore, overestimation of K u

..

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מצגת שניה 18.3 ושיעור שני.pdf