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Module 2 # 3. Pharmacokinetics. how to make sense of the previous lesson or why do I have to learn this stuff?. 15. 12. 9. Cp tox. plasma conc. (mg/L). 6. Cp eff. 3. 0. 0. 10. 20. 30. time (hr). d rug dosing calculations. Concepts: target plasma concentration

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module 2 3

Module 2# 3

Pharmacokinetics

slide2

how to make sense of the previous lesson

  • or
  • why do I have to learn this stuff?
slide3

15

12

9

Cptox

plasma conc. (mg/L)

6

Cpeff

3

0

0

10

20

30

time (hr)

drug dosing calculations

Concepts: target plasma concentration

therapeutic window

After distribution:

Cp (free) approximates [D]

slide4
Therapeutic window: Difference between the minimum effective concentrations (MEC) required for a desired response and one that produces an adverse effect.
  • For some drugs it is small (only two- to three-fold difference)
  • E.g. digoxin, theophylline, lidocaine, aminoglycosides, cyclosporine, anticonvulsants.
the loading dose
the loading dose

If we know the target plasma concentration, and the Vd, we can calculate the i.v.loading dose:

L = Cp.Vd

For the oral loading dose we have to take the fraction bioavailable into account (0-1)

L =Cp.Vd/F

slide6
Loading Dose (DL) = a dose of drug sufficient to produce a plasma concentration of drug that would fall within the therapeutic window after only one or very few doses over a very short interval. It is larger than the dose rate needed to maintain the concentration within the window and would produce toxic concentrations if given in repeated doses.

Information needed to calculate a DL

  • Volume of distribution, usually in L or L/Kg - from literature - For example: Goodman & Gilman’ Pharmacology, 2001,  Table A-II-1
  • Desired concentration - from literature
slide7

Loading dose

  • DL = target Cp Vd/ F

e.g. lidocaine t1/2 = 1-2 h

Post MI arrythmias – life threatening – can’t wait 4-8 h – DL is used

  • Another example: Digoxin for heart failure
  • If only maintenanace dose is given it takes 10 days t1/2 = 61 h

DL = 1.5 ng/ml x 580 / 0.7 = 1243 micg ~ 1 mg

Can be given iv divided – 0.5 mg, aft 6-8 h 0.25 mg, aft 6h 0.125 mg and then 0.125 mg (to avoid overdigitalization and toxicity)

slide9

Drug with high Vd

high tissue

binding

slide10

So now if we want an equation,

W

C

W

=

V

C = --------

or

i.e

V

D = Cp Vd

where C = concentration (plasma), W = weight of drug

or dose of drug, and V = volume of solution or vol. of

distribution

slide11

Reference books say the Vd of theophylline

is 0.5 L per kg of body weight.

What is the IV loading dose to give a serum

theophylline concentration of 12 mg/L in

a 62 kg man?

V

C

or LD = Cp Vd

W

=

LD = (12 mg/L) (0.5 L/kg) (62 kg)

= 372 mg

we usually use mcg/mL (or ug/mL), which is the same thing as mg/L

slide12

Actually, theophylline for injection is avail-

able only as Aminophylline

(theophylline+ethylenediamine), which contains

80% theophylline. In this problem, we need

to give a larger weight of Aminophylline to

get the same weight of theophylline.

372 mg Theo

= 465 mg Amino

0.80

This gives us a loading dose

slide13

Half-life ( t 1/2 )

time interval after which the concentration

is half that at the beginning of the time

interval

has real-life meaning only in first-order

kinetics!

half life t 1 2

iv bolus

100

Cp (mg/L)

slope=-kel (k)

10

2

4

6

8

10

time (hr)

half life (t1/2)

the time taken for the Cp to fall by half

mathematically after i.v. bolus: Cpt = Cp0e-kt

where k is the rate constant (in hrs-1) so:

ln Cpt = ln Cp0 -kt

when Cpt/ Cp0 = 1/2 = t1/2 = ln2/k = 0.693/k

most drugs follow first order kinetics
most drugs follow first order kinetics:

rate of change of drug in the body

= -kel.Amount of Drug in the body .

where kel(k) is the rate constant of elimination

a few drugs (e.g. ethanol) follow 0 order kinetics:

rate of change = k

a few drugs (e.g. phenytoin) follow 1st order kinetics:

at low conc. and 0 order at high conc.

slide16

Rate processes (mainly elimination)

zero-order

constant amount of drug eliminated per

unit of time

elimination rate constant has units of weight/time:

Ke = 50 mg/h, for example

typical drug: ethanol

slide17

Rate processes (mainly elimination)

first-order

constant fraction or percent of drug eliminated per

unit of time

elimination rate constant has units of weight/time:

Ke = 0.25/h or 0.25 h-1 , for example.

typical drug: theophylline, most other drugs

slide18

first-order elimination following

single IV bolus dose

log

Cp

Cp

Time

Time

slide19

Special properties of first-order kinetics

Cp is 90% of the way to new steady-state

level after change of dose in 3 t1/2,

96% in 5 t1/2

this is true regardless of

initial concentration

final concentration

dose

but only for first-order kinetics !

slide20

First-order is also known as linear kinetics:

there is a linear relationship between

dose and steady-state concentration

Cpss

Dose

slide21

A patient’s peak serum phenobarbital level is

12.0 ug/mL at steady state on a dose of 60 mg/day.

If we want a level of 20 ug/mL, what new dose

should we order if we know that phenobarbital

has linear kinetics?

60 mg/day

12.0 ug/mL

=

20 ug/mL

X

X = 100 mg/day

slide22

If we change the dose to 90 mg/day, when can

we get another plasma level to check our calculations?

In adults, the half-life averages 100 hours. In

3 half-lives (300 h, or 12.5 days) we will be 90%

of the way from 12.2 ug/mL to the new steady-

state level (we hope it’s 20 ug/mL). This is close

enough to get a level. Reschedule the patient

to come back in 2 weeks for this level.

slide23

change

dose

level OK

level

not OK

Cp

Time

clearance
clearance

defined as the volume of blood from which drug is irreversibly removed per unit time.

ml/min

can calculate: UV/P = excretion rate/plasma concentration

at steady state: rate of excretion = rate in (the dose, M)

M = rate of excretion = Cldrug. Cp

clearance25
Clearance
  • Clearance is the VOLUME of plasma that can be freed of drug per unit time, i.e., gives estimate of function of organs of elimination and rate of removal of drug from the body
  • Rate of Elimination (mg/hr)      CL = -------------------------------- = vol/time      Concentration (mg/L)                
clearance calculations
Clearance calculations

From renal physiology:  

CL  =  (Ux * V)   /  Px 

L/h   =  ( mg/L  *  L/h)  /  mg/L

Where: Ux = urine concentration of "x" (mg/L)            V  = urine flow rate (L/h)            Px  = Systemic venous plasma concn. of "x" (mg/L)

            CL = renal clearance (L/hr)

clearance27
Clearance
  • Whole body clearance is simply the sum of all organ clearances

CL(body) = CL(renal) + CL(hepatic) + CL(pulmonary) + CL(etc)

relation between cl k and vd
relation between Cl, k and Vd
  • the rate of drug removal (drug out) depends on the amount of drug in the body and the rate constant of excretion (kel, k)
  • drug out = k*Amount of drug in the body (A)
  • A = Cp*Vd
  • drug out = k*Cp*Vd
  • but drug out = Cl*Cp
  • Cl*Cp = k*Cp*Vd
  • therefore: Cl = k*Vd
calculating clearance

100

80

60

plasma conc. (mg/L)

40

20

0

0

10

20

30

time (hr)

calculating clearance

Cldrug = Dose

AUC

t1/2 = 0.693

k

or

k = 0.693

t1/2

t1/2 = 0.693 Vd

Cl

Cldrug = k.Vd = (0.693/t1/2 ).Vd

Cltotal = Cl renal + Clnon-renal

slide30

Relationship between CL(body) and Ke

    • Whole body clearance and the elimination rate constant are related, but emphasize different aspects of the same processes.
    • Elimination Rate constant (Ke) is the FRACTION of drug in the body that is eliminated each unit of time, e.g., "fraction per hour".
    • Calculate Ke from CL & Vd: If one knows the Vd of a drug and the CL (body), one can calculate the FRACTION of the drug in the body that is removed per unit time. This is the same fraction as the fraction of "plasma equivalent" that is completely cleared of drug per unit time. This fraction is the Ke.
the maintenance dose
the maintenance dose
  • Maintenance Dose (DM) = The dose needed to maintain the concentrationwithin the therapeutic window when given repeatedly at a constant interval

M = Cldrug.Cp

for oral dosing

M = Cldrug.Cp/F

why clearance and vd
Why clearance and Vd?
  • clearance and volume of distribution are independent variables.
  • they determine the half-life (Cl = k*Vd)

t1/2 = 0.693

k

or

k = 0.693

t1/2

t1/2 = 0.693 Vd

Cl

bioavailability
bioavailability

absolute bioavailability = F

= AUCoral / AUCi.v.

= a fraction between 0 and 1

150

iv

100

plasma conc. (mg/L)

50

po

0

0

5

10

15

time (hrs)

Some drugs with low "F" in humans (F < .5) - Alprenolol , 5-fluorouracil, Lidocaine, Morphine, Nifedipine, Propranolol, Salicylamide

oral dosing
oral dosing

the time taken to reach steady state depends only on t1/2

(= 4-5 x t1/2)

Css = F.dose

CL.T

M (average Cp) =Cl.Cp/F

for drugs with short t1/2, must dose frequently

slide35

Time to steady-state or elimination is independent of dosage

  • Time required to reach steady state
    • Depends on elimination rate
    • Requires 5 elimination half-lives to reach 97% of steady state
    • Requires 5 elimination half-lives to eliminate 97% of drug
summary
summary
  • drug dosing should aim for the target plasma concentration.
  • the volume of distribution is useful in calculating the loading dose
  • the clearance is useful in calculating the maintenance dose
  • the time to reach steady state depends only on the
  • half life
slide39

About the linear model....

It is based on the exponential function

y = e x

where e is the base of natural logarithms ( ln )

on your calculator, e x is

or

e x

ln x

inv

slide40

C1= C2 e kt

C1/C2 = e kt

2/1 = e kt

ekt = 2

kt = 0.693

t = 0.693 / k

( C1, t1 )

log

Cp

( C2, t2 )

t

Time

C2 = C1 e - kt

slide41

A patient’s acetaminophen level at 2:00 PM is

86.2 ug/mL and at 6:00 the same day is 27.8

ug/mL. If we know that ke of acetaminophen

in this patient is 0.283 h -1 , what was the level

at 10:00 AM?

C1= C2 e kt

( C1, t1 )

log

Cp

C1= 86.2 e ( 0.283 ) ( 4 )

( C2, t2 )

t

C1= 267.3 ug/mL

Time

slide42

How did we know that ke of acetaminophen

in this patient is 0.283 h -1 ?

Either of the two previous equations can also be

stated as

C1

ln

C2

k e

=

t 2 - t 1

slide43

If the patient’s acetaminophen level at 2:00 PM (1400)

is 86.2 ug/mL and at 6:00 (1800) the same day

is 27.8 ug/mL,

C1

ln

C2

k e

=

t 2 - t 1

86.2 ug/mL

ln

27.8 ug/mL

k e

( 1800 - 1400 ) h

k e

=

0.283 h - 1

slide44

What is the half-life of acetaminophen in this

patient?

One last equation:

0.693

t 1/2 =

k e

So

0.693

t 1/2 =

0.283 h - 1

t 1/2 =

2.45 h

slide45

What good is all this?

Well, we started with the bottom line

and worked backward so that the importance

of each equation might be clear.

The next slide is the problem

as it would occur in real life. Try working

through it toward the bottom line, starting

with information existing at the start of the

case.

slide46

A patient is admitted following a suicide attempt in

which he took an unknown quantity of acetaminophen.

After appropriate gastric lavage and supportive ther-

apy, he is being considered for therapy with the spe-

cific antidote (N-acetylcysteine). This antidote is rec-

ommended if the serum acetaminophen concentration

4 h after the ingestion is > 200 ug/mL. The overdose

is reported to have occurred at 10 AM (1000); at

2:00 PM (1400) his serum acetaminophen level was

86.2 ug/mL and at 6:00 (1800) the same day it is

27.8 ug/mL. Should he receive N-acetylcysteine? Is

this patient’s half-life consistent with the population

average (2.5 h) or is there something unusual about

his metabolism? Work it out.

slide47

A patient who is taking 400 mg of carbamazepine

every 12 hours has a trough serum carbamazepine

level of 4.2 ug/mL. Her seizure control is

inadequate; if we want to achieve a serum drug

level of 8.0 ug/mL, what should the new dose be?

We know the drug usually has linear kinetics.

Hint: we usually use trough levels with this drug.

With linear kinetics, the trough levels are pro-

portional to the dose just like peak levels are.

761 mg every 12 h; you would probably do this

increase in two steps a week apart.

slide48

A patient weighs 88 pounds. He needs a

loading dose of phenytoin for status epilepticus.

If the average Vd is 0.65 L/kg, what is the correct

loading dose? (Although this drug has nonlinear

kinetics, this fact only influences accumulation

during repeated dosing; it has no effect on loading

doses.) The therapeutic range of this drug is

10-20 ug/mL, and we would like to achieve a level

of 12 ug/mL.

312 mg - Did you forget to convert pounds

to kilograms?

slide49

A 76-kg patient has received an intravenous bolus

of 400 mg of a drug with the following population

averages: Vd 2.3 L/kg, t 1/212 h, minimum effective

concentration 1 ug/mL. How long will it be before

another dose is needed? Assume linear kinetics.

(Hint: first calculate ke , then calculate the drug

concentration after the bolus dose. Then use the

C2equation and try various values for the elapsed

time to see the largest value that keeps C2above

1.0 ug/mL. [If you like math, you may want to solve

this equation for t ; first take the logarithm of both

sides and then solve.]

I get about 14 hours.

slide50

First-order: steady state following

repeated IV bolus dosing

all peaks same

Cp

all troughs same

Time