general principles of pharmacology and toxicology ii.pdf · general principles of pharmacology and...

Post on 16-May-2020

30 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

General Principles of Pharmacology and Toxicology

Parisa Gazerani, Pharm D, PhDAssistant Professor

Center for Sensory-Motor Interaction (SMI)Department of Health Science and TechnologyAalborg University

2

Pharmacokinetics II

3

Pharmacokinetics made easy!

• In pharmacokinetics the body is represented as a single or multiple compartments in to which the drug is distributed.

Volume of Distribution = Dose/Plasma Concentration

Clearance = Elimination Rate Constant x Volume of Distribution (CL = kV)

Loading dose = Cp(Target) x VD

Maintenance Dose = CL x CpSSav (target average steady state drug concentration)

4

Question 1

What Is the is the loading dose required for drug A if:

Target concentration is 10 mg/LVD is 0.75 L/kgPatients weight is 75 kg

Answer is on the next slide

5

Answer – Loading Dose of Drug A

VD = 0.75 L/kg x 75 kg = 56.25 LTarget Conc. = 10 mg/L

Loading Dose = Target Concentration x VD

Loading Dose = 10 mg/L x 56.25 L = 565 mg

This would probably be rounded to 560 or even 500 mg.

6

Question 2

What maintenance dose is required for drug B if:

Target average SS concentration is 10 mg/LCL of drug A is 0.015 L/kg/hrPatient weighs 75 kg

Answer is on the next slide

7

Answer – Maintenance Dose of Drug B

Maintenance Dose = CL x CpSSav

CL = 0.015 L/hr/kg x 75kg = 1.125 L/hr

Dose = 1.125 L/hr x 10 mg/L = 11.25 mg/hr

So will need 11.25 x 24 mg per day = 270 mg

8

• Half-life is the time taken for the drug concentration to fall to half its original value

• The elimination rate constant (k) is the fraction of drug in the body which is removed per unit time.

Dru

g C

once

ntra

tion C1

C2

Time

9

t1/2 = 0.693/kC0

Log

Con

cn.

C0/2 t1/2t1/2

t1/2Time

Time to eliminate ~ 4 t1/2

10

Steady-state (SS)

• Steady-state occurs after a drug has been given for approximately five elimination half-lives.

• At steady-state the rate of drug administration equals the rate of elimination and plasma concentration - time curves found after each dose should be approximately superimposable.

• Rate in = Rate Out

• Reached in 4 – 5 half-lives (linear kinetics)

• Important when interpreting drug concentrations in TDM or assessing clinical response

11

Accumulation to Steady State 100 mg given every half-life

100100

187.5187.5194194

175175

150150

757587.587.5 9494 9797

5050

200200……

…… 100100

12

C

t

Cpav

Four half lives to reach steady state

13

Test Yourself!

Elimination of a drug refers to:

a) Excretion of its breakdown products in the urineb) Renal excretion of the unchanged drugc) Uptake of drug from the blood into the liverd) Metabolism of the drug in the livere) Distribution of the drug into fat

Correct answers: b, d

14

Test Yourself!

The elimination rate of a drug is:

a) A constant for a particular drug and patientb) The extent to which it is excreted in urinec) Directly proportional to the plasma drug concentrationd) Directly proportional to the clearancee) The extent to which the drug is excreted in the faeces

Correct answers: c, d

15

Test Yourself!

Clearance:

a) Depends on the elimination rateb) Refers to the efficiency of elimination of drug by an organ or the

whole bodyc) Cannot be greater than blood flow to an organd) Determines the steay state drug concentration during constant

dosinge) Is determined by the half-life

Correct answers: b, c, d

16

Test Yourself!

Two definition of clearance are:

a) The volume of blood or plasma irreversibly cleared of drug per unit time

b) The time taken to reduce the plasma concentration by halfc) The constant relating the rate of elimination of a drug to the plasma

drug concentrationd) The amount of drug metabolized per unit timee) The amount of drug excreted in urine per unit time

Correct answers: a, c

17

Test Yourself!

Volume of distribution is:

a) The total volume of the bodyb) The volume of the extracellular fluidc) Equal to the volume of total body waterd) The constant relating amount of drug in the body to the plasma drug

concentratione) The volume of the body minus the blood volume

Correct answer: d

18

Test Yourself!

The loading dose of a drug is determined by:

a) The clearanceb) The elimination ratec) The target plasma drug concentrationd) The volume of distributione) The molecular weight of drug

Correct answers: c, d

19

Test Yourself!

The rate of distribution of a drug can determine:

a) The volume of distributionb) The onset of drug effectc) The rate of drug eliminationd) The duration of drug effecte) The clearance

Correct answers: b, d

20

Test Yourself!

Volume of distribution can be measured using:

a) The plasma drug concentration extrapolated back to zero time after administration

b) The rate of elimination at a particulatr time after the dosec) The clearanced) The rate of onset of drug effecte) The duration of drug effect

Correct answer: a

21

Test Yourself!

Half-life:

a) Is the time taken for the plasma concentration to fall by half b) Has units of ”per hour”c) Is the time taken for the amount of drug in the body to fall by halfd) Decreases as elimination constant increasese) Increases as the elimination constant increases

Correct answers: a, c, d

22

Test Yourself!

After a single dose of a drug which has a half life of 12 hours, what percentage of the dose is still in the body after 1 day?

a) 87.5%b) 75%c) 50%d) 25%e) 12.5%

Correct answer: d

23

Test Yourself!

During a constant rate intravenous infusion of a drug with an elimination rate constant of 0.173 per hour, the plasma drug concentration will be what percentage of steady state after 16 hours?

a) 25%b) 50%c) 75%d) 87.5%e) 93.75%

Correct answer: e

24

Test Yourself!

Half-life determines:

a) The loading doseb) The time to reach steady statec) The drug concentration at steady state during constant dosingd) The duration of action after a single dosee) The flactuation in plasma drug concentration during a dosing interval

Correct answers: b, d, e

25

Test Yourself!

Bioavailability:

a) Is the extent to which a drug is absorbed from the gutb) Is the fraction of the drug metabolized first pass by liverc) Is the fraction of the dose reaching the systemic circulation intactd) Is a measure of both first-pass metabolism and absorption from the

gute) Only refers to intravenous drug administration

Correct answers: c, d

26

Test Yourself!

A new generic drug is tested in a bioavailability study against the innovator brand. The AUC oral for the generic is 1200 mg*hour/L and that for the innovator brand is 1000 mg*hour/L

a) The relative bioavailability of the generic formulation compared to the innovator brand is 1.2

b) The absolute bioavailability of the innovator brand is 1.0c) The generic formulation would usually be regarded as bioequivalent

to the innovator brandd) The absolue bioavailability of the innovator brand is 0.83e) The generic brand is better than the innovator brand

Correct answers: a, c

27

Test Yourself!

Which of the following routs of administration completely avoid first-pass clearance?

a) Buccalb) Sublingualc) Rectald) Orale) Transdermal

Correct answers: a, b, e

28

Test Yourself!

With respect to the renal clearance of drugs by glomerular filtration:

a) Glumerular filtration rate is about 40% of renal blood flowb) Only unbound drug is filteredc) Drug clearance by glumerular filtration is equal to glomerular filtration

rated) Both bound and unbound drug can be filtered

Correct answer: b

29

Test Yourself!

The proportion of drug re-absorbed from the renal tubule depends on:

a) The glomerular filtration rateb) The urine flow ratec) The extent of drug secretion into the renal tubuled) The liphophilicity of the non-ionized druge) The urine pH

Correct answers: b, d, e

30

Test Yourself!

The renal clearance of a lipophilic drug which is a weak acid with a pKa of 5.0 is likely to be:

a) Reduced by cimetidineb) Reduced by probenecidc) Increased by acidifying the urined) Increase by alkalinizing the urinee) Increased by an increase in urine flow rate

Correct answers: b, d, e

31

Test Yourself!

The fraction of a drug unbound in plasma is:

a) Usually not affected by the drug concentrationb) Independent of the concentration of binding proteinc) Dependent on the affinity of drug for the binding proteind) Usually increased as drug concentration increasese) Usually decreases as drug concentration increases

Correct answers: a, c

32

Test Yourself!

The AUC for the plasma concentration versus time curve of a drug after a single intravenous dose is determined by:

a) Volume of distributionb) Dosec) Half-lifed) Clearancee) bioavailability

Correct answers: b, d

33

Test Yourself!

Calculation of the volume of distribution of a drug after a single oral dose requires which of the following parameters?

a) The time of max plasma concentrationb) The clearancec) The elimination rate constant or half-lifed) The bioavailabilitye) The absorption rate constant

Correct answers: b, c, d

34

Explore it!

Table gives plasma drug concentrations (Cp) obtained following an intravenous bolus administration of a 250 mg dose of a drug that exhibited the characteristics of a one-compartment model and was eliminated exclusively by urinary excretion.

Time (h) Plasma Con. (ug/mL)

0.5 68.0

1.0 54.0

2.0 30.0

3.0 18.5

5.0 6.0

7.0 1.8

35

Plotting the data revealed the following:

a) The elimination half life (t1/2) =1.275 hb) The overal elimination rate constant (K) = 0.543/hc) The intial plasma concentration (Cp)0 = 90 ug/mL

Slope= -K/2.303

100 Intercept = initial plasma conc.

Dru

g co

nc

10

Time (h)

t1/2 5

1

36

Questions

a) The apparent volume of distribution (Vd)?

V = Dose/(Cp)0

b) The drug plasma concentration at 75 min following the administration of a 2.5 mg/kg dose to a subject weighting 70 kg?

c) The time at which the plasma concentration of the drug will fall below 20 ug/mL, following the administration of a 275 mg dose?

Cp = (Cp)0 e –Kt or Cp/(Cp)0 = e –Kt or Ln (Cp/(Cp)0)/-K = t

37

Answer

V = Dose/(Cp)0 V= 250/90= 2.77 L

(Cp)0 = Dose/V (Cp)0 = 175/2.77 = 63.176

Cp = (Cp)0 e –Kt Cp = 63.176 e 0.543 x 75min or 1.25 h = 31.956

(Cp)0 = Dose/V (Cp)0 = 275000ug/2770 mL = 99.277 ug/mL

t = Ln (Cp/(Cp)0)/-K t = Ln (20/99.277) 0.543 = 2.94 h

38

Take a closer look!

Find more problem sets with answers in this book.

39

Take a closer look!

Find more self-test questions with answers in this book.

top related