General Principles of Pharmacology and Toxicology
Parisa Gazerani, Pharm D, PhD Assistant Professor
Center for Sensory-Motor Interaction (SMI) Department of Health Science and Technology Aalborg University Pharmacokinetics II
2 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)
3 Question 1
What Is the is the loading dose required for drug A if:
Target concentration is 10 mg/L VD is 0.75 L/kg Patients weight is 75 kg
Answer is on the next slide
4 Answer – Loading Dose of Drug A
VD = 0.75 L/kg x 75 kg = 56.25 L Target 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.
5 Question 2
What maintenance dose is required for drug B if:
Target average SS concentration is 10 mg/L CL of drug A is 0.015 L/kg/hr Patient weighs 75 kg
Answer is on the next slide
6 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
7 • 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.
C1 tion
C2 Drug Concentra
Time
8 t1/2 = 0.693/k C0
C0/2 t1/2 t1/2 Log Concn. t1/2
Time
Time to eliminate ~ 4 t1/2
9 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
10 Accumulation to Steady State 100 mg given every half-life
194 … 200 187.5 175 150
100 97 … 100 87.5 94 97 75 50
11 C
Cpav
t
Four half lives to reach steady state
12 Test Yourself!
Elimination of a drug refers to: a) Excretion of its breakdown products in the urine b) Renal excretion of the unchanged drug c) Uptake of drug from the blood into the liver d) Metabolism of the drug in the liver
e) Distribution of the drug into fat Correct answers: b, d b, answers: Correct
13 Test Yourself!
The elimination rate of a drug is: a) A constant for a particular drug and patient b) The extent to which it is excreted in urine c) Directly proportional to the plasma drug concentration d) Directly proportional to the clearance
e) The extent to which the drug is excreted in the faeces
d c, answers: Correct
14 Test Yourself!
Clearance: a) Depends on the elimination rate b) Refers to the efficiency of elimination of drug by an organ or the whole body c) Cannot be greater than blood flow to an organ d) Determines the steay state drug concentration during constant dosing
e) Is determined by the half-life Correct answers: b, c, d c, b, answers: Correct
15 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 half c) The constant relating the rate of elimination of a drug to the plasma drug concentration d) The amount of drug metabolized per unit time
e) The amount of drug excreted in urine per unit time Correct answers: a, c a, answers: Correct
16 Test Yourself!
Volume of distribution is: a) The total volume of the body b) The volume of the extracellular fluid c) Equal to the volume of total body water d) The constant relating amount of drug in the body to the plasma drug concentration
e) The volume of the body minus the blood volume Correct answer: d answer: Correct
17 Test Yourself!
The loading dose of a drug is determined by: a) The clearance b) The elimination rate c) The target plasma drug concentration d) The volume of distribution
e) The molecular weight of drug
d c, answers: Correct
18 Test Yourself!
The rate of distribution of a drug can determine: a) The volume of distribution b) The onset of drug effect c) The rate of drug elimination d) The duration of drug effect
e) The clearance Correct answers: b, d b, answers: Correct
19 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 dose c) The clearance d) The rate of onset of drug effect
e) The duration of drug effect Correct answer: a answer: Correct
20 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 half d) Decreases as elimination constant increases
e) Increases as the elimination constant increases Correct answers: a, c, d c, a, answers: Correct
21 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 answer: Correct
22 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 answer: Correct
23 Test Yourself!
Half-life determines: a) The loading dose b) The time to reach steady state c) The drug concentration at steady state during constant dosing d) The duration of action after a single dose
e) The flactuation in plasma drug concentration during a dosing interval Correct answers: b, d, e d, b, answers: Correct
24 Test Yourself!
Bioavailability: a) Is the extent to which a drug is absorbed from the gut b) Is the fraction of the drug metabolized first pass by liver c) Is the fraction of the dose reaching the systemic circulation intact d) Is a measure of both first-pass metabolism and absorption from the gut
e) Only refers to intravenous drug administration
d c, answers: Correct
25 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 1.0 c) The generic formulation would usually be regarded as bioequivalent to the innovator brand d) The absolue bioavailability of the innovator brand is 0.83
e) The generic brand is better than the innovator brand Correct answers: a, c a, answers: Correct
26 Test Yourself!
Which of the following routs of administration completely avoid first-pass clearance? a) Buccal b) Sublingual c) Rectal d) Oral
e) Transdermal Correct answers: a, b, e b, a, answers: Correct
27 Test Yourself!
With respect to the renal clearance of drugs by glomerular filtration: a) Glumerular filtration rate is about 40% of renal blood flow b) Only unbound drug is filtered c) Drug clearance by glumerular filtration is equal to glomerular filtration rate
d) Both bound and unbound drug can be filtered Correct answer: b answer: Correct
28 Test Yourself!
The proportion of drug re-absorbed from the renal tubule depends on: a) The glomerular filtration rate b) The urine flow rate c) The extent of drug secretion into the renal tubule d) The liphophilicity of the non-ionized drug
e) The urine pH Correct answers: b, d, e d, b, answers: Correct
29 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 cimetidine b) Reduced by probenecid c) Increased by acidifying the urine d) Increase by alkalinizing the urine
e) Increased by an increase in urine flow rate Correct answers: b, d, e d, b, answers: Correct
30 Test Yourself!
The fraction of a drug unbound in plasma is: a) Usually not affected by the drug concentration b) Independent of the concentration of binding protein c) Dependent on the affinity of drug for the binding protein d) Usually increased as drug concentration increases
e) Usually decreases as drug concentration increases Correct answers: a, c a, answers: Correct
31 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 distribution b) Dose c) Half-life d) Clearance
e) bioavailability Correct answers: b, d b, answers: Correct
32 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 concentration b) The clearance c) The elimination rate constant or half-life d) The bioavailability
e) The absorption rate constant Correct answers: b, c, d c, b, answers: Correct
33 Explore it!
Table gives plasma drug Time (h) Plasma Con. concentrations (Cp) obtained following (ug/mL) an intravenous bolus administration of 0.5 68.0 a 250 mg dose of a drug that exhibited the characteristics of a one- 1.0 54.0 compartment model and was eliminated exclusively by urinary 2.0 30.0 excretion. 3.0 18.5
5.0 6.0
7.0 1.8
34 Plotting the data revealed the following: a) The elimination half life (t1/2) =1.275 h b) The overal elimination rate constant (K) = 0.543/h c) The intial plasma concentration (Cp)0 = 90 ug/mL
100 Intercept = initial plasma conc.
Slope= -K/2.303
10 Drug conc
5 t1/2
1 Time (h)
35 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?
–Kt –Kt Cp = (Cp)0 e or Cp/(Cp)0 = e or Ln (Cp/(Cp)0)/-K = t
36 Answer
V = Dose/(Cp)0 V= 250/90= 2.77 L
(Cp)0 = Dose/V (Cp)0 = 175/2.77 = 63.176
–Kt 0.543 x 75min or 1.25 h Cp = (Cp)0 e Cp = 63.176 e = 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
37 Take a closer look!
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38 Take a closer look!
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