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Lecture 9

Applications of pharmacokinetic models of drug action

Pharmacokinetics – applications – Exercise I.

Tse and Szeto (1982) published a study on bioavailability of theophylline after intrave-nous administration and oral administration to dogs (Tab. 1). Concentration of theophylline in plasma was described by a one-compartment model. The following doses were used- intravenous dose: 50 mg of aminophylline (85% theophylline) - oral dose A: elixophylline (theophylline, 100 mg capsule);- oral dose C: aminophylline (aminophylline, 200 mg tablet = 170 mg theophyl.)

Plot a graph of the concentration in plasma vs. time, and determine: a) half-life of elimination (t1/2) of theophylline after intravenous administration and

oral administration of a capsule and tablet,

b) rate constant of elimination (kel) of theophylline after intravenous administration,

oral administration of a capsule and tablet,

c) apparent distribution volume (Vd) of theophylline for i. v. administration,

d) half-life (t1/2) and rate constant (ka) of absorption for oral administration of theophylline,

e) determine the area (AUC)024 under the curve cp(t) with help of trapezoidal rule for

integration for all doses,

f) for the area (AUC)024 calculate the total area (AUC)0

under the curve cp(t) for all doses and routes of administration,

g) using the (AUC)0 determine the absolute bioavailability (F) of doses administered

p. o.,

h) determine relative bioavailability (Frel) of dose administered p. o . in the form of 200 mg tablet as compared to capsule.

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33

Pharmacokinetics – applications – Exercise I.

Tab. 1.

Plasma

Cp, Vd

i.v. / p.o. admin.

Ddose of

drug

central compartment

elimination

mk

absorption

ak

ek

metabolisation

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Pharmacokinetics – applications – Exercise I.

Solution:a) from graph: half-life of elimination - i. v. admin: 42,5 mg theophylline: t1/2 = 5,75

h; p. o. admin., 100 mg capsule: t1/2 = 5,75 h; p. o. admin, 200 mg tablet: t1/2 = 5,75 h. Rate of elimination of drug is independent of the route of administration and drug formulation.

b) rate constant of elimination: kel = ln2/t1/2 = 0,1205 h for kinetics of 1st order

c) apparent distribution volume: Vd = D/cp0 (dose/init. conc.)

D = 50 mg.0,85 = 42,5 mg theophylline, cp0 = 4,8 g.dm-3, Vd = 8,854 dm3

d) half-life (t1/2)abs and rate constant (kabs) of absorption can be determined graphically from the dependence log cp vs. time:

- p. o. admin, 100 mg capsule: (t1/2)abs = 0.1875 h, kabs = 3,696 h-1

- p. o. admin, 200 mg tablet: (t1/2)abs = 0.200 h, kabs = 3,465 h-1

Values of kabs are fairly high, maximum on the curves appears in short time tmax ~ 1,5 h

0

2

4

6

8

10

12

14

16

0 2 4 6 8 10 12 14 16 18 20 22 24

Intravenóznepodanie

Perorálnepodanie A

t1/2

-0,6

-0,4

-0,2

0

0,2

0,4

0,6

0,8

1

1,2

1,4

0 2 4 6 8 10 12 14 16 18 20 22 24

time [h]

log c

p0

time [h]

cp0

[g d

m-3

]

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Pharmacokinetics – applications – Exercise I.

e,f) (AUC)024 in the time from t = 0 till t = 24 we calculate from the graph with help

of the trapezoid rule, AUC)24 = cp,24/kel, (AUC)0

= (AUC)024 + (AUC)24

:- i. v. admin: 42,5 mg theophylline:

(AUC)024 = 33,380 μg.ml-1.h, (AUC)24

= 2,074 μg.ml-1 h, (AUC)0 = 35,454

μg.ml-1.h- p. o. admin: 100 mg theophylline capsule: (AUC)0

24 = 96,293 μg.ml-1.h, (AUC)24 = 8,298 μg.ml-1.h, (AUC)0

= 104,591μg.ml-1.h

- p. o. admin: 200 mg aminophylline tablet: (AUC)0

24 = 137,212 μg.ml-1.h, (AUC)24 = 7,054 μg.ml-1.h, (AUC)0

= 144,266μg.ml-1.h

g) alternatively: (AUC)0 = D0/Vd.kel where dose D0 for i. v. bolus 50 mg of

aminophylline (42,5 mg theophylline), Vd = 8854,16 ml, kel = 1066.926 ml.h-1 will be (AUC)0

= 39,834 μg ml-1 h(AUC)0

= [kaFD0/Vd(ka-kel)]. (1/kel – 1/ka)- p. o. admin: 100 mg theophylline capsule: (AUC)0

= 104,366 μg.ml-1.h- p. o. admin: 200 mg aminophylline (170 mg theophylline) tablet: (AUC)0

= 160,202 μg.ml-1.h- p. o. admin: 100 mg theophylline in tablet: (AUC)0

= 94,236 μg.ml-1.habsolute bioavailability: F = {[(AUC)0

]capsule/Dcapsule}/ {[(AUC)0 ]i.v.bolus/Di.v.bolus} =

{104,366/100 mg theoph.}/{39,834/42,5 mg theoph.} = 1,11 (F 1)uncertainty in AUC determination

F = {[(AUC)0 ]tablet/Dtablet}/ {[(AUC)0

]i.v.bolus/Di.v.bolus} = 0,905 (90,5%)(90,5% theophylline entered systemic circulation)

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Pharmacokinetics – applications – Exercise I.

h) relative bioavailability: Frel = {[(AUC)0

]tablet/Dtablet}/{[(AUC)0 ]capsule/Dcapsule} =

= {160,202/170 mg}/{104,366/100 mg} = 0,903 (90,3%)Bioequivalence: bioavailability (F) + time, in which the drug reaches max. conc.

(tmax) + maximal conc. in plasma (cp,max)- p. o. admin.: 100 mg theophylline in capsule:

observation: tmax = between 1,0 h and 1,5 h cp,max = 11,15 μg.ml-1

- p. o. admin: 200 mg aminophylline in tablet: observation: tmax = around 1,0 h cp,max = 15,70 μg.ml-1

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Pharmacokinetics – applications – Exercise II.

Man (70 kg) diagnosed with asthma was admitted to hospital with s asthmatic attack, he was administered i.v. bolus of 300 mg aminophylline dihydrate (79% theophylline)

every 8 h (dosing interval ). According to patient’s record :

- half-life of elimination of theophylline t1/2 = 7 h, - apparent distribution volume for theophylline: Vd = 30 l

Calculate:

a) maximal concentration of theophylline in plasma in steady state (cp,max),

b) minimal concentration of theophylline in plasma in steady state (cp,min),

c) mean concentration of theophylline in plasma in steady state (ĉp) for dosing:

- i.v. bolus of 300 mg aminophylline dihydrate (79% drug) every 8 h,

- i.v. bolus of 150 mg aminophylline dihydrate (79% drug) every 4 h,

d) concentration of theophylline in plasma before 4th dose,

e) concentration of theophylline in plasma shortly after 4th dose,

f) accumulation factor of drug (R) for dosing according to point c),

g) fluctuations () for dosing according to point c),

h) what initial dose is needed to cause a steady state of infusion,

i) How long will it take until half of the steady state concentration will be reached (i.e. fss = 0,5)

j) patient with body mass of 90 kg receives 520 mg of theophylline per day in tablets. 100% od the drug is absorbed from tablets, half-life of elimination t1/2 = 4,5 h and apparent volume of distribution Vd = 0,48 dm3.kg-1. What average concentration of theophylline in plasma (ĉp) will be reached in steady state ?

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Pharmacokinetics – applications – Exercise II.

Solution:a) content of the drug in aminophylline is 79%, therefore 300 mg of aminophylline

comprises 237 mg of theophylline, rate constant of elimination kel = 0,099 h-1

initial concentration in plasma: cp0 = D/Vd = 237000 μg/ 30000 ml = 7,90 μg.ml-1

in case of intravenous admin. will the maximum plasma conc. in steady state be:cp,max = cp0/(1 – e-kel) = 7,90 μg.ml-1 /(1 – e-0,099 h-1.8 h) = 14,44 μg.ml-1

b) in case of intravenous administration will the minimum plasma concentration in steady state be: cp,min = [cp0/(1 – e-kel)]. e-kel = 6,540 μg.ml-1

c) in case of intravenous administration will the mean plasma concentration in steady state for dosage of 300 mg of aminophylline every 8 h be:

ĉp = D0/(Vdkel) = 237000 μg/(30000 ml.0,099 h-1.8 h) = 9,975 μg.ml-1

for a dosage of 150 mg of aminophylline (118,5 mg theophylline) every 4 h:

ĉp = D0/(Vdkel) = 118500 μg/(30000 ml.0,099 h-1.4 h) = 9,975 μg.ml-1

(half dose, half dosing interval = same mean concentration of the drug)

d) concentration of theophylline just before the 4th dose will be the minimal concentration after the 3rd dose

cp3,min = {cp0(1 – e-3kel)/(1 – e-kel)}. e-kel = = {7,90 μg.ml-1(1 – e-3.0,099.8)/(1 – e-0,099.8)}. e-0,099.8 = 5,933 μg.ml-1

e) concentration of theophylline immediately after the 4th dose will be the maximal concentration after the 4th dose:

cp4,max = {cp0(1 – e-4kel)/(1 – e-kel)} = = {7,90 μg.ml-1(1 – e-4.0,099.8)/(1 – e-0,099.8)} = 13,832 μg.ml-1

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Pharmacokinetics – applications – Exercise II.

Solution:f) accumulation factor of theophylline (R) can be calculated e.g. with help of max.

conc. in plasma in steady state cp,max and after the 1st dose cp1,max. For dosage of 237 mg every 8 h we get:

cp,max = 14,440 mg.ml-1

cp,min = 6,540 mg.ml-1

ĉp = 9,975 mg.ml-1

maximal conc. after 1st dose: cp1,max = D0/Vd = 237 mg/ 30000 ml = 7,90 μg.ml-1

minimal conc. in plasma after 1st dose:cp1,min = cp0.e

-kel. = 7,90 μg.ml-1 .e-0.099.8 = 3,578 mg.ml-1

accumulation factor R = cp,max /cp1,max = 1/(1 – e-kel.) = 1/(1 – e-0,099.8) = 1,828for half dose of 118,5 mg every 4 h we get higher accumulation (shorter interval: R = cp,max /cp1,max = 1/(1 – e-kel) = 1/(1 – e-0,099.4) = 3,058

g) Fluctuations can be obtained with help of half-life (t1/2) and dosage interval (): = 1/e-kel. = 1/e-0,099.8 = 2,207

for half dosage of 118,5 mg every 4 h we get:

= 1/e-kel. = 1/e-0,099.4 = 1,485

smaller dose administered more frequently leads to higher accumulation and lowerfluctuations during infusion

h) Administration of the dose DM = 300 mg of aminophylline (i.e. 237 mg theophylline) every 8 h leads to steady state concentration of 14,44 μg.ml-1. Initial dose, which leads to a steady state can be determined as:

DL = DM./(1 – e-kel.) = 300 mg/(1 – e-0,099.8) = 548,37 mg aminophylline

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Pharmacokinetics – applications – Exercise II.

Solution:i) time needed to reach the fraction fss of steady state concentration can be

determined as: fss = 1 – e-kel.t = 0,5 where kel = ln2/t1/2 number of elimination half-lifes

needed to reach 0,5 of steady state concentration: N = t/t1/2 = ln0,5/-0,693 = 1i.e. 1 half-life t1/2 = 7 h to reach 50% of steady state concentration

j) patient with body mass of 90 kg obtains daily dose D = 520 mg of theophylline,

i.e. 5,720 mg.kg-1 , dosing interval = 24 h, absorbed fraction (bioavailability) F

= 100%, t1/2 = 4,5 h, kel = 0.154 h-1, Vd = 0,48 dm3.kg-1

Systemic clearance: Cls = Vd.kel = 6,72 dm3.h-1

Average concentration of theophylline in plasma in steady state for extra vascular administration:ĉp = F D/(Vd.kel. ) = 1.520000 μg/(43636 ml.0,154 h-1.24 h) = 3,224 μg.ml-1

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Pharmacokinetics – applications – Exercise III.

30 year old patient with Pseudomonas sp. infection was given infusion of 125 mg tobramycine with concentration of 6,5 μg.ml-1 in the duration of 1 h. 3 h after this infusion dropped the plasma conc. to cp = 5,0 μg.ml-1 and after 7 h to cp = 3,5 μg.ml-1. Linear regression logcp vs. time (t) gave the rate constant of elimination kel

= 0,088 h-1.

Calculate:a) distribution volume of the patient,b) maximal concentration of tobramycine in plasma in steady state (cpmax) and

minimal concentration of tobramycine in plasma in steady state (cpmin) if the patient was given maintenance dose of 70 mg during 1 h every 12 h,

c) Will the maintenance dose secure the same maximal concentration as the initial dose?

d) if the initial dose would be given continuously (serious error) what theoretical concentration of tobramycine would be reached (should the patient survive this treatment)?

e) If the initial dose would be given repeatedly for 1 h every 12 h (next error), what theoretical conc. of tobramycine would be reached?

f) what average steady state concentration (ĉp) would be reached in the case e) g) what minimal steady state concentration (cpmin) would be reached in the case e) h) what rate of continuous infusion (Q) would lead to final concentration of

tobramycine in plasma of ĉp = 6,5 mg.dm-3?

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Pharmacokinetics – applications – Exercise III.

Solution:

a) Distribution volume: Vd = QL/(cp.kel).[1 – e-kel.t’] =

= 125 mg.h-1/(6,5 mg.dm-3 .0,088 h-1).[1 – e-0,088 h-1.1 h] = 18,4 dm3

b) For the maintenance dose: cpmax = Q/(Vd.kel){[1 – e-kek.t]/[1 – e-kel.]} =

= 70 mg.h-1/(18,4 dm3.0,088 h-1).{[1 – e-0,088.1]/[1 – e-0,088.12]} = = 5,6 mg.dm-3

cpmin = cpmax .e-kel( - t) = 5,6 mg.dm-3 .e-0,088(12-1) = 2,1 mg.dm-3

c) maximal concentration of tobramycine in plasma after initial dose of 6,5 mg.dm-3

is different from the concentration in steady state of 5,6 mg.dm-3

d) continuous administration of the initial infusion dose would lead to: ĉp = QL/(Vd.kel) = 125 mg.h-1/(18,4 dm3.0,088 h=1) = 77,2 mg.dm-3

which represent highly toxic concentration

e) repeated administration of initial infusion dose would lead to maximal conc.:cpmax = Q/(Vd.kel){[1 – e-kek.t]/[1 – e-kel.]} =

= 125 mg.h-1/(18,4 dm3.0,088 h-1).{[1 – e-0,088.1]/[1 – e-0,088.12]} = = 9,97 mg.dm-3

which is also too high concentration

f) Repeated administration of initial infusion dose would lead to average steady state concentration:

ĉp = FD/( Vd.kel) = 1.125 mg/(12 h.0,088 h-1.18,4 dm3) = 6,43 mg.dm-3

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Pharmacokinetics – applications – Exercise III.

Solution:

g) repeated administration of initial infusion dose would lead to minimal concentration :

cpmin = cpmax .e-kel.(-t) = 9,97 mg.dm-3 .e-0,088(12-1) = 3,79 mg.dm-3

which is also too high concentration

h) for the continuous infusion the required flow rate will be: Q = Vd.kel.cp = 18,4 dm3.0,088 h-1.6,5 mg.dm-3 = 10,5 mg.h-1

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Literature

Recommended literature

• ATKINS, Peter W. – DE PAULA, Julio: Physical Chemistry: Thermodynamics, Structure, and Change, 10th Ed., Oxford University Press, Oxford, UK, 2014.

• BOROUJERDI, Mehdi: Pharmacokinetics and Toxicokinetics, CRC Press, Taylor & Francis Group, Boca Raton, FL, U.S.A., 2015.

• JAMBHEKAR, Sunil S. - BREEN, Philip J.: Basic Pharmacokinetics, 2nd Ed., Pharmaceutical Press, London, UK, 2012.

• KERNS, Edward H. - DI, Li: Drug-like Properties: Concepts, Structure Design and Methods, Elsevier, Burlington, MA, U.S.A., 2008.

• PATRICK, Graham L.: An Introduction to Medicinal Chemistry, 5th Ed., Oxford University Press, Oxford, UK, 2013.

• VAN HOLDE, Kensal E. – JOHNSON, W. Curtis - HO, P. Shing.: Principles of Physical Biochemistry, 2nd Ed., Pearson Prentice Hall, Upper Saddle River, NJ, U.S.A., 2006.