Pharmacokinetic Models

One Compartment Model–IV Bolus–Absorption

One Compartment Model

• Simplest compartmental model• Body is assumed to behave as if it were a

single, well stirred fluid.

I.V. Bolus Dose

-dX/dt = CL•Cp

-d(X/V)/dt = (CL/V)•Cp = -dCp/dt

Cp = Cp,oe-(CL/V)t

v

CL

Example: Dose = 300 mg, as iv bolusV = 35 L and CL = 2 L/h

Cp,o = Dose/V = 300 mg ÷ 35 L = 8.57 mg/L

I.V. Bolus , 300 mgV = 35 L; CL = 2 L/h

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

I.V. Bolus , 300 mgV = 35 L; CL = 2 L/h

1.00

10.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

I.V. Bolus , 300 mgV = 35 L; CL = 2 L/h

1.00

10.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]One Compartment Model

Cp = Cp,oe-(CL/V)t

log Cp = log Cp,o – (CL/V)t/2.3

Slope = -(CL/V)/2.3

-2.3Slope = CL/V = KE

V = Dose ÷ Cp,o

V = 300 mg ÷ 8.56 mg/L = 35 L

CL = KEV

I.V. Bolus , 300 mgV = 35 L; CL = 2 L/h

1.00

10.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]One Compartment Model

-2.303 x Slope = KE

Slope = (y2 – y1)/(x2 – x1)

= (log Cp2 – log Cp1)/

( t2 – t1)

= log (Cp2/Cp1)/(t2 – t1)

= log (7.65/1.10)/(2 – 36)

= -0.0248KE = (-2.303)(-0.0248) = 0.0571 h-1

One Compartment ModelI.V. Bolus , 200 mg

V = 35 L; CL = 2 L/h; F = 0.77

1.00

10.00

0 10 20 30 40TIME [h]

Cp

[mg/

L]

Half Life:

t1/2 = ln 2/KE = 0.693/0.0571 h-1

= 12 h

t1/2 = 0.693 V/CL

= 0.693•35L/2L h-1

= 12 h

CL = KEV = (0.0571 h-1)(35 L)

= 2 L/h

One Compartment ModelKEY CONCEPT!

• Half Life and KE depend on both CL and V; a change in either CL or V will cause a change in t1/2 and KE.

KE

t1/2

CL

CL

V

V

Area Under the Curve

• AUC = the area under the Cp,t profile, from time = 0 to time = , usually for a single dose.

• AUCt1-t2 = the area under the curve from time = t1 to time = t2.

EE

ptKpp KV

DoseKC

dteCdtCAUC E

0,

00,

0

CLDoseAUC

AUCDoseCL

Summary of Wednesday

v

CL

I.V. Bolus Dose Cp = Cp,oe-(CL/V)t

I.V. Bolus , 300 mgV = 35 L; CL = 2 L/h

1.00

10.00

0 10 20 30 40

TIME [h]C

p [m

g/L]

- 2.3 x Slope = CL/V = KE

V = Dose ÷ Cp,o

CL = KEV

KE

t1/2

CL

CL

V

V

CLDoseAUC

Calculation of AUCTrapezoidal Method, R&T, p.469

Cp

mg/L

1 2 4 8Time [h]

2 3 2.4 1.8

12

21

22

1ttCCAUC ppt

t

1012

201

0

AUC

5.2122

322

1

AUC

4.5242

4.234

2

AUC

4.8482

8.14.28

4

AUC

63.08.18

Elast KCAUC

AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L

CL for individual pathways

DB

MB

Durine

Murine

Expired air

CLH

CLR

CLP

CL = CLH + CLR + CLP

CLDoseMCL u

H

CLDoseXCL u

R

CLDoseXCL P

P

One Compartment Model

• A two-fold change in CL:I.V. Bolus , 300 mg

V = 35 L; CL = 2 & 4 L/h

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

I.V. Bolus , 300 mgV = 35 L; CL = 2 & 4 L/h

0.10

1.00

10.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

One Compartment Model

• A two-fold change in V:I.V. Bolus , 300 mg

V = 35 & 70 L; CL = 2 L/h

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

I.V. Bolus , 300 mgV = 35 & 70 L; CL = 2 L/h

1.00

10.00

0 10 20 30 40

TIME [h]

Cp

[mg/

L]

One Compartment ModelAbsorption Input

• Drug enters body by a first-order, monoexponential process.

dX/dt = kaXg - CL•Cp

v

CL

ka

tkt

VCL

a

ap

aee

VCLkV

DoseFkC

tktK

Ea

ap

aE eeKkV

DoseFkC

Absorption

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 10 20 30Time [h]

Cp

[mg/

L]

0.1

1.0

0 10 20 30Time [h]

Cp

[mg/

L]

The slope of the log-linear phase reflects the smaller of ka and KE.

Absorption

• Shape parameters– Cmax

– Tmax

– AUC– t1/2

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 10 20 30Time [h]

Cp

[mg/

L]Use shape parameters to deduce changes in PK parameters: ka, CL, V, F

Tmax

At the peak Cp, dCp/dt = 0

tktK

Ea

ap

aE eeKkV

DoseFkC

0maxmax

TKE

Tka

Ea

ap Ea eKekKkV

DoseFkdt

dC

Ea

Ea

KkKkT

ln

max

vCL

ka

Cmax

0 pEGap CK

VXk

dtdC

tkG

aeDoseFX

maxmax CKe

VDoseFk

ETka a

Ea

E

a

a KkKk

k

E

a eVKDoseFkC

ln

max

EaE

KkK

a

E

kK

VDoseFC

max

vCL

ka

AUC

0

dteeKkV

DoseFkdtCAUC tktK

Ea

ap

aE

CLDoseFAUC

ka

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 10 20 30Time [h]

Cp

[mg/

L]

ka1 1/hr 0.25

ka2 1/hr 0.75

ka

___Cmax

___Tmax

___t1/2

___AUC

CL

CL

___Cmax

___Tmax

___t1/2

___AUC

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 10 20 30Time [h]

Cp

[mg/

L]

CL1 L/hr 20CL2 L/hr 40

F•Dose

F•Dose

___Cmax

___Tmax

___t1/2

___AUC

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 10 20 30Time [h]

Cp

[mg/

L]

F1 1F2 0.66

V

V

___Cmax

___Tmax

___t1/2

___AUC

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.0 10.0 20.0 30.0Time [h]

Cp

[mg/

L]

V1 L 250V2 L 50

Shape parameters as functions of PK parameters

Ea

Ea

KkKkT

ln

max

EaE

E

KkK

a

ETK

kK

VDoseFe

VDoseFC

/

maxmax

CLDoseFAUC

CL

Vt 2ln2/1

Peak Shape Analysis

Peak Shape Analysis

PK Parameters from single-dose plasma concentration profile

CLDoseFAUC

AUCDose

FCL

Uncertainty in ‘F’ is transmitted to CL. The ratio Dose/AUC gives the true value of CL only when F=1. If F1, then the calculation gives a CL value that is larger than the true value.

PK Parameters from single-dose plasma concentration profile

tktK

Ea

ap

aE eeKkV

DoseFkC

0.1

1.0

0 10 20 30Time [h]

Cp

[mg/

L]

-2.3 x slope = KE (usually)

FV

KF

CLss

E

Pharmacokinetic Models

One Compartment ModelAbsorption RateBioavailability

Absorption Rate

v

CL

ka

tktK

Ea

ap

aE eeKkV

DoseFkC

XG = Dose•e-kat

Chloramphenicol500 mg po

0.00.51.01.52.02.53.03.5

0 10 20 30

Time [h]

Cp

[mg/

L]

How can the value of ka be determined from the Cp,t profile?

Determination of ka

XGI XB XE

vCL

ka

Chloramphenicol500 mg po

0.00.51.01.52.02.53.03.5

0 10 20 30

Time [h]

Cp

[mg/

L]

Chloramphenicol 500 mg po

0

100

200

300

400

500

0 5 10 15

Time [h]

Xg [m

g]

XGI cannot be measured; ka must come from Cp,t profile

X = amount

Determination of ka

1. Computer fit of equation using software such as WinNonlin.

vCL

ka

tktK

Ea

ap

aE eeKkV

DoseFkC

2. Graphical analysis; aka “method of residuals”, “feathering”, “peeling”

Method of Residuals

When ka > 4KE, e-kat goes to 0 before e-KEt does. After e-kat goes to 0:

tK

Ea

ap

EeKkV

DoseFkC

3.2tK

KkVDoseFkLogCLog E

Ea

ap

Chloramphenicol500 mg po

0.1

1.0

10.0

0 10 20 30Time [h]

Cp

[mg/

L]

Method of ResidualsSubtract the Cp,t profile from the line:

tktK

Ea

ap

aE eeKkV

DoseFkC

tK

Ea

ap

EeKkV

DoseFkC

tktK

Ea

atK

Ea

ap

aEE eeKkV

DoseFkeKkV

DoseFkC

tk

Ea

ap

aeKkV

DoseFkC

Chloramphenicol500 mg po

0.1

1.0

10.0

0 10 20 30Time [h]

Cp

[mg/

L]

Method of ResidualsWhat if KE > 4ka?

tktK

Ea

ap

aE eeKkV

DoseFkC

Chloramphenicol500 mg po

0.1

1.0

10.0

0 10 20 30Time [h]

Cp

[mg/

L]

tk

Ea

ap

aeKkV

DoseFkC

tktK

Ea

atk

Ea

ap

aEa eeKkV

DoseFke

KkVDoseFk

C

tK

Ea

ap

EeKkV

DoseFkC

Bioavailability - F

0

dteeKkV

DoseFkdtCAUC tktK

Ea

ap

aE

CLDoseFAUC

std

stdstd

test

testtest

std

test

CLDoseF

CLDoseF

AUCAUC

test

std

std

test

std

test

DoseDose

AUCAUC

FF

Bioavailability - F

test

std

std

test

std

test

DoseDose

AUCAUC

FF

When AUCstd is from an i.v. dose, Fstd = 1.00 and the “absolute bioavailability” of the test is determined.

When AUCstd is somethingelse such as the innovator’s product or a solution, “relative bioavailability” is determined.