Dosing Regimen Design

Infusion regimen

Why is a dosing regimen necessary?

• To replace the drug that the body eliminates.

How can drug be replaced?

• Continuously or intermittently.

Continuous Input

• Learning Objectives– input rate to achieve a desired plasma

concentration.– kinetics of accumulation; i.e., how long to

steady state.– loading dose.

– determination of CL, V, KE and t1/2.

Examples of continuous input

• i.v. drip

• i.v. infusion

• transdermal patch

• sustained release oral dosage forms

• Ocusert

• Norplant

Kinetics of continuous input

A = amount of drug in body

Cp = plasma concentration

Ko = input rate (amt/time)

KE = elimination rate constant, (CL/V)

v

CL

Ko

dA/dt = rate in - rate out

dA/dt = Ko - KEA

A = V x Cp and KEV = CL

dA/dt = Ko - CL•Cp

dA/dt = 0 at a plateau Cp

Rate In = Rate OutKo = CL •Cp,ss

Cp,ss = Ko/CL

Example: Diazepam Cp,t profile in young and old:

Ln C

p

Time

young

old

What’s different?

Vss CL t1/2 fup

[L/kg] [L/h/kg] [h] [%]

0.88 0.0174 44.5 2.49

1.39 0.0156 71.5 2.76

Herman and Wilkinson. Br. J. Clin. Pharmacol. 42:147,1996. #2919

DR adjustment?

Principles

• When infused at the same rate (one compartment model assumed):– all drugs with the same half life will have the

same steady-state amount of drug in the body.– all drugs with the same clearance will have the

same steady-state plasma concentration.

Accumulation Kinetics

tKss

tK

E

o EE eAeK

KA 11inf

tKssp

tKop

EE eCeCL

KC 11 ,inf,

Time to steady state

3.3 t1/2 is time to 90% of Cp,ss

t = nt1/2 where n = no. of half lives that have passed

n

ntttK ee E

21

2/12/12ln

nssptK

sspp CeCC E 2111 ,,inf,

Example

This drug has a V = 45 L and a CL = 12 L/h. What infusion rate is needed to achieve a Cp,ss of 25 mg/L?

Ko = CL x Cp,ss = 12 L/h x 25 mg/L = 300 mg/h

How long will it take to get to 90% of steady state?

t1/2 = ln 2 V/CL = (0.693)(45)/(12) = 2.6 h

t90% = 3.3 t1/2 = (3.3)(2.6) = 8.6 h

How much drug is in the body at steady state?

Ass = Cp,ssV = 45 L x 25 mg/L = 1,125 mg

Summary

Cp,ss Ass Time to SS

Ko

CL

V

Diagram

Ass

Css

Ko CL

t90%

V

VP, VE, VR

RE/I, fur, fup

QH, fup, CLint,u

GFR, etc.

Review

With a constant rate of input, Ko

Rate In = _______ Rate Out

Rate In = Css x ___ CL

Rate In = Ass x ___ KE

All drugs with same CL will have same ____

All drugs with same t1/2 will have same ____

Css

Ass

Post-Infusion Cp Profile

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0 20 40 60 80 100 120

Time

Infusion

Post-Infusion Cp Profile

0.00001

0.00010

0.00100

0.01000

0.10000

0 20 40 60 80 100 120

Time

Infusion

Changing to a new Cp,ss

CL by 25%

Ko by 25%

V by 25%

Cp

Time

Bolus and Infusion

A “loading dose” may be used to start at steady state immediately.

Loading Dose = Ass = Ko/KE = CssV

Rowland and Tozer, Figure 6-5. p. 74

Time to Steady State

3.3 t1/2 is time to 90% of Cp,ss.

When Cp,0 is 0, this is within ±10% of Cp,ss.

When Cp,0 is 0, the time to ±10% of Cp,ss differs from the t90%.

What is the appropriate endpoint for calculation of the time to steady state?

Calculation of time to ±10% Cp,ss

tKss

tKpp

EE eCeCC 10,

Plasma concentration at any time after bolus + infusion:

Given Cp,0 and Css, the time to reach any Cp can be calculated from:

ntK

pss

pss EeCC

CC

2

1

0,

Example

Cp,0 = 500 mg/L and Css = 100 mg/L, how long does it take to reach 110 mg/L? (i.e., 110% of Css)

ntK

pss

pss EeCC

CC

2

1

0,

(1/2)n = (100 – 110)/(100 – 500) = -10/-400 = 0.025

(n)[ln (0.5)] = ln (0.025)

n = ln (0.025) / ln (0.5) = 5.32 half lives

What would be the Cp after 3.3 half lives? 140 mg/L

Assessment of PK parameters

1. CL = Ko / Cp,ss

TimeL

og (

Cp,

ss -

Cp)

-2.3 slope = KE

2. Get KE from the slope of a semilog plot of (Cp,ss – Cp) vs. t.

3. V = CL / KE tKsspp

EeCC 1,inf,