Dosing Regimen Individualization Body Size Variability in Body Size Does one size fit all ? If body size is outside the range 25% of normal (for 70.

Dosing Regimen Individualization

Body Size

Variability in Body Size

Does one size fit all ?

If body size is outside the range 25% of normal (for 70 kg, less than 52.5 kg [115 lb] or greater than 87.5 kg [193 lb]), an adjustment in the DR should be considered.

LD: Proportional to V

How to make the adjustment depends on the distribution characteristics of the drug:

Polar and distributes in total body water:

LD = LDusual x (W/70)

If patient is obese, adipose tissue has little water, so base LD on what the patient’s weight would be if they were normal weight.

Adjustment for W in the obese patient.

1. Calculate lean body mass (LBM) and use 55 kg as the normal LBM:

LD = LDusual x (LBM/55)

LBMmale = 1.10 W - 128 (W/H)2 W, kg; H, cm

LBMfemale = 1.07 W - 148 (W/H)2

2. Calculate ideal body weight (IBW) and use 70 kg as the normal IBW:

LD = LDusual x (IBW/70)

IBWmale = 0.0023(H)2 W, kg; H, cm

IBWfemale = 0.0021(H)2

Fat-soluble drugs

Use the same guideline as for polar drug.

Due to slow distribution into adipose tissue, only load the nonadipose tissue and let the maintenance dose fill the adipose tissue.

Drugs that distribute only into ECW

LD should be based upon the BSA of the patient.

No special allowance is needed for obesity.

LD = LDusual x (BSA/1.87)

BSA (m2) = 0.02350 * H0.42246 * W0.51456

H, cm; W, kg

Maintenance Dose

fup

CLint,u

QH

QR

GFRCLRS,int

Key Physiological Variables:

Allometric Equation

For many physiological variables, the allometric equation describes their relationship to body weight, W, among a group of animals of different body weight:

baWVariablePhysiol .

liver weight (kg) = 0.0370 W0.849 W, kg

log W

log C

L a = CL @ W = 1

b = slope

DM adjustment: b value

(DM/)new = (DM/)std X [W/70]0.75

For W = 35 kg

[35/70]0.75 = 0.60

FDM/

Ko unusual

CLunusual

CLusual

FDM/

Ko usual

=

CLnew = CLstd X [W/70]0.75

Drug Clearance: b values

Allometric exponent

0.15

0.25

0.35

0.45

0.55

0.65

0.75

0.85

0.95

1.05

1.15

1.25

1.35

1.45

1.55

Fre

qu

en

cy

0

5

10

15

20

25

30

n = 110

mean = 0.77 0.19

median = 0.75

T.-M. Hu, W.L. Hayton. AAPS PharmSci 2001; 3(4) article 29 (http://www.pharmsci.org/)

slope = b = 0.74

DM adjustment

For all but renally eliminated drugs:(DM/)new = (DM/)std X [W/70]0.75

For W = 35 kg

[35/70]0.75 = 0.60 [35/70]0.67 = 0.63

For renally eliminated drugs:(DM/)new = (DM/)std X [W/70]0.67

Allometric Equation Approach

Uses CL and V values in at least two subject groups that have different body weights.

Example: Cefotiam in Sumo Wrestlers

Height[cm]

Weight[kg]

Vss

[L]

CL[L/h]

t1/2

[h]

Sumo 181 124 30.2 38.3 0.91

Control 169 58 17.9 23.5 0.74

K Chiba et al., Antimicrob. Agents Chemother. 33:1188,1989.

What is a Sumo Wrestler ?

DR for Sumo Wrestler

Usual regimen: 2 g stat (LD) and 1 g q 6 h (DM / ).

Calculate b value for Vss and for CL

Vss = aWb 30.2 = a 124b & 17.9 = a 58b

(30.2/17.9) = (124/58)b a values cancel

b = ln (30.2/17.9) / ln (124/58) = 0.688

LDsumo = (124/70)0.688 X 2 g = 2.96 g

DR for Sumo Wrestler, cont.

Calculate b value for CL

CL = aWb 38.3 = a 124b & 23.5 = a 58b

(38.3/23.5) = (124/58)b

b = ln (38.3/23.5) / ln (124/58) = 0.643

DMsumo = (124/70)0.643 X 1 g = 1.44 g