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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.
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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
fup CLint,u QH QR GFR CLRS,int Maintenance Dose Key Physiological Variables:
log CL log W 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: liver weight (kg) = 0.0370 W0.849 W, kg a = CL @ W = 1 b = slope
FDM/ Ko FDM/ Ko CLunusual CLusual = usual unusual DM adjustment: b value CLnew = CLstd X [W/70]0.75 (DM/)new = (DM/)std X [W/70]0.75 For W = 35 kg [35/70]0.75 = 0.60
Drug Clearance: b values 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/)
DM adjustment For all but renally eliminated drugs: (DM/)new = (DM/)std X [W/70]0.75 For renally eliminated drugs: (DM/)new = (DM/)std X [W/70]0.67 For W = 35 kg [35/70]0.75 = 0.60 [35/70]0.67 = 0.63
Allometric Equation Approach Uses CL and V values in at least two subject groups that have different body weights. Example: Cefotiam in Sumo Wrestlers K Chiba et al., Antimicrob. Agents Chemother. 33:1188,1989.
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