Pharmacokinetics Mnemonics

The most important pharmacokinetic parameters from a dosing point of view are:

  1. The clearance (CL) – determines the maintenance dose-rate
  2. The volume of distribution (Vd) – determines the loading dose (LD)
  3. The half-life (t½) – determines the time to steady state and the dosing interval

Vd = Volume of distribution
D = Dose
Co = Concentration
CL = Clearance
kE = Elimination constant
t1/2 = Half life

pharmacokinetics equations

In the mnemonics below, equations can be created by insertin “=” after the 1st word and “/” before the last word.

Mnemonic: ViDeo = DOwnloader/CONverter

Vd = D/Co

Also, Loading dose = Vd X Concentration targeted

E.g. To achieve a target Cp of 1.5 µg/L for digoxin (Vd = 500 L) –

LD (µg) = 500 (L) X 1.5 (µg/L) = 750 µg

Mnemonic: ViDeo = CLear/KlEar

Vd = CL/kE

E.g. A 2,000 mg dose with a concentration of 600 mg/L has a clearance of 0.05 L/hr. What is the elimination rate constant (KE)?

Vd = CL/kE = 0.05/kE
Also, Vd = D/Co = 2000/600

i.e. 0.05/kE = 2000/600
kE = 0.05 X 600/2000 = 0.015 hr

Mnemonic: HalfLife = Learnt To make ViDeo CLear

A very common assumption is that it takes about 5.5 half lives to completely clear a drug from the body.

t1/2 = ln(2) Vd/CL = 0.693 Vd/CL

E.g. In the above case, if we need to calculate t1/2 –

t1/2 = 0.693 X Vd/CL or 0.693/kE = 0.693/0.015 = 46.2 hr

Maintenance dose calculation:

Clearance of theophylline is 2.8 L/hr and Target concentration is 10 mg/L. What is the TDS maintenance dose?

Dosing rate = CL X Target concentration = 2.8 X 10 = 28 mg/hr

Maintenance dose = Dosing rate X Dosing interval = 28 X 8 = 224 mg

If the bio-availability is <1, divide by bioavailability (F). Suppose, if the bioavailability of theophylline is 0.8. Then,

Maintenance dose = 224/0.8 = 280 mg

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