Question

13. Aspirin is a common anti-inflammatory medication. A drug dose was administered orally and initial concentration of Aspirin in plasma was 33.2 mg/L. Knowing that Aspirin elimination follows zero order kinetics and drug half-life is 15 min. Determine: How long will it take for a drug concentration to decrease to 5% of its initial value? What is the elimination rate constant?

Answer 1

• Therapeutic drug concentrations in anxiety are 5 to 30 µg/mL after recommended doses (400–600 mg). A single 400 mg oral dose to healthy men produced average plasma concentrations of 7.7 mg/L at 2 hours, 4.4 mg/L at 4 hours and 1.6 mg/L at 24 hours. After 800 mg, a plasma concentration of 16 mg/L was achieved 2 hours after ingestion.

  Meprobamate is especially prone to induce sedation. This effect can be sought in DFC. A long half-life permits the drug to be detected for a long time in biological samples.

• Elimination Rate Constant

  The elimination rate constant (k_el) is the fraction of drug eliminated per unit time. It is not an independent pharmacokinetic parameter because it depends both on clearance and V_d:

  Eq. 11.10kel=ClearanceVd

  k_el can also be calculated directly from concentration data. As seen in Fig. 11.5, the concentration–time curve is linear on the log scale and the k_el is the slope of the line given by

  

  The concentration–time curve plotted on the log scale is linear for a drug that follows a one-compartment model. The elimination rate constant (k_el) is the slope of the straight line on the log scale.

  Eq. 11.11kel=lnC1−lnC2t2−t1

  where C₁ and C₂ are the concentrations measured at time one (t₁) and time two (t₂), respectively. When the elimination rate constant is known, the concentration after a single intravenous dose can be calculated at any time (t) by,

  Eq. 11.12C(t)=DoseVd×e−kel·t

  Although useful for calculations, k_el is more difficult to relate from a clinical perspective. The half-life (t_1/2) is easier to conceptualize and is calculated from k_el:

  Eq. 11.13t1/2=0.693kel=0.693×VdClearance

  The t_1/2 represents the time needed to eliminate 50% of the drug in the body. After 3.3 half-lives, 90% of the drug will be eliminated from the body (Table 11.2). The relationship between t_1/2 (= 0.693/k_el) and dosing interval (tau) will determine the amount of drug accumulated at steady-state with multiple doses by the accumulation