Medical Pharmacology Question Bank

Chapter 2: Pharmacokinetics — Module 3: Metabolism and Excretion
Tier: Tier 3 — Clinical Vignettes

1. A 52-year-old man with a seizure disorder has been stable on phenytoin 300 mg daily with plasma levels consistently between 12-15 mg/L (therapeutic range 10-20 mg/L) for five years. His neurologist increases the dose to 350 mg daily — a 17% increase. Three weeks later the patient presents with nystagmus, ataxia, and confusion. His plasma phenytoin level is 38 mg/L. Which pharmacokinetic explanation best accounts for the disproportionate concentration increase?

A) The dose increase triggered CYP2C9 enzyme induction, paradoxically increasing
B) The patient began taking an over-the-counter CYP2C9 inhibitor between visits,
C) Phenytoin's primary metabolic enzyme (CYP2C9) is operating near its maximum
D) The 17% dose increase raised plasma concentration by exactly 17%, but laboratory
E) Phenytoin switches abruptly from first-order to zero-order kinetics at a

ANSWER: C

Rationale:

Phenytoin is the paradigmatic example of capacity-limited (Michaelis- Menten) elimination — its CYP2C9-mediated metabolism operates near saturation at therapeutic plasma concentrations. The key concept is that at concentrations near the Km of CYP2C9 for phenytoin (approximately 5-10 mg/L), the drug operates in the transition zone between first-order and zero-order kinetics. At 12-15 mg/L, the enzyme is already working at a substantial fraction of its maximum velocity (Vmax) and cannot proportionally increase elimination in response to a dose increase. When the dose rises from 300 to 350 mg/day (a 17% increase in drug input), the elimination system — already near capacity — cannot match the new input rate at the old concentration. Drug accumulates until plasma concentration rises to a new level where the remaining incremental capacity of CYP2C9 can finally match the new dosing rate. This new steady state happens to be at 38 mg/L — more than double the previous level — from a 17% dose increase. The clinical lesson: phenytoin dose adjustments must be made in small increments (25 mg at a time), with months allowed between changes and frequent plasma level monitoring.

2. A 68-year-old woman with atrial fibrillation is started on amiodarone 200 mg three times daily (loading phase). Her cardiologist explains that without a loading dose, it would take 5-9 months to achieve therapeutic plasma concentrations. Amiodarone has a half-life of approximately 40-55 days. Which pharmacokinetic principle explains why loading doses are necessary for amiodarone?

A) Amiodarone has very poor oral bioavailability, and a loading dose compensates
B) Steady-state plasma concentration is reached after 4-5 half-lives — for amiodarone
C) Amiodarone undergoes autoinduction of its own metabolism during the first
D) The loading dose is required to saturate plasma protein binding sites before
E) Amiodarone's half-life is actually only 24-48 hours — the published 40-55 day

ANSWER: B

Rationale:

The time to reach steady state depends exclusively on the drug's elimination half-life and equals approximately 4-5 half-lives. For amiodarone with t½ = 40-55 days: 4-5 × 40-55 days = 160-275 days (5.3-9.2 months). Starting at a maintenance dose of 200 mg daily without loading, plasma and tissue amiodarone concentrations would remain subtherapeutic for months — unacceptable for a drug intended to suppress an active arrhythmia. The loading dose strategy addresses this by administering a higher daily dose (600 mg/day during loading) — the dosing rate far exceeds the elimination rate, rapidly accumulating amiodarone in the body to concentrations associated with antiarrhythmic efficacy. The loading dose does NOT make steady state arrive faster — the time to true pharmacokinetic steady state (4-5 half-lives) is unchanged by loading. What it achieves is that drug concentrations quickly reach therapeutically effective levels even before steady state. This is the core pharmacokinetic rationale for loading doses: clinically necessary when the time to steady state at maintenance dosing is longer than the patient can wait for therapeutic benefit.

3. A patient with normal renal function is started on warfarin 5 mg daily. After 7 days his INR is 2.5 — within the therapeutic range of 2-3. His physician increases the dose to 6 mg daily. The physician plans to check the INR again in 3 days to see whether the new dose is working appropriately. Warfarin has a half-life of approximately 36 hours. A clinical pharmacist advises that checking the INR at 3 days after the dose change is pharmacokinetically premature. Which explanation best supports the pharmacist's concern?

A) Warfarin requires 7 days to exert any pharmacodynamic effect because it must
B) The new steady-state warfarin plasma concentration will not be achieved until
C) Warfarin should only be monitored weekly because INR is highly variable
D) The pharmacist is incorrect — for drugs with a 36-hour half-life, steady state
E) Three days is too long to wait — warfarin toxicity can develop within 24 hours

ANSWER: B

Rationale:

The pharmacokinetic principle is that after any dose change, the new steady state is achieved after 4-5 half-lives. For warfarin with t½ ≈ 36 hours: 4-5 × 36 hours = 144-180 hours = 6-7.5 days. At 3 days (72 hours) after the dose change, only approximately 2 half-lives have elapsed — warfarin plasma concentrations are at approximately 75% of their eventual new steady-state value and are still rising. The INR measured at 72 hours reflects this intermediate, still-rising warfarin concentration. If the physician adjusts the dose further based on this premature measurement (which may appear to show insufficient anticoagulation), they risk over-anticoagulating the patient once true steady state is reached at 6-7 days. Option D contains the critical pharmacokinetic error: 2 half-lives (72 hours) achieves approximately 75% of steady state, not 100%. The 4-5 half-lives rule for ~97% of steady state is the pharmacokinetically appropriate threshold for interpretable monitoring. The correct monitoring schedule for warfarin dose changes is INR at 5-7 days after any dose adjustment.

4. A patient with chronic kidney disease (eGFR 30 mL/min) requires treatment with an antibiotic. The antibiotic has the following properties: total clearance (normal renal function) = 120 mL/min; fraction eliminated renally unchanged (fe) = 0.80; non-renal clearance = 24 mL/min. Which approach to dose adjustment is pharmacokinetically correct?

A) Reduce the dose by 75% because eGFR is 75% reduced (from 120 to 30 mL/min)
B) No dose adjustment is needed — drugs with normal renal clearance are adequately
C) Reduce the dose proportionally to the reduction in total clearance: renal CL
D) Reduce the dose by 80% because fe = 0.80 means 80% of the drug is affected
E) Increase the dose in CKD because reduced renal excretion means the drug

ANSWER: C

Rationale:

Renal dose adjustment requires separating total clearance into its renal and non-renal components, then adjusting only the renal component for the patient's reduced GFR — non-renal clearance (hepatic metabolism, biliary excretion) is assumed unchanged unless hepatic disease is also present. Step-by-step calculation: Normal renal CL = fe × total CL = 0.80 × 120 = 96 mL/min. Non-renal CL = total CL - renal CL = 120 - 96 = 24 mL/min. Adjusted renal CL (at eGFR 30) = 96 × (30/120) = 96 × 0.25 = 24 mL/min. Adjusted total CL = adjusted renal CL + non-renal CL = 24 + 24 = 48 mL/min. Dose reduction ratio = adjusted CL / normal CL = 48/120 = 0.40. The dose should be reduced to 40% of normal — or the dosing interval extended by the reciprocal (1/0.40 = 2.5-fold). Option D — reducing by the fe fraction — is a common error that also does not account for the actual degree of GFR reduction (it would mean the same dose reduction at any level of renal impairment, which is clearly incorrect).

5. A 45-year-old man receiving vancomycin for MRSA bacteremia has blood cultures drawn and two vancomycin levels measured: a peak of 32 mg/L drawn 1 hour after infusion completion, and a trough of 10 mg/L drawn immediately before the next dose. The vancomycin is given every 12 hours. The MRSA isolate has a vancomycin MIC of 1 mg/L. The target AUC/MIC ratio for MRSA treatment is 400-600 mg·h/L. Using the trapezoidal approximation AUC per interval ≈ (Cpeak + Ctrough)/2 × τ, is the current regimen achieving the target?

A) AUC per 12-hour interval = (32 + 10)/2 × 12 = 252 mg·h/L; AUC24 = 252 × 2 = 504
B) AUC per 12-hour interval = (32 + 10)/2 × 12 = 252 mg·h/L; this is the 24-hour
C) AUC24 = Cpeak × 24 = 32 × 24 = 768 mg·h/L; AUC/MIC = 768; above target —
D) AUC/MIC cannot be calculated from two time points — a full 12-point
E) AUC24 = Ctrough × τ = 10 × 24 = 240 mg·h/L; AUC/MIC = 240 — below target;

ANSWER: A

Rationale:

The linear trapezoidal approximation for AUC within one dosing interval at steady state uses the average of peak and trough concentrations multiplied by the interval duration: AUC per interval = (Cpeak + Ctrough)/2 × τ = (32 + 10)/2 × 12 hours = 21 × 12 = 252 mg·h/L. Since vancomycin is dosed every 12 hours (two doses per 24 hours), the 24-hour AUC is two dosing-interval AUCs: AUC24 = 252 × 2 = 504 mg·h/L. The AUC/MIC ratio = 504 / 1 = 504 — within the target range of 400-600 mg·h/L for MRSA bacteremia treatment per current ASHP/IDSA/SIDP guidelines. Current guidelines recommend AUC/MIC-guided vancomycin monitoring over traditional trough-only monitoring because trough alone is a poor predictor of treatment outcomes and nephrotoxicity risk.

6. A patient is started on a new antiepileptic drug with a volume of distribution of 60 liters and a clearance of 3 liters per hour. The drug is given orally twice daily with a bioavailability of 75%. The target average steady-state concentration is 8 mg/L. What maintenance dose should be prescribed for each twice-daily administration?

A) Dose = Vd × Css = 60 × 8 = 480 mg twice daily — using the loading dose formula
B) Dose = (CL × Css × τ) / F = (3 L/hr × 8 mg/L × 12 hr) / 0.75 = 288/0.75
C) Dose = CL × Css = 3 × 8 = 24 mg per dose — omitting both the dosing interval
D) Dose = (CL × Css) / F = (3 × 8) / 0.75 = 32 mg per dose — omitting the
E) Dose = Css × τ / F = 8 × 12 / 0.75 = 128 mg per dose — omitting clearance

ANSWER: B

Rationale:

The maintenance dose equation is derived from the steady-state condition where rate of drug input equals rate of drug elimination: (F × Dose) / τ = CL × Css. Rearranging for the dose: Dose = (CL × Css × τ) / F. Substituting the values: CL = 3 L/hr; Css = 8 mg/L; τ = 12 hours (for twice-daily dosing); F = 0.75. Dose = (3 × 8 × 12) / 0.75 = 288 / 0.75 = 384 mg per dose. This dose is given twice daily (every 12 hours). Verification: daily drug entering systemic circulation = 384 mg × 2 doses × 0.75 bioavailability = 576 mg/day. Daily drug eliminated at Css = CL × Css × 24 hr = 3 L/hr × 8 mg/L × 24 hr = 576 mg/day. Balance confirmed. Option A uses the loading dose formula (Vd × Css), which calculates the amount of drug that must be in the body to immediately achieve the target concentration — this is different from the maintenance dose, which must replace ongoing elimination. The loading dose for this drug would be 60 L × 8 mg/L = 480 mg — coincidentally close to the maintenance dose, but derived by a completely different equation. END OF FOUNDATION AND APPLIED QUESTIONS — MODULE 3 pharmacology2000.com | Chapter 2: General Principles — Pharmacokinetics For advanced quantitative pharmacokinetics and complex clinical reasoning cases, see: PK-Module3-Clinical-Reasoning