A 60-year-old man is admitted to the intensive care unit for a severe exacerbation of chro… — USMLE Step 3

A 60-year-old man is admitted to the intensive care unit for a severe exacerbation of chronic obstructive pulmonary disease. He is started on a continuous intravenous infusion of theophylline at a rate of $40 \text{ mg/hr}$ . Forty-eight hours later, his steady-state serum theophylline concentration is measured at $10 \text{ mg/L}$ (therapeutic range: $10-20 \text{ mg/L}$ ). His estimated volume of distribution ( $V_d$ ) for theophylline is $40 \text{ L}$ .

On hospital day 4, the patient develops a catheter-associated urinary tract infection, and empiric intravenous ciprofloxacin is added to his regimen. On hospital day 8, the patient complains of intractable nausea, palpitations, and tremors. An electrocardiogram shows sinus tachycardia. A repeat serum theophylline concentration is found to be $20 \text{ mg/L}$ . The theophylline infusion rate had remained constant at $40 \text{ mg/hr}$ .

The physician immediately stops the theophylline infusion to allow the drug concentration to fall to $10 \text{ mg/L}$ before restarting at a lower rate.

Assuming first-order elimination kinetics and that the volume of distribution remains unchanged, approximately how many hours will it take from the time the infusion is stopped for the serum theophylline concentration to decrease to $10 \text{ mg/L}$ ?

A
3.5 hours
B
7.0 hours
C
10.0 hours
14.0 hoursAnswer
E
28.0 hours

Answer

14.0 hours

The correct answer is 14.0 hours. In a continuous intravenous infusion, the steady-state concentration (

C_{ss}

) is determined by the infusion rate divided by the drug clearance (

Cl = \frac{\text{Rate}}{C_{ss}}

). Initially, the patient's clearance was

\frac{40 \text{ mg/hr}}{10 \text{ mg/L}} = 4 \text{ L/hr}

. The initial elimination half-life was

t_{1/2} = \frac{0.7 \times V_d}{Cl} = \frac{0.7 \times 40 \text{ L}}{4 \text{ L/hr}} = 7 \text{ hours}

. When ciprofloxacin, a strong CYP1A2 inhibitor, was added, theophylline metabolism was impaired. The new steady-state concentration of

20 \text{ mg/L}

indicates a new clearance of

\frac{40 \text{ mg/hr}}{20 \text{ mg/L}} = 2 \text{ L/hr}

. Because half-life is inversely proportional to clearance, the new half-life is

\frac{0.7 \times 40 \text{ L}}{2 \text{ L/hr}} = 14 \text{ hours}

. Since the goal is to reduce the concentration from

20 \text{ mg/L}

10 \text{ mg/L}

(a 50% reduction), exactly one half-life must pass. Because the inhibitor is still present, the drug will be eliminated according to the new, prolonged half-life of 14 hours.

Step-by-Step Solution

Determine the new systemic clearance (

Cl

) of theophylline.

Cl = 2 \text{ L/hr}

The new steady-state concentration (

C_{ss}

) is

20 \text{ mg/L}

while the infusion rate remained

40 \text{ mg/hr}

. Using the formula

Cl = \frac{\text{Infusion Rate}}{C_{ss}}

, the new clearance is

\frac{40}{20} = 2 \text{ L/hr}

Calculate the current elimination half-life (

t_{1/2}

t_{1/2} = 14 \text{ hours}

Half-life is calculated as

t_{1/2} = \frac{0.7 \times V_d}{Cl}

(using 0.7 to approximate

\ln(2)

). Using the stable

V_d

40 \text{ L}

and the new clearance of

2 \text{ L/hr}

, the calculation is

\frac{0.7 \times 40}{2} = 14 \text{ hours}

. The baseline half-life was 7 hours, but clearance was halved by the CYP1A2 inhibitor.

Determine the time required for the concentration to fall to the target level.

14 \text{ hours}

The concentration must fall from

20 \text{ mg/L}

10 \text{ mg/L}

, which is exactly a 50% reduction. Under first-order kinetics, this requires exactly one half-life. Since ciprofloxacin is still present and inhibiting metabolism, the current prolonged half-life of 14 hours applies.

Key Concept

The steady-state concentration of a continuous infusion is inversely proportional to clearance. When clearance is reduced by an enzyme inhibitor, the elimination half-life increases proportionately. Subsequent drug wash-out follows the newly prolonged half-life.

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