One-compartment: IV infusion MCQs With Answer

Introduction

One-compartment: IV infusion MCQs With Answer

This question set is designed for M.Pharm students studying Advanced Biopharmaceutics & Pharmacokinetics (MPH 202T). It focuses on one-compartment intravenous (IV) continuous infusion — covering key principles, governing equations, steady-state concepts, loading doses, relationships among clearance, volume and elimination rate constant, and practical calculation problems. Questions range from conceptual understanding to quantitative problems typical in pharmacy exams and therapeutic drug monitoring. Working through these MCQs will reinforce how infusion rate, clearance and volume determine steady-state concentration, time to steady state, and strategies like loading doses to reach therapeutic levels rapidly.

Q1. In a one-compartment model during a constant IV infusion, the steady-state plasma concentration (Css) is determined by which relationship?

• Css = V / R0
• Css = R0 / CL
• Css = R0 × V
• Css = CL / R0

Correct Answer: Css = R0 / CL

Q2. Which expression correctly describes the plasma concentration C(t) at time t during a constant-rate IV infusion in a one-compartment model?

• C(t) = (R0 / CL) × (1 – e^{-k t})
• C(t) = (R0 / V) × e^{-k t}
• C(t) = (CL / R0) × (1 – e^{-k t})
• C(t) = (R0 / CL) × e^{-k t}

Correct Answer: C(t) = (R0 / CL) × (1 – e^{-k t})

Q3. Approximately how many half-lives are required to reach about 95% of steady-state concentration during a continuous IV infusion?

• 1 half-life
• 2 half-lives
• 3 half-lives
• 5 half-lives

Correct Answer: 3 half-lives

Q4. What is the correct formula for a loading dose (LD) intended to immediately achieve a target steady-state concentration Css in a one-compartment model?

• LD = CL × Css
• LD = V × Css
• LD = (R0 / CL) × V
• LD = R0 × Css

Correct Answer: LD = V × Css

Q5. Which equation correctly relates the elimination rate constant (k), clearance (CL), and volume of distribution (V) in a one-compartment model?

• k = V / CL
• k = CL × V
• k = CL / V
• k = CL + V

Correct Answer: k = CL / V

Q6. If the infusion rate R0 of a drug is doubled while clearance (CL) remains constant, what happens to the steady-state concentration (Css)?

• Css is halved
• Css remains unchanged
• Css doubles
• Css increases by the square root of 2

Correct Answer: Css doubles

Q7. Calculate the infusion rate R0 required to maintain a steady-state concentration of 10 mg/L if clearance (CL) is 5 L/h.

• R0 = 2 mg/h
• R0 = 50 mg/h
• R0 = 5 mg/h
• R0 = 100 mg/h

Correct Answer: R0 = 50 mg/h

Q8. A drug has a half-life of 4 hours. Approximately how long will it take to reach 90% of steady-state concentration during continuous infusion?

• ≈8.3 hours
• ≈13.3 hours
• ≈23.3 hours
• ≈2.3 hours

Correct Answer: ≈13.3 hours

Q9. For a drug infused at R0 = 100 mg/h with CL = 10 L/h and V = 20 L, what is the plasma concentration after 2 hours? (Use C(t) = Css (1 – e^{-k t}), k = CL/V)

• ≈6.32 mg/L
• ≈3.16 mg/L
• ≈9.50 mg/L
• 10.00 mg/L

Correct Answer: ≈6.32 mg/L

Q10. After stopping a continuous IV infusion at steady state, which expression describes how concentration declines with time?

• C(t) = Css × (1 – e^{-k t})
• C(t) = Css × e^{-k t}
• C(t) = (R0 / CL) × e^{k t}
• C(t) = CL × V × e^{-k t}

Correct Answer: C(t) = Css × e^{-k t}

Q11. If a patient’s clearance is reduced by 50% but the infusion rate R0 is unchanged, what is the expected change in steady-state concentration?

• Css decreases by 50%
• Css remains the same
• Css increases by 50%
• Css doubles

Correct Answer: Css doubles

Q12. The time constant (τ = 1/k) in a one-compartment infusion model corresponds to approximately what fraction of steady-state concentration achieved at t = τ?

• 36.8% of Css
• 50% of Css
• 63.2% of Css
• 86.5% of Css

Correct Answer: 63.2% of Css

Q13. For an infusion rate of 30 mg/h and clearance of 2 L/h, the steady-state plasma concentration is:

• 13 mg/L
• 15 mg/L
• 60 mg/L
• 0.067 mg/L

Correct Answer: 15 mg/L

Q14. A drug has V = 40 L and CL = 4 L/h. What is its elimination half-life (t1/2)? (Use t1/2 = 0.693 × V / CL)

• ≈6.93 hours
• ≈0.173 hours
• ≈10 hours
• ≈0.693 hours

Correct Answer: ≈6.93 hours

Q15. To immediately achieve a target steady-state concentration of 5 mg/L in a patient with V = 20 L, what loading dose should be administered?

• 15 mg
• 100 mg
• 5 mg
• 400 mg

Correct Answer: 100 mg

Q16. At the very start of an IV infusion (t = 0, C = 0), the initial rate of change in drug concentration (dC/dt) equals which of the following?

• R0 / CL
• R0 / V
• k × Css
• Zero

Correct Answer: R0 / V

Q17. True or false: The time required to reach steady state during a continuous IV infusion depends on the infusion rate R0.

• True
• False
• Depends only on V
• Depends only on CL

Correct Answer: False

Q18. For therapeutic drug monitoring, when is the most appropriate time to obtain a steady-state concentration during a continuous infusion?

• Immediately after starting the infusion
• After 3–5 half-lives (when steady state is approximated)
• Before the infusion is started
• One hour after starting regardless of t1/2

Correct Answer: After 3–5 half-lives (when steady state is approximated)

Q19. If a loading dose equal to V × Css is administered at the same time a maintenance infusion (R0 consistent with Css) is started, what is the expected plasma concentration profile?

• Concentration will be immediately at Css and then decline
• Concentration will gradually rise to Css over several half-lives
• Concentration will be immediately at Css and then be maintained at Css by the infusion
• Concentration will overshoot to twice Css and slowly decline

Correct Answer: Concentration will be immediately at Css and then be maintained at Css by the infusion

Q20. Consider two drugs infused at rates achieving the same Css. Drug A has V = 20 L and CL = 4 L/h; Drug B has V = 10 L and CL = 1 L/h. Which statement about time to reach steady state is correct?

• Drug A reaches steady state faster because it has larger V
• Drug B reaches steady state faster because it has smaller CL
• Both reach steady state at the same rate
• Drug A reaches steady state faster because its V/CL (≈5 h) is smaller than Drug B’s V/CL (≈10 h)

Correct Answer: Drug A reaches steady state faster because its V/CL (≈5 h) is smaller than Drug B’s V/CL (≈10 h)

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