Home Books Katzung & Trevor's Pharmacology: Examination & Board Review, 12e

Chapter 3: Pharmacokinetics

Mr Jones has zero kidney function and is undergoing hemodialysis 3 days per week while awaiting a kidney transplant. He takes metformin for type 2 diabetes mellitus and was previously stabilized (while his kidney function was adequate) at a dosage of 500 mg twice daily, given orally. The plasma concentration at this dosage with normal kidney function was found to be 1.4 mg/L. He has had 6 dialysis procedures and metformin toxicity is suspected. A blood sample now shows a metformin concentration of 4.2 mg/L. What was Mr Jones’ clearance of metformin while his kidney function was normal?

(A) 238 L/d

(B) 29.8 L/h

(C) 3 L/d

(D) 238 L/h

(E) 30 L/min

Examination questions often provide more information than is needed—to test the student’s ability to classify and organize data. In question 1, the data provided for Mr Jones on dialysis is irrelevant, even though choice A, 238 L/d, is the correct clearance while on dialysis. By definition, clearance is calculated by dividing the rate of elimination by the plasma concentration:

The answer is B.

Ms Smith, a 65-year-old woman with pneumonia, was given tobramycin, 150 mg, intravenously. After 20 minutes, the plasma concentration was measured and was found to be 3 mg/L. Assuming no elimination of the drug in 20 minutes, what is the apparent volume of distribution of tobramycin in Ms Smith?

(A) 3 L/min

(B) 3 L

(C) 50 L

(D) 7 L

(E) 0.1 mg/min

The volume of distribution (V_d) is the apparent volume into which the loading dose is distributed. It is calculated by dividing the dose by the resulting plasma concentration, C_p:

The answer is C.

St. John’s Wort, a popular botanical remedy, is a potent inducer of hepatic phase I CYP3A4 enzymes. Verapamil and phenytoin are both eliminated from the body by metabolism in the liver. Verapamil has a clearance of 1.5 L/min, approximately equal to liver blood flow, whereas phenytoin has a clearance of 0.1 L/min. Based on this fact, which of the following is most correct?

(A) St. John’s Wort will increase the half-life of phenytoin and verapamil

(B) St. John’s Wort will decrease the volume of distribution of CYP3A4 substrates

(C) St. John’s Wort will decrease the hepatic extraction of phenytoin

(D) St. John’s Wort will decrease the first-pass effect for verapamil

(E) St. John’s Wort will increase the clearance of phenytoin

Induction of phase I metabolizing enzymes will decrease the half-life of substrates of these enzymes. P450 enzyme induction has no effect on volume of distribution. Hepatic extraction, the first-pass effect, and clearance for CYP3A4 substrates will be increased by inducers. However, the extraction of verapamil is already equal to the hepatic blood flow, so further increase in metabolism will not increase clearance of this drug. The answer is E.

A 45-year-old woman with small cell lung cancer has elected to participate in the trial of a new chemotherapeutic agent. It is given by constant intravenous infusion of 10 mg/h. Plasma concentrations (C_p) are measured with the results shown in the following table.

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Time After Start of Infusion (h)

Plasma Concentration (mg/L)

0.7

3.0

3.6

3.84

4.0

What conclusion can be drawn from these data?

(A) Clearance is 2.0 L/h

(B) Doubling the rate of infusion would result in a plasma concentration of 16 mg/L at 40 h

(C) Elimination follows zero-order kinetics

(D) Half-life is 2 h

(E) Volume of distribution is 30 L

By inspection of the data in the table, it is clear that the steady-state plasma concentration is approximately 4 mg/L. None of the measured concentrations is equal to one half of the steady state value; so the half-life is not immediately apparent. However, according to the constant infusion principle (Figure 3–3), 2 half-lives are required to reach 75% of the final concentration; 75% (3.0 mg/L) of the final steady-state concentration was reached at 4 h. If 4 h equals 2 half-lives, the half-life must be 2 h. Rearranging the equation for maintenance dosing (dosing rate = CL × C_p), it can be determined that the clearance (CL) = dosing rate/plasma concentration (C_p), or 2.5 L/h. The volume of distribution (V_d) can be calculated from the half-life equation (t_1/2 = 0.693 × V_d/CL) and is equal to 7.2 L. This drug follows first-order kinetics, as indicated by the progressive approach to the steady-state plasma concentration. The answer is D.

Figure 3–3. Plasma concentration (plotted as percentage of maximum) of a drug given by constant intravenous infusion for 8 half-lives and then stopped. The concentration rises smoothly with time and always reaches 50% of steady state after 1 half-life, 75% after 2 half-lives, 87.5% after 3 half-lives, and so on. The decline in concentration after stopping drug administration follows the same type of curve: 50% is left after 1 half-life, 25% after 2 half-lives, and so on. The asymptotic approach to steady state on both increasing and decreasing limbs of the curve is characteristic of drugs that have first-order kinetics.

You are the only physician in a clinic that is cut off from the outside world by violent storms, flooding, and landslides. A 15-year-old girl is brought to the clinic with severe asthmatic wheezing. Because of the lack of other drugs, you decide to use intravenous theophylline for treatment. The pharmacokinetics of theophylline include the following average parameters: V_d 35 L; CL 48 mL/min; half-life 8 h. If an intravenous infusion of theophylline is started at a rate of 0.48 mg/min, how long would it take to reach 93.75% of the final steady-state concentration?

(A) Approximately 48 min

(B) Approximately 7.4 h

(C) Approximately 8 h

(D) Approximately 24 h

(E) Approximately 32 h

For a drug with first-order kinetics, the approach of the drug plasma concentration to steady-state concentration during continuous infusion follows a stereotypical curve (Figure 3–3) that rises rapidly at first and gradually reaches a plateau. It reaches 50% of steady state at 1 half-life, 75% at 2 half-lives, 87.5% at 3, 93.75% at 4, and progressively halves the difference between its current level and 100% of steady state with each half-life. The answer is E, 32 h, or 4 half-lives.

A 74-year-old retired mechanic is admitted with a myocardial infarction and a severe acute cardiac arrhythmia. You decide to give lidocaine to correct the arrhythmia. A continuous intravenous infusion of lidocaine, 1.92 mg/min, is started at 8 AM. The average pharmacokinetic parameters of lidocaine are: V_d 77 L; clearance 640 mL/min; half-life 1.4 h. What is the expected steady-state plasma concentration?

(A) 40 mg/L

(B) 3.0 mg/L

(C) 0.025 mg/L

(D) 7.2 mg/L

(E) 3.46 mg/L

The drug is being administered continuously and the steady-state concentration (C_p(ss)) for a continuously administered drug is given by the equation in question 1. Thus,

Rearranging:

The answer is B.

A new drug is under study in phase 1 trials. It is found that this molecule is avidly taken up by extravascular tissues so that the final total amount in the extravascular compartment at steady state is 100 times the amount remaining in the blood plasma. What is the probable volume of distribution in a hypothetical person with 8 L of blood and 4 L of plasma?

(A) Insufficient data to calculate

(B) 8 L

(C) 14.14 L

(D) 100 L

(E) 404 L

Let Z be the amount in the blood plasma. If the amount in the rest of the body is 100 times greater, then the total amount in the body is 101Z. The concentration in the blood plasma (C_p) is Z/4 L. According to the definition:

The answer is E.

A 63-year-old woman in the intensive care unit requires an infusion of procainamide. Its half-life is 2 h. The infusion is begun at 9 AM. At 1 PM on the same day, a blood sample is taken; the drug concentration is found to be 3 mg/L. What is the probable steady-state drug concentration after 16 or more hours of infusion?

(A) 3 mg/L

(B) 4 mg/L

(C) 6 mg/L

(D) 9.9 mg/L

(E) 15 mg/L

According to the curve that relates plasma concentration to infusion time (Figure 3–3), a drug reaches 50% of its final steady-state concentration in 1 half-life, 75% in 2 half-lives, etc. From 9 AM to 1 PM is 4 h, or 2 half-lives. Therefore, the measured concentration at 1 PM is 75% of the steady-state value (0.75 × C_p(ss)). The steady-state concentration is 3 mg/L divided by 0.75, or 4 mg/L. The answer is B.

A 30-year-old man is brought to the emergency department in a deep coma. Respiration is severely depressed and he has pinpoint pupils. His friends state that he self-administered a large dose of morphine 6 h earlier. An immediate blood analysis shows a morphine blood level of 0.25 mg/L. Assuming that the V_d of morphine in this patient is 200 L and the half-life is 3 h, how much morphine did the patient inject 6 h earlier?

(A) 25 mg

(B) 50 mg

(C) 100 mg

(D) 200 mg

(E) Not enough data to predict

According to the curve that relates the decline of plasma concentration to time as the drug is eliminated (Figure 3–3), the plasma concentration of morphine was 4 times higher immediately after administration than at the time of the measurement, which occurred 6 h, or 2 half-lives, later. Therefore, the initial plasma concentration was 1 mg/L. Since the amount in the body at any time is equal to V_d × plasma concentration (text Equation 1), the amount injected was 200 L × 1 mg/L, or 200 mg. The answer is D.

Gentamicin, an aminoglycoside antibiotic, is sometimes given as a single large intravenous bolus dose of once a day to achieve a highly active peak plasma concentration. Gentamicin’s volume of distribution is about 20 L in a 70 kg patient and, in your patient, the half-life is 4 h. If your patient is given an IV bolus dose of 360 mg, what will the trough concentration of gentamicin be 24 hours later just before the next intravenous bolus?

(A) 18 mg/L

(B) 4.5 mg/L

(C) 1.13 mg/L

(D) 0.56 mg/L

(E) 0. 28 mg/L

This problem requires, first, the calculation of the initial peak plasma concentration, and then the decline in concentration as the drug is eliminated by 50% in each half-life. The peak concentration = dose/V_d, or 360 mg/20 L or 18 mg/L. The trough concentration will be measured 24 hours or 6 half-lives later. After 1 half-life C_p will be 9; after 2, 4.5; after 3, 2.25; after 4 half-lives, 1.125 mg/L; after 5, 0.56; and after 6 half-lives, 0.28 mg/L. The answer is E.

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Chapter 3: Pharmacokinetics

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