Pharmacokinetics (ADME principles)

Pharmacokinetics (ADME principles) US Medical PG Question 1: What is the primary mechanism for iron absorption in the duodenum?

• A. Simple diffusion
• B. Passive paracellular transport
• C. Endocytosis
• D. DMT1 transporter (Correct Answer)

Pharmacokinetics (ADME principles) Explanation: ***DMT1 transporter*** - The **divalent metal transporter 1 (DMT1)** is the primary mechanism for absorbing **non-heme iron (ferrous iron, Fe2+)** into duodenal enterocytes. - This active transport process is pH-dependent and drives iron uptake against a concentration gradient. *Simple diffusion* - Applies to the movement of substances down their concentration gradient without the aid of membrane proteins, which is not the main mechanism for iron due to its ionic nature. - While some highly lipid-soluble substances can cross membranes this way, metal ions like iron require specific transporters. *Passive paracellular transport* - Involves substances moving *between* cells, rather than *through* them, often occurring in leaky epithelia. - While some fluid and electrolytes may use this route, it is not the primary or regulated pathway for iron absorption. *Endocytosis* - A process where cells engulf substances by forming vesicles from the plasma membrane. - While some macromolecules are absorbed via endocytosis, it is not the major mechanism for absorbing dietary iron in the duodenum.

Pharmacokinetics (ADME principles) US Medical PG Question 2: A 65-year-old female patient with a past medical history of diabetes mellitus and an allergy to penicillin develops an infected abscess positive for MRSA on the third day of her hospital stay. She is started on an IV infusion of vancomycin at a dose of 1000 mg every 12 hours. Vancomycin is eliminated by first-order kinetics and has a half life of 6 hours. The volume of distribution of vancomycin is 0.5 L/kg. Assuming no loading dose is given, how long will it take for the drug to reach 94% of its plasma steady state concentration?

• A. 30 hours
• B. 12 hours
• C. 6 hours
• D. 18 hours
• E. 24 hours (Correct Answer)

Pharmacokinetics (ADME principles) Explanation: ***24 hours*** - For a drug eliminated by **first-order kinetics**, it takes approximately **4 half-lives** to reach **93.75%** of steady state concentration, which is conventionally rounded to **94%**. - Since the half-life of vancomycin is **6 hours**, reaching 94% of steady state requires: 4 × 6 hours = **24 hours**. - This follows the pharmacokinetic principle that each half-life brings the drug closer to steady state: 1 t½ = 50%, 2 t½ = 75%, 3 t½ = 87.5%, 4 t½ = 93.75%. *30 hours* - This duration represents **five half-lives** (5 × 6 hours), at which point approximately **96.875%** (often rounded to 97%) of steady state would be reached. - This exceeds the 94% target specified in the question. *18 hours* - This duration represents **three half-lives** (3 × 6 hours), at which point approximately **87.5%** of steady state concentration would be reached. - This falls short of the 94% target. *12 hours* - This duration represents **two half-lives** (2 × 6 hours), at which point approximately **75%** of steady state concentration would be reached. - This is insufficient time to reach 94% of plasma steady state concentration. *6 hours* - This duration represents **one half-life**, at which point approximately **50%** of steady state concentration would be reached. - This is far too short to achieve near-steady state levels.

Pharmacokinetics (ADME principles) US Medical PG Question 3: A scientist is studying the excretion of a novel toxin X by the kidney in order to understand the dynamics of this new substance. He discovers that this new toxin X has a clearance that is half that of inulin in a particular patient. This patient's filtration fraction is 20% and his para-aminohippuric acid (PAH) dynamics are as follows: Urine volume: 100 mL/min Urine PAH concentration: 30 mg/mL Plasma PAH concentration: 5 mg/mL Given these findings, what is the clearance of the novel toxin X?

• A. 1,500 mL/min
• B. 600 mL/min
• C. 300 mL/min
• D. 60 mL/min (Correct Answer)
• E. 120 mL/min

Pharmacokinetics (ADME principles) Explanation: ***60 ml/min*** - First, calculate the **renal plasma flow (RPF)** using PAH clearance: RPF = (Urine PAH conc. × Urine vol.) / Plasma PAH conc. = (30 mg/mL × 100 mL/min) / 5 mg/mL = 600 mL/min. - Next, calculate the **glomerular filtration rate (GFR)**, which is the clearance of inulin. GFR = RPF × Filtration Fraction = 600 mL/min × 0.20 = 120 mL/min. Toxin X clearance is half of inulin clearance, so 120 mL/min / 2 = **60 mL/min**. *1,500 ml/min* - This value is likely obtained if an incorrect formula or conversion was made, possibly by misinterpreting the units or the relationship between GFR, RPF, and filtration fraction. - It significantly overestimates the clearance for a substance that is cleared at half the rate of inulin. *600 ml/min* - This value represents the **renal plasma flow (RPF)**, calculated using the PAH clearance data. - It does not account for the filtration fraction or the fact that toxin X clearance is half of inulin clearance (GFR). *300 ml/min* - This value would be obtained if the renal plasma flow (RPF) was incorrectly halved, or if an intermediate calculation was misinterpreted as the final answer. - It does not align with the given filtration fraction and the relationship between toxin X and inulin clearance. *120 ml/min* - This value represents the **glomerular filtration rate (GFR)**, which is equal to the clearance of inulin (RPF × Filtration Fraction = 600 mL/min × 0.20 = 120 mL/min). - The question states that the clearance of toxin X is **half** that of inulin, so this is an intermediate step, not the final answer.

Pharmacokinetics (ADME principles) US Medical PG Question 4: A research group wants to assess the safety and toxicity profile of a new drug. A clinical trial is conducted with 20 volunteers to estimate the maximum tolerated dose and monitor the apparent toxicity of the drug. The study design is best described as which of the following phases of a clinical trial?

• A. Phase 0
• B. Phase III
• C. Phase V
• D. Phase II
• E. Phase I (Correct Answer)

Pharmacokinetics (ADME principles) Explanation: ***Phase I*** - **Phase I clinical trials** involve a small group of healthy volunteers (typically 20-100) to primarily assess **drug safety**, determine a safe dosage range, and identify side effects. - The main goal is to establish the **maximum tolerated dose (MTD)** and evaluate the drug's pharmacokinetic and pharmacodynamic profiles. *Phase 0* - **Phase 0 trials** are exploratory studies conducted in a very small number of subjects (10-15) to gather preliminary data on a drug's **pharmacodynamics and pharmacokinetics** in humans. - They involve microdoses, not intended to have therapeutic effects, and thus cannot determine toxicity or MTD. *Phase III* - **Phase III trials** are large-scale studies involving hundreds to thousands of patients to confirm the drug's **efficacy**, monitor side effects, compare it to standard treatments, and collect information that will allow the drug to be used safely. - These trials are conducted after safety and initial efficacy have been established in earlier phases. *Phase V* - "Phase V" is not a standard, recognized phase in the traditional clinical trial classification (Phase 0, I, II, III, IV). - This term might be used in some non-standard research contexts or for post-marketing studies that go beyond Phase IV surveillance, but it is not a formal phase for initial drug development. *Phase II* - **Phase II trials** involve several hundred patients with the condition the drug is intended to treat, focusing on **drug efficacy** and further evaluating safety. - While safety is still monitored, the primary objective shifts to determining if the drug works for its intended purpose and at what dose.

Pharmacokinetics (ADME principles) US Medical PG Question 5: A 26-year-old man comes to the emergency department because of a 1-week history of worsening fatigue, nausea, and vomiting. Six weeks ago, he was diagnosed with latent tuberculosis and appropriate low-dose pharmacotherapy was initiated. Physical examination shows right upper quadrant tenderness and scleral icterus. Laboratory studies show elevated aminotransferases. Impaired function of which of the following pharmacokinetic processes is the most likely explanation for this patient's symptoms?

• A. Acetylation (Correct Answer)
• B. Glucuronidation
• C. Hydrolysis
• D. Sulfation
• E. Reduction

Pharmacokinetics (ADME principles) Explanation: ***Acetylation*** - This patient is exhibiting symptoms of **hepatotoxicity** (fatigue, nausea, vomiting, RUQ tenderness, scleral icterus, elevated aminotransferases) after starting low-dose pharmacotherapy for latent tuberculosis. The most common drug used for latent TB is **isoniazid**, which is primarily metabolized by **N-acetylation**. - Impaired acetylation, particularly in **slow acetylators**, can lead to higher plasma concentrations of isoniazid and its toxic metabolites, increasing the risk of **drug-induced liver injury**. *Glucuronidation* - **Glucuronidation** is a Phase II metabolic pathway that conjugates drugs with **glucuronic acid** to increase water solubility and facilitate excretion. - While important for the metabolism of many drugs and endogenous substances (e.g., bilirubin), it is not the primary mechanism of metabolism or the main pathway implicated in the hepatotoxicity of **isoniazid**. *Hydrolysis* - **Hydrolysis** is a chemical reaction in which water is used to break down a compound, often involving ester or amide bonds. - This process is not the primary metabolic pathway for **isoniazid**, nor is impaired hydrolysis a common cause of its hepatotoxicity. *Sulfation* - **Sulfation** is a Phase II metabolic reaction that conjugates drugs with a **sulfate group**, typically for detoxification and excretion. - While various drugs undergo sulfation, it is not the dominant metabolic pathway for **isoniazid**, and impaired sulfation is not typically associated with isoniazid-induced hepatotoxicity. *Reduction* - **Reduction** reactions involve the gain of electrons or hydrogen atoms, or the loss of oxygen, and are part of drug metabolism for certain compounds. - However, reduction is not the primary clearance mechanism for **isoniazid**, and abnormal reduction is not commonly implicated in its hepatotoxic effects.

Pharmacokinetics (ADME principles) US Medical PG Question 6: In a healthy patient with no renal abnormalities, several mechanisms are responsible for moving various filtered substances into and out of the tubules. Para-aminohippurate (PAH) is frequently used to estimate renal blood flow when maintained at low plasma concentrations. The following table illustrates the effect of changing plasma PAH concentrations on PAH excretion: Plasma PAH concentration (mg/dL) | Urinary PAH concentration (mg/dL) 0 | 0 10 | 60 20 | 120 30 | 150 40 | 180 Which of the following mechanisms best explains the decreased rate of increase in PAH excretion observed when plasma PAH concentration exceeds 20 mg/dL?

• A. Decreased glomerular filtration of PAH
• B. Increased rate of PAH reabsorption
• C. Increased flow rate of tubular contents
• D. Saturation of PAH transport carriers (Correct Answer)
• E. Increased diffusion rate of PAH

Pharmacokinetics (ADME principles) Explanation: ***Saturation of PAH transport carriers*** - PAH is primarily cleared by **tubular secretion** via organic anion transporters (OATs) in the proximal tubule, which have a **finite transport maximum (Tm)**. - When plasma PAH concentration exceeds the capacity of these carriers (as seen above 20 mg/dL), the transporters become saturated, leading to a **decreased incremental excretion** despite rising plasma levels. *Decreased glomerular filtration of PAH* - **Glomerular filtration rate (GFR)** for PAH is proportional to its plasma concentration and is typically constant in a healthy kidney, so it would not decrease with increasing plasma PAH. - A decrease in GFR would lead to a *reduced* overall excretion, but not specifically explain the *decreased rate of increase* at higher plasma concentrations. *Increased rate of PAH reabsorption* - PAH is **minimally reabsorbed** in the renal tubules; its primary mechanism of removal from the blood is active secretion. - An increase in reabsorption would lead to *less* PAH in the urine, but there's no physiological basis for increased reabsorption as plasma concentration rises. *Increased flow rate of tubular contents* - While an increased flow rate can sometimes affect solute reabsorption or secretion, it would generally lead to a more, not less, efficient clearance of secreted substances. - This mechanism does not explain the **saturation kinetics** observed with PAH at higher plasma concentrations. *Increased diffusion rate of PAH* - PAH is a charged organic anion, and its movement across tubular membranes is primarily mediated by **active transport** rather than simple diffusion. - Even if diffusion played a minor role, an increased diffusion rate would generally lead to *more* excretion, not the observed plateau in the rate of increase.

Pharmacokinetics (ADME principles) US Medical PG Question 7: A researcher is investigating the effects of a new antihypertensive medication on renal physiology. She gives a subject a dose of the new medication, and she then collects plasma and urine samples. She finds the following: Hematocrit: 40%; Serum creatinine: 0.0125 mg/mL; Urine creatinine: 1.25 mg/mL. Urinary output is 1 mL/min. Renal blood flow is 1 L/min. Based on the above information and approximating that the creatinine clearance is equal to the GFR, what answer best approximates filtration fraction in this case?

• A. 10%
• B. 17% (Correct Answer)
• C. 33%
• D. 50%
• E. 25%

Pharmacokinetics (ADME principles) Explanation: ***17%*** - First, calculate **GFR** using the creatinine clearance formula: GFR = (Urine creatinine × Urinary output) / Serum creatinine = (1.25 mg/mL × 1 mL/min) / 0.0125 mg/mL = **100 mL/min**. - Next, calculate **Renal Plasma Flow (RPF)** from Renal Blood Flow (RBF) and Hematocrit: RPF = RBF × (1 - Hematocrit) = 1000 mL/min × (1 - 0.40) = **600 mL/min**. - Finally, calculate **Filtration Fraction (FF)** = GFR / RPF = 100 mL/min / 600 mL/min = 0.1667 = **16.7%, which approximates to 17%**. - This is the correct answer based on the physiological calculations and represents a normal filtration fraction. *10%* - This would correspond to a filtration fraction of 0.10, which would require either a GFR of 60 mL/min (lower than calculated) or an RPF of 1000 mL/min (higher than calculated). - This value is too low given the provided parameters and doesn't match the calculation from the given data. *25%* - This value would suggest FF = 0.25, requiring a GFR of 150 mL/min with the calculated RPF of 600 mL/min. - This is higher than the calculated GFR of 100 mL/min and doesn't match the given creatinine values. *33%* - This would imply FF = 0.33, requiring a GFR of approximately 200 mL/min with RPF of 600 mL/min. - This is significantly higher than the calculated GFR and would represent an abnormally elevated filtration fraction. *50%* - A filtration fraction of 50% is unphysiologically high and would indicate severe pathology. - This would require a GFR of 300 mL/min with the calculated RPF, which is impossible given the provided creatinine clearance data.

Pharmacokinetics (ADME principles) US Medical PG Question 8: A researcher is investigating the behavior of two novel chemotherapeutic drugs that he believes will be effective against certain forms of lymphoma. In order to evaluate the safety of these drugs, this researcher measures the concentration and rate of elimination of each drug over time. A partial set of the results is provided below. Time 1: Concentration of Drug A: 4 mg/dl Concentration of Drug B: 3 mg/dl Elimination of Drug A: 1 mg/minute Elimination of Drug B: 4 mg/minute Time 2: Concentration of Drug A: 2 mg/dl Concentration of Drug B: 15 mg/dl Elimination of Drug A: 0.5 mg/minute Elimination of Drug B: 4 mg/minute Which of the following statements correctly identifies the most likely relationship between the half-life of these two drugs?

• A. The half-life of drug A is always longer than that of drug B
• B. The half-life of both drug A and drug B are constant
• C. The half-life of both drug A and drug B are variable
• D. The half-life of drug A is variable but that of drug B is constant
• E. The half-life of drug A is constant but that of drug B is variable (Correct Answer)

Pharmacokinetics (ADME principles) Explanation: ***The half-life of drug A is constant but that of drug B is variable*** - Drug A shows a **constant fraction** eliminated per unit time (1 mg/minute from 4 mg/dl, then 0.5 mg/minute from 2 mg/dl), indicating **first-order kinetics** and thus a constant half-life. - Drug B's elimination rate remains constant (4 mg/minute) despite varying concentrations (3 mg/dl then 15 mg/dl), which suggests **zero-order kinetics** and a variable half-life dependent on concentration. *The half-life of drug A is always longer than that of drug B* - This statement is incorrect because Drug B exhibits **zero-order kinetics**, meaning its **half-life changes** with concentration, making a constant comparison invalid. - At very high concentrations, Drug B's half-life could actually be longer than Drug A's if the elimination rate is slow relative to the large amount of drug. *The half-life of both drug A and drug B are constant* - This is incorrect because Drug B demonstrates **zero-order kinetics**, where the elimination rate is constant, but the **half-life is variable** and directly depends on the drug concentration. - For zero-order kinetics, a constant amount of drug is eliminated per unit time, not a constant fraction, which causes the half-life to change. *The half-life of both drug A and drug B are variable* - This is incorrect because Drug A exhibits **first-order kinetics**, where a **constant proportion** of the drug is eliminated per unit time, resulting in a **constant half-life**. - Its elimination rate is directly proportional to its concentration (1 mg/min from 4 mg/dl, 0.5 mg/min from 2 mg/dl), which defines first-order kinetics. *The half-life of drug A is variable but that of drug B is constant* - This statement is the opposite of what the data indicates for Drug A; Drug A's elimination is **proportional to its concentration**, signifying **first-order kinetics** and a constant half-life. - Drug B's elimination rate is constant regardless of concentration, which points to **zero-order kinetics** and thus a variable half-life.

Pharmacokinetics (ADME principles) US Medical PG Question 9: A 65-year-old man with a history of myocardial infarction is admitted to the hospital for treatment of atrial fibrillation with rapid ventricular response. He is 180 cm (5 ft 11 in) tall and weighs 80 kg (173 lb). He is given an intravenous bolus of 150 mg of amiodarone. After 20 minutes, the amiodarone plasma concentration is 2.5 mcg/mL. Amiodarone distributes in the body within minutes, and its elimination half-life after intravenous administration is 30 days. Which of the following values is closest to the volume of distribution of the administered drug?

• A. 60 L (Correct Answer)
• B. 80 L
• C. 150 L
• D. 17 L
• E. 10 L

Pharmacokinetics (ADME principles) Explanation: ***60 L*** - The **volume of distribution (Vd)** is calculated using the formula: **Vd = Dose / Plasma Concentration**. - Given: Dose = 150 mg (150,000 mcg), Plasma concentration = 2.5 mcg/mL - Calculation: Vd = 150,000 mcg / 2.5 mcg/mL = 60,000 mL = **60 L** - Note: This calculation represents a simplified scenario. In clinical practice, amiodarone has an extremely large volume of distribution (60-100 L/kg or ~4,800-8,000 L in this patient) due to extensive tissue distribution, but the question tests the ability to apply the basic pharmacokinetic formula. *80 L* - This value would result if the plasma concentration were 1.875 mcg/mL (150,000 mcg / 80,000 mL), not the given 2.5 mcg/mL. - This represents a common calculation error when working with pharmacokinetic parameters. *150 L* - This value would require a plasma concentration of 1 mcg/mL (150,000 mcg / 150,000 mL), which is lower than the measured 2.5 mcg/mL. - This error might occur if the dose value were confused with the volume of distribution. *17 L* - This value would be obtained with a plasma concentration of approximately 8.8 mcg/mL (150,000 mcg / 17,000 mL), significantly higher than the measured 2.5 mcg/mL. - This represents a significant underestimation of Vd and would suggest limited drug distribution. *10 L* - This value would require a plasma concentration of 15 mcg/mL (150,000 mcg / 10,000 mL), which is 6-fold higher than the given 2.5 mcg/mL. - Such a small Vd would suggest drug confined primarily to plasma, which is inappropriate for lipophilic drugs with extensive tissue distribution.

Pharmacokinetics (ADME principles) US Medical PG Question 10: An experimental infusable drug, X729, is currently being studied to determine its pharmacokinetics. The drug was found to have a half life of 1.5 hours and is eliminated by first order kinetics. What is the minimum number of hours required to reach a steady state concentration of >90%?

• A. 6 (Correct Answer)
• B. 3
• C. 7.5
• D. 1.5
• E. 4.5

Pharmacokinetics (ADME principles) Explanation: ***6*** - For a drug eliminated by **first-order kinetics**, approximately **4 to 5 half-lives** are required to reach **steady-state concentration**. - To reach >90% of steady-state, at least **4 half-lives** are needed, where **93.75%** of the steady state is achieved. - The time taken would be **4 half-lives × 1.5 hours/half-life = 6 hours**, making this the **minimum time** to exceed 90%. *3* - This represents only **2 half-lives** (2 × 1.5 hours = 3 hours), which would achieve roughly **75%** of the steady-state concentration. - This is insufficient to reach >90% of the steady-state concentration. *7.5* - This time point represents **5 half-lives** (5 × 1.5 hours = 7.5 hours), which would achieve approximately **97%** of the steady-state concentration. - While this does exceed 90%, the question asks for the **minimum** number of hours required, and 90% is already exceeded at 6 hours (4 half-lives). *1.5* - This is only **1 half-life**, which would achieve approximately **50%** of the steady-state concentration. - This is far too early to reach a >90% steady-state concentration. *4.5* - This represents **3 half-lives** (3 × 1.5 hours = 4.5 hours), achieving approximately **87.5%** of the steady-state concentration. - While close to 90%, it does not quite reach "greater than 90%".

Pharmacokinetics (ADME principles) Flashcard 1 of 7

Question: What compartment is the majority of a drug with medium Vd found in? _____

Answer: Extracellular fluid

Pharmacokinetics (ADME principles) Flashcard 2 of 7

Question: How is Warfarin transported in the blood?_____

Answer: Bound to albumin

Pharmacokinetics (ADME principles) Flashcard 3 of 7

Question: What phase of drug metabolism includes methylation, glucuronidation, acetylation, and sulfation (i.e. conjugation)? _____

Answer: Phase II

Pharmacokinetics (ADME principles) Flashcard 4 of 7

Question: Phase _____ of clinical trials assesses optimal dosing and treatment efficacy

Answer: II

Pharmacokinetics (ADME principles) Flashcard 5 of 7

Question: Phase _____ of clinical trials assesses safety, toxicity, pharmacokinetics, and pharmacodynamics

Answer: I

Pharmacokinetics (ADME principles) Flashcard 6 of 7

Question: Which drug elimination is characterized by a constant amount of drug eliminated per unit time?_____

Answer: Zero-order elimination

Pharmacokinetics (ADME principles) Flashcard 7 of 7

Question: Phase _____ drug metabolism usually yields very polar, water-soluble metabolites (inactive)

Answer: II