Scientific Foundations

Recall Questions

1. Adding a reversible competitive antagonist typically causes which change to the agonist’s graded dose–response curve?

   1. Left shift; E_Max increases
   2. Right shift; E_Max unchanged
   3. Right shift; E_Max decreases
   4. No shift; EC₅₀ unchanged

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2. Clearance (Cl) is best defined as:

   1. The fraction of drug absorbed
   2. The volume of plasma cleared of drug per unit time
   3. The amount of drug in the body divided by plasma concentration
   4. Time required for the drug to reach steady state

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3. Phase I metabolism most commonly involves:

   1. Glucuronidation and sulfation
   2. Direct renal excretion of unchanged drug
   3. Protein binding
   4. Oxidation/reduction/hydrolysis often via CYP enzymes

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4. A pharmaceutical company is developing a new drug for type 2 diabetes that mimics insulin signalling. The target receptor, when activated, undergoes autophosphorylation of its intracellular tyrosine residues, triggering downstream glucose uptake pathways.

   1. G-protein coupled receptor (GPCR)
   2. Transmembrane ion channel
   3. Intracellular nuclear receptor
   4. Transmembrane receptor with intrinsic tyrosine kinase activity
   5. Ligand-gated nicotinic acetylcholine receptor

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5. In a laboratory experiment, an agonist produces a dose-response curve with an EC₅₀ of 10 nM and an E_max of 100%. A competitive antagonist is then added. A new dose-response curve is produced.

   1. EC₅₀ is unchanged; E_max is reduced to 60%.
   2. EC₅₀ is shifted to 100 nM; E_max remains at 100%.
   3. Both EC₅₀ and E_max are reduced proportionally.
   4. EC₅₀ is reduced to 1 nM; E_max remains at 100%.
   5. EC₅₀ is unchanged; E_max increases to 120%.

   Show Explanation

   A competitive antagonist produces a parallel rightward shift of the dose-response curve EC₅₀ increases (less potent apparent agonist effect) but E_max is preserved because the antagonism is surmountable. Reduction in E_max (A, C) is characteristic of non-competitive antagonism. A leftward shift (D) would indicate increased potency. E_max cannot exceed 100% in a standard system

6. You are comparing two drugs:

   • Drug A: Weak acid, pK_a = 4.0
   • Drug B: Weak based, pK_a = 9.0

   Both drugs are in the stomach, where the luminal pH = 2.0. Which of the following best describes the ionisation state and absorption of these two drugs in the stomach?

   1. Both Drug A and Drug B are predominantly uncharged and are well absorbed from the stomach
   2. Drug A is predominantly uncharged and well absorbed; Drug B is predominantly charged and poorly absorbed from the stomach
   3. Drug A is predominantly charged and poorly absorbed; Drug B is predominantly uncharged and well absorbed from the stomach
   4. Both Drug A and Drug B are predominantly charged and poorly absorbed from the stomach.
   5. Drug A and Drug B have identical absorption profiles because both are passively diffused

   Show Explanation

   Use the Henderson-Hasselbalch equation to determine the fraction of each drug in protonated and unprotonated form: $$ pH = pK_a + \log_{10}\frac{[U]}{[P]} $$ $$ \frac{[U]}{[P]} = 10 ^{pH - pK_a} $$ Drug A is mostly in the protonated form ([U]:[P] is 1:100) and weak acids are uncharged in the unprotonated form. Almost all of Drug B is the protonated form ([U]:[P] is 1:10⁷) and weak bases are charged in the protonated form. Uncharged molecules are more readily absorbed because they can diffuse across the cell membrane.

Application Questions

1. A 65-year old patient presents with a blood test indicating hypercholesterolemia. You decide to prescribe a statin which have been found to lower serum cholesterol levels. You are considering Atorvastatin, which is lipophilic, and Pravastatin, which is hydrophilic. Which statin would have the larger volume of distribution in the patient?

   1. Atorvastatin
   2. Pravastatin
   3. Neither, they would have similar volumes of distribution

   Show Explanation

   Because lipophilic statins more easily cross cell membranes they would be more readily absorbed and distributed into tissues and have a greater volume of distribution compared to a drug which is hydrophilic.

2. A 60 kg woman has a bacterial infection. You decide to administer an antibiotic that has a volume of distribution of 0.2 L/kg. The drug has a clearance of 200 ml/min. If you administer the drug by constant intravenous infusion, approximately how long will it take for the plasma concentration of the antibiotic to reach steady-state levels?

   1. 30 min
   2. 1 hr
   3. 3 hr
   4. 24 hr

   Show Explanation

   The volume of distribution (Vd) is 12 L and the clearance is 12 L/hr. We can use this formula to determine the half-life of the antibiotic: $$ t_{1/2} = \frac{0.693 \times V_d}{Cl} $$ $$ t_{1/2} = \frac{0.693 \times 12 L}{12\frac{L}{hr}} $$ $$ t_{1/2} = 0.693 hr $$ Since it takes 3 to 4 half-lives to reach steady-state, the best choice is 3 hr (160 min).

3. A patient presents with an aspirin overdose. The treating team decides to administer intravenous sodium bicarbonate. Which mechanism BEST explains how sodium bicarbonate enhances aspirin elimination?

   1. Sodium bicarbonate directly binds aspirin in the plasma, forming an inactive complex
   2. Alkalinisation of the urine ionises the weak acid aspirin in the tubular lumen, trapping it and reducing reabsorption
   3. Sodium bicarbonate activates CYP450 enzymes to increase aspirin metabolism
   4. Alkalinisation of the urine converts aspirin to its uncharged form, enhancing its passive reabsorption
   5. Sodium bicarbonate acts as a physiological antagonist, counteracting aspirin's antiplatelet effect

   Show Explanation

   Aspirin is a weak acid. Alkalinising the urine (raising urinary pH above aspirin's pK_a) causes aspirin to become ionised (charged) in the tubular lumen. Charged molecules cannot cross lipid membranes by passive diffusion → ion trapping occurs → aspirin is excreted rather than reabsorbed. This is the Henderson-Hasselbalch principle applied clinically. Sodium bicarbonate does not bind aspirin directly (A), does not activate CYP450 (C), alkalinisation does not enhance reabsorption (D), and it is not a physiological antagonist for antiplatelet effects (E).

4. A 45-year-old woman with chronic pain is switched from intravenous morphine to oral morphine at discharge. Her physician explains that a higher oral dose will be required to achieve the same analgesic effect. Which of the following BEST explains why the oral dose must be higher than the IV dose?

   1. Oral drugs are absorbed more rapidly than IV drugs, leading to faster elimination
   2. Oral drugs must pass through the hepatoportal circulation before reaching systemic circulation, where hepatic metabolism reduces the amount of active drug available
   3. Oral bioavailability is always exactly 50% of the IV dose for all drugs
   4. Oral administration increases the volume of distribution, requiring higher doses to achieve target plasma concentrations
   5. IV morphine binds more extensively to plasma proteins, making it more potent than oral morphine

   Show Explanation

   The first-pass effect (hepatic presystemic metabolism) is the key principle here. Drugs absorbed from the GI tract enter the portal circulation and pass through the liver before reaching systemic circulation. The liver metabolises a proportion of the drug, reducing the amount reaching the target. IV administration bypasses this entirely, giving F = 1.0. Option (A) is incorrect — oral absorption is generally slower, not faster. Option (C) is false — bioavailability varies widely between drugs. Options (D) and (E) incorrectly attribute the difference to Vd or protein binding.

5. A 70 kg patient is prescribed a drug with the following properties:

   • V_d = 1.5L/kg
   • Bioavailability (F) = 0.75
   • Target plasma concentration = 10 mg/L

   What loading dose should be administered orally to rapidly achieve the target plasma concentration?

   1. 525 mg
   2. 700 mg
   3. 1050 mg
   4. 1400 mg
   5. 2100 mg

   Show Explanation

   Using the loading dose equation: $$\text{Dose} = \frac{[Drug]_{target} \times V_d}{F}$$ $$\text{Dose} = \frac{10 \text{ mg/L} \times (1.5 \text{ L/kg} \times 70 \text{ kg})}{0.75}$$ $$= \frac{10 \times 105}{0.75} = \frac{1050}{0.75} = \mathbf{1400 \text{ mg}}$$

6. A 67 kg patient requires treatment with a drug that has the following properties:

   • V_d = 1.5L/kg
   • Clearance = 6 L/hr
   • Bioavailability (F) = 0.75
   • Target steady-state plasma concentration = 15 mg/L
   • Oral dosing inverval = every 8 hours

   What oral maintenance dose should be prescribed every 8 hours?

   1. 480 mg
   2. 640 mg
   3. 960 mg
   4. 1280 mg
   5. 1500 mg

   Show Explanation

   Using the oral maintenance dosing equation: $$\frac{\text{Dose} \times F}{\text{Interval}} = Cl \times [Drug]_{ss}$$ $$\text{Dose} = \frac{Cl \times [Drug]_{ss} \times \text{Interval}}{F}$$ $$= \frac{0.1 \text{ L/min} \times 15 \text{ mg/L} \times 480 \text{ min}}{0.75}$$ $$= \frac{720}{0.75} = \mathbf{960 \text{ mg}}$$

7. A 24-year-old woman with epilepsy is prescribed phenytoin for seizure control. She is also taking the oral contraceptive pill (OCP). Three months later, she presents to her GP having discovered she is pregnant. Which of the following BEST explains the mechanism by which phenytoin reduced contraceptive efficacy?

   1. Phenytoin competitively inhibits CYP3A4, reducing conversion of OCP hormones to active metabolites
   2. Phenytoin induces CYP450 enzymes, accelerating metabolism and reducing plasma levels of OCP hormones below therapeutic concentrations
   3. Phenytoin displaces OCP hormones from plasma protein binding sites, increasing free drug levels and causing rapid renal excretion
   4. Phenytoin inhibits gastric absorption of OCP hormones, reducing their bioavailability
   5. Phenytoin and OCP hormones compete for the same Phase II conjugation pathways, reducing OCP efficacy

   Show Explanation

   Phenytoin is a CYP450 enzyme inducer. By upregulating CYP450 expression (principally CYP3A4), phenytoin accelerates the metabolism of OCP oestrogens and progestogens, reducing their plasma concentrations below therapeutic levels — resulting in contraceptive failure.

8. A 55-year-old man on warfarin, an anticoagulant medication, for atrial fibrillation presents to his GP with a new complaint of dyspepsia. He is started on cimetidine, to reduce acid production in his stomach. Two weeks later, a blood test shows his blood clots too slowly, increasing his risk of dangerous bleeding. Which mechanism BEST explains the dangerous rise in INR following cimetidine initiation?

   1. Cimetidine induces CYP450 enzymes, increasing warfarin production from its prodrug form
   2. Cimetidine alkalinises gastric pH, increasing warfarin absorption and bioavailability
   3. Cimetidine inhibits CYP450 enzymes responsible for warfarin metabolism, leading to warfarin accumulation and increased anticoagulation
   4. Cimetidine competes with warfarin for renal tubular secretion, reducing warfarin clearance
   5. Cimetidine induces Phase II glucuronidation, converting warfarin to a more potent active metabolite

   Show Explanation

   Cimetidine is a CYP450 inhibitor. By blocking CYP-mediated metabolism of warfarin (primarily CYP2C9), cimetidine causes warfarin accumulation in plasma. Since warfarin has a narrow therapeutic index, even small increases in its plasma level can tip a patient from anticoagulation into dangerous bleeding.

9. A 58-year-old on long-term atorvastatin for hypercholesterolaemia is prescribed a 5-day course of Paxlovid (nirmatrelvir/ritonavir) for a COVID-19 infection. Atorvastatin has an oral bioavailability (F) of only 14% under normal circumstances. After ritonavir co-administration, its bioavailability increases substantially. Which of the following BEST explains why atorvastatin normally has such low oral bioavailability (F = 14%)?

   1. Atorvastatin is poorly absorbed from the GI tract due to its large molecular size
   2. Atorvastatin undergoes extensive first-pass metabolism via CYP3A4 in the intestinal wall and liver before reaching systemic circulation
   3. Atorvastatin is actively secreted back into the gut lumen by P-glycoprotein, preventing absorption
   4. Atorvastatin is highly protein-bound in plasma, reducing its measured bioavailability
   5. Atorvastatin is a prodrug that requires renal conversion to its active form, accounting for its low apparent oral bioavailability

   Show Explanation

   Atorvastatin's very low bioavailability (F = 14%) is primarily due to extensive first-pass metabolism via CYP3A4 enzymes present in both the intestinal wall and the liver. Despite good GI absorption, the majority of drug is metabolised before it can reach the systemic circulation. Ritonavir is an inhibitor of CYP3A4 which increases the bioavailability of atorvastatin.

10. A 100 kg patient presents with phenytoin toxicity. His current plasma phenytoin level is 50 mg/L. Phenytoin has a volume of distribution (V_d) of 0.8 L/kg. The patient's plasma volume is 3 L. The clinical team considers plasmapheresis (plasma exchange) to rapidly reduce the drug level. What will the plasma phenytoin concentration be IMMEDIATELY after a single plasma exchange procedure?

    1. 0 mg/L
    2. 25 mg/L
    3. 35 mg/L
    4. 48 mg/L
    5. 50 mg/L

    Show Explanation

    • Step 1 — Calculate total drug in body: $$ \text{Total drug} = [Drug]_{plasma} \times V_d = 50 \text{ mg/L} \times (0.8 \text{ L/kg} \times 100 \text{ kg}) = 50 \times 80 = 4{,}000 \text{ mg} $$
    • Step 2 — Calculate drug removed in plasma exchange: $$ \text{Drug removed} = [Drug]_{plasma} \times V_{plasma} = 50 \text{ mg/L} \times 3 \text{ L} = 150 \text{ mg} $$
    • Step 3 — Calculate remaining drug: $$ \text{Drug remaining} = 4{,}000 - 150 = 3{,}850 \text{ mg} $$
    • Step 4 — Calculate new plasma concentration: $$ [Drug]_{new} = \frac{3{,}850 \text{ mg}}{80 \text{ L}} \approx \mathbf{48 \text{ mg/L}} $$

    Plasma exchange removes only **150 mg of 4,000 mg total** — just **3.75%** of total body drug content. The plasma level barely changes.

11. A patient requires IV theophylline therapy. The following pharmacokinetic parameters are known:

    • Target steady-state plasma concentration: 15 mg/L
    • Half-life: 4 hours
    • Volume of distribution: 25 L

    What IV maintenance infusion rate is required to maintain the target steady-state concentration?

    1. 15.0 mg/hr
    2. 32.3 mg/hr
    3. 64.5 mg/hr
    4. 93.8 mg/hr
    5. 375.0 mg/hr

    Show Explanation

    • Step 1 — Calculate clearance: $$ Cl = \frac{0.693 \times V_d}{t_{1/2}} = \frac{0.693 \times 25 \text{ L}}{4 \text{ hr}} = \frac{17.325}{4} = 4.33 \text{ L/hr} $$
    • Step 2 — Calculate IV maintenance infusion rate: $$ \text{Rate}_{in} = Cl \times [Drug]_{ss} = 4.33 \text{ L/hr} \times 15 \text{ mg/L} = \mathbf{64.5 \text{ mg/hr}} $$

12. Using the same theophylline patient from the question above, the physician decides to administer a loading dose before starting the maintenance infusion to rapidly achieve the target plasma concentration of 15 mg/L. What loading dose should be administered, and approximately how long would it take to reach steady state WITHOUT a loading dose?

    1. Loading dose = 64.5 mg; steady state without loading dose = 4 hours
    2. Loading dose = 375 mg; steady state without loading dose = 12–16 hours
    3. Loading dose = 375 mg; steady state without loading dose = 4–8 hours
    4. Loading dose = 250 mg; steady state without loading dose = 12–16 hours
    5. Loading dose = 750 mg; steady state without loading dose = 24–48 hours

    Show Explanation

    • Loading dose: $$ \text{Loading dose} = [Drug]_{target} \times V_d = 15 \text{ mg/L} \times 25 \text{ L} = \mathbf{375 \text{ mg}} $$ Note: For IV administration, bioavailability F = 1.0, so no correction is needed.
    • Time to steady state without loading dose: $$ \text{Time to steady state} = 4-5 \times t_{1/2} = 4-5 \times 4 \text{ hr} = \mathbf{16-20 \text{ hr}} $$

13. A 70 kg patient on long-term lithium (300 mg twice daily, every 12 hours) for bipolar disorder presents with toxicity. His previous steady-state level was 1.1 mmol/L; his current level is 3.2 mmol/L (toxic range: >1.5 mmol/L). The molecular weight of lithium carbonate relevant to this calculation gives 6.9 mg per mmol of lithium. Using these values, what is this patient's lithium clearance?

    1. 0.34 L/hr
    2. 0.69 L/hr
    3. 1.10 L/hr
    4. 3.20 L/hr
    5. 6.90 L/hr

    Show Explanation

    • Step 1 — Convert dose to mmol/hr: $$ \text{Rate}_{in} = \frac{300 \text{ mg}}{12 \text{ hr}} \div 6.9 \text{ mg/mmol} = \frac{25 \text{ mg/hr}}{6.9 \text{ mg/mmol}} = 3.62 \text{ mmol/hr} $$
    • Step 2 — Calculate clearance using steady-state or maintenance by continuous infusion equation: $$ Cl = \frac{\text{Rate}_{in}}{[Drug]_{ss}} = \frac{3.62 \text{ mmol/hr}}{3.2 \text{ mmol/L}} \approx \mathbf{1.1 \text{ L/hr}} $$

    Note: We use the **current toxic level** (3.2 mmol/L) as the steady-state concentration because this represents the new equilibrium achieved on the current dosing regimen

14. Using the lithium patient from the question above, the team withholds lithium and monitors the patient. Lithium has a V_d of 0.6 L/kg in this patient. Approximately how long will it take for the lithium level to fall from 3.2 mmol/L to below 1.0 mmol/L (the lower end of the therapeutic range)?

    1. ~13 hours
    2. ~27 hours
    3. ~53 hours
    4. ~80 hours
    5. ~106 hours

    Show Explanation

    • Step 1 — Calculate half-life: $$ t_{1/2} = \frac{0.693 \times V_d}{Cl} = \frac{0.693 \times (0.6 \text{ L/kg} \times 70 \text{ kg})}{1.1 \text{ L/hr}} = \frac{0.693 \times 42}{1.1} = \frac{29.1}{1.1} \approx \mathbf{26.5 \text{ hr}} $$
    • Step 2 — Determine how many half-lives needed (starting level = 3.2 mmol/L, target < 1.0 mmol/L): $$ 3.2 \xrightarrow{t_{1/2} = 26.5\text{hr}} 1.6 \xrightarrow{t_{1/2} = 26.5\text{hr}} 0.8 \text{ mmol/L} $$ Two half-lives are required: $$ 2 \times 26.5 \text{ hr} = \mathbf{53 \text{ hours} \approx \text{just over 2 days}} $$

15. An 85 kg patient with epilepsy is prescribed levetiracetam 500 mg twice daily (BID). The following pharmacokinetic parameters apply:

    • V_d = 0.6L/kg
    • Clearance = 5L/hr
    • Bioavailability (F) = 1.0
    • Therapeutic range: 5 - 30 mg/L

    Which of the following BEST describes the expected peak and trough plasma concentrations at steady state for the 500 mg BID regimen?

    1. Peak ~5 mg/L; Trough ~1 mg/L — subtherapeutic throughout
    2. Peak ~13 mg/L; Trough ~3 mg/L — within the therapeutic range at peak but subtherapeutic at trough
    3. Peak ~20 mg/L; Trough ~10 mg/L — within therapeutic range throughout
    4. Peak ~26 mg/L; Trough ~6 mg/L — upper therapeutic range at peak, low-therapeutic at trough
    5. Peak ~40 mg/L; Trough ~9 mg/L — toxic at peak

    Show Explanation

    • Step 1 — Calculate Vd: $$ V_d = 0.6 \text{ L/kg} \times 85 \text{ kg} = 51 \text{ L} $$
    • Step 2 — Calculate steady-state concentration: $$ [Drug]_{ss} = \frac{\text{Daily dose}}{Cl \times 24\text{hr}} = \frac{1000 \text{ mg/day}}{5 \text{ L/hr} \times 24\text{hr}} \approx 8.3 \text{ mg/L} $$
    • Step 3 — Calculate bolus concentration per dose: $$ [Drug]_{bolus} = \frac{500 \text{ mg}}{51 \text{ L}} \approx 10 \text{ mg/L} $$
    • Step 4 — Calculate peak and trough:
      1. Peak = [Drug]_steady-state + 1/2 [Drug]_Bolus = 8.3 mg/L + 5 mg/L = 13.3 mg/L
      2. Trough = [Drug]_steady-state - 1/2 [Drug]_Bolus = 8.3 mg/L - 5 mg/L = 3.3 mg/L