Chapter 33 — Anti-Cancer Drugs Part I — Module 2 — Alkylating Agents
1. Cyclophosphamide requires hepatic cytochrome P450 oxidation to its active form. Which CYP isoform is the predominant catalyst of this 4-hydroxylation step?
A) CYP3A4
B) CYP2B6
C) CYP2D6
D) CYP1A2
E) CYP2E1
ANSWER: B
Rationale:
Option B is correct. The 4-hydroxylation of cyclophosphamide that initiates its activation is catalyzed predominantly by CYP2B6, with lesser contributions from CYP3A4 and CYP2C9. Because CYP2B6 is the principal activating isoform, agents that induce or inhibit it (along with CYP3A4) shift the balance between active metabolite formation and toxicity.
Option A: Option A is incorrect because CYP3A4 contributes to cyclophosphamide activation but is not the predominant isoform; CYP2B6 leads.
Option C: Option C is incorrect because CYP2D6 is central to metabolism of many cardiovascular and psychiatric drugs but is not the major cyclophosphamide-activating enzyme.
Option D: Option D is incorrect because CYP1A1/1A2 activate dacarbazine, not cyclophosphamide.
Option E: Option E is incorrect because CYP2E1 metabolizes small molecules such as ethanol and acetaminophen and is not the predominant cyclophosphamide activator.
2. Ifosfamide and cyclophosphamide are structural isomers, yet ifosfamide is distinctly associated with neurotoxic encephalopathy. Which metabolic difference accounts for this?
A) Ifosfamide generates substantially more acrolein per gram, which crosses into the central nervous system
B) Ifosfamide is activated by monoamine oxidase rather than cytochrome P450, producing a neurotoxic amine
C) Ifosfamide bypasses aldehyde dehydrogenase entirely, allowing aldophosphamide to accumulate in the brain
D) Ifosfamide undergoes a higher rate of side-chain oxidation (N-dechloroethylation), generating chloroacetaldehyde
E) Ifosfamide is more lipophilic, so a greater fraction of phosphoramide mustard enters the central nervous system
ANSWER: D
Rationale:
Option D is correct. Ifosfamide undergoes a substantially higher rate of side-chain oxidation (N-dechloroethylation) than cyclophosphamide. This pathway releases chloroacetaldehyde, the neurotoxic metabolite responsible for ifosfamide encephalopathy, which cyclophosphamide produces to a much lesser degree.
Option A: Option A is incorrect because, although ifosfamide does produce a greater acrolein load, acrolein causes urothelial injury (hemorrhagic cystitis), not the encephalopathy; the neurotoxin is chloroacetaldehyde.
Option B: Option B is incorrect because ifosfamide is activated by cytochrome P450, like cyclophosphamide, not by monoamine oxidase.
Option C: Option C is incorrect because the encephalopathy is driven by chloroacetaldehyde from side-chain oxidation, not by aldophosphamide accumulation from bypassing aldehyde dehydrogenase.
Option E: Option E is incorrect because the neurotoxicity reflects a specific metabolite (chloroacetaldehyde), not greater central penetration of phosphoramide mustard.
3. Cyclophosphamide activation yields two distinct products with separate roles. Which pairing correctly assigns each product to its function?
A) Phosphoramide mustard is the antitumor DNA-cross-linking species; acrolein is the urotoxic byproduct
B) Acrolein is the antitumor DNA-cross-linking species; phosphoramide mustard is the urotoxic byproduct
C) 4-hydroxycyclophosphamide is the antitumor species; carboxyphosphamide is the urotoxic byproduct
D) Phosphoramide mustard is the urotoxic byproduct; chloroacetaldehyde is the antitumor species
E) Aldophosphamide is the antitumor species; phosphoramide mustard is an inert excretion product
ANSWER: A
Rationale:
Option A is correct. When aldophosphamide decomposes, it yields phosphoramide mustard, the bifunctional alkylating species that produces the antitumor DNA (deoxyribonucleic acid) interstrand and intrastrand cross-links, and acrolein, the reactive aldehyde responsible for urothelial injury and hemorrhagic cystitis. The two products have distinct roles: one kills tumor cells, the other injures the bladder.
Option B: Option B is incorrect because the roles are reversed: phosphoramide mustard is the antitumor alkylator, and acrolein is the urotoxin.
Option C: Option C is incorrect because 4-hydroxycyclophosphamide is an upstream intermediate and carboxyphosphamide is the inactive detoxification product, not the antitumor and urotoxic pair.
Option D: Option D is incorrect because phosphoramide mustard is the antitumor species, not the urotoxin, and chloroacetaldehyde is the ifosfamide neurotoxin, not the antitumor species.
Option E: Option E is incorrect because aldophosphamide is a transient intermediate, and phosphoramide mustard is the active alkylator rather than an inert product.
4. Aldehyde dehydrogenase (ALDH) expression influences which cells are protected from cyclophosphamide. Which statement correctly describes this relationship?
A) ALDH converts phosphoramide mustard to acrolein, so high-ALDH cells suffer more bladder injury
B) ALDH activates aldophosphamide to phosphoramide mustard, so high-ALDH cells are preferentially killed
C) ALDH inactivates aldophosphamide to carboxyphosphamide, so high-ALDH normal cells are relatively protected
D) ALDH is expressed only in tumor cells, conferring selective tumor resistance to cyclophosphamide
E) ALDH inhibits cytochrome P450 activation, so high-ALDH cells generate less active metabolite overall
ANSWER: C
Rationale:
Option C is correct. Aldehyde dehydrogenase (ALDH) oxidizes aldophosphamide to the non-toxic carboxyphosphamide, diverting it away from forming phosphoramide mustard. Normal hematopoietic stem cells and hepatocytes express high ALDH and are therefore relatively protected, whereas tumor cells with low ALDH activity are comparatively more vulnerable to the active metabolite.
Option A: Option A is incorrect because ALDH detoxifies aldophosphamide; it does not convert phosphoramide mustard to acrolein.
Option B: Option B is incorrect because ALDH inactivates the precursor rather than activating it; high ALDH protects cells rather than sensitizing them.
Option D: Option D is incorrect because ALDH is expressed at high levels in normal hematopoietic and hepatic cells, conferring protection there, not selectively in tumor cells.
Option E: Option E is incorrect because ALDH acts downstream on aldophosphamide and does not inhibit the upstream CYP activation step.
5. Cisplatin forms several types of platinum-DNA lesions. Which lesion predominates quantitatively?
A) Interstrand cross-links, which account for the majority of cisplatin-DNA adducts
B) Monofunctional N3-adenine adducts, which represent most cisplatin-DNA lesions
C) 1,2-intrastrand d(GpG) adducts at the N7 position of adjacent guanines
D) Protein-DNA cross-links linking platinum to histone lysine residues
E) O6-guanine methyl adducts identical to those formed by temozolomide
ANSWER: C
Rationale:
Option C is correct. The aquated cisplatin species attacks N7 of guanine, and the predominant lesion is the 1,2-intrastrand d(GpG) adduct between adjacent guanines (roughly 65% of lesions), followed by 1,2-d(ApG) intrastrand adducts (about 25%), with interstrand cross-links a minority (approximately 5 to 8%).
Option A: Option A is incorrect because interstrand cross-links, though biologically important, are a small minority of cisplatin lesions, not the majority.
Option B: Option B is incorrect because the predominant lesions are intrastrand GpG adducts, not monofunctional N3-adenine adducts.
Option D: Option D is incorrect because protein-DNA cross-links are not the predominant cisplatin lesion; intrastrand GpG adducts are.
Option E: Option E is incorrect because O6-guanine methylation is the temozolomide lesion; cisplatin forms platinum adducts at N7-guanine, not methyl adducts at O6.
6. Cisplatin transport across the tumor cell membrane involves specific copper-handling proteins. Which assignment of transporters to function is correct?
A) CTR1 mediates cisplatin efflux; ATP7A and ATP7B mediate cisplatin uptake
B) ATP7B mediates uptake; CTR1 mediates efflux and is upregulated in resistant cells
C) CTR1 and ATP7A both mediate uptake; efflux occurs only by passive diffusion
D) ATP7A mediates uptake; ATP7B is unrelated to platinum handling
E) CTR1 mediates cisplatin uptake; ATP7A and ATP7B mediate efflux and contribute to resistance
ANSWER: E
Rationale:
Option E is correct. Cisplatin enters cells primarily through the copper transporter CTR1 (copper transporter 1), and efflux through the copper-exporting ATPases ATP7A and ATP7B is a recognized resistance mechanism. The direction is uptake by CTR1, export by ATP7A/ATP7B.
Option A: Option A is incorrect because the roles are reversed: CTR1 is the uptake transporter, and ATP7A/ATP7B perform efflux.
Option B: Option B is incorrect because CTR1 mediates uptake, not efflux, and ATP7B is an exporter, not an importer.
Option C: Option C is incorrect because ATP7A is an efflux ATPase, not an uptake transporter, and efflux is carrier-mediated rather than purely passive.
Option D: Option D is incorrect because ATP7A is an efflux transporter (not uptake), and ATP7B is very much involved in platinum efflux and resistance.
7. Carboplatin's reduced reactivity relative to cisplatin is a direct consequence of its leaving group. Which structural feature distinguishes carboplatin?
A) Its two chloride leaving groups are replaced by ammine ligands, eliminating aquation entirely
B) Its two chloride leaving groups are replaced by a bidentate cyclobutanedicarboxylate, which hydrolyzes slowly
C) Its platinum center is replaced by a diaminocyclohexane carrier ligand that resists mismatch repair
D) Its leaving groups are two water molecules, making it pre-activated and highly reactive in plasma
E) Its leaving group is a glucose moiety that targets the drug to tissues expressing GLUT2
ANSWER: B
Rationale:
Option B is correct. In carboplatin, the two chloride leaving groups of cisplatin are replaced by a single bidentate cyclobutanedicarboxylate (CBDCA) ligand. This ligand is hydrolyzed (aquated) much more slowly than chloride, producing a less reactive platinum species and accounting for carboplatin's lower nephrotoxicity, neurotoxicity, and emetogenicity.
Option A: Option A is incorrect because the ammine ligands are retained in carboplatin; the chlorides are replaced by cyclobutanedicarboxylate, and aquation still occurs, only slowly.
Option C: Option C is incorrect because the diaminocyclohexane (DACH) carrier ligand is the defining feature of oxaliplatin, not carboplatin.
Option D: Option D is incorrect because carboplatin is not pre-aquated; its slow-hydrolyzing CBDCA ligand makes it less reactive, not more.
Option E: Option E is incorrect because a glucose moiety directing uptake via GLUT2 describes streptozocin, not carboplatin.
8. Oxaliplatin retains activity in certain tumors that are resistant to cisplatin. What property of its DNA adducts explains this?
A) Its bulky DACH-platinum adducts are not recognized by mismatch repair proteins, so it remains active in MMR-deficient tumors
B) Its adducts are repaired exclusively by base excision repair, which is absent in colorectal cancer
C) Its adducts are reversible, allowing tumor cells to tolerate them without triggering resistance pathways
D) Its adducts form only interstrand cross-links, which cisplatin cannot produce at all
E) Its adducts are removed by MGMT, making MGMT-methylated tumors uniquely sensitive to oxaliplatin
ANSWER: A
Rationale:
Option A is correct. Oxaliplatin carries a 1,2-diaminocyclohexane (DACH) carrier ligand, producing bulkier DNA adducts than cisplatin. These DACH-platinum adducts are not recognized by the mismatch repair (MMR) proteins hMSH2 and hMSH6 that normally trigger MMR-mediated apoptosis in response to platinum damage, which is why oxaliplatin remains active in MMR-deficient (microsatellite-unstable) tumors that are cisplatin-resistant.
Option B: Option B is incorrect because oxaliplatin adducts are handled chiefly by nucleotide excision repair, not exclusively by base excision repair, and the explanation here is evasion of MMR recognition.
Option C: Option C is incorrect because oxaliplatin adducts are covalent and not reversible; the basis for retained activity is MMR evasion.
Option D: Option D is incorrect because cisplatin can form interstrand cross-links; the distinguishing feature of oxaliplatin is the bulky DACH adduct that evades MMR.
Option E: Option E is incorrect because MGMT reverses O6-guanine alkylation from methylating agents, not platinum adducts; oxaliplatin is not an MGMT substrate.
9. Mechlorethamine is the most reactive and vesicant of the nitrogen mustards, making extravasation a serious hazard. What is the antidote for mechlorethamine extravasation?
A) Dexrazoxane
B) Mesna
C) Leucovorin
D) Sodium thiosulfate
E) Amifostine
ANSWER: D
Rationale:
Option D is correct. Sodium thiosulfate is the antidote for mechlorethamine extravasation; it reacts with and neutralizes the reactive alkylating species in the tissue, limiting local injury from this highly vesicant agent.
Option A: Option A is incorrect because dexrazoxane is used for anthracycline extravasation and anthracycline cardioprotection, not for mechlorethamine.
Option B: Option B is incorrect because mesna binds urinary acrolein to prevent hemorrhagic cystitis and is not an extravasation antidote.
Option C: Option C is incorrect because leucovorin rescues normal cells from methotrexate (an antimetabolite) and has no role in mechlorethamine extravasation.
Option E: Option E is incorrect because amifostine is a cytoprotectant for cisplatin nephrotoxicity, not an antidote for mechlorethamine extravasation.
10. Melphalan is used for high-dose conditioning before autologous stem cell transplantation in multiple myeloma. Why is the intravenous route preferred for high-dose conditioning?
A) Oral melphalan is inactivated by gastric acid before it can be absorbed
B) Oral melphalan undergoes extensive first-pass conversion to an inactive metabolite
C) Oral melphalan has highly variable absorption, making intravenous dosing necessary for reliable high-dose exposure
D) Oral melphalan cannot cross the intestinal mucosa because it is too lipophilic
E) Oral melphalan is contraindicated because it causes severe direct intestinal alkylation injury
ANSWER: C
Rationale:
Option C is correct. Oral melphalan (L-phenylalanine mustard) has highly variable bioavailability (roughly 15 to 90 percent), so the intravenous route is mandated for high-dose conditioning, where reliable and predictable drug exposure is essential for myeloablation.
Option A: Option A is incorrect because the limitation is erratic absorption, not acid inactivation in the stomach.
Option B: Option B is incorrect because the issue is variable absorption rather than extensive first-pass metabolism to an inactive product.
Option D: Option D is incorrect because melphalan is absorbed orally (albeit unpredictably); it is not blocked by excessive lipophilicity.
Option E: Option E is incorrect because oral melphalan is used at standard doses for myeloma; the reason for IV conditioning is reliability of exposure, not direct intestinal injury.
11. Spontaneous decomposition of carmustine and lomustine in aqueous solution generates two reactive species. Which pair correctly identifies them and their actions?
A) An aziridinium ion that alkylates DNA and a sulfhydryl that scavenges free radicals
B) A methyl diazonium ion that methylates O6-guanine and a nitrite that nitrosylates proteins
C) An aquated platinum species that binds N7-guanine and a chloride that suppresses aquation
D) A phosphoramide mustard that cross-links DNA and an acrolein that injures the urothelium
E) A chloroethyl carbonium ion that alkylates DNA and an isocyanate that carbamylates proteins
ANSWER: E
Rationale:
Option E is correct. Nitrosoureas decompose non-enzymatically to a chloroethyl carbonium ion, the DNA (deoxyribonucleic acid) alkylating species that ultimately forms interstrand cross-links, and an isocyanate, which carbamylates lysine residues on proteins (including DNA repair enzymes), inhibiting their function and contributing to cytotoxicity.
Option A: Option A is incorrect because the aziridinium ion is the reactive intermediate of nitrogen mustards, not nitrosoureas, and nitrosoureas do not generate a radical-scavenging sulfhydryl.
Option B: Option B is incorrect because the methyl diazonium ion is the temozolomide/dacarbazine methylating species, not the nitrosourea decomposition product.
Option C: Option C is incorrect because aquated platinum species describe cisplatin chemistry, not nitrosoureas.
Option D: Option D is incorrect because phosphoramide mustard and acrolein are cyclophosphamide products, not nitrosourea decomposition species.
12. Carmustine and streptozocin are both nitrosoureas, yet their dose-limiting and cumulative toxicities differ. Which statement correctly discriminates between them?
A) Both share pulmonary fibrosis as the dose-limiting toxicity, differing only in the threshold dose
B) Carmustine causes cumulative pulmonary fibrosis at high total doses, whereas streptozocin's dose-limiting toxicity is nephrotoxicity
C) Streptozocin causes pulmonary fibrosis, whereas carmustine's dose-limiting toxicity is nephrotoxicity
E) Both share nephrotoxicity as the dose-limiting toxicity, with neither causing pulmonary injury
ANSWER: B
Rationale:
Option B is correct. Carmustine produces cumulative pulmonary fibrosis, characteristically after high total doses (above roughly 1,200 mg/m^2), which becomes an absolute contraindication to further treatment. Streptozocin is the nitrosourea whose dose-limiting toxicity is nephrotoxicity (tubular injury) rather than the marrow or pulmonary toxicity of the others.
Option A: Option A is incorrect because the two do not share pulmonary fibrosis as the dose-limiting toxicity; streptozocin is dose-limited by nephrotoxicity.
Option C: Option C is incorrect because the assignments are reversed: carmustine causes the pulmonary fibrosis and streptozocin the nephrotoxicity.
Option D: Option D is incorrect because carmustine's signature cumulative toxicity is pulmonary fibrosis, not hemorrhagic cystitis, and streptozocin causes nephrotoxicity, not ototoxicity.
Option E: Option E is incorrect because carmustine causes pulmonary injury; the two do not both spare the lung.
13. Temozolomide methylates DNA at several positions, but its cytotoxicity is concentrated in one lesion. Which statement correctly describes the chemistry and the key lesion?
A) The ultimate reactive species is the methyl diazonium ion, and O6-methylguanine is the principal cytotoxic lesion despite being a minority of adducts
B) The ultimate reactive species is a chloroethyl carbonium ion, and N7-guanine adducts account for the cytotoxicity
C) The ultimate reactive species is phosphoramide mustard, and interstrand cross-links account for the cytotoxicity
D) The ultimate reactive species is an aquated platinum complex, and 1,2-d(GpG) intrastrand adducts are the lethal lesion
E) The ultimate reactive species is acrolein, and N3-adenine adducts account for the majority of cell death
ANSWER: A
Rationale:
Option A is correct. Temozolomide hydrolyzes to MTIC, which decomposes to the methyl diazonium ion, the ultimate methylating species. Although O6-methylguanine accounts for only about 5 percent of adducts (N7-guanine adducts are far more numerous), it is responsible for most of the cytotoxic effect because the resulting persistent mismatches drive futile mismatch repair, double-strand breaks, and apoptosis.
Option B: Option B is incorrect because the chloroethyl carbonium ion is the nitrosourea species; temozolomide acts through a methyl diazonium ion, and N7-guanine adducts are the less cytotoxic majority.
Option C: Option C is incorrect because phosphoramide mustard and interstrand cross-links are cyclophosphamide chemistry, not temozolomide's.
Option D: Option D is incorrect because aquated platinum and GpG intrastrand adducts describe cisplatin, not temozolomide.
Option E: Option E is incorrect because acrolein is the cyclophosphamide urotoxin and is not the methylating species; O6-methylguanine, not N3-adenine, drives temozolomide cytotoxicity.
14. Dacarbazine and temozolomide both generate the same active methylating intermediate (MTIC), yet only temozolomide is orally administered. What accounts for this difference?
A) Dacarbazine is destroyed by stomach acid, whereas temozolomide is acid-stable, allowing oral use
B) Dacarbazine is a vesicant requiring central venous access, whereas temozolomide is non-irritant
C) Dacarbazine requires hepatic CYP1A1/CYP1A2 activation and is given intravenously, whereas temozolomide is activated by spontaneous hydrolysis and is oral
D) Dacarbazine must be given intrathecally to reach its target, whereas temozolomide is absorbed enterally
E) Dacarbazine is a prodrug activated only in tumor tissue, whereas temozolomide is active as administered
ANSWER: C
Rationale:
Option C is correct. Dacarbazine requires enzymatic activation by hepatic CYP1A1 and CYP1A2 to generate MTIC, so it is administered intravenously and does not achieve meaningful central nervous system concentrations. Temozolomide produces the same MTIC by spontaneous, non-enzymatic hydrolysis at physiological pH, allowing oral administration and effective central nervous system penetration.
Option A: Option A is incorrect because the route difference reflects the mode of activation (enzymatic versus spontaneous), not acid stability.
Option B: Option B is incorrect because the reason dacarbazine is intravenous is its requirement for hepatic activation, not vesicant properties requiring central access.
Option D: Option D is incorrect because dacarbazine is given intravenously, not intrathecally.
Option E: Option E is incorrect because dacarbazine is activated by hepatic CYP enzymes, not selectively in tumor tissue, and temozolomide is a prodrug activated by spontaneous hydrolysis rather than active as administered.
15. Busulfan is used as a myeloablative conditioning agent. Which statement correctly pairs its most serious non-hematologic toxicity with the drug used to treat it once established?
A) Pulmonary fibrosis, treated with high-dose corticosteroids
B) Hemorrhagic cystitis, treated with mesna
C) Ifosfamide-type encephalopathy, treated with methylene blue
D) Hepatic sinusoidal obstruction syndrome, treated with defibrotide
E) Nephrotoxicity, treated with amifostine
ANSWER: D
Rationale:
Option D is correct. The most serious non-hematologic toxicity of high-dose busulfan is hepatic sinusoidal obstruction syndrome (SOS, formerly veno-occlusive disease). Therapeutic drug monitoring of busulfan AUC reduces the risk by avoiding over-exposure, and defibrotide is the agent approved to treat established severe SOS.
Option A: Option A is incorrect because cumulative pulmonary fibrosis is the carmustine toxicity; busulfan's serious non-hematologic toxicity is hepatic SOS.
Option B: Option B is incorrect because hemorrhagic cystitis and mesna belong to the cyclophosphamide/ifosfamide pairing, not busulfan.
Option C: Option C is incorrect because methylene blue treats ifosfamide encephalopathy, not a busulfan toxicity.
Option E: Option E is incorrect because nephrotoxicity with amifostine prophylaxis is the cisplatin pairing, not busulfan SOS.
16. Amifostine and mesna are both thiol-based cytoprotectants, but they protect against different toxicities by different mechanisms. Which statement correctly describes amifostine?
A) It binds urinary acrolein to prevent hemorrhagic cystitis, filling the same role as mesna
B) It chelates circulating platinum in plasma, preventing any drug from reaching the tumor
C) It inhibits aldehyde dehydrogenase, blocking formation of the cyclophosphamide urotoxin at its source
D) It reverses ifosfamide encephalopathy by serving as a mitochondrial electron acceptor
E) It is a thiophosphate prodrug activated by alkaline phosphatase to the free thiol WR-1065, used for cisplatin nephroprotection
ANSWER: E
Rationale:
Option E is correct. Amifostine is an organic thiophosphate prodrug that is selectively dephosphorylated by alkaline phosphatase in normal tissues to its active free-thiol metabolite WR-1065, which scavenges free radicals and binds platinum in normal cells. It is approved for cisplatin nephroprotection, distinguishing it mechanistically from mesna, which binds urinary acrolein.
Option A: Option A is incorrect because binding urinary acrolein to prevent hemorrhagic cystitis is the function of mesna, not amifostine.
Option B: Option B is incorrect because amifostine acts as a cytoprotectant in normal tissue via WR-1065; it does not chelate all circulating platinum and abolish antitumor delivery.
Option C: Option C is incorrect because amifostine does not inhibit aldehyde dehydrogenase; it is a free-radical-scavenging thiol generated from a phosphatase-activated prodrug.
Option D: Option D is incorrect because the mitochondrial electron acceptor that reverses ifosfamide encephalopathy is methylene blue, not amifostine.
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