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FMGE biochemistry questions

41 original FMGE-style questions on biochemistry, each with the correct answer, an explanation, why the other options are wrong and the educational objective. Solve them in the quiz or read them below, grouped by section.

Questions41 with answers
Sections9
Options per question4
Time≈ 62 min
FMGE Biochemistry: practice quizPick an answer to see the correct option and the explanation at once. Your best score is saved in this browser.

Enzymes & Clinical Enzymology

5 questions
1

A 60-year-old woman is brought to the emergency department 2 hours after sudden onset of severe chest pain. Her ECG is inconclusive. Which serum marker is expected to rise earliest and also return to normal earliest if she has sustained a myocardial infarction?

  1. ALactate dehydrogenase
  2. BMyoglobin
  3. CTroponin I
  4. DCK-MB
Show answer
Answer: B. Myoglobin

Myoglobin is a small cytoplasmic heme protein that diffuses rapidly into blood after myocardial injury, making it detectable within 1-2 hours of infarction, earlier than CK-MB or troponin. Because of its small size it is also cleared rapidly by the kidneys, returning to baseline within 24 hours. This combination of early rise and early fall makes it useful for very early detection but poor for later diagnosis due to lack of cardiac specificity. Troponin and CK-MB rise slightly later but stay elevated much longer, making them better for confirming diagnosis over a wider time window.

  • A. LDH rises latest (6-12 hours) and remains elevated for up to 10-14 days.
  • C. Troponin I rises later (3-4 hours) and stays elevated for 7-10 days, so it is not the earliest to normalize.
  • D. CK-MB rises by 4-6 hours and normalizes by 2-3 days, slower than myoglobin on both counts.
Educational objective: Myoglobin is the earliest-rising and earliest-normalizing marker after myocardial infarction, though it lacks cardiac specificity.
2

A 45-year-old man with a history of chronic alcohol use presents with severe epigastric pain radiating to the back, nausea and vomiting for 12 hours. Examination shows epigastric tenderness and guarding. Which serum enzyme estimation is most specific for confirming the suspected diagnosis?

  1. ASerum gamma-glutamyl transferase
  2. BSerum amylase
  3. CSerum alkaline phosphatase
  4. DSerum lipase
Show answer
Answer: D. Serum lipase

Serum lipase is more specific for pancreatic injury than amylase because it is produced predominantly by the pancreas, whereas amylase also originates from salivary glands and other tissues, causing more false positives. Lipase also rises slightly later but stays elevated longer (up to 8-14 days) than amylase, giving a wider diagnostic window in acute pancreatitis. A level three times the upper limit of normal, combined with characteristic abdominal pain, supports the diagnosis per revised Atlanta criteria. ALP and GGT reflect hepatobiliary and cholestatic pathology rather than pancreatic acinar injury.

  • A. GGT indicates hepatobiliary disease or alcohol-induced enzyme induction, not pancreatitis specifically.
  • B. Amylase is less specific since salivary glands, fallopian tubes and other organs also contribute to serum levels.
  • C. Alkaline phosphatase is a marker of cholestasis and bone turnover, not pancreatic injury.
Educational objective: Serum lipase is the most specific enzyme marker for diagnosing acute pancreatitis.
3

A 50-year-old woman has an isolated elevated serum alkaline phosphatase on a routine health checkup. Serum calcium, phosphate, bilirubin, AST and ALT are all within normal limits. Which additional test would best help determine whether the elevated enzyme originates from liver or bone?

  1. ASerum CK-MB
  2. BSerum LDH
  3. CSerum 5'-nucleotidase
  4. DSerum GGT
Show answer
Answer: D. Serum GGT

GGT is present in high concentration in hepatobiliary tissue but is absent in bone, so a raised GGT alongside raised ALP points to a hepatobiliary source, while a normal GGT with raised ALP suggests a bony origin such as Paget's disease, osteomalacia or metastasis. 5'-nucleotidase is also used for this purpose but is less widely available and less commonly tested in Indian practice compared to GGT. CK-MB and LDH are not used to differentiate the origin of ALP since they are not specific to either liver or bone pathology in this context. This stepwise approach avoids unnecessary imaging when the biochemical pattern already clarifies the source.

  • A. CK-MB is a cardiac marker and gives no information about the source of ALP.
  • B. LDH is a nonspecific marker released from multiple tissues and does not help localize ALP elevation.
  • C. 5'-nucleotidase also localizes ALP to liver but is less commonly used in routine Indian clinical practice than GGT.
Educational objective: A concomitantly raised GGT with elevated ALP indicates a hepatobiliary source, while a normal GGT suggests a bony cause.
4

A 40-year-old man with long-standing alcohol use presents with jaundice and tender hepatomegaly. His liver function tests show AST 180 U/L and ALT 70 U/L. In a patient with acute viral hepatitis, which pattern of these two enzymes is typically expected instead?

  1. AIsolated rise in AST with normal ALT
  2. BAST/ALT ratio greater than 2
  3. CEqual and markedly elevated AST and ALT with ratio near 1
  4. DAST/ALT ratio less than 1
Show answer
Answer: D. AST/ALT ratio less than 1

In viral hepatitis, hepatocyte injury releases both enzymes, but ALT, being more liver-specific and having a longer half-life, rises higher than AST, giving an AST/ALT ratio typically less than 1. In contrast, alcoholic liver disease classically shows an AST/ALT ratio greater than 2, due to mitochondrial AST release, pyridoxine (vitamin B6) deficiency impairing ALT synthesis, and alcohol-induced mitochondrial injury. This ratio is a useful bedside clue to distinguish the two common causes of hepatitis. Absolute enzyme levels are usually much higher in viral hepatitis (often in the thousands) compared to alcoholic hepatitis.

  • A. Isolated AST rise with normal ALT is not typical of either viral or alcoholic hepatitis, both of which raise both enzymes.
  • B. A ratio greater than 2 is characteristic of alcoholic liver disease, not viral hepatitis.
  • C. A ratio near 1 with both enzymes markedly elevated is seen in some cases but the classic teaching point is ALT exceeding AST in viral hepatitis.
Educational objective: An AST/ALT ratio greater than 2 suggests alcoholic liver disease, whereas a ratio less than 1 is typical of viral hepatitis.
5

A 30-year-old man is brought to the emergency department after being trapped under rubble for 10 hours following a building collapse. His urine appears dark brown, and he complains of severe muscle pain in both legs. Which enzyme is most useful for monitoring the extent and progression of his muscle injury?

  1. ASerum amylase
  2. BTotal creatine kinase
  3. CAlkaline phosphatase
  4. DLactate dehydrogenase
Show answer
Answer: B. Total creatine kinase

Total creatine kinase, predominantly the CK-MM isoform, is released in large quantities from damaged skeletal muscle and correlates well with the severity of muscle injury in conditions like rhabdomyolysis from crush injury. Levels can rise to many times the upper limit of normal and are used both to establish the diagnosis and to track resolution or worsening of muscle damage, which also guides monitoring for acute kidney injury from myoglobin-induced tubular damage. LDH and AST may also rise but are far less specific and less sensitive for quantifying skeletal muscle damage. CK remains the standard enzyme for this purpose in Indian and international practice.

  • A. Serum amylase reflects pancreatic or salivary gland pathology, unrelated to skeletal muscle injury.
  • C. Alkaline phosphatase reflects hepatobiliary or bone pathology, not skeletal muscle breakdown.
  • D. LDH is nonspecific and released from many tissues, making it a poor choice for monitoring muscle-specific injury.
Educational objective: Serum creatine kinase (CK-MM) is the enzyme of choice for diagnosing and monitoring rhabdomyolysis and skeletal muscle injury.

Lipid Metabolism & Lipoproteins

7 questions
6

A 52-year-old man with hypercholesterolemia is started on a drug that competitively inhibits the rate-limiting enzyme of cholesterol biosynthesis in the liver. Which enzyme is inhibited by this drug?

  1. ACholesterol 7-alpha-hydroxylase
  2. BHMG-CoA reductase
  3. CHMG-CoA synthase
  4. DAcetyl-CoA carboxylase
Show answer
Answer: B. HMG-CoA reductase

HMG-CoA reductase catalyzes the conversion of HMG-CoA to mevalonate, the committed and rate-limiting step of cholesterol synthesis. Statins are structural analogues of HMG-CoA that competitively inhibit this enzyme, lowering intracellular cholesterol. This upregulates LDL receptor expression on hepatocytes, increasing clearance of LDL from the blood. The enzyme is regulated by sterol-mediated feedback and is the target of the most effective class of lipid-lowering drugs.

  • A. Cholesterol 7-alpha-hydroxylase is the rate-limiting enzyme of bile acid synthesis, not cholesterol synthesis.
  • C. HMG-CoA synthase acts upstream, forming HMG-CoA, but is not the rate-limiting or statin target enzyme.
  • D. Acetyl-CoA carboxylase is the rate-limiting enzyme of fatty acid synthesis, not cholesterol synthesis.
Educational objective: HMG-CoA reductase is the rate-limiting enzyme of cholesterol synthesis and the target of statin drugs.
7

A 10-year-old boy presents with multiple yellowish nodules over the extensor tendons of both hands and a thick, grayish-white ring around the cornea. His father died of myocardial infarction at the age of 34 years. Fasting lipid profile shows total cholesterol 420 mg/dL, triglycerides 110 mg/dL, and HDL-cholesterol 40 mg/dL. Which lipoprotein abnormality best explains this presentation?

  1. AElevated IDL with elevated cholesterol and triglycerides
  2. BElevated LDL with normal triglycerides
  3. CElevated chylomicrons with normal LDL
  4. DElevated VLDL with elevated triglycerides
Show answer
Answer: B. Elevated LDL with normal triglycerides

This child has the classic phenotype of homozygous familial hypercholesterolemia: tendon xanthomas, corneal arcus, and premature coronary artery disease in a parent. Type IIa hyperlipoproteinemia results from LDL receptor gene mutations, causing markedly elevated LDL cholesterol with normal triglyceride levels since VLDL and chylomicron metabolism are unaffected. This pattern distinguishes it from the combined hyperlipidemias (Type IIb, III) where triglycerides are also raised. Early diagnosis is important because homozygotes develop atherosclerosis and myocardial infarction even in childhood.

  • A. Elevated IDL with both cholesterol and triglyceride elevation describes Type III (broad beta disease), not isolated hypercholesterolemia.
  • C. Elevated chylomicrons (Type I) cause eruptive xanthomas and severe hypertriglyceridemia, not tendon xanthomas with isolated hypercholesterolemia.
  • D. Elevated VLDL with high triglycerides describes Type IV hyperlipoproteinemia, which does not match this normal triglyceride profile.
Educational objective: Type IIa familial hypercholesterolemia shows isolated marked elevation of LDL cholesterol with normal triglycerides, presenting with tendon xanthomas and premature atherosclerosis.
8

A biochemistry researcher is studying the hydrolysis of triglycerides present in circulating chylomicrons and VLDL by an enzyme located on the luminal surface of capillary endothelium in adipose tissue and skeletal muscle. Which apolipoprotein is required as a cofactor to activate this enzyme?

  1. AApolipoprotein A-I
  2. BApolipoprotein B-48
  3. CApolipoprotein C-II
  4. DApolipoprotein E
Show answer
Answer: C. Apolipoprotein C-II

Apolipoprotein C-II, present on the surface of chylomicrons and VLDL, is an essential cofactor that activates lipoprotein lipase anchored on capillary endothelium. Activated lipoprotein lipase hydrolyzes the triglyceride core of these particles, releasing free fatty acids and glycerol for uptake by adipose and muscle tissue. A deficiency of apo C-II causes severe hypertriglyceridemia similar to lipoprotein lipase deficiency itself, because the enzyme remains inactive despite being present. This distinguishes genetic causes of chylomicronemia syndrome.

  • A. Apo A-I is the major structural protein of HDL and activates LCAT, not lipoprotein lipase.
  • B. Apo B-48 is a structural protein of chylomicrons required for their assembly and secretion, not for activating lipoprotein lipase.
  • D. Apo E mediates receptor-mediated uptake of chylomicron and VLDL remnants by the liver, not lipoprotein lipase activation.
Educational objective: Apolipoprotein C-II is the essential cofactor that activates lipoprotein lipase to hydrolyze triglycerides in chylomicrons and VLDL.
9

A 6-month-old infant presents with hypoketotic hypoglycemia, lethargy, and hepatomegaly during a febrile illness with poor feeding. Enzyme studies reveal a defect in the transport system that carries long-chain fatty acyl groups from the cytosol into the mitochondrial matrix for beta-oxidation. Which of the following enzymes forms the acyl-carnitine conjugate needed for this transport?

  1. ACarnitine palmitoyltransferase I
  2. BThiolase
  3. CAcyl-CoA dehydrogenase
  4. DEnoyl-CoA hydratase
Show answer
Answer: A. Carnitine palmitoyltransferase I

Carnitine palmitoyltransferase I (CPT-I), located on the outer mitochondrial membrane, converts long-chain fatty acyl-CoA to fatty acyl-carnitine, enabling translocation across the mitochondrial membranes via carnitine-acylcarnitine translocase. The acyl group is transferred back to CoA by CPT-II on the inner mitochondrial membrane for beta-oxidation to proceed. CPT-I is the rate-limiting and regulated step of long-chain fatty acid oxidation, inhibited by malonyl-CoA during fatty acid synthesis to prevent a futile cycle. Defects in the carnitine shuttle present with fasting-induced hypoketotic hypoglycemia and hepatic dysfunction because fatty acids cannot be oxidized to supply ketones and energy.

  • B. Thiolase cleaves the beta-ketoacyl-CoA to release acetyl-CoA during beta-oxidation, unrelated to mitochondrial transport.
  • C. Acyl-CoA dehydrogenase catalyzes the first oxidation step of beta-oxidation inside the mitochondrial matrix, after fatty acids have already been transported in.
  • D. Enoyl-CoA hydratase catalyzes the hydration step within the beta-oxidation spiral, not the carnitine transport step.
Educational objective: Carnitine palmitoyltransferase I forms acyl-carnitine to shuttle long-chain fatty acids into mitochondria for beta-oxidation, and is the rate-limiting step of this process.
10

A 19-year-old known diabetic is brought to the emergency department with vomiting, abdominal pain, and rapid deep breathing. Blood glucose is 420 mg/dL and arterial blood gas shows metabolic acidosis with a high anion gap. Insulin deficiency in this patient leads to unrestrained lipolysis and hepatic ketogenesis primarily because of increased availability of which substrate combined with reduced activity of a key regulatory enzyme?

  1. AGlycerol-3-phosphate with high malonyl-CoA levels
  2. BAcetyl-CoA with low malonyl-CoA levels
  3. CCitrate with high acetyl-CoA carboxylase activity
  4. DPyruvate with low oxaloacetate levels
Show answer
Answer: B. Acetyl-CoA with low malonyl-CoA levels

In insulin deficiency, unrestrained lipolysis in adipose tissue releases large amounts of free fatty acids that reach the liver and undergo beta-oxidation, generating excess acetyl-CoA. Insulin deficiency also reduces acetyl-CoA carboxylase activity, lowering malonyl-CoA levels; since malonyl-CoA normally inhibits CPT-I, its fall permits unrestricted entry of fatty acyl-CoA into mitochondria for oxidation. The resulting surplus of acetyl-CoA exceeds the capacity of the citric acid cycle (further limited by low oxaloacetate due to increased gluconeogenesis) and is diverted to ketone body synthesis in hepatic mitochondria. This produces acetoacetate and beta-hydroxybutyrate, causing the high anion gap metabolic acidosis seen in diabetic ketoacidosis.

  • A. High malonyl-CoA would inhibit CPT-I and fatty acid entry into mitochondria, which would reduce rather than promote ketogenesis.
  • C. High acetyl-CoA carboxylase activity would increase malonyl-CoA and inhibit fatty acid oxidation, opposite to what occurs in insulin deficiency.
  • D. Low oxaloacetate contributes to reduced acetyl-CoA entry into the TCA cycle but is not the primary trigger initiating increased ketogenic substrate flow.
Educational objective: Insulin deficiency lowers malonyl-CoA, relieving inhibition of CPT-I and allowing excess acetyl-CoA from fatty acid oxidation to be diverted into hepatic ketogenesis.
11

A 35-year-old man presents with grossly enlarged, orange-yellow tonsils and hepatosplenomegaly. Lipid profile shows plasma HDL-cholesterol of 4 mg/dL (normal 40-60 mg/dL) with mildly elevated triglycerides. Peripheral nerve biopsy shows lipid-laden foam cells. This condition results from a defect in which protein responsible for cholesterol efflux from peripheral cells to apolipoprotein A-I?

  1. ALipoprotein lipase
  2. BCholesteryl ester transfer protein (CETP)
  3. CLecithin-cholesterol acyltransferase (LCAT)
  4. DATP-binding cassette transporter A1 (ABCA1)
Show answer
Answer: D. ATP-binding cassette transporter A1 (ABCA1)

Tangier disease is caused by mutations in the ABCA1 transporter gene, which normally mediates efflux of free cholesterol and phospholipids from peripheral cells onto apolipoprotein A-I to form nascent HDL. Loss of ABCA1 function prevents HDL maturation, so apo A-I is rapidly degraded, resulting in markedly low plasma HDL-cholesterol. Cholesterol instead accumulates within tissue macrophages, producing the characteristic orange-yellow enlarged tonsils, hepatosplenomegaly, and foam cells in peripheral nerves. Despite very low HDL, paradoxically the risk of premature coronary artery disease is only mildly increased in this disorder.

  • A. Lipoprotein lipase deficiency causes severe hypertriglyceridemia and eruptive xanthomas, not the tonsillar findings described.
  • B. CETP deficiency causes elevated, not decreased, HDL-cholesterol due to reduced transfer of cholesteryl esters to other lipoproteins.
  • C. LCAT deficiency causes impaired esterification of cholesterol on HDL, leading to corneal opacities and renal disease, not the orange tonsils of Tangier disease.
Educational objective: Tangier disease results from ABCA1 transporter defects that impair cholesterol efflux to apo A-I, causing severely low HDL and cholesterol-laden foam cells in tissues.
12

A lipid profile is drawn from a 48-year-old man with established coronary artery disease: total cholesterol 240 mg/dL, triglycerides 150 mg/dL, HDL-cholesterol 35 mg/dL, and calculated LDL-cholesterol 175 mg/dL. Oxidative modification and uptake by scavenger receptors on macrophages of which particle is primarily responsible for foam cell formation and atherosclerotic plaque development in this patient?

  1. AChylomicrons
  2. BVery low-density lipoprotein (VLDL)
  3. CLow-density lipoprotein (LDL)
  4. DHigh-density lipoprotein (HDL)
Show answer
Answer: C. Low-density lipoprotein (LDL)

LDL is the principal atherogenic lipoprotein because it carries the bulk of circulating cholesterol and, upon oxidative modification within the arterial intima, is taken up by macrophage scavenger receptors that are not subject to feedback downregulation. This unregulated uptake leads to intracellular cholesterol accumulation, forming foam cells that are the hallmark of early atherosclerotic lesions. Small, dense LDL particles are particularly prone to oxidation and arterial wall penetration. This mechanism underlies the direct correlation between elevated LDL-cholesterol and cardiovascular risk, and the rationale for LDL-lowering therapy.

  • A. Chylomicrons are too large to penetrate the arterial endothelium and are rapidly cleared, making them non-atherogenic under normal conditions.
  • B. VLDL is a precursor of LDL and contributes modestly to atherogenesis, but it is less atherogenic than its oxidized LDL remnant.
  • D. HDL removes cholesterol from peripheral tissues and transports it to the liver, making it anti-atherogenic rather than atherogenic.
Educational objective: Oxidized LDL taken up by macrophage scavenger receptors drives foam cell formation and is the most atherogenic lipoprotein particle.
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Amino Acid & Protein Metabolism

4 questions
13

A 4-month-old infant, born of a non-consanguineous marriage, presents with developmental delay, a musty body odor, and hypopigmented skin and hair compared to the parents. Newborn screening by the Guthrie test had shown markedly elevated blood phenylalanine. Which dietary modification is most appropriate for long-term management of this condition?

  1. ARestriction of dietary methionine with cystine supplementation
  2. BRestriction of dietary branched-chain amino acids
  3. CHigh-protein diet with phenylalanine supplementation
  4. DRestriction of dietary phenylalanine with tyrosine supplementation
Show answer
Answer: D. Restriction of dietary phenylalanine with tyrosine supplementation

This infant has phenylketonuria caused by deficiency of phenylalanine hydroxylase, which normally converts phenylalanine to tyrosine in the liver. Accumulated phenylalanine and its metabolites (phenylpyruvate, phenylacetate causing the musty odor) are neurotoxic and impair melanin synthesis, causing the hypopigmentation and intellectual disability if untreated. Management requires a lifelong diet low in phenylalanine, started as early as possible after diagnosis, to prevent irreversible neurological damage, combined with tyrosine supplementation since it becomes a conditionally essential amino acid. Regular monitoring of blood phenylalanine levels guides dietary adjustments to maintain levels within the safe therapeutic range.

  • A. Restriction of methionine with cystine supplementation is used in homocystinuria due to cystathionine beta-synthase deficiency.
  • B. Restriction of branched-chain amino acids is the dietary management for maple syrup urine disease, not PKU.
  • C. A high-protein diet with phenylalanine supplementation would worsen the neurotoxic accumulation of phenylalanine and is contraindicated.
Educational objective: Phenylketonuria due to phenylalanine hydroxylase deficiency is managed with a lifelong phenylalanine-restricted, tyrosine-supplemented diet to prevent intellectual disability.
14

A 35-year-old man notices that his urine turns black after being left standing for a few hours. He also complains of low back pain, and examination reveals bluish-black discoloration of the ear cartilage. Deficiency of which enzyme best explains this presentation?

  1. APhenylalanine hydroxylase
  2. BHomogentisate 1,2-dioxygenase
  3. CFumarylacetoacetase
  4. DTyrosinase
Show answer
Answer: B. Homogentisate 1,2-dioxygenase

Alkaptonuria results from deficiency of homogentisate 1,2-dioxygenase, which normally breaks down homogentisic acid in the tyrosine degradation pathway. Accumulated homogentisic acid is excreted in urine, where it oxidizes and polymerizes on exposure to air, turning the urine black. Chronic deposition of this pigment in connective tissue causes ochronosis, seen as bluish-black discoloration of cartilage and sclera, along with degenerative arthropathy. The condition is a classic example of an autosomal recessive inborn error of amino acid metabolism.

  • A. Phenylalanine hydroxylase deficiency causes phenylketonuria.
  • C. Fumarylacetoacetase deficiency causes tyrosinemia type I, not black urine.
  • D. Tyrosinase deficiency causes albinism, not urine darkening.
Educational objective: Alkaptonuria is caused by homogentisate 1,2-dioxygenase deficiency, leading to homogentisic acid accumulation and urine darkening on standing.
15

A 6-day-old neonate presents with poor feeding, lethargy, and a sweet, burnt-sugar odor to the urine. Blood tests show markedly elevated levels of leucine, isoleucine, and valine. Which enzyme deficiency is responsible for this condition?

  1. ABranched-chain alpha-ketoacid dehydrogenase
  2. BMedium-chain acyl-CoA dehydrogenase
  3. CPropionyl-CoA carboxylase
  4. DMethylmalonyl-CoA mutase
Show answer
Answer: A. Branched-chain alpha-ketoacid dehydrogenase

Maple syrup urine disease is caused by deficiency of the branched-chain alpha-ketoacid dehydrogenase complex, a mitochondrial enzyme that requires thiamine as a cofactor. This enzyme normally catalyzes oxidative decarboxylation of the alpha-ketoacids derived from the branched-chain amino acids leucine, isoleucine, and valine. Its deficiency causes accumulation of these amino acids and their corresponding ketoacids, producing the characteristic sweet, burnt-sugar odor of urine and sweat. Untreated disease leads to severe neurological damage, and management requires a diet restricted in branched-chain amino acids.

  • B. Medium-chain acyl-CoA dehydrogenase deficiency affects fatty acid oxidation, not branched-chain amino acids.
  • C. Propionyl-CoA carboxylase deficiency causes propionic acidemia, not branched-chain amino acid accumulation.
  • D. Methylmalonyl-CoA mutase deficiency causes methylmalonic acidemia.
Educational objective: Maple syrup urine disease is due to branched-chain alpha-ketoacid dehydrogenase deficiency, causing accumulation of leucine, isoleucine, and valine.
16

A newborn has pale white skin, white hair, and pink irides, with no family history of similar features. Melanocytes are present in normal numbers on skin biopsy, but pigment granules are absent. Deficiency of which enzyme best explains the lack of pigmentation?

  1. ADopamine beta-hydroxylase
  2. BHomogentisate dioxygenase
  3. CTyrosinase
  4. DTyrosine hydroxylase
Show answer
Answer: C. Tyrosinase

Albinism results from deficiency or dysfunction of tyrosinase, the copper-containing enzyme in melanocytes that catalyzes the rate-limiting conversion of tyrosine to dopaquinone, the first step in melanin synthesis. Although melanocytes are present in normal numbers, they fail to produce melanin pigment due to this enzymatic block, resulting in the characteristic hypopigmentation of skin, hair, and eyes. This condition contrasts with vitiligo, where melanocytes themselves are destroyed rather than being functionally deficient in enzyme activity.

  • A. Dopamine beta-hydroxylase converts dopamine to norepinephrine in catecholamine synthesis, not melanin synthesis.
  • B. Homogentisate dioxygenase deficiency causes alkaptonuria, unrelated to pigmentation.
  • D. Tyrosine hydroxylase is involved in catecholamine synthesis, not melanin production.
Educational objective: Albinism is caused by tyrosinase deficiency, blocking the conversion of tyrosine to melanin despite normal melanocyte numbers.

Nucleotide Metabolism & Gout

3 questions
17

A 3-year-old boy presents with choreoathetoid movements, intellectual disability, and repeated self-biting of his lips and fingers requiring protective restraints. Serum uric acid is markedly elevated. Which enzyme deficiency underlies this condition?

  1. AAdenosine deaminase
  2. BXanthine oxidase
  3. CHypoxanthine-guanine phosphoribosyltransferase
  4. DPurine nucleoside phosphorylase
Show answer
Answer: C. Hypoxanthine-guanine phosphoribosyltransferase

Lesch-Nyhan syndrome is an X-linked recessive disorder caused by deficiency of hypoxanthine-guanine phosphoribosyltransferase (HGPRT), an enzyme essential for the purine salvage pathway. Its absence prevents reutilization of hypoxanthine and guanine, shunting them toward degradation into uric acid and causing severe hyperuricemia. The clinical triad consists of hyperuricemia, neurological dysfunction with choreoathetosis and intellectual disability, and characteristic compulsive self-mutilating behavior such as lip and finger biting. Management includes allopurinol for hyperuricemia, though it does not improve the neurobehavioral features.

  • A. Adenosine deaminase deficiency causes severe combined immunodeficiency, not this triad.
  • B. Xanthine oxidase deficiency causes xanthinuria with low, not high, uric acid levels.
  • D. Purine nucleoside phosphorylase deficiency causes a T-cell immunodeficiency, not self-mutilation and hyperuricemia.
Educational objective: Lesch-Nyhan syndrome is caused by HGPRT deficiency, producing hyperuricemia, neurological deficits, and compulsive self-injurious behavior.
18

A 48-year-old man presents with acute onset of severe pain, redness, and swelling of the first metatarsophalangeal joint. Joint fluid aspiration under polarized light microscopy reveals needle-shaped, negatively birefringent crystals. These crystals are composed of the final breakdown product of which metabolic pathway?

  1. AHeme degradation
  2. BBranched-chain amino acid catabolism
  3. CPyrimidine nucleotide degradation
  4. DPurine nucleotide degradation
Show answer
Answer: D. Purine nucleotide degradation

Gout results from deposition of monosodium urate crystals, which are the end product of purine nucleotide degradation. Purines (adenine and guanine) are degraded through hypoxanthine and xanthine to uric acid by xanthine oxidase, and humans lack uricase, the enzyme that would further break down uric acid into the more soluble allantoin in most other mammals. This limited solubility predisposes to crystal formation in joints, especially the first metatarsophalangeal joint, when serum urate levels are high. The negatively birefringent needle-shaped crystals under polarized light confirm the diagnosis of gout.

  • A. Heme degradation produces bilirubin, not uric acid crystals.
  • B. Branched-chain amino acid catabolism produces propionyl-CoA and other intermediates, unrelated to urate crystals.
  • C. Pyrimidine degradation produces beta-alanine and beta-aminoisobutyrate, not uric acid.
Educational objective: Uric acid, the end product of purine degradation, deposits as monosodium urate crystals in joints, causing gout.
19

A 6-month-old boy presents with recurrent severe pneumonia, oral thrush, and chronic diarrhea since birth. Investigations show absolute lymphopenia affecting both T and B cell lineages with low serum immunoglobulins. Thymic shadow is absent on chest X-ray. Blood levels of deoxyadenosine and its nucleotides are markedly elevated, and these metabolites are found to be toxic to lymphocytes. Which enzyme deficiency best explains this presentation?

  1. APurine nucleoside phosphorylase
  2. BAdenine phosphoribosyltransferase
  3. CAdenosine deaminase
  4. DHypoxanthine-guanine phosphoribosyltransferase
Show answer
Answer: C. Adenosine deaminase

Adenosine deaminase (ADA) converts adenosine and deoxyadenosine to inosine and deoxyinosine. Its deficiency causes accumulation of deoxyadenosine, which is phosphorylated to dATP, a potent inhibitor of ribonucleotide reductase that blocks DNA synthesis and is especially toxic to lymphocytes. This leads to severe combined immunodeficiency with absence of both T and B cells. ADA deficiency accounts for a significant proportion of autosomal recessive SCID cases and is treatable with enzyme replacement or gene therapy.

  • A. Purine nucleoside phosphorylase deficiency causes a milder immunodeficiency affecting mainly T cells, with intact B cell function.
  • B. Adenine phosphoribosyltransferase deficiency causes 2,8-dihydroxyadenine urolithiasis, not immunodeficiency.
  • D. HGPRT deficiency causes Lesch-Nyhan syndrome with hyperuricemia and neurological features, not immunodeficiency.
Educational objective: Adenosine deaminase deficiency causes SCID through toxic accumulation of dATP that inhibits ribonucleotide reductase in lymphocytes.

Vitamins

6 questions
20

A 48-year-old man with chronic alcohol dependence is brought to the emergency department with confusion, unsteady gait, and inability to move his eyes laterally. On examination, he has horizontal nystagmus, bilateral lateral rectus palsy, and a wide-based ataxic gait. He is started on intravenous dextrose before any vitamin supplementation, after which his confusion worsens. Deficiency of which vitamin best explains this clinical picture?

  1. AVitamin B12 (cobalamin)
  2. BVitamin B1 (thiamine)
  3. CVitamin B3 (niacin)
  4. DVitamin B6 (pyridoxine)
Show answer
Answer: B. Vitamin B1 (thiamine)

Thiamine is a cofactor for pyruvate dehydrogenase, alpha-ketoglutarate dehydrogenase, and transketolase, enzymes essential for cerebral glucose metabolism. Deficiency, common in chronic alcoholics due to poor intake and impaired absorption, causes Wernicke encephalopathy, presenting with ophthalmoplegia, ataxia, and confusion. Giving glucose without thiamine precipitates or worsens the encephalopathy by increasing demand on thiamine-dependent enzymes. If untreated, it progresses to Korsakoff psychosis with anterograde amnesia and confabulation.

  • A. B12 deficiency causes subacute combined degeneration with sensory ataxia, but not ophthalmoplegia or this acute presentation.
  • C. Niacin deficiency causes pellagra with dermatitis, diarrhea, and dementia, not ophthalmoplegia or ataxia.
  • D. Pyridoxine deficiency causes peripheral neuropathy and sideroblastic anemia, not this triad.
Educational objective: Thiamine deficiency causes Wernicke-Korsakoff syndrome, and glucose must never be given before thiamine in at-risk patients.
21

A 55-year-old strict vegetarian woman presents with progressive weakness, tingling in both feet, and difficulty walking in the dark. Examination shows loss of vibration and position sense in the lower limbs with a positive Romberg sign. Peripheral smear shows macrocytic red cells with hypersegmented neutrophils. Serum vitamin B12 is low. Which mechanism best explains the combination of anemia and neurological findings in this patient?

  1. ADeficient hydroxylation of proline and lysine residues in collagen
  2. BImpaired conversion of folate to its active tetrahydrofolate form
  3. CDecreased absorption of dietary iron from the proximal jejunum
  4. DImpaired conversion of homocysteine to methionine affecting myelin synthesis and DNA synthesis
Show answer
Answer: D. Impaired conversion of homocysteine to methionine affecting myelin synthesis and DNA synthesis

Vitamin B12 is a cofactor for methionine synthase, which converts homocysteine to methionine while regenerating tetrahydrofolate from methyltetrahydrofolate for DNA synthesis. B12 is also required for the conversion of methylmalonyl-CoA to succinyl-CoA; its deficiency causes accumulation of methylmalonyl-CoA, which disrupts normal fatty acid synthesis in neuronal membranes and contributes to demyelination. The combined defect in DNA synthesis causes megaloblastic anemia, while the lipid synthesis defect causes subacute combined degeneration of the spinal cord. This explains why B12 deficiency affects both the bone marrow and nervous system, unlike isolated folate deficiency.

  • A. This describes the mechanism of vitamin C deficiency affecting collagen, unrelated to this presentation.
  • B. This describes folate deficiency, which causes megaloblastic anemia without neurological disease.
  • C. Iron deficiency causes microcytic anemia, not the macrocytic picture described.
Educational objective: Vitamin B12 deficiency causes megaloblastic anemia and subacute combined degeneration through impaired methionine synthase and methylmalonyl-CoA mutase activity.
22

A 24-year-old woman plans to conceive in the next few months. Her elder sibling's child was born with meningomyelocele. She has no other medical illness and takes no regular medications. Which strategy is most appropriate to reduce the risk of a neural tube defect in her future pregnancy?

  1. AVitamin B12 supplementation throughout pregnancy
  2. BFolic acid supplementation starting at least one month before conception and continued through the first trimester
  3. CFolic acid supplementation starting only after the pregnancy test is confirmed positive
  4. DIron and folic acid supplementation starting in the second trimester
Show answer
Answer: B. Folic acid supplementation starting at least one month before conception and continued through the first trimester

Folate is essential for one-carbon transfer reactions required for purine and thymidylate synthesis, which are critical during neural tube closure occurring by the 28th day after conception, often before pregnancy is recognized. Periconceptional folic acid supplementation, started at least four weeks before conception and continued through the first trimester, significantly reduces the risk of neural tube defects, especially in women with a family history. Starting supplementation only after a positive pregnancy test is too late because neural tube closure is already complete by then. National programs recommend higher doses in women with a prior affected pregnancy or family history.

  • A. Vitamin B12 deficiency is not the primary preventable cause of neural tube defects; folate is the key nutrient.
  • C. By the time pregnancy is confirmed, the neural tube has already closed, so this timing fails to prevent the defect.
  • D. Second trimester supplementation is far too late since neural tube closure occurs in the first month of gestation.
Educational objective: Periconceptional folic acid supplementation before and during early pregnancy prevents neural tube defects by supporting one-carbon metabolism during neural tube closure.
23

A 32-year-old man on antitubercular therapy for the past three months presents with burning pain and tingling sensation in both feet, worse at night. Examination shows reduced vibration sense and absent ankle reflexes bilaterally. There is no history of diabetes or alcohol use. Which vitamin supplementation would have most likely prevented this complication?

  1. AVitamin B1 (thiamine)
  2. BVitamin B2 (riboflavin)
  3. CVitamin B9 (folic acid)
  4. DVitamin B6 (pyridoxine)
Show answer
Answer: D. Vitamin B6 (pyridoxine)

Pyridoxal phosphate, the active form of vitamin B6, is an essential cofactor for transamination reactions and for the synthesis of neurotransmitters and heme. Isoniazid forms a hydrazone complex with pyridoxal phosphate and also inhibits pyridoxine phosphokinase, leading to functional B6 deficiency. This results in peripheral neuropathy, which is why prophylactic pyridoxine is co-administered with isoniazid, especially in malnourished patients, pregnant women, and those with diabetes or HIV infection. The neuropathy is predominantly sensory and dose related.

  • A. Thiamine deficiency causes beriberi and Wernicke encephalopathy but is not specifically depleted by isoniazid.
  • B. Riboflavin deficiency causes angular stomatitis and glossitis, not isoniazid-related neuropathy.
  • C. Folic acid deficiency causes megaloblastic anemia, not the peripheral neuropathy seen with isoniazid.
Educational objective: Isoniazid causes peripheral neuropathy by inducing functional vitamin B6 deficiency, preventable by co-administering pyridoxine.
24

A 7-year-old boy from a low-income family, who eats mostly powdered milk with almost no fresh fruits or vegetables, presents with swollen, bleeding gums and bruising over both legs. Examination shows perifollicular hemorrhages and swelling over the costochondral junctions. Which biochemical defect best explains these findings?

  1. AImpaired hydroxylation of proline and lysine residues in procollagen
  2. BImpaired synthesis of glycosaminoglycans in connective tissue
  3. CImpaired cross-linking of elastin fibers by lysyl oxidase
  4. DImpaired carboxylation of glutamate residues in clotting factors
Show answer
Answer: A. Impaired hydroxylation of proline and lysine residues in procollagen

Vitamin C is a cofactor for prolyl hydroxylase and lysyl hydroxylase, enzymes that hydroxylate proline and lysine residues in procollagen. This hydroxylation is essential for stable triple helix formation and cross-linking of collagen fibers. Deficiency leads to weak, poorly cross-linked collagen, causing capillary fragility and the bleeding gums, perifollicular hemorrhages, and subperiosteal hemorrhages characteristic of scurvy. The swollen costochondral junctions, described as the scorbutic rosary, reflect poor collagen support at these sites.

  • B. Glycosaminoglycan synthesis defects characterize mucopolysaccharidoses, unrelated to this presentation.
  • C. Lysyl oxidase cross-links collagen and elastin using copper as a cofactor, not vitamin C.
  • D. Carboxylation of clotting factor glutamate residues depends on vitamin K, not vitamin C.
Educational objective: Vitamin C deficiency impairs proline and lysine hydroxylation in collagen synthesis, causing the capillary fragility and bleeding seen in scurvy.
25

A biochemistry researcher is studying vitamin D metabolism. Cholecalciferol is first hydroxylated in the liver to form 25-hydroxycholecalciferol. This metabolite then undergoes a second hydroxylation to form the most biologically active form of vitamin D, which has the highest affinity for the vitamin D receptor. At which site does this second, activating hydroxylation occur?

  1. AKidney, catalyzed by 1-alpha-hydroxylase
  2. BSkin, catalyzed by 7-dehydrocholesterol reductase
  3. CLiver, catalyzed by 25-hydroxylase
  4. DIntestine, catalyzed by calcium-binding protein synthesis
Show answer
Answer: A. Kidney, catalyzed by 1-alpha-hydroxylase

25-hydroxycholecalciferol, formed in the liver, travels to the kidney where 1-alpha-hydroxylase in the proximal tubule converts it to 1,25-dihydroxycholecalciferol, calcitriol, the most active form of vitamin D. Calcitriol binds the vitamin D receptor with much higher affinity than 25-hydroxycholecalciferol and promotes intestinal calcium absorption, bone resorption, and renal calcium reabsorption. Parathyroid hormone, low serum phosphate, and low calcium stimulate renal 1-alpha-hydroxylase activity. This renal step is why chronic kidney disease leads to deficient calcitriol production and secondary hyperparathyroidism.

  • B. The skin step converts 7-dehydrocholesterol to cholecalciferol using UV light, unrelated to hydroxylation activation.
  • C. 25-hydroxylation in the liver produces the storage and transport form, not the most active form of vitamin D.
  • D. The intestine is a target organ for calcitriol action, not a site of vitamin D hydroxylation.
Educational objective: The kidney performs the final activating 1-alpha-hydroxylation of vitamin D to produce calcitriol, the most biologically active form.

Acid-Base & Electrolyte Biochemistry

3 questions
26

A 19-year-old man with type 1 diabetes mellitus is brought to the emergency department with vomiting and abdominal pain for 2 days. His respiratory rate is 32/min with deep breathing. Arterial blood gas shows: pH 7.21, PaCO2 20 mmHg, HCO3- 9 mEq/L. Which of the following best describes this acid-base disturbance?

  1. ARespiratory alkalosis with metabolic compensation
  2. BMetabolic alkalosis with respiratory compensation
  3. CMixed metabolic and respiratory acidosis
  4. DMetabolic acidosis with appropriate respiratory compensation
Show answer
Answer: D. Metabolic acidosis with appropriate respiratory compensation

The low pH confirms acidemia, and the low HCO3- indicates a primary metabolic acidosis, consistent with diabetic ketoacidosis. Using Winter's formula, expected PaCO2 = 1.5(9) + 8 ± 2 = 21.5 ± 2 mmHg, which matches the measured PaCO2 of 20 mmHg, confirming appropriate respiratory compensation by hyperventilation (Kussmaul breathing). If the PaCO2 were higher or lower than predicted, a mixed disorder would be suspected. This pattern is classic for uncompensated metabolic acidosis undergoing compensation, not a primary respiratory process.

  • A. Respiratory alkalosis would show a high pH, not a low pH as seen here.
  • B. Metabolic alkalosis would present with a high pH and high HCO3-, which is not seen here.
  • C. The PaCO2 falls within the predicted compensatory range, so this is not a mixed disorder.
Educational objective: Winter's formula predicts the expected respiratory compensation in metabolic acidosis; a PaCO2 matching the predicted range confirms simple compensation rather than a mixed disorder.
27

A 45-year-old man presents with confusion and rapid breathing after ingesting an unknown substance. Serum sodium is 140 mEq/L, chloride is 96 mEq/L, and bicarbonate is 10 mEq/L. His serum osmolality is elevated with a large osmolar gap. Which of the following is the most likely cause of his acid-base disorder?

  1. AHyperchloremic acidosis due to normal saline infusion
  2. BDiarrhea
  3. CMethanol poisoning
  4. DRenal tubular acidosis
Show answer
Answer: C. Methanol poisoning

The anion gap is calculated as Na+ - (Cl- + HCO3-) = 140 - (96+10) = 34 mEq/L, which is markedly elevated above the normal range of 8-12 mEq/L. A high anion gap metabolic acidosis with an elevated osmolar gap strongly suggests ingestion of a toxic alcohol such as methanol, which is metabolized to formic acid. Methanol toxicity classically causes visual disturbances and a high anion gap acidosis with osmolar gap, unlike the other causes listed. Diarrhea and renal tubular acidosis cause a normal anion gap (hyperchloremic) acidosis, and saline-induced acidosis is also a normal anion gap state.

  • A. Saline-induced acidosis is a dilutional hyperchloremic acidosis with a normal anion gap, not an elevated osmolar gap.
  • B. Diarrhea causes bicarbonate loss resulting in a normal anion gap hyperchloremic acidosis, not a high anion gap with osmolar gap.
  • D. Renal tubular acidosis is a normal anion gap acidosis due to impaired renal acid handling.
Educational objective: A high anion gap metabolic acidosis with an elevated osmolar gap points to toxic alcohol ingestion such as methanol or ethylene glycol.
28

A biochemistry lecturer explains how the blood maintains a stable pH despite continuous production of carbon dioxide by tissues. Which of the following best explains the primary mechanism by which the bicarbonate-carbonic acid buffer system resists changes in blood pH?

  1. ABicarbonate directly binds excess hydrogen ions to form water and oxygen, which are excreted by the kidney
  2. BHemoglobin in red blood cells converts all excess acid directly into ammonium ions for renal excretion
  3. CPlasma proteins denature in the presence of excess acid, releasing stored bicarbonate into circulation
  4. DCarbonic anhydrase rapidly interconverts carbon dioxide and carbonic acid, allowing excess hydrogen ions to be buffered by bicarbonate and the resulting carbon dioxide to be exhaled
Show answer
Answer: D. Carbonic anhydrase rapidly interconverts carbon dioxide and carbonic acid, allowing excess hydrogen ions to be buffered by bicarbonate and the resulting carbon dioxide to be exhaled

The bicarbonate buffer system works because carbonic anhydrase catalyzes the rapid reaction CO2 + H2O to form carbonic acid, which dissociates into H+ and HCO3-. When excess H+ enters the blood, it combines with HCO3- to form carbonic acid, which is then converted back to CO2 and water and eliminated by the lungs through ventilation. This open system is highly effective because the lungs can rapidly adjust CO2 elimination, unlike a closed buffer system. The other options describe reactions that do not occur or misattribute the mechanism to hemoglobin or plasma proteins.

  • A. Bicarbonate buffers H+ by forming carbonic acid and ultimately CO2 and water, not water and oxygen.
  • B. Ammonium excretion is a renal mechanism for acid elimination but is separate from the bicarbonate buffer reaction in blood.
  • C. Plasma proteins act as buffers through their own amino acid side chains, not by releasing stored bicarbonate.
Educational objective: The bicarbonate buffer system is effective because carbonic anhydrase links it to pulmonary CO2 excretion, making it an open and highly efficient buffering system.

Molecular Biology & Genetics

7 questions
29

During DNA replication in a eukaryotic cell, an enzyme detects a mismatched nucleotide incorporated at the growing 3' end of the newly synthesized strand and removes it before synthesis continues. Which enzymatic activity is primarily responsible for this proofreading function?

  1. A3' to 5' exonuclease activity of DNA polymerase
  2. BLigase activity joining Okazaki fragments
  3. CEndonuclease activity of topoisomerase
  4. D5' to 3' exonuclease activity of DNA polymerase
Show answer
Answer: A. 3' to 5' exonuclease activity of DNA polymerase

DNA polymerases possess an intrinsic 3' to 5' exonuclease activity that recognizes mismatched base pairs immediately after their incorporation and excises the incorrect nucleotide before polymerization resumes. This proofreading activity greatly increases the fidelity of DNA replication, reducing the error rate from about 1 in 10^4-10^5 bases to about 1 in 10^7 bases. The 5' to 3' exonuclease activity, in contrast, is used for removing RNA primers during Okazaki fragment processing, not for proofreading newly added bases. Topoisomerase relieves supercoiling, and ligase seals nicks between fragments, neither of which corrects base mismatches.

  • B. Ligase seals the nick between adjacent DNA fragments and plays no role in correcting mismatched nucleotides.
  • C. Topoisomerase relieves torsional strain ahead of the replication fork and does not correct base-pairing errors.
  • D. 5' to 3' exonuclease activity removes RNA primers and damaged DNA ahead of the replication fork, not mismatched bases just added.
Educational objective: The 3' to 5' exonuclease activity of DNA polymerase provides proofreading during replication, ensuring high fidelity of newly synthesized DNA.
30

A diagnostic laboratory wants to detect a very small quantity of HIV viral RNA in a patient's blood sample when antibody-based tests are negative during the early window period. Which of the following techniques is most appropriate, and what is its underlying principle?

  1. ASouthern blotting, which separates DNA fragments by size using gel electrophoresis and hybridization
  2. BPolymerase chain reaction, which exponentially amplifies specific nucleic acid sequences using repeated cycles of denaturation, primer annealing, and extension
  3. CELISA, which detects antigen-antibody complexes using an enzyme-linked colorimetric reaction
  4. DKaryotyping, which visualizes the number and structure of chromosomes under a microscope
Show answer
Answer: B. Polymerase chain reaction, which exponentially amplifies specific nucleic acid sequences using repeated cycles of denaturation, primer annealing, and extension

Polymerase chain reaction (PCR) amplifies a specific target nucleic acid sequence exponentially through repeated cycles of denaturation at high temperature, primer annealing, and extension by a heat-stable DNA polymerase, typically Taq polymerase. This allows detection of extremely small quantities of viral genetic material, making it the method of choice for early diagnosis of HIV infection during the window period before antibodies develop. Southern blotting and karyotyping are used for different applications such as detecting specific DNA fragments or chromosomal abnormalities, and ELISA detects antibodies or antigens rather than nucleic acids directly. For an RNA virus like HIV, reverse transcriptase PCR is used to first convert RNA to cDNA before amplification.

  • A. Southern blotting identifies specific DNA sequences in a sample but lacks the sensitivity to detect trace amounts of viral RNA without prior amplification.
  • C. ELISA detects antibodies or antigens and would be negative during the early window period before seroconversion occurs.
  • D. Karyotyping is used to study chromosome number and structure and has no role in detecting viral nucleic acids.
Educational objective: PCR detects minute quantities of specific nucleic acid sequences through exponential amplification, making it ideal for early diagnosis of viral infections before antibody formation.
31

A 55-year-old woman with metastatic ovarian cancer is started on a chemotherapeutic agent that stabilizes the covalent complex between DNA and an enzyme that normally relaxes supercoiled DNA by making transient single-strand breaks. This stabilization prevents re-ligation of the DNA strand, leading to accumulation of DNA breaks and cell death. Which of the following drugs acts by this mechanism?

  1. AVincristine
  2. BTopotecan
  3. CCyclophosphamide
  4. DMethotrexate
Show answer
Answer: B. Topotecan

Topotecan is a topoisomerase I inhibitor that binds to the topoisomerase I-DNA cleavable complex and prevents re-ligation of the single-strand break, leading to accumulation of DNA strand breaks and subsequent cell death during replication. This mechanism distinguishes it from drugs acting on folate metabolism, microtubules, or DNA cross-linking. Topoisomerase I normally relieves torsional stress ahead of the replication fork by creating transient single-strand nicks and resealing them, and camptothecin derivatives like topotecan and irinotecan exploit this mechanism for their cytotoxic effect. Methotrexate inhibits dihydrofolate reductase, vincristine binds tubulin to inhibit mitotic spindle formation, and cyclophosphamide causes DNA cross-linking as an alkylating agent.

  • A. Vincristine binds tubulin and inhibits microtubule polymerization, arresting cells in mitosis rather than acting on topoisomerase.
  • C. Cyclophosphamide is an alkylating agent that cross-links DNA strands, a different mechanism from topoisomerase inhibition.
  • D. Methotrexate inhibits dihydrofolate reductase, blocking folate-dependent nucleotide synthesis, not topoisomerase function.
Educational objective: Camptothecin derivatives such as topotecan inhibit topoisomerase I by stabilizing the DNA-enzyme cleavable complex, preventing strand re-ligation and causing lethal DNA damage.
32

A 42-year-old man presents with progressive involuntary jerky movements of the limbs and face, personality changes, and cognitive decline over the past year. His father had similar symptoms beginning at age 50 and died 10 years after onset. Genetic testing reveals an expanded CAG trinucleotide repeat in a gene on chromosome 4. Which of the following best describes the inheritance pattern and typical feature of this disorder?

  1. AAutosomal recessive inheritance requiring both parents to be carriers
  2. BMitochondrial inheritance transmitted exclusively through the maternal line
  3. CAutosomal dominant inheritance with anticipation, showing earlier age of onset in successive generations
  4. DX-linked recessive inheritance affecting primarily male offspring
Show answer
Answer: C. Autosomal dominant inheritance with anticipation, showing earlier age of onset in successive generations

Huntington disease is caused by an expanded CAG trinucleotide repeat in the huntingtin gene on chromosome 4 and follows autosomal dominant inheritance, so an affected parent has a 50 percent chance of transmitting the disease allele to each offspring. The disorder characteristically shows genetic anticipation, where the repeat expansion tends to increase in successive generations, particularly through paternal transmission, leading to earlier age of onset and often more severe disease in children compared to their affected parent. Clinically, it presents with chorea, psychiatric disturbances, and progressive dementia, matching the presentation described. The other inheritance patterns do not fit the family history of an affected father transmitting disease to his son with earlier onset in the next generation.

  • A. Autosomal recessive inheritance would require both parents to carry the mutant allele, which does not match a disease transmitted from an affected father.
  • B. Mitochondrial inheritance is transmitted only through the maternal line, but this patient's affected parent was his father.
  • D. X-linked recessive disorders are transmitted from carrier mothers to sons, not from an affected father to a son as described here.
Educational objective: Huntington disease is an autosomal dominant trinucleotide repeat disorder (CAG expansion) that shows genetic anticipation with earlier onset in successive generations.
33

A researcher wants to confirm a suspected deletion in a specific gene by analyzing patient genomic DNA extracted from peripheral blood leukocytes. Which technique is best suited for this purpose?

  1. ASouthern blot
  2. BELISA
  3. CNorthern blot
  4. DWestern blot
Show answer
Answer: A. Southern blot

Southern blotting detects and characterizes DNA fragments, including deletions, insertions and rearrangements, by hybridizing labeled probes to DNA separated by gel electrophoresis after restriction digestion. It is the classic method for diagnosing gene deletions, such as in fragile X syndrome or alpha-thalassemia deletions. Northern blot analyzes RNA to study gene expression, not structural DNA changes. Western blot detects proteins using antibodies, and ELISA is an immunoassay for antigen-antibody reactions, neither of which examines genomic DNA directly.

  • B. ELISA detects antigens or antibodies in a sample and cannot analyze genomic DNA structure.
  • C. Northern blot detects RNA and is used to study gene expression levels, not DNA deletions.
  • D. Western blot identifies proteins using specific antibodies, not DNA sequences.
Educational objective: Southern blot is the test of choice for detecting DNA-level abnormalities such as gene deletions, while Northern and Western blots analyze RNA and protein respectively.
34

A 30-year-old woman with a known mitochondrial DNA point mutation causing myopathy wants to know the risk of transmitting the disease to her children. Her husband is unaffected and has no mitochondrial mutation. Which statement best describes the expected inheritance pattern?

  1. AEach child has a 50% chance of inheriting the mutation, following an autosomal dominant pattern
  2. BOnly her sons can inherit the mutation and transmit it to the next generation
  3. CThe mutation skips generations and appears only in alternate generations, similar to autosomal recessive inheritance
  4. DAll of her children, both sons and daughters, can inherit the mutation, but only her daughters can transmit it further
Show answer
Answer: D. All of her children, both sons and daughters, can inherit the mutation, but only her daughters can transmit it further

Mitochondrial DNA is inherited exclusively through the maternal line because the ovum contributes essentially all mitochondria to the zygote, while sperm mitochondria are degraded after fertilization. Therefore, an affected mother transmits mitochondrial mutations to all her children, both male and female, but only her daughters can pass the mutation on to subsequent generations since sons do not pass mitochondria to their offspring. This pattern is distinct from Mendelian autosomal or sex-linked inheritance.

  • A. Mitochondrial inheritance does not follow the 50% segregation ratio seen in autosomal dominant Mendelian inheritance.
  • B. Sons can also inherit the mutation since all children receive maternal mitochondria, but sons cannot transmit it further.
  • C. Mitochondrial inheritance does not skip generations; it is transmitted continuously through the maternal line.
Educational objective: Mitochondrial DNA mutations are transmitted exclusively through the mother to all offspring, but only daughters can pass the trait to the next generation.
35

A 28-year-old man presents for voluntary HIV screening at a counseling center. A test is performed that uses a plate coated with HIV antigens, patient serum, an enzyme-linked secondary antibody, and a chromogenic substrate, with color intensity measured by spectrophotometry. Which test is being described?

  1. APolymerase chain reaction
  2. BRapid immunochromatographic test
  3. CWestern blot
  4. DELISA
Show answer
Answer: D. ELISA

ELISA (enzyme-linked immunosorbent assay) is the standard screening test for HIV infection. It detects antibodies against HIV antigens coated on a solid-phase plate; a secondary antibody linked to an enzyme produces a color change proportional to antibody concentration, read spectrophotometrically. ELISA is highly sensitive, making it suitable as an initial screening test, with confirmatory testing done using Western blot or similar supplemental assays. The described plate-based, enzyme-substrate color reaction format is characteristic of ELISA rather than confirmatory or point-of-care tests.

  • A. PCR detects viral nucleic acid and does not use antigen-coated plates or chromogenic antibody-based detection.
  • B. Rapid immunochromatographic tests use lateral flow strips, not microtiter plates with spectrophotometric reading.
  • C. Western blot is a confirmatory test that separates viral proteins by electrophoresis and detects antibodies against specific bands, not a plate-based colorimetric assay.
Educational objective: ELISA is the standard initial screening test for HIV, detecting antibodies via antigen-coated plates and enzyme-linked color reactions.

Hemoglobin & Porphyrin Metabolism

2 questions
36

A 45-year-old man presents with yellowish discoloration of the eyes and dark-colored urine for one week. Laboratory tests show total bilirubin 8 mg/dL, direct bilirubin 6 mg/dL, and urine is positive for bilirubin but negative for excess urobilinogen. Which pattern best describes this biochemical picture?

  1. AConjugated hyperbilirubinemia, suggesting hepatobiliary obstruction
  2. BUnconjugated hyperbilirubinemia, suggesting defective hepatic uptake
  3. CUnconjugated hyperbilirubinemia, suggesting hemolysis
  4. DMixed hyperbilirubinemia, suggesting hemolytic anemia with liver failure
Show answer
Answer: A. Conjugated hyperbilirubinemia, suggesting hepatobiliary obstruction

A predominantly direct (conjugated) bilirubin fraction, with direct bilirubin comprising more than half of total bilirubin, indicates conjugated hyperbilirubinemia, typically due to hepatocellular or post-hepatic obstructive causes. Conjugated bilirubin is water-soluble and appears in urine, explaining the dark urine and positive urine bilirubin. Reduced urobilinogen reflects decreased bilirubin reaching the intestine due to biliary obstruction, consistent with cholestasis. Unconjugated bilirubin, in contrast, is not water-soluble, does not appear in urine, and is associated with increased urobilinogen in hemolytic states due to increased bilirubin delivery to the gut.

  • B. Defective hepatic uptake, as in Gilbert syndrome, causes mild unconjugated hyperbilirubinemia, not a direct-predominant pattern with bilirubinuria.
  • C. Hemolysis causes unconjugated hyperbilirubinemia with increased urobilinogen, not a predominant direct fraction with absent urine urobilinogen.
  • D. There is no evidence of hemolysis or liver failure here; the pattern is purely conjugated and obstructive.
Educational objective: A direct bilirubin fraction exceeding half of total bilirubin with bilirubinuria and reduced urobilinogen indicates conjugated hyperbilirubinemia from hepatobiliary obstruction.
37

A newborn develops severe jaundice with serum total bilirubin of 22 mg/dL, almost entirely unconjugated, and does not respond to phenobarbital therapy. Liver biopsy shows complete absence of enzyme activity in hepatocytes. Which enzyme defect best explains this presentation compared to a milder, phenobarbital-responsive form of the same disorder?

  1. AComplete absence of UDP-glucuronosyltransferase, as in Crigler-Najjar syndrome type 1
  2. BPartial reduction of UDP-glucuronosyltransferase activity, as in Gilbert syndrome
  3. CDeficiency of bilirubin UDP-glucuronosyltransferase substrate uptake protein
  4. DDeficiency of canalicular multidrug resistance protein 2
Show answer
Answer: A. Complete absence of UDP-glucuronosyltransferase, as in Crigler-Najjar syndrome type 1

Crigler-Najjar syndrome type 1 results from a complete absence of hepatic UDP-glucuronosyltransferase (UGT1A1) activity, leading to severe unconjugated hyperbilirubinemia that does not respond to enzyme-inducing drugs like phenobarbital and carries a risk of kernicterus. In contrast, Gilbert syndrome involves only a partial reduction in UGT1A1 activity, typically around 30 percent of normal, causing mild, fluctuating unconjugated hyperbilirubinemia that is usually asymptomatic and does not require treatment. Crigler-Najjar type 2 has partial enzyme deficiency and does respond to phenobarbital, distinguishing it from type 1.

  • B. Gilbert syndrome causes only mild hyperbilirubinemia with partial enzyme activity, not the severe levels and biopsy findings described here.
  • C. This describes a transporter defect, not the glucuronyltransferase enzyme deficiency responsible for this unconjugated hyperbilirubinemia.
  • D. MRP2 deficiency causes Dubin-Johnson syndrome, which presents with conjugated, not unconjugated, hyperbilirubinemia.
Educational objective: Crigler-Najjar syndrome type 1 is caused by complete UDP-glucuronosyltransferase deficiency causing severe phenobarbital-unresponsive unconjugated hyperbilirubinemia, unlike the partial deficiency in Gilbert syndrome.

Nutrition & Energy Metabolism

4 questions
38

A 2-year-old child from a low-income family presents with generalized pitting edema, a distended abdomen, sparse depigmented hair, and skin showing flaky, peeling patches. Despite the edema, subcutaneous fat is relatively preserved over the limbs. Serum albumin is 1.8 g/dL. Which condition best fits this presentation?

  1. AMarasmus
  2. BKwashiorkor
  3. CMarasmic kwashiorkor
  4. DVitamin A deficiency
Show answer
Answer: B. Kwashiorkor

Kwashiorkor results from severe protein deficiency with relatively preserved total caloric intake, leading to hypoalbuminemia that causes generalized edema, a protuberant abdomen from hepatomegaly and ascites, skin depigmentation and flaking, and sparse brittle hair, often with a reddish discoloration. Subcutaneous fat is typically preserved, distinguishing it from marasmus, which results from combined protein-calorie deficiency and presents with severe muscle wasting, loss of subcutaneous fat, and a wizened, old-man appearance without edema. Serum albumin is markedly low in kwashiorkor due to inadequate protein intake despite adequate calories.

  • A. Marasmus presents with severe wasting of fat and muscle without edema, unlike this child who has preserved fat and prominent edema.
  • C. Marasmic kwashiorkor would show both severe wasting and edema together, whereas this child has preserved subcutaneous fat with isolated edema typical of kwashiorkor.
  • D. Vitamin A deficiency causes night blindness and xerophthalmia, not generalized edema and hypoalbuminemia.
Educational objective: Kwashiorkor is caused by protein deficiency with adequate calories, producing edema, hypoalbuminemia, and skin and hair changes with preserved subcutaneous fat.
39

A 30-year-old man weighing 70 kg undergoes indirect calorimetry after an overnight fast, lying awake and physically and mentally at rest in a thermoneutral room 12 hours after his last meal. His oxygen consumption is used to calculate his basal metabolic rate. Which of the following factors is the major determinant of the value obtained in this healthy adult?

  1. AAmbient room humidity
  2. BTotal body fat percentage
  3. CLean body mass
  4. DDietary protein intake over the preceding week
Show answer
Answer: C. Lean body mass

Basal metabolic rate (BMR) reflects the energy expended to maintain essential physiological functions at rest, and it correlates most closely with lean (fat-free) body mass because muscle and visceral organs are metabolically far more active than adipose tissue. This is why BMR is higher in muscular individuals and in men compared to women of the same body weight, and why it declines with age as lean mass falls. Thyroid hormone, sex, age, and body surface area also influence BMR, but lean body mass is the single strongest predictor. Recent protein intake affects the thermic effect of food, not basal conditions measured after an overnight fast.

  • A. Humidity affects heat loss and comfort but is not a determinant of basal metabolic rate.
  • B. Fat tissue is metabolically less active, so adiposity does not drive BMR the way lean mass does.
  • D. Diet-induced thermogenesis affects postprandial energy expenditure, not the basal state measured after fasting.
Educational objective: Lean body mass is the principal determinant of basal metabolic rate because metabolically active tissue, not fat, drives resting energy expenditure.
40

In a metabolic ward, a subject's expired air is analyzed and the ratio of carbon dioxide produced to oxygen consumed is calculated. When this subject is metabolizing only carbohydrate as fuel, the ratio equals 1.0, whereas during a period of exclusive fat oxidation the value measured is approximately:

  1. A0.8
  2. B1.2
  3. C0.9
  4. D0.7
Show answer
Answer: D. 0.7

The respiratory quotient (RQ) is the ratio of CO2 produced to O2 consumed during metabolism of a given substrate. Fatty acids are relatively hydrogen-rich and oxygen-poor compared to carbohydrates, so their complete oxidation requires proportionally more O2 relative to the CO2 generated, giving an RQ of about 0.7. Carbohydrate oxidation, where the molecule already contains O and H in the same ratio as water, gives an RQ of 1.0. A mixed diet typically gives an RQ around 0.8, and protein oxidation gives a value near 0.8-0.9.

  • A. 0.8 corresponds to a mixed diet or protein oxidation, not pure fat metabolism.
  • B. An RQ above 1.0 suggests lipogenesis from excess carbohydrate, not fat oxidation.
  • C. 0.9 is closer to the value for protein oxidation, not fat.
Educational objective: Pure fat oxidation gives a respiratory quotient of about 0.7, compared to 1.0 for carbohydrate, because fat is relatively deficient in oxygen and requires more O2 per CO2 produced.
41

A factory worker is brought to the emergency department after accidental exposure to a cyanide-containing compound. He is cyanotic despite adequate ventilation, and his venous blood is bright red due to impaired tissue oxygen extraction. Cyanide produces this effect by binding to which component of the mitochondrial electron transport chain?

  1. AFerric (Fe3+) ion of cytochrome a3 in complex IV
  2. BFAD cofactor of complex II
  3. CIron-sulfur cluster of complex I
  4. DUbiquinone binding site of complex III
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Answer: A. Ferric (Fe3+) ion of cytochrome a3 in complex IV

Cyanide binds tightly to the ferric (Fe3+) form of cytochrome a3, the terminal component of complex IV (cytochrome c oxidase), blocking the transfer of electrons to molecular oxygen. This halts the entire electron transport chain, collapses the proton gradient, and stops ATP synthesis despite normal oxygen delivery, producing histotoxic hypoxia. Tissues cannot use the oxygen present in blood, so venous blood remains oxygenated and bright red, and cellular respiration shifts to anaerobic glycolysis with lactic acidosis. Treatment includes hydroxocobalamin or sodium nitrite plus sodium thiosulfate to detoxify cyanide.

  • B. Complex II is not the primary site of cyanide toxicity.
  • C. Complex I is inhibited by agents such as rotenone and amytal, not cyanide.
  • D. Complex III inhibitors such as antimycin A act at this site, not cyanide.
Educational objective: Cyanide causes histotoxic hypoxia by binding the ferric ion of cytochrome a3 in complex IV, blocking electron transfer to oxygen and halting oxidative phosphorylation.

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