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NEET PG · India

NEET PG biochemistry questions

50 original NEET PG-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.

Questions50 with answers
Sections9
Options per question4
Time≈ 75 min
NEET PG Biochemistry: practice quizPick an answer to see the correct option and the explanation at once. Your best score is saved in this browser.

Enzymology & Vitamins

8 questions
1

A 28-year-old man is started on a new anticonvulsant for seizure control. Two weeks later he develops severe abdominal pain, constipation and dark-coloured urine. Urine porphobilinogen is markedly elevated. The enzyme whose activity is increased by induction of its synthesis in this condition is regulated by negative feedback from the end product of the pathway it initiates. Which enzyme is this?

  1. AALA synthase
  2. BUroporphyrinogen decarboxylase
  3. CPorphobilinogen deaminase
  4. DFerrochelatase
Show answer
Answer: A. ALA synthase

ALA synthase (ALAS) catalyzes the first and rate-limiting step of heme synthesis, condensing glycine and succinyl-CoA in a reaction requiring pyridoxal phosphate. Heme, the end product, represses ALAS gene transcription and inhibits its transport into mitochondria, forming a classic negative feedback loop. Drugs that induce cytochrome P450 (such as many anticonvulsants) deplete the free heme pool, derepressing ALAS and precipitating acute attacks in susceptible individuals with porphyria. This mechanism explains drug-precipitated acute intermittent porphyria.

  • B. This enzyme acts later in the cytoplasmic pathway and is not feedback-regulated by heme.
  • C. This enzyme is deficient, not regulatory, in acute intermittent porphyria and is not the rate-limiting or feedback-regulated step.
  • D. Ferrochelatase catalyzes the final step, inserting iron into protoporphyrin IX, and is not the rate-limiting enzyme.
Educational objective: ALA synthase is the rate-limiting enzyme of heme synthesis and is feedback-inhibited by heme, explaining drug-induced porphyric attacks.
2

A 45-year-old man with a history of alcohol use presents with severe epigastric pain radiating to the back, nausea and vomiting for 12 hours. Serum amylase is 650 U/L (reference 30-110 U/L). To confirm pancreatic origin rather than salivary gland inflammation, which enzyme test should be ordered next?

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

Serum lipase is more specific for pancreatic injury than amylase because lipase is produced almost exclusively by the pancreas, whereas amylase is also secreted by the salivary glands and can be elevated in parotitis, macroamylasemia, and other conditions. Lipase also rises earlier and remains elevated longer than amylase in acute pancreatitis, improving diagnostic sensitivity. Current guidelines recommend lipase, or amylase plus lipase together, over amylase alone for diagnosing acute pancreatitis. A lipase level more than three times the upper limit of normal strongly supports the diagnosis.

  • A. ALT reflects hepatocellular injury and is used to suggest a biliary cause of pancreatitis, not to confirm pancreatic origin of hyperamylasemia.
  • C. GGT is a marker of hepatobiliary disease and alcohol use, not specific for pancreatitis.
  • D. Alkaline phosphatase reflects biliary obstruction or bone turnover, not pancreatic acinar injury specifically.
Educational objective: Serum lipase is more specific than amylase for diagnosing acute pancreatitis because it is not produced by the salivary glands.
3

A 60-year-old man presents 4 days after an episode of severe chest pain. Troponin has already normalized, but the clinician wants retrospective biochemical confirmation of myocardial injury using lactate dehydrogenase isoenzymes. Serum LDH isoenzyme electrophoresis shows the LDH1 fraction exceeding the LDH2 fraction. What does this pattern indicate?

  1. AHemolytic anemia without organ damage
  2. BSkeletal muscle injury
  3. CAcute hepatocellular necrosis
  4. DMyocardial infarction
Show answer
Answer: D. Myocardial infarction

LDH1 (H4), composed predominantly of heart-type subunits, normally runs lower than LDH2 (H3M1) in healthy serum, giving an LDH2 greater than LDH1 pattern. In myocardial infarction, myocardial cell necrosis releases large amounts of LDH1, reversing the normal ratio so that LDH1 exceeds LDH2, a pattern called "flipped LDH." This flip typically appears by 12-24 hours and can persist up to 7-10 days, making it useful for late presentations when troponin has normalized. Liver and skeletal muscle injury instead raise LDH5, the M4 isoenzyme.

  • A. Hemolysis releases LDH1 and LDH2 from red cells but does not by itself reverse the LDH1:LDH2 ratio to the degree seen with myocardial necrosis.
  • B. Skeletal muscle injury raises LDH5, not LDH1, since skeletal muscle is rich in M subunits.
  • C. Hepatocellular necrosis predominantly elevates LDH5 (muscle-type, M4 subunit), not LDH1.
Educational objective: A reversed LDH1:LDH2 ratio ("flipped LDH"), with LDH1 exceeding LDH2, indicates myocardial necrosis.
4

A 50-year-old man with chronic alcohol dependence and poor dietary intake develops nystagmus, ataxia and confusion. His condition improves after thiamine administration. Thiamine pyrophosphate is a cofactor for the mitochondrial enzyme most directly responsible for the neuronal energy failure underlying this presentation, which catalyzes oxidative decarboxylation of a Krebs cycle intermediate. Which enzyme is this?

  1. ASuccinate dehydrogenase
  2. BIsocitrate dehydrogenase
  3. CAlpha-ketoglutarate dehydrogenase
  4. DCitrate synthase
Show answer
Answer: C. Alpha-ketoglutarate dehydrogenase

Alpha-ketoglutarate dehydrogenase requires thiamine pyrophosphate as an essential cofactor to catalyze the oxidative decarboxylation of alpha-ketoglutarate to succinyl-CoA in the citric acid cycle. In thiamine deficiency, reduced activity of this enzyme impairs mitochondrial ATP generation, and neurons in highly oxidative, thiamine-dependent regions such as the mammillary bodies and periaqueductal gray are especially vulnerable, producing the ophthalmoplegia, ataxia and confusion of Wernicke encephalopathy. Pyruvate dehydrogenase is also thiamine-dependent but is not a Krebs cycle enzyme itself. Prompt thiamine replacement reverses the enzymatic block and can reverse early symptoms.

  • A. Succinate dehydrogenase uses FAD, not thiamine pyrophosphate, as its cofactor.
  • B. Isocitrate dehydrogenase uses NAD+ as a cofactor and does not require thiamine pyrophosphate.
  • D. Citrate synthase does not require thiamine pyrophosphate and does not catalyze an oxidative decarboxylation.
Educational objective: Thiamine pyrophosphate is an essential cofactor for alpha-ketoglutarate dehydrogenase, and its deficiency impairs this enzyme to cause Wernicke encephalopathy.
5

A 35-year-old man with a diet consisting almost exclusively of polished rice presents with bilateral pitting pedal edema, elevated jugular venous pressure, and a resting heart rate of 118/min with a wide pulse pressure. There is no significant peripheral neuropathy. Which clinical diagnosis best fits this presentation?

  1. APellagra
  2. BWet beriberi
  3. CSubacute combined degeneration
  4. DDry beriberi
Show answer
Answer: B. Wet beriberi

Wet beriberi is the cardiovascular form of thiamine deficiency, characterized by high-output heart failure with peripheral edema, tachycardia, and a wide pulse pressure due to peripheral vasodilation from impaired oxidative metabolism and compensatory arteriovenous shunting. Dry beriberi instead predominantly affects the peripheral nerves, producing symmetric sensorimotor polyneuropathy without significant cardiac involvement. The polished rice diet is a classic risk factor since thiamine is concentrated in the rice husk, which is removed during polishing. Distinguishing wet from dry beriberi guides clinical suspicion, though both respond to thiamine repletion.

  • A. Pellagra is caused by niacin deficiency and presents with dermatitis, diarrhea and dementia, not high-output heart failure.
  • C. Subacute combined degeneration results from vitamin B12 deficiency and affects the dorsal columns and corticospinal tracts, not the cardiovascular system.
  • D. Dry beriberi presents with peripheral neuropathy rather than high-output cardiac failure and edema.
Educational objective: Wet beriberi is thiamine deficiency presenting as high-output cardiac failure with edema and tachycardia, distinct from the neuropathic dry beriberi.
6

A 24-year-old woman from a low-income household presents with painful cracks and fissures at the corners of her mouth, a magenta-colored tongue with atrophic papillae, and scaly greasy skin lesions around the nose and ears. She reports a diet poor in dairy products, eggs and green vegetables. Deficiency of which vitamin best explains this presentation?

  1. ANiacin (vitamin B3)
  2. BPyridoxine (vitamin B6)
  3. CRiboflavin (vitamin B2)
  4. DFolate (vitamin B9)
Show answer
Answer: C. Riboflavin (vitamin B2)

Riboflavin deficiency classically causes angular stomatitis (cracks at the mouth corners), magenta glossitis, and seborrheic dermatitis affecting the nasolabial folds, scrotum or vulva. Riboflavin is a precursor of the coenzymes FAD and FMN, which are essential for numerous oxidation-reduction reactions, and deficiency impairs mucosal and epithelial cell turnover, producing these characteristic mucocutaneous findings. Dietary sources include milk, eggs, and leafy green vegetables, and deficiency is more common in populations with limited intake of these foods. Treatment is oral riboflavin supplementation with dietary correction.

  • A. Niacin deficiency causes pellagra with dermatitis, diarrhea and dementia, not the characteristic magenta tongue and angular stomatitis of riboflavin deficiency.
  • B. Pyridoxine deficiency causes peripheral neuropathy, seizures and sideroblastic anemia rather than this mucocutaneous pattern.
  • D. Folate deficiency causes megaloblastic anemia and neural tube defects, not angular stomatitis and magenta glossitis.
Educational objective: Riboflavin deficiency causes angular stomatitis, magenta glossitis, and seborrheic dermatitis due to impaired FAD/FMN-dependent oxidation-reduction reactions.
7

A 40-year-old man from a region where maize (corn) is the dietary staple presents with a symmetric, hyperpigmented, scaly rash on sun-exposed skin, chronic watery diarrhea, and progressive memory impairment with disorientation. Which biochemical mechanism best explains this deficiency state?

  1. AReduced conversion of methionine to cysteine impairing glutathione synthesis
  2. BReduced conversion of tryptophan to niacin impairing NAD+/NADP+ synthesis
  3. CReduced conversion of phenylalanine to tyrosine impairing catecholamine synthesis
  4. DReduced conversion of histidine to histamine impairing mast cell function
Show answer
Answer: B. Reduced conversion of tryptophan to niacin impairing NAD+/NADP+ synthesis

Pellagra results from niacin deficiency and classically presents with the triad of dermatitis, diarrhea and dementia. Maize is deficient in both niacin and the essential amino acid tryptophan, which is normally a minor endogenous source of niacin through a pathway requiring vitamin B6-dependent enzymes; maize protein (zein) is also poor in tryptophan, compounding the deficiency. Niacin is required to synthesize NAD+ and NADP+, coenzymes central to oxidative energy metabolism, and their deficiency impairs energy-dependent processes in rapidly dividing epithelium, gut mucosa and neurons, producing the classic triad. Treatment with niacin or a tryptophan-rich, balanced protein diet reverses the clinical features.

  • A. This pathway relates to cystathionine metabolism and homocystinuria, not pellagra.
  • C. This pathway relates to phenylketonuria and catecholamine synthesis, unrelated to the pellagra triad.
  • D. This pathway is unrelated to niacin metabolism or the pellagra triad.
Educational objective: Pellagra results from niacin deficiency, compounded by impaired tryptophan-to-niacin conversion, causing dermatitis, diarrhea and dementia.
8

A 30-year-old woman started on a four-drug regimen for pulmonary tuberculosis two months ago presents with symmetric burning paresthesias and numbness in both feet. Her tuberculosis treatment includes a drug that interferes with the activation of a B-vitamin required as a cofactor for glutamate decarboxylase and transaminase reactions. Which vitamin deficiency is responsible for her neuropathy?

  1. ACobalamin (vitamin B12)
  2. BPyridoxine (vitamin B6)
  3. CThiamine (vitamin B1)
  4. DFolate (vitamin B9)
Show answer
Answer: B. Pyridoxine (vitamin B6)

Isoniazid forms a hydrazone complex with pyridoxal phosphate and also inhibits the enzyme pyridoxine kinase, which is required to convert pyridoxine to its active cofactor form, pyridoxal phosphate. Pyridoxal phosphate is an essential cofactor for transaminases, glutamate decarboxylase (synthesizing GABA), and delta-aminolevulinic acid synthase in heme synthesis, so its depletion impairs neurotransmitter synthesis and nerve function, producing a symmetric peripheral neuropathy. This adverse effect is preventable with prophylactic pyridoxine supplementation, routinely co-administered with isoniazid in at-risk patients such as pregnant women, malnourished individuals, and those with diabetes or HIV. Isoniazid-induced neuropathy is a key dose-related adverse effect taught alongside isoniazid-induced hepatotoxicity.

  • A. Cobalamin deficiency causes subacute combined degeneration and megaloblastic anemia, unrelated to isoniazid metabolism.
  • C. Thiamine deficiency is not caused by isoniazid and classically presents with beriberi or Wernicke encephalopathy rather than this specific drug-induced neuropathy.
  • D. Folate deficiency causes megaloblastic anemia, not the peripheral neuropathy characteristic of isoniazid therapy.
Educational objective: Isoniazid causes peripheral neuropathy by inducing pyridoxine (vitamin B6) deficiency, preventable with prophylactic pyridoxine supplementation.

Carbohydrate Metabolism

7 questions
9

In a biochemistry practical, students measure the activity of glycolytic enzymes under varying ATP and citrate concentrations in a liver homogenate. The enzyme catalyzing the rate-limiting and most highly regulated step of glycolysis is allosterically inhibited by high ATP and citrate, and activated by AMP and fructose-2,6-bisphosphate. Which enzyme is being studied?

  1. APyruvate kinase
  2. BHexokinase
  3. CAldolase
  4. DPhosphofructokinase-1
Show answer
Answer: D. Phosphofructokinase-1

Phosphofructokinase-1 (PFK-1) catalyzes the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate and is the key rate-limiting, irreversible, and most tightly regulated step of glycolysis. It is allosterically inhibited by high ATP and citrate (signaling adequate energy and citric acid cycle intermediates) and activated by AMP (signaling low energy) and fructose-2,6-bisphosphate (the key signal for fed-state glycolysis generated by phosphofructokinase-2 in response to insulin). Hexokinase and pyruvate kinase are also regulated but are not the primary rate-limiting step of the pathway.

  • A. Pyruvate kinase catalyzes the final step and is regulated by fructose-1,6-bisphosphate and ATP, but it is not the primary control point described here.
  • B. Hexokinase catalyzes the first committed step of glucose metabolism and is inhibited by its product glucose-6-phosphate, but it is not the rate-limiting enzyme of glycolysis itself.
  • C. Aldolase catalyzes a reversible cleavage step and is not allosterically regulated by ATP, citrate, AMP, or fructose-2,6-bisphosphate.
Educational objective: Phosphofructokinase-1 is the rate-limiting enzyme of glycolysis, allosterically inhibited by ATP and citrate and activated by AMP and fructose-2,6-bisphosphate.
10

A 4-month-old infant is brought with a protuberant abdomen and poor weight gain. Examination shows marked hepatomegaly, a doll-like round face, and thin limbs. The infant becomes irritable and tremulous if a feed is delayed beyond 3 hours. Investigations show venous glucose 32 mg/dL, lactate 48 mg/dL (normal 4.5-19.8 mg/dL), triglycerides 420 mg/dL, and uric acid 9.8 mg/dL. Administration of glucagon fails to raise the blood glucose. Which of the following enzyme defects best explains this presentation?

  1. AMuscle phosphorylase
  2. BDebranching enzyme (amylo-1,6-glucosidase)
  3. CGlucose-6-phosphatase
  4. DFructose-1,6-bisphosphatase
Show answer
Answer: C. Glucose-6-phosphatase

Von Gierke disease (GSD type I) is caused by deficiency of glucose-6-phosphatase, the enzyme that catalyzes the final step of both glycogenolysis and gluconeogenesis by converting glucose-6-phosphate to free glucose. Loss of this step causes severe fasting hypoglycemia unresponsive to glucagon, along with accumulation of glucose-6-phosphate that is shunted into glycolysis (lactic acidosis), the pentose phosphate pathway (hyperuricemia from increased purine degradation), and lipogenesis (hypertriglyceridemia). The massive hepatomegaly results from glycogen and fat accumulation in hepatocytes, and the doll-like facies with thin limbs and truncal fat is a classic described phenotype. Management includes frequent feeds and uncooked cornstarch to prevent hypoglycemia.

  • A. Muscle phosphorylase deficiency (McArdle disease, GSD V) causes exercise intolerance without fasting hypoglycemia, since it does not affect hepatic glucose output.
  • B. Debranching enzyme deficiency (GSD III, Cori disease) causes milder hypoglycemia with normal or mildly elevated lactate and responds partially to glucagon after a short fast.
  • D. Fructose-1,6-bisphosphatase deficiency impairs gluconeogenesis only, sparing glycogenolysis, so hypoglycemia is typically precipitated by fasting or illness rather than being this severe and glycogen-storage-predominant.
Educational objective: Von Gierke disease (GSD type I, glucose-6-phosphatase deficiency) presents with glucagon-unresponsive fasting hypoglycemia, hepatomegaly, lactic acidosis, hyperuricemia, and hyperlipidemia.
11

A 19-year-old man complains of painful cramping in his calves and thighs whenever he sprints or lifts heavy weights, forcing him to stop after a few minutes. He notices that if he slows down and rests briefly at the onset of pain, he can then continue the same activity with less discomfort, a phenomenon he calls his 'second wind.' After a vigorous workout, his urine turns dark brown. Serum creatine kinase is markedly elevated at rest. Which of the following best explains the biochemical basis of his exercise-induced symptoms?

  1. ADeficient acid alpha-glucosidase causing lysosomal glycogen accumulation in muscle
  2. BDeficient muscle carnitine palmitoyltransferase II impairing fatty acid oxidation
  3. CDeficient hepatic glucose-6-phosphatase impairing glucose release during exertion
  4. DInability to mobilize muscle glycogen due to deficient myophosphorylase activity
Show answer
Answer: D. Inability to mobilize muscle glycogen due to deficient myophosphorylase activity

McArdle disease (GSD type V) results from deficiency of myophosphorylase, the muscle-specific isoform of glycogen phosphorylase, which prevents glycogenolysis in skeletal muscle during anaerobic, high-intensity exercise. This causes early muscle fatigue, cramps, and myoglobinuria due to rhabdomyolysis, with a resting elevated CK reflecting chronic muscle membrane leakage. The 'second wind' phenomenon occurs because, after brief rest, increased blood flow delivers free fatty acids and glucose from the circulation that can be used aerobically, bypassing the defective glycogenolytic pathway. Liver glycogen metabolism and blood glucose maintenance are unaffected, so fasting hypoglycemia does not occur.

  • A. Acid alpha-glucosidase deficiency (Pompe disease) causes a lysosomal storage myopathy with progressive weakness and cardiomegaly in infancy, not isolated exercise-induced cramping with a second-wind phenomenon.
  • B. Carnitine palmitoyltransferase II deficiency impairs long-chain fatty acid oxidation and classically causes myalgia with prolonged, low-intensity exercise or fasting, not the second-wind phenomenon seen with glycogenolytic defects.
  • C. Hepatic glucose-6-phosphatase deficiency causes fasting hypoglycemia (Von Gierke disease), not exercise-induced muscle cramps with a normal fasting glucose.
Educational objective: McArdle disease (GSD V, myophosphorylase deficiency) causes exercise-induced cramps, myoglobinuria, and a characteristic 'second wind' phenomenon due to inability to break down muscle glycogen anaerobically.
12

A 5-month-old infant presents with generalized hypotonia ('floppy baby'), macroglossia, and feeding difficulty. Examination reveals massive cardiomegaly with signs of heart failure, and chest X-ray shows a markedly enlarged cardiac silhouette. Skeletal muscle biopsy shows vacuoles filled with glycogen that stain positively with periodic acid-Schiff and are not digested by diastase, consistent with lysosomal accumulation. Which of the following enzyme defects is responsible for this condition?

  1. ABranching enzyme (amylo-1,4 to 1,6-transglucosidase)
  2. BAlpha-1,4-glucosidase (acid maltase)
  3. CLiver phosphorylase
  4. DPhosphoglucomutase
Show answer
Answer: B. Alpha-1,4-glucosidase (acid maltase)

Pompe disease (GSD type II) is caused by deficiency of lysosomal acid alpha-glucosidase (acid maltase), the enzyme responsible for degrading glycogen that has been taken up into lysosomes by autophagy. Because the defect is lysosomal rather than cytoplasmic, glycogen accumulates within lysosomes throughout the body, producing a distinctive vacuolar, diastase-resistant PAS-positive pattern on biopsy. The infantile form presents with severe hypotonia, macroglossia, and massive cardiomegaly leading to early heart failure and death if untreated, distinguishing it from other GSDs that mainly affect liver or muscle cytoplasmic glycogen metabolism. Enzyme replacement therapy with recombinant alglucosidase alfa is the mainstay of treatment.

  • A. Branching enzyme deficiency causes GSD type IV (Andersen disease), characterized by progressive liver cirrhosis, not a lysosomal storage pattern with cardiomegaly.
  • C. Liver phosphorylase deficiency causes GSD type VI, a mild hepatic glycogenosis with hepatomegaly and mild hypoglycemia, without lysosomal storage or cardiomegaly.
  • D. Phosphoglucomutase deficiency is an extremely rare cause of glycogen storage disease affecting glycolysis/glycogenolysis interconversion, not a lysosomal enzyme defect.
Educational objective: Pompe disease (GSD II) is due to lysosomal acid alpha-glucosidase deficiency, causing lysosomal glycogen accumulation with infantile hypotonia and massive cardiomegaly.
13

A 10-day-old breastfed neonate develops vomiting, poor feeding, lethargy, and jaundice. Examination reveals hepatomegaly and bilateral cataracts. Blood culture grows Escherichia coli, raising suspicion of neonatal sepsis. Urine testing shows a strongly positive reducing substance test but a negative glucose oxidase strip test. Which of the following enzyme deficiencies is most likely responsible for this infant's condition?

  1. AUDP-galactose-4-epimerase
  2. BAldose reductase
  3. CGalactokinase
  4. DGalactose-1-phosphate uridyltransferase
Show answer
Answer: D. Galactose-1-phosphate uridyltransferase

Classic galactosemia is caused by deficiency of galactose-1-phosphate uridyltransferase (GALT), which leads to accumulation of galactose-1-phosphate in the liver, kidney, brain, and lens. This causes hepatomegaly, jaundice, cataracts, vomiting, failure to thrive, and a predisposition to E. coli sepsis, classically attributed to impaired neutrophil function and the favorable growth of E. coli in a galactose-rich environment. The urine reducing substance test is positive (galactose reduces copper sulfate) while the glucose-specific oxidase test is negative, confirming a non-glucose reducing sugar. Because galactosemia is a treatable cause of neonatal illness and blindness, it is included in many newborn screening programs, with treatment being lifelong exclusion of galactose and lactose from the diet.

  • A. UDP-galactose-4-epimerase deficiency is usually a rare, often clinically mild or peripheral form of galactosemia and is not the classic severe presentation described here.
  • B. Aldose reductase catalyzes glucose to sorbitol conversion in the polyol pathway and is unrelated to galactose metabolism or this clinical picture.
  • C. Galactokinase deficiency causes isolated cataracts without hepatomegaly, sepsis risk, or significant systemic toxicity, since galactose-1-phosphate does not accumulate.
Educational objective: Classic galactosemia results from galactose-1-phosphate uridyltransferase deficiency, presenting with hepatomegaly, cataracts, and E. coli sepsis risk in neonates, detectable on newborn screening.
14

A 7-month-old infant, previously well on breast milk alone, is started on fruit juices and commercial weaning foods. Within days, the parents notice recurrent vomiting, poor feeding, sweating, and episodes of lethargy after these feeds, along with progressive abdominal distension. Investigations during a symptomatic episode show blood glucose 38 mg/dL and significant hypophosphatemia. The infant improves markedly when fructose- and sucrose-containing foods are withdrawn. Deficiency of which enzyme best explains this presentation?

  1. AFructokinase
  2. BAldolase B
  3. CSucrase-isomaltase
  4. DFructose-1,6-bisphosphatase
Show answer
Answer: B. Aldolase B

Hereditary fructose intolerance is caused by deficiency of aldolase B, the liver enzyme that cleaves fructose-1-phosphate in the second step of fructose metabolism. Fructose-1-phosphate accumulates after ingestion of fructose or sucrose (which is hydrolyzed to fructose and glucose), trapping hepatic phosphate and causing severe hypoglycemia, hypophosphatemia, and inhibition of both glycogenolysis and gluconeogenesis. Symptoms classically appear when fructose-containing foods such as fruit juices or sweetened weaning foods are introduced, as seen in this previously asymptomatic breastfed infant. Chronic exposure leads to vomiting, failure to thrive, hepatomegaly, and renal tubular dysfunction, and management requires strict dietary exclusion of fructose, sucrose, and sorbitol.

  • A. Fructokinase deficiency (essential fructosuria) is a benign, asymptomatic condition with fructose appearing in urine but no toxic metabolite accumulation.
  • C. Sucrase-isomaltase deficiency causes carbohydrate malabsorption with osmotic diarrhea after sucrose ingestion, not hypoglycemia or hypophosphatemia.
  • D. Fructose-1,6-bisphosphatase deficiency impairs gluconeogenesis and causes fasting hypoglycemia with lactic acidosis, but it is not specifically triggered by fructose/sucrose ingestion in the same way and does not cause marked hypophosphatemia from fructose-1-phosphate trapping.
Educational objective: Hereditary fructose intolerance is due to aldolase B deficiency, causing fructose-1-phosphate accumulation with hypoglycemia and hypophosphatemia after fructose or sucrose ingestion.
15

A 52-year-old man with type 2 diabetes mellitus on metformin undergoes routine follow-up. His self-monitored blood glucose readings over the past 2 months have consistently ranged between 180 and 240 mg/dL. However, his HbA1c is reported as 5.8%, suggesting good glycemic control. He was diagnosed 3 weeks ago with hereditary spherocytosis presenting as hemolytic anemia, for which he has required two blood transfusions. Which of the following best explains the discrepancy between his HbA1c and his actual glycemic control?

  1. ATransfused blood contains hemoglobin with a higher oxygen affinity
  2. BReduced red cell lifespan limits the time available for hemoglobin glycation
  3. CHemolysis causes increased renal clearance of glycated hemoglobin
  4. DMetformin directly inhibits the non-enzymatic glycation of hemoglobin
Show answer
Answer: B. Reduced red cell lifespan limits the time available for hemoglobin glycation

HbA1c reflects the average blood glucose concentration over the preceding 8-12 weeks because glucose binds irreversibly and non-enzymatically to hemoglobin over the lifespan of the red blood cell, roughly 120 days. In conditions with increased red cell turnover, such as hemolytic anemias (e.g., hereditary spherocytosis) or after blood transfusion, red cells are destroyed and replaced prematurely, so there is less time for glycation to occur, leading to a falsely low HbA1c despite poor actual glycemic control. This is a well-recognized limitation of HbA1c, and in such patients fructosamine (reflecting glycation of serum proteins over 2-3 weeks) or direct glucose monitoring is preferred for assessing control. Clinicians must recognize this discordance to avoid under-treating hyperglycemia.

  • A. Oxygen affinity of transfused hemoglobin is unrelated to the measurement or interpretation of glycated hemoglobin levels.
  • C. Glycated hemoglobin is not renally cleared; it is removed from circulation only when the red cell itself is destroyed.
  • D. Metformin acts mainly by reducing hepatic gluconeogenesis and improving insulin sensitivity; it does not inhibit hemoglobin glycation chemistry.
Educational objective: HbA1c underestimates true glycemic control in conditions of increased red cell turnover (hemolysis, recent transfusion, blood loss) because glycation requires time proportional to red cell lifespan.
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Lipid Metabolism

7 questions
16

A 32-year-old man presents with yellowish nodular deposits over both Achilles tendons and the extensor tendons of his hands, along with a grayish-white ring around the corneal periphery. He reports that his father died of a myocardial infarction at age 38. His fasting lipid profile shows total cholesterol of 410 mg/dL and LDL cholesterol of 340 mg/dL, with normal triglycerides. He is heterozygous for a mutation affecting lipoprotein metabolism. Which of the following best describes the underlying molecular defect in this disorder?

  1. ADeficiency of apolipoprotein C-II needed for lipoprotein lipase activation
  2. BDefective apolipoprotein B-48 impairing chylomicron assembly in the intestine
  3. CDefective LDL receptor leading to impaired hepatic clearance of LDL particles
  4. DDeficiency of lipoprotein lipase impairing chylomicron and VLDL catabolism
Show answer
Answer: C. Defective LDL receptor leading to impaired hepatic clearance of LDL particles

Familial hypercholesterolemia is an autosomal dominant disorder most commonly caused by mutations in the gene encoding the LDL receptor, which impair hepatic uptake and clearance of circulating LDL particles. This leads to markedly elevated LDL cholesterol from birth, with heterozygotes typically showing total cholesterol in the 300-500 mg/dL range and homozygotes showing even higher levels with onset of disease in childhood. The clinical hallmarks are tendon xanthomas (classically Achilles and extensor tendons of the hand), corneal arcus, and premature coronary artery disease, often with a strong family history of early myocardial infarction. Diagnosis is clinical and supported by lipid profile and family history, with statins being first-line therapy, and PCSK9 inhibitors or LDL apheresis reserved for severe or homozygous cases.

  • A. Apolipoprotein C-II deficiency similarly causes a chylomicronemia syndrome with severe hypertriglyceridemia, not the isolated hypercholesterolemia pattern seen here.
  • B. Apolipoprotein B-48 defects affect chylomicron formation and cause abetalipoproteinemia-related fat malabsorption, not elevated LDL cholesterol with tendon xanthomas.
  • D. Lipoprotein lipase deficiency causes type I hyperlipoproteinemia with massive chylomicronemia, severe hypertriglyceridemia, and eruptive xanthomas, not isolated high LDL with tendon xanthomas.
Educational objective: Familial hypercholesterolemia is caused by LDL receptor mutations impairing hepatic LDL clearance, producing markedly elevated LDL cholesterol, tendon xanthomas, and premature coronary artery disease.
17

A 58-year-old man is started on a lipid-lowering drug after a myocardial infarction. Six weeks later he complains of diffuse muscle aching in both thighs. Serum creatine kinase is 1850 U/L (normal up to 190 U/L). Which mechanism best explains the lipid-lowering action of the drug responsible for this adverse effect?

  1. AIrreversible binding to bile acids in the intestinal lumen
  2. BActivation of PPAR-alpha leading to increased lipoprotein lipase activity
  3. CInhibition of intestinal cholesterol absorption via NPC1L1 transporter blockade
  4. DCompetitive inhibition of HMG-CoA reductase, reducing hepatic cholesterol synthesis
Show answer
Answer: D. Competitive inhibition of HMG-CoA reductase, reducing hepatic cholesterol synthesis

Statins competitively inhibit HMG-CoA reductase, the rate-limiting enzyme of hepatic cholesterol synthesis, which upregulates LDL receptor expression and lowers plasma LDL-cholesterol. This same class is the most common cause of drug-induced myopathy, ranging from myalgia to rhabdomyolysis, due to reduced synthesis of isoprenoid intermediates needed for normal muscle cell function. Risk is increased with higher doses, hypothyroidism, renal impairment, and concomitant fibrates or CYP3A4 inhibitors. Markedly elevated CK with myalgia in a patient on this therapy is a classic presentation of statin-induced myopathy.

  • A. This describes bile acid sequestrants (e.g., cholestyramine), which are not associated with myopathy.
  • B. This describes fibrates, which can cause myopathy but act through PPAR-alpha, not HMG-CoA reductase inhibition.
  • C. This describes ezetimibe, which does not cause myopathy of this severity.
Educational objective: Statins lower cholesterol by inhibiting HMG-CoA reductase and are the most common cause of drug-induced myopathy with elevated creatine kinase.
18

A 10-year-old boy presents with recurrent episodes of abdominal pain and eruptive xanthomas over the trunk. Fasting serum triglycerides are 2800 mg/dL with lactescent (milky) plasma, total cholesterol is mildly elevated, and the plasma shows a creamy supernatant layer with a clear infranatant on standing overnight at 4°C. Lipoprotein lipase activity is markedly reduced. Which Fredrickson phenotype does this profile represent?

  1. AType IIa
  2. BType IV
  3. CType III
  4. DType I
Show answer
Answer: D. Type I

Type I hyperlipoproteinemia (familial chylomicronemia) results from lipoprotein lipase deficiency or deficiency of its cofactor ApoC-II, causing massive accumulation of chylomicrons in plasma. This produces markedly elevated triglycerides with a creamy chylomicron layer on top and a clear infranatant on refrigeration, along with eruptive xanthomas and recurrent pancreatitis-like abdominal pain in childhood. Total cholesterol is only mildly raised because chylomicrons carry mostly triglyceride. This pattern is distinct from the isolated LDL elevation of Type IIa or the mixed patterns of Type III and IV. This is a mechanism-based explanation correlating lipoprotein biochemistry with clinical and gross plasma findings.

  • A. Type IIa shows isolated LDL elevation with a clear plasma sample, not chylomicronemia.
  • B. Type IV shows elevated VLDL with turbid but not creamy plasma and normal lipoprotein lipase activity.
  • C. Type III (dysbetalipoproteinemia) shows broad beta band on electrophoresis with palmar xanthomas, not a creamy chylomicron layer.
Educational objective: A creamy supernatant with clear infranatant on refrigerated plasma and markedly reduced lipoprotein lipase activity indicates Type I hyperlipoproteinemia (chylomicronemia).
19

A 6-month-old infant is brought with lethargy, hypoglycemia (blood glucose 32 mg/dL), and hepatomegaly after a prolonged fasting period due to a febrile illness. Urine ketones are absent despite the hypoglycemia. Serum free fatty acids are elevated, and plasma acylcarnitine profile shows markedly reduced free carnitine with elevated long-chain acylcarnitines. Which mechanism best explains the hypoketotic hypoglycemia in this infant?

  1. ADeficiency of malonyl-CoA decarboxylase in the cytosol
  2. BDeficiency of acetyl-CoA carboxylase in hepatocytes
  3. CImpaired transport of long-chain fatty acids into the mitochondrial matrix
  4. DReduced activity of hormone-sensitive lipase in adipose tissue
Show answer
Answer: C. Impaired transport of long-chain fatty acids into the mitochondrial matrix

Carnitine is required to shuttle long-chain fatty acyl groups across the inner mitochondrial membrane via the carnitine-acylcarnitine translocase system (carnitine palmitoyltransferase I and II). Deficiency of carnitine or of CPT enzymes blocks beta-oxidation of long-chain fatty acids, so during fasting the liver cannot generate acetyl-CoA for ketogenesis, producing hypoketotic hypoglycemia despite high circulating free fatty acids. This is confirmed by low free carnitine with accumulation of long-chain acylcarnitines upstream of the block. Hepatomegaly results from fat accumulation in hepatocytes.

  • A. Malonyl-CoA decarboxylase deficiency is a rare distinct disorder affecting malonyl-CoA metabolism, not the primary carnitine shuttle defect described.
  • B. Acetyl-CoA carboxylase deficiency would not explain the specific acylcarnitine accumulation pattern seen in this infant.
  • D. Reduced hormone-sensitive lipase would decrease free fatty acid release, not cause the elevated free fatty acids seen here.
Educational objective: Carnitine or CPT deficiency blocks mitochondrial entry of long-chain fatty acids, causing hypoketotic hypoglycemia with low free carnitine and elevated long-chain acylcarnitines during fasting.
20

A 24-year-old man with type 1 diabetes mellitus is brought to the emergency department with vomiting and altered sensorium after omitting his insulin doses for two days. Arterial blood gas shows pH 7.12 and bicarbonate 8 mEq/L, serum glucose is 480 mg/dL, and urine ketones are strongly positive. Which hepatic metabolic change is primarily responsible for the markedly increased ketone body production in this patient? </br>

  1. AIncreased activity of glucose-6-phosphatase releasing free glucose from glycogen
  2. BIncreased conversion of fatty acid-derived acetyl-CoA to acetoacetate due to low hepatic malonyl-CoA levels
  3. CIncreased activity of glycogen synthase in hepatocytes
  4. DIncreased conversion of pyruvate to oxaloacetate via pyruvate carboxylase
Show answer
Answer: B. Increased conversion of fatty acid-derived acetyl-CoA to acetoacetate due to low hepatic malonyl-CoA levels

Insulin deficiency and glucagon excess activate hormone-sensitive lipase, releasing free fatty acids that reach the liver in large amounts. Low insulin also lowers malonyl-CoA (via reduced acetyl-CoA carboxylase activity), relieving inhibition of carnitine palmitoyltransferase I, which allows unrestricted entry of fatty acyl-CoA into mitochondria for beta-oxidation. The resulting surge of acetyl-CoA exceeds the oxidative capacity of the TCA cycle and is diverted to ketone body synthesis (acetoacetate and beta-hydroxybutyrate), producing the anion-gap metabolic acidosis seen in diabetic ketoacidosis. This mechanism explains why ketogenesis, not glycogenolysis or gluconeogenesis directly, drives the acidosis.

  • A. Glycogenolysis contributes to hyperglycemia but does not directly drive ketone body formation.
  • C. Glycogen synthase activity is suppressed, not increased, in insulin deficiency and is unrelated to ketogenesis.
  • D. Increased pyruvate carboxylase activity supports gluconeogenesis and worsens hyperglycemia but is not the direct cause of ketogenesis.
Educational objective: In insulin deficiency, low hepatic malonyl-CoA disinhibits carnitine palmitoyltransferase I, driving fatty acid oxidation and excess acetyl-CoA into ketone body synthesis, causing diabetic ketoacidosis.
21

Which apolipoprotein acts as the essential cofactor that activates lipoprotein lipase on the capillary endothelium, enabling hydrolysis of triglycerides carried within chylomicrons and VLDL particles?

  1. AApoC-II
  2. BApoA-I
  3. CApoB-100
  4. DApoE
Show answer
Answer: A. ApoC-II

ApoC-II is a small apolipoprotein found on chylomicrons and VLDL that serves as the obligatory activating cofactor for lipoprotein lipase, the endothelial enzyme that hydrolyzes triglycerides into free fatty acids and glycerol for tissue uptake. Deficiency of ApoC-II or lipoprotein lipase itself causes severe hypertriglyceridemia (Type I hyperlipoproteinemia) despite normal enzyme protein being present. In contrast, ApoB-100 is the structural ligand for the LDL receptor, ApoA-I activates lecithin-cholesterol acyltransferase (LCAT) in HDL metabolism, and ApoE mediates receptor-mediated clearance of chylomicron and VLDL remnants.

  • B. ApoA-I is the major HDL apolipoprotein that activates LCAT for cholesterol esterification, not lipoprotein lipase.
  • C. ApoB-100 is the structural apolipoprotein of LDL that binds the LDL receptor, not the lipoprotein lipase activator.
  • D. ApoE mediates hepatic uptake of chylomicron and VLDL remnants via the LDL receptor and LRP, not lipoprotein lipase activation.
Educational objective: ApoC-II is the essential cofactor that activates lipoprotein lipase for hydrolysis of triglyceride-rich lipoproteins.
22

A 7-year-old boy presents with massive splenomegaly, bone pain in the long bones, and easy bruising. Complete blood count shows pancytopenia, and X-ray of the femur shows an Erlenmeyer flask deformity of the distal femur. Bone marrow aspirate reveals large cells with a crumpled-tissue-paper appearance of the cytoplasm on light microscopy. Which enzyme deficiency is responsible for this condition?

  1. AAcid sphingomyelinase
  2. BAlpha-galactosidase A
  3. CGlucocerebrosidase
  4. DHexosaminidase A
Show answer
Answer: C. Glucocerebrosidase

Gaucher disease is caused by deficiency of glucocerebrosidase (acid beta-glucosidase), the lysosomal enzyme that cleaves glucocerebroside into glucose and ceramide. Glucocerebroside accumulates in macrophages, producing characteristic Gaucher cells with a crumpled-tissue-paper cytoplasmic appearance, seen in bone marrow, spleen, and liver. Clinical features include massive hepatosplenomegaly, bone pain with Erlenmeyer flask deformity of the femur, pathologic fractures, and pancytopenia from marrow infiltration and hypersplenism. Type 1 (non-neuronopathic) is the most common form and is treatable with enzyme replacement therapy (imiglucerase).

  • A. Acid sphingomyelinase deficiency causes Niemann-Pick disease, with foam cells showing a zebra body pattern rather than crumpled-tissue-paper cytoplasm.
  • B. Alpha-galactosidase A deficiency causes Fabry disease, presenting with angiokeratomas and neuropathic pain, not Gaucher cells.
  • D. Hexosaminidase A deficiency causes Tay-Sachs disease, a neurodegenerative disorder with a cherry-red spot and no hepatosplenomegaly.
Educational objective: Gaucher disease results from glucocerebrosidase deficiency, causing glucocerebroside accumulation in macrophages that appear as crumpled-tissue-paper Gaucher cells, with bone pain and Erlenmeyer flask deformity.

Amino Acid & Protein Metabolism

6 questions
23

A 3-month-old infant is found on newborn screening to have a markedly elevated blood phenylalanine level with normal tyrosine. The infant is clinically well at present but is at risk of severe intellectual disability if untreated. The deficient enzyme normally converts phenylalanine to tyrosine using tetrahydrobiopterin as a cofactor. Lifelong dietary restriction of which amino acid, with supplementation of its hydroxylated product, is the mainstay of management?

  1. ATryptophan, with serotonin supplementation
  2. BTyrosine, with phenylalanine supplementation
  3. CPhenylalanine, with tyrosine supplementation
  4. DMethionine, with cysteine supplementation
Show answer
Answer: C. Phenylalanine, with tyrosine supplementation

Phenylketonuria results from deficiency of phenylalanine hydroxylase, which normally converts phenylalanine to tyrosine in the liver using tetrahydrobiopterin as a cofactor. Loss of this enzyme causes accumulation of phenylalanine and its toxic metabolites (phenylpyruvate, phenylacetate), which impair myelination and brain development. Since tyrosine synthesis from phenylalanine is blocked, tyrosine becomes a conditionally essential amino acid that must be supplied in the diet. Management is a low-phenylalanine diet with tyrosine supplementation, started as early as possible and continued lifelong.

  • A. Tryptophan metabolism is unrelated to the phenylalanine hydroxylase defect.
  • B. Tyrosine is the product, not the toxic substrate; it must be supplemented, not restricted.
  • D. Methionine/cysteine metabolism is affected in homocystinuria, not PKU.
Educational objective: Phenylketonuria is caused by phenylalanine hydroxylase deficiency and is managed with a phenylalanine-restricted, tyrosine-supplemented diet.
24

A 5-day-old neonate, born of a non-consanguineous marriage, develops poor feeding, lethargy, and a sweet caramel-like odor in the urine. Blood investigations show elevated leucine, isoleucine, and valine levels with their corresponding ketoacids. Which enzyme deficiency best explains this presentation?

  1. ABranched-chain alpha-ketoacid dehydrogenase
  2. BMedium-chain acyl-CoA dehydrogenase
  3. CMethylmalonyl-CoA mutase
  4. DPropionyl-CoA carboxylase
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 requiring thiamine pyrophosphate that normally decarboxylates the ketoacids derived from leucine, isoleucine, and valine. Its deficiency leads to accumulation of these three branched-chain amino acids and their ketoacids in blood and urine, producing the characteristic sweet, maple syrup-like odor. Untreated infants develop poor feeding, vomiting, lethargy, and encephalopathy within the first week of life. Diagnosis is confirmed by elevated plasma branched-chain amino acids and urinary ketoacids, and management involves dietary restriction of these amino acids.

  • B. This causes a fatty acid oxidation defect presenting with hypoketotic hypoglycemia, unrelated to branched-chain amino acids.
  • C. This causes methylmalonic acidemia, a different organic acidemia without the characteristic odor.
  • D. Deficiency of this enzyme causes propionic acidemia, presenting with metabolic acidosis, not a sweet urine odor.
Educational objective: Maple syrup urine disease is due to branched-chain alpha-ketoacid dehydrogenase deficiency, causing accumulation of leucine, isoleucine, and valine with a characteristic sweet urine odor.
25

A 40-year-old man presents with dark brownish-black discoloration of the sclerae and ear cartilage along with early-onset degenerative arthritis of the large joints. He mentions that his urine turns black after being left standing for a few hours, though it appears normal when freshly passed. Which enzyme deficiency is responsible for this condition?

  1. AHomogentisate 1,2-dioxygenase
  2. BTyrosinase
  3. CFumarylacetoacetate hydrolase
  4. D4-Hydroxyphenylpyruvate dioxygenase
Show answer
Answer: A. Homogentisate 1,2-dioxygenase

Alkaptonuria is an autosomal recessive disorder caused by deficiency of homogentisate 1,2-dioxygenase, an enzyme in the tyrosine degradation pathway that normally converts homogentisic acid to maleylacetoacetate. Accumulated homogentisic acid is excreted in urine, where it oxidizes and polymerizes on exposure to air, turning the urine black on standing. Chronic deposition of oxidized homogentisate pigment in connective tissue produces ochronosis, seen as bluish-black pigmentation of cartilage, sclera, and skin, along with premature degenerative arthropathy of large joints. Diagnosis is confirmed by detecting homogentisic acid in urine.

  • B. Tyrosinase deficiency causes albinism due to absent melanin synthesis, not black urine.
  • C. This enzyme deficiency causes tyrosinemia type I, presenting with liver failure and renal tubular dysfunction, not ochronosis.
  • D. Deficiency of this enzyme causes tyrosinemia type III, not the black urine and ochronosis of alkaptonuria.
Educational objective: Alkaptonuria results from homogentisate 1,2-dioxygenase deficiency, causing urine that darkens on standing and ochronotic arthropathy.
26

A 14-year-old boy presents with sudden-onset left hemiparesis. On examination he is tall and thin with long fingers, and slit-lamp examination reveals downward lens dislocation. Laboratory studies show markedly elevated plasma homocysteine and methionine. Deficiency of which enzyme is most likely responsible for the thrombotic event in this patient?

  1. AMethylmalonyl-CoA mutase
  2. BCystathionine beta-synthase
  3. CMethionine synthase
  4. DCystathionase
Show answer
Answer: B. Cystathionine beta-synthase

Classic homocystinuria is most commonly caused by cystathionine beta-synthase deficiency, a pyridoxal phosphate-dependent enzyme that converts homocysteine to cystathionine. Its deficiency causes accumulation of homocysteine and methionine, producing the clinical triad of marfanoid habitus, downward lens dislocation (ectopia lentis), and thromboembolic events. Elevated homocysteine damages vascular endothelium and promotes a prothrombotic state by increasing tissue factor expression and impairing nitric oxide-mediated vasodilation, predisposing to arterial and venous thrombosis including stroke. Management includes pyridoxine, folate, and vitamin B12 supplementation along with dietary methionine restriction.

  • A. This causes methylmalonic acidemia, not the ectopia lentis and marfanoid habitus seen here.
  • C. Methionine synthase deficiency causes homocystinuria with low, not elevated, methionine levels.
  • D. Cystathionase deficiency causes cystathioninuria, a benign condition without thrombotic risk.
Educational objective: Cystathionine beta-synthase deficiency causes homocystinuria, in which elevated homocysteine predisposes to vascular endothelial damage and thromboembolism.
27

A 55-year-old man with cirrhosis due to chronic alcohol use presents with confusion, asterixis, and a flapping tremor. Serum ammonia is elevated at 140 µmol/L. In astrocytes, excess ammonia is detoxified predominantly by combining with glutamate to form a compound that causes astrocyte swelling and contributes to cerebral edema. Which compound is this?

  1. AGlutamine
  2. BAlpha-ketoglutarate
  3. CGamma-aminobutyric acid
  4. DGlutathione
Show answer
Answer: A. Glutamine

In the brain, astrocytes lack a complete urea cycle and instead detoxify ammonia primarily through glutamine synthetase, which combines ammonia with glutamate to form glutamine. In hepatic failure, chronically elevated ammonia drives excessive glutamine synthesis, and the resulting osmotic accumulation of glutamine within astrocytes causes cellular swelling, a key contributor to low-grade cerebral edema in hepatic encephalopathy. Glutamine also acts as a Trojan horse, being transported into mitochondria where it is hydrolyzed back to ammonia, perpetuating mitochondrial dysfunction and oxidative stress. This mechanism explains the correlation between hyperammonemia and the neuropsychiatric manifestations of hepatic encephalopathy.

  • B. Alpha-ketoglutarate is consumed, not formed, in the glutamate dehydrogenase reaction that generates glutamate for ammonia fixation.
  • C. GABA is an inhibitory neurotransmitter whose activity is altered in hepatic encephalopathy but is not the direct ammonia-detoxifying product.
  • D. Glutathione is an antioxidant tripeptide unrelated to ammonia detoxification in astrocytes.
Educational objective: Astrocytes detoxify ammonia via glutamine synthetase, and resulting glutamine accumulation causes astrocyte swelling in hepatic encephalopathy.
28

A 6-year-old child has generalized hypopigmentation of skin and hair since birth, along with photophobia, nystagmus, and reduced visual acuity. Fundus examination reveals a hypopigmented retina. Skin biopsy shows melanocytes present in normal numbers but unable to produce pigment. Deficiency of which enzyme is the most common cause of this condition?

  1. ATyrosinase
  2. BDopa decarboxylase
  3. CHomogentisate oxidase
  4. DPhenylalanine hydroxylase
Show answer
Answer: A. Tyrosinase

Oculocutaneous albinism type 1, the most common and severe form, results from deficiency or absence of tyrosinase, the rate-limiting enzyme that converts tyrosine to dopaquinone in the melanin synthesis pathway within melanocytes. Melanocytes are present in normal numbers but cannot synthesize melanin, resulting in generalized hypopigmentation of skin, hair, and eyes. Ocular findings include photophobia, nystagmus, reduced visual acuity, and a hypopigmented fundus due to lack of melanin in the retinal pigment epithelium and iris. The clinical spectrum ranges from complete absence of pigment (tyrosinase-negative) to partial pigment production (tyrosinase-positive), depending on residual enzyme activity.

  • B. Dopa decarboxylase is involved in catecholamine synthesis, not the primary defect in albinism.
  • C. This enzyme deficiency causes alkaptonuria, unrelated to pigmentation.
  • D. Phenylalanine hydroxylase deficiency causes phenylketonuria, which can cause mild hypopigmentation but not the classic albinism phenotype described.
Educational objective: Oculocutaneous albinism most commonly results from tyrosinase deficiency, blocking melanin synthesis despite normal melanocyte numbers.

Hemoglobin & Porphyrin Disorders

7 questions
29

A 28-year-old woman presents with severe colicky abdominal pain, vomiting, constipation, and new-onset hypertension and tachycardia. She was recently started on phenytoin for a seizure disorder following a febrile illness. Urine turns dark red-brown on exposure to sunlight. Which enzyme deficiency underlies this presentation, and the use of which drug most likely precipitated this acute attack?

  1. AUroporphyrinogen decarboxylase; phenytoin caused photosensitization
  2. BALA dehydratase; phenytoin caused direct enzyme inhibition
  3. CFerrochelatase; phenytoin inhibited heme synthesis directly
  4. DPorphobilinogen deaminase; phenytoin induced hepatic ALA synthase
Show answer
Answer: D. Porphobilinogen deaminase; phenytoin induced hepatic ALA synthase

Acute intermittent porphyria is caused by a partial deficiency of porphobilinogen deaminase (hydroxymethylbilane synthase), the third enzyme of the heme synthesis pathway. Drugs like phenytoin that induce hepatic cytochrome P450 enzymes increase the demand for heme, which upregulates ALA synthase, the rate-limiting enzyme of the pathway, through loss of heme-mediated negative feedback. This drives further accumulation of the neurotoxic precursors ALA and porphobilinogen proximal to the partial enzyme block, precipitating an acute attack with abdominal pain, autonomic instability, and neuropsychiatric symptoms. Classic precipitants include barbiturates, phenytoin, sulfonamides, alcohol, and starvation, and management includes intravenous hemin and glucose to suppress ALA synthase induction.

  • A. Uroporphyrinogen decarboxylase (uroporphyrinogen decarboxylase/cosynthase) deficiency causes porphyria cutanea tarda with skin photosensitivity, not acute neurovisceral attacks.
  • B. ALA dehydratase deficiency porphyria is extremely rare, and phenytoin acts by inducing ALA synthase, not by directly inhibiting an enzyme.
  • C. Ferrochelatase deficiency causes erythropoietic protoporphyria, characterized by painful photosensitivity, not acute abdominal attacks.
Educational objective: Acute intermittent porphyria results from porphobilinogen deaminase deficiency, and enzyme-inducing drugs like phenytoin precipitate attacks by upregulating hepatic ALA synthase.
30

A 4-year-old boy from a slum area presents with abdominal pain, irritability, and regression of acquired language milestones. He is found to have a habit of eating paint flakes from the walls. Peripheral smear shows basophilic stippling of erythrocytes. Blood lead level is markedly elevated. Which of the following enzymes is inhibited early and most severely, leading to accumulation of its substrate in blood and urine?

  1. ACoproporphyrinogen oxidase
  2. BALA dehydratase
  3. CUroporphyrinogen decarboxylase
  4. DPorphobilinogen deaminase
Show answer
Answer: B. ALA dehydratase

Lead has a high affinity for sulfhydryl groups and inhibits two key enzymes of heme synthesis: ALA dehydratase (most sensitive) and ferrochelatase. Inhibition of ALA dehydratase causes accumulation of delta-aminolevulinic acid (ALA) in blood and urine, a hallmark biochemical finding in lead poisoning. Ferrochelatase inhibition causes accumulation of protoporphyrin IX, which combines with zinc to give elevated zinc protoporphyrin, another diagnostic marker. Basophilic stippling results from residual ribosomal RNA aggregates due to disrupted RNA degradation.

  • A. Not the principal target of lead; its deficiency causes hereditary coproporphyria.
  • C. Not significantly affected by lead; its deficiency causes porphyria cutanea tarda.
  • D. Not the primary enzyme inhibited by lead; it is inhibited by acute intermittent porphyria mutations, not lead toxicity.
Educational objective: Lead poisoning inhibits ALA dehydratase and ferrochelatase, causing accumulation of ALA and zinc protoporphyrin with basophilic stippling on peripheral smear.
31

A 48-year-old man with a history of chronic alcohol use presents with painful blistering and fragility of the skin over the dorsum of both hands after sun exposure, along with hyperpigmentation and hypertrichosis of the face. Urine sent to the laboratory turns dark on standing. Which enzyme deficiency is responsible for this presentation?

  1. AFerrochelatase
  2. BUroporphyrinogen decarboxylase
  3. CPorphobilinogen deaminase
  4. DALA synthase
Show answer
Answer: B. Uroporphyrinogen decarboxylase

Porphyria cutanea tarda is caused by deficiency of uroporphyrinogen decarboxylase (uroporphyrinogen III decarboxylase), leading to accumulation of uroporphyrin, which is photosensitizing and causes blistering skin lesions on sun-exposed areas. It is the most common porphyria and is often precipitated or unmasked by alcohol, estrogen use, hepatitis C, or iron overload. Unlike the acute porphyrias, there is no significant abdominal pain or neuropsychiatric involvement because porphobilinogen and ALA levels are not markedly elevated. Urine darkens on exposure to light and air due to oxidation of porphyrins.

  • A. Ferrochelatase deficiency causes erythropoietic protoporphyria, presenting with acute painful photosensitivity without blistering, typically in childhood.
  • C. Deficiency causes acute intermittent porphyria, which presents with abdominal pain and neuropsychiatric symptoms but no photosensitivity.
  • D. ALA synthase is upregulated, not deficient, in porphyrias; its deficiency causes a rare sideroblastic anemia, not porphyria.
Educational objective: Porphyria cutanea tarda results from uroporphyrinogen decarboxylase deficiency, causing photosensitive blistering skin lesions without acute neurovisceral symptoms.
32

A 6-year-old boy of African descent presents with severe bone pain and a hemoglobin of 7.2 g/dL. Hemoglobin electrophoresis confirms homozygous HbS disease. Which of the following best describes the molecular basis for polymerization of deoxygenated hemoglobin in this condition?

  1. ASubstitution of glutamate by valine at position 6 of the beta chain creates a hydrophobic patch that binds a complementary site on another HbS molecule
  2. BA point mutation increases the affinity of hemoglobin for oxygen, promoting tissue hypoxia
  3. CA frameshift mutation in the beta-globin gene causes absence of beta-chain synthesis
  4. DA nonsense mutation in the beta-globin gene produces a truncated, unstable globin chain
Show answer
Answer: A. Substitution of glutamate by valine at position 6 of the beta chain creates a hydrophobic patch that binds a complementary site on another HbS molecule

Sickle cell disease results from a single point mutation (GAG to GTG) in the sixth codon of the beta-globin gene, substituting hydrophobic valine for glutamate at position 6. This creates a hydrophobic patch on the surface of deoxygenated HbS that fits into a complementary hydrophobic pocket on an adjacent deoxy-HbS molecule, promoting polymerization into long fibers that distort the red cell into a sickle shape. Sickling is reversible with reoxygenation initially but repeated cycles cause irreversible membrane damage and hemolysis. This leads to vaso-occlusion, pain crises, and chronic hemolytic anemia.

  • B. HbS has normal or slightly decreased oxygen affinity; the pathology arises from polymerization on deoxygenation, not from increased oxygen affinity.
  • C. This describes the mechanism of beta-thalassemia, not sickle cell disease.
  • D. This describes a mechanism seen in some unstable hemoglobin variants, not HbS, which results from a missense mutation.
Educational objective: Sickle cell disease arises from a glutamate-to-valine substitution at position 6 of beta-globin, creating a hydrophobic site that drives HbS polymerization upon deoxygenation.
33

A 24-year-old asymptomatic woman undergoes a routine antenatal check-up. Her hemoglobin is 10.8 g/dL with a mean corpuscular volume of 68 fL and mean corpuscular hemoglobin of 21 pg. Serum iron and ferritin are normal. Hemoglobin electrophoresis shows HbA2 of 5.8% (normal less than 3.5%). Which of the following best explains the elevated HbA2 level in this patient?

  1. AIncreased synthesis of gamma-globin chains compensates for absent beta-globin production
  2. BReduced synthesis of beta-globin chains shifts alpha-chain pairing preferentially toward delta-globin chains
  3. CReduced iron incorporation into protoporphyrin IX increases delta-chain synthesis
  4. DDeletion of both alpha-globin gene loci reduces competition for delta chains
Show answer
Answer: B. Reduced synthesis of beta-globin chains shifts alpha-chain pairing preferentially toward delta-globin chains

In beta-thalassemia trait, one beta-globin allele has reduced output, decreasing beta-chain synthesis. Free alpha chains, which are produced in relative excess, then pair preferentially with the available delta chains to form more HbA2 (alpha2-delta2) and, to a lesser extent, with gamma chains to form HbF. This compensatory shift results in the characteristic elevated HbA2 (greater than 3.5%) seen in beta-thalassemia trait, a key diagnostic marker that distinguishes it from iron deficiency anemia, which also causes microcytosis but with normal or low HbA2.

  • A. Increased HbF synthesis does occur mildly but does not explain the specific rise in HbA2, which is the diagnostic hallmark.
  • C. Iron metabolism is unrelated to delta-chain globin gene expression or HbA2 levels.
  • D. Alpha-thalassemia reduces alpha-chain production and does not raise HbA2; it may even lower it.
Educational objective: Beta-thalassemia trait is diagnosed by an elevated HbA2 (greater than 3.5%) resulting from preferential pairing of excess alpha chains with delta chains.
34

A 35-year-old man presents with yellowish discoloration of the sclera for three days, along with dark urine and pale stools. Laboratory evaluation shows total bilirubin 8.4 mg/dL, direct (conjugated) bilirubin 6.1 mg/dL, alanine aminotransferase 620 U/L, and aspartate aminotransferase 540 U/L. Urine is positive for bilirubin but negative for urobilinogen on dipstick testing. Which of the following best explains this pattern of urine findings?

  1. AHemolysis increases unconjugated bilirubin, which is filtered unchanged by the glomerulus into urine
  2. BUnconjugated bilirubin is water-soluble and is excreted directly into urine without intestinal conversion
  3. CConjugated bilirubin is water-soluble and filtered by the kidney, while reduced bile flow to the intestine decreases urobilinogen formation
  4. DConjugated bilirubin is reabsorbed from the intestine and converted to urobilinogen entirely within the renal tubule
Show answer
Answer: C. Conjugated bilirubin is water-soluble and filtered by the kidney, while reduced bile flow to the intestine decreases urobilinogen formation

Conjugated bilirubin is water-soluble (unlike unconjugated bilirubin, which is bound to albumin and lipid-soluble) and can be filtered by the glomerulus and excreted in urine, producing bilirubinuria, a feature only of conjugated hyperbilirubinemia. Reduced bile flow into the intestine, due to hepatocellular or obstructive pathology, decreases the amount of bilirubin converted by gut bacteria into urobilinogen, which normally gets partly reabsorbed and excreted in urine; hence urine urobilinogen is low or absent. This pattern, with predominantly conjugated hyperbilirubinemia, bilirubinuria, and absent urobilinogen, points toward a hepatocellular or obstructive cause rather than hemolysis.

  • A. This describes hemolytic (unconjugated) hyperbilirubinemia, which does not cause bilirubinuria since unconjugated bilirubin is not filtered by the kidney.
  • B. Unconjugated bilirubin is not water-soluble and is never excreted in urine regardless of its serum level.
  • D. Urobilinogen is formed by gut bacterial action on bilirubin in the intestine, not within the renal tubule.
Educational objective: Bilirubinuria with low urine urobilinogen indicates conjugated hyperbilirubinemia from hepatocellular or obstructive liver disease, not hemolysis.
35

A newborn develops progressive jaundice in the first week of life with a total serum bilirubin of 24 mg/dL, almost entirely unconjugated, and develops lethargy with opisthotonic posturing. Liver biopsy shows no structural abnormality, and there is no evidence of hemolysis. The infant's enzyme activity for bilirubin conjugation is found to be completely absent. Which of the following conditions is most consistent with this presentation, as distinguished from a milder disorder of the same enzyme pathway?

  1. ADubin-Johnson syndrome
  2. BCrigler-Najjar syndrome type 1
  3. CGilbert syndrome
  4. DRotor syndrome
Show answer
Answer: B. Crigler-Najjar syndrome type 1

Crigler-Najjar syndrome type 1 results from complete absence of UDP-glucuronosyltransferase (UGT1A1) activity, causing severe unconjugated hyperbilirubinemia that can lead to kernicterus (manifesting as lethargy and opisthotonus) if untreated. Gilbert syndrome, in contrast, is a much milder condition with 20-30% residual UGT1A1 activity, causing mild unconjugated hyperbilirubinemia usually under 3 mg/dL that becomes apparent during stress, fasting, or illness, without neurological complications. Crigler-Najjar type 2 has partial enzyme deficiency and responds to phenobarbital, unlike type 1, which requires phototherapy and often liver transplantation.

  • A. Dubin-Johnson syndrome causes conjugated hyperbilirubinemia due to a canalicular transport defect, not unconjugated hyperbilirubinemia.
  • C. Gilbert syndrome causes only mild unconjugated hyperbilirubinemia and does not cause kernicterus.
  • D. Rotor syndrome also causes conjugated hyperbilirubinemia due to impaired hepatic storage and excretion, unrelated to UGT1A1 deficiency.
Educational objective: Crigler-Najjar syndrome type 1 is complete UGT1A1 deficiency causing severe unconjugated hyperbilirubinemia with risk of kernicterus, unlike the mild partial deficiency of Gilbert syndrome.

Acid-Base & Electrolytes

5 questions
36

A 19-year-old man is brought to the emergency department with vomiting and abdominal pain for 2 days. He is a known type 1 diabetic who ran out of insulin a week ago. His blood pressure is 100/64 mmHg, pulse is 112/min, and respiratory rate is 28/min with deep breathing. Arterial blood gas and serum electrolytes show: pH 7.18, HCO3- 8 mEq/L, Na+ 138 mEq/L, Cl- 96 mEq/L, K+ 5.2 mEq/L. What is the anion gap and its interpretation?

  1. AAnion gap 14 mEq/L, normal anion gap metabolic acidosis
  2. BAnion gap 42 mEq/L, high anion gap metabolic acidosis
  3. CAnion gap 34 mEq/L, high anion gap metabolic acidosis
  4. DAnion gap 20 mEq/L, high anion gap metabolic acidosis
Show answer
Answer: C. Anion gap 34 mEq/L, high anion gap metabolic acidosis

Anion gap is calculated as Na+ - (Cl- + HCO3-) = 138 - (96 + 8) = 34 mEq/L, which is markedly elevated above the normal range of 8-16 mEq/L. This confirms a high anion gap metabolic acidosis, consistent with diabetic ketoacidosis, where accumulation of beta-hydroxybutyrate and acetoacetate anions widens the gap. The low HCO3- and compensatory tachypnea (Kussmaul breathing) support this diagnosis. Other causes of high anion gap acidosis include lactic acidosis, uremia, and toxic alcohol ingestion (the mnemonic MUDPILES).

  • A. Incorrect arithmetic; this value would suggest a normal anion gap acidosis, which does not fit this clinical picture.
  • B. Incorrect arithmetic; this overestimates the gap.
  • D. Incorrect arithmetic; this underestimates the true gap.
Educational objective: Anion gap = Na+ - (Cl- + HCO3-); a value above 16 mEq/L indicates high anion gap metabolic acidosis, as seen in diabetic ketoacidosis.
37

A 6-year-old boy presents with failure to thrive and bony deformities. Investigations reveal: serum pH 7.30, serum HCO3- 16 mEq/L, serum K+ 3.0 mEq/L, urine pH 5.2 despite systemic acidosis, and generalized aminoaciduria, glucosuria, and phosphaturia on urine analysis. Serum uric acid is low. Which biochemical abnormality best differentiates this condition from distal renal tubular acidosis?

  1. AImpaired aldosterone-mediated potassium secretion with hyperkalemia
  2. BReduced ammonium excretion due to decreased renal mass
  3. CImpaired proximal tubular bicarbonate reabsorption with associated glucosuria, aminoaciduria, and phosphaturia
  4. DInability to secrete hydrogen ions in the distal tubule with urine pH persistently above 5.5
Show answer
Answer: C. Impaired proximal tubular bicarbonate reabsorption with associated glucosuria, aminoaciduria, and phosphaturia

Proximal (type II) RTA results from a defect in proximal tubular bicarbonate reabsorption, leading to bicarbonate wasting along with other proximal tubular functions such as glucose, amino acid, and phosphate reabsorption, producing a Fanconi-like picture. Because distal acidification remains intact, the urine can still be acidified to a pH below 5.5 once serum bicarbonate falls below the reduced renal threshold, as seen here with urine pH 5.2. In contrast, distal (type I) RTA shows an inability to acidify urine despite systemic acidosis, with urine pH persistently above 5.5. The presence of generalized proximal tubular dysfunction (glucosuria, aminoaciduria, phosphaturia) points specifically to type II RTA.

  • A. This describes type IV RTA, which is associated with hyperkalemia, not the hypokalemia seen in this case.
  • B. This is a feature of type IV RTA related to aldosterone deficiency or resistance, not proximal RTA.
  • D. This describes distal (type I) RTA, which is inconsistent with the urine pH of 5.2 and the proximal tubular wasting pattern seen here.
Educational objective: Proximal (type II) RTA causes bicarbonate wasting with generalized proximal tubular dysfunction and a urine pH that can fall below 5.5, unlike distal (type I) RTA where urine pH remains inappropriately above 5.5.
38

A 60-year-old man with small cell lung carcinoma is evaluated for lethargy and confusion. He is euvolemic on examination with no edema or signs of dehydration. Investigations show: serum Na+ 118 mEq/L, serum osmolality 250 mOsm/kg, urine osmolality 450 mOsm/kg, urine Na+ 45 mEq/L, normal renal, adrenal, and thyroid function, and he is not on diuretics. Which of the following best explains this biochemical profile?

  1. APrimary polydipsia causing dilutional hyponatremia
  2. BCerebral salt wasting syndrome
  3. CDiuretic-induced natriuresis with hypovolemic hyponatremia
  4. DSyndrome of inappropriate antidiuretic hormone secretion from the tumor
Show answer
Answer: D. Syndrome of inappropriate antidiuretic hormone secretion from the tumor

SIADH is characterized by euvolemic hyponatremia with inappropriately concentrated urine (urine osmolality greater than 100 mOsm/kg) relative to low serum osmolality, along with urine sodium typically above 40 mEq/L, in the absence of renal, adrenal, or thyroid dysfunction or diuretic use. Small cell lung carcinoma is a classic paraneoplastic cause due to ectopic ADH secretion, which causes water retention and dilutional hyponatremia without volume overload. The combination of low serum osmolality with an inappropriately high urine osmolality confirms failure to suppress ADH despite hypotonicity. Cerebral salt wasting is excluded because that condition causes hypovolemia with renal sodium loss, not euvolemia.

  • A. Primary polydipsia typically causes dilute urine (low urine osmolality), not concentrated urine as seen here.
  • B. Cerebral salt wasting causes hypovolemia with volume depletion, unlike the euvolemic state described here.
  • C. This patient is euvolemic, not hypovolemic, and is not on diuretics.
Educational objective: SIADH is diagnosed by euvolemic hyponatremia with inappropriately high urine osmolality (>100 mOsm/kg) and urine sodium (>40 mEq/L) relative to low serum osmolality, in the absence of renal, adrenal, or thyroid disease.
39

A 55-year-old man with chronic kidney disease on irregular dialysis presents with weakness. His ECG shows tall, peaked T waves, widened QRS complex, and loss of P waves. Serum potassium is 7.8 mEq/L. Which of the following is the most likely underlying mechanism for this biochemical abnormality in this patient?

  1. AReduced renal excretion of potassium due to decreased glomerular filtration rate
  2. BIncreased aldosterone secretion causing renal potassium retention
  3. CRedistribution of potassium from intracellular to extracellular compartment due to insulin deficiency
  4. DExcessive dietary potassium intake exceeding normal renal clearance
Show answer
Answer: A. Reduced renal excretion of potassium due to decreased glomerular filtration rate

In chronic kidney disease, the primary mechanism of hyperkalemia is decreased glomerular filtration rate leading to impaired renal excretion of potassium, since the kidney is the main route of potassium elimination. As GFR falls below approximately 10-15 mL/min/1.73m2, the kidneys can no longer adequately excrete the daily potassium load, especially when dialysis sessions are missed. The characteristic ECG changes of hyperkalemia progress from peaked T waves to widened QRS, flattened or absent P waves, and eventually a sine-wave pattern as potassium rises. Aldosterone levels are typically appropriately elevated as a compensatory response in CKD, not the primary cause of hyperkalemia.

  • B. Aldosterone would be expected to rise as a compensatory response to hyperkalemia and would promote potassium excretion, not retention; primary hyperaldosteronism causes hypokalemia.
  • C. Insulin deficiency is a recognized cause of hyperkalemia but is not the primary mechanism in this patient with CKD and missed dialysis.
  • D. Dietary excess alone rarely causes severe hyperkalemia in a person with normal renal function; here the primary defect is reduced excretory capacity.
Educational objective: In chronic kidney disease, hyperkalemia results primarily from reduced renal excretion due to decreased GFR, manifesting on ECG as peaked T waves progressing to QRS widening and P wave loss.
40

A 48-year-old woman undergoes routine health check-up and is found to have an incidentally elevated serum calcium. She denies bone pain, renal stones, or abdominal symptoms. Investigations show: serum calcium 11.6 mg/dL, serum phosphate 2.1 mg/dL, serum intact PTH 110 pg/mL (reference 10-65 pg/mL), serum alkaline phosphatase mildly elevated, and 24-hour urinary calcium elevated. Which of the following best explains this biochemical pattern?

  1. AVitamin D toxicity from excessive supplementation
  2. BPrimary hyperparathyroidism due to a parathyroid adenoma
  3. CFamilial hypocalciuric hypercalcemia due to calcium-sensing receptor mutation
  4. DHumoral hypercalcemia of malignancy due to PTH-related peptide secretion
Show answer
Answer: B. Primary hyperparathyroidism due to a parathyroid adenoma

Primary hyperparathyroidism, most commonly caused by a solitary parathyroid adenoma, presents with inappropriately elevated or high-normal intact PTH in the setting of hypercalcemia, along with hypophosphatemia due to PTH-mediated phosphate wasting in the proximal tubule. PTH increases bone resorption and renal calcium reabsorption while promoting urinary calcium excretion overall because the filtered load is increased, so urinary calcium is typically elevated rather than low. Mild elevation of alkaline phosphatase reflects increased osteoblastic activity secondary to PTH-driven bone turnover. This classic combination of high calcium, low phosphate, high PTH, and hypercalciuria distinguishes primary hyperparathyroidism from other causes of hypercalcemia.

  • A. Vitamin D toxicity causes hypercalcemia with suppressed PTH and typically elevated serum phosphate, not hypophosphatemia.
  • C. Familial hypocalciuric hypercalcemia presents with low urinary calcium excretion and only mildly elevated or normal PTH, unlike the elevated urinary calcium seen here.
  • D. In malignancy-associated hypercalcemia from PTHrP, intact PTH levels are suppressed, not elevated, because PTHrP does not cross-react in the standard intact PTH assay.
Educational objective: Primary hyperparathyroidism shows hypercalcemia with elevated intact PTH, hypophosphatemia, and increased urinary calcium excretion, distinguishing it from PTH-independent causes of hypercalcemia.

Hormones & Signal Transduction

3 questions
41

A researcher is studying signal transduction in hepatocytes exposed to two different hormones. Hormone A acts through a Gs-protein coupled receptor leading to adenylyl cyclase activation, while Hormone B acts through a Gq-protein coupled receptor leading to phospholipase C activation. Which pair of hormones correctly matches Hormone A and Hormone B respectively based on their classic second messenger pathways?

  1. AAngiotensin II and glucagon
  2. BGlucagon and angiotensin II
  3. CGlucagon and insulin
  4. DInsulin and angiotensin II
Show answer
Answer: B. Glucagon and angiotensin II

Glucagon binds a Gs-coupled receptor that activates adenylyl cyclase, raising intracellular cAMP, which activates protein kinase A to stimulate glycogenolysis and gluconeogenesis in the liver. Angiotensin II acts through the Gq-coupled AT1 receptor, activating phospholipase C, which hydrolyzes PIP2 into IP3 and diacylglycerol, releasing intracellular calcium and activating protein kinase C. This distinction is a classic example of how different G-protein subtypes couple receptors to distinct second messenger cascades. Insulin, in contrast, works through a receptor tyrosine kinase mechanism, not a G-protein coupled pathway.

  • A. This reverses the correct hormone-pathway pairing for Hormone A and Hormone B.
  • C. Insulin does not signal through the IP3/DAG pathway; it uses receptor tyrosine kinase signaling.
  • D. Insulin does not act through a Gs-coupled cAMP pathway; it signals via receptor tyrosine kinase.
Educational objective: Gs-coupled receptors (e.g., glucagon) signal via adenylyl cyclase and cAMP, while Gq-coupled receptors (e.g., angiotensin II) signal via phospholipase C generating IP3 and DAG.
42

A 35-year-old woman is administered a synthetic glucocorticoid for a severe allergic reaction. Which of the following best describes the primary mechanism by which this hormone exerts its biological effect at the cellular level?

  1. AIt binds a cell surface receptor coupled to a G-protein, activating adenylyl cyclase to generate cAMP
  2. BIt binds a receptor tyrosine kinase, triggering autophosphorylation and downstream MAP kinase activation
  3. CIt diffuses across the plasma membrane and binds a cytoplasmic or nuclear receptor, and the hormone-receptor complex then binds DNA to alter gene transcription
  4. DIt binds a ligand-gated ion channel, causing rapid influx of calcium ions into the cell
Show answer
Answer: C. It diffuses across the plasma membrane and binds a cytoplasmic or nuclear receptor, and the hormone-receptor complex then binds DNA to alter gene transcription

Glucocorticoids, being lipophilic steroid hormones, diffuse freely across the plasma membrane and bind to cytoplasmic or nuclear glucocorticoid receptors. The hormone-receptor complex undergoes a conformational change, dissociates from heat shock proteins, dimerizes, and translocates to the nucleus where it binds glucocorticoid response elements on DNA. This binding modulates the transcription of target genes, upregulating anti-inflammatory proteins such as lipocortin-1 and downregulating pro-inflammatory cytokines. This genomic mechanism explains the characteristic delayed onset but prolonged duration of action of steroid hormones compared to peptide hormones acting through membrane receptors.

  • A. This describes peptide hormone signaling (e.g., glucagon), not the mechanism of lipophilic steroid hormones.
  • B. This describes growth factor or insulin signaling through receptor tyrosine kinases, not steroid hormone action.
  • D. This describes neurotransmitter action at ligand-gated ion channels, not the genomic mechanism of steroid hormones.
Educational objective: Steroid hormones like glucocorticoids act via intracellular nuclear receptors that bind DNA response elements to directly regulate gene transcription.
43

A 30-year-old healthy man eats a carbohydrate-rich meal, leading to insulin release from pancreatic beta cells. In skeletal muscle and adipose tissue, which sequence of events best describes how insulin promotes increased cellular glucose uptake?

  1. AInsulin binds its receptor, activating adenylyl cyclase to increase cAMP, which phosphorylates and activates GLUT4 transporters already present in the membrane
  2. BInsulin binds its receptor tyrosine kinase, triggering autophosphorylation and activation of the PI3-kinase/Akt pathway, which stimulates translocation of GLUT4 transporters to the plasma membrane
  3. CInsulin binds a G-protein coupled receptor, activating phospholipase C, which increases intracellular calcium to stimulate GLUT4 synthesis
  4. DInsulin diffuses into the cell and binds a cytoplasmic receptor, which directly upregulates GLUT4 gene transcription
Show answer
Answer: B. Insulin binds its receptor tyrosine kinase, triggering autophosphorylation and activation of the PI3-kinase/Akt pathway, which stimulates translocation of GLUT4 transporters to the plasma membrane

Insulin binds the alpha subunits of its receptor, a receptor tyrosine kinase, causing autophosphorylation of the beta subunits and activation of intrinsic tyrosine kinase activity. This phosphorylates insulin receptor substrate (IRS) proteins, which recruit and activate PI3-kinase, generating PIP3 and activating Akt (protein kinase B). Activated Akt triggers translocation of GLUT4-containing vesicles from intracellular storage sites to the plasma membrane, increasing glucose uptake in skeletal muscle and adipose tissue within minutes. This rapid, non-genomic mechanism is distinct from the slower transcriptional effects insulin also has on metabolic enzyme gene expression.

  • A. Insulin receptor signaling does not involve adenylyl cyclase or cAMP; GLUT4 translocation is driven by the PI3-kinase/Akt pathway.
  • C. Insulin does not signal through a G-protein coupled receptor or phospholipase C pathway; this describes a different class of hormone receptor.
  • D. Insulin acts through a cell surface receptor tyrosine kinase, not an intracellular cytoplasmic receptor like steroid hormones.
Educational objective: Insulin stimulates glucose uptake by activating its receptor tyrosine kinase and the PI3-kinase/Akt pathway, which drives translocation of GLUT4 transporters to the plasma membrane in muscle and fat cells.

Molecular Biology & Genetics

5 questions
44

A 28-year-old man presents with fever, cough and weight loss. Sputum smear for acid-fast bacilli is negative, but the clinician wants rapid molecular confirmation of Mycobacterium tuberculosis and rifampicin resistance within two hours. The technique used amplifies a specific DNA segment exponentially using repeated cycles of denaturation, primer annealing and extension catalyzed by a thermostable enzyme. Which enzyme is essential for this amplification process?

  1. ARNA polymerase II
  2. BReverse transcriptase
  3. CTaq DNA polymerase
  4. DDNA ligase
Show answer
Answer: C. Taq DNA polymerase

Polymerase chain reaction (PCR) relies on a heat-stable DNA polymerase, classically Taq polymerase from Thermus aquaticus, to extend primers at high temperatures (around 72°C) across repeated thermal cycles without denaturing. This enzyme tolerates the 94-95°C denaturation step repeated for 25-35 cycles, allowing exponential doubling of the target DNA sequence each cycle. This principle underlies CBNAAT/Xpert MTB-RIF, which also uses real-time PCR to detect rpoB gene mutations for rifampicin resistance. Reverse transcriptase is used in RT-PCR to convert RNA to cDNA first, not in standard DNA amplification.

  • A. RNA polymerase II transcribes DNA into mRNA in eukaryotic cells and plays no role in PCR.
  • B. Reverse transcriptase converts RNA to DNA and is used only when the starting template is RNA (RT-PCR), not for amplifying existing DNA.
  • D. DNA ligase joins DNA fragments by forming phosphodiester bonds and is used in cloning, not in PCR amplification.
Educational objective: PCR amplifies specific DNA sequences exponentially using a thermostable DNA polymerase (Taq) and is the basis of rapid molecular diagnostics like CBNAAT for tuberculosis.
45

A 32-year-old female blood donor is tested and found reactive on a fourth-generation antigen-antibody combination assay for HIV during initial screening. According to NACO guidelines for confirmatory diagnosis before informing the patient, which test should be performed next?

  1. AA second and third rapid/ELISA test using different antigen or principle
  2. BCD4 cell count by flow cytometry
  3. CWestern blot for HIV antibodies
  4. DHIV-1 RNA viral load by real-time PCR
Show answer
Answer: A. A second and third rapid/ELISA test using different antigen or principle

NACO's HIV testing strategy for diagnosis requires a sample reactive on the first ELISA/rapid test to be retested with two additional assays based on different antigens or test principles, following the three-test strategy for diagnosis. A sample is labelled HIV-positive only when it is reactive on all three different assays. Western blot, though historically used as confirmatory in some countries, is not part of the standard NACO algorithm due to high cost and limited availability. Viral load and CD4 count are used for monitoring and staging, not for establishing initial diagnosis.

  • B. CD4 count assesses immune status for staging and treatment initiation, not for confirming HIV infection.
  • C. Western blot is not part of the NACO three-test diagnostic strategy in India; it is costly and not used for routine confirmation here.
  • D. Viral load testing is used for monitoring disease progression and treatment response, not for initial diagnostic confirmation.
Educational objective: NACO recommends a sequential three-test strategy using assays of different antigen/principle for confirming HIV diagnosis, not Western blot.
46

A researcher wants to detect and quantify a specific messenger RNA transcript in a tissue sample to study gene expression levels under hypoxic conditions. Which laboratory technique is most appropriate for this purpose?

  1. ASouthern blot
  2. BNorthern blot
  3. CWestern blot
  4. DSouthwestern blot
Show answer
Answer: B. Northern blot

Northern blot is used to detect and analyze RNA, particularly mRNA, making it the appropriate technique for studying gene expression at the transcript level. The sample RNA is separated by electrophoresis, transferred to a membrane, and hybridized with a labelled complementary probe. Southern blot detects DNA fragments (used in genetic mutation analysis, paternity testing), while Western blot detects proteins using antibodies. Southwestern blot is a specialized technique detecting DNA-binding proteins, not mRNA.

  • A. Southern blot is used to detect and analyze DNA fragments, such as in restriction fragment length polymorphism analysis, not RNA.
  • C. Western blot detects proteins using specific antibodies, not RNA transcripts.
  • D. Southwestern blot identifies DNA-binding proteins, not mRNA expression levels.
Educational objective: Northern blot is the technique of choice for detecting and quantifying specific RNA transcripts to study gene expression.
47

A 45-year-old man presents with chorea, personality changes and progressive cognitive decline. His father had similar symptoms starting at age 55, but the patient's symptoms began a decade earlier and are more severe. Genetic testing reveals an expanded CAG trinucleotide repeat in the HTT gene on chromosome 4, with a greater repeat number than in his father. This phenomenon of worsening severity and earlier onset in successive generations is best explained by which mechanism?

  1. AGenomic imprinting silencing the maternal allele in each generation
  2. BMitochondrial heteroplasmy increasing with each generation
  3. CInstability and progressive expansion of the trinucleotide repeat during paternal gametogenesis
  4. DRandom somatic mutation accumulating independently in each generation
Show answer
Answer: C. Instability and progressive expansion of the trinucleotide repeat during paternal gametogenesis

Huntington disease is an autosomal dominant trinucleotide (CAG) repeat disorder showing genetic anticipation, where repeat length tends to expand further during spermatogenesis due to instability of the repeat sequence during DNA replication. Larger repeat numbers correlate with earlier age of onset and greater disease severity in subsequent generations, especially with paternal transmission. The CAG repeat codes for glutamine, and expanded repeats produce a toxic mutant huntingtin protein with expanded polyglutamine tracts. This mechanism of anticipation is also seen in myotonic dystrophy and fragile X syndrome.

  • A. Huntington disease is not an imprinted disorder; the anticipation effect is linked to CAG repeat instability, not allele silencing.
  • B. Huntington disease is a nuclear gene disorder, not a mitochondrial disorder; heteroplasmy is irrelevant here.
  • D. Anticipation in Huntington disease results from a specific, predictable mechanism (repeat instability), not random independent somatic mutation.
Educational objective: Anticipation in trinucleotide repeat disorders like Huntington disease results from progressive expansion of unstable repeats during gametogenesis, causing earlier onset and greater severity in successive generations.
48

A 30-year-old woman has progressive external ophthalmoplegia, bilateral ptosis and proximal muscle weakness. Her mother and maternal aunt have similar symptoms, but her father is unaffected, and none of her paternal relatives show the disease. Her brother is also affected. Muscle biopsy shows ragged red fibers on Gomori trichrome stain. Which of the following best describes the pattern of inheritance in this family?

  1. AMitochondrial inheritance, transmitted exclusively through the mother
  2. BX-linked recessive inheritance
  3. CAutosomal recessive inheritance with consanguinity
  4. DAutosomal dominant inheritance with variable penetrance
Show answer
Answer: A. Mitochondrial inheritance, transmitted exclusively through the mother

The pattern described, affected mother transmitting disease to all her children (both sons and daughters) while an affected father transmits to none of his children, is characteristic of mitochondrial inheritance. This occurs because mitochondria and their DNA are inherited almost exclusively from the ovum cytoplasm, not sperm. Ragged red fibers on muscle biopsy are a classic histological finding in mitochondrial myopathies due to the accumulation of abnormal mitochondria beneath the sarcolemma. Conditions like mitochondrial encephalomyopathy and chronic progressive external ophthalmoplegia follow this maternal transmission pattern.

  • B. X-linked recessive disease typically affects males whose mothers are carriers, not daughters, and is transmitted via maternal carriers, not matching this family pattern directly here.
  • C. Autosomal recessive inheritance requires two carrier parents and is unrelated to the strict maternal line transmission seen in this family.
  • D. Autosomal dominant inheritance would show transmission from either affected parent to roughly half of offspring of both sexes; here only the mother transmits the disease.
Educational objective: Mitochondrial disorders show maternal inheritance, affected mothers transmit disease to all children, while affected fathers transmit to none, and ragged red fibers on biopsy are a key diagnostic clue.

Nutrition & Biochemical Assessment

2 questions
49

A 2-year-old child from a low-income family is brought with generalized edema, a distended abdomen, sparse depigmented hair and a rash with areas of hyperpigmentation and desquamation on the limbs. The child is apathetic but has retained significant subcutaneous fat. Serum albumin is 2.0 g/dL (normal 3.5-5.0). Which of the following best explains the pathogenesis of the edema seen in this child?

  1. APrimary deficiency of dietary fat causing impaired lipoprotein synthesis
  2. BPure caloric deficiency causing depletion of subcutaneous fat and muscle wasting
  3. CSevere dietary protein deficiency causing hypoalbuminemia and reduced plasma oncotic pressure
  4. DExcessive dietary carbohydrate intake causing osmotic fluid retention
Show answer
Answer: C. Severe dietary protein deficiency causing hypoalbuminemia and reduced plasma oncotic pressure

Kwashiorkor results from severe dietary protein deficiency despite relatively adequate caloric (carbohydrate) intake, leading to hypoalbuminemia, which reduces plasma oncotic pressure and causes generalized edema. The characteristic features, edema, flaky-paint dermatosis, hair changes, hepatomegaly and preserved subcutaneous fat, distinguish it from marasmus. Marasmus, in contrast, results from combined protein-calorie deficiency with severe wasting of fat and muscle but without edema, as albumin levels are usually relatively preserved. The retained subcutaneous fat and low albumin in this child point specifically to protein deficiency as the predominant defect.

  • A. Fat deficiency is not the primary defect in kwashiorkor; the hallmark is protein deficiency causing hypoalbuminemia.
  • B. Pure caloric deficiency describes marasmus, which presents with severe wasting and loss of subcutaneous fat, not edema with preserved fat as seen here.
  • D. Edema in kwashiorkor is due to hypoalbuminemia, not excess carbohydrate causing osmotic retention.
Educational objective: Kwashiorkor results from predominant protein deficiency causing hypoalbuminemia and edema with preserved subcutaneous fat, distinguishing it from the pure caloric deficiency and wasting seen in marasmus.
50

A 62-year-old man presents with painless jaundice, significant weight loss over three months and a palpable, non-tender gallbladder. Contrast-enhanced CT abdomen shows a hypodense mass in the head of the pancreas with dilated intrahepatic and extrahepatic biliary radicles. Which serum tumor marker is most useful for supporting the diagnosis and monitoring treatment response in this condition?

  1. ACA 19-9
  2. BCarcinoembryonic antigen (CEA)
  3. CAlpha-fetoprotein (AFP)
  4. DProstate-specific antigen (PSA)
Show answer
Answer: A. CA 19-9

CA 19-9 is the most useful and widely used tumor marker for pancreatic adenocarcinoma, both for supporting diagnosis in the appropriate clinical context and for monitoring response to treatment and detecting recurrence. It is a sialylated Lewis blood group antigen found elevated in over 70-80% of pancreatic cancer cases, though it lacks specificity and can be elevated in biliary obstruction and other malignancies. CEA is more commonly associated with colorectal carcinoma, AFP with hepatocellular carcinoma and germ cell tumors, and PSA with prostate carcinoma. The clinical picture here, painless jaundice with a palpable gallbladder (Courvoisier sign) and a pancreatic head mass, is classic for pancreatic cancer.

  • B. CEA is primarily associated with colorectal carcinoma and is less specific or sensitive than CA 19-9 for pancreatic cancer.
  • C. AFP is the marker for hepatocellular carcinoma and germ cell tumors, not pancreatic adenocarcinoma.
  • D. PSA is specific to prostate carcinoma and is irrelevant to this clinical presentation.
Educational objective: CA 19-9 is the tumor marker of choice for diagnosis support and monitoring of pancreatic adenocarcinoma.

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