INI-CET 2025 — Biochemistry

Q: A child presents with fatigue and hepatomegaly. Liver enzymes (ALT, AST) are elevated. Ketosis was significant. Liver biopsy shows excess glycogen accumulation. After feeding, blood glucose levels rise, but there is no rise in glucose after overnight fasting. Which of the following enzyme deficiencies is most likely responsible for this presentation?

Glucose-6-phosphatase. ***Glucose-6-phosphatase*** - This deficiency is characteristic of Glycogen Storage Disease type I (**GSD I** or **Von Gierke disease**), which prevents the final step of both glycogenolysis and gluconeogenesis (conversion of **Glucose-6-Phosphate** to free glucose). - The inability to release free glucose from the liver, especially during fasting, causes severe **fasting hypoglycemia**, significant **ketosis**, and massive **hepatomegaly** due to trapped glycogen and fat accumulation. *Liver phosphorylase* - Deficiency in **Liver phosphorylase** (GSD VI or Hers disease) impairs the breakdown of glycogen but does not affect the gluconeogenesis pathway. - While it leads to hepatomegaly and hypoglycemia, the symptoms are generally **milder** than GSD I, and the profound metabolic derangements (severe ketosis, lactic acidosis) seen in this case are typically less pronounced. *Muscle phosphorylase* - This enzyme deficiency, known as **GSD V** or **McArdle disease**, primarily affects the skeletal muscle, leading to muscle pain, cramps, and **exercise intolerance**. - It does not cause hepatomegaly or issues with hepatic glucose release and therefore is an **unlikely** cause of fasting hypoglycemia in a child. *Phosphofructokinase* - Phosphofructokinase (GSD VII or **Tarui disease**) deficiency affects both muscle and erythrocytes, presenting similar to McArdle disease with **exercise intolerance** and often mild hemolysis. - PFK deficiency primarily affects glycolysis and does not directly impair the liver's ability to release glucose via **gluconeogenesis** or the final G6Pase step during fasting. [Practice more Biochemistry PYQs on OnCourse]

Q: Cancer cells take up excess glucose because?

Lactate is produced even in the presence of Oxygen. ***Lactate is produced even in the presence of Oxygen*** - This phenomenon is known as the **Warburg effect** (or aerobic glycolysis). Cancer cells preferentially ferment glucose to lactate, even when adequate **oxygen** is available, bypassing efficient oxidative phosphorylation. - This inefficient use of glucose allows rapid generation of metabolic intermediates (e.g., carbon backbones) required for the synthesis of **lipids**, proteins, and nucleic acids needed for rapid cell proliferation. *High NADH/NAD ratio* - A **high NADH/NAD+ ratio** signals abundance of reducing equivalents, which would typically inhibit glycolysis and favor oxidative phosphorylation. - Rapid glycolysis, as seen in cancer (Warburg effect), requires the constant regeneration of **NAD+** from NADH via lactate dehydrogenase for the pathway to continue. *High GLUT2* - While cancer cells increase glucose uptake by overexpressing glucose transporters, the typically overexpressed transporter in many solid tumors is **GLUT1**, not GLUT2. - **GLUT2** is primarily found in the liver, kidney, and pancreatic beta cells and is less commonly the primary high-affinity transporter driving the intense uptake seen in malignant cells. *Absence of Oxygen* - The characteristic metabolic change in cancer is that glucose uptake and lactate production occur despite the **presence of oxygen** (aerobic glycolysis). - If oxygen were truly absent (anaerobic conditions), all cell types would produce lactate; hence, the defining feature of cancer is the metabolic shift occurring in an **oxygenated environment**. [Practice more Biochemistry PYQs on OnCourse]

Q: How many PCR cycles are needed to produce half the amplified DNA copies produced by 27 cycles of PCR?

26. ***26 (Correct)*** - PCR amplification follows an **exponential growth** pattern, where the template DNA is theoretically doubled in each cycle. The product quantity is proportional to 2^n (where n is the cycle number). - If 27 cycles produce 2^27 copies, half of this amount is calculated as 2^27 / 2, which mathematically simplifies to 2^(27-1). Therefore, exactly **26 cycles** are needed. *25 (Incorrect)* - 25 cycles would produce 2^25 copies. Since 2^27 = 4 × 2^25, 25 cycles yield only **one-quarter** (25%) of the DNA produced by 27 cycles, not half. - In exponential amplification, a difference of two cycles corresponds to a **four-fold change** in product amount. *13 (Incorrect)* - 13 cycles would produce 2^13 copies, which represents an extremely small fraction of the total DNA generated after 27 cycles (2^27). - This amount belongs to the early phase of the reaction and is mathematically too low to be relevant to the calculation for half the **final product quantity**. *12 (Incorrect)* - 12 cycles produce 2^12 copies, which is 2^15 times less than 2^27 copies. - This number is irrelevant as it doesn't align with the required calculation based on the principles of **geometric progression** in PCR. [Practice more Biochemistry PYQs on OnCourse]

Q: Mechanism of inhibition caused by cyanide and carbon monoxide poisoning involves inhibition of which enzyme?

Cytochrome C oxidase. ***Cytochrome C oxidase*** - Cyanide and carbon monoxide are **powerful inhibitors of Cytochrome C oxidase (Complex IV)** in the electron transport chain. - Cyanide binds to the **ferric iron (Fe³⁺)** in Complex IV, while carbon monoxide also binds to Complex IV, preventing oxygen from binding. - Inhibition of Complex IV **stops the transfer of electrons to oxygen**, halting the entire process of oxidative phosphorylation and cellular respiration, leading to **cellular hypoxia and energy deficit**. *Incorrect: NADH dehydrogenase* - This enzyme, also known as Complex I, is primarily inhibited by compounds like **Rotenone** and **Amytal**. - While crucial for the ETC, it is not the target of carbon monoxide or cyanide. *Incorrect: Succinate dehydrogenase* - This enzyme, known as Complex II, is an integral part of both the ETC and the Krebs cycle. - It is specifically inhibited by compounds like **Malonate** and is not the primary target in cyanide or carbon monoxide poisoning. *Incorrect: Cytochrome C oxidoreductase* - This enzyme represents Complex III (also called the Cytochrome bc1 complex). - It transfers electrons from ubiquinone to cytochrome C, but its inhibition is not the primary mechanism of action for cyanide or carbon monoxide, which directly target Complex IV. [Practice more Biochemistry PYQs on OnCourse]

10 Previous Year Questions with Answers & Explanations

Questions

A child presents with fatigue and hepatomegaly. Liver enzymes (ALT, AST) are elevated. Ketosis was significant. Liver biopsy shows excess glycogen accumulation. After feeding, blood glucose levels rise, but there is no rise in glucose after overnight fasting. Which of the following enzyme deficiencies is most likely responsible for this presentation?

Cancer cells take up excess glucose because?

How many PCR cycles are needed to produce half the amplified DNA copies produced by 27 cycles of PCR?

Mechanism of inhibition caused by cyanide and carbon monoxide poisoning involves inhibition of which enzyme?

A deficiency of Glucose-6-Phosphatase is associated with which of the following bilirubin patterns?

What is the pH of the zwitterion as shown in the titration curve of glycine?

The PRIMARY mechanism of toxicity of which substance is NOT through Cytochrome C oxidase inhibition?

A 90 kg obese man is taking a carbohydrate-rich diet. Which of the following enzymes/intermediates will be elevated in him? 1. Malonyl CoA 2. Acetyl CoA Carboxylase 3. PDH 4. Citrate Lyase

Match the following blotting techniques with the type of biomolecule they detect: Blotting Technique Detects A. Southern blot 1. RNA B. Northern blot 2. Lipids C. Western blot 3. DNA D. Eastern blot 4. Protein

Q10

IGF-1 and IGF-2 are structurally most similar to which of the following molecules?

INI-CET 2025 - Biochemistry INI-CET Practice Questions and MCQs

Question 1: A child presents with fatigue and hepatomegaly. Liver enzymes (ALT, AST) are elevated. Ketosis was significant. Liver biopsy shows excess glycogen accumulation. After feeding, blood glucose levels rise, but there is no rise in glucose after overnight fasting. Which of the following enzyme deficiencies is most likely responsible for this presentation?

A. Glucose-6-phosphatase (Correct Answer)
B. Liver phosphorylase
C. Muscle phosphorylase
D. Phosphofructokinase

Explanation: ***Glucose-6-phosphatase*** - This deficiency is characteristic of Glycogen Storage Disease type I (**GSD I** or **Von Gierke disease**), which prevents the final step of both glycogenolysis and gluconeogenesis (conversion of **Glucose-6-Phosphate** to free glucose). - The inability to release free glucose from the liver, especially during fasting, causes severe **fasting hypoglycemia**, significant **ketosis**, and massive **hepatomegaly** due to trapped glycogen and fat accumulation. *Liver phosphorylase* - Deficiency in **Liver phosphorylase** (GSD VI or Hers disease) impairs the breakdown of glycogen but does not affect the gluconeogenesis pathway. - While it leads to hepatomegaly and hypoglycemia, the symptoms are generally **milder** than GSD I, and the profound metabolic derangements (severe ketosis, lactic acidosis) seen in this case are typically less pronounced. *Muscle phosphorylase* - This enzyme deficiency, known as **GSD V** or **McArdle disease**, primarily affects the skeletal muscle, leading to muscle pain, cramps, and **exercise intolerance**. - It does not cause hepatomegaly or issues with hepatic glucose release and therefore is an **unlikely** cause of fasting hypoglycemia in a child. *Phosphofructokinase* - Phosphofructokinase (GSD VII or **Tarui disease**) deficiency affects both muscle and erythrocytes, presenting similar to McArdle disease with **exercise intolerance** and often mild hemolysis. - PFK deficiency primarily affects glycolysis and does not directly impair the liver's ability to release glucose via **gluconeogenesis** or the final G6Pase step during fasting.

Question 2: Cancer cells take up excess glucose because?

A. Lactate is produced even in the presence of Oxygen (Correct Answer)
B. High NADH/NAD ratio
C. High GLUT2
D. Absence of Oxygen

Explanation: ***Lactate is produced even in the presence of Oxygen*** - This phenomenon is known as the **Warburg effect** (or aerobic glycolysis). Cancer cells preferentially ferment glucose to lactate, even when adequate **oxygen** is available, bypassing efficient oxidative phosphorylation. - This inefficient use of glucose allows rapid generation of metabolic intermediates (e.g., carbon backbones) required for the synthesis of **lipids**, proteins, and nucleic acids needed for rapid cell proliferation. *High NADH/NAD ratio* - A **high NADH/NAD+ ratio** signals abundance of reducing equivalents, which would typically inhibit glycolysis and favor oxidative phosphorylation. - Rapid glycolysis, as seen in cancer (Warburg effect), requires the constant regeneration of **NAD+** from NADH via lactate dehydrogenase for the pathway to continue. *High GLUT2* - While cancer cells increase glucose uptake by overexpressing glucose transporters, the typically overexpressed transporter in many solid tumors is **GLUT1**, not GLUT2. - **GLUT2** is primarily found in the liver, kidney, and pancreatic beta cells and is less commonly the primary high-affinity transporter driving the intense uptake seen in malignant cells. *Absence of Oxygen* - The characteristic metabolic change in cancer is that glucose uptake and lactate production occur despite the **presence of oxygen** (aerobic glycolysis). - If oxygen were truly absent (anaerobic conditions), all cell types would produce lactate; hence, the defining feature of cancer is the metabolic shift occurring in an **oxygenated environment**.

Question 3: How many PCR cycles are needed to produce half the amplified DNA copies produced by 27 cycles of PCR?

A. 26 (Correct Answer)
B. 12
C. 25
D. 13

Explanation: ***26 (Correct)*** - PCR amplification follows an **exponential growth** pattern, where the template DNA is theoretically doubled in each cycle. The product quantity is proportional to 2^n (where n is the cycle number). - If 27 cycles produce 2^27 copies, half of this amount is calculated as 2^27 / 2, which mathematically simplifies to 2^(27-1). Therefore, exactly **26 cycles** are needed. *25 (Incorrect)* - 25 cycles would produce 2^25 copies. Since 2^27 = 4 × 2^25, 25 cycles yield only **one-quarter** (25%) of the DNA produced by 27 cycles, not half. - In exponential amplification, a difference of two cycles corresponds to a **four-fold change** in product amount. *13 (Incorrect)* - 13 cycles would produce 2^13 copies, which represents an extremely small fraction of the total DNA generated after 27 cycles (2^27). - This amount belongs to the early phase of the reaction and is mathematically too low to be relevant to the calculation for half the **final product quantity**. *12 (Incorrect)* - 12 cycles produce 2^12 copies, which is 2^15 times less than 2^27 copies. - This number is irrelevant as it doesn't align with the required calculation based on the principles of **geometric progression** in PCR.

Question 4: Mechanism of inhibition caused by cyanide and carbon monoxide poisoning involves inhibition of which enzyme?

A. Succinate dehydrogenase
B. Cytochrome C oxidase (Correct Answer)
C. NADH dehydrogenase
D. Cytochrome C oxidoreductase

Explanation: ***Cytochrome C oxidase*** - Cyanide and carbon monoxide are **powerful inhibitors of Cytochrome C oxidase (Complex IV)** in the electron transport chain. - Cyanide binds to the **ferric iron (Fe³⁺)** in Complex IV, while carbon monoxide also binds to Complex IV, preventing oxygen from binding. - Inhibition of Complex IV **stops the transfer of electrons to oxygen**, halting the entire process of oxidative phosphorylation and cellular respiration, leading to **cellular hypoxia and energy deficit**. *Incorrect: NADH dehydrogenase* - This enzyme, also known as Complex I, is primarily inhibited by compounds like **Rotenone** and **Amytal**. - While crucial for the ETC, it is not the target of carbon monoxide or cyanide. *Incorrect: Succinate dehydrogenase* - This enzyme, known as Complex II, is an integral part of both the ETC and the Krebs cycle. - It is specifically inhibited by compounds like **Malonate** and is not the primary target in cyanide or carbon monoxide poisoning. *Incorrect: Cytochrome C oxidoreductase* - This enzyme represents Complex III (also called the Cytochrome bc1 complex). - It transfers electrons from ubiquinone to cytochrome C, but its inhibition is not the primary mechanism of action for cyanide or carbon monoxide, which directly target Complex IV.

Question 5: A deficiency of Glucose-6-Phosphatase is associated with which of the following bilirubin patterns?

A. Direct bilirubin normal, indirect bilirubin increased, and urobilinogen normal (Correct Answer)
B. Indirect bilirubin increased, direct bilirubin normal, and raised urine urobilinogen
C. Direct bilirubin increased, indirect bilirubin decreased, and urine urobilinogen/urobilin increased
D. Raised direct bilirubin, normal indirect bilirubin, and absent urine bilirubin

Explanation: ***Direct bilirubin normal, indirect bilirubin increased, and urobilinogen normal*** - Glucose-6-Phosphatase (G6Pase) deficiency causes **Glycogen Storage Disease Type I (von Gierke disease)**, which is primarily characterized by severe **hypoglycemia**, hepatomegaly, lactic acidosis, and hyperuricemia. - While jaundice is NOT a classical feature of von Gierke disease, some patients may show mild elevation of **unconjugated (indirect) bilirubin**. - The proposed mechanism involves accumulation of **glucuronic acid** (a metabolic byproduct that cannot be converted to glucose due to G6Pase deficiency), which may competitively inhibit **UDP-glucuronyltransferase (UGT1A1)**, the enzyme responsible for bilirubin conjugation. - This results in increased **indirect (unconjugated) bilirubin** with normal direct bilirubin and normal urobilinogen, similar to the pattern seen in Gilbert syndrome. *Indirect bilirubin increased, direct bilirubin normal, and raised urine urobilinogen* - This pattern with raised urobilinogen is more characteristic of **hemolytic conditions** where excessive red blood cell breakdown overwhelms the liver's conjugation capacity. - The increased urobilinogen suggests increased enterohepatic circulation of bilirubin metabolites, which is not a typical feature of G6Pase deficiency. - While indirect bilirubin may be elevated in von Gierke disease, the raised urobilinogen makes this pattern less specific. *Direct bilirubin increased, indirect bilirubin decreased, and urine urobilinogen/urobilin increased* - This pattern is **internally contradictory** and does not match any known pathophysiological mechanism in G6Pase deficiency. - If G6Pase deficiency affects bilirubin metabolism, it would impair conjugation (leading to **increased** indirect bilirubin), not decrease it. - Additionally, decreased indirect bilirubin with increased direct bilirubin would require enhanced conjugation, which is opposite to the proposed mechanism of UGT1A1 inhibition. *Raised direct bilirubin, normal indirect bilirubin, and absent urine bilirubin* - This pattern represents **conjugated hyperbilirubinemia** with defective biliary excretion, characteristic of **Dubin-Johnson syndrome** or **Rotor syndrome**. - These conditions involve defects in hepatocyte transport proteins (MRP2 in Dubin-Johnson), not impaired conjugation. - This pattern is inconsistent with the proposed mechanism in G6Pase deficiency, where impaired conjugation would increase **unconjugated** bilirubin.

Question 6: What is the pH of the zwitterion as shown in the titration curve of glycine?

A. 6 (Correct Answer)
B. 2
C. 9
D. 4.4

Explanation: ***Correct Option: 6*** - The pH of the zwitterion corresponds to the **isoelectric point (pI)**, the pH at which the amino acid has a net charge of zero. According to the curve, this occurs at pH 5.97, which is approximately 6. - The pI is calculated as the average of the two pKa values: **pI = (pKa1 + pKa2) / 2 = (2.34 + 9.60) / 2 = 5.97**. *Incorrect Option: 4.4* - At a pH of 4.4, which is between pKa1 and pI, the glycine molecule would have a net **positive charge** as the concentration of the cationic form ([NH3+-CH2-COOH]) would still be significant. - This value does not represent any specific inflection or equivalence point on the titration curve for glycine. *Incorrect Option: 2* - A pH of approximately 2 is near the **pKa1 (2.34)** of the carboxyl group, which is the point where the concentration of the fully protonated form equals the concentration of the zwitterion. - At this pH, the molecule has a net **positive charge**, as the amino group is fully protonated and the carboxyl group is only half-deprotonated. *Incorrect Option: 9* - A pH of approximately 9 is near the **pKa2 (9.60)** of the amino group, representing the buffering region for this group. - At this pH, the molecule has a net **negative charge**, as there is a significant amount of the fully deprotonated form ([NH2-CH2-COO-]) present.

Question 7: The PRIMARY mechanism of toxicity of which substance is NOT through Cytochrome C oxidase inhibition?

A. Nitric oxide (NO)
B. Cyanide (CN)
C. Oligomycin
D. Carbon monoxide (CO) (Correct Answer)

Explanation: ***Carbon monoxide (CO)*** - **Primary mechanism of toxicity** is through binding to hemoglobin forming **carboxyhemoglobin**, preventing oxygen transport - While CO can bind to **cytochrome oxidase**, its **dominant clinical effect** occurs at the oxygen delivery level, not cellular respiration *Hydrogen sulfide (H₂S)* - **Direct inhibitor** of cytochrome C oxidase by binding to the **heme iron center** - Functions similarly to cyanide, causing **histotoxic hypoxia** by blocking cellular oxygen utilization *Nitric oxide (NO)* - **Potent reversible inhibitor** of cytochrome C oxidase competing with oxygen at the active site - Physiological regulator of **cellular respiration** and important in hypoxia signaling pathways *Cyanide (CN⁻)* - **Classic inhibitor** of cytochrome C oxidase, binding with high affinity to the **oxidized cytochrome a₃** - Causes rapid **metabolic failure** by completely blocking the electron transport chain at Complex IV

Question 8: A 90 kg obese man is taking a carbohydrate-rich diet. Which of the following enzymes/intermediates will be elevated in him? 1. Malonyl CoA 2. Acetyl CoA Carboxylase 3. PDH 4. Citrate Lyase

A. $1,2,3,4$ (Correct Answer)
B. $1,2,3$
C. $1,3,4$
D. $2,3,4$

Explanation: ***Correct Answer: 1,2,3,4 (All of them)*** In an obese person consuming a carbohydrate-rich diet, **high insulin levels** drive maximum flux through **lipogenesis (fatty acid synthesis)**. This results in elevation of ALL enzymes and intermediates in the pathway: - **PDH (Pyruvate Dehydrogenase)**: Activated by insulin to convert pyruvate → Acetyl CoA in mitochondria - **Citrate Lyase**: Elevated to cleave citrate → cytosolic Acetyl CoA (substrate for fatty acid synthesis) - **Acetyl CoA Carboxylase (ACC)**: The **rate-limiting enzyme** of fatty acid synthesis, activated by insulin (dephosphorylation) and citrate - **Malonyl CoA**: The **first committed intermediate** in fatty acid synthesis, product of ACC action on Acetyl CoA All four components work sequentially in the pathway from glucose → fatty acids, and all are upregulated in this metabolic state. *Incorrect Option: 1,2,3* This incorrectly excludes **Citrate Lyase**, which is essential for providing cytosolic Acetyl CoA from mitochondrial citrate. Without elevated Citrate Lyase activity, fatty acid synthesis cannot proceed efficiently despite high carbohydrate intake. *Incorrect Option: 2,3,4* This incorrectly excludes **Malonyl CoA**, the direct product of Acetyl CoA Carboxylase and the committed intermediate for fatty acid synthesis. When ACC is highly active (as it would be with high insulin), Malonyl CoA concentration must be elevated. *Incorrect Option: 1,3,4* This incorrectly excludes **Acetyl CoA Carboxylase (ACC)**, the **rate-limiting enzyme** of the entire fatty acid synthesis pathway. In a high carbohydrate/high insulin state, ACC is maximally activated by dephosphorylation and allosteric activation by citrate, making this a critical error.

Question 9: Match the following blotting techniques with the type of biomolecule they detect: Blotting Technique Detects A. Southern blot 1. RNA B. Northern blot 2. Lipids C. Western blot 3. DNA D. Eastern blot 4. Protein

A. A-3, B-1, C-4, D-2 (Correct Answer)
B. A-4, B-2, C-1, D-3
C. A-2, B-4, C-3, D-1
D. A-1, B-3, C-2, D-4

Explanation: ***A-3, B-1, C-4, D-2*** - **Southern blot** is named after Edwin Southern and is the foundational technique used to detect specific **DNA** sequences. - **Northern blot** detects specific **RNA** sequences, while **Western blot** targets specific **Proteins** using antibodies. ***A-1, B-3, C-2, D-4*** - This option incorrectly matches Southern blot (A) with RNA (1) and Northern blot (B) with DNA (3). - The standard nomenclature links Southern with **DNA** and Northern with **RNA** due to their evolutionary development from the original technique. ***A-4, B-2, C-1, D-3*** - This option incorrectly assigns Southern blot (A) to Protein (4) and Northern blot (B) to Lipids (2). - Western blot (C) detects **Protein**, not RNA (1), as suggested here. ***A-2, B-4, C-3, D-1*** - This option incorrectly matches Southern blot (A) with Lipids (2) and Western blot (C) with DNA (3). - **Eastern blot** (D) is a technique designed to detect **post-translational modifications** (like lipids and carbohydrates) on proteins, making the D-2 match plausible, but the other matches are incorrect (i.e., **Southern blot** detects DNA).

Question 10: IGF-1 and IGF-2 are structurally most similar to which of the following molecules?

A. Preproinsulin
B. Insulin (Correct Answer)
C. Proinsulin
D. C-peptide

Explanation: ***Insulin*** - **Insulin-like growth factors (IGFs)**, including IGF-1 and IGF-2, belong to the same peptide growth factor superfamily as insulin, sharing notable **sequence homology** (~50% amino acid similarity) and a similar three-dimensional structure. - Both IGFs and insulin are small peptides stabilized by **disulfide bonds** forming A and B domains, and act via similar receptor tyrosine kinases (the insulin receptor and IGF-1 receptor can cross-react). - Mature insulin represents the **structurally closest molecule** to the secreted, functional IGFs. *Preproinsulin* - **Preproinsulin** is the earliest precursor and includes an N-terminal **signal peptide** (the 'pre' sequence) which is cleaved off upon entry into the endoplasmic reticulum. - Since IGFs mature into secreted proteins without this transient signal sequence in their final structure, preproinsulin is structurally less similar than mature insulin. *Proinsulin* - **Proinsulin** consists of the A chain and B chain linked by the connecting **C-peptide** domain. - Although IGFs are derived from a single precursor chain like proinsulin, **mature insulin** (two chains, A and B) is structurally closer to the secreted IGFs than proinsulin (which has three domains: A, B, and C). *C-peptide* - The **C-peptide** is the connecting segment linking the A and B chains of proinsulin, which is cleaved and removed before insulin becomes mature. - It is a short, linear peptide with little structural resemblance to the complex, disulfide-bonded domains of functional IGFs or insulin.