Enzymes — MCQs

Enzymes Indian Medical PG Practice Questions and MCQs

Question 341: NAD acts as a cofactor for all except?

A. Isocitrate dehydrogenase
B. Alpha ketoglutarate dehydrogenase
C. Malate dehydrogenase
D. Succinyl thiokinase (Correct Answer)

Explanation: **Explanation:** The correct answer is **Succinyl thiokinase** (also known as Succinyl-CoA synthetase). This enzyme catalyzes the conversion of Succinyl-CoA to Succinate in the TCA cycle. Unlike the other enzymes listed, it does not involve a redox reaction requiring NAD+; instead, it is responsible for **substrate-level phosphorylation**, where the energy from the thioester bond is used to synthesize GTP (or ATP). **Analysis of Options:** * **Isocitrate dehydrogenase:** This is the rate-limiting enzyme of the TCA cycle. It catalyzes the oxidative decarboxylation of isocitrate to $\alpha$-ketoglutarate, reducing **NAD+ to NADH**. * **Alpha-ketoglutarate dehydrogenase:** This multi-enzyme complex requires five cofactors (Thiamine, Lipoic acid, CoA, FAD, and **NAD+**). It reduces NAD+ to NADH during the conversion of $\alpha$-ketoglutarate to Succinyl-CoA. * **Malate dehydrogenase:** This enzyme catalyzes the final step of the TCA cycle, converting malate to oxaloacetate. This is a dehydrogenation reaction that requires **NAD+** as an electron acceptor. **High-Yield Clinical Pearls for NEET-PG:** 1. **Substrate-Level Phosphorylation:** In the TCA cycle, this occurs only at the **Succinyl thiokinase** step. 2. **NADH Production:** Three steps in the TCA cycle produce NADH: Isocitrate DH, $\alpha$-ketoglutarate DH, and Malate DH. 3. **FADH2 Production:** Occurs only at the **Succinate dehydrogenase** step (which is also Complex II of the Electron Transport Chain). 4. **Mnemonic for TCA Cofactors:** $\alpha$-ketoglutarate DH and Pyruvate DH both require "**T**ender **L**oving **C**are **F**or **N**ancy" (Thiamine, Lipoate, CoA, FAD, NAD).

Question 342: Cytochrome oxidase is a:

A. Hemoprotein (Correct Answer)
B. Flavin mononucleotide
C. Flavin adenine dinucleotide
D. Flavin adenine trinucleotide

Explanation: **Explanation:** **Cytochrome oxidase (Complex IV)** is the terminal enzyme of the mitochondrial electron transport chain (ETC). It is classified as a **hemoprotein** because it contains two heme groups (**heme a and heme a3**) as essential prosthetic groups. These heme groups contain iron atoms that cycle between the ferrous ($Fe^{2+}$) and ferric ($Fe^{3+}$) states to facilitate the transfer of electrons to molecular oxygen, reducing it to water. **Analysis of Options:** * **Option A (Correct):** Cytochrome oxidase contains two heme moieties and two copper centers ($Cu_A$ and $Cu_B$). The presence of heme makes it a classic hemoprotein. * **Option B & C (Incorrect):** Flavin mononucleotide (FMN) and Flavin adenine dinucleotide (FAD) are prosthetic groups for **Complex I** (NADH dehydrogenase) and **Complex II** (Succinate dehydrogenase), respectively. Cytochrome oxidase does not utilize flavin nucleotides. * **Option D (Incorrect):** Flavin adenine trinucleotide is a non-existent molecule in biological systems. **High-Yield Clinical Pearls for NEET-PG:** * **Inhibitors:** Cytochrome oxidase is the target of lethal toxins like **Cyanide, Carbon Monoxide (CO), Hydrogen Sulfide ($H_2S$), and Azide**. They bind to the iron in heme a3, halting cellular respiration. * **Copper Requirement:** It is one of the few enzymes requiring copper. Copper deficiency can impair its function, contributing to the neurological symptoms seen in **Menkes disease**. * **Final Electron Acceptor:** It catalyzes the final step of the ETC where oxygen acts as the terminal electron acceptor.

Question 343: Which of the following enzymes is responsible for the transfer of amino groups from an amino acid to an alpha-keto acid?

A. Transaminase (Correct Answer)
B. Transketolase
C. Deaminase
D. Lyase

Explanation: **Explanation:** **Transaminases (Aminotransferases)** are the enzymes responsible for **transamination**, the first step in the catabolism of most amino acids. This process involves the reversible transfer of an amino group (–NH₂) from an amino acid to an α-keto acid (typically α-ketoglutarate), resulting in the formation of a new amino acid (Glutamate) and a new α-keto acid. This reaction requires **Pyridoxal Phosphate (PLP)**, a derivative of Vitamin B6, as an essential cofactor. **Analysis of Incorrect Options:** * **Transketolase:** An enzyme of the Hexose Monophosphate (HMP) shunt that transfers two-carbon units. It requires Thiamine pyrophosphate (TPP) as a cofactor. * **Deaminase:** These enzymes catalyze **deamination**, which is the total removal of an amino group from a molecule (releasing it as free ammonia), rather than transferring it to another substrate. * **Lyase:** A class of enzymes that catalyze the cleavage of bonds (C-C, C-O, C-N) by means other than hydrolysis or oxidation, often resulting in the formation of a double bond. **High-Yield Clinical Pearls for NEET-PG:** * **Diagnostic Markers:** Serum Glutamate Oxaloacetate Transaminase (**SGOT/AST**) and Serum Glutamate Pyruvate Transaminase (**SGPT/ALT**) are critical markers for liver injury. ALT is more specific to the liver, while AST is also found in cardiac and skeletal muscle. * **Cofactor Dependency:** Always remember that **Vitamin B6 (PLP)** is the mandatory cofactor for all transamination reactions. * **Exceptions:** Lysine, Threonine, Proline, and Hydroxyproline do not undergo transamination.

Question 344: Which enzyme is responsible for the postprandial utilization of glucose?

A. Fructokinase
B. Glucokinase (Correct Answer)
C. Hexokinase
D. All of the above

Explanation: **Explanation:** The correct answer is **Glucokinase (Hexokinase IV)**. **Why Glucokinase is the correct answer:** Glucokinase is primarily located in the liver and pancreatic beta cells. Its unique kinetic properties make it ideal for the **postprandial (fed) state**: 1. **High $K_m$ (Low affinity):** It only becomes significantly active when blood glucose levels are high (e.g., after a meal). 2. **High $V_{max}$ (High capacity):** It can rapidly phosphorylate large amounts of glucose, allowing the liver to "clear" glucose from the portal blood and store it as glycogen, preventing postprandial hyperglycemia. 3. **Lack of Product Inhibition:** Unlike hexokinase, it is not inhibited by its product (Glucose-6-Phosphate), allowing continuous glucose uptake even when energy levels are high. **Why other options are incorrect:** * **Hexokinase (Types I-III):** These are found in extrahepatic tissues. They have a **low $K_m$ (high affinity)**, meaning they are already saturated at fasting glucose levels. Their role is to ensure tissues like the brain get glucose even during starvation, not to manage a postprandial surplus. * **Fructokinase:** This enzyme is specific to fructose metabolism (converting fructose to fructose-1-phosphate) and does not utilize glucose. **High-Yield Clinical Pearls for NEET-PG:** * **Molecular Sensor:** Glucokinase acts as the "glucose sensor" in the pancreas for insulin release. * **MODY Type 2:** Mutations in the glucokinase gene lead to Maturity-Onset Diabetes of the Young (MODY) type 2. * **Localization:** Glucokinase is regulated by the **Glucokinase Regulatory Protein (GKRP)**, which sequesters it in the nucleus during fasting. * **Inducibility:** Glucokinase is induced by **Insulin**, further enhancing its role in the fed state.

Question 345: A substance that binds to an enzyme at a site other than the catalytic site is known as which of the following?

A. Competitive inhibitor
B. None of the above
C. Reversible inhibitor
D. Non-competitive inhibitor (Correct Answer)

Explanation: ### Explanation **Correct Answer: D. Non-competitive inhibitor** **1. Why it is correct:** In **non-competitive inhibition**, the inhibitor binds to an **allosteric site** (a site other than the active/catalytic site) of the enzyme. Because it does not compete for the active site, it can bind to both the free enzyme (E) and the enzyme-substrate (ES) complex. This binding induces a conformational change that reduces the enzyme's catalytic activity. * **Kinetics:** The $V_{max}$ is decreased (because the enzyme is effectively "poisoned"), but the $K_m$ remains unchanged (because the affinity for the substrate at the active site is not directly affected). **2. Why other options are incorrect:** * **A. Competitive inhibitor:** These substances are structural analogs of the substrate and bind **directly to the active site**. They "compete" with the substrate; thus, $V_{max}$ remains the same (can be overcome by increasing substrate concentration), but $K_m$ increases. * **C. Reversible inhibitor:** This is a broad category that includes both competitive and non-competitive inhibitors. While a non-competitive inhibitor is often reversible, the question specifically asks for the term defining the **site of binding**, making "Non-competitive" the more specific and accurate choice. **3. NEET-PG High-Yield Pearls:** * **Irreversible Inhibition:** Also known as "Suicide Inhibition" (e.g., Aspirin inhibiting COX, Allopurinol inhibiting Xanthine Oxidase). * **Lineweaver-Burk Plot:** In non-competitive inhibition, the plots intersect on the **negative X-axis** (same $K_m$), whereas in competitive inhibition, they intersect on the **Y-axis** (same $V_{max}$). * **Uncompetitive Inhibition:** The inhibitor binds **only** to the ES complex (rare; $V_{max}$ and $K_m$ both decrease). * **Classic Example:** Heavy metal poisoning (e.g., Lead, Mercury) often acts via non-competitive inhibition by binding to -SH groups on enzymes.

Question 346: Sodium fluoride inhibits which enzyme in glycolysis?

A. Hexokinase
B. Pyruvate kinase
C. Aconitase
D. Enolase (Correct Answer)

Explanation: ### Explanation **Correct Answer: D. Enolase** **Mechanism of Action:** Sodium fluoride (NaF) is a potent inhibitor of **Enolase**, the enzyme responsible for the ninth step of glycolysis. Enolase catalyzes the dehydration of 2-phosphoglycerate to phosphoenolpyruvate (PEP). The inhibition occurs because fluoride ions, in the presence of inorganic phosphate, form a complex with magnesium ions (**Magnesium-Fluorophosphate complex**). Since Enolase requires $Mg^{2+}$ as a cofactor, this complex displaces the magnesium, effectively inactivating the enzyme and halting glycolysis. **Analysis of Incorrect Options:** * **A. Hexokinase:** This is the first regulatory enzyme of glycolysis. It is inhibited by its product, Glucose-6-Phosphate, but not by fluoride. * **B. Pyruvate Kinase:** This is the final enzyme of glycolysis. While it also requires $Mg^{2+}$, it is not the primary target of fluoride inhibition. * **C. Aconitase:** This is an enzyme of the **TCA cycle** (not glycolysis). It is inhibited by **Fluoroacetate** (via conversion to fluorocitrate), not sodium fluoride. **Clinical Pearls for NEET-PG:** 1. **Blood Glucose Estimation:** NaF is added to "Grey-top" vacutainers (along with Potassium Oxalate) to prevent **ex vivo glycolysis** by RBCs and WBCs. This ensures that the glucose level measured in the lab reflects the patient's actual blood sugar at the time of collection. 2. **Anticoagulant vs. Preservative:** In the grey-top tube, Potassium Oxalate acts as the anticoagulant, while Sodium Fluoride acts as the **antiglycolytic agent/preservative**. 3. **Fluoride and Teeth:** In dentistry, fluoride prevents dental caries by inhibiting the enolase of oral bacteria (like *S. mutans*), preventing acid production that erodes enamel.

Question 347: In non-competitive enzyme action, what occurs?

A. Vmax is increased
B. Apparent Km is increased
C. Apparent Km is unchanged (Correct Answer)
D. Concentration of active enzyme molecules is reduced

Explanation: In **non-competitive inhibition**, the inhibitor binds to a site other than the active site (the allosteric site). It can bind to either the free enzyme (E) or the enzyme-substrate complex (ES). ### Why the Correct Answer is Right: * **Apparent $K_m$ is unchanged:** Since the inhibitor does not compete with the substrate for the active site, the affinity of the enzyme for its substrate remains unaffected. Therefore, the concentration of substrate required to reach half of the maximum velocity ($K_m$) remains the same. * **Note on Option D:** While the inhibitor effectively "inactivates" a portion of the enzyme population, the standard biochemical description of non-competitive inhibition focuses on the kinetic parameters ($V_{max}$ and $K_m$). ### Why Incorrect Options are Wrong: * **A. $V_{max}$ is increased:** Incorrect. In non-competitive inhibition, $V_{max}$ **decreases**. Because the inhibitor reduces the catalytic efficiency of the enzyme, increasing substrate concentration cannot overcome the inhibition. * **B. Apparent $K_m$ is increased:** Incorrect. This occurs in **competitive inhibition**, where the inhibitor mimics the substrate and competes for the active site, requiring more substrate to reach $1/2 V_{max}$. * **D. Concentration of active enzyme molecules is reduced:** While technically true in a physical sense, in the context of enzyme kinetics questions, the primary answer sought is the effect on $K_m$ and $V_{max}$. ### NEET-PG High-Yield Pearls: 1. **Competitive Inhibition:** $V_{max}$ Unchanged, $K_m$ Increased (e.g., Statins, Methotrexate). 2. **Non-Competitive Inhibition:** $V_{max}$ Decreased, $K_m$ Unchanged (e.g., Cyanide on Cytochrome Oxidase, Fluoride on Enolase). 3. **Uncompetitive Inhibition:** Both $V_{max}$ and $K_m$ Decrease (e.g., Lithium on Inositol Monophosphatase). 4. **Lineweaver-Burk Plot:** In non-competitive inhibition, the plots intersect on the **negative X-axis** ($-1/K_m$).

Question 348: All of the following are covalent modifications of enzyme regulation EXCEPT?

A. Phosphorylation
B. ADP Ribosylation
C. Acetylation
D. Glycosylation (Correct Answer)

Explanation: **Explanation:** Enzyme regulation via **covalent modification** involves the reversible addition or removal of specific chemical groups to an enzyme’s amino acid residues, thereby altering its activity (turning it "on" or "off"). **Why Glycosylation is the Correct Answer:** While glycosylation is a covalent attachment of carbohydrates to proteins, it is primarily a **post-translational modification** involved in protein folding, stability, and cell-surface signaling (e.g., ABO blood groups or lysosomal enzyme tagging). It is generally **not** a mechanism used for the rapid, reversible regulation of enzyme catalytic activity in response to metabolic signals. **Analysis of Incorrect Options:** * **Phosphorylation (Option A):** The most common covalent modification. Enzymes like *Glycogen Phosphorylase* are activated by phosphorylation, while *Glycogen Synthase* is inactivated. * **ADP-Ribosylation (Option B):** Involves the transfer of ADP-ribose from NAD+. This is a key regulatory mechanism and is also the pathogenetic mechanism for toxins like **Cholera** and **Diphtheria** (inhibiting Elongation Factor-2). * **Acetylation (Option C):** Common in histones and metabolic enzymes. For example, acetylation of histones regulates gene expression by altering DNA binding. **High-Yield Clinical Pearls for NEET-PG:** * **Zymogen Activation:** Another form of covalent modification, but it is **irreversible** (e.g., Trypsinogen to Trypsin). * **Key Enzyme:** *HMG-CoA Reductase* (rate-limiting for cholesterol) is inactivated by phosphorylation. * **Toxin Link:** *Pertussis toxin* causes ADP-ribosylation of the Gi protein, leading to increased cAMP levels.

Question 349: Glyceraldehyde-3-phosphate dehydrogenase is inhibited by iodoacetate. This is an example of which type of enzyme inhibition?

A. Noncompetitive (Correct Answer)
B. Uncompetitive
C. Competitive
D. Allosteric

Explanation: **Explanation:** The inhibition of **Glyceraldehyde-3-phosphate dehydrogenase (GAPDH)** by **iodoacetate** is a classic example of **irreversible noncompetitive inhibition**. 1. **Why Noncompetitive is Correct:** GAPDH is a key glycolytic enzyme that contains a critical **sulfhydryl (-SH) group** at its active site (cysteine residue). Iodoacetate acts as an alkylating agent that covalently binds to this -SH group. Because it forms a stable covalent bond, it permanently inactivates the enzyme regardless of the substrate concentration. In medical biochemistry, irreversible inhibitors are categorized under the broad umbrella of noncompetitive inhibition because they decrease the $V_{max}$ and cannot be overcome by adding more substrate. 2. **Why Other Options are Wrong:** * **Competitive:** Competitive inhibitors bind reversibly to the active site and can be displaced by increasing substrate concentration ($K_m$ increases, $V_{max}$ unchanged). Iodoacetate’s binding is covalent and irreversible. * **Uncompetitive:** These inhibitors bind only to the enzyme-substrate (ES) complex. Iodoacetate binds to the free enzyme. * **Allosteric:** Allosteric inhibition involves binding at a site distant from the active site to induce a conformational change. Iodoacetate directly targets the functional group within the active site. **High-Yield Clinical Pearls for NEET-PG:** * **Glycolysis Blockade:** By inhibiting GAPDH, iodoacetate stops glycolysis, leading to a depletion of ATP and NADH. * **Fluoride vs. Iodoacetate:** While iodoacetate inhibits GAPDH, **Fluoride** (used in blood collection tubes) inhibits **Enolase** by chelating magnesium. * **Suicide Inhibition:** A related concept is "suicide inhibition" (e.g., Allopurinol inhibiting Xanthine Oxidase), where the enzyme converts a substrate analogue into a reactive inhibitor. * **Arsenite:** Note that **Arsenite** inhibits the Pyruvate Dehydrogenase complex by binding to the -SH groups of lipoic acid, a mechanism similar to iodoacetate's action on GAPDH.

Question 350: Which of the following does NOT allosterically inhibit phosphofructokinase-1 (PFK-1)?

A. Citrate
B. ATP
C. H+
D. AMP/ADP (Correct Answer)

Explanation: **Explanation:** Phosphofructokinase-1 (PFK-1) is the **rate-limiting and key committed step** of glycolysis, converting Fructose-6-phosphate to Fructose-1,6-bisphosphate. Its regulation is crucial for balancing cellular energy needs. **Why AMP/ADP is the correct answer:** AMP and ADP are indicators of a **low-energy state** in the cell. When ATP is consumed, AMP levels rise. AMP acts as a potent **allosteric activator** of PFK-1, signaling the cell to increase glycolytic flux to generate more ATP. Therefore, it does not inhibit the enzyme; it stimulates it. **Analysis of Incorrect Options (Inhibitors):** * **ATP:** Although a substrate, high levels of ATP act as an allosteric inhibitor. It binds to a regulatory site to decrease the enzyme's affinity for Fructose-6-phosphate, signaling that the cell has sufficient energy. * **Citrate:** An intermediate of the TCA cycle. High citrate levels indicate that the cycle is "saturated" and precursors for energy production are abundant, leading to the feedback inhibition of PFK-1. * **H+ (Low pH):** Accumulation of protons (acidosis) inhibits PFK-1. This is a protective mechanism, particularly in skeletal muscle, to prevent excessive lactic acid production and subsequent tissue damage during anaerobic glycolysis. **High-Yield NEET-PG Pearls:** * **Most Potent Activator:** Fructose-2,6-bisphosphate is the most powerful allosteric activator of PFK-1 (overcomes ATP inhibition). * **Insulin vs. Glucagon:** Insulin increases PFK-1 activity (via F-2,6-BP), while glucagon decreases it. * **Reciprocal Regulation:** Factors that activate PFK-1 (like F-2,6-BP and AMP) typically inhibit Fructose-1,6-bisphosphatase (gluconeogenesis), preventing a futile cycle.

Practice by Chapter

Enzyme Classification and Nomenclature

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Enzyme Kinetics and Michaelis-Menten Equation

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Enzyme Inhibition: Competitive and Non-competitive

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Allosteric Regulation

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Coenzymes and Cofactors

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Isoenzymes and Clinical Significance

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Enzyme Regulation: Covalent Modification

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Enzyme Regulation: Zymogen Activation

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Enzyme Induction and Repression

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Ribozymes and Catalytic RNA

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Enzyme Diagnostic Applications

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Enzyme Therapy and Inhibitors as Drugs

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Enzymes — MCQs

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