Enzymes — MCQs

Enzymes Indian Medical PG Practice Questions and MCQs

Question 171: Fluoride causes inhibition of which of the following enzymes?

A. Pyruvate dehydrogenase (PDH)
B. Enolase (Correct Answer)
C. Glucose-6-phosphate dehydrogenase (G6PD)
D. Pyruvate kinase

Explanation: **Explanation:** **Correct Option: B. Enolase** Fluoride is a potent inhibitor of **Enolase**, the ninth enzyme in the glycolytic pathway. Enolase catalyzes the dehydration of 2-phosphoglycerate to phosphoenolpyruvate (PEP). The inhibition occurs because fluoride ions combine with magnesium (Mg²⁺) and phosphate to form a **magnesium-fluorophosphate complex**. This complex displaces the essential Mg²⁺ ions from the enzyme's active site, effectively halting glycolysis. **Analysis of Incorrect Options:** * **A. Pyruvate dehydrogenase (PDH):** This multi-enzyme complex converts pyruvate to Acetyl-CoA. It is primarily inhibited by high ratios of ATP/ADP and NADH/NAD⁺, or by Arsenite (which binds to lipoic acid), but not by fluoride. * **C. Glucose-6-phosphate dehydrogenase (G6PD):** This is the rate-limiting enzyme of the Pentose Phosphate Pathway (HMP Shunt). It is regulated by the availability of NADP⁺ and inhibited by high levels of NADPH. * **D. Pyruvate kinase:** This enzyme catalyzes the final step of glycolysis. It is inhibited by ATP, Alanine, and Glucagon (via phosphorylation), but is not the target of fluoride. **Clinical Pearls for NEET-PG:** * **Blood Glucose Estimation:** In clinical practice, blood samples for glucose estimation are collected in **Grey-topped vials** containing **Sodium Fluoride (NaF)**. This prevents "in vitro" glycolysis by RBCs, ensuring the measured glucose level reflects the patient's actual blood sugar at the time of draw. * **Anticoagulant Pairing:** NaF is usually paired with **Potassium Oxalate** (an anticoagulant) because fluoride alone is a poor anticoagulant. * **Water Fluoridation:** At low concentrations, fluoride prevents dental caries by inhibiting bacterial enolase in oral plaque.

Question 172: Glutathione synthetase requires which cofactor?

A. Copper
B. Selenium
C. Magnesium (Correct Answer)
D. Iron

Explanation: **Explanation:** The correct answer is **Magnesium (Mg²⁺)**. **1. Why Magnesium is correct:** Glutathione synthetase is the second enzyme in the synthesis of Glutathione (GSH), catalyzing the condensation of γ-glutamylcysteine and glycine. This reaction is **ATP-dependent**. In biochemistry, almost all enzymes that utilize ATP (kinases and synthetases) require **Magnesium** as a mandatory cofactor. Magnesium ions complex with ATP (Mg-ATP) to shield the negative charges of the phosphate groups, facilitating the nucleophilic attack required for the reaction to proceed. **2. Why other options are incorrect:** * **Selenium (B):** This is a common distractor. Selenium is the essential cofactor for **Glutathione Peroxidase**, the enzyme responsible for neutralizing hydrogen peroxide using reduced glutathione. It is not required for the synthesis of glutathione itself. * **Copper (A) & Iron (D):** These are involved in redox reactions (e.g., Cytochrome c oxidase, Catalase, Superoxide Dismutase) but do not play a role in the ligase activity of glutathione synthetase. **3. Clinical Pearls & High-Yield Facts for NEET-PG:** * **Glutathione Components:** It is a tripeptide made of **Glutamate, Cysteine, and Glycine**. Cysteine is the rate-limiting amino acid. * **Two-Step Synthesis:** 1. γ-glutamylcysteine synthetase (requires Mg²⁺/ATP). 2. Glutathione synthetase (requires Mg²⁺/ATP). * **Glutathione Peroxidase vs. Reductase:** Peroxidase requires **Selenium**; Reductase requires **NADPH** (derived from the HMP Shunt) and **FAD**. * **Clinical Correlation:** Deficiency of glutathione synthetase leads to **5-oxoprolinuria** (pyroglutamic aciduria), characterized by metabolic acidosis, hemolytic anemia, and neurological symptoms.

Question 173: What is an abzyme?

A. Isoenzyme
B. Allosteric enzyme
C. Abnormal enzyme
D. Antibody with a catalytic activity (Correct Answer)

Explanation: ### Explanation **Correct Answer: D. Antibody with a catalytic activity** **Concept:** The term **Abzyme** is a portmanteau of "Antibody" and "Enzyme." These are monoclonal antibodies produced against a **transition-state analog** of a specific reaction. According to the transition-state theory, enzymes work by binding most tightly to the high-energy transition state of a substrate. By creating an antibody that mimics this binding site, the antibody can stabilize the transition state of a chemical reaction, thereby lowering the activation energy and acting as a catalyst. **Why Incorrect Options are Wrong:** * **A. Isoenzyme:** These are physically distinct forms of the same enzyme (different amino acid sequences) that catalyze the same chemical reaction but may have different kinetic properties (e.g., LDH1 vs. LDH5). * **B. Allosteric enzyme:** These are enzymes whose activity is regulated by the binding of an effector molecule at a site other than the active site (the allosteric site), causing a conformational change (e.g., Phosphofructokinase-1). * **C. Abnormal enzyme:** This is a non-specific term usually referring to enzymes with genetic mutations (enzymopathies) or altered function due to denaturation, not a specific class of catalytic antibodies. **High-Yield Clinical Pearls for NEET-PG:** * **Mechanism:** Abzymes are generated by immunizing animals with a stable molecule that resembles the **transition state** of a reaction, not the substrate itself. * **Clinical Potential:** Abzymes are being researched for targeted prodrug activation, clearing toxins (e.g., cocaine esterase abzymes for overdose), and viral inactivation (e.g., HIV gp120). * **Natural Occurrence:** Autoimmune diseases (like SLE or Hashimoto’s) can sometimes produce natural abzymes that can cleave DNA or proteins. * **Synonym:** Also known as **Catmabs** (Catalytic Monoclonal Antibodies).

Question 174: What effect does a competitive inhibitor have on an enzyme?

A. It alters the Vmax of the reaction.
B. It binds to the same site as the substrate. (Correct Answer)
C. It decreases the apparent Km for the substrate.
D. It decreases the turnover number.

Explanation: ### Explanation **1. Why Option B is Correct:** Competitive inhibition occurs when an inhibitor molecule closely resembles the substrate in structure. Because of this structural similarity, the inhibitor competes directly with the substrate for the **active site** of the enzyme. When the inhibitor occupies the active site, the substrate cannot bind. This competition can be overcome by increasing the substrate concentration ($[S]$), which "outcompetes" the inhibitor. **2. Analysis of Incorrect Options:** * **Option A (Vmax):** In competitive inhibition, **$V_{max}$ remains unchanged**. Since high concentrations of substrate can displace the inhibitor, the enzyme can still reach its maximum velocity. * **Option C (Km):** Competitive inhibitors **increase the apparent $K_m$**. $K_m$ is the substrate concentration required to reach $1/2 V_{max}$. Because the inhibitor interferes with substrate binding, more substrate is needed to achieve the same rate, signifying a decreased affinity. * **Option D (Turnover Number/$k_{cat}$):** The turnover number represents the maximum number of substrate molecules converted to product per active site per unit time. Since $V_{max}$ is unchanged, the $k_{cat}$ remains constant in competitive inhibition. **3. NEET-PG High-Yield Clinical Pearls:** * **Lineweaver-Burk Plot:** On a double-reciprocal plot, competitive inhibition lines intersect at the **y-axis** ($1/V_{max}$ is constant). * **Classic Clinical Examples:** * **Statins** (e.g., Atorvastatin) are competitive inhibitors of HMG-CoA reductase. * **Methanol poisoning** is treated with **Ethanol** or **Fomepizole**, which competitively inhibit Alcohol Dehydrogenase. * **Methotrexate** is a competitive inhibitor of Dihydrofolate Reductase (DHFR). * **Sulfonamides** compete with PABA for Dihydropteroate synthase.

Question 175: What is the Michaelis Menten constant (Km)?

A. The substrate concentration at which the reaction velocity is half of the maximum velocity (Vmax). (Correct Answer)
B. The substrate concentration at which the reaction velocity is maximum.
C. Both of the above.
D. None of the above.

Explanation: ### Explanation **1. Why Option A is Correct:** The Michaelis-Menten constant ($K_m$) is a fundamental parameter in enzyme kinetics. It is defined as the **specific substrate concentration $[S]$ at which the reaction velocity ($v$) is exactly half of the maximum velocity ($V_{max}$)**. Mathematically, from the Michaelis-Menten equation: $v = \frac{V_{max}[S]}{K_m + [S]}$. When $[S] = K_m$, the equation simplifies to $v = \frac{V_{max}}{2}$. Biochemically, $K_m$ reflects the **affinity** of an enzyme for its substrate. A **low $K_m$** indicates high affinity (the enzyme reaches half-saturation at low substrate levels), while a **high $K_m$** indicates low affinity. **2. Why Other Options are Incorrect:** * **Option B:** The substrate concentration at which velocity is maximum is theoretically infinite. In a hyperbolic curve, the reaction velocity approaches $V_{max}$ asymptotically but never truly reaches it at a finite concentration. * **Option C & D:** These are incorrect because $K_m$ has a singular, specific definition related to $1/2 V_{max}$. **3. NEET-PG High-Yield Clinical Pearls:** * **Glucokinase vs. Hexokinase:** This is a classic exam favorite. **Hexokinase** has a **low $K_m$** (high affinity), allowing it to trap glucose even at low blood levels. **Glucokinase** (in the liver/pancreas) has a **high $K_m$** (low affinity), functioning only when glucose levels are high (postprandial). * **Enzyme Inhibitors:** * **Competitive Inhibition:** $K_m$ increases (affinity decreases), but $V_{max}$ remains unchanged. * **Non-competitive Inhibition:** $K_m$ remains unchanged, but $V_{max}$ decreases. * **Lineweaver-Burk Plot:** On a double-reciprocal plot, the x-intercept is $-1/K_m$.

Question 176: Which enzyme is diagnostic of myocardial infarction in a case of hypothyroidism?

A. SGOT
B. LDH
C. Aldolase
D. CPK-MB (Correct Answer)

Explanation: **Explanation:** The correct answer is **CPK-MB** (Creatine Phosphokinase-MB). **Why CPK-MB is the correct answer:** In patients with **hypothyroidism**, there is a baseline elevation of total CPK (specifically the CPK-MM isoform) due to increased muscle membrane permeability and decreased clearance of the enzyme. Because total CPK is already high in hypothyroid patients, it cannot be used to diagnose a Myocardial Infarction (MI). However, the **CPK-MB isoenzyme** is specific to cardiac muscle. Even in the presence of high total CPK, a rise in the MB fraction (typically >5% of total CPK) remains a highly specific and diagnostic marker for acute myocardial injury in these patients. **Analysis of Incorrect Options:** * **A. SGOT (AST):** While AST rises in MI, it is found in the liver, skeletal muscle, and RBCs. It lacks the cardiac specificity required, especially when skeletal muscle involvement is suspected in hypothyroidism. * **B. LDH:** LDH is a late marker for MI. Like CPK, total LDH can be elevated in various conditions (including hypothyroidism and hemolysis), making it less specific than CPK-MB for acute diagnosis. * **C. Aldolase:** This enzyme is a marker for skeletal muscle diseases (like muscular dystrophy) and has no diagnostic role in myocardial infarction. **High-Yield Clinical Pearls for NEET-PG:** * **Gold Standard:** While CPK-MB is the answer here, **Cardiac Troponins (I and T)** are the overall most sensitive and specific markers for MI. * **Time Frame:** CPK-MB rises within 4–6 hours, peaks at 24 hours, and returns to normal within 48–72 hours. It is the marker of choice for detecting **re-infarction**. * **Hypothyroidism Fact:** Always remember that **elevated CPK-MM** is a common biochemical finding in hypothyroid myopathy.

Question 177: In competitive inhibition, what happens to the substrate affinity to the enzyme?

A. Decreases (Correct Answer)
B. Increases
C. Any of the above
D. Does not change

Explanation: In competitive inhibition, the inhibitor structurally resembles the substrate and competes for the same **active site** on the enzyme. ### Why the Correct Answer is Right (Option A) The term **"affinity"** refers to how strongly an enzyme binds to its substrate, which is inversely represented by the **Michaelis constant ($K_m$)**. * In competitive inhibition, the presence of the inhibitor makes it harder for the substrate to bind to the active site. * To achieve the same reaction velocity ($1/2 V_{max}$), a much higher concentration of substrate is required to "outcompete" the inhibitor. * Since $K_m$ increases, the **affinity decreases**. ### Why Other Options are Wrong * **Option B (Increases):** Affinity never increases in the presence of an inhibitor. An increase in affinity would mean a decrease in $K_m$, which occurs in certain types of allosteric activation, not inhibition. * **Option D (Does not change):** This occurs in **Non-competitive inhibition**. In that case, the inhibitor binds to an allosteric site, not the active site; therefore, the binding ability (affinity/$K_m$) of the substrate remains unchanged, but the maximal velocity ($V_{max}$) decreases. ### NEET-PG High-Yield Pearls 1. **The $V_{max}$ Rule:** In competitive inhibition, $V_{max}$ remains **unchanged** because the inhibition can be completely reversed by increasing substrate concentration. 2. **Lineweaver-Burk Plot:** The lines for inhibited and uninhibited reactions intersect on the **Y-axis** ($1/V_{max}$ is constant). 3. **Clinical Example:** **Statins** (e.g., Atorvastatin) are competitive inhibitors of HMG-CoA reductase. **Methanol poisoning** is treated with Ethanol because they compete for the active site of Alcohol Dehydrogenase.

Question 178: The activity of which of the following enzymes is directly affected by citrate?

A. Fructose-2,6-bisphosphatase
B. Isocitrate dehydrogenase
C. Phosphofructokinase I (Correct Answer)
D. Pyruvate carboxylase

Explanation: ### Explanation **Correct Answer: C. Phosphofructokinase I (PFK-1)** **Why it is correct:** Phosphofructokinase-1 (PFK-1) is the **rate-limiting enzyme** of glycolysis. It is subject to complex allosteric regulation to ensure that glucose is only broken down when the cell requires energy. **Citrate** acts as a potent **allosteric inhibitor** of PFK-1. * **The Mechanism:** High levels of citrate in the cytosol indicate that the TCA cycle is "saturated" and energy stores (ATP) are high. Citrate enhances the inhibitory effect of ATP on PFK-1, effectively slowing down glycolysis to prevent the unnecessary buildup of metabolic intermediates. **Why the other options are incorrect:** * **A. Fructose-2,6-bisphosphatase:** This enzyme is part of the bifunctional enzyme complex (with PFK-2) regulated primarily by **cAMP-dependent phosphorylation** (via Glucagon/Insulin ratio), not directly by citrate. * **B. Isocitrate dehydrogenase:** This is the rate-limiting enzyme of the TCA cycle. It is primarily regulated by the **ADP/ATP ratio** and **NAD+/NADH ratio**, and stimulated by **Ca²⁺**. While citrate is a product of the preceding step, it does not directly regulate this enzyme. * **D. Pyruvate carboxylase:** This gluconeogenic enzyme is obligatorily activated by **Acetyl-CoA**, not citrate. This ensures that when Acetyl-CoA levels are high, pyruvate is diverted toward oxaloacetate for gluconeogenesis. **High-Yield Clinical Pearls for NEET-PG:** * **Most potent activator of PFK-1:** Fructose-2,6-bisphosphate (overcomes citrate inhibition). * **Dual Role of Citrate:** Citrate inhibits glycolysis (PFK-1) but **activates fatty acid synthesis** by stimulating **Acetyl-CoA Carboxylase (ACC)**. This coordinates carbohydrate breakdown with lipid storage. * **Citrate Shuttle:** Citrate is the form in which acetyl groups leave the mitochondria to enter the cytosol for de novo lipogenesis.

Question 179: Active sites of serine proteases contain which amino acid residue?

A. Histidine
B. Lysine or threonine
C. Arginine
D. Serine (Correct Answer)

Explanation: **Explanation:** **Serine proteases** are a class of enzymes that cleave peptide bonds in proteins. The correct answer is **Serine** because these enzymes utilize a uniquely reactive serine residue in their active site to perform a nucleophilic attack on the carbonyl carbon of the substrate's peptide bond. The hallmark of serine proteases is the **"Catalytic Triad,"** a coordinated structure consisting of three specific amino acids: **Serine (Ser 195), Histidine (His 57), and Aspartate (Asp 102)**. In this mechanism, Histidine acts as a base to withdraw a proton from Serine, making the Serine oxygen highly nucleophilic and capable of breaking the peptide bond. **Analysis of Incorrect Options:** * **Option A (Histidine):** While Histidine is part of the catalytic triad, the question asks for the primary residue that defines the class and performs the nucleophilic attack. Serine is the definitive residue for this group. * **Option B (Lysine/Threonine):** Threonine is found in the active sites of "Threonine proteases" (e.g., proteasomes), but not serine proteases. Lysine is often involved in binding but not as the primary nucleophile in this class. * **Option C (Arginine):** Arginine is frequently involved in substrate recognition (e.g., Trypsin cleaves after Arg/Lys), but it does not function as the catalytic residue in the active site. **High-Yield Clinical Pearls for NEET-PG:** * **Examples of Serine Proteases:** Trypsin, Chymotrypsin, Elastase, Thrombin, Plasmin, and C1 esterase. * **Inhibitors:** Serine proteases are irreversibly inhibited by **Diisopropylfluorophosphate (DFP)** and nerve gases (Sarin), which bind to the active site Serine. * **Clinical Correlation:** **Alpha-1 Antitrypsin deficiency** leads to uncontrolled activity of Neutrophil Elastase (a serine protease), resulting in emphysema and liver cirrhosis.

Question 180: Aspartate transaminase, an enzyme, is most abundant in which of the following organs?

A. Brain
B. Heart (Correct Answer)
C. Spleen
D. Retina

Explanation: **Explanation:** **Aspartate Transaminase (AST)**, formerly known as Serum Glutamic Oxaloacetic Transaminase (SGOT), is a pyridoxal phosphate (Vitamin B6)-dependent enzyme. It catalyzes the reversible transfer of an amino group between aspartate and alpha-ketoglutarate to form oxaloacetate and glutamate. **Why Heart is Correct:** AST is found in high concentrations in tissues with high metabolic activity. The **heart (myocardium)** contains the highest concentration of AST per unit of tissue, followed closely by the liver and skeletal muscle. Historically, AST was the first cardiac biomarker used to diagnose Myocardial Infarction (MI). Following an MI, AST levels rise within 6–8 hours, peak at 24–48 hours, and return to baseline within 4–6 days. **Why Other Options are Incorrect:** * **Brain:** While AST is present in the brain, its concentration is significantly lower than in cardiac or hepatic tissues. * **Spleen:** The spleen contains minimal amounts of AST; it is not a primary source for this enzyme. * **Retina:** The retina relies on specific metabolic enzymes, but AST is not found there in clinically significant or "most abundant" quantities. **High-Yield Clinical Pearls for NEET-PG:** 1. **Tissue Distribution:** Heart > Liver > Skeletal Muscle > Kidney > Pancreas > RBCs. 2. **De Ritis Ratio:** The AST/ALT ratio. A ratio **>2:1** is highly suggestive of **Alcoholic Liver Disease**, as alcohol is a mitochondrial toxin and AST has both mitochondrial and cytosolic isoenzymes. 3. **Hemolysis:** Since AST is present in RBCs, a hemolyzed blood sample will show falsely elevated AST levels. 4. **Current Status:** In modern practice, AST has been replaced by **Troponins (I and T)** for MI diagnosis due to their superior cardiac specificity.

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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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