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BI2.1-5 | Enzyme — Gate Quiz
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Hexokinase catalyzes the phosphorylation of glucose using ATP. To which IUB class of enzymes does hexokinase belong?
Correct! Hexokinase transfers a phosphoryl group from ATP to glucose, making it a transferase (EC class 2).
IUB classification: 1-Oxidoreductases, 2-Transferases, 3-Hydrolases, 4-Lyases, 5-Isomerases, 6-Ligases. Hexokinase is a transferase (kinases are all transferases).
Incorrect. Enzymes that transfer functional groups from one molecule to another are transferases.
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The active site of an enzyme is best described as:
Correct! The active site is a small, three-dimensionally shaped cleft or pocket that binds substrate with specificity and contains catalytic residues.
Active sites are formed by amino acid residues that may be far apart in primary sequence but brought together in 3D structure. The lock-and-key and induced fit models describe substrate binding.
Incorrect. The active site is a specific region (not the whole enzyme) where substrate binds and catalysis occurs.
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In Michaelis-Menten kinetics, the Km represents the substrate concentration at which:
Correct! Km is the substrate concentration at which v = Vmax/2. It reflects enzyme-substrate affinity.
Low Km means high affinity. Hexokinase Km for glucose is approximately 0.1 mM vs. glucokinase approximately 10 mM, underlying their different physiological roles.
Incorrect. Km is defined as the [S] at which v = Vmax/2.
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On a Lineweaver-Burk (double reciprocal) plot, where does the x-intercept lie?
Correct! The x-intercept is -1/Km. Setting 1/v = 0 in the equation 1/v = (Km/Vmax)(1/[S]) + 1/Vmax gives 1/[S] = -1/Km.
Lineweaver-Burk plot: y-intercept = 1/Vmax, x-intercept = -1/Km, slope = Km/Vmax. Inhibitor effects on this plot distinguish competitive from non-competitive inhibition.
Incorrect. Setting 1/v = 0 gives x-intercept = -1/Km.
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Malonate inhibits succinate dehydrogenase by competing with succinate for the active site. Adding excess succinate reverses this inhibition. This is an example of:
Correct! Malonate is a structural analogue of succinate that competes for the active site. High substrate concentration overcomes this inhibition.
Competitive inhibitors increase apparent Km, Vmax unchanged. Malonate-succinate is the classical textbook example of competitive inhibition.
Incorrect. The ability to reverse inhibition with excess substrate defines competitive inhibition.
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A non-competitive inhibitor binds to the enzyme at a site other than the active site. Which of the following best describes its effect?
Correct! Non-competitive inhibition: substrate binding is unaffected (Km unchanged) but catalytic efficiency is reduced (Vmax decreases).
Pure non-competitive inhibitors bind to allosteric sites on both free enzyme and enzyme-substrate complex. Km unchanged, Vmax decreased. Example: heavy metals inhibiting many enzymes.
Incorrect. Non-competitive inhibitors reduce Vmax but do not affect Km.
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Phosphofructokinase-1 (PFK-1) is inhibited by ATP and citrate and activated by AMP and ADP. Binding of these metabolites at sites other than the active site is called:
Correct! Allosteric regulation involves modulators binding to regulatory sites that alter enzyme conformation and activity.
PFK-1 integrates energy status signals: high ATP/citrate (energy surplus) = inhibition; high AMP/ADP (energy deficit) = activation. Allosteric enzymes show sigmoidal (not hyperbolic) kinetics.
Incorrect. Binding of regulatory molecules at non-active sites constitutes allosteric regulation.
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During fasting, glucagon activates a cascade leading to phosphorylation of glycogen synthase. What happens to glycogen synthase activity?
Correct! Phosphorylation INACTIVATES glycogen synthase. During fasting, glucagon triggers cAMP-PKA-mediated phosphorylation of glycogen synthase, making it inactive.
Phosphorylation has OPPOSITE effects on glycogen synthase vs. phosphorylase. It activates phosphorylase (breakdown) but inactivates synthase (synthesis). This reciprocal control ensures glucose homeostasis.
Incorrect. Phosphorylation by PKA inactivates glycogen synthase.
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Lactate dehydrogenase (LDH) exists as 5 isoforms. The isoform predominantly found in cardiac muscle is:
Correct! LDH-1 (H4) predominates in cardiac muscle. Its elevation in blood is a marker of myocardial infarction.
In MI, LDH-1 > LDH-2 (the flip) indicates myocardial damage. LDH-1 favours conversion of lactate to pyruvate (aerobic tissue).
Incorrect. LDH-1 (H4) is the cardiac isoform; LDH-5 (M4) is the skeletal muscle/liver isoform.
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A 55-year-old man presents with chest pain. Serum markers show elevated CK-MB. In which tissue does CK-MB predominantly originate?
Correct! CK-MB (MB dimer) is predominantly found in cardiac muscle. Elevation indicates myocardial damage.
CK isoforms: CK-BB (brain), CK-MB (heart), CK-MM (skeletal muscle). In MI, CK-MB rises within 4-6 hours, peaks at 24 hours, returns to normal by 48-72 hours.
Incorrect. CK-MB is the cardiac isoform of creatine kinase.
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