Question 1

Catalysis by serine proteases — Identify the primary nucleophile that initiates peptide bond cleavage in the catalytic cycle

Accepted Answer

Introduction: Serine proteases (e.g., chymotrypsin, trypsin, elastase) catalyze peptide bond hydrolysis through a conserved catalytic triad. This question focuses on the nucleophile that initiates the acylation step of catalysis. Given Data / Assumptions: The catalytic triad comprises Ser, His, and Asp (or Glu in some families). Nucleophilic attack on the carbonyl carbon forms a tetrahedral intermediate. Hydrolysis proceeds in two stages: acylation and deacylation. Concept / Approach: During acylation, the Ser hydroxyl is deprotonated by His to generate a reactive alkoxide, which attacks the peptide carbonyl. Water becomes the nucleophile only in the second stage (deacylation) to release the acyl-enzyme intermediate and regenerate free enzyme. Step-by-Step Solution: 1) His abstracts a proton from Ser-OH, assisted by Asp that stabilizes His.2) Ser-O⁻ (alkoxide) attacks the substrate carbonyl → first tetrahedral intermediate.3) Collapse releases the amine leaving group and forms the acyl-enzyme.4) Water then acts as the nucleophile in deacylation, attacking the acyl-enzyme to release the carboxyl product. Verification / Alternative check: Covalent inhibition by fluorophosphonates selectively labels the active-site Ser, supporting its role as the primary nucleophile in acylation. Kinetic solvent isotope effects further distinguish steps involving water in the second half-reaction. Why Other Options Are Wrong: b) Water is the nucleophile in deacylation, not the initiating step.c) There is temporal separation; not simultaneous co-nucleophiles.d) Asparagine is not part of the catalytic triad nucleophile.e) Proline's imino group does not perform nucleophilic attack here. Common Pitfalls: Conflating acylation and deacylation; assuming water initiates all protease reactions; overlooking the role of His and Asp in activating Ser. Final Answer: Serine (active-site Ser as an alkoxide) is the primary initiating nucleophile.

Question 2

Metabolic pathway organization — Which cellular strategies commonly enhance the efficiency of enzyme-driven pathways?

Accepted Answer

Introduction: Cells organize metabolism spatially and structurally to accelerate flux, minimize diffusion losses, and prevent side reactions. This question examines three widely used strategies that make pathways more efficient. Given Data / Assumptions: Sequential reactions benefit from substrate channeling. Compartments create specialized chemical environments. Membrane anchoring places enzymes in proximity to substrates or each other. Concept / Approach: Bringing enzymes together reduces transit time between steps and can protect unstable intermediates. Compartmentalization also enables distinct pH, redox, or cofactor conditions that match pathway needs, improving yield and regulation. Step-by-Step Solution: 1) Compartmentalization: organelles like mitochondria or peroxisomes confine pathways, enhancing coordination and avoiding interference.2) Multienzyme complexes: physical assemblies promote substrate channeling directly from one active site to the next.3) Membrane localization: fixing enzymes in order along a membrane (e.g., electron transport) ensures rapid handoff of intermediates. Verification / Alternative check: Experimental reconstitution of enzyme scaffolds often yields higher pathway flux. Natural examples include pyruvate dehydrogenase complex and fatty acid synthase, both of which exemplify multienzyme channeling. Why Other Options Are Wrong: e) Spatial organization is a well-documented determinant of metabolic efficiency; claiming no effect contradicts abundant biochemical evidence. Common Pitfalls: Assuming diffusion is always fast enough; overlooking metabolite leakage; ignoring that scaffolding can also aid regulation and cofactor recycling. Final Answer: All of the above strategies enhance efficiency.

Question 3

Biochemistry—Metabolic Pathway Logic In the enzyme-catalyzed pathway below, predict the effect of inactivating enzyme E2 on the appearance of intermediates and final product. Assume a unidirectional metabolic sequence under steady supply of A. A ---…

Accepted Answer

Introduction / Context: Enzyme-catalyzed pathways convert substrates into products through a series of discrete, ordered steps. Each step is typically catalyzed by a specific enzyme. This question examines pathway logic: what happens upstream and downstream when one enzyme (E2) is specifically inactivated, while the rest remain functional and substrate A is continuously available. Given Data / Assumptions: Linear pathway: A --(E1)--> B --(E2)--> C --(E3)--> D. Only E2 is inactivated; E1 and E3 retain full activity. Substrate A input continues; no alternative bypass routes exist. Products accumulate according to flux permitted by available enzymes. Concept / Approach: In a linear pathway, blocking a step prevents formation of all downstream intermediates and products beyond that step, but upstream reactions may continue and cause accumulation of intermediates just before the block. Therefore, activity prior to E2 (E1) proceeds, while any reaction requiring C cannot occur. E3 remains idle due to lack of its substrate C. Step-by-Step Solution: Step 1: With E1 active, A is converted to B. Thus A and B are present (A as feed, B as product of E1).Step 2: E2 is inactive, so B cannot be converted to C. Therefore C cannot form.Step 3: Without C, E3 lacks substrate and cannot produce D. Therefore D cannot form.Step 4: The net effect is presence of A and B only; C and D are absent. Verification / Alternative check: Consider steady-state flux J. If a step is completely blocked, J downstream becomes zero. Upstream steps may still run, leading to B accumulation until regulatory feedback or substrate limitation reduces E1 flux. Why Other Options Are Wrong: A, B, C, and D all present: impossible without E2 to make C. A, B, C present but not D: C cannot form without E2. A only: E1 is still active, so B forms. None present: feed A is still present and E1 still forms B. Common Pitfalls: Assuming E3 can act on B directly; assuming spontaneous conversion of B to C without catalysis; overlooking that upstream reactions can continue despite a downstream block. Final Answer: A and B will still be produced, but not C or D

Question 4

Biochemistry—Reaction Kinetics and Transition State Before most molecules can react, they must be destabilized to reach the transition state. Which factor typically achieves this requirement in biochemical reactions?

Accepted Answer

Introduction / Context: Chemical reactions proceed when reactants surmount an energy barrier to form an activated complex (transition state). This barrier is the activation energy, which must be supplied, either thermally or by catalytic mechanisms that reduce the required amount. The question probes understanding of how reactions are initiated at the molecular level in cells. Given Data / Assumptions: Molecules require sufficient energy to reach a high-energy transition state. Enzymes lower, but do not eliminate, the activation energy. Reaction spontaneity (negative free energy change) does not imply instantaneous reaction. Cellular conditions are mild (aqueous, near-neutral pH, moderate temperatures). Concept / Approach: Identify the fundamental requirement for bond breaking/forming: energy to distort bonds into the transition-state geometry. Catalysts provide alternative pathways with lower activation energy, often by stabilizing the transition state and properly orienting substrates, thereby reducing the energy needed to initiate reaction. Step-by-Step Solution: Step 1: Recognize that progress over the energy barrier requires activation energy input.Step 2: Enzymes lower the barrier so that only a small energy input is necessary under physiological conditions.Step 3: Conclude that providing activation energy (or lowering it via catalysis) is the key requirement for reaction initiation. Verification / Alternative check: Arrhenius behavior shows reaction rates depend exponentially on activation energy. Lowering this energy dramatically increases rate, confirming the central role of activation energy rather than arbitrary shape changes or pathway labels. Why Other Options Are Wrong: Random 3D shape change: does not necessarily generate the proper activated geometry. Oxidation: a chemical transformation, not a universal prerequisite for all reactions. Biosynthetic vs catabolic label: does not determine activation barrier. Unlimited concentration: increases collision frequency but cannot replace the need to cross the energy barrier. Common Pitfalls: Confusing thermodynamics (free energy change) with kinetics (activation energy); assuming catalysts change overall free energy, which they do not. Final Answer: Input of a small amount of activation energy to reach the transition state

Question 5

Biochemistry—Core Properties of Enzymes Which of the following statements correctly describes enzymes and their catalytic function?

Accepted Answer

Introduction / Context: Enzymes are nature’s catalysts, enabling rapid, specific chemical transformations in living systems. Understanding what enzymes change (rate, pathway) and what they do not change (overall thermodynamic favorability) is foundational to biochemistry and cell physiology. Given Data / Assumptions: Reactions have an intrinsic free energy change (ΔG) governed by reactants and products. Activation energy (Ea) determines the kinetic barrier to reaction. Protein conformation dictates active-site geometry and specificity. Concept / Approach: Evaluate each statement against enzyme principles: enzymes lower Ea by stabilizing the transition state, orienting substrates, and providing catalytic groups. They do not change ΔG, equilibrium constant, or reaction stoichiometry; they simply allow equilibrium to be reached faster. Their function depends critically on native 3D structure (fold) that forms the active site. Step-by-Step Solution: Step 1: Assess A: Enzymes do not alter ΔG or equilibrium; correct.Step 2: Assess B: Most enzymes are proteins, and structure determines activity; correct.Step 3: Assess C: Lowering activation energy is the principal kinetic effect; correct.Step 4: Since A–C are correct, choose D (All of the above). Verification / Alternative check: Thermodynamic cycles show ΔG depends only on initial and final states; catalysts change the pathway, not the endpoints. Kinetic data demonstrate exponential sensitivity of rate to Ea. Why Other Options Are Wrong: E excludes statement B, which is a central truth about protein enzymes. Common Pitfalls: Thinking enzymes change equilibrium position; ignoring that some catalysts can be RNA (ribozymes), though the general protein statement remains valid for most enzymes. Final Answer: All of the above

Question 6

Biochemistry—Catalysis and Energetics Chymotrypsin increases the rate of peptide bond hydrolysis by approximately 10^10. At physiological temperature, this rate enhancement corresponds most closely to what reduction in activation free energy (ΔΔG‡)?

Accepted Answer

Introduction / Context: Rate enhancements by enzymes can be translated into decreases in activation free energy using the relationship between rate constants and activation barriers. This question connects kinetic observables (fold-increase in rate) with thermodynamic barriers (ΔG‡) at a given temperature. Given Data / Assumptions: Rate enhancement ≈ 10^10 relative to uncatalyzed reaction. Temperature near 298 K for estimation. Relationship: ΔΔG‡ ≈ −RTln(k_cat/k_uncat). R ≈ 8.314 J/mol·K; RT at 298 K ≈ 2.48 kJ/mol. Concept / Approach: Use the exponential dependence of rate on activation energy. A 10^10 enhancement implies a substantial lowering of the barrier. Compute ΔΔG‡ with RT and the natural logarithm of the rate ratio to estimate the energetic advantage provided by chymotrypsin’s catalytic strategy (transition-state stabilization, covalent catalysis, general acid-base catalysis, and precise positioning). Step-by-Step Solution: Step 1: Compute ln(10^10) = 10 * ln(10) ≈ 10 * 2.3026 ≈ 23.026.Step 2: Multiply by RT: 2.48 kJ/mol * 23.026 ≈ 57.1 kJ/mol.Step 3: Round to the closest listed value: approximately 58 kJ/mol. Verification / Alternative check: Sensitivity analysis: if T varies modestly around 298 K, RT shifts slightly (about 2.5 kJ/mol), giving ΔΔG‡ still near 57–59 kJ/mol, confirming that 58 kJ/mol is the best match among options. Why Other Options Are Wrong: 40 or 49 kJ/mol: underestimate the barrier reduction implied by a 10^10 rate boost. 88 kJ/mol: overestimates the barrier reduction. 25 kJ/mol: far too small for 10 orders of magnitude in rate. Common Pitfalls: Using log base 10 directly without converting to natural log; forgetting to multiply by RT in kJ/mol rather than J/mol; mixing enthalpic and entropic interpretations. Final Answer: 58 kJ/mol

Question 7

Biochemistry—Serine Protease Architecture What is the common functional feature shared by all classical serine proteases (e.g., trypsin, chymotrypsin, elastase)?

Accepted Answer

Introduction / Context: Serine proteases are a major protease family characterized by a conserved catalytic apparatus. Despite differing substrate preferences, their catalytic machinery follows a shared design logic that defines the family and explains their powerful hydrolytic activity against peptide bonds. Given Data / Assumptions: Serine proteases include trypsin, chymotrypsin, and elastase. Catalytic residues typically form a triad: Ser, His, Asp. Specificity pockets differ between family members to recognize side chains. Concept / Approach: Identify the invariant catalytic hallmark of the serine protease fold: a single nucleophilic serine whose reactivity is enhanced by histidine (as a base) and aspartate (to orient and stabilize His). This arrangement supports covalent catalysis via an acyl-enzyme intermediate followed by hydrolytic deacylation. Step-by-Step Solution: Step 1: Recognize that each enzyme uses one key serine Oγ as the nucleophile.Step 2: Note that the histidine deprotonates serine to generate the alkoxide nucleophile; aspartate stabilizes histidine.Step 3: Understand that specificity pockets vary (e.g., deep hydrophobic for chymotrypsin; acidic for trypsin), so “universally hydrophobic/hydrophilic pocket” is not a common feature. Verification / Alternative check: Structural superposition of serine proteases shows conserved catalytic triad geometry despite variations in the S1 pocket that confer substrate specificity. Why Other Options Are Wrong: Universally hydrophobic or hydrophilic pockets: pocket properties vary by enzyme and substrate preference. Cluster of many reactive serines: only one serine is catalytic; others are structural. Metal dependence: serine proteases are not metalloenzymes. Common Pitfalls: Assuming specificity determinants are conserved identically across the family; confusing catalytic serine with noncatalytic serines. Final Answer: A single, catalytically essential serine residue in the active site (within a catalytic triad)

Question 8

Biochemistry—Protease Specificity Rules What peptide bond specificity is characteristic of the protease clostripain?

Accepted Answer

Introduction / Context: Different proteases display distinct primary specificities, recognizing particular side chains near the scissile bond. Knowing these rules allows prediction of fragment patterns after digestion and informs protein sequencing strategies. Given Data / Assumptions: Clostripain is a protease from Clostridium species. Primary specificity distinguishes it from trypsin and other proteases. We focus on the residue on the P1 side of the cleavage site. Concept / Approach: Recall canonical specificities: trypsin cleaves after lysine or arginine (with context effects), while clostripain shows strong preference for cleavage after arginine specifically. Therefore, among the options, the correct description is cleavage on the carboxyl side of arginine residues. Step-by-Step Solution: Step 1: Identify the distinctive feature of clostripain—arginine selectivity.Step 2: Determine the side of cleavage: carboxyl side (after Arg) is the usual convention for these proteases.Step 3: Match the phrasing exactly with the option indicating “after arginine.” Verification / Alternative check: Empirical digests and proteomic mapping consistently show Arg-specific cleavage patterns for clostripain under standard conditions. Why Other Options Are Wrong: After His or after Lys: not the hallmark specificity of clostripain. “None of the above”: incorrect because Arg specificity is well established. Before Arg: incorrect side of cleavage for the described enzyme. Common Pitfalls: Confusing clostripain with trypsin; mixing up “before” versus “after” nomenclature when describing cleavage sites. Final Answer: It cleaves on the carboxyl side (after) arginine residues

Question 9

Biochemistry—Multienzyme Complexes Tryptophan synthase of Escherichia coli is a classic bifunctional oligomeric enzyme. What is its subunit composition?

Accepted Answer

Introduction / Context: Many biosynthetic pathways use multienzyme complexes that channel intermediates between active sites. Tryptophan synthase is a well-studied example that couples two sequential reactions to produce tryptophan from indole-3-glycerol phosphate and serine, minimizing diffusion losses of the reactive intermediate indole. Given Data / Assumptions: Tryptophan synthase is described as bifunctional and oligomeric. The canonical E. coli enzyme is known to have distinct subunits often labeled A (α) and B (β). Subunits assemble into a higher-order complex for efficient catalysis. Concept / Approach: Identify the minimal accurate description: two different subunits (A and B) associate to form the complete enzyme. This composition underlies substrate channeling: the α-subunit generates indole, which is directly delivered to the β-subunit active site for condensation with serine to form tryptophan. Step-by-Step Solution: Step 1: Recognize “bifunctional” implies two catalytic activities housed in distinct subunits.Step 2: Recall the established α2β2 architecture in many organisms; the essential point is the presence of A and B subunits.Step 3: Select the option that explicitly states “two proteins designated A and B.” Verification / Alternative check: Structural and kinetic studies show inter-subunit tunnels guiding indole from A to B, consistent with the AB composition. Why Other Options Are Wrong: Only A or only B: would delete half the catalytic function. “A plus one-subunit a”: nonstandard and incorrect nomenclature. Three proteins A, B, C: not supported for the classic bacterial enzyme. Common Pitfalls: Confusing the two-subunit composition with other multienzyme assemblies that include additional regulatory subunits; conflating stoichiometry details with the essential identity of A and B. Final Answer: Two proteins designated A and B forming a functional complex

Question 10

Biochemistry—Comparing Protease Mechanisms Which statement is FALSE when comparing classical serine proteases with the HIV protease (an aspartyl protease)?

Accepted Answer

Introduction / Context: Proteases employ diverse catalytic strategies to hydrolyze peptide bonds. Serine proteases and aspartyl proteases (such as HIV protease) represent two mechanistic archetypes. Recognizing what is shared and what is distinct is crucial for understanding inhibitor design and drug specificity. Given Data / Assumptions: Serine proteases use a catalytic Ser-His-Asp triad and proceed via a covalent acyl-enzyme intermediate. Aspartyl proteases use two aspartates to activate a water molecule for direct attack; no covalent acyl intermediate with the enzyme is formed. Both enzyme classes require water for hydrolysis and exhibit sequence preferences. Concept / Approach: Identify the statement that incorrectly claims a shared mechanistic detail. The presence or absence of a covalent acyl-enzyme intermediate is the critical discriminator: present in serine proteases; absent in aspartyl proteases, which use general acid-base catalysis to activate water for direct nucleophilic attack on the peptide carbonyl. Step-by-Step Solution: Step 1: Confirm that serine proteases form an acyl-enzyme intermediate via the catalytic serine.Step 2: Recall that HIV protease activates water between two aspartates; nucleophile is water, not an enzyme residue, so no acyl-enzyme forms.Step 3: Conclude that the statement claiming both form an acyl intermediate is false. Verification / Alternative check: Mechanistic studies and inhibitor structures (e.g., transition-state mimics) for HIV protease show tetrahedral intermediates without enzyme-bound acyl covalent species. Why Other Options Are Wrong: Nucleophilic attack and water requirement: both true in their respective mechanisms. Sequence specificity: both classes exhibit contextual preferences. Metal independence: neither class is a metallo-protease. Common Pitfalls: Assuming all proteases use identical covalent strategies; overlooking that many proteases are noncovalent hydrolases utilizing activated water. Final Answer: Both form a covalent acyl-enzyme intermediate during catalysis

Question 11

Biochemistry—Cyclooxygenase-1 (COX-1) and Physiology How is the enzyme COX-1 important for human health under normal conditions?

Accepted Answer

Introduction / Context: Cyclooxygenases (COX-1 and COX-2) convert arachidonic acid into prostaglandin H2, the precursor of diverse prostanoids. COX-1 is constitutively expressed in many tissues and supports homeostatic functions, including gastric mucosal protection. Understanding COX-1 roles explains both therapeutic benefits and side effects of nonsteroidal anti-inflammatory drugs (NSAIDs). Given Data / Assumptions: COX-1 is a constitutive cyclooxygenase active under basal conditions. Prostanoids derived from COX-1 support mucus and bicarbonate secretion and regulate gastric blood flow. Aspirin and many NSAIDs inhibit COX activity, reducing prostaglandin synthesis. Concept / Approach: Identify the specific physiological function linked to COX-1 rather than general cellular processes. The maintenance of gastric mucosa via prostaglandins is the classical COX-1 function; blocking COX-1 may compromise the stomach lining and increase ulcer risk, particularly with chronic NSAID use. Step-by-Step Solution: Step 1: Recognize that COX-1 catalyzes early steps in prostaglandin biosynthesis from arachidonic acid.Step 2: Link prostaglandins to gastric protective functions (mucus, bicarbonate, mucosal blood flow).Step 3: Conclude that COX-1 activity is essential for stomach lining maintenance under normal physiology. Verification / Alternative check: Clinical observations: nonselective NSAIDs that inhibit COX-1 and COX-2 can cause gastric irritation and ulcers, consistent with loss of COX-1-derived protective prostaglandins. Why Other Options Are Wrong: CO2 transport is mediated by carbonic anhydrase and hemoglobin buffering, not COX-1. DNA synthesis involves polymerases and nucleotide metabolism, not COX-1. Aspirin is an acetylating inhibitor of COX; COX-1 is not a derivative of aspirin. Heme catabolism to biliverdin involves heme oxygenase, not COX-1. Common Pitfalls: Conflating COX-1 and COX-2 roles; assuming all eicosanoid pathways are interchangeable without tissue-specific functions. Final Answer: It catalyzes formation of prostaglandins that help maintain the stomach lining

Question 12

Biochemistry—Shared Features of Protease Classes Which common mechanistic features are shared by serine proteases and aspartate (aspartyl) proteases?

Accepted Answer

Introduction / Context: Protease families achieve peptide bond hydrolysis using different active-site chemistries, yet they share several mechanistic themes. Recognizing these shared features aids in understanding catalysis, substrate recognition, and inhibitor design across protease classes. Given Data / Assumptions: Serine proteases use a Ser-His-Asp triad and form a transient acyl-enzyme intermediate; water is required for deacylation. Aspartate proteases (e.g., pepsin family, HIV protease) use two Asp residues to activate a water molecule for direct nucleophilic attack. Both enzyme classes recognize specific sequence contexts around the scissile bond (S1, S1′, etc.). Concept / Approach: Check each statement for validity across both families. Water involvement is universal for hydrolysis; base-assisted activation of the nucleophile occurs via His (serine proteases) or via Asp residues that polarize and deprotonate water (aspartate proteases). Sequence specificity arises from substrate-binding pockets that accommodate side chains selectively. Step-by-Step Solution: Step 1: Confirm water usage. Serine proteases require water to hydrolyze the acyl-enzyme, and aspartate proteases use water as the nucleophile.Step 2: Confirm base activation. Histidine acts as a base for serine proteases; a pair of aspartates activates water in aspartate proteases.Step 3: Confirm specificity. Binding pockets select residues, conferring sequence preferences for both families. Verification / Alternative check: Active-site structures and kinetic isotope effects support base-assisted proton transfers in both classes; substrate profiling defines sequence preferences. Why Other Options Are Wrong: Each of A–C is correct; therefore D is correct and E is incorrect. Common Pitfalls: Assuming that only serine proteases use a base or that only one class shows sequence specificity; ignoring the different identities of the base (His vs Asp pair). Final Answer: All of the above

Question 13

In serine protease catalysis (e.g., trypsin), what specific catalytic role do Asp 102 and His 57 play within the catalytic triad (Ser 195–His 57–Asp 102)?

Accepted Answer

Introduction: Serine proteases such as trypsin, chymotrypsin, and elastase share a conserved catalytic triad composed of Ser 195, His 57, and Asp 102. This question probes the precise role of Asp 102 and His 57 in activating Ser 195 for nucleophilic attack on the peptide bond carbonyl. Understanding this charge-relay (proton-shuttle) system is central to appreciating how proteases accelerate peptide bond hydrolysis by many orders of magnitude. Given Data / Assumptions: The active site contains Ser 195, His 57, and Asp 102 in a specific hydrogen-bonded geometry. Hydrolysis requires nucleophilic attack by Ser 195 Oγ on the substrate carbonyl carbon. Proton transfers must occur efficiently during acylation and deacylation steps. Concept / Approach: Asp 102 orients and stabilizes the protonated form of His 57. His 57 acts as a general base to abstract a proton from Ser 195, generating a powerful alkoxide nucleophile. During catalysis, His 57 also serves as a general acid to donate a proton to the leaving group and later to water, thereby shuttling protons between participants. This cooperative behavior is the classical charge-relay or proton-shuttle mechanism. Step-by-Step Solution: Positioning: Asp 102 hydrogen bonds to His 57, increasing His 57’s basicity and fixing its orientation.Activation: His 57 abstracts a proton from Ser 195 Oγ to form Ser O− (nucleophile).Acylation: Ser O− attacks the peptide carbonyl → tetrahedral intermediate stabilized in the oxyanion hole.Collapse: His 57 donates a proton to the amide nitrogen → leaving group departs → acyl-enzyme forms.Deacylation: Water is activated by His 57, attacks the acyl-enzyme, and the acyl bond is hydrolyzed, regenerating Ser 195. Verification / Alternative check: Mutational and kinetic studies show large loss of catalytic efficiency when Asp 102 or His 57 is altered, confirming their roles in proton transfer and Ser 195 activation rather than mere structural support. Why Other Options Are Wrong: Neutralize each other’s charges: does not explain nucleophile formation or observed kinetics. Keep specificity pocket open: substrate selectivity is determined by pocket size/charge, not by Asp 102–His 57. Clamp substrate: binding subsites do this; the triad’s function is chemical catalysis. Recruit a metal ion: serine proteases are not metalloenzymes. Common Pitfalls: Confusing the oxyanion hole (stabilizes intermediates) with the catalytic triad (performs proton transfers), or thinking Asp 102 directly attacks the substrate. Final Answer: Function together as a proton shuttle (charge-relay) that activates Ser 195

Question 14

In Escherichia coli, what reaction is catalyzed by the pyruvate (pyruvic acid) dehydrogenase complex at the junction between glycolysis and the TCA cycle?

Accepted Answer

Introduction: The pyruvate dehydrogenase complex (PDC) links glycolysis to the tricarboxylic acid (TCA) cycle by converting pyruvate into acetyl-CoA. This question checks recognition of the oxidative decarboxylation step that produces acetyl-CoA, CO2, and reduced cofactors crucial for aerobic metabolism. Given Data / Assumptions: Substrate: pyruvate (end product of glycolysis). Products: acetyl-CoA and CO2; NAD+ is reduced to NADH. Multienzyme assembly: E1 (pyruvate dehydrogenase), E2 (dihydrolipoyl transacetylase), E3 (dihydrolipoyl dehydrogenase). Concept / Approach: PDC performs oxidative decarboxylation. Thiamine pyrophosphate (TPP) decarboxylates pyruvate, lipoamide accepts the acyl group, CoA forms acetyl-CoA, and FAD/NAD+ reoxidize the lipoamide. The net effect is oxidation of pyruvate coupled to the formation of acetyl-CoA and CO2 with generation of NADH. Step-by-Step Solution: E1: TPP-dependent decarboxylation of pyruvate → hydroxyethyl-TPP.E2: Transfer of the two-carbon unit to lipoamide, then to CoA → acetyl-CoA.E3: Reoxidation of reduced lipoamide via FAD and NAD+ → NADH + H+.CO2 is released during decarboxylation, establishing the link to TCA entry. Verification / Alternative check: Stoichiometry: pyruvate + CoA + NAD+ → acetyl-CoA + CO2 + NADH + H+, matching oxidative decarboxylation rather than reduction. Why Other Options Are Wrong: Decatalyze/retard: not meaningful biochemical actions. Reduction to acetyl-CoA: the reaction is oxidative, not reductive. Direct oxaloacetate formation: that is catalyzed by pyruvate carboxylase (anaplerosis), not PDC. Common Pitfalls: Confusing PDC with pyruvate carboxylase or lactate dehydrogenase; forgetting NADH production. Final Answer: Catalyze the oxidation of pyruvate to acetyl-CoA and CO2

Question 15

For trypsin and chymotrypsin, the peptide-bond cleavage specificity depends primarily on what property of the substrate-binding specificity pocket?

Accepted Answer

Introduction: Protease specificity arises from how the substrate side chain at the P1 position fits into the enzyme’s specificity pocket. This question focuses on trypsin and chymotrypsin, which share the same catalytic triad but cut different substrates because their binding pockets differ in physical and electrostatic properties. Given Data / Assumptions: Trypsin, chymotrypsin, and elastase are serine proteases with identical catalytic chemistry. Substrate recognition differs mainly at the specificity pocket that binds the side chain N-terminal to the scissile bond (P1). Catalytic residues (Ser-His-Asp) are conserved and do not dictate which substrates are chosen. Concept / Approach: Trypsin’s pocket contains a negatively charged Asp that attracts Lys/Arg; chymotrypsin has a large hydrophobic pocket accommodating bulky aromatics; elastase has a shallow, occluded pocket favoring small residues. Thus, size, shape, and charge of the pocket determine cleavage preference, not the mere presence of Ser 195 or the water content per se. Step-by-Step Solution: Identify the P1 side chain as the key determinant.Map enzyme pockets: trypsin → Asp at bottom; chymotrypsin → hydrophobic cavity.Infer electrostatics and sterics guide binding and specificity.Conclude pocket properties (size/shape/charge) control cleavage site choice. Verification / Alternative check: Mutational swaps of pocket residues interconvert specificities (e.g., trypsin↔chymotrypsin), demonstrating causality of pocket chemistry rather than catalytic triad identity. Why Other Options Are Wrong: Ser 195 proximity: present in all serine proteases and does not confer specificity. Low-barrier H-bond: influences catalysis rate, not residue preference. Absence of water: unrealistic; water participates in deacylation. Covalent pre-modification: not a general requirement for these enzymes. Common Pitfalls: Equating catalytic efficiency with specificity or confusing P1 vs other subsites (P2, P3, etc.). Final Answer: Size, shape, and charge (chemical environment) of the specificity pocket

Question 16

Which of the following enzymes are classified as serine proteases based on their catalytic mechanism?

Accepted Answer

Introduction: Enzyme families are often grouped by their catalytic residues and mechanisms. Serine proteases utilize an activated serine (with His and Asp) to perform nucleophilic attack on peptide bonds. This question asks you to recognize classical members of this family. Given Data / Assumptions: Serine proteases possess the catalytic triad Ser-His-Asp. They proceed via acyl-enzyme and deacylation steps. Trypsin, chymotrypsin, and elastase are textbook examples. Concept / Approach: Although their specificity pockets differ, trypsin, chymotrypsin, and elastase share the same core chemistry centered on Ser 195 activation through a charge-relay system. Therefore, all listed enzymes fall under the serine protease class. Step-by-Step Solution: Identify the catalytic triad presence.Confirm acylation/deacylation intermediates with an oxyanion hole.Note differing substrate preferences due to pocket variations, not mechanism.Conclude that all three are serine proteases. Verification / Alternative check: Active-site serine can be specifically inhibited by diisopropyl fluorophosphate (DFP) and related reagents across these enzymes, confirming a shared mechanism. Why Other Options Are Wrong: Each single enzyme alone misses the completeness of the set. None of these: contradicts well-established classifications. Common Pitfalls: Assuming that different specificities imply different catalytic classes; here, specificity arises from the binding pocket, not the chemistry. Final Answer: All of these

Enzymes Questions