Q: Metalloenzymes: Which of the following enzymes contains a Zinc (Zn) ion as a catalytic cofactor?

Introduction: Many enzymes require metal ions for catalytic activity or structural stability. Recognizing which enzymes are Zn-dependent helps in understanding catalytic mechanisms and inhibitor design. Given Data / Assumptions: Classical biochemistry of metalloenzymes. Focus on presence of Zn at the catalytic center. Concept / Approach: Zinc commonly functions as a Lewis acid to polarize water or substrate functional groups. Proteases such as carboxypeptidase A and matrix metalloproteinases are well-known Zn enzymes. Zn coordinates with histidine, glutamate, or aspartate side chains in the active site, activating water for nucleophilic attack on peptide bonds. Step-by-Step Solution: 1) Identify known Zn proteases: carboxypeptidase A is archetypal.2) Evaluate alternatives: phosphorylase b kinase is regulated by Ca^2+ and phosphorylation, not Zn at its catalytic core.3) Tyrosine hydroxylase uses a non-heme Fe^2+ cofactor.4) Many PDEs are Mg^2+/Mn^2+ dependent rather than Zn-centered at catalysis.5) Pyruvate dehydrogenase complex uses thiamine pyrophosphate, lipoamide, FAD, NAD^+, and Mg^2+; Zn is not central. Verification / Alternative check: Structural studies of carboxypeptidase A show tetrahedral Zn coordination essential for peptide bond hydrolysis. Why Other Options Are Wrong: B, D: These rely primarily on other metal ions (Ca^2+, Mg^2+/Mn^2+). C: Requires Fe^2+. E: PDH uses Mg^2+ with TPP; no Zn at the catalytic core. Common Pitfalls: Assuming all hydrolases are Zn-dependent; many are serine-, cysteine-, or metallo-enzymes with diverse cofactors. Final Answer: Carboxypeptidase A

Q: Non-competitive Inhibition: What is the primary kinetic effect on an enzyme-catalyzed reaction?

Introduction: Non-competitive inhibition reflects inhibitor binding to an allosteric site on E and often also on ES, reducing catalytic turnover independent of substrate binding at the active site. This question probes the hallmark kinetic signature. Given Data / Assumptions: Classical non-competitive model (special case of mixed inhibition with identical Ki on E and ES). Michaelis–Menten single-substrate behavior. Concept / Approach: Because the inhibitor decreases the fraction of active enzyme capable of catalysis, Vmax drops. Since substrate affinity is not primarily altered in the true non-competitive limit, Km remains roughly the same. On Lineweaver–Burk plots, 1/Vmax increases (y-intercept rises), while x-intercept (−1/Km) stays near constant; slope increases accordingly. Step-by-Step Solution: 1) Inhibitor binds E and ES at a distinct site ⇒ catalytic turnover number kcat is effectively reduced.2) Vmax = kcat * [E]_total ⇒ Vmax decreases as active fraction falls.3) Km remains approximately constant in the true non-competitive case since substrate binding affinity is not selectively impaired.4) Therefore, velocity cannot increase under ordinary conditions as inhibitor concentration rises. Verification / Alternative check: Graphically, lines for different inhibitor concentrations intersect on the x-axis but have higher y-intercepts, confirming lower Vmax with similar Km. Why Other Options Are Wrong: B: Exclusive ES binding describes uncompetitive inhibition, not non-competitive. C: B is incorrect, so both cannot be true. D and E: Contradict the defining reduction of Vmax under inhibition. Common Pitfalls: Confusing non-competitive with uncompetitive (both lower Vmax, but effects on Km differ), or assuming inhibitors can enhance velocity under typical steady-state conditions. Final Answer: It decreases Vmax while Km (for the true non-competitive case) remains essentially unchanged.

Q: Lineweaver–Burk Signature of Non-competitive Inhibition: Which change is characteristic?

Introduction: Interpreting Lineweaver–Burk plots is central to distinguishing inhibition modes. Non-competitive inhibition has a specific graphical fingerprint that reflects a reduction in catalytic capacity (Vmax). Given Data / Assumptions: True non-competitive case (inhibitor binds E and ES equally well). Michaelis–Menten kinetics; plotting 1/v vs 1/[S]. Concept / Approach: In the non-competitive limit, Vmax decreases; Km remains essentially unchanged. On a double-reciprocal plot, this means 1/Vmax (y-intercept) increases, x-intercept (−1/Km) is nearly constant, and slope = Km/Vmax increases accordingly. Lines for different inhibitor levels intersect on or near the x-axis. Step-by-Step Solution: 1) Recognize parameter mapping: y-intercept = 1/Vmax; x-intercept = −1/Km.2) Non-competitive inhibition lowers Vmax ⇒ y-intercept rises.3) Km unaffected ⇒ x-intercept about constant.4) Slope increases since denominator Vmax is reduced. Verification / Alternative check: Families of LB lines under non-competitive inhibition commonly intersect on the x-axis; kinetic fitting to v = (Vmax/(1+I/Ki)) * S / (Km + S) reproduces the same geometry. Why Other Options Are Wrong: A: No uniform lateral shift occurs. C: x-intercept alone does not change; y-intercept does. D: They are not equally correct; only one statement matches the canonical pattern. E: Vmax does not increase; slope does not decrease. Common Pitfalls: Confusing non-competitive with competitive (y-intercept constant in competitive) or uncompetitive (both intercepts shift). Final Answer: The y-intercept (1/Vmax) increases while the x-intercept (−1/Km) remains approximately unchanged.

Q: International Enzyme Classification (EC Numbers): Which class is EC 2?

Introduction: The Enzyme Commission (EC) numbering system classifies enzymes into major classes based on the reactions they catalyze. Knowing the mapping helps interpret literature and databases. Given Data / Assumptions: Standard EC classes: 1–6 (and extensions). Question asks specifically for class EC 2. Concept / Approach: EC classes: 1 Oxidoreductases, 2 Transferases, 3 Hydrolases, 4 Lyases, 5 Isomerases, 6 Ligases. Transferases move a functional group (e.g., methyl, glycosyl, phosphoryl) from a donor to an acceptor, central in metabolism and regulation. Step-by-Step Solution: 1) Recall class mapping list.2) Identify EC 2 in the mapping.3) Conclude EC 2 corresponds to Transferases. Verification / Alternative check: Cross-checking with enzyme databases (e.g., BRENDA/ExplorEnz) confirms EC 2 = Transferases. Why Other Options Are Wrong: A: EC 3, not 2. B: EC 6. D: EC 5. E: EC 1. Common Pitfalls: Mixing up EC 1 and EC 2 due to both being common in pathways; remember 1 = redox (oxidation–reduction), 2 = group transfer. Final Answer: Transferases

Q: Effect of Changing Kinetic Parameters: For an enzyme with Km = 10 mM and Vmax = 100 mmol/min, at [S] = 100 mM, which statements about velocity changes are correct?

Introduction: This problem applies the Michaelis–Menten equation to assess how changes in Vmax and Km affect reaction velocity at a fixed substrate concentration. It tests quantitative intuition at saturating and near-saturating conditions. Given Data / Assumptions: Km = 10 mM. Vmax = 100 mmol/min. [S] = 100 mM. Single-substrate Michaelis–Menten kinetics. Concept / Approach: Use v = Vmax * [S] / (Km + [S]). At [S] ≫ Km, velocity approaches Vmax, so changes in Vmax have roughly proportional effects. Decreasing Km at fixed [S] increases the fraction [S]/(Km + [S]), raising v but with diminishing returns as [S] greatly exceeds Km. Step-by-Step Solution: 1) Baseline: v0 = 100 * 100 / (10 + 100) = 10000 / 110 ≈ 90.91 mmol/min.2) Increase Vmax 10-fold: Vmax' = 1000 ⇒ v' = 1000 * 100 / 110 = 100000 / 110 ≈ 909.09 mmol/min (≈ 10× increase).3) Decrease Km 10-fold: Km' = 1 mM ⇒ v' = 100 * 100 / (1 + 100) = 10000 / 101 ≈ 99.01 mmol/min (increase relative to 90.91).4) Therefore, (a) and (b) are both true; (d) is false. Verification / Alternative check: Limiting case logic: as [S] → ∞, v → Vmax; thus scaling Vmax scales v. Reducing Km increases saturation fraction toward 1, raising velocity until limited by Vmax. Why Other Options Are Wrong: D: Increasing Vmax cannot decrease v; it raises the upper limit. E: Contradicts calculations showing both effects increase v. Common Pitfalls: Assuming that once [S] ≫ Km, changing Km has no effect; it still slightly increases v unless saturation is complete. Final Answer: Both (a) and (b)

Q: Most enzymes work by

decreasing energy of activation

Question 1

Michaelis–Menten mechanism: What is the rate-determining step under the usual assumption k2 ≪ k−1?

Accepted Answer

Introduction: Classic Michaelis–Menten kinetics assumes a mechanism E + S ⇌ ES → E + P with the chemical conversion (k2) slower than ES dissociation (k−1). Under this common condition, the product formation step is rate-determining. Given Data / Assumptions: Steady-state approximation: d[ES]/dt ≈ 0 after a brief pre-steady-state period. k2 ≪ k−1, which makes chemistry slower than ES breakdown to E + S. No product inhibition and single-substrate reaction. Concept / Approach: When k2 is the smallest forward rate constant, the overall turnover is gated by ES → E + P. Thus Vmax = kcat * [E]total where kcat ≈ k2 under this mechanism. Step-by-Step Solution: 1) Mechanism: E + S ⇌ ES (k1, k−1) followed by ES → E + P (k2).2) If k2 is slow, each catalytic cycle’s duration is dominated by the chemistry step.3) Therefore, the product formation step determines the maximal rate; increasing [S] saturates ES but cannot exceed the limit set by k2. Verification / Alternative check: At saturation, V = Vmax = k2 * [E]total. Experimentally, turnover numbers (kcat) measured at saturating substrate reflect the rate-determining chemical step, corroborating that ES → E + P is limiting when k2 is smallest. Why Other Options Are Wrong: (a) ES formation is usually fast under typical MM assumptions. (b) ES dissociation to E + S is faster than chemistry in the stated regime. (d) Only one step (chemistry) is rate-determining here. (e) Inhibitor binding is not part of the uninhibited MM mechanism. Common Pitfalls: Confusing pre-steady-state binding steps with overall steady-state turnover; assuming ES formation is rate-limiting even at high [S]. Final Answer: Product formation step (ES → E + P).

Question 2

Enzyme Active Site: What remains true about its nature and position? (In biochemistry and enzyme kinetics context) — choose the most accurate statement.

Accepted Answer

Introduction: Enzymes accelerate biochemical reactions by binding substrates at a specialized region called the active site. This question tests understanding of the active site’s location, flexibility, and molecular complementarity to the substrate. Given Data / Assumptions: Discussion limited to protein enzymes. Focus on structural features (shape, position) and dynamic behavior upon substrate binding. Standard induced-fit and lock-and-key models apply conceptually. Concept / Approach: Active sites are typically pockets or grooves formed by residues that may be distant in the primary sequence but are brought together in the folded tertiary structure. Binding relies on geometric and chemical complementarity, including hydrogen bonding, ionic interactions, hydrophobic effects, and van der Waals forces. Many enzymes exhibit induced fit, where subtle conformational adjustments improve catalytic alignment of catalytic residues with substrate reactive groups. Step-by-Step Solution: 1) Position: Active sites are usually localized clefts/grooves, not uniformly spread and not necessarily at the exact geometric center of the protein.2) Complementarity: Binding depends on complementary shape and electrostatics between the active site and the substrate, not identity of amino acid sequence to substrate.3) Dynamics: Induced fit allows small conformational changes after initial binding, improving transition state stabilization rather than remaining completely rigid.4) Catalysis: Proper orientation of catalytic residues lowers activation energy for the reaction. Verification / Alternative check: Classic examples include hexokinase, which closes around glucose, and serine proteases, whose specificity pocket matches side-chain geometry of substrates. Both illustrate localized pockets and induced-fit/selection mechanisms. Why Other Options Are Wrong: A: Not necessarily central; many active sites are near surfaces or clefts. B: Active sites are not perfectly rigid; induced fit is common. D: Binding is not by sequence identity to substrate. E: Active site is localized, not uniform across the surface. Common Pitfalls: Confusing lock-and-key (static) with induced fit (dynamic), assuming the site must be central, or believing sequence identity drives binding rather than chemical complementarity. Final Answer: It is a three-dimensional cleft/groove whose shape and chemistry are complementary to the substrate; it can adjust by induced fit.

Question 3

Lineweaver–Burk Plot: What is the characteristic effect of a competitive inhibitor?

Accepted Answer

Introduction: Competitive inhibition is a fundamental concept in enzyme kinetics. On a Lineweaver–Burk (double-reciprocal) plot, competitive inhibitors alter specific parameters in a predictable way. This question checks recognition of which intercept and slope are affected. Given Data / Assumptions: Michaelis–Menten single-substrate kinetics. Classic competitive inhibitor competes with substrate at the active site. Lineweaver–Burk plotting: 1/v versus 1/[S]. Concept / Approach: In competitive inhibition, Vmax is unchanged because sufficiently high substrate concentrations can outcompete the inhibitor. However, the apparent affinity decreases, so Km,app increases. On the Lineweaver–Burk plot, the y-intercept equals 1/Vmax (unchanged) and the x-intercept equals −1/Km,app (moves toward zero as Km,app increases). The slope equals Km,app/Vmax, which therefore increases. Step-by-Step Solution: 1) Identify parameters: y-intercept = 1/Vmax; x-intercept = −1/Km.2) Competitive inhibition leaves Vmax unchanged ⇒ y-intercept unchanged.3) Km increases to Km,app ⇒ x-intercept magnitude decreases (moves toward zero).4) Slope = Km/Vmax increases because Km increases, Vmax constant. Verification / Alternative check: Graphical families of lines for competitive inhibition intersect at the same y-intercept but have different slopes, confirming y-intercept constancy and x-intercept shift. Why Other Options Are Wrong: A: Both intercepts do not shift; y-intercept stays constant. B: There is no uniform horizontal shift with intercepts fixed. D: Slope increases; it does not remain unchanged. E: y-intercept (1/Vmax) does not decrease since Vmax is unchanged. Common Pitfalls: Confusing competitive with non-competitive or uncompetitive patterns; forgetting that only apparent Km changes in competitive inhibition. Final Answer: It changes the x-intercept (moves it toward zero) while the y-intercept remains unchanged.

Question 4

Enzyme Inhibition: Which types can occur in biochemical reactions?

Accepted Answer

Introduction: Enzyme inhibition regulates metabolic flux, enables pharmacological control, and reveals mechanisms of catalysis. Inhibition can be temporary or permanent, and this question tests recognition of the main categories. Given Data / Assumptions: Standard biochemical classification of inhibitors. Focus on whether inhibition can be reversible, irreversible, or both. Concept / Approach: Reversible inhibitors bind non-covalently (competitive, non-competitive, uncompetitive, mixed), allowing dissociation and restoration of activity. Irreversible inhibitors usually form covalent bonds with catalytic residues (e.g., serine in serine proteases) or bind so tightly that dissociation is negligible on experimental timescales. Step-by-Step Solution: 1) Define reversible: non-covalent, equilibrium binding; effect depends on inhibitor concentration and dissociation constant.2) Define irreversible: covalent modification or quasi-irreversible tight binding that permanently inactivates the enzyme molecules affected.3) Recognize both categories are well-documented in enzymology and pharmacology. Verification / Alternative check: Classic examples include competitive reversible inhibition by methotrexate on dihydrofolate reductase, and irreversible inhibition by aspirin acetylating cyclooxygenase active-site serine. Why Other Options Are Wrong: A: Excludes irreversible inhibitors. B: Excludes reversible inhibitors. D: Contradicts abundant evidence. E: Allosteric activation is not inhibition and can be reversible; it does not preclude inhibition. Common Pitfalls: Equating irreversible with only covalent (some near-irreversible tight binding exists), or assuming reversible equals only competitive (there are multiple reversible modes). Final Answer: Both reversible and irreversible inhibition

Question 5

Determining the Degree of Cooperativity in Enzymes: Which graphical method is appropriate?

Accepted Answer

Introduction: Cooperativity describes how binding of one ligand molecule affects affinity for additional molecules, typical in multimeric proteins. This question asks for the plot used to quantify cooperativity. Given Data / Assumptions: Multisubunit enzyme or protein exhibiting sigmoidal kinetics. Hill formalism applies for estimating the Hill coefficient nH. Concept / Approach: The Hill plot linearizes cooperative binding data by plotting log[θ/(1−θ)] versus log[S] (or log[L]), where θ is fractional saturation. The slope gives the Hill coefficient nH, indicating positive cooperativity (nH > 1), non-cooperative behavior (nH ≈ 1), or negative cooperativity (nH < 1). Step-by-Step Solution: 1) Collect saturation or velocity data over substrate concentrations spanning sub- and supra-K0.5 ranges.2) Convert to fractional saturation or use v/Vmax to approximate θ for enzymes.3) Plot log[θ/(1−θ)] versus log[S] to obtain a straight line over the cooperative range.4) Determine slope = nH, intercept provides apparent affinity parameter. Verification / Alternative check: Non-linear regression with the Hill equation v = Vmax * S^nH / (K0.5^nH + S^nH) provides nH directly and serves as a numerical cross-check. Why Other Options Are Wrong: B: Koshland’s model explains mechanism but the suggested curve alone does not yield a simple numeric nH. C: Michaelis–Menten hyperbola applies to non-cooperative enzymes. D: Graphical determination is possible. E: Eadie–Hofstee is for MM kinetics, not specifically for cooperativity degree. Common Pitfalls: Over-interpreting nH as the exact number of binding sites; it is an empirical index of cooperativity strength. Final Answer: Hill plot

Question 6

Metalloenzymes: Which of the following enzymes contains a Zinc (Zn) ion as a catalytic cofactor?

Accepted Answer

Introduction: Many enzymes require metal ions for catalytic activity or structural stability. Recognizing which enzymes are Zn-dependent helps in understanding catalytic mechanisms and inhibitor design. Given Data / Assumptions: Classical biochemistry of metalloenzymes. Focus on presence of Zn at the catalytic center. Concept / Approach: Zinc commonly functions as a Lewis acid to polarize water or substrate functional groups. Proteases such as carboxypeptidase A and matrix metalloproteinases are well-known Zn enzymes. Zn coordinates with histidine, glutamate, or aspartate side chains in the active site, activating water for nucleophilic attack on peptide bonds. Step-by-Step Solution: 1) Identify known Zn proteases: carboxypeptidase A is archetypal.2) Evaluate alternatives: phosphorylase b kinase is regulated by Ca^2+ and phosphorylation, not Zn at its catalytic core.3) Tyrosine hydroxylase uses a non-heme Fe^2+ cofactor.4) Many PDEs are Mg^2+/Mn^2+ dependent rather than Zn-centered at catalysis.5) Pyruvate dehydrogenase complex uses thiamine pyrophosphate, lipoamide, FAD, NAD^+, and Mg^2+; Zn is not central. Verification / Alternative check: Structural studies of carboxypeptidase A show tetrahedral Zn coordination essential for peptide bond hydrolysis. Why Other Options Are Wrong: B, D: These rely primarily on other metal ions (Ca^2+, Mg^2+/Mn^2+). C: Requires Fe^2+. E: PDH uses Mg^2+ with TPP; no Zn at the catalytic core. Common Pitfalls: Assuming all hydrolases are Zn-dependent; many are serine-, cysteine-, or metallo-enzymes with diverse cofactors. Final Answer: Carboxypeptidase A

Question 7

Non-competitive Inhibition: What is the primary kinetic effect on an enzyme-catalyzed reaction?

Accepted Answer

Introduction: Non-competitive inhibition reflects inhibitor binding to an allosteric site on E and often also on ES, reducing catalytic turnover independent of substrate binding at the active site. This question probes the hallmark kinetic signature. Given Data / Assumptions: Classical non-competitive model (special case of mixed inhibition with identical Ki on E and ES). Michaelis–Menten single-substrate behavior. Concept / Approach: Because the inhibitor decreases the fraction of active enzyme capable of catalysis, Vmax drops. Since substrate affinity is not primarily altered in the true non-competitive limit, Km remains roughly the same. On Lineweaver–Burk plots, 1/Vmax increases (y-intercept rises), while x-intercept (−1/Km) stays near constant; slope increases accordingly. Step-by-Step Solution: 1) Inhibitor binds E and ES at a distinct site ⇒ catalytic turnover number kcat is effectively reduced.2) Vmax = kcat * [E]_total ⇒ Vmax decreases as active fraction falls.3) Km remains approximately constant in the true non-competitive case since substrate binding affinity is not selectively impaired.4) Therefore, velocity cannot increase under ordinary conditions as inhibitor concentration rises. Verification / Alternative check: Graphically, lines for different inhibitor concentrations intersect on the x-axis but have higher y-intercepts, confirming lower Vmax with similar Km. Why Other Options Are Wrong: B: Exclusive ES binding describes uncompetitive inhibition, not non-competitive. C: B is incorrect, so both cannot be true. D and E: Contradict the defining reduction of Vmax under inhibition. Common Pitfalls: Confusing non-competitive with uncompetitive (both lower Vmax, but effects on Km differ), or assuming inhibitors can enhance velocity under typical steady-state conditions. Final Answer: It decreases Vmax while Km (for the true non-competitive case) remains essentially unchanged.

Question 8

Lineweaver–Burk Signature of Non-competitive Inhibition: Which change is characteristic?

Accepted Answer

Introduction: Interpreting Lineweaver–Burk plots is central to distinguishing inhibition modes. Non-competitive inhibition has a specific graphical fingerprint that reflects a reduction in catalytic capacity (Vmax). Given Data / Assumptions: True non-competitive case (inhibitor binds E and ES equally well). Michaelis–Menten kinetics; plotting 1/v vs 1/[S]. Concept / Approach: In the non-competitive limit, Vmax decreases; Km remains essentially unchanged. On a double-reciprocal plot, this means 1/Vmax (y-intercept) increases, x-intercept (−1/Km) is nearly constant, and slope = Km/Vmax increases accordingly. Lines for different inhibitor levels intersect on or near the x-axis. Step-by-Step Solution: 1) Recognize parameter mapping: y-intercept = 1/Vmax; x-intercept = −1/Km.2) Non-competitive inhibition lowers Vmax ⇒ y-intercept rises.3) Km unaffected ⇒ x-intercept about constant.4) Slope increases since denominator Vmax is reduced. Verification / Alternative check: Families of LB lines under non-competitive inhibition commonly intersect on the x-axis; kinetic fitting to v = (Vmax/(1+I/Ki)) * S / (Km + S) reproduces the same geometry. Why Other Options Are Wrong: A: No uniform lateral shift occurs. C: x-intercept alone does not change; y-intercept does. D: They are not equally correct; only one statement matches the canonical pattern. E: Vmax does not increase; slope does not decrease. Common Pitfalls: Confusing non-competitive with competitive (y-intercept constant in competitive) or uncompetitive (both intercepts shift). Final Answer: The y-intercept (1/Vmax) increases while the x-intercept (−1/Km) remains approximately unchanged.

Question 9

International Enzyme Classification (EC Numbers): Which class is EC 2?

Accepted Answer

Introduction: The Enzyme Commission (EC) numbering system classifies enzymes into major classes based on the reactions they catalyze. Knowing the mapping helps interpret literature and databases. Given Data / Assumptions: Standard EC classes: 1–6 (and extensions). Question asks specifically for class EC 2. Concept / Approach: EC classes: 1 Oxidoreductases, 2 Transferases, 3 Hydrolases, 4 Lyases, 5 Isomerases, 6 Ligases. Transferases move a functional group (e.g., methyl, glycosyl, phosphoryl) from a donor to an acceptor, central in metabolism and regulation. Step-by-Step Solution: 1) Recall class mapping list.2) Identify EC 2 in the mapping.3) Conclude EC 2 corresponds to Transferases. Verification / Alternative check: Cross-checking with enzyme databases (e.g., BRENDA/ExplorEnz) confirms EC 2 = Transferases. Why Other Options Are Wrong: A: EC 3, not 2. B: EC 6. D: EC 5. E: EC 1. Common Pitfalls: Mixing up EC 1 and EC 2 due to both being common in pathways; remember 1 = redox (oxidation–reduction), 2 = group transfer. Final Answer: Transferases

Question 10

Effect of Changing Kinetic Parameters: For an enzyme with Km = 10 mM and Vmax = 100 mmol/min, at [S] = 100 mM, which statements about velocity changes are correct?

Accepted Answer

Introduction: This problem applies the Michaelis–Menten equation to assess how changes in Vmax and Km affect reaction velocity at a fixed substrate concentration. It tests quantitative intuition at saturating and near-saturating conditions. Given Data / Assumptions: Km = 10 mM. Vmax = 100 mmol/min. [S] = 100 mM. Single-substrate Michaelis–Menten kinetics. Concept / Approach: Use v = Vmax * [S] / (Km + [S]). At [S] ≫ Km, velocity approaches Vmax, so changes in Vmax have roughly proportional effects. Decreasing Km at fixed [S] increases the fraction [S]/(Km + [S]), raising v but with diminishing returns as [S] greatly exceeds Km. Step-by-Step Solution: 1) Baseline: v0 = 100 * 100 / (10 + 100) = 10000 / 110 ≈ 90.91 mmol/min.2) Increase Vmax 10-fold: Vmax' = 1000 ⇒ v' = 1000 * 100 / 110 = 100000 / 110 ≈ 909.09 mmol/min (≈ 10× increase).3) Decrease Km 10-fold: Km' = 1 mM ⇒ v' = 100 * 100 / (1 + 100) = 10000 / 101 ≈ 99.01 mmol/min (increase relative to 90.91).4) Therefore, (a) and (b) are both true; (d) is false. Verification / Alternative check: Limiting case logic: as [S] → ∞, v → Vmax; thus scaling Vmax scales v. Reducing Km increases saturation fraction toward 1, raising velocity until limited by Vmax. Why Other Options Are Wrong: D: Increasing Vmax cannot decrease v; it raises the upper limit. E: Contradicts calculations showing both effects increase v. Common Pitfalls: Assuming that once [S] ≫ Km, changing Km has no effect; it still slightly increases v unless saturation is complete. Final Answer: Both (a) and (b)

Question 11

Most enzymes work by

Accepted Answer

decreasing energy of activation

Question 12

When [S] = 0.1 *KM, the velocity of an enzyme catalyzed reaction is about:

Accepted Answer

0.1 * Vmax

Question 13

Which of the following refers to pseudo steady state?

Accepted Answer

d(CES)/dt = 0

Question 14

In the steady state the material balance equation for any component of a system is

Accepted Answer

rate of addition - rate of removal + rate of formation = 0

Question 15

The equation for the rate of product formation for simple enzyme reaction is given by (Where rmax, maximum reaction rate, Cs substrate concentration, Cp product concentration ES, CES enzyme-substrate concentration)

Accepted Answer

rp = rmax Cs/(Km+Cs)

Question 16

Which of the following statement(s) regarding enzymes, is/are false?

Accepted Answer

Enzymes provide activation energy for reactions

Question 17

The slope of Lineweaver Burk plot for Michaelis Menten equation is

Accepted Answer

Km/Vmax

Question 18

An enzyme has a Km of 4.7 x 10-5M. If the Vmax of the preparation is 22m moles liter-1 min-1, what velocity would be observed in the presence of 2.0 x 10-4M substrate and 5.0 x 10-5M of a competitive inhibitor?

Accepted Answer

13.54μ moles liter-1min-1

Question 19

For an enzyme that displays Michaelis-Menten kinetics, the reaction velocity (as a fraction of Vmax) observed at [S] = 2 KM will be

Accepted Answer

0.66

Question 20

The usual method(s) to solve rate equation of simple enzyme kinetics is/are

Accepted Answer

all of these

Enzymes and Kinetics Questions