Enzyme practice

Revised Enzymes Practice Exam

Multiple Choice Questions

Which statement best describes an enzyme?
- A) It increases ∆G
- B) It lowers ∆G‡
- C) It is consumed in reaction
- D) It raises activation barrier
Ribozymes are:
- A) carbohydrate catalysts
- B) RNA catalysts
- C) lipid catalysts
- D) protein inhibitors
Enzymes accelerate reactions mainly by:
- A) increasing temperature
- B) stabilizing the transition state
- C) altering equilibrium
- D) raising reactant energy
Which class catalyzes oxidation-reduction?
- A) Transferase
- B) Ligase
- C) Oxidoreductase
- D) Lyase
The ES complex is also called:
- A) Lineweaver complex
- B) Michaelis complex
- C) Allosteric complex
- D) Transition complex
Enzymes DO NOT:
- A) lower ∆G‡
- B) speed reaction rate
- C) change equilibrium
- D) bind substrate
In the proximity model, enzymes improve rate by reducing:
- A) enthalpy
- B) entropy cost
- C) hydrophobicity
- D) redox potential
A reaction showing saturation versus [S] follows:
- A) First-order kinetics always
- B) Michaelis-Menten behavior
- C) Zero-order only
- D) Non-enzymatic kinetics
KM is:
- A) product affinity
- B) measure of substrate affinity
- C) turnover number
- D) equilibrium constant
kcat is:
- A) molecules per enzyme per second
- B) substrate affinity
- C) inhibition constant
- D) pH optimum
Enzyme efficiency is often limited by:
- A) pKa
- B) diffusion
- C) KM
- D) ∆H
Lineweaver-Burk plots are useful because:
- A) they eliminate noise
- B) they linearize data
- C) they require no kinetics assumptions
- D) they remove inhibitors
In competitive inhibition, Vmax:
- A) increases
- B) decreases
- C) unchanged
- D) becomes zero
In competitive inhibition, apparent KM:
- A) decreases
- B) unchanged
- C) increases
- D) becomes undefined
Uncompetitive inhibition binds:
- A) free enzyme only
- B) ES complex only
- C) both E and ES
- D) active site only
Mixed inhibition results in:
- A) no change in Vmax
- B) lines intersect left of y-axis
- C) parallel lines
- D) increased Vmax
Noncompetitive inhibition is a subtype of:
- A) mixed
- B) competitive
- C) uncompetitive
- D) irreversible
General acid-base catalysis uses:
- A) metal ions
- B) transient covalent bonds
- C) proton donation/acceptance
- D) hydrophobic collapse

Additional Concepts

Covalent catalysis requires:
- A) metal cofactor
- B) nucleophile on enzyme
- C) denatured enzyme
- D) irreversible inhibitor
Metal ion catalysis stabilizes:
- A) positive charges
- B) negative charges
- C) hydrophobic pockets
- D) peptide bonds
Chymotrypsin uses:
- A) only acid-base catalysis
- B) only metal catalysis
- C) multiple catalytic mechanisms
- D) none of these
Lysozyme acts by:
- A) oxidizing substrate
- B) cleaving peptidoglycan
- C) phosphorylating sugars
- D) reducing disulfides
The blood coagulation cascade mostly uses:
- A) allosteric regulation
- B) reversible inhibition
- C) irreversible covalent modification
- D) competitive inhibition
Zymogens are activated by:
- A) reversible phosphorylation
- B) irreversible cleavage
- C) dimerization
- D) pH changes
Allosteric enzymes usually:
- A) follow MM kinetics exactly
- B) exhibit sigmoidal curves
- C) cannot be inhibited
- D) use metal ions only

Short Answer Questions

Explain how enzymes lower the activation energy (∆G‡) of a reaction.
- Enzymes lower ∆G‡ by stabilizing the transition state, orienting substrates, reducing entropy cost, and providing catalytic residues.
Describe the evidence supporting the transition-state stabilization model.
- Stronger binding to transition state (TS) analogs than substrates; structural data showing TS complementarity.
Define kcat and KM and explain their physical meaning.
- kcat = turnover number (the number of substrate molecules converted to product per enzyme molecule per second).
- KM ≈ substrate affinity (lower KM = higher affinity for the substrate).
Compare competitive and uncompetitive inhibition in terms of KM and Vmax.
- Competitive Inhibition: Vmax unchanged, KM increases (because more substrate is needed to reach half Vmax).
- Uncompetitive Inhibition: Vmax decreases, KM decreases.
Outline the catalytic steps of chymotrypsin.
- Binding → nucleophilic attack by Ser195 → tetrahedral intermediate → cleavage → water attack → regeneration of enzyme.
Describe the mechanism of lysozyme cleavage of peptidoglycan.
- Asp52 performs nucleophilic attack (SN2), Glu35 facilitates acid/base catalysis, resulting in covalent intermediate hydrolyzed to cleave the substrate.
Explain why enzyme activity shows saturation at high substrate concentration.
- ES complex formation saturates enzyme; at high substrate concentration, all active sites are occupied, preventing further increases in reaction rate.
Describe the difference between sequential and ping-pong two-substrate mechanisms.
- Sequential Mechanism: both substrates bind to the enzyme before product release.
- Ping-Pong Mechanism: one product is released before the second substrate binds, creating a transient intermediate.

Answer Key

MCQ Answers:
1: B
2: B
3: B
4: C
5: B
6: C
7: B
8: B
9: B
10: A
11: B
12: B
13: C
14: C
15: B
16: B
17: A
18: C
19: B
20: B
21: C
22: B
23: C
24: B
25: B
SA Answers:
1: Enzymes lower ∆G‡ by stabilizing the transition state, orienting substrates, reducing entropy cost, and providing catalytic residues.
2: Stronger binding to TS analogs than substrates; structural data showing TS complementarity.
3: kcat = turnover number; KM ≈ substrate affinity (lower KM = higher affinity).
4: Competitive: Vmax unchanged, KM increases. Uncompetitive: Vmax decreases, KM decreases.
5: Binding → nucleophilic attack by Ser195 → tetrahedral intermediate → cleavage → water attack → regeneration of enzyme.
6: Asp52 performs nucleophilic attack (SN2), Glu35 acid/base catalysis, covalent intermediate hydrolyzed.
7: ES complex formation saturates enzyme; at high [S], all active sites occupied.
8: Sequential: both substrates bind before product release. Ping-pong: one product released before second substrate binds.