All flashcards
Flashcard 1: What is the relationship between equilibrium constant Keq and enzyme catalysis?
Answer: Enzymes do not change Keq. Since enzymes do not affect ΔG, they leave the ratio of products to reactants at equilibrium unchanged.
Flashcard 2: What is the induced-fit model of enzyme-substrate binding?
Answer: Substrate binding causes enzyme conformational change to fit better. This model explains how enzymes adapt their shape upon substrate binding to enhance catalytic efficiency.
Flashcard 3: What is the lock-and-key model of enzyme-substrate binding?
Answer: Active site is preformed and complementary to the substrate. This rigid model describes perfect geometric complementarity between enzyme and substrate without conformational changes.
Flashcard 4: What is the definition of Vmax in enzyme kinetics?
Answer: Maximum rate when enzyme is saturated with substrate. Vmax represents the enzyme's turnover rate when all active sites are occupied by substrate.
Flashcard 5: What is the primary effect of an enzyme on a reaction energy diagram?
Answer: It lowers the activation energy Ea. By stabilizing the transition state, enzymes reduce the energy barrier required for the reaction to proceed.
Flashcard 6: What is the relationship between ΔG and enzyme catalysis for a given reaction?
Answer: Enzymes do not change ΔG of the reaction. Catalysts like enzymes accelerate reactions without altering the overall free energy change between reactants and products.
Flashcard 7: Which option best describes how enzymes increase reaction rate?
Answer: They stabilize the transition state and provide a lower-Ea pathway. Enzymes accelerate reactions by lowering the activation energy through transition state stabilization.
Flashcard 8: What is the Michaelis-Menten equation for initial rate v0?
Answer: v0=Km+[S]Vmax[S]. This equation models the hyperbolic relationship between initial velocity and substrate concentration in enzyme kinetics.
Flashcard 9: What is the difference between an apoenzyme and a holoenzyme?
Answer: Apoenzyme lacks cofactor; holoenzyme is the active complete enzyme. The apoenzyme requires the cofactor to form the functional holoenzyme for catalytic activity.
Flashcard 10: Which option best defines a prosthetic group in enzyme function?
Answer: A tightly bound cofactor that remains associated with the enzyme. Prosthetic groups enhance enzyme function by remaining covalently or tightly bound during catalysis.
Flashcard 11: What is the catalytic triad commonly found in serine proteases?
Answer: Serine, histidine, and aspartate. These residues work together to facilitate nucleophilic attack and proton transfer in peptide bond hydrolysis.
Flashcard 12: Which option best defines the transition state in enzyme-catalyzed reactions?
Answer: The highest-energy, unstable configuration between reactants and products. The transition state represents the peak of the energy barrier that must be overcome for the reaction to occur.
Flashcard 13: What is the definition of Km in Michaelis-Menten kinetics?
Answer: [S] at which v0=2Vmax. Km reflects the enzyme's affinity for the substrate, where half-maximal velocity is achieved.
Flashcard 14: Which option best defines a cofactor in enzyme catalysis?
Answer: A nonprotein helper, often a metal ion, required for activity. Cofactors assist in catalysis by participating in redox reactions or stabilizing intermediates.
Flashcard 15: Which option best defines a coenzyme in enzyme catalysis?
Answer: An organic cofactor, often vitamin-derived, required for activity. Coenzymes act as carriers of chemical groups or electrons during enzymatic reactions.
Flashcard 16: At [S]=Km, what is v0 expressed in terms of Vmax?
Answer: v0=2Vmax. At [S]=Km, the enzyme operates at half its maximum capacity according to Michaelis-Menten kinetics.
Flashcard 17: If [S]≫Km, what is the approximate value of v0?
Answer: v0≈Vmax. High substrate concentrations saturate the enzyme, making the reaction rate independent of [S].
Flashcard 18: If [S]≪Km, what is the approximate form of v0?
Answer: v0≈KmVmax[S]. Low substrate levels make the reaction rate directly proportional to [S], following first-order kinetics.
Flashcard 19: What is the Lineweaver-Burk (double-reciprocal) form of Michaelis-Menten?
Answer: v01=VmaxKm[S]1+Vmax1. This linear transformation allows determination of Km and Vmax from intercepts on a double-reciprocal plot.
Flashcard 20: On a Lineweaver-Burk plot, what is the y-intercept equal to?
Answer: Vmax1. The y-intercept corresponds to the reciprocal of the maximum velocity in the Lineweaver-Burk equation.
Flashcard 21: On a Lineweaver-Burk plot, what is the x-intercept equal to?
Answer: −Km1. The x-intercept is the negative reciprocal of Km, indicating substrate affinity in the plot.
Flashcard 22: Which option best defines competitive inhibition in enzyme kinetics?
Answer: Inhibitor binds active site; increases Km; Vmax unchanged. Competitive inhibitors compete for the active site, requiring higher [S] to achieve the same velocity.
Flashcard 23: Which option best defines pure noncompetitive inhibition in enzyme kinetics?
Answer: Inhibitor binds allosteric site; decreases Vmax; Km unchanged. Noncompetitive inhibitors reduce enzyme activity by binding elsewhere, without affecting substrate binding affinity.
Flashcard 24: Which option best defines uncompetitive inhibition in enzyme kinetics?
Answer: Inhibitor binds ES; decreases Km and decreases Vmax. Uncompetitive inhibitors bind only to the enzyme-substrate complex, altering both apparent affinity and maximum rate.
Flashcard 25: Which option best defines the active site of an enzyme?
Answer: The specific region where substrate binds and catalysis occurs. The active site facilitates catalysis by providing a microenvironment that stabilizes the transition state through specific substrate interactions.