All flashcards
Flashcard 1: State the Lineweaver nten (double-reciprocal) form of the Michaelis nten equation.
Answer: rac{1}{v}= rac{K_m}{V_{max}} rac{1}{[S]}+ rac{1}{V_{max}}. The double-reciprocal plot linearizes Michaelis-Menten kinetics, allowing determination of Km and V_max from slope and intercepts.
Flashcard 2: What is the approximate rate when [S] K_m?
Answer: v V_{max}. Enzyme saturation at high [S] limits the rate to V_max, independent of further increases in substrate concentration.
Flashcard 3: What is the approximate rate law when [S] K_m?
Answer: v rac{V_{max}}{K_m}[S]. This linear approximation derives from the Michaelis-Menten equation when [S] is much less than Km, simplifying to a first-order rate.
Flashcard 4: Identify where an uncompetitive inhibitor binds relative to substrate binding.
Answer: Only to the ES complex (not to free enzyme). Uncompetitive inhibition requires substrate-bound enzyme, forming a dead-end complex that lowers apparent Km and V_max.
Flashcard 5: Identify where a competitive inhibitor binds relative to the substrate binding site.
Answer: Active site (mutually exclusive with substrate). Competitive inhibitors mimic substrate structure, binding reversibly to the same site and preventing substrate access.
Flashcard 6: Which inhibition type is best overcome by increasing [S] (substrate concentration)?
Answer: Competitive inhibition. Increasing [S] can outcompete competitive inhibitors for the active site, restoring V_max unlike other inhibition types.
Flashcard 7: What change in Km and Vmax is caused by uncompetitive inhibition?
Answer: Km decreases; Vmax decreases. Uncompetitive inhibitors bind only to ES complex, decreasing both Km and V_max by stabilizing the complex.
Flashcard 8: What change in Km and Vmax is caused by pure noncompetitive inhibition?
Answer: Vmax decreases; Km unchanged. Noncompetitive inhibitors bind elsewhere, reducing effective enzyme concentration and thus V_max, without altering Km.
Flashcard 9: What change in Km and Vmax is caused by competitive inhibition?
Answer: Km increases; Vmax unchanged. Competitive inhibitors compete for the active site, increasing apparent Km but not affecting V_max at high [S].
Flashcard 10: In a Lineweaver nten plot, what is the slope equal to?
Answer: rac{K_m}{V_{max}}. The slope reflects the ratio of Km to V_max, indicating how quickly velocity approaches maximum as [S] increases.
Flashcard 11: In a Lineweaver nten plot, what is the x-intercept equal to?
Answer: - rac{1}{K_m}. The x-intercept occurs where 1/v = 0, solving to -1/Km in the Lineweaver-Burk equation.
Flashcard 12: In a Lineweaver nten plot, what is the y-intercept equal to?
Answer: rac{1}{V_{max}}. The y-intercept in the Lineweaver-Burk plot corresponds to infinite [S], where velocity equals V_max.
Flashcard 13: Which condition on [S] makes the rate approximately zero-order in substrate?
Answer: [S] K_m. At high [S] relative to Km, the enzyme is saturated, making velocity independent of [S] and zero-order in substrate.
Flashcard 14: What is the definition of the turnover number kcat?
Answer: Catalytic cycles per enzyme per second at saturation. k_cat quantifies an enzyme's efficiency by measuring product formation rate per active site when fully saturated.
Flashcard 15: What is the definition of Km in Michaelis–Menten kinetics?
Answer: Substrate concentration where v=V_{max}. Km represents enzyme-substrate affinity, as it is the [S] at which half-maximal velocity is achieved in the Michaelis-Menten model.
Flashcard 16: State the Michaelis nten equation for initial velocity v.
Answer: v= rac{V_{max}[S]}{K_m+[S]}. The equation describes hyperbolic kinetics, where velocity approaches V_max as [S] increases, based on enzyme-substrate binding equilibrium.
Flashcard 17: State the relationship between Vmax, kcat, and total enzyme [E]T.
Answer: Vmax=kcat[E]T. V_max scales with total enzyme concentration, as k_cat is the turnover rate per enzyme molecule at saturation.
Flashcard 18: What is the definition of Vmax in Michaelis–Menten kinetics?
Answer: Maximum initial rate at saturating [S]. V_max is achieved when all enzyme active sites are saturated with substrate, reflecting the enzyme's maximum catalytic capacity.
Flashcard 19: What is catalytic efficiency, and how is it written mathematically?
Answer: rac{k_{cat}}{K_m}. Catalytic efficiency measures how well an enzyme converts substrate to product at low [S], combining turnover and affinity.
Flashcard 20: Which condition on [S] makes the rate approximately first-order in substrate?
Answer: [S] K_m. At low [S] relative to Km, velocity is directly proportional to [S], following first-order kinetics in the Michaelis-Menten approximation.
Flashcard 21: What is the effect of phosphorylation on enzyme activity in general terms?
Answer: Covalent modification that can increase or decrease activity. Phosphorylation adds a phosphate group via kinases, modulating enzyme conformation to either activate or inhibit catalysis.
Flashcard 22: Which option describes a feedback inhibitor in a metabolic pathway?
Answer: End product inhibits an earlier enzyme, often the first committed step. Feedback inhibition regulates pathways by allosterically inhibiting upstream enzymes to prevent overproduction of end products.
Flashcard 23: What does a Hill coefficient nH>1 indicate about substrate binding?
Answer: Positive cooperativity. A Hill coefficient greater than 1 signifies enhanced binding affinity after initial substrate attachment in multimeric enzymes.
Flashcard 24: What is the key kinetic hallmark of allosteric enzymes versus Michaelis nten enzymes?
Answer: Sigmoidal v versus [S] (cooperativity). Allosteric enzymes exhibit cooperative binding, yielding a sigmoidal curve unlike the hyperbolic Michaelis-Menten kinetics.
Flashcard 25: What is the defining feature of irreversible inhibition in enzyme kinetics?
Answer: Covalent or extremely tight binding permanently inactivates enzyme. Irreversible inhibitors form stable bonds with the enzyme, reducing active enzyme concentration permanently.