3. Enzyme Kinetics & Michaelis–Menten

What is enzyme kinetics primarily concerned with?

A. Protein color

B. Rates of enzyme-catalyzed reactions

C. DNA base pairing

D. Chromatography column packing

B. Rates of enzyme-catalyzed reactions

Explanation:1. B — Kinetics measures reaction rates and how conditions affect them.

For E + S ⇌ ES → E + P, what does ES represent?

A. Equilibrium solvent

B. Enzyme standard

C. Enzyme-substrate complex

D. Elution sample

C. Enzyme-substrate complex

Explanation:2. C — ES is the enzyme bound to substrate.

In the scheme E + S ⇌ ES → E + P, what does kcat describe?

A. Substrate binding

B. Product-forming conversion of bound substrate

C. Protein staining

D. Inhibitor dissociation only

B. Product-forming conversion of bound substrate

Explanation:3. B — kcat is the catalytic turnover rate for conversion of ES to product.

Why are initial rates measured?

A. Product has not accumulated appreciably and substrate concentration is approximately its starting value

B. The enzyme is inactive at later times

C. The reaction is always at equilibrium initially

D. Absorbance cannot be measured later

A. Product has not accumulated appreciably and substrate concentration is approximately its starting value

Explanation:4. A — At initial time, [P] is approximately zero and substrate depletion is minimal.

What is the steady-state assumption?

A. Enzyme concentration changes continuously

B. [S] is always zero

C. [ES] is approximately constant during the measured initial-rate interval

D. Product formation stops

C. [ES] is approximately constant during the measured initial-rate interval

Explanation:5. C — The formation and breakdown rates of ES approximately balance.

Which mass balance is used for total enzyme?

A. ET = [S] + [ES]

B. ET = [S] + [P]

C. ET = [E] + [P]

D. ET = [E] + [ES]

D. ET = [E] + [ES]

Explanation:6. D — Enzyme is either free or substrate-bound in the simple model.

Why is substrate commonly used in large excess over enzyme?

A. To stop product formation

B. To make enzyme concentration unknown

C. To make free substrate approximately equal to total substrate

D. To force Km to zero

C. To make free substrate approximately equal to total substrate

Explanation:7. C — When [S] greatly exceeds [ES], substrate depletion into ES is negligible.

Which equation is the Michaelis–Menten equation?

A. v0 = Km/(Vmax + [S])

B. v0 = Vmax[S]/(Km + [S])

C. v0 = Vmax + Km[S]

D. v0 = [S]/Vmax

B. v0 = Vmax[S]/(Km + [S])

Explanation:8. B — This is the standard initial-rate relationship derived in the lab manual.

What does Vmax represent?

A. Rate at enzyme saturation

B. Substrate concentration at half rate

C. Rate constant for binding

D. Equilibrium dissociation constant

A. Rate at enzyme saturation

Explanation:9. A — At saturating substrate, enzyme active sites are effectively occupied and the rate approaches Vmax.

At what substrate concentration is v0 equal to one-half Vmax?

A. [S] = 0

B. [S] = 2Km

C. [S] = Km

D. [S] = Vmax

C. [S] = Km

Explanation:10. C — By the Michaelis–Menten equation, [S] = Km gives v0 = Vmax/2.

A lower Km is commonly interpreted in this manual as indicating:

A. Irreversible inhibition

B. Lower Vmax necessarily

C. More enzyme present necessarily

D. Greater apparent affinity/sensitivity to substrate

D. Greater apparent affinity/sensitivity to substrate

Explanation:11. D — Lower Km means half-maximal velocity is reached at a lower substrate concentration.

If enzyme concentration is doubled while substrate remains sufficiently abundant, what happens to initial rate?

A. It doubles

B. It halves

C. It remains unchanged

D. It becomes zero

A. It doubles

Explanation:12. A — At fixed substrate conditions, v0 is proportional to total enzyme concentration.

At very low substrate concentration, the rate is approximately:

A. Zero-order with respect to substrate

B. First-order with respect to substrate

C. Independent of enzyme

D. Always equal to Vmax

B. First-order with respect to substrate

Explanation:13. B — When [S] is much smaller than Km, v0 is approximately (Vmax/Km)[S].

At saturating substrate concentration, the rate becomes approximately:

A. Zero-order with respect to substrate

B. First-order with respect to substrate

C. Negative

D. Equal to Km

A. Zero-order with respect to substrate

Explanation:14. A — Once enzyme is saturated, increasing substrate has little effect on rate.

What is the relationship between Vmax, kcat, and total enzyme concentration?

A. ET = Vmax × Km

B. Vmax = Km/ET

C. kcat = Km × ET

D. Vmax = kcat × ET

D. Vmax = kcat × ET

Explanation:15. D — The manual defines Vmax as kcat multiplied by total enzyme concentration.

What is the unit of kcat?

A. Concentration

B. Volume

C. Time⁻¹

D. Absorbance

C. Time⁻¹

Explanation:16. C — kcat is a turnover frequency, commonly expressed in s⁻¹ or min⁻¹.

If Vmax = 120 µM/min and ET = 2 µM, what is kcat?

A. 0.0167 min⁻¹

B. 60 min⁻¹

C. 122 min⁻¹

D. 240 min⁻¹

B. 60 min⁻¹

Explanation:17. B — kcat = Vmax/ET = 120/2 = 60 min⁻¹.

What is the shape of a typical Michaelis–Menten plot of v0 versus [S]?

A. Exponential decay

B. Straight line through all concentrations

C. Bell-shaped curve

D. Hyperbola approaching Vmax

D. Hyperbola approaching Vmax

Explanation:18. D — The rate rises and approaches a saturation asymptote.

If two enzymes have the same Vmax but different Km values, what differs?

A. The substrate concentration needed to reach half-maximal rate

B. Their maximum rates

C. The y-intercept of the Michaelis plot

D. The reaction's product identity

A. The substrate concentration needed to reach half-maximal rate

Explanation: Km controls how quickly the curve approaches Vmax.

If two enzymes have the same Km but different Vmax values, what differs?

A. The substrate identity

B. The half-maximal substrate concentration

C. The maximum catalytic rate

D. The x-intercept of the Michaelis curve

C. The maximum catalytic rate

Explanation: 20. C — Vmax changes the vertical asymptote while Km remains the half-maximal [S].