Questions: Michaelis-Menten Kinetics and Enzyme Catalysis

Question 1

Enzyme A has Km = 0.05 mM and Vmax = 1 μM/s. Enzyme B has Km = 2 mM and Vmax = 100 μM/s. In a cell where substrate concentration is maintained at 0.01 mM, which enzyme produces product faster?

• A. Enzyme A, because its lower Km means it has higher affinity and will be more active at low substrate
• B. Enzyme B, because its higher Vmax means it can process substrates faster when bound
• C. Enzyme B, because its higher Vmax/Km (specificity constant) makes it more efficient at sub-Km concentrations
• D. They produce identical rates because both are operating well below their respective Km values

Answer: (C)

At [S] << Km for both enzymes, rate ≈ (Vmax/Km) × [S]. Enzyme A: Vmax/Km = 1/0.05 = 20 s⁻¹ (in equivalent units). Enzyme B: Vmax/Km = 100/2 = 50 s⁻¹. Enzyme B's specificity constant is 2.5× higher, so it produces product faster despite having a higher Km. This destroys the 'lower Km = better enzyme' misconception: what matters at low substrate is Vmax/Km, not Km alone. An enzyme with modest affinity but very fast catalysis can outperform one with tight binding but slow turnover.

Question 2

Why does enzyme-catalyzed reaction rate level off and reach a maximum (Vmax) at high substrate concentrations, rather than continuing to increase as in an ordinary bimolecular reaction?

• A. High substrate concentrations inhibit the enzyme by product-like feedback
• B. All enzyme active sites become occupied (saturated), leaving no free enzyme to bind additional substrate
• C. At high substrate, the reaction becomes thermodynamically unfavorable and slows
• D. Substrate molecules begin competing with each other for enzyme, reducing the effective concentration

Answer: (B)

Enzyme active sites are finite and specific binding pockets. Once every enzyme molecule has substrate bound (fully saturated), the rate is limited entirely by how fast the ES complex converts to product (k_cat × [E]_total = Vmax). Adding more substrate cannot increase the rate further because there are no free active sites to bind it. This saturation behavior is the defining feature of enzyme kinetics and is absent in simple bimolecular reactions, which have no binding capacity limit.

Question 3

The Michaelis constant Km is equal to the dissociation constant (Kd) of the enzyme-substrate complex, so a lower Km usually indicates tighter enzyme-substrate binding.

• T. True
• F. False

Answer: False

Km = (k₋₁ + k_cat)/k₁, while Kd = k₋₁/k₁. These are equal only when k_cat << k₋₁ — that is, when product formation is slow compared to substrate release. For many efficient enzymes, k_cat is significant relative to k₋₁, and Km > Kd. Furthermore, even if Km did equal Kd, tighter binding (lower Kd) is not always beneficial: an enzyme that holds substrate too tightly may release product slowly, reducing turnover. True catalytic efficiency is best measured by k_cat/Km (the specificity constant), not by Km or Kd alone.

Question 4

At substrate concentrations far above Km (say [S] = 100 × Km), doubling the substrate concentration will approximately double the reaction rate.

• T. True
• F. False

Answer: False

At [S] >> Km, the Michaelis-Menten equation simplifies to v ≈ Vmax — the rate is essentially at its maximum and is zero-order in substrate. Doubling [S] from 100 Km to 200 Km changes the rate from 100/101 × Vmax to 200/201 × Vmax — a change of less than 0.5%. Doubling substrate doubles the rate only in the first-order regime ([S] << Km), where v ≈ (Vmax/Km)[S]. The order of the reaction with respect to substrate transitions from first-order (at low [S]) to zero-order (at high [S]) — this transition is the hallmark of saturation kinetics.

Question 5

What is the physical meaning of the Michaelis constant Km, and why is the specificity constant Vmax/Km a better single measure of enzyme efficiency than Km alone?

Model answer: Km is the substrate concentration at which the reaction rate is exactly half of Vmax. Physically, it reflects the balance between substrate binding (k₁), substrate release back to free enzyme (k₋₁), and catalytic conversion to product (k_cat). Km alone measures something close to binding affinity but is not true affinity. Vmax/Km (the specificity constant, equal to k_cat/Km) measures how fast the enzyme converts substrate to product when substrate is scarce ([S] << Km) — the regime where most enzymes actually operate in cells. Two enzymes can have the same Km but very different Vmax values, making one far more productive. The specificity constant captures both binding and catalysis in a single number and is the correct efficiency metric for comparing enzymes.

An enzyme with low Km (tight binding) but low k_cat (slow catalysis) can be outperformed by one with higher Km but much higher k_cat. The specificity constant rewards the combination of finding substrate quickly (high k₁, low Km) and converting it quickly (high k_cat). The theoretical upper limit of ~10⁹ M⁻¹s⁻¹ is set by diffusion — 'perfect' enzymes like catalase approach this limit.