Biochem: Enzymes, Kinetics, and Catalysis

How do enzymes catalyze reactions, and what is the biological significance?

• They lower the activation energy by binding the transition state in a complementary pocket

• The cell’s enzymes select a specific subset of all possible reactions, and that subset defines what the cell does.

Why does life require high activation energies (slow uncatalyzed chemistry)? (HW 1.2, Mock 3 Q7)

• Catalysis is a selection principle. If every reaction were fast without a catalyst, no reaction could be favored by accelerating it.

• High barriers also make biopolymers kinetically stable even though they are thermodynamically unstable, and let systems sit far from equilibrium, which energy coupling needs.

Two ways to accelerate a reaction; why does life prefer one? Is there an upper limit? (HW 1.4, 1.5)

• Raise T or lower Δ‡G°

• Raising T speeds up all reactions

• lowering Δ‡G° by enzymatic catalysis speeds up one, so speed becomes a selection principle. T

• he upper limit is k⁺ ≤ k_BT/h, reached at zero activation energy.

Show that a catalyst cannot change the equilibrium position (HW 1.6, Mock 2 Q5)

• Forward and reverse reactions pass through the same transition state, so a catalyst lowers Δ‡G°_fwd and Δ‡G°_rev by the same amount.

• Keq = k⁺/k⁻ = exp(−(Δ‡G°_fwd − Δ‡G°_rev)/k_BT) depends only on the difference, which is unchanged.

One-sentence reason enzymes promote both directions equally (HW 3.8, my answer)

• Forward and reverse transitions share one transition state, so stabilizing it speeds both equally

• a catalyst that sped up only one direction would violate the second law (it would turn heat into work).

How can an enzyme accelerate a reaction without changing ΔG°? Distinguish activation free energy from reaction free energy (Mock 3 Q5).

• The enzyme lowers the barrier (activation free energy, which sets the rate) but leaves the free-energy difference between reactants and products (which sets equilibrium) untouched.

• It does that by stabilizing only the transition state, not the substrates or products.

If [ES] = [E], what is v₀? (HW 3.4, Mock 1 Q1)

v₀ = Vmax/2. Justification: p_ES = [S]/([S] + K_M) = ½ gives [S] = K_M, so v₀ = Vmax·K_M/(K_M + K_M) = Vmax/2. Equivalently, half the enzyme is bound, so v₀ = kcat[ES] = kcat[E]tot/2.

Why does v₀ fall as K_M rises? (HW 3.5, Mock 3 Q8)

When kcat ≪ k⁻, K_M ≈ k⁻/k⁺, the dissociation constant of ES. A large K_M means the ES complex falls apart easily, so at a given [S] less enzyme is bound and many binding cycles are needed per product molecule. Formally v₀ = Vmax[S]/([S] + K_M) shrinks as K_M grows.

Competitive inhibitor: K_M rises, Vmax unchanged (HW 3.6, my answer)

• The inhibitor competes with substrate for the same active site

• Extra substrate is needed to reach half-saturation, so the apparent K_M increases.

• At very high [S] the substrate outcompetes the inhibitor and all enzyme can be driven into ES, so Vmax is unchanged.

Non-competitive inhibitor: effect on Vmax (HW 3.7, my answer)

• Vmax decreases.

• The inhibitor binds E and ES alike and inactivates them, and no amount of substrate undoes that.

• It effectively removes active enzyme, and Vmax = kcat[E]active.

Lineweaver–Burk (HW 2.16).

• Inverting Michaelis–Menten gives 1/v₀ = (K_M/Vmax)(1/[S]) + 1/Vmax, a straight line in 1/v₀ versus 1/[S]. Slope K_M/Vmax; y-intercept 1/Vmax; x-intercept −1/K_M.

Catalytic efficiency.

• kcat/K_M ≤ k⁺.

• Some enzymes approach the diffusion limit (10⁸–10⁹ M⁻¹s⁻¹), meaning nearly every enzyme–substrate encounter is productive (acetylcholinesterase, fumarase, catalase in the Lecture 4 table).