Homework 3 — Enzyme Kinetics

Problem 1 — Full dynamics vs Michaelis-Menten

From the data table: V₀ = V_maxs₀/(K_m+s₀) fits exactly with V_max = 5 mM/min (the one correct statement) and K_m = 3 mM. With e₀ = 10 mM: k₂ = V_max/e₀ = 0.5/min, and using k_f = 10k_r in K_m = (k_r+k₂)/k_f: k_f ≈ 0.172/min, k_r ≈ 0.0172/min. The three ODEs: ds/dt = −k_f e s + k_rc,   dc/dt = k_f e s − (k_r+k₂)c,   dp/dt = k₂c, with e = e₀−c.

Why the curves differ: Michaelis-Menten assumes the complex C is in quasi-steady state and that s ≈ s₀ initially. During the first fraction of a minute (left plot) the complex is still forming, so the MM product curve leads the true one; late in the reaction substrate depletion makes both converge.

Problem 2 — Determine K_m by inspection, then simulate

V_max = 6.0 µM/s. The data row V₀ = 3.0 µM/s (= ½V_max) occurs at [S]₀ = 100 mM — so K_m = 100 mM with no calculation needed. Drag the K_m slider and watch the curve pass through the data.

Problem 4 — Lineweaver-Burk plot with an inhibitor

1/V = (K_m/V_max)(1/[S]) + 1/V_max. Both datasets are fitted live by least squares. The two lines come out nearly parallel (slope ratio ≈1.3 — for real experimental data that difference is unlikely to be significant), while V_max and K_m both fall by about the same factor (≈3×). Parallel Lineweaver-Burk lines with proportionally reduced V_max and K_m are the signature of uncompetitive inhibition: the inhibitor binds only the ES complex, so adding more substrate produces more complex to inhibit — the inhibition cannot be overridden by extra substrate.

Problem 5 — PFK-1: Michaelis-Menten or Hill?

The experimental curve is sigmoidal (slow start, then steep rise) — that is Hill dynamics, i.e. cooperative binding: V = V_max[S]ⁿ/(Kⁿ+[S]ⁿ). Adjust the sliders to fit the data; the best MM fit (dashed) can never reproduce the sigmoid foot at low [S].