Numerical Problems: Enzyme Kinetics

Q1: An enzyme has a Vmax of 100 µM/s and a Km of 2mM. What is the velocity of the reaction when the substrate concentration is 2mM?

Solution: According to the Michaelis-Menten equation: v = (Vmax * [S]) / (Km + [S]) Given: Vmax = 100 µM/s, Km = 2mM, [S] = 2mM. v = (100 * 2) / (2 + 2) = 200 / 4 = 50 µM/s. This is consistent with the definition of Km as the substrate concentration at which the reaction velocity is half of Vmax.

Q2: The Vmax for an enzyme-catalyzed reaction is 60 µM/min. The Km is 1 mM. What is the substrate concentration when the velocity is 20 µM/min?

Solution: Using the Michaelis-Menten equation: v = (Vmax * [S]) / (Km + [S]) 20 = (60 * [S]) / (1 + [S]) 20 * (1 + [S]) = 60 * [S] 20 + 20[S] = 60[S] 20 = 40[S] [S] = 20 / 40 = 0.5 mM.

Q3: A Lineweaver-Burk plot gives a y-intercept of 0.05 (µM/min)⁻¹ and an x-intercept of -0.2 (mM)⁻¹. Calculate Vmax and Km.

Solution: The y-intercept = 1/Vmax. 0.05 = 1/Vmax Vmax = 1 / 0.05 = 20 µM/min. The x-intercept = -1/Km. -0.2 = -1/Km Km = 1 / 0.2 = 5 mM.

Q4: An enzyme has a Km of 5x10⁻⁵ M. A competitive inhibitor is present at a concentration of 2x10⁻⁵ M, and the inhibitor constant Ki is 4x10⁻⁵ M. Calculate the apparent Km (Km_app).

Solution: For competitive inhibition, the apparent Km is calculated as: Km_app = Km * (1 + [I]/Ki) Km_app = 5x10⁻⁵ * (1 + (2x10⁻⁵ / 4x10⁻⁵)) Km_app = 5x10⁻⁵ * (1 + 0.5) Km_app = 5x10⁻⁵ * 1.5 = 7.5x10⁻⁵ M.