An enzyme kinetics assay reveals that V_max = 100 nmol/s and K_m = 10 μM. For this assay, what substrate concentration is required for a reaction rate V₀ of 75 nmol/s?

Question

Quizplus · Accepted Answer

11ecba2d_0f08_9cd3_a3bf_6dc037acdea6_MD0008_00 The Michaelis-Menten equation describes the velocity V₀ of any enzyme-catalyzed reaction as a function of substrate concentration [S]. The general form of the equation is 11ecba2d_0f08_c3e4_a3bf_75c138a8629d_MD0008_00 V0=VmaxSKm+[S] where V_max is the maximum velocity and K_m is the [S] at which V₀ = ½V_max. When any three of the four values (V₀, V_max, K_m, and [S]) are known, the remaining value can be determined mathematically. The question states that V_max = 100 nmol/s, K_m = 10 μM, and V₀ = 75 nmol/s. Each of these values can be substituted into the Michaelis-Menten equation: 11ecba2d_0f08_c3e5_a3bf_f1368c957a4c_MD0008_00 75 nmol/s=100 nmol/s×S10 µM+S and [S] can be determined. To solve for [S], first both sides of the equation should be multiplied by (10 μM + [S]) to yield 11ecba2d_0f08_eaf6_a3bf_d1658bc24c9f_MD0008_00 75 nmol/s×10 µM+S=100 nmol/s×S The left side of the equation can then be expanded to 11ecba2d_0f08_eaf7_a3bf_9733b8fb0120_MD0008_00 750nmol/s µM+75 nmol/s×S=100 nmol/s×S Then (75 nmol/s) × [S] is subtracted from both sides, giving 11ecba2d_0f09_1208_a3bf_8bb3f8763c8d_MD0008_00 750nmol/s µM=25 nmol/s×S Finally, both sides of the equation are divided by 25 nmol/s, resulting in 11ecba2d_0f09_1209_a3bf_d712228fa7a6_MD0008_00 750 nmol/s µM25 nmol/s=S=30 µM Therefore, a substrate concentration of 30 μM would yield a reaction rate of 75 nmol/s. (Choice A) The Michaelis-Menten equation shows that if [S] were 20 μM, V₀ would be 66.7 nmol/s. (Choice C) The Michaelis-Menten equation shows that if [S] were 50 μM, V₀ would be 83.3 nmol/s. (Choice D) The Michaelis-Menten equation shows that if [S] were 75 μM, V₀ would be 88.2 nmol/s. Educational objective: The Michaelis-Menten equation 11ecba2d_0f09_391a_a3bf_eff7bcfd7e45_MD0008_00 V0=Vmax [S]Km+[S] relates the observed rate V₀ of an enzyme-catalyzed reaction to the substrate concentration [S] by the constants V_max and K_m. If any three of the parameters V₀, V_max, [S], and K_m are known, the fourth can be determined by solving the equation.

An Enzyme Kinetics Assay Reveals That Vmax = 100 Nmol/s

An Enzyme Kinetics Assay Reveals That V_max = 100 Nmol/s