Question

16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve on the graph. 0 0 5 10 15 20 25 30 35 [Substrate] (mm) Apparent km: Apparent Vmax:

Answer 1

16, a) v vana =4.5 4M/S Max Maz = 2.25 MM/ 2 k = 3 MM b) Adding a non-competitive inbibitor reduces the Vagay, so the apparent Curve will be below the Original eurve Non-compelitive inbibitor reduces the Vmax but does not alter kan Khan Apparent km - 30 M Apparent Vmax = 2.4 4 M/s. Apparent v max ₃ 1.2 you/s 2 If the inbibitor concentration (1) and kI is given. Vmax Apparent can be calculated from the ean apparest Vmax Vmax [I] + KI

5 4.5 4 3.5 3 2.5 Vmax/2 V. (MM/s) 2 1.5 Apparent Vmax/2 1 0.5 0 0 Km 5 10 15 20 25 30 35 [Substrate] (mm)