Page 5 22 A Km and Vmax for an enzyme is determined in the a

Page 5 22. A Km and Vmax for an enzyme is determined in the absence and presence of an inhibitor () The LB plots resulting from these experiments are shown in the graph below. Calculate the Kmapp and Vmax.\"both in the absence and presence of inhibitor. If the inhibitor is used at a concentration of 250 uM, what is its Ki? What type of inhibition is observed? How did you decide? 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 y=2.5348x + 0.0386 y 1 5.298x + 0.0283 eno l with 1 -0.02-0.01 0.01 0.02 0.03 1substtrate] (1/ uM)

Solution

From the above shown plot,

Vmax = 1/intercept on y-axis

without Inhibitor Vmax = 1/0.0386 = 25.910 umol/min

with inhibitor Vmax = 1/0.0283 = 35.336 umol/min

Kmax = slope x Vmax

without Inhibitor Km = 2.5348 x 25.910 = 65.68 uM

with inhibitor Km = 15.298 x 35.336 = 540.57 uM

With [I] = 250 uM

Km,app = Km[1 + [I]/Ki]

540.57 = 65.68[1 + 250/Ki]

Ki = 0.138 uM

Type of inhibition = both Vmax and Km have increased upon addition of inhibitor, so it is a mixed inhibition.