An enzyme has a Km of 10 mM and a Vm of 100 mmol/min If [S]=10 mM, which will increase the velocity more◦ 10-fold decrease in Km or 10-fold increase in V m? If [S]=100 mM, which will increase the velocity more◦ 10-fold decrease in Km or 10-fold increase in V m? Use both qualitative and quantitative analysis

The Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =

Which of the following statements are true about the relationships of [S], KM, and Vmax?  (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.

Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitor competitive or noncompetitive? [S] (M) 10,000 1 x 104 2.000-5 x 104 666673 1.5 x 10-3 400- 2.5 x 10-³ 2005x103 V, No Inhibitor (Arbitrary units) 0.026 38,461 0.092 10.87 37100 0.136 7.35 Wak 0.150 -67 0.165 6.06 L slope: 0033 yint=5057 полесте um.x=0.198 km 6.5x 10-4 0.010 0.040 25 10.086 11.63 0.120 8.33 0.142 7.04 2500 V, Inhibitor Present (Arbitrary Units) 100 ↓ $op.= .009 yint: 5.21 900x12 km: 1.73 y/o conf. V10-3 321000

The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →