An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22...
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
A simple noncompetitive inhibitor of acetylcholinesterase binds to the enzyme to affect Vmax only; it does not affect KM. Part A For an inhibition constant of KI=2.9×10?4M, what concentration of inhibitor is needed to give a 90% inhibition of the enzyme? Express your answer using two significant figures and include the appropriate units. [I] =
4) (5 points) What fraction of Vmax is observed at [S] = 5 KM? 5) (20 points) For the following data: [S] (μM) V0 (no inhibitor) V0 (2.45 μM inhibitor present) 2.1 0.031 0.020 4.2 0.06 0.045 13 0.138 0.09 20 0.153 0.13 52 0.170 0.135 a) Construct a 1/v (y-axis) versus 1/[S] (x-axis) plot in the space below. b) Is the inhibition competitive, noncompetitive, or uncompetitive? c) Calculate KM, KMapp, Vmax, and Vmaxapp....
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
Which is the most inappropriate description of enzyme inhibition? Select one: a. A competitive inhibitor increases Km only. b. A noncompetitive inhibitor decreases Vmax only. c. An uncompetitive inhibitor decreases both Km and Vmax. d. All of the above e. None of the above
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.(a)At what inhibitor concentration will 75% of the enzyme be bound to the inhibitor if there is no substrate present? (b) This enzyme has a Km of 4.0 × 10-5 M and a Vmax of 50 μM/s. At a substrate concentration of 3.0 × 10-4 M, calculate (i) the velocity of reaction in the presence of the inhibitor at 4.8 x 10-5 M (ii) the degree of...
In a competitive inhibition, Km is increased while Vmax is unchanged. An enzyme is being assayed in the presence of a fixed amount of a competitive inhibitor. How could the rate be increased in this reaction?
The following observations come from Lineweaver-Burke plots, based on kinetic data generated from a Michaelis/Menton-type enzyme (E) that catalyzes the hydrolysis of a peptide substrate (S). All data were generated in the presence of 18.0 μM total enzyme. The enzyme-catalyzed reaction has a Km of 3.00 μM and a Vmax of 2.00 μM/sec. The enzyme-catalyzed reaction in the presence of 15.0 μM of Inhibitor A has an apparent Km of 2.25 μM and an apparent Vmax of 1.50 μM/sec. The...
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. Identify the type of inhibitor with explanation.