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....
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 .
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
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
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...
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?
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.
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...