
How do I calculate the apparent vmax?
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How do I calculate the apparent vmax? 16. At right is a graph obtained from a...
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. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 KM: 3mm V. (mM/s) 1.5 1 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...
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...
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...
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2. i) (10 points) The...
b. Look at the graph below of how a competitive inhibitor affects the kinetics of an enzyme C. Rate of reaction is the Vmax of the enzyme affected? Why or why not: explain in terms of substrate concentration and enzyme active site saturation) Without inhibitor With competitive inhibitor d. is Vmax/2 affected? Why or why not: explain in terms of Vmax. Substrate concentration e. Is Km affected? Explain in terms of the active site. Hint a competitive inhibitor is competing...
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
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
10.What type of inhibitor is this? How do you know? (2)
11.For your assigned inhibitor 1, what are the apparent Km &
Vmax? (NOTE: apparent Km& Vmax are just the Km & Vmax in
presence of inhibitor, at a given concentration.) (2)
Kinetics experiments were performed on PGI. Enzyme activity
(initial velocity, Vo) was measured at varying concentrations of
Glucose-6-phosphate (G6P). The enzyme concentration used in all
experiments was 1.5 μM.
12.What will be the reaction rate with 0.500 mM...
The following data was obtained for an enzyme in the absence of an inhibitor, and in the presence of two different inhibitors. The concentration of each inhibitor was 10 mM. The total concentration of enzyme was the same for each experiment. [S] {mM} without inhibitor v, {umol/(ml*s)} with inhibitor A v, {umol/(ml*s)} With inhibitor B v, {umol/(ml*s)} 0.0 0.0 0.0 0.0 1.0 3.6 3.2 2.6 2.0 6.3 5.3 4.5 4.0 10.0 7.8 7.1 8.0 14.3 10.1 10.2 12.0 16.7 11.3...
please graph all 3 lines and explain the
vmax&km
How to: Lineweaver Burke 1. The following data was determined for an enzyme in the absence of an inhibitor and in the presence of two different inhibitors (V2 and V3). Determine the V. and K for the enzyme (1) Plot the data and determine the type of inhibition for each inhibitor (S) mm 1 V2 4.3 5.5 V1 12 20 29 2 relliate 150b