1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vma...
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...
4. The following data were obtained from an enzyme kinetics experiment. Graph the data using a Lineweaver-Burk plot and determine, by inspection of the graph, the values for Km and Vmax. ISI (M) V (nmol/min) 0.20 0.26 0.33 1.00 1.43 1.67 2.08 3.33 5. You measured the kinetics of an enzyme activity as a function of substrate concentration (see Table). The enzyme concentration was maintained constant at a level of 1 M. [S] AM Vopmol/min 2.9 3.8 4.4 Plot the...
The following data set was collected from an experiment conducted in the lab where a new enzyme is being characterized. Use the Lineweaver-Burk method in order to determine the values of KM and Vmax. Complete your work on a separate piece of paper and upload the excel file or picture of your work. Your work must include the following: Table of reciprocals Reciprocal plot (straight-line graph) Calculations and results for KM and Vmax In order to get full credit for...
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
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...
Name Page Number Date Vmax 1. Estimate Vmax and Km using the velocity vs. substrate concentration plot - KM 2. Calculate the Vmax and Km using the Lineweaver-Burk plot. KM Vmax (S), uM 0.08 0.12 0.54 1.23 1.82 2.72 Reaction Velocity (UM/min) 0.15 0.21 0.70 1.1 1.3 15 4.94 10.00 1.8
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the enzyme- catalized reactions are explained by Michaelis-Menten equation. IMPORTANT: Show the calculations and indicate the units for all your answers. a. For an enzyme which follows the Michaelis-Menten enzyme kinetics, Km is 50 mmol L. Calculate the substrate concentration required to obtain the initial velocity (V.) equivalent to 90% of the maximum velocity (Vmax). b. The Vmax of the above reaction is 250 mmol...
After collecting enzyme kinetics data using substrates in mM and reaction velocities in mM/min, you make a Lineweaver-Burk plot. Your line of best fit has the equation: y = 0.00160 x + 0.00759. Calculate the Vmax of the enzyme using the equation: 1 KM 1 - = v Vmax [S] Vmax +
11. In Excel, prepare Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available: V, umol/min umol/min (No Inhibitor) S], mM (Inhibitor Present) 3.66 5.12 6.18 6.98 7.60 4.58 6.40 7.72 8.72 9.50 3.0 5.0 7.0 9.0 11.0 a. What are the KM and Vmax values for the inhibited and uninhibited reaction 5 pts. each reaction) b. Is the inhibitor competitive or noncompetitive? (5 pts.) Micheli-Menten) EQUATIONS: VV
9. The following data were obtained from an enzyme kinetics experiment. Graph the data using a Lineweaver-Burk plot and determine the values for Km and Vmax. [S] (UM) V(nmol/min) 0.20 0.26 1.43 1.67 2.08 3.33 0.33 1.00