S | 1/S | V | 1/V |
0.1 | 10 | 3.33 | 0.3 |
0.2 | 5 | 5 | 0.2 |
0.5 | 2 | 7.14 | 0.14 |
0.8 | 1.25 | 8 | 0.125 |
1 | 1 | 8.33 | 0.12 |
2 | 0.5 | 9.09 | 0.11 |
Enzyme concentration = 10^-6 M = 10^-6 × 10^3 mM
Kcat is turnover number.
Please rate.
3.33 3. Suppose that the following data are obtained for an enzyme-catalyzed reaction: [S(mm) V (mmol...
Suppose that the following data are obtained for an enzyme-catalyzed reaction: [S] (mM) V (mmol ml-1min-1) 0.1 3.33 0.2 5.00 0.5 7.14 0.8 8.0 1.0 8.33 2.0 9.09 a.) From a double-reciprocal plot of the data, determine Km and Vmax. b.) Assuming that the enzyme present in the system had a concentration of 10-6 M, calculate its turnover number.
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...
Consider the data collected for an enzyme-catalyzed reaction [S] (mM) vo (mM s1 0.25 2.00 0.50 3.33 1.00 5.00 2.00 6.67 Determine Vmax and K,,m for this reaction. mM s- Vmax mM
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 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...
s- (3 pts) The flowing rates have been an enzyme- catalyzed reaction at various substrate concentrations: Run no. 103 [S]M Rate, v/ (Ms') 2.41 3.33 4.78 6.17 7.41 9.52 0.0 12.5 0.4 0.6 4 2.0 4.0 a- From Line-Weaver double reciprocal plot, obtain and Mechaelis-Menten constant, max b- If the enzyme concentration is 1.00 x 10-11 M calculate kz.
] a. The equation of Lineweaver-Burk double-reciprocal plot of caffeine dehydrogenase-catalyzed reaction is y = 12x + 3. Calculate Vmax (mmol/s) and Km (mmol/L). b. [5 points] Estimate V for caffeine concentration of 400 mmol/L. c. [10 points] The enzyme caffeine dehydrogenase (Cdh) is an inducible quinone-dependent oxidoreductase. Describe how the addition of caffeine into the culture medium will be detected and transcriptionally regulated by the two-component system in Pseudomonas sp. CBB1.
Consider the data collected for an enzyme-catalyzed reaction. [S] (MM) 2 (MM :5-1) 0.20 0.86 0.33 1.20 2.00 2.42 4.00 2.69 Determine Vmax and Km for this reaction. mMs-1 Km = mM
The value of Km for the shown data for a hexokinase-catalyzed reaction is with the unit of . The value of Vmax for. the same reaction is with the unit of . Be sure to give the values with the correct number of significant figures. You might have to construct a kinetic plot. For units, choose one answer from (uM, 1/ UM, HM/second, uM x second, mM, 1/mM, second, 1/second, mM/second, mM x second) vo (mM/sec) Glucose concentration (mm) 0.10...
2. The table shows the kinetic data for a reaction catalyzed by an enzyme under the following conditions: in the absence of an inhibitor, and in the presence of two different inhibitors, (1) and (2) each at a concentration of 10 mM. Assume the total enzyme concentration, [Elo, is the same for all reactions and the enzyme obeys the Michaelis-Menten mechanism In the presence of presence of 10 mM inhibitor 1 inhibitor2 In the 10 mM No inhibitor mM 2.5...