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16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax...
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...
How do I calculate the apparent vmax? 16. At right is a graph obtained from a...
How do I calculate the apparent vmax?
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 M/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 1.5 KM: 3mm ve (MM/s) 1 0.5 0 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...
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5...
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 catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The...
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
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. T...
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...
The following observations come from Lineweaver-Burke plots, based on kinetic data generated from a Michaelis/Menton-type enzyme...
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...
You perform a series of enzyme activity assays and then graph the data using a Lineweaver-Burk...
You perform a series of enzyme activity assays and then graph the data using a Lineweaver-Burk plot. You determine the X-intercept is at -0.02 mM-1 and the Y-intercept is at 5.0 (mM/sec)-1. Calculate the Vmax and Km for this enzyme. A. Vmax = 0.20 mM/sec; Km = 50.0 mM B. Vmax = 0.20 mM/sec; Km = ‒50.0 mM C. Vmax = 5.0 mM/sec; Km = 0.02 mM D. Vmax = 5.0 mM/sec; Km = ‒0.02 mM
b. Look at the graph below of how a competitive inhibitor affects the kinetics of an...
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...
9. The following data were obtained from an enzyme kinetics experiment. Graph the data using a...
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
Please do everything like you were answering a test. Don’t attempt if youre not going to do all p...
Please do everything like you were answering a test. Don’t attempt
if youre not going to do all parts. Do it ASAP and I will give you
a good rating. If not I will report you. Thank you so much for
being the best. Show work if necessary and be concise. There’s no
way for me to separate so do all parts.
do all parts please. I cant seperate it because it will be
refunded.
2. i) (10 points) The...
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...
How do I calculate the apparent vmax?
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 M/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 1.5 KM: 3mm ve (MM/s) 1 0.5 0 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...
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 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
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...
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...
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
Please do everything like you were answering a test. Don’t attempt
if youre not going to do all parts. Do it ASAP and I will give you
a good rating. If not I will report you. Thank you so much for
being the best. Show work if necessary and be concise. There’s no
way for me to separate so do all parts.
do all parts please. I cant seperate it because it will be
refunded.
2. i) (10 points) The...
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