a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When...
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 enzyme did the inhibitor inactivate?
Homework Answers
What is the rate of an enzyme catalyzed reaction if the Vmax is 100µmol S→P/min and...
What is the rate of an enzyme catalyzed reaction if the Vmax is 100µmol S→P/min and the Km is 7 mM and the substrate concentration is 11mM? Is the enzyme working at Vmax? What if the substrate concentration is raised to 25mM?
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where...
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
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...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. kcat kcat = 2.73×104 min−1 Part B Calculate kcat/KM for the enzyme reaction.
At an enzyme concentration of 1microM the Vmax is 100microM/min and the Km is 20nm. At...
At an enzyme concentration of 1microM the Vmax is 100microM/min and the Km is 20nm. At an enzyme concentration of 2microM, what would the V0 be at a substrate concentration of 40mM?
1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vma...
1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vmax and the amount of enzyme present in this experiment. Use excel to obtain the lineweaver burk plot. Substrate concentration (MM) 1 Initial velocity (umol/min) 167 250 334 376 498 499 100 1,000
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration...
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
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
In a competitive inhibition, Km is increased while Vmax is unchanged. An enzyme is being assayed...
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?
Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ....
Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ. What is the velocity Vo of the enzyme at the following substrate concentrations? 1 uM 10 uM 100 HM Question3 For an enzyme, the following measurements have been made: Substrate concentration [S] Initial velocity Vo 10 20 40 90 120 180 300 500 10,000 50,000 0.12 0.20 0.30 0.42 0.45 0.39 0.53 0.56 0.60 0.60
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...
1. The turnover number for an enzyme is known to be 5000 min. From the following set of data, calculate the Km, Vmax and the amount of enzyme present in this experiment. Use excel to obtain the lineweaver burk plot. Substrate concentration (MM) 1 Initial velocity (umol/min) 167 250 334 376 498 499 100 1,000
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
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
Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ. What is the velocity Vo of the enzyme at the following substrate concentrations? 1 uM 10 uM 100 HM Question3 For an enzyme, the following measurements have been made: Substrate concentration [S] Initial velocity Vo 10 20 40 90 120 180 300 500 10,000 50,000 0.12 0.20 0.30 0.42 0.45 0.39 0.53 0.56 0.60 0.60
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