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?
Homework Answers
![from that enzyme kinetics we know Vmax [E.] So, if at [eo] = 1 MM, Vmax = 100 MM/min then at [E] 22 MM , Donor = 200 MM /min](http://img.homeworklib.com/questions/893d6480-734a-11ea-bf44-b9522ecd2c63.png?x-oss-process=image/resize,w_560)
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At an enzyme concentration of 1microM the Vmax is
100microM/min and the Km is 20nm. At...
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...
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?
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.
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95...
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations: A. 1 mM B. 9 mM C. 11 mM
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation...
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
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)
What is the velocity of a Michaelis-Menten enzyme reaction (in terms of vmax) when the concentration...
What is the velocity of a Michaelis-Menten enzyme reaction (in terms of vmax) when the concentration of substrate is 4 times the value of KM? Show your work.
How do competitive inhibitors affect the KM and Vmax of an enzyme? Draw a plot of...
How do competitive inhibitors affect the KM and Vmax of an enzyme? Draw a plot of velocity as a function of substrate concentration, both with and without inhibitor added.
Name Page Number Date Vmax 1. Estimate Vmax and Km using the velocity vs. substrate concentration...
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
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
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
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)
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
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
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