54. What is the catalyze reaction rate? Km=2 mM Kcat= 3 s-1 At the enzyme concentration=10...
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54. What is the catalyze reaction rate?
Km=2 mM
Kcat= 3 s-1
At the enzyme concentration=10...
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...
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 has a Km for substrate of 10 mM and Vmax of 5 mol L-1...
An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1 sec-1 at a total enzyme concentration of 1 nM. At [S] = 10 mM, kcat is: A) 2500 per M per sec. B) 5000 per M per sec. C) 1250 per M per sec. D) 2500 per sec. E) 5000 per sec.
Kcat= 30.0sec^-1 Km= 0.0050M Total enzyme Concentration= 1uM What is the max rate of the enzyme?
Kcat= 30.0sec^-1 Km= 0.0050M Total enzyme Concentration= 1uM What is the max rate of the enzyme?
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km...
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km of 0.0030 M have an initial velocity that is 25% of the maximum velocity? a. 0.0010M b 0.0090 M c. 0.0015 M d. 0.0060 M e.Cannot be determined
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km...
At what substrate concentration would an enzyme with a kcat of 30.0 sec-1 and a Km of 0.0030 M have an initial velocity that is 25% of the maximum velocity? a. 0.0010M b 0.0090 M c. 0.0015 M d. 0.0060 M e.Cannot be determined
An enzyme that follows Michaelis-Menten kinetics has a KM value of 20.0 μM and a kcat...
An enzyme that follows Michaelis-Menten kinetics has a KM value of 20.0 μM and a kcat value of 211 s−1. At an initial enzyme concentration of 0.0100 μM, the initial reaction velocity was found to be 1.07×10−6 μM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures.
1.What will be the reaction rate at 0.0550 mM [G6P]? (2) 2. If the enzyme concentration...
1.What will be the reaction rate at 0.0550 mM [G6P]? (2)
2. If the enzyme concentration is unchanged, at what G6P
concentration will the reaction rate equal 1.650 mM/s? (2)
3. What PGI concentration is required to give a reaction rate of
4.000 mM/s at 0.0300 mM [G6P]? (2)
Calculated Vmax=1.440mM/s
Km=0.0330 mM
The enzyme phosphoglucose isomerase (PGI) catalyzes the following reaction in glycolysis сH,ОРО,2- Phosphoglucose 2-03POH2C сH,он о. н Н н isomerase но он н Н Н он он...
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
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 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
1.What will be the reaction rate at 0.0550 mM [G6P]? (2)
2. If the enzyme concentration is unchanged, at what G6P
concentration will the reaction rate equal 1.650 mM/s? (2)
3. What PGI concentration is required to give a reaction rate of
4.000 mM/s at 0.0300 mM [G6P]? (2)
Calculated Vmax=1.440mM/s
Km=0.0330 mM
The enzyme phosphoglucose isomerase (PGI) catalyzes the following reaction in glycolysis сH,ОРО,2- Phosphoglucose 2-03POH2C сH,он о. н Н н isomerase но он н Н Н он он...
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
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