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Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was 5x10-4...
3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4...
3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5 minutes?...
3. Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was...
3. Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, pNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1 st order of reaction? (3 points) b) What will be the product concentration at the end of...
1. An enzyme with a Km of 1x10 % M was assayed using an initial substrate...
1. An enzyme with a Km of 1x10 % M was assayed using an initial substrate concentration of 3x10-- M. After 2 min, 5 percent of the substrate was converted. How much substrate will be converted after 10 min. 30 min. 60 min? How long must the reaction be run to achieve 99% conversion? (Assume that the enzyme follows Michaelis-Menten kinetics.)
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22...
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22 micromoles X liters^-1 Xmin^-1). What velocity would be observed in the presence of 2X10^-4 M substrate and 5X10^-4M of a. a competitive inhibitor b. a noncompetitive inhibitor c. an uncompetitive inhibiter Ki in all three cases is 3X10^-4M. What is the degree of inhibition in all three cases?
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36...
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. Identify the type of inhibitor with explanation.
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50...
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
The results for enzyme activity analysis are presented in Table 5 (below). In this case, do...
The results for enzyme activity analysis are presented in Table 5 (below). In this case, do not use the graph to determine: (1.5 pts each) (Hint- use M-M equation) a) Vmax; b) Km; c) initial velocity at [S] 1 x 10-1 M; d) the amount of product formed during the first 5 minutes at [S] 2x10 3M; at a [S of 2x10 6M? e) what is Km and Vmax if the free [E is increased 3-fold? 3. v (umol/min 0.25...
The Km of an enzymatic reaction is 10^-4 M. What substrate concentration is needed in order...
The Km of an enzymatic reaction is 10^-4 M. What substrate concentration is needed in order to attain 60% if the maximum velocity of the reaction?
Refer to the below Enzyme activity data (Table 4) to determine the following: (1.5 pts each) a) V...
Refer to the below Enzyme activity data (Table 4) to determine the following: (1.5 pts each) a) Vmax b) why is the velocity v constant at [S] greater than 2 x 10-3 M? c) what is the free [E] at [S] = 2 x 10-2 M? 2. SI (mol/L 2 x 10-l 2 x 10-2 2 x 10-3 2 x 104 5 x 10-4 1,3 x 10-5 ol/min 60.00 60.00 60.00 48.00 45.00 2.00 Table 4. Enzyme activity assay data...
4) (5 points) What fraction of Vmax is observed at [S] = 5 KM? 5) (20...
4) (5 points) What fraction of Vmax is
observed at [S] = 5 KM?
5) (20 points) For the following data:
[S] (μM)
V0 (no inhibitor)
V0 (2.45 μM inhibitor present)
2.1
0.031
0.020
4.2
0.06
0.045
13
0.138
0.09
20
0.153
0.13
52
0.170
0.135
a) Construct a 1/v (y-axis) versus 1/[S]
(x-axis) plot in the space below.
b) Is the inhibition competitive, noncompetitive, or
uncompetitive?
c) Calculate KM, KMapp,
Vmax, and Vmaxapp....
3. Acid Phosphatase was assayed at [S] =2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, PNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1st order of reaction? (3 points) b) What will be the product concentration at the end of 3.5 minutes?...
3. Acid Phosphatase was assayed at [S] = 2x10-5 M. The Km for the substrate was 5x10-4 M. At the end of one minute 2.5% of the substrate, pNPP, had been converted to pNP (a product). Answer the following questions using this information: a) What percent of PNPP will be converted at the end of 3.5 minutes assuming product formation is in 1 st order of reaction? (3 points) b) What will be the product concentration at the end of...
1. An enzyme with a Km of 1x10 % M was assayed using an initial substrate concentration of 3x10-- M. After 2 min, 5 percent of the substrate was converted. How much substrate will be converted after 10 min. 30 min. 60 min? How long must the reaction be run to achieve 99% conversion? (Assume that the enzyme follows Michaelis-Menten kinetics.)
The following data were recorded for the enzyme-catalyzed reaction. Substrate concentration (M) 6.25 x 100 7.50 x 10 1.00 x 10-4 1.00 x 10-3 Reaction velocity (nM/min) 15 56 60 75 (1) Estimate Km and Vmax- (2) What would V be at S=2.5 x 10-5 ?
The results for enzyme activity analysis are presented in Table 5 (below). In this case, do not use the graph to determine: (1.5 pts each) (Hint- use M-M equation) a) Vmax; b) Km; c) initial velocity at [S] 1 x 10-1 M; d) the amount of product formed during the first 5 minutes at [S] 2x10 3M; at a [S of 2x10 6M? e) what is Km and Vmax if the free [E is increased 3-fold? 3. v (umol/min 0.25...
Refer to the below Enzyme activity data (Table 4) to determine the following: (1.5 pts each) a) Vmax b) why is the velocity v constant at [S] greater than 2 x 10-3 M? c) what is the free [E] at [S] = 2 x 10-2 M? 2. SI (mol/L 2 x 10-l 2 x 10-2 2 x 10-3 2 x 104 5 x 10-4 1,3 x 10-5 ol/min 60.00 60.00 60.00 48.00 45.00 2.00 Table 4. Enzyme activity assay data...
4) (5 points) What fraction of Vmax is
observed at [S] = 5 KM?
5) (20 points) For the following data:
[S] (μM)
V0 (no inhibitor)
V0 (2.45 μM inhibitor present)
2.1
0.031
0.020
4.2
0.06
0.045
13
0.138
0.09
20
0.153
0.13
52
0.170
0.135
a) Construct a 1/v (y-axis) versus 1/[S]
(x-axis) plot in the space below.
b) Is the inhibition competitive, noncompetitive, or
uncompetitive?
c) Calculate KM, KMapp,
Vmax, and Vmaxapp....
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