Determine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax 140.85 micromol/minute) HO.HO [S] (M) 1/[S] vo (μmol/min) 1/vo OHO (M) (μmol/min) 2.50E-06 28 4.00E+05 3.57E-02 4.00E-06 40 2.50E+05 2.50E-02 1.00E-05 70 1.00E+05 1.43E-02 2.00E-05 95 5.00E+04 1.05E-02 4.00E-05 112 2.50E+04 8.93E-03 1.00E-04 128 1.00E+04 7.81E-03 2.00E-03 139 5.00E+02 7.19E-03 1.00E-02 140 1.00E+02 7.14E-03 y = 7.0 × 10-8 x + 0.0071
Determine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax 140.85 micromol/minute) HO.HO [S] (M) 1/[S] vo (μmol/min) 1/vo OHO (M) (μmol/min) 2.50E-06 28 4.00E+05 3.57E-02 4.00E-06 40 2.50E+05 2.50E-02 1.00E-05 70 1.00E+05 1.43E-02 2.00E-05 95 5.00E+04 1.05E-02 4.00E-05 112 2.50E+04 8.93E-03 1.00E-04 128 1.00E+04 7.81E-03 2.00E-03 139 5.00E+02 7.19E-03 1.00E-02 140 1.00E+02 7.14E-03 y = 7.0 × 10-8 x + 0.0071
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter7: Reaction Rates And Chemical Equilibrium
Section: Chapter Questions
Problem 7.42P
Related questions
Question
![Determine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax
140.85 micromol/minute)
HO.HO
[S] (M)
1/[S]
vo (μmol/min)
1/vo OHO
(M)
(μmol/min)
2.50E-06
28
4.00E+05
3.57E-02
4.00E-06
40
2.50E+05
2.50E-02
1.00E-05
70
1.00E+05
1.43E-02
2.00E-05
95
5.00E+04
1.05E-02
4.00E-05
112
2.50E+04
8.93E-03
1.00E-04
128
1.00E+04
7.81E-03
2.00E-03
139
5.00E+02
7.19E-03
1.00E-02
140
1.00E+02
7.14E-03
y = 7.0 × 10-8 x + 0.0071](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16403921-ce7c-46c3-b6d3-82ff9d7b2bd5%2F533827d0-3461-4231-88b9-15509e2d61c1%2Fi69dfo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Determine the kcat of the reaction given [E] = 2.5E-7 and the enzyme has a mass of kDa. (Km = 9.86E-6 micromol; Vmax
140.85 micromol/minute)
HO.HO
[S] (M)
1/[S]
vo (μmol/min)
1/vo OHO
(M)
(μmol/min)
2.50E-06
28
4.00E+05
3.57E-02
4.00E-06
40
2.50E+05
2.50E-02
1.00E-05
70
1.00E+05
1.43E-02
2.00E-05
95
5.00E+04
1.05E-02
4.00E-05
112
2.50E+04
8.93E-03
1.00E-04
128
1.00E+04
7.81E-03
2.00E-03
139
5.00E+02
7.19E-03
1.00E-02
140
1.00E+02
7.14E-03
y = 7.0 × 10-8 x + 0.0071
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