1. If Vmax for an E catalyzed reaction is 3.0 μmole/min in thepresence of 1.5 micrograms of enzyme (mol. Wt 30,000) What is theturnover number? Express answer in sec-1
2. An enzyme catalyzed reaction whereby substrate S isisomerized to product P, using 2.5 mg of enzyme, mol wt 125 000,was found to have a Km of 3 x 10-3 M and a maximumvelocity of 275 mmole/min. What is the turnover number?
1. If Vmax for an E catalyzed reaction is 3.0 μmole/min in thepresence of 1.5 micrograms of enzyme (mol. Wt 30,000) What is theturnover number? Express answer in sec-1
2. An enzyme catalyzed reaction whereby substrate S isisomerized to product P, using 2.5 mg of enzyme, mol wt 125 000,was found to have a Km of 3 x 10-3 M and a maximumvelocity of 275 mmole/min. What is the turnover number?
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Related questions
1) The kinetic parameter of an enzyme is:
Km = 0.15 mM for substrate A
Km = 1.2 mM for substrate B
Vmax is same for both the substrate
a) Calculate the initial velocity (Vo) for each substrate in terms of Vmax when:
i) [S] = 0.15 mM
ii) [S] = 1.5 mM
iii) [S] = 15 mM
b) Which substrate has greater affinity for the given enzyme?
2) The following data was obtained for an enzyme-catalyzed reaction. Estimate the Km and Vmax for this enzyme and find what will be the Vo at [S] = 2.5 x 10-5 M?
[S], M | Vo (nmoles/min/mg/protein) |
6.25 x 10-6 | 15 |
7.50 x 10-5 | 56.3 |
1.50 x 10-4 | 60 |
1.00 x 10-3 | 74.5 |
1.00 x 10-2 | 75 |