1.) the turnover number for an enzyme is known to be 5000 min-1. From the follow
ID: 55187 • Letter: 1
Question
1.) the turnover number for an enzyme is known to be 5000 min-1. From the following set of data, compute the Km and the total amount of enzyme present int these experiments.
1 167
2 250
4 334
6 376
100 498
1000 499
a.) Vmax for the enzyme is _____________. briefly explain how you determined Vmax
b.) Km for the enzyme is _______________. brielfy explan how you determined Km.
Explanation / Answer
Calculate 1/S and 1/V values (convert initial velocity in micro mol to milli mol before starting calculation. 167 micromol = 0.167 milli mol
1/S
1/V
1
6
0.5
4
0.25
3
0.167
2.65
0.01
2.0
0.001
2.0
* Plot 1/S value on X-axis and 1/V values on Y-axis
* Draw a line connecting the points. It gives a straight line
* The Y intercept is the 1/Vmax value
* The y-intercept for this values, when plotted on a graph is 1.6
The Vmax for this enzyme is 0.625 mM/sec
Km of the enzyme is 2.5 mM
The total enzyme present (ET) is 0.125 micromole (0.000125 milli mole)
1/S
1/V
1
6
0.5
4
0.25
3
0.167
2.65
0.01
2.0
0.001
2.0
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