An enzyme has velocities of 0.11 nmol/min at a substrate concentration of 1 μM, 0.34 nmol/min at a substrate concentration of 5 μM, and 0.45 nmol/min at a substrate concentration of 10 μM. What is the Vmax for the enzyme if the KM is 5 × 10–6 M?
0.22 nmol/min
0.68 nmol/min
0.90 nmol/min
1.12 nmol/min
None of the answers is correct.

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  1. By definition, 2*.34 nmol/min

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  2. So the answer would be none of the above? I had gotten C?

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  3. Sorry I just realized what you meant. It's 2 times .34 which equals .68

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