Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration...

Question

Question

Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L.

Part A

If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute?

Express your answer to three significant figures.

kcat

=

2.73×10⁴

min−1

Part B

Calculate kcat/KM for the enzyme reaction.

Answer 1

Answer #1

Answer 2

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