Kcat= 30.0sec^-1 Km= 0.0050M Total enzyme Concentration= 1uM What is the max rate of the enzyme?

Question

Question

Kcat= 30.0sec^-1

Km= 0.0050M

Total enzyme Concentration= 1uM

What is the max rate of the enzyme?

Answer 1

Answer #1

Kcat = Vmax/Et

Kcat = 30

Vmax = ?

Et = 1 × 10^-6

30 = Vmax/10^-6

Vmax = 30 × 10^6 = 3 × 10^7 M per second

Vmax = maximum velocity of an enzyme catalysed reaction

Km = substrate concentration at which the velocity of the reaction is half of Vmax

Kcat is turnover number, number of substrate molecules converted to products by one active site of enzyme at a given time

Please rate high.

Answer 2

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