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Problem 36.16 11 of 14 > Part A The enzyme fumarase catalyzes the hydrolysis of fumarate: Fumarate(aq) + H2O(1) L - malate(aq). The turnover number for this enzyme is 2.5 x 10 s-4, and the Michaelis constant is 4.2 x 10" M. What is the rate of fumarate conversion if the initial enzyme concentration is 1.0x10-6 M and the fumarate concentration is 1.0x10-4 M ? Express your answer with the appropriate units. Ro = Value Units Submit Request Answer Provide Feedback Next >

Answer 1

consider the catalyzed hydrolysis of fumarate which is shown as below:

Fumarate + H₂O ------> L-malate

the rate expression for the emzyme catalyzed reaction is

R = K₂ [S]_o[E]_o / ([S]_o + K_m)

here, K₂  is the turnover number having value (2.5 X 10³ s^-1) K_m is the Michaelis constant having value (4.2 X10^-6 M) [E]_o is the initial enzyme concentration having value (1 X 10^-6 M) and [S]_o is the initial concentation having value (1 X 10^-4 M)

by substitute all the required values

R = (2.5 X10³ s^-1 X 1 X 10^-6 M X 1 X 10^-4 M) / (1 X 10^-4 M + 4.2 X10^-6 M)

R = (2.5 X10³ s^-1 X 1 X 10^-6 M X 1 X 10^-4 M) / ( 2.402 X 10^-4 M)

R = 2.488 X 10^-3 M s^-1

hence the rate of fumarate conversion is 2.45X 10^-3 M s^-1

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