Biochemical Engineering - Enzymes and Kinetics - Discussion
Discussion Forum : Enzymes and Kinetics - Section 1 (Q.No. 22)
22.
An enzyme is assayed at an initial substrate concentration of 2 x 10-5M. In 6 minute, half of the substrate is used. The Km for the substrate is 2 x 10-3M. The value of k in minute is
Discussion:
5 comments Page 1 of 1.
Jhon said:
2 years ago
I Didn't understand. Please explain to me.
Neha yadav said:
9 years ago
Solution to this question:
Tthe enzymatic reaction is mainly first order reaction.
From the first order of reaction, k = 0.693/half time (t1/2).
Half time is 6 min as given in the question.
So, 0.693/6 = 0.1155.
Tthe enzymatic reaction is mainly first order reaction.
From the first order of reaction, k = 0.693/half time (t1/2).
Half time is 6 min as given in the question.
So, 0.693/6 = 0.1155.
(2)
Akku said:
10 years ago
Please explain this.
(1)
Saipraveen said:
10 years ago
Sir please do the solution I didn't get the solution to this question.
(1)
Samuels Austin said:
1 decade ago
Can I get an explanation of the calculation on how the answer came to be?
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