An enzyme-catalyzed reaction was carried out with the substrate concentration in
ID: 817924 • Letter: A
Question
An enzyme-catalyzed reaction was carried out with the substrate concentration initially 1,000 times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to the product and the amount of product formed was 12 umol. If in a separate experiment , 1/3 as much enzyme and twice as much substrate had been combined , how long would it take for the same amount of product to be formed?
I know that the answer is 27 minutes, but can someone please show me how to solve the problem?
Explanation / Answer
(when [S] >>>Km, nearly all enzyme will be substrate bound, and [ES] approaches [E]total.Then Vmax = k2[E]total
Since only 1% of substrate has been converted, the concentration of substrate is still pretty high.
Now k2 is also called kcat, termed as the turn over number of the enzyme. This is a fundamental/intrinsic property of the enzyme and should not change with enzyme or substrate concentration. Now from the equation above, you can see that the velocity or rate of the reaction would only be dependent on the total enzyme concentration in this scenario. Since there is only 1/3rd amount of the enzyme, it would take three times the amount of time it used to previously to turn over the same amount of substrate and produce the same amount of product. Hence time = 9 x 3 = 27 min.
Hope this makes sense. Good luck!
Sources:
Has had straight A's in Enzyme Kinetics courses during Masters' and Doctoral level!