Question

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 mmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 mmol) of product to be formed?

Answer 1

Given, substrate concentration initially is 1000 times greater than the Km for that substrate. Now, when [S] >>> Km, initial velocity (Vo) will be equal to maximum velocity (Vm) reaction will be of zeroth order.

So, when [S] >>> Km, all of the enzyme will be bound in the enzyme-substrate complex & no free enzyme left. As a result, Vm = k2 [E]Total where, k2 = turnover number

From the above equation, we see that rate will be dependent on the concentration of enzyme (As turnover number is constant) & substrate concentration will have no effect on it. Now, initially 9 minutes require to produce 12 mmol of the product. So, if we reduce the enzyme concentration to 1/3, we will require 3x time to produce the same amount of product.

So, required time = (9 x 3) minutes = 27 minutes