An enzyme-catalyzed reaction was carried out with the initial substrate concentration 1,000 times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been conveed to the product, and the amount of product was 12 mmol. If, in a separate experiment, one-third as much enzyme and twice as much of the substrate is combined, how long it would to take for the same amount (12 mmol) of product to be formed?
Since we know that Enzyme is more impoant than substrate, so enzyme concentration we have to see first if that is reduced then it doesn't matter whether we increase substrate concentration or not time will increase only. So, since enzyme is decreased to 1/3 time will increase by 3 times irrespective of increase in substrate concentration By using Michaelis-Menten equation, that is, V0 = Vmax x / Km + and by definition Vmax = Kcat x Where is enzyme concentration, Kcat is turnover number So, Similarly, As velocity is 3 times slower than first velocity so same amount of product formation will require 3x more time. Hence, Time taken will be = 3 x t = 3 x 9 = 27 minutes
13.5 mins
27 mins
8 mins
9 mins
Enzyme kinetics