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 umol. 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 micro-moles) of product to be formed?

Answer 1

27 min

Step-by-step explanation:

By the Michaelis-Menten Kinetics Model, the initial velocity of the enzymatic reaction is given by:

`v0 = (vmax [S])/(Km +[S])`

where vmax is the maximum velocity, [S] is the substrate concentration and Km is the equilibrium constant. The maximum velocity is directly proportional to the enzyme concentration.

So, for 12μmol of the product formed in 9 min, the velocity is also:

v0 = 12/9 = 1.33 μmol/min

For [S] 1,000 times higher then Km, Km can be unconsidered in the equation, so:

`v0 = (vmax[S])/([S])`

v0 = vmax

vmax = 1.33 μmol/min

If the enzyme concentration decreases by three, the maximum velocity will also decrease by three, so it will be: 0.443μmol/min. Km continues the same, and [S] will be multiplied by 2, so Km can still be unconsidered. So:

v0 = vmax

v0 = 0.443 μmol/min

Then,

0.443 = 12/t

t = 12/0.443

t = 27 min