Question

Determine the value of the turnover number of the enzyme invertase, given that Rmax (Vmax) for invertase = 15.8 mmol·L–1·min–1 and [E]t = 4.2 μmol·L–1. Invertase has a single active site.

Answer 1

Given:

Vmax = 15.8 mmol·L–1·min–1 = 15800 μmol·L–1·min–1

[E]t = 4.2 μmol·L–1.

Turnover number kcat is defined as the maximum number of chemical conversions per second that a single catalytic site will perform for a given enzyme concentration. It can be calculated from the maximum reaction rate Vmax and catalyst site concentration [E]T.

We have already changed units to be same in Given data above.

kcat = Vmax/[E]T

         = 15800 μmol·L–1·min–1 / 4.2 μmol·L–1

         = 3761/min

To convert per second, we have to divide by 60.

         = 62.68/sec

Turnover number for enzyme is 62.68/sec