Answer:
99203s⁻¹
Explanation:
Turnover number is defined as the maximum number of chemical conversions of substrate molecules per second that a single catalytic site will execute for a given enzyme concentration.
It is possible to calculate this number using the maximum reaction rate (Vmax) and catalyst site concentration, thus:
[tex]k_{cat}=\frac{V_{max}}{[E_{T}}[/tex]
As umol are 1x10⁻⁶ moles and nmol are 1x10⁻⁹ moles:
[tex]k_{cat}=\frac{249x10^{-6}molL^{-1}s^{-1}}{2,51x10^{-9}molL^{-1}}[/tex]
[tex]k_{cat}=99203s^{-1}[/tex]
I hope it helps!