Answer:
The value of turnover number is [tex]1.06\times10^5 s^{-1}[/tex]
Explanation:
As the given data
[tex]R_{max}=249 \mu\, mol/Ls=249 \times 10^{-6} mol/Ls[/tex]
[tex][E]_t=235 n \, mol\L=235 \times 10^{-9} mol/L[/tex]
As per the given formula of turnover number for a single active site enzyme is given as
[tex]n_{turnover}=\frac{R_{max}}{[E]_t}[/tex]
By substituting values
[tex]n_{turnover}=\frac{R_{max}}{[E]_t}\\n_{turnover}=\frac{249 \times 10^{-6} mol/Ls}{2.35 \times 10^{-9} mol/L}\\n_{turnover}=1.06\times10^5 s^{-1}[/tex]
So the value of turnover number is [tex]1.06\times10^5 s^{-1}[/tex]