Explanation:
The given data is as follows.
[tex]V_{max} = 6.8 \times 10^{-10} \mu mol/min[/tex]
[tex]K_{m} = 5.2 \times 10^{-6} M[/tex]
Now, according to Michaelis-Menten kinetics,
[tex]V_{o} = V_{max} \times [\frac{S}{(S + Km)}][/tex]
where, S = substrate concentration = [tex]10.4 \times 10^{-6}[/tex] M
Now, putting the given values into the above formula as follows.
[tex]V_{o} = V_{max} \times [\frac{S}{(S + Km)}][/tex]
[tex]V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}][/tex]
[tex]V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667[/tex]
= [tex]4.5 \times 10^{-10} \mu mol/min[/tex]
This means that [tex]V_{o}[/tex] would approache [tex]V_{max}[/tex].