Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
[tex]V_0=\frac{V_{max\ [S]}}{k_M+[S]}[/tex]
Here, [tex]K_m[/tex] is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
[tex]V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}[/tex]
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.