(Solved): The Michaelis-Menten rate equation for an enzyme subject to competitive inhibition is V0=Km ...

The Michaelis-Menten rate equation for an enzyme subject to competitive inhibition is $V_0=\frac{V_{\max }[\mathrm{S}]}{?K_{\mathrm{m}}+[\mathrm{S}]}$ Beginning with the same Michaelis-Menten starting point, $V_0=k_2[\mathrm{ES}]$ a new definition of total enzyme as $\left[{\mathrm{E}}_{\mathrm{t}}\right]=[\mathrm{E}]+[\mathrm{ES}]+[\mathrm{EI}]$ and the definitions of $?$ and $K_{\mathrm{I}}$ provided, derive the rate equation for competitive inhibition. Use the derivation of the Michaelis-Menten equation as a guide, developing a term for [ES]. In the previous steps, you derived the equations $\begin{array}{l}\left[{\mathrm{E}}_{\mathrm{t}}\right]=[\mathrm{E}]+[\mathrm{ES}]+\frac{[\mathrm{E}][\mathrm{I}]}{K_{\mathrm{I}}}\\ \left[{\mathrm{E}}_{\mathrm{t}}\right]=[\mathrm{ES}]+[\mathrm{E}]\left(1+\frac{[\mathrm{I}]}{K_{\mathrm{I}}}\right)\end{array}$ Now define the $1+\frac{[\mathrm{I}]}{K_{\mathrm{I}}}$ term in equation 4 as $?$, and write the new equation. $\left[{\mathrm{E}}_{\mathrm{t}}\right]=$ The substituted version of equation 4 will become equation 5 in the derivation.