You have engineered a protease to catalytically break the backbone amine bond of a peptide $($histidine$-$alanine$-$glutamine$)$ substrate. The substrate binds to the enzyme active site only if the substrate's histidine is positively charged. Conversely, an aspartic acid residue in the enzyme's active site must be negatively charged to enable binding.

a$.$ Draw a schematic of the reaction mechanisms.

b$.$ Derive an expression for the rate of product formation as a function of rate constants and concentrations of total protease, unbound peptide $($in either protonation state$),$ and $H+.$

c$.$ Plot the initial reaction rate as a function of $pH$ for the following parameter values:

Protease$-$substrate association rate constant $={10}^6{M}^{-1}{S}^{-1}$

Protease$-$substrate dissociation rate constant $=2s^{-1}$

Catalytic turnover number $=10s^{-1}$

Total enzyme concentration $=1nM$

Unbound peptide concentration $=2M$

d$.$ What $pH$ yields the maximum reaction rate?

$($Hint: $($a$)$

The binding event occurs only between $E^{-}($carboxyl group deprotonated$)$ and $S{H}^{+}($histidine group protonated$).$ You can write the usual enzymatic reaction between these an enzyme and a substrate except that it involves these two species.

$($b$)$ Solve for $E^{-}S{H}^{+}$using QSSA. But you should remember that $E^{-}$is in equilibrium with $EH$ and $H^{+},$ and that $S{H}^{+}$is in equilibrium with $S$ and $H^{+}.$ Then you can write an equation for $\frac{d[P]}{dt}=k_{\text{c}\text{a}\text{t}}[ES{]}_{\text{Q}\text{S}\text{S}\text{A}}.$

$($c$)$ You will need pKa values for aspartic acid and histidine. Use $3\mathrm{.}90$ and $6\mathrm{.}04,$ respectively.

$($d$)$ You should differentiate $\frac{d[P]}{dt}$ with respect to $H^{+}$and set that to zero and solve for $H^{+}.)$