Chapter 9, Problem 9.13P

The hydrogen ion, H⁺, binds with the enzyme (E⁻) to activate it in the form EH. The hydrogen ion, H⁺, also binds with EH to deactivate it by forming ${\text{EH}}_2^{+}$

$\begin{array}{l}\begin{array}{cc}{\text{H}}^{\text{+}}+{\text{E}}^{-}\stackrel{}{\rightleftarrows }\text{ EH}& K_{\text{1}}\text{=}\frac{\text{(EH)}}{{\text{(H}}^{\text{+}}{\text{)(E}}^{-}\text{)}}\\ {\text{H}}^{\text{+}}+\text{EH }\stackrel{}{\rightleftarrows }{\text{ EH}}_{\text{2}}^{\text{+}}& K_{\text{2}}\text{=}\frac{{\text{(EH}}_{\text{2}}^{\text{+}}\text{)}}{{\text{(H}}^{\text{+}}\text{)(EH)}}\end{array}\hfill \\ \text{EH}+\text{S }\stackrel{K_{\text{M}}}{\rightleftarrows }\text{ EHS}\to \text{EH}+\text{P,}{K}_{\text{M}}\text{=}\frac{\text{(EHS)}}{\text{(EH)(S)}}\hfill \end{array}$

Figure P9-13_B Enzyme pH dependence.

where E⁻ and ${\text{EH}}_{\text{2}}^{\text{+}}$ are inactive.

1. (a) Determine if the preceding sequence can explain the optimum in enzyme activity with pH shown in Figure P9-13_B.
2. (b) List ways you can work this problem incorrectly.
3. (c) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.