= Solution
For the <Hanes-Woolf plot>, the slope is $1/V_{\max}$, the vertical intercept is $K_m/V_{\max}$, and the horizontal intercept is $-K_m$. The lines have approximately equal slopes, $0.50\,\mathrm{min\,\mu M^{-1}}$. The dashed <TEM-1> line intercepts the vertical axis at about $13\,\mathrm{min}$, while the dotted <TEM-52> line intercepts it at about $2.5\,\mathrm{min}$. Thus
$$
\boxed{V_{\max,1}\simeq V_{\max,52}\simeq2.0\,\mathrm{\mu M\,min^{-1}},\quad K_{m,1}\simeq26\,\mathrm{\mu M},\quad K_{m,52}\simeq5\,\mathrm{\mu M}.}
$$
The negative-axis intercepts independently give approximately $-26$ and $-5\,\mathrm{\mu M}$. Allow normal graph-reading uncertainty, especially at the smaller intercept. \b[The limiting rates are similar; the <Michaelis constant> is substantially lower for <TEM-52>.] A <Michaelis constant> is not generally a binding <dissociation constant>, and calculating a <turnover number> additionally requires the concentration of active <enzyme>.
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