 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, $0.50\min \mu {\mathrm{M}}^{-1}$. The dashed TEM-1 line intercepts the vertical axis at about $13\min$, while the dotted TEM-52 line intercepts it at about $2.5\min$. ThusThe negative-axis intercepts independently give approximately $-26$ and $-5\mu \mathrm{M}$. Allow normal graph-reading uncertainty, especially at the smaller intercept. 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.