Chapter 12 - Enzyme Kinetics, Inhibition, and Control - Exercises - Page 398: 9

We know that $$v_{0}=V_{\max }[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ Therefore $$v_{0} / V_{\max }=[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ $$0.95=[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ $$[\mathrm{S}]=0.95 K_{M}+0.95[\mathrm{S}]$$ $$0.05[\mathrm{S}]=0.95 K_{M}$$ $$[\mathrm{S}]=(0.95 / 0.05) K_{M}=19 K_{M}$$

We know that $$v_{0}=V_{\max }[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ Therefore $$v_{0} / V_{\max }=[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ $$0.95=[\mathrm{S}] /\left(K_{M}+[\mathrm{S}]\right)$$ $$[\mathrm{S}]=0.95 K_{M}+0.95[\mathrm{S}]$$ $$0.05[\mathrm{S}]=0.95 K_{M}$$ $$[\mathrm{S}]=(0.95 / 0.05) K_{M}=19 K_{M}$$