Firefly luciferase is often used as a reporter enzyme in cell and molecular biology. It catalyzes the oxidation of the substrate luciferin, with the simultaneous emission of a photon. The kinetic parameters for luciferase are $k_{\text{c}\text{a}\text{t}}=0\mathrm{.}04$mol oxyluciferin $\frac{?}{\text{m}\text{o}\text{l}}$ luciferase and $K_m=100M.$ Consider an assay wherein ${10}^5$ cells are suspended in a $100L$ microplate well, each cell is expressing ${10}^5$ molecules of luciferase, and $1mM$ luciferin is added to the well at time zero. At what time will the light$-$generation intensity drop to $90\%$ of its maximum value? At what time will it drop to $50\%$ of its maximum value?

$($Hint: The light intensity is proportional to $\frac{d[P]}{dt}.$ You have to first find the maximum value of $\frac{d[P]}{dt}=\frac{k_{\text{c}\text{a}\text{t}}[E{]}_0[S]}{K_m+[S]}.$ At which $[$S$]$ will this be maximum? Then, you can find $S$ at which $\frac{d[P]}{dt}=0\mathrm{.}9(\frac{d[P]}{dt}{)}_{\text{m}\text{a}\text{x}}.$ If you found $[$S$],$ then you can use the same approach as problem $1($solve the differential equation$)$ and find $t$ for that $[$S$].$ You can repeat this procedures for $50\%$ maximum value$)$