Problem $1$

In uncompetitive inhibition, the inhibitor

reversibly binds to the enzyme$-$substrate complex

after it has been formed, as shown in the reaction

scheme. Show that the rate of product generation

can be described by the following rate expression:

$v=\frac{V_m[(S)]}{K_m^{\text{'}}+[S](1+\frac{[I]}{K_{\text{I}}})},V_m=k_2[E_t]$

Problem $2$

In substrate inhibition, the substrate itself acts as

an inhibitor, as shown in the reaction scheme. Show

that in this case the rate of product generation can

be described by the following rate expression:

$v=\frac{V_m[(S)]}{K_m^{\text{'}}+[S]+\frac{[S{]}^2}{K_{S_l}}},V_m=k_2[E_t]$

$E+S{}^{{}_m^{\text{'}\text{'}}}ES{k}_{2}E+P$

$+$

$|{|}_1$

$K_1$

ESI

$E+S{K}_m^{\text{'}}ES{k}_{2}E+P$

$+$

$S^{\text{l}\text{o}\text{n}\text{g}\text{r}\text{i}\text{g}\text{h}\text{t}\text{a}\text{r}\text{r}\text{o}\text{w}}{K}_{S_1}$

$E{S}_2$