User:Hannes Röst/Article3

Intermediate

 * $$ \begin{align}

E + S \underset{k_{-1}}{\overset{k_{1}} {\begin{smallmatrix}\displaystyle\longrightarrow \\ \displaystyle\longleftarrow \end{smallmatrix}}} ES  \overset{k_2} {\longrightarrow} EI  \overset{k_3} {\longrightarrow} E + P \end{align}$$ So we can write down the following kinetic equations where we use the quasi-steady-state assumptions for both intermediate enzyme complexes:
 * $$ \begin{align}

\frac{d{[}P{]}}{dt} &= k_3 {[}EI{]} \\ \frac{d{[}EI{]}}{dt} &= k_2{[}ES{]} - k_3 {[}EI{]} \; \overset{!} = 0 \;   \\ \frac{d{[}ES{]}}{dt} &= k_1{[}E{]}{[}S{]} - {[}ES{]} (k_{-1} + k_2) \; \overset{!} = 0 \;  \\ \end{align}$$ Again we can write down the total enzyme concentration and assume it to be constant over the time scale in question
 * $$ \begin{align}

{[}E{]}_0 &= {[}E{]} + {[}ES{]} + {[}EI{]} \; \overset{!} = \; \text{const} \end{align}$$ Now again we express [ES] in terms of the other variables
 * $$ \begin{align}

0 &= k_1{[}S{]}({[}E{]}_0 - {[}ES{]} -{[}EI{]} ) - {[}ES{]} (k_{-1} + k_2) \\ k_1{[}S{]}({[}E{]}_0 -{[}EI{]}) &= k_1{[}S{]}{[}ES{]} + {[}ES{]} (k_{-1} + k_2) \\ {[}S{]}({[}E{]}_0 -{[}EI{]}) &= {[}S{]}{[}ES{]} + {[}ES{]} \underbrace{\frac{(k_{-1} + k_2) }{k_{-1}}}_{K_M}\\ {[}S{]}({[}E{]}_0 -{[}EI{]}) &= ( K_M + {[}S{]}) {[}ES{]}\\ {[}ES{]} &= \frac{{[}S{]}({[}E{]}_0 -{[}EI{]})}{K_M + {[}S{]}} \end{align}$$ Now we plug this into the equation for $$d{[}EI{]}/dt $$ and we can express [EI] in terms of the other variables
 * $$ \begin{align}

0 &= k_2{[}ES{]} - k_3 {[}EI{]} \\ 0 &= k_2\frac{{[}S{]}({[}E{]}_0 -{[}EI{]})}{K_M + {[}S{]}} - k_3 {[}EI{]} \\ 0 &= k_2\frac{{[}S{]}{[}E{]}_0}{K_M + {[}S{]}}- k_2\frac{{[}S{]}{[}EI{]}}{K_M + {[}S{]}} - k_3 {[}EI{]} \\ k_2\frac{{[}S{]}{[}E{]}_0}{K_M + {[}S{]}} &= (k_2\frac{{[}S{]}}{K_M + {[}S{]}} + k_3 ){[}EI{]} \\ {[}EI{]} &= k_2\frac{{[}S{]}{[}E{]}_0}{(K_M + {[}S{]})(k_2\frac{{[}S{]}}{K_M + {[}S{]}} + k_3)} \\ {[}EI{]} &= k_2\frac{{[}S{]}{[}E{]}_0}{k_2 {[}S{]} + k_3(K_M + {[}S{]})} \end{align}$$ Finally we can find again an expression for the rate of substrate conversion to product
 * $$ \begin{align}

\frac{d{[}P{]}}{dt} &= v_0 = k_3 {[}EI{]} = k_3 k_2 \frac{{[}E{]}_0 {[}S{]}}{k_3 K_M  + {[}S{]}(k_2 + k_3)}\\ \frac{d{[}P{]}}{dt} &= \underbrace{\dfrac{k_3 k_2}{k_2 + k_3}}_{k_{cat}} \cdot \frac{{[}E{]}_0 {[}S{]}}{\underbrace{\frac{k_3}{k_2 + k_3} K_M}_{K_M^{\prime}}  + {[}S{]}}\\ v_0 &= k_{cat}\frac{{[}S{]} {[}E{]}_0}{K_M^{\prime}+ {[}S{]}} \end{align}$$ In the last equation we have written a Michaelis-Menten like formula with modified constants, thus in this case we have for the measured constants the following functions
 * $$ \begin{align}

K_M^{\prime} \ &\stackrel{\mathrm{def}}{=}\ \frac{k_3}{k_2 + k_3} K_M = \frac{k_3}{k_2 + k_3} \cdot \frac{k_{2} + k_{-1}}{k_{1}}\\ k_{cat} \ &\stackrel{\mathrm{def}}{=}\ \dfrac{k_3 k_2}{k_2 + k_3} \end{align} $$ We see that for the limiting case $$k_3 \gg k_2$$, thus when the last step from EI to E + P is much faster than the previous step, we get again the original equation. Mathematically we have then $$ K_M^{\prime} \approx K_M $$ and $$k_{cat} \approx k_2$$.