Team:Imperial College London/Drylab/Enzyme/Analysis

Assumptions

Enzymatic assumptions:

• The enzyme is specific only for the substrate and not for any other chemicals.
• Only one enzyme, our enzyme of interest is present and participating in the reaction.
• There is negligible formation of product without the enzyme.
• The rate of enzymatic activity remains constant over time because:
  • There is no co-operativity of the system. Binding of substrate to one enzyme binding site doesn't influence the affinity or activity of an adjacent site.
  • No allosteric regulations from either the product or the substrate.
  • There is no product inhibition of the enzyme.

• Total [E] does not change with regards to time. Enzyme will not be used up nor degraded over time, thus [E] at the beginning of the reaction will be the same as [E] at the end of reaction.
• The reaction catalysed is irreversible
• [S] >> [E], such that the free concentration of substrate is very close to the concentration I added. This also ensures a constant substrate concentration throughout the assay. This allows easy determination of [E].
• For which

<math> \begin{align} E + S \underset{k_{2}}{\overset{k_{1}} {\begin{smallmatrix}\displaystyle\longrightarrow \\ \displaystyle\longleftarrow \end{smallmatrix}}} ES \overset{k_3} {\longrightarrow} E + P \end{align}</math>

, k1>>k2

Law of mass action assumptions:

• Free diffusion
• Free unrestricted molecular motion

Michaelis-Menten assumption:

• There is a quasi-steady state of [ES] , ie: , during which the enzyme is used for the reaction. This means that the rate of formation of ES complex is equal to the rate of dissociation of ES complex.
• <math> \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} </math>

• from the above assumptions, we know that