Question

The rate law for the enzyme catalyzed reaction $ E+SES\xrightarrow{{{k}_{b}}}P+E, $ is given by;
(A) $ rate=\dfrac{{{k}_{b}}\left[ S \right]\left[ E \right]}{{{K}_{M}}+\left[ S \right]}. $
(B) $ rate=\dfrac{{{k}_{b}}\left[ S \right]}{{{K}_{M}}+\left[ S \right]}. $
(C) $ rate=\dfrac{{{k}_{b}}\left[ E \right]}{{{K}_{M}}+\left[ S \right]}. $
(D) $ rate={{k}_{b}}\left[ S \right]\left[ E \right]. $

Answer 1

Hint: We know that in a chemical reaction, the enzyme acts as a biological catalyst. The role of the catalyst is to speed up the reaction rate but they do not involve chemical reaction. The catalyst lowers the activation energy for the reaction.

Complete step by step solution:
Enzymes are the biological catalysts which are used in different chemical reactions. The role of the catalyst is to lower the activation energy for a reaction. The lower is the activation for the reaction, the faster is the rate. The enzyme speeds up the reaction by lowering the activation energy. The enzyme contains an active site which is known as an enzyme active site. It is the location on the enzyme surface where the substrate binds and the chemical reaction takes place, catalyzed by the enzyme.
The binding of the substrate to an enzyme active site is known as enzyme substrate complex. Many enzymes change their shape when the substrate binds to it. This process is known as “induced fit”. This term means that the precise orientation of the enzyme is needed for catalytic activity which can be induced by the binding of the substance.
The equation for the formation of complex is shown as; $ E+SES\xrightarrow{{{k}_{b}}}P+E; $
Where, E is the enzyme; S is the substrate and P is the product.
Here the formation can be rewritten as: $ E+SES\xrightarrow{{{k}_{b}}}P+E; $
Thus the predefined enzyme and substrate is given as; $ [ES]=\dfrac{\left[ E \right]\left[ S \right]}{\left[ S \right]+{{K}_{M}}} $ ………(this equation is predefined formula for enzyme and substrate).
Thus now just substituting the value of $ [ES] $ in rate formula;
 $ rate={{k}_{b}}\times [ES]={{k}_{b}}\times \left\{ \dfrac{\left[ E \right]\left[ S \right]}{\left[ S \right]+{{K}_{M}}} \right\} $
Therefore, correct answer is option A i.e. $ rate=\dfrac{{{k}_{b}}\left[ S \right]\left[ E \right]}{{{K}_{M}}+\left[ S \right]}. $

Note:
Remember that the enzyme does not change the equilibrium constant of the reaction which is denoted by $ Keq $ . The equilibrium constant depends only on the difference in the energy level of the reactant and product. The enzyme forms complex with the substrate.