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Biology LibreTexts

More Complicated Kinetics

A More Complicated Reaction

Not all reactions can be characterized so simply as a simple substrate interacting with an enzyme to form an \(ES\) complex, which then turns over to form product. Sometimes, intermediates form. For example, a substrate \(S\) might interact with \(E\) to form a complex, which then is cleaved to products \(P\) and \(Q\). \(Q\) is released from the enzyme, but \(P\) might stay covalently attached. This happens often in the hydrolytic cleavage of a peptide bond by a protease, when an activated nucleophile like Ser reacts with the sessile peptide bond in a nucleophilic substitution reaction, releasing the amine end of the former peptide bond as the leaving group, while the carboxy end of the peptide bond remains bonded to the Ser as an Ser-acyl intermediate. Water then enters and cleaves the acyl intermediate, freeing the carboxyl end of the original peptide bond. This is shown in the written reaction below:

18 covalentintmech.gif

To simplify the derivation of the kinetic equation, let's assume that \(E\), \(S\), and \(ES\) are in rapid equilibrium defined by the dissociation constant, \(K_s\). Assume \(Q\) has a visible absorbance that is easy to monitor. Assume from the steady state assumption that:

\[ \dfrac{d[E-P]}{dt} = k_2[ES] - k_3[E-P] = 0 \]

assuming \(k_3\) is a pseudo-first order rate constant and [H2O] does not change. The velocity depends on which step is rate limiting. If \(k_3 \ll k_2\), then the \(k_3\) step is rate-limiting and

\[v = k_3[E-P]\]

If \(k_2 \ll k_3\), then the \(k_2\) step is rate-limiting. Then

\[v = k_2[ES]\]

For a homework assignment, you derived the following kinetic equation for this reaction, assuming \(v = k_2[ES]\).

\[ v= \dfrac{\dfrac{k_2k_3}{k_2+k_3}E_0S}{\left[ \dfrac{k_3}{k_2+k_3}\right]K_s + S}=\dfrac{k_{cat}E_oS}{K_M+S} \tag{12}\]

You can verify that you get the same equation if you assume that \(v = k_3[E-P]\) (\( k_{cat}\) will be defined below.) This equation looks quite complicated, especially if you substitute for \(K_s\), \(k_{-1}/k_1\). All the kinetic constants can be expressed as functions of the individual rate constants. However, this equation can be simplified by realizing the following:


\[S \gg [K_s(k_3)/(k_2+k_3)]\]

\[v = [(k_2k_3)/(k_2 + k_3)]E_o = V_m\]

\[ \dfrac{K_s(k_3)}{(k_2+k_3)} = constant = K_m \]

Substituting these into equation 7 gives

\[  v = \dfrac{V_m(S)}{K_m+ S} \tag{13}\]

This again is the general form of the Michaelis-Menten equation. Note in the first bulleted item above that for this reaction,

\[ V_m = \dfrac{k_2k_3}{k_2 + k_3}E_o.\]

This is more complicated than our earlier definition of \(V_m = k_3E_o\). They are similar in that the term \(E_o\) is multiplied by a constant which is itself a function of rate constant(s). The rate constants are generally lumped together into a generic constant called \(k_{cat}\). For the simple reaction \(k_{cat} = k_3\), but for the more complicated reaction with a covalent intermdiate which we just derived, \(k_{cat} = (k_2k_3)/(k_2 + k_3)\). For all reactions,

\[ V_m = k_{cat}E_o. \tag{14}\]

20 MMequations.gif

Figure: Summary of \(v_o\) vs. \(S\) equations - Meanings of \(K_m\), \(k_{cat}\) and \(V_m\).