Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 438 - Introductory Biochemistry - Spring 2013

Lecture Notes:


Introduction To Enzymes

Enzymes are the heart of Biochemistry

Enzymes generally have a cleft for active site, generally <5%of surface: look like pac man. Need large structure to maintain shape etc. with many weak bonds.

Six Classes of enzymes

  1. Oxidoreductases - redox reactions [Rxn 1, p136]
  2. Transferases - group transfer reactions [Rxn 2, p137]
  3. Hydrolases - hydrolysis reactions [Rxn 3, p137]
    that these first three comprize the majority of biologically catalyzed reactions [pie graph, p137]
  4. Lyases - lysis generating a double bonds (non-oxidative elimination reactions) [Rxn 4, p137]
  5. Isomerases - isomerization reactions in a single molecule [Rxn 5, p138]
  6. Ligases - join (ligate) two substrate molecules [Rxn 6, p138]

Look at major aspects of enzyme study:


Models for Enzyme Specificity:

Enzyme Kinetics


Gives information on dynamic systems.

Sets the parameters for catalytic mechanisms such as:

A right arrow C + X;

B + X right arrow D etc.

Review some Kinetics from General Chemistry:

We have now reviewed kinetics as tools. Before we go to enzymes a few comments:


Plots of vi = d[P]/dt vs. [S] for 0 - 3rd order

reaction order plot

Look at simple, one-substrate enzymes:

For simple enzyme, S right arrow P get rectangular hyperbola type plot for vi vs [S], similar to Mb binding curve. [Figure 5.4a]

M-M plot

Let's look at a mathematical model and attempt to generate curve. This was first done by Michaelis and Menten for an equilibrium model. Better is the steady state model of Haldane and Briggs (more general), which we will derive.

For S right arrow P assume

chemical equation for  E +S in equilibrium with [ES] (showing rate constants k1 and k2) in equilibrium with E + P  (with rate constants k3 and k4)

And for initial reaction conditions [P] = 0 & therefore k4 = 0, so have

chemical equation for  E +S in equilibrium with [ES] (showing rate constants k1 and k2) decaying to E + P  (with rate constants k3)

Now vi = d[P]/dt = k3[ES] (Note that kcat is often used instead of k3);

Assume steady state (steady state assumption: d[ES]/dt= 0):

d[ES]/dt= 0; Thus: 0 = d[ES]/dt= k1[E][S] - k2[ES] - k3[ES].



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© R. A. Paselk 2010;

Last modified 20 February 2013