Doc Brown's Advanced A Level Chemistry

Advanced A Level Chemistry

Kinetics-Rates Part 7

Selected Case Studies of a variety of chemical reactions and their rate expressions

7.3 Enzyme kinetics - biological catalyzed reactions

(enzymes being biological catalysts)


Advanced A Level Kinetics Index


Case study 4.3 Enzyme kinetics – Biological Catalysts

Explanation and derivation of orders of reactants and how to write the rate expression

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  • KEY in sequence: E = free enzyme, S = free substrate reactant molecule, ES = enzyme–reactant complex, EP = enzyme–product complex, E = free enzyme, P = free product

  • BIOLOGICAL CATALYSIS – HOW DOES AN ENZYME WORK?

  • WHY IS THE REACTION PROFILE QUITE 'COMPLICATED' COMPARED TO MANY OTHER REACTION?

  • The mechanism of enzyme catalysed:

    • Step (1) E + S ==> ES

      • The 'docking in' of the substrate S of the onto the 'active site' of E to form the enzyme–substrate complex ES. This is often quoted as the rate determining step (rds), but it isn't (see step (2) below!).

    • Step (2) ES ==> EP

      • The conversion of the substrate–enzyme complex ES into the enzyme–product EP, is the rate determining step. The docking in of the substrate to be held e.g. by inter molecular forces is likely to have a low activation energy and a relatively high probability initially,  so E + S ==> ES cannot be the rds. However, ES ==> EP involves bond breaking and will have a much higher activation energy giving a much lower probability of ES  ==> EP transformation.

      • However, [ES] cannot be measured directly, so a simplified kinetics approach is to treat the formation and transformation of ES/EP through to P as a function of the concentrations of E and S (see the rate expressions below, which is a big bad fudge seen here and in most textbooks!). The situation should be dealt with via the Michaelis–Menten equation, but this goes way above AS–A2 chemistry demands.

    • Step (3) EP ==> E + P

      • The breakdown of the EP complex leading to the departure of the product P and leave the enzyme E, as a free 'active site' completing the catalytic cycle.

topA possible detailed reaction profile is shown below.

The uncatalysed profile would be a single, and much higher 'hump' (see 2nd simplified diagram below).

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  • Here Ea1 is small for the formation of the enzyme–substrate complex or intermediate ES. The energy of ES is lower than the reactants because it involves an energy releasing interaction whether it be via intermolecular forces or chemical bonding.

  • Ea2 is much larger for the transformation of the ES intermediate into the EP intermediate because it involves bond breaking in the substrate molecule.

    • Ea2 is still much smaller than Ea for the uncatalysed reaction, which is not shown here, but is shown in the simplified diagram below e.g. the uncatalysed decomposition of hydrogen peroxide solution has an Ea of 76kJmol–1, but with an enzyme like catalase/peroxidase, the Ea is reduced to 30kJmol–1 AND the rate of reaction is increased by a factor of 108, impressive biochemistry!!!

    • Ea3 is relatively small for the breakdown of the EP intermediate into the free product and free enzyme.

  • However, the simplified reaction profile diagram below should be good enough in an exam and shows the relevant activation energies.

    • Here, Ea1 is for E+S ==> ES, Ea2 is for ES ==> EP ==> E + P

      • (Ea for EP ==> E + P isn't shown separately)

    • and Ea3 for the uncatalysed reaction, which with no enzyme is significantly greater.

      • My thanks to Professor Martin Chaplin of London South Bank University for his most helpful emails concerning reaction profiles.

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  • Treating step (2) as the rds, gives the rate expression ...

    • rate = kES[ES],

    • that is the rate of reaction = rate of transformation of ES into EP .

    • However, since you cannot measure the concentration of the enzyme–substrate complex, the rate is treated as a function of the concentrations of the enzyme and substrate, [E] and [S], which form the ES intermediate, assuming a steady state situation, giving

      • rate = k1[E][S]

      • (Which is a considerably simplified approach if you care to research further in a thick undergraduate textbook of biochemistry and researching the Michaelis–Menten equation!)

    • So we are treating the reaction as 1st order with respect to both enzyme and substrate, 2nd order overall.

    • For a set of experiments, using a fixed amount of enzyme, the initial rate should be proportional to the concentration of substrate at relatively low concentrations, so the effective rate expression is pseudo 1st order ... 

      • rate = k2[S]

    • If you keep the substrate concentration constant and vary the quantity of enzyme, the effective rate expression is also pseudo 1st order ...

      • rate = k3[E]

  • topHowever, if the substrate concentration is very high, the maximum number of active centres on the enzymes are occupied and the rate is completely controlled by the transformation of the ES complex at its maximum possible concentration and independently of the high substrate concentration. This means you are now dealing with the maximum possible rate, which only depends on the amount of enzyme present,

    • so the rate expression becomes zero order with respect to the substrate,

      •  rate = k3[E]

    • and for a constant [E], and varying [S] at high concentrations of S the rate = k4,

    • The two situations are summed up by graph (1) below for a constant amount of enzyme. The graph moves from 1st order kinetics to zero order.

    • Graph (2) is a 'simple' comparison of the effects of no inhibitor, a constant amount of a competitive or a non–competitive inhibitor for a constant amount of enzyme.

      • Vo is the initial reaction rate for a constant amount of enzyme and varying the substrate concentration.

      • Vmax is the maximum possible rate when the maximum number of active centres are occupied with S or P.

      • Competitive inhibition: Although the inhibitor initially decreases the reaction rate, the substrate can increasingly compete with the inhibitor to occupy the 'active sites' as the substrate concentration increases (Le Chatelier's equilibrium concentration principle) so that the maximum reaction rate Vmax can eventually equal Vmax for the non inhibited enzyme.

      • Non–competitive inhibition: Since the inhibitor molecule does not bind to the 'active site' of the enzyme it cannot be 'replaced' by increasing the substrate concentration. However, because it constantly interferes with a fraction of the enzymes, the reaction rate will be reduced giving the lower the Vmax which can never reach Vmax however high the substrate concentration.

      • See also enzyme structure and function notes for more details on inhibition

(1) doc b, (2) doc b

  • The remarkable efficiency of enzymes - biological catalysts are superb!

    • For example the decomposition of hydrogen peroxide - an unwanted harmful chemical produced in the body!

    • 2H2O2(aq)  ==> 2H2O(l)  +  O2(g)

    • Three activation energies (Ea) are quoted

      • (i) uncatalysed: 75 kJmol-1

      • (ii) catalysed with colloidal platinum: 49 kJmol-1

      • (iii) catalysed by the enzyme catalase in you body: 23 kJmol-1

    • The numbers speak for themselves!

    • Both catalysts are effective in considerably reducing the activation energy and hence greatly increasing the speed of the decomposition reaction.

    • BUT, the enzyme catalase is far superior to a class transition metal catalyst!

  • On the (c) doc b GCSE page describing enzyme uses and rates graphs for concentration, pH and temperature changes.

  • On the Isomerism and Stereochemistry Part II page there is a discussion of enzyme structure and function.

Advanced A Level Kinetics Index


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