Bio307 Lecture 5 (enzyme Kinetics I).ppt

Enzymes alter only the reaction rate and not the reaction equilibrium Enzymes accelerate the forward and reverse reactions by precisely the same factor

kF forward rate constant kR reverse rate constant

Equilibrium concentration of B is 100 times more than A, whether an enzyme is present or not, but with an enzyme the equilibrium is reached faster

Enzymes accelerate reactions by facilitating the formation of a transition state

Enzymes lower the activation energy Enzymes accelerate reaching the equilibrium

One approach to understanding how enzymes achieve this facilitation is to assume that the substrate in transition state (S‡) and the substrate (S) are in equilibrium.

in which K‡ is the equilibrium constant for the formation of S‡, and v is the rate of formation of product from S‡.

The rate of the reaction is proportional to the concentration of S‡:

because only S‡ can be converted into product. The concentration of S‡ is in turn related to the free energy difference between S‡ and S, because these two chemical species are assumed to be in equilibrium; the greater the difference between these two states, the smaller the amount of S‡.

Because the reaction rate is proportional to the concentration of S‡, and the concentration of S‡ depends on ΔG‡, the rate of reaction V depends on ΔG‡. Specifically,

In this equation, k is Boltzmann's constant, and h is Planck's constant. The value of kT/h at 25°C is 6.2 × 1012 s-1. Suppose that the free energy of activation is 6.82 kcal mol-1 (28.53 kJ mol-1). The ratio [S‡]/[S] (=K’eq) is then 10-5 (see Table 8.4). If we assume for simplicity's sake that [S] = 1 M, then the reaction rate V is 6.2 × 107 s-1. If ΔG‡ were lowered by 1.36 kcal mol-1 (5.69 kJ mol-1), the ratio [S‡]/[S] is then 10-4, and the reaction rate would be 6.2 × 108 s-1. As Table 8.4 shows, a decrease of 1.36 kcal mol-1 in ΔG‡ yields a tenfold larger V. A relatively small decrease in ΔG‡ (20% in this particular reaction) results in a much greater increase in V.

Table 8.4. Relation between ΔG°′ and K′eq (at 25°C)

ΔG°′ K′eq

kcal mol-1

kJ/mol-1

10-5

6.82

28.53

10-4

5.46

22.84

10-3

4.09

17.11

10-2

2.73

11.42

10-1

1.36

5.69

1

0

0

10

-1.36

-5.69

102

-2.73

-11.42

103

-4.09

-17.11

104

-5.46

-22.84

105

-6.82

-28.53

The formation of an enzyme-substrate complex is the first step in enzymatic catalysis

Spectroscopic characteristics of many enzymes and substrates change on formation of an ES complex Tryptophan synthase – Pyridoxal phosphate as co-factor (Tryptophan from serine and an indole derivative)

The active sites of enzymes have some common features Active site: region in enzyme that binds substrates and cofactors if any Catalytic groups: residues that participate in bond-formation Interaction of enzyme and substrate at the active site promotes the formation of the transition state, The active site is the region of the enzyme where the free energy is lowered.

The active site is a 3-dimesional cleft formed by groups that come from different parts of the aa - sequence

Lysozyme

Degrades the cell wall of bacteria

• The active site takes up a relatively small part of the total volume of an enzyme

Myoglobin

• Active sites are located in clefts/crevices

Substrates are bound to enzymes by multiple weak attractions H-bonds Electrostatic interactions Van der Waals forces Hydrophobic interactions

ES-complexes: Equilibrium constants 10-2 – 10-8 Free energies of interaction -3 to -12 kcal/mol Covalent bonds: Energies -50 to -110 kcal/mol

Ribonuclease: part of active site