topic 13, chemical kinetics and qssa, rev

1

Chemical Kinetics, Mechanisms

• We will derive the rate of an overall reaction from a system of elementary reaction rates, a mechanism.

• The stoichiometry of an elementary reaction defines the concentration dependence of the rate expression.

• The quasi-steady state assumption and equilibrium assumption will minimize the number of rate constants in a rate expression.

2

3

4

5

6

7

Idea is to find some linear combination of the elementary reactions to produce the stoichiometry of the overall observed reaction. If this cannot be done, the proposed reaction mechanism is not valid.

8

9

10

11

•Proceeds as written with no identifiable intermediates. This implies a collision.

•Rate depends on the concentrations of the reactants…the probability of collision.

• Inherently reversible, Tolman’s principle. Reverse reaction must be possible based on energy, symmetry and steric considerations.

•Transition state theory – if during a vibration the transition state falls apart, it may become either reactant or product. The step to transition state is so fast that we assume equilibration. ( e.g. molecular motion happens on a scale of 1013/sec.)

Elementary reactions - characteristics

12

• Do we have to solve a system of ODE’s for a given mechanism to know something about the reaction kinetics, or is there a way to “estimate” kinetic information which can be handled offline?

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

Michaelis Menten Kinetics, Example

Verify that the proposed mechanism can give a kinetic expression in which, when CS is very

large, rate is linear with CE0

E= enzymeS=substrateES= bound substrateP=product

• QSSA: ES is a steady state intermediate

• Mass balance

38

3P

P ESdCR k Cdt

= =

1 2 30ESE S ES ES

dC k C C k C k Cdt

= = − −

0E E ESC C C= +

39

From QSSA equation:k1(CE)(CS) = (k2 + k3)(CES)

Substitute mass balance:k1(CEo-CES)(CS) = (k2 + k3)(CES)

Distribute:k1(CEo)(CS) - k1(CES)(CS) = (k2 + k3)(CES)

Collect ES terms:k1(CEo)(CS) = (k2 + k3)(CES) + k1(CES)(CS)k1(CEo)(CS) = CES (k2 + k3 + k1S)

40

01

2 3 1

E SES

S

k C CC

k k k C=

+ +

0

2 3

1

E SES

S

C CC k k C

k

=+

+

Now we can return to the original production equation for P, and express in terms that do not involve concentrations of intermediates.

03 2 33

1

; where E SPES M

M S

k C C k kdC k C Kdt K C k

+= = =

+

Verification of Michaelis-Menten Kinetics

• When S is very large, rate goes to a maximal velocity given by…

• If k2>>k3; then KM has a value = dissociation constant.

41

03~P ER k C

More complicated biochemical 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.

42

H. Jacubowski, St. Johns Univ. MN

• 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.

43

• 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.

44

• Assume rapid equilibrium in first step, characterized by equilibrium constant KS.

• Assume E-P is an intermediate at steady state.

• The second reaction is the rate limiting step.

• Verify:

45