prof. r. shanthini 09 nov 2012 enzyme kinetics and associated reactor design: introduction to...

Prof. R. Shanthini 09 Nov 2012

Enzyme kinetics and associated reactor design:

Introduction to enzymes, enzyme catalyzed reactions and

simple enzyme kinetics

- learn about enzymes

- learn about enzyme catalyzed reactions

- study the kinetics of simple enzyme catalyzed reactions

CP504 – ppt_Set 02


What is an Enzyme?

Enzymes are mostly proteins, and hence they consists of amino acids.

Enzymes are present in all living cells, where they help converting nutrients into energy and fresh cell material.

Enzymes breakdown of food materials into simpler compounds.

Examples: - pepsin, trypsin and peptidases break down proteins

into amino acids - lipases split fats into glycerol and fatty acids- amylases break down starch into simple sugars


Enzymes are very efficient (biological) catalysts.

Enzyme catalytic function is very specific and effective.

Enzymes bind temporarily to one or more of the reactants of the reaction they catalyze.

Enzymes does not get consumed in the reaction that it catalyses.

What is an Enzyme?


How does an Enzyme help?

Enzymes speed up reactions enormously. To understand how they do this, examine the concepts of activation energy & the transition state.

In order to react, the molecules involved are distorted, strained or forced to have an unlikely electronic arrangement. That is the molecules must pass through ahigh energy state.

This high energy state is called the transition statetransition state.

The energy required to achieve it is called the activation activation energyenergy for the reaction.



The higher the free energy change for the transition barrier,the slower the reaction rate.



Enzymes lower energy barrier by forcing the reacting molecules through a different transition state. This transition state involves interactions with the enzyme.

EnzymeEnzyme


Oxidoreductase: transfer oxygen atoms or electron

Transferase: transfer a group (amine, phosphate, aldehyde,

oxo, sulphur, etc)

Hydrolase: hydrolysis

Lyase: transfer non-hydrolytic group from substrate

Isomerase: isomerazion reactions

Ligase: bonds synthesis, using energy from ATPs

Enzyme classification


Examples of Enzyme Catalysed Reactions

CO2+ H2O H2CO3 Carbonic anhydrase

Carbonic anhydrase is found in red blood cells.

It catalyzes the above reaction enabling red blood cells to transport carbon dioxide from the tissues (high CO2) to the lungs (low CO2).

One molecule of carbonic anhydrase can process millions of molecules of CO2 per second.

Example 1:

Examples of enzyme catalyzed reactions


2H2O2 2H2O + O2 Catalase

Catalase is found abundantly in the liver and in the red blood cells.

One molecule of catalase can breakdown millions of molecules of hydrogen peroxide per second.

Hydrogen peroxide is a by-product of many normal metabolic processes.

It is a powerful oxidizing agent and is potentially damaging to cells which must be quickly converted into less dangerous substances.

Example 2:

Examples of enzyme catalyzed reactions


- in the food industry for removing hydrogen peroxide from milk prior to cheese production

- in food-wrappers to prevent food from oxidizing

- in the textile industry to remove hydrogen peroxide from fabrics to make sure the material is peroxide-free

- to decompose the hydrogen peroxide which is used (in some cases) to disinfect the contact lens

Industrial use of catalase


See the hand out on the same topic

Examples of Industrial Enzymes


Enzymes are very specific.

Absolute specificity - the enzyme will catalyze only one reaction

Group specificity - the enzyme will act only on molecules that have specific functional groups, such as amino, phosphate or methyl groups

Linkage specificity - the enzyme will act on a particular type of chemical bond regardless of the rest of the molecular structure

Stereochemical specificity - the enzyme will act on a particular steric or optical isomer

More on enzymes


Source: http://waynesword.palomar.edu/molecu1.htm



E + S ES



Lock & Key Theory Of Enzyme Specificity(postulated in 1894 by Emil Fischer)

E + S ES E + P


Active Site Of Enzyme Blocked By Poison Molecule



Induced Fit Model(postulated in 1958 by Daniel Koshland )

Source: http://www.mun.ca/biology/scarr/Induced-Fit_Model.html

Binding of the first substrate induces a conformational shift that helps binding of the second substrate with far lower energy than otherwise required. When catalysis is complete, the product is released, and the enzyme returns to its uninduced state.

E + S ES E + P


E + S ES E + Pk1

k2

k3

which is equivalent to

S

P[E]

S for substrate (reactant)

E for enzyme

ES for enzyme-substrate complex

P for product

Simple Enzyme Kinetics


Michaelis-Menten approach to the rate equation:

Assumptions:

1. Product releasing step is slower and it determines the reaction rate

2. ES forming reaction is at equilibrium

3. Conservation of mass (CE0 = CE + CES)

E + S ES E + Pk1

k2

k3

Initial concentration of E

Concentration of E at time t

Concentration of ES at time t



E + S ES E + Pk1

k2

k3

rP = - rS = k3 CES

Product formation (= substrate utilization) rate:

k1 CE CS = k2 CES

Since ES forming reaction is at equilibrium, we get

(1)

(2)



E + S ES E + Pk1

k2

k3

k1 (CE0 – CES) CS = k2 CES

Using CE0 = CE + CES in (2) to eliminate CE, we get

which is rearranged to give

CE0CSCES = k2/k1 + CS

(3)



E + S ES E + Pk1

k2

k3

k3CE0CSrP =

rmaxCS = KM + CS

(4)

Using (3) in (1), we get

k2/k1 + CS

- rS =

where rmax = k3CE0

and KM = k2 / k1 (6)

(5)


E + S ES E + Pk1

k2

k3

Substrate binding step

Other terminology used

Catalytic step

k3 = kcat

rmax = k3CE0 = kcatCE0

KM = k2 / k1

(6)

(5a)


Assumptions:

1. Steady-state of the intermediate complex ES

2. Conservation of mass (CE0 = CE + CES)

E + S ES E + Pk1

k2

k3

Initial concentration of E

Concentration of E at time t

Concentration of ES at time t

Briggs-Haldane approach to the rate equation:


E + S ES E + Pk1

k2

k3


rP = k3 CES

Product formation rate:

(7)

rs = - k1 CECS + k2 CES

Substrate utilization rate:

(8)

k1 CECS = k2 CES + k3 CES

Since steady-state of the intermediate complex ES is assumed, we get

(9)


E + S ES E + Pk1

k2

k3


rP = - rS = k3 CES

Combining (7), (8) and (9), we get

(10)

k1 (CE0 - CES)CS = (k2 + k3)CES

Using CE0 = CE + CES in (9) to eliminate CE, we get

which is rearranged to give

CE0CSCES = (k2+k3)/k1 + CS

(11)


E + S ES E + Pk1

k2

k3


where rmax = k3CE0

and KM = (k2 + k3) / k1(13)

Combining (10) and (11), we get

k3CE0CS- rS =(k2+k3)/k1 +CS

rmaxCS =

KM + CS

(5)

(12)rP =

When k3 << k2 (i.e. product forming step is slow),

KM = k2 / k1(6)


where rmax = k3CE0 = kcatCE0

and KM = f(rate constants)

- rS rmaxCS =

KM + CS rP =

Simple Enzyme Kinetics (in summary)

S

P[E]

rmax is proportional to the initial concentration of the enzyme

KM is a constant

prof. r. shanthini 09 nov 2012 enzyme kinetics and associated reactor design: introduction to...

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