BAHAN KULIAH BIOKIMIA POWER POINT BAGIAN 1 /BIOCHEMISTRY POWER POINT LECTURES PART 1 | Karya Tulis Ilmiah

Enzyme Kinetics and Catalysis

Serine proteases

•_{Diverse and widespread proteolytic enzymes}

•_{Involved in digestion, development, clotting,} inflammation…

Use of an Artificial Substrate

P-Nitrophenolate is very yellow while the acetate is

The kinetics show

1. A “burst phase” where the product is rapidly formed with amounts stoichiometric with the enzyme.

A covalent bond between a Serine and the substrate suggests an “active Serine”. These Serines can be labeled with inhibitors such as diidopropyl

phosphofluoridate specifically killing the enzyme.

DIPF is extremely toxic because other active Serines can be labeled. Such as acetylcholine esterase.

Affinity labeling

His 57 is a second important catalytic residue. A substrate containing a reactive group binds at the active site of the enzyme and reacts with a nearby reactive amino acid group. A Trojan horse effect.

Catalytic mechanism

1. After the substrate binds Ser 195 nucleophilically

attacks the scissile peptide bond to form a transition state complex called the tetrahedral intermediate (covalent

catalysis) the imidazole His 52 takes up the proton Asp 102 is hydrogen bonded to His 57. Without Asp 102 the rate of catalysis is only 0.05% of wild-type.

2. Tetrahedral intermediate decomposes to the

acyl-enzyme intermediate. His 57 acts as an acid donating a proton.

1. Conformational distortion forms the tetrahedral

intermediate and causes the carboxyl to move close to the oxyanion hole

2. Now it forms two hydrogen bonds with the enzyme that

cannot form when the carbonyl is in its normal conformation.

Enzyme Kinetics

Rates of Enzyme Reactions

How fast do reactions take place

•_{Reaction rates}

Thermodynamics says I know the difference between state 1 and state 2 and G = (G_f - G_i)

But

Changes in reaction rates in response to differing conditions is related to path followed by the reaction

and

Enzyme kinetics are important for many

reasons

1. Substrate binding constants can be measured as well as inhibitor strengths and maximum catalytic rates.

2. Kinetics alone will not give a chemical mechanism but combined with chemical and structural data

mechanisms can be elucidated.

3. Kinetics help understand the enzymes role in metabolic pathways.

Chemical kinetics and Elementary

Reactions

A simple reaction like A  B may proceed through several

elementary reactions like A  I₁  I₂  B Where I₁ and I₂ are

intermediates.

The characterization of elementary reactions comprising an overall reaction process constitutes its mechanistic

description.

Rate Equations

Consider aA + bB + • • • + zZ. The rate of a reaction is proportional to the frequency with which the reacting molecules simultaneously bump into each other

   

a b

 

z

Z

B

A

k

The order of a reaction = the sum of exponents

Generally, the order means how many molecules have to bump into each other at one time for a reaction to occur.

A first order reaction one molecule changes to

another

A



B

A second order reaction two molecules react

A + B



P + Q

or

3rd order rates A + B + C  P + Q + R rarely occur

and higher orders are unknown.

Let us look at a first order rate

A  B

 

 

dt

P

d

dt

A

d







v

 = velocity of the reaction

in Molar per min. or

moles per min per volume

k = the rate constant of the reaction

 

 

A

dt

A

d

Instantaneous rate: the rate of reaction at any specified time point that is the definition of the derivative.

We can predict the shape of the curve if we know the order of the reaction.

A second order reaction: 2A  P

 

 

2

A

A

k

dt

d

v







Or for A + B  P + Q

 

 

  

B

A

B

A

k

dt

d

dt

d

It is difficult to determine if the reaction is either first or second order by directly plotting changes in

concentration.

 

 

A

dt

A

d

k





d

 

 

k

dt

A

A











t 0 A A

dt

k

-A

A

o

d

 

 

t

o

k

A

ln

A

ln





 

 

-kt

o

e

A

However, the natural log of the concentration is

directly proportional to the time. - for a first order

reaction-The rate constant for the first order reaction has units of s-1 or min-1 since

velocity = molar/sec

and v = k[A] : k = v/[A]

The half-life of a first order reaction

   

2

A

A



o _{Plugging in}

to rate equation

 

 

1₂ o

A

2

A

ln





kt

























k

k

693

.

0

2

ln

t

2

The half-life of a first order reaction can be used to determine the amount of material left after a length of time.

The time for half of the reactant which is initially present to decompose or change.

32P, a common radioactive isotope, emits an

energetic  particle and has a half-life of 14 days.

A second order reaction such like 2A  P

 

 

 

 







t

dt

k

0 A o A 2 o

A

A

d

-  - 



A

_o



kt

1

A

1

When the reciprocal of the concentration is plotted verses time a second order reaction is characteristic of a straight line.

The half-life of a second order reaction is

and shows a dependents on the initial concentration 2

 

o

The Transition State

A bimolecular reaction A + B C A B + C at some point in the reaction coordinate an intermediate ternary complex will exist

A B C

This forms in the process of bond formation and bond breakage and is called a transition state.

H_a + H_b H_c H_a H_b + Hc

An energy contour of the hydrogen reaction as the three molecules approach the transition state at location c.

This is called a saddle point and has a higher energy than the

Energy diagrams for the transition state using the hydrogen molecule

Transition state diagram for a spontaneous reaction. X‡ is the symbol for the

 

k

  

A

B

k'

 

X

dt

P

d





Q P





B



A

For the reaction

‡ Where [X] is the

concentration of the transition state species

 

  

A

B

X

K

‡



‡

G

RTlnK

-

‡





‡



G

‡

is the Gibbs free energy of the activated

complex.

 

  

B

A

e

k'

t

P

-



RT

G



d

d

‡

The greater the G‡, the more unstable the transition

state and the slower the reaction proceeds.

This hump is the activation barrier or kinetic barrier for a reaction.

The activated complex is held together by a weak bond that would fly apart during the first vibration of the bond and can be

expressed by k' =where is the vibrational frequency of the

bond that breaks the activated complex and  is the probability

Now we have to define . E = h and  = E/h where h is Planks constant relating frequency to Energy. Also through a statistical treatment of a classical

oscillator E= K_bT where K_b is Boltzmann constant.

By putting the two together

h

T

K

k





b

RT G b

h

T

K

k

 



e

And

The rate of reaction decreases as its free energy of activation, G‡ increases

or

the reaction speeds up when thermal energy is added

Multi-step reactions have rate determining steps

Consider

A



 

k1

I



 

k2

P

Catalysts act to lower the activation barrier of the reaction being catalyzed by the enzyme.

Where G‡

cat = G‡uncat- G‡cat

The rate of a reaction is increased by

RT

G

_cat

e



G‡

cat = 5.71 kJ/mol is a ten fold increase in rate.

This is half of a hydrogen bond!!

G‡

cat = 34.25 kJ/mol produces a million fold

increase in rate!!