physbiochem enzyme kinetics

8/2/2019 PhysBiochem Enzyme Kinetics

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Physical Biochemistry

Enzyme kinetics



Catalysis of chemical reactions

• Catalysts are defined as species that increasethe rate of a chemical reaction without alteringthe position of the equilibrium: – Equilibrium constant K is not affected.

• Catalysts may directly participate in the reactionbut are regenerated at the end of the reaction.

CProductsJ

JBIICA

+→

→+

→+

CProductsCBA +→++

Catalyst

Overall reaction



Enzymes

• Very high rate enhancements : – Typically 10 6 – 10 14 compared to uncatalyzed rate. – Diffusion-controlled/limited : Enhancement is so high,

the rate essentially depends on fast a substrate canreach the active site.

• High selectivity : – Only particular substrates can bind.

• High specificity : – Only particular reaction is catalyzed:• E.g. Glucosyl transferase will only add glucose on to 2

position of another glucose, but not at the other positions 3,4and 6.



Enzymes

• Have ability to respond to external signalsor their environment: – Activity of enzymes can be controlled or

altered. – Very important prerequisite to control

metabolism.



Free energy of binding

• Enzymes use the free energy of binding ∆bG to

change the free energy of activation ∆ ‡G.• Rate of reaction is determined by the free

energy difference between transition state andreactants: – Reduction of this difference increases rate.

• For single-substrate reactions, two ways: – Destabilize substrate / stabilize transition state. – Altering the reaction mechanism.



Destabilize substrate / Stabilize transition state.

• Binding usuallyfavorable, butincreases freeenergy of activation.

• Favorable freeenergy of binding for distorted substrates: – Distortion of

substrate towardstransition state.

– Lower free energy of activation.



Altering reaction mechanism

• Enzyme provides functional groups that aremore effective than the normal reactants : – Glu of enzyme acts as proton donor instead of water

of solvent.• This changes the reaction mechanism towards a

path with a lower free energy of activation: – Increases rate.

• Free energy of binding is used to position thesubstrate in the best possible orientation withrespect to the functional groups.



Single-substrate enzymekinetics

• Derivation of the Michaelis-Mentenequation.

• Two ways: – Steady state approximation of enzyme-

substrate complex ES. – Equilibrium approximation:

• ES in equilibrium with E and S slightly altered bythe breakdown to products.

ProductsEESSE 21

1

+ → ⇔+−

k k

k



Steady-state approximation for ES

ProductsEESSE 21

1

+ → ⇔+−

k k

k

[ ] [ ] [ ] [ ] [ ]0ESESSEES 211 =−−= − k k k dt

d Steady-state approximation:

[ ] [ ] [ ]ESEE −=total

Free enzyme concentration:

Concentration of enzyme present at the start of the reaction.

[ ] [ ] [ ]( )[ ] [ ] [ ]0211=−−−=

−ES k ES k S ES E k

dt

ES d total



Steady-state approximation for ES

[ ] [ ] [ ]( )[ ] [ ] [ ]0211=−−−=

−ES k ES k S ES E k

dt

ES d total

[ ] [ ] [ ]

[ ]S

SEES

1

21

+

+=

−

k

k k total

[ ] [ ] [ ]

[ ]S k

k k

S E k ES k v total

++

==−

1

21

22

Rate of the reaction:

[ ][ ]SS

max+

=m

K

V v

1

21

k

k k K

m

+= − [ ]total

E k V 2max

= Michaelis-MentenMichaelis-Menten

equation.equation.



Michaelis-Menten equation

• Relates the rate v of the enzyme-catalyzed reaction to the concentration of the free substrate [S].

• Approximation when [S] total >> [ES]:

[ ][ ]SSmax

+=

m K

V v

[ ] [ ] [ ] [ ]total total S ES S S ≈−=



Equilibrium approximation

ProductsEESSE 21

1

+ → ⇔+−

k k

k

[ ] [ ][ ] [ ]

[ ] [ ]m

m K K

SEESES

SE=⇒=Dissociation constant:

[ ][ ]

[ ][ ] [ ]

[ ] [ ]

[ ][ ] [ ]

[ ][ ]S

SSE

E

SE

ESEES

EES

+=

+

=+

==m

m

m

total K

K

K θ Define:

10 ≤≤ θ Fraction of enzyme thatis active or has a boundreacting substrate.

Divide by [E], multiply by K m



Equilibrium approximation

[ ][ ]S

S+

=m

K θ 10 ≤≤ θ

Rate of the reaction:[ ][ ]SSmax

max +==

m K

V V v θ

Michaelis-Menten equationMichaelis-Menten equation

V max is the limiting (maximum) rate attainable, when all enzyme

is present as ES.

[ ][ ]total E

ES=θ




• [S] >> K m: – K m + [S] → [S] – v = V max = k 2 [ES]

• [S] < < K m: – K m + [S] → K m – v = V max [S] / K m – Straight line with slope

V max / K m• [S] = K m:

– v = V max / 2.[ ]

[ ]S

Smax

+

=m K

V v

[ ] m K > >S

[ ] m K < <S




• V max is the limiting velocity of the reaction at a givenconcentration of the enzyme:

– Enzyme fully saturated withsubstrate S.

• K m is the Michaelis constant : – Steady state: Measure of life

time of ES complex. – Equilibrium approximation: K m

is a dissociation constant.

[ ]total E k V

2max=

1

1

1

21

:mEquilibriu

:stateSteady

k

k K

k

k k

K

m

m

−

−

=

+

=




• k cat is catalytic constant : – First order rate constant of the

decomposition reaction into products.

– Same as k 2 for single stepdecomposition. – Also rate constant at high

concentration of S:• v = k cat [E]total .

[ ]total

cat

V k

Emax=

[ ]total E k V 2max =

ProductsEESSE 21

1

+ → ⇔+−

k k

k




• k A is specificity constant or catalyticefficiency :

– Measures degree of selectivity.

– Maximum value isdiffusion rate constant,

when diffusion is theslowest step.

– Also second order rateconstant at small

concentration of S

m

cat A

K

k k =

[ ][ ]

[ ] [ ] [ ] [ ]SESE

SSmax

total A

m

total cat

m

k K

k

K

V v =→

+=

[ ] m K < <S




• Turnover number: – Moles of substrate consumed or product

formed per mole enzyme per unit time. – Same as k cat .



Analysis of kinetic data

• Fit experimental data to Michael-Mentenequation with non-linear regression.

• Rearrange rate to obtain a straight line.

[ ][ ]SSmax

+=

m K

V v

[ ] maxmax

1S11

V V

K

vm +=

[ ]S1

versus1v

Lineweaver-Burk plot:x-intercept: – K

m -1

Y-intercept (x=0): V max

-1

[ ] mmK

v K V

S v −= max

[ ]v

S v versus Eadie-Hofstee plot:Slope: -K

m -1

X-intercept (y=0): V max

[ ] [ ]maxmax

SS

V

K

V v

m+= [ ] [ ]SversusSv

Hanes plot:X-intercepts (y=0): - K

m

Slope: V max

-1



Worked example

• The activity of the enzyme urease, whichcatalyses the reaction CO(NH 2)2 + H 2O ⇔ CO 2 +2NH 3 was studied as a function of urea

concentration with the following results. What arethe values of K m and V max .

[Urea] mM 30 60 100 150 250 400

v mmol ureaconsumed (mgenzyme) -1 min -1

3.37 5.53 7.42 8.94 10.70 12.04 Microsoft ExcelWorksheet

http://z/Teaching/PhysBiochem/enzyme-kinetics-urease.xls





Limitations• Pre-steady-state period:

– Reaction reaches not immediately equilibrium:• Can be used to obtain better information about reaction.

– Rate equations much more complex.• Approach to equilibrium:

– Assumption: Forward reaction towards products is irreversible. – Practise: Reversible reaction if product concentration is high enough.• Product inhibition:

– Product has an ability to bind to active site to become an competitiveinhibitor.

• Substrate inhibition:

– Some enzymes are inhibited by high substrate concentration. – Usually indicates a second binding site for substrate.• Enzymes with more than one active site:

– Active sites may affect each other. – Positive, negative cooperativity.






































Temperature affect kinetics

• Reaction usually follows Arrhenius equation for T -dependency.

• Enzyme: – Unfolds above meltingtemperature rendering itless active.

– Multistep reaction:• Different steps may be rate

determining at differenttemperatures.

RT

E a

Aek −

=±±±

∆−∆=∆ S T H G










































Temperature affect kinetics

• Enzyme: – May exists in more than one active

interconvertible conformation.

– Each active form has different free energy of activation.













Effects of p H

• Changes in pH affects rates of enzyme-catalyzed reactions by: – H+ and/or OH - appear in rate equation(s).

– Changes in ionization state of substrate:• Additional acid-base catalysis.• Changes in substrate binding free energy:

– Affects K m.

– pH affect stability of enzyme:• Changes in ionization state of titrating residues: – May affect binding and therefore K m.

– May V max if titrating residues directly participate in reaction.










































One titrating residue

• Example: – Active site containing one titrating site. – Enzyme is active when this titrating site is

deprotonated . – Enzyme is activated when p H is increased – H+ acts as non-competitive inhibitor.

++ +⇔ HS:ES:EHm K

Inactive Active



































One titrating residue

• Typical plots• V max is higher at

higher pH.

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 3.6 3.8 4 4.2 4.4 5 6 8 10 12 14

pH

Occupancy

Charge state























Two titrating sites

• Active site contains two titrating residues: – One with low p K a1 .

– One with high p K a2

.

−−−

+

−−

+

−−⇔−−⇔−−

+

+

+

+YEXYEHXYHEHX

H

H

H

H

ActiveInactive Inactive

21 p p aaK K <




























































Two titrating sites

• Typical plot for well separatedpK a’s.

• Maximumnarrows whenpK a areoverlapping.




















Two titrating sites

• Effective V max may not reachideal V max , whenpK a’s are tooclose.

−−−

+

−−

+

−−⇔−−⇔−−

+

+

+

+YEXYEHXYHEHX

H

H

H

H


















































physbiochem enzyme kinetics

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