4. ENZYME KINETICSEnzymekinetics

4. ENZYME KINETICS

Enzyme kinetics

Investigation of enzymatic reaction rate, identification of pa- rameters.

E + S ↔ E + P

For stoichiometric calculations all components should be gi- ven in moles or grams. But: enzymes are not pure proteins!

→ amount of enzymes is measured through their catalytic effect → ACTIVITY

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Enzyme kinetics

One UNIT is the amount of the enzyme which consumes 1 µmol substrate or forms 1 µmol product during 1 minute at given reac- tion circumstances.

SI: 1 Katal: 1 mol substrate (product) during 1 s.

(too huge!!) → nKat = 10^-9Kat (nanoKatal)

1 Kat = 6*10⁷U, 1U =1.666*10^-8Kat, 1U= 1/60 µKat

Specific activity: U/mass or U/volume → U/mg, U/ml

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Michaelis-Menten kinetics

Conditions:

k_-2= 0 (the second step is irreversible)

the first step reaches the equilibrium quickly = RAPID EQUILIBRIUM:

Dissociation constant of (ES):

stable ES complex, EP complex negligible

k₁SE = k_-1(ES)

(ES) S.E k

K k

1 1

s = ⁻ =

E + S ES E + P

k₁ k_-1

k₂ k_-2

one active centre, one substrate

concentration can be applied (instead of activity) (S) >> (E₀) i.e. E₀ / S << 1

Reaction rate:

Mass balance for E:

Divide these equations!

V dP

dt k (ES)₂

= =

E

( ES )

E

Michaelis-Menten kinetics

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E + S ES E + P

k₁ k_-1

k₂ k_-2

Michaelis-Menten kinetics

Divide the two equations:

substitute:

Rearrange:

because

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V E

k (ES) E (ES)

o

= 2

+

V E

k S K E E S

K E

o 2

s

s

= +

1 s

1

k S .E

K k ( E S )

= − =

S K

S K

1 S K

S E

k V

s s s o

2 = +

= +

V

= k E

dt k (ES)₂

= =

The rate equation:

or

V V

S K 1 S

K

max

s

s

= +

V V S

K S

max s

= +

Michaelis-Menten kinetics

7

Leonor Michaelis 1875-1949 Maud Menten

1879-1960

Michaelis, L., Menten, M. (1913) Die kinetik der invertinwirkung, Biochemische Zeitung 49, 333-369

M és M

Briggs-Haldane kinetics

The same differential equtions but the condition:

(quasi) steady state:

(S) >> (E₀) i.e. E₀/S << 1 k₁ES > k_-1(ES) ill. k₁ES > k₂(ES)

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E + S ES E + P

k₁ k_-1

k₂ k_-2

( )

( ) ( ) ( )

( )

1 1

1 1 2

2

dS k ES k ES

dt

d ES k ES k ES k ES

dt

dP k ES dt

−

−

= − +

= − −

=

d(ES)/dt =0

Briggs, G. E., and Haldane, J. B. (1925) A Note on the Kinetics of Enzyme Action, Bio-chem J 19, 338-339.

Briggs-Haldane kinetics

After a short transition peri- od (pre-steady state) the rate is almost constant (quasi-steady state).

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Quasi st-st

time Pre- st-st

( )

( )

( )

d ES k E S k ES k ES 0

dt

K_m= (k_-1+ k₂) / k₁

E+(ES)= E_o

( )( )

( ) ( )

1 1 2

1

1 2

k E S k k ES

k E S

ES k k

−

−

⋅ ⋅ = +

= ⋅ ⋅ +

Michaelis constant

S K

V S S K

S E V k

m max m

o 2

= +

= +

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Briggs-Haldane kinetics

Michaelis-Menten Briggs-Haldane

V V S

K S

max s

= +

V V S

K S

max m

= +

1 2

m

1

k k

K k

− +

=

if (k₁) >> (k₂) the two constants are equal!

Discussion

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m s

1

K K k

= + k

1 s

1

K k k

= −

1^thorder range

S E k S K E

V k _{0 0}

S

2 = ′

≈

catalytic effectivity = specifity constant

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S >>K_s

S <<K_s

Discussion

V_max=k₂E₀

rectangular hiperbole:

V=(V_max/K_s)S

V V S

K S

max s

= +

zero order range

Hiperbole

V

S V_max

-K_m 0

K_m

m m m s m

m s m s

m V

K S

V K K

S

V K V K S V V

x y a

+ +

− + =

−

= +

=

Experimental area

In M-M and B-H equations V means initial reaction rate (V₀ → extrapolated to t=0).

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How to measure reaction rate?

Parameter estimation

Linearised diagrams are used:

Calculation of nonlinear regression was complicated without computers

It provides additional info about enzyme inhibition 1. Lineweaver-Burk plot

1/v – 1/S

1 1

1

max max

K

V = V + V

S

Linearised forms

2. Hanes-Langmuir plot S/v – S

3. Eady-Hofstee plot v/S – v

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V

V = V max − K m S 1

S K m S

V V V

max max

= + ⋅

tg α= k₂

Effect of enzyme concentration

If v_max = k₂.E₀, then:

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V_max: its not a climax, but limit → border of rate Its not an enzyme feature, it depends on E₀:

V_max= k₂ . E₀ → = ACTIVITY

k₂ is the real enzyme feature = turnover number [s^-1] → transformation frequency

Extending to every enzymes and every kinetics:

k_cat[ s^-1 ]: Turnover frequency of one enzyme molecule (at S-saturation): how many substra- te molecules are transformed in one second by one enzyme molecule.

V_max= k_cat. E₀

Interpretation of kinetic parameters

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Kinetic parameters: K

, K

Affinity of az enzyme to substrate

Usually the S concentration in a living cell – easy adap- tion to changes

K_Shas changed → Inhibitor? Activator?

Enzyme analytics:

- activity measurement:

S>>K_S v=v_max

- substrate measurement:

S<<K_S linear range

unsensitive

very sensitive

k₁ 10⁷-10¹⁰ dm³mol^-1min^-1[max. value (~10¹¹) limited by diffusivity of small molecules] k_-1 10²-10⁶min^-1

k₂ 50-10⁷min^-1

K_m 10^-6- 10^-2 mol/dm³

Interpretation of kinetic parameters

Molecular switches Metabolic enzymes

Restriction enzymes

Many enzyme catalysed reactions - mainly biopolymer hyd- rolysis - are highly shifted to the right hand side, practically k_-2may really be neglected.

But conversions like

glucose fructose

(glucose isomerase)

~50 : 50 %

are of reversible character.

Reversible reactions

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K k k

k

K k k

k

ms

2 1

1

mp

2 1

2

= +

= +

−

−

−

V k E

V k E

maxs 2 o

maxp 1 o

=

=

2 1

2 1 2 1 um) eq(uilibri

2 2 2 1

1 1

k k

k K k

K K

k K k k

K k

−

−

−

−

=

=

=

=

1/K_S K_P

Reversible reactions

While k_-2= 0 in both kinetic models reactions seems to be irre- versible. Models for reversible (equilibrium) reactions are built up from models of two countercurrent irreversible reaction.

E + S ES E + P k₁

k_-1

k₂ k_-2

E

WHAT WILL HAPPEN?

S → P or P → S

?

S P

V

= V

- V

= k

(ES) - k

(EP)

V

V S P

K

K 1 P

K S

V S

K V P

K

1 S

K

P K

netto

maxs

eq

ms

mp

maxs ms

maxP mP

ms mP

=

 −

 



 +

 

 +

=

−

+ +

What does it depend on?

Reversible M-M equation

Reversible reactions

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K

, S , P value!