• Nem Talált Eredményt

4. ENZYME KINETICS

N/A
N/A
Protected

Academic year: 2022

Ossza meg "4. ENZYME KINETICS"

Copied!
5
0
0

Teljes szövegt

(1)

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

2

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

3

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

4 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 )=Eo

Michaelis-Menten kinetics

5

E + S ES E + P k1

k-1

k2

k-2

Michaelis-Menten kinetics

Divide the two equations:

substitute:

Rearrange:

because

6

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 = +

+

=

Vmax=k E2 o V=dPdt =k (ES)2

(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

8

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)

9

E + S ES E + Pk1 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).

10 Quasi st-st

time Pre- st-st

( )

1 1

( )

2

( )

d ES k E S k ES k ES 0

dt = ⋅ ⋅ − − =

Km= (k-1+ k2) / k1

E+(ES)=Eo

( )( ) ( ) ( )

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

= +

= +

11

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!

12

Discussion

2

m s

1

K K k

= +k

1 s

1

K k k

=

(3)

1thorder range

S E k S K E

V k 0 0

S

2 = ′

catalytic effectivity = specifity constant

13

S >>Ks

S <<Ks

Discussion

Vmax=k2E0

rectangularhiperbole:

V=(Vmax/Ks)S

V V S

K S

max s

= +

zero order range

Hiperbole

V

S Vmax

-Km 0

Km

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).

15

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

16

1 1

m

1

max max

K V = V + V

S

Linearised forms

2. Hanes-Langmuir plot S/v – S

3. Eady-Hofstee plot v/S – v

17

V

V = V maxK m S 1

S K m S

V V V

max max

= + ⋅

tg α= k2

Effect of enzyme concentration

If vmax= k2.E0, then:

18

(4)

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

Vmax= k2. E0 = ACTIVITY k2 is the real enzyme feature = turnover number [s-1] →

transformation frequency

Extending to every enzymes and every kinetics:

kcat[ 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.

Vmax= kcat. E0

Interpretation of kinetic parameters

19

Kinetic parameters: K

s

, K

m

Affinity of az enzyme to substrate

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

KShas changed→ Inhibitor? Activator?

Enzyme analytics:

- activity measurement:

S>>KS v=vmax - substrate measurement:

S<<KS linear range

20 unsensitive

very sensitive

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

k-1 102-106min-1 k2 50-107min-1 Km 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

23

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/KS KP

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.

24

E + S ES E + P k1

k-1 k2 k-2

(5)

E

WHAT WILL HAPPEN?

S → P or P → S

?

S P

V

netto

= V

foreward

- V

backward

= k

2

(ES) - k

-2

(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

25

K

eq

, S , P value!

Hivatkozások

KAPCSOLÓDÓ DOKUMENTUMOK

[r]

[r]

A segment s ∈ S is said to be shielded if there are at least 5k4 k segments of the same type, belonging to different edges of E, preceding s and at least 5k4 k such edges coming after

Whether the reaction is instantaneous or takes place in the intermediate regime with partial depletion ofNaOCl, is checked by calculating the experimental enhancement factor for a

Temperature Reaction rate (v ).. At a certain temperature the rate increases by many orders of magnitude... The background of an explosion is very often a chain reaction with

The kinetics and inhibitory effects of cyanide on the granular starch hydrolyzing enzyme (GSHE) activity during hydrolysis of cassava (Manihot esculenta Crantz) starch at low

A new method for determining of reaction mechanism and kinetics, catalyst pellet model and adsorption kinetics was developed on the basis of a computer analysis

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