# Enzymes and Enzyme Kinetics - CMBE

Enzymes and Enzyme Kinetics - CMBE

G. Other Enzyme Models

Multisubstrate models – many enzymatic

reactions have more than one substrate

that influences its kinetics

A + B

P + Q

1. Compulsory Order System

A must react first with the enzyme, and P only

is formed after Q is released.

Model:

E + A

k 1

EA

k -1

EA + B

k 2

EAB

k -2

EAB

k 3

EPQ

k -3

EPQ

k 4

EP + Q

EP

k 5

E + P

Note:

E + B

… etc.

1

2

3

4

5

9 Equations can be written (one for each component):

___ dA

dt

___ dE

dt

____ dEA

dt

____ dB

dt

_____ dEAB

dt

= -k 1 [E][A] + k -1 [EA]

= -k1 [E][A] + k -1 [EA] + k 5 [EP]

= k 1 [E][A] - k -1 [EA] - k 2 [EA][B] + k -2 [EAB]

= -k 2 [EA][B] + k -2 [EAB]

= k 2 [EA][B] - k -2 [EAB] - k 3 [EAB] + k -3 [EPQ]

etc.

Assumptions

_____ dEA

dt

_____ dEAB _____ dEPQ _____ dEP _____ dE

= = = = = 0

dt dt dt dt

2 Equilibrium

Reactions 1, 2, and 3 remain at equilibrium

Result

where:

____ dP

dt

0

= _________________

[E] 0

____ 1 _____ K 3 _______ K 2 K 3

+ +

K 1 [B] kobs [A][B]

k obs

K i = dissociation constant

for reaction i

k obs = observed first order

rate constant

Note

____ dP

dt

0

= _________________

[E] 0

____ 1 _____ K 3 _______ K 2 K 3

+ +

K 1 [B] kobs [A][B]

k obs

If [A]

2. Random Order System

Either A or B can react first with enzyme, and

complex formed can further react with other

substrate to form enzyme:substrate 1 :substrate 2

complex.

Model:

E + A

EA + B

EPQ

EQ

k 1

EB

k -1

Q

EA E + B

k -1 ′

k 4

EQ + P EPQ

k 4 ′

EP +

k 2

k 2 ′

EAB EB + A EAB

k -2 k -2 ′

EAB

k 3

EPQ

k -3

k 5 k

E + Q

5 ′

EP E + P

k 1 ′

Usually one product is always easiest to release.

3. Ping-Pong Model

Also known as the enzyme-substituted

mechanism or double displacement mechanism.

Both substrates need not be bound together at

the active site. This mechanism is common for

phosphate-transferring enzymes.

Model:

E + A

k 1

EA

k -1

EA

k 2

E* + P

k 3

E* + B E*B

k -3

E*B

k 4

E + Q

____ dP

dt

= ______________

[E] 0

K′

where:

____ K″ ____ K′′′

+ +

[A] [B]

K’s depend on rate

constants

Note that if [A]

H. pH and temperature effects

1. pH

pH has an obvious influence on enzyme

structure because:

1) the charge of amino acids (as zwitterions)

is a function of pH

2) the charge of several residues is a function

of pH….

Residue pK R

Aspartate 3.86

Glutamate 4.25

Histidine 6.0

Cysteine 8.33

Tyrosine 10.07

Lysine 10.53

Arginine 12.48

These values for pK a change slightly when the residue is on a protein.

H

3

Consider glutamate-lysine:

2 )

C O

O

+ (CH 2) 2 H O

_

N C C

H

N

H

C

(CH

C

4

O

O _

N H

+

3

H 2 N

H 3 N +

R 1 H

pK ≈ 4.3

H 3 N + R 1 –

COO –

R 2 H +

pK ≈ 9.8

R 1

COO –

R 2 H +

H 3 N +

R 2 H +

R 1 H

R 2 H +

pK ≈ 2.2

COO –

COOH

H 2 N

pK ≈ 10.5

R 1

COO –

R 2

A change in the charge on a residue

may alter the charge in the active site

_

+

+

_

_

+

+

At “medium” pH

At “low” pH

A change in the charge on a residue

may alter the charge internally, reducing

the ion pairing which is holding the

protein together.

_

+

_

_

+ +

+ ◦

At “medium” pH

At “low” pH

One (usually not particularly good) way to consider

mathematically the effect of pH is to treat hydrogen

ion (or lack of ion) as a competitive inhibitor. In

other words, the substrate will bind with enzyme

having an altered active site, but this binding will

EH + S

EH

k 1

k -1

k 3

EHS

E – + H +

k -3

k 4

EH

+

2

k 2

EH + P

Such reactions are

usually at equilibria

K 1 =

[EH][H + ]

[EH 2+ ]

EH + H + k -4

K 2 =

[E – ][H + ]

[EH]

____ dP

dt

= ___________ V MAX [S]

K

APP

M + [S]

Where:

K M APP = K M

1 + ____ K 2

[H + ]

+

____ [H + ]

K 1

Note similarity with

competitive inhibition...

K M APP = K M

1 + ___ [I]

K I

Example of what this model predicts:

Assume K 1 = 10 -5 or pK 1 = 5

K 2 = 10 -8 or pK 2 = 8

Then

__________________________

pH [H + ] K

APP

M /K M

4 10 -4 11.0

5 10 -5 2.0

6 10 -6 1.11

6.5 10 -6.5 1.06

7 10 -7 1.11

8 10 -8 2.0

9 10 -9 11.0

Enzyme Kinetics at Different pH

12

10

8

K M

APP /KM

6

4

2

0

3 4 5 6 7 8 9 10

pH

...model predicts no change in V MAX !

Enzyme Kinetics at Different pH

10

8

6

V

4

2

0

0 2 4 6 8

[S]

6.5

7.4

8.0

8.5

Problem with model is prediction that at high substrate

concentration, maximum reaction rate is restored.

Model prediction (V MAX = 10):

10

Reality (for example)....

8

[S] = 10K M

V

6

[S] = 5K M

4

2

[S] = K M

0

4 5 6 7 8 9

pH

2. Temperature

Temperature has two effects on enzyme kinetics:

1) activation due to the hastening of the

release of product from enzyme-substrate

complex (just like the temperature effect in

chemical reactions).

2) reversible inactivation and irreversible

denaturation – loss of enzyme because of

destabilization of bonds or more

significant loss of structural integrity (such

Note: textbook only considers irreversible damage.

Model:

E + S

k 1

k 2

ES E + P

k -1

E

k 1 ″

k -1 ″

E I

k 2 ″

E D

Reversible

Irreversible

Note: textbook only considers second step and uses k D instead of k 2 ″.

k 2 ″ and k 2 depend on temperature according to the

Arrhenius equation.

k 2 = A e –E/RT

k 2 ″ = A″ e –E″/RT

E = Energy of activation

E″ = Energy of denaturation

A, A″ (min -1 )

E, E″ (cal/mol)

Write 6 temperature-dependent differential equations.

___ dS

dt

___ dE

dt

____ dES

dt

____ dP

dt

_____ dE I

dt

_____ dE D

dt

= -k 1 [E][S] + k -1 [ES]

= -k1 [E][S] + k -1 [ES] - k 1 ″[E] + k -1 ″[E I ]+ Ae -E/RT [ES]

=

=

=

=

k 1 [E][S] - k -1 [ES] - Ae -E/RT [ES]

Ae -E/RT [ES]

k 1 ″[E] - k -1 ″[E I ]-A″e -E″/RT [E I ]

A″e -E″/RT [E I ]

Notes:

1) the amount of enzyme lost depends on

temperature and duration of exposure. (see Fig.

3-15 p. 77 “The nature of the plot will depend on

the length of time the reaction mixture is exposed

to the test temperature.” This plot is only valid for

“10-min exposure”!)

2) other rate constants (e.g., k 1 , k -1 ) are also affected

by temperature. However, the forward and reverse

rate constants tend to be influenced similarly.

Over a small temperature range, an equilibrium is

not altered as much as irreversible rate constants.

A specific example (simulated by the solution to the

differential equations):

activation

A = 9.1 10 8 min -1

denaturation

E = 10,900 cal/mol

A″ = 6.6 10 52 min -1 E″ = 77,900 cal/mol

__________________________

T(°C) k k 2 ″ (min -1 2 (min -1 )

)

30 12.5 0.000429

35 16.8 0.00350

40 22.3 0.0268

45 29.4 0.192

50 38.4 1.29

55 49.7 8.22

A specific example (simulated by the solution to the

differential equations):

activation

A = 9.1 10 8 min -1

denaturation

E = 10,900 cal/mol

A″ = 6.6 10 52 min -1 E″ = 77,900 cal/mol

S 0 = 8.0 g/L

E 0 = 0.03 g/L

k 1 = 30 L/g min

k -1 = 160 min -1

k 1 ″ = 15 min -1

k -1 ″ = 12 min -1

6

Figure 5a

Enzyme Kinetics at Different Temperatures

5

4

P (g/L)

3

40C

35C

30C

45C

2

1

50C

0

0 10 20 30 40

Time (minutes)

Figure 5b

Enzyme Kinetics at Different Temperatures

0.30

0.25

dP/dt

0.20

0.15

40C

35C

0.10

0.05

50C

45C

30C

0.00

0 10 20 30 40

Time (minutes)

Figure 5c

Enzyme Kinetics at Different Temperatures

0.020

0.015

30C

35C

E + ES

0.010

40C

0.005

50C

45C

0.000

0 10 20 30 40

Time (minutes)

Figure 5d

Enzyme Kinetics at Different Temperatures

6

5

40 min

P (g/L)

4

3

20 min

2

8 min

1

0

30 35 40 45 50

Temperature (C)

Figure 5e

Enzyme Kinetics at Different Temperatures

0.7

0.6

50C

0.5

v = dP/dt

0.4

0.3

0.2

40C

30C

0.1

0.0

0 2 4 6 8 10

S

What is effect of activation, inactivation, denaturation?

activation

V MAX k 2

As T increases, k 2 increases, V MAX increases

___________________

T(°C) k 2 V MAX

30 12.5 0.375

40 22.3 0.669

50 38.4 1.152

(Green/dashed curves on Figure 5e show increase in

reaction rate with temperature due to activation ONLY

without inactivation and without denaturation)

inactivation

E I

E

=

k 1 ″

k -1 ″

=

15

12

= 1.25

[E][S]

[ES]

=

k -1

k 1

=

160

30

= 5.33 g/L

For S 0 = 8 g/L, [E]/[ES] = 0.667

E 0 = E + ES + E I = 0.667ES + ES + 1.25(0.667)ES

= 0.667ES + ES + 0.833ES

useful

enzyme

(67%)

inactive

enzyme

(33%)

Equilibrium causes 33% of the enzyme to be in the inactive form

(Yellow/solid curves on Figure 5e show reduction in activity due to inactivation)

denaturation

Effective useful enzyme concentration decreases with time because

a portion becomes (irreversibly) denatured.

(Trajectories with “dots” on Figure 5e show reduction in activity with

time due to denaturation)

0.7

0.6

0.5

0.4

v

0.3

0.2

0.1

0.0

0 2 4 6 8 10

S

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