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Michaelis-Menten Kinetics

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Michaelis-Menten Kinetics
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Michaelis-Menten Kinetics

Enzyme Kinetics

Enzymatic reaction

k1 k2

E+S k-1

ES E+P

Rate expression for product formation

v = dP/dt = k2(ES)

d(ES)/dt = k1(E)(S)-k-1(ES)-k2(ES)



Conservation of enzyme

(E) = (E0) – (ES)

Two Methods to Proceed

• Rapid equilibrium assumption: define

equilibrium coefficient

K’m = k-1/k1 = [E][S]/[ES]



• Quasi-steady state assumption

[ES] = k1[E][S]/(k-1+k2)



• Both methods yield the same final equation

Michaelis- Menten Kinetics

Michaelis-Menten Kinetics

• When v= 1/2 Vmax, [S]= Km so Km is

sometimes called the half-saturation

constant and sometimes the Michaelis

constant



Vmax [S ] k 2 [E 0 ][S ]

v 

K m  [S ] K m  [S ]

Michaelis-Menten Kinetics



Vmax [S ] k 2 [E 0 ][S ]

v 

K m  [S ] K m  [S ]



• units on k2 are amount product per amount

of enzyme per unit time (also called the

“turnover number”). Units on E0 are

amount of enzyme (moles, grams, units,

etc.) per unit volume

• Km has the same units as [S] (mole/liter,

etc.)

Experimentally Determining Rate

Parameters for Michaelis-Menten

Kinetics



Lineweaver-Burk

Eadie-Hofstee

Hanes- Woolf

Batch Kinetics

Determining Parameters



• Rearrange the equation into a linear form.

• Plot the data.

• What kind of data would we have for an

experiment examining enzyme kinetics?

• Describe an experiment.

• The intercept and slope are related to the

parameter values.

Enzyme Kinetics Experiment

Place enzyme and substrate (reactants) in a

constant temperature, well stirred vessel.

Measure disappearance of reactant or

formation of product with time.

Why constant temperature?





Why well stirred?





What about the medium? Buffer?

Lineweaver-Burk

(double reciprocal plot)

– Rewrite Michaelis-Menten rate expression







1 Km 1 1

 

v Vmax [S ] Vmax

– Plot 1/v versus 1/[S]. Slope is Km/Vmax,

intercept is 1/Vmax

Graphical Solution



intercepts

1/ V

1 Km 1 1 Slope = Km/ Vmax

 

v Vmax [S] Vmax

1/ Vmax







-1/ Km 1/ [S]

Example: Lineweaver-Burk

-5 -5

[S] x 10 M V, M/min x 10

1.0 1.17

1.5 1.50

2.0 1.75

2.5 1.94

3.0 2.10

3.5 2.23

4.0 2.33

4.5 2.42

5.0 2.50

Resulting Plot









slope = Km/ Vmax= 0.5686

y intercept = 1/ Vmax= 2.8687

Michaelis-Menten Kinetics





Vmax [S ] k 2 [E 0 ][S ]

v 

K m  [S ] K m  [S ]

Vmax = 1/2.8687 x 10-4 = 3.49 x 10-5 M/min

Km= 0.5686 x Vm = 1.98 x 10-5 M

Other Methods

• Eadie-Hofstee plot



v

v  Vmax  Km

[S ]



• Hanes- Woolf



[S ] K m 1

  [S ]

v Vmax Vmax

Comparison of Methods

• Lineweaver-Burk: supposedly gives good

estimate for Vmax, error is not symmetric

about data points, low [S] values get more

weight

• Eadie-Hofstee: less bias at low [S]

• Hanes-Woolf: more accurate for Vmax.

• When trying to fit whole cell data – I don’t

have much luck with any of them!

Batch Kinetics

d[S ] Vmax [S ]

v 

dt K m  [S ]

integrate

[S 0 ]

Vmaxt  [S 0 ]  [S ]  K m ln

[S ]

rearrange

[S 0 ]  [S ] K m [S 0 ]

 ln Vmax

t t [S ]

Inhibited Enzyme Kinetics

• Competitive Inhibition

• Noncompetitive Inhibition

• Uncompetitive Inhibition

• Substrate Inhibition

Effects of Temperature and pH

Experiments: Initial rate at

different substrate concentrations



E S1= 20 E S2=10 E S3=6.7 E S4=5 E S5=4









Measure S for a short time period. Calculate v from:

v = [S(time 0) – S(time 1)]/delta time

Time (min) S (g/L)

Experiment 0 20

Using S1 0.5 19.43



v= (20-19.3)g/L]/0.5 min = 1.14 g/L/min





Time (min) S (g/L )

Experiment 0 10

Using S2 0.5 9.565



v= (10-9.565)g/L]/0.5 min = 0.87 g/L/min

Experimental Data

S (mmol/L) v (mmol/L/min)

20 1.14

10 0.87

6.7 0.70

5.0 0.59

4.0 0.50



Problems with this method?

Rate is not measured at a constant substrate

concentration – substrate decreasing. Must have

sensitive assay for substrate to measure initial rates.

20

18

16 regression

14 S/v = 0.6S + 5.6

12

S/v (min)









10

8

6 experimental data

regression

4

2

0

0 5 10 15 20 25

S (g/L)

Allosteric Enzyme Kinetics

In an enzyme with more than one substrate

binding site, binding of one substrate

molecule affects the binding of another.

n

d[S ] V max[S]

v  n

dt K m  [S ]n









n>1, cooperation; n0



I=0







Vmax is unchanged

1/Vmax



-1/Km -1/Km,app 1/[S]

Practice deriving kinetic

expressions



Derive competitive inhibition equation (3.22

in your text)? Write down all assumptions.

Noncompetitive Inhibition

Inhibitor binds to the enzyme, but not at the active

site. However, the enzyme affinity for substrate is

reduced.

E  S  ES  P [E ][S ] [EI ][S ]



Km  

  [ES ] [ESI ]

I I

[E ][I ] [ES ][I ]

KI  

EI S  ESI [EI ] [ESI ]



[E 0 ]  [E ]  [ES ]  [EI ]  [ESI ]



v  k 2 [ES]

Noncompetitive Inhibition

Cofactors and Coenzymes

Holoenzymes- three parts

• Apoenzyme- Protein portion

• Cofactor- inorganic ion (ex: metal ions),

improve the fit of enzyme with substrate

• Coenzyme- nonprotein organic molecule

(ex: NAD- nicotinamide adenine

dinucleotide), many synthesized from

vitamins (why vitamins are essential)

Noncompetitive Inhibition

Rate is given by:



Vmax Vmax,app

v 

 [I ]  K m 

  

 K m 

1 1 1

 K I 

  [S ] 



   [S ] 

 





Question: What is the magnitude of Vmax,app

relative to Vmax, and what will be the effect of

v? How can you moderate the effects of this

type of inhibition.

Noncompetitive Inhibition

1/v I>0





I=0





1/Vmax,app



1/Vmax



-1/Km 1/[S]

Km is unchanged

Uncompetitive Inhibition

Inhibitor binds only to ES complex, and not

to E alone.

E  S  ES  P [E ][S ]



Km 

 [ES ]

I

[E ][I ]

KI 

[EI ]

ESI

[E 0 ]  [E ]  [ES ]  [ESI ]



v  k 2 [ES]

Uncompetitive Inhibition

Rate is given by:

Vmax

[S ]

 [I ] 

1 

 K I  Vmax,app [S ]

v  

 

 [S ] K m ,app  [S ]

Km

 [I ] 

1

 K I 

 





What is the magnitude of Vmax,app relative to Vmax?

What is the magnitude of Km,app relative to Km?

Uncompetitive Inhibition

1/v

I>0



I=0

1/Vmax,app





1/Vmax



-1/Km,app -1/Km 1/[S]

Substrate Inhibition

No substrate inhibition

v



Substrate inhibition

Vmax [S ]

v 

[S]2



K m  [S ] 

KSi



[S ]max. rate  K mK Si

 S

Enzyme Deactivation

• Enzymes are denatured by

– Temperature

– pH

– Radiation

– Irreversible binding by inhibitors

• Temperature can both increase

(thermal activation) and decrease

(thermal denaturation) rate

Temperature effects

At moderate temperatures, higher

temperatures give higher rates.

E a

v  k 2 [E ], where k 2  Ae RT









At higher temperatures, rate starts to

decrease as enzyme denatures faster

d[E ] E d

k d t

  k d [E ], or [E]  [E 0 ]e , where kd  Ad e d RT

dt

Temperature Effects

Effect on rate is a combination of the two effects





E a

k d t

v  Ae RT

[E 0 ]e



Activation energy  10

kcal/mol

Deactivation energy  100

kcal/mol


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