enzymes

Enzymes

Fall 2007

Lecture 2

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Enzymes• High molecular weight proteins

15,000 < MW <4,000,000

• Catalysts• Nomenclature - end with “ase”• Holoenzyme - enzyme containing a non-protein group like a metal

– (apoenzyme + cofactor)– (protein + metal)

• Isoenzymes - catalyze same rxn• Classified based on type of rxn catalyzed (Table 3.1)


Examples(enzyme - use - source)

• Trypsin - anti-inflammatory, meat tenderizers - animal pancreas

• amylase - syrup, glucose production - Bacillus subtilis• protease - detergents, silver recovery - B. subilis• invertase - confectionaries- Sacharomyces cerevisiae• cellulase - breaks down cellulose - bacteria/yeast/mold• penicillinase - remove penicillin from allergic individuals

- bacteria


Enzyme Function• Lower activation energy of a reaction by binding to the

substrate and forming a substrate-enzyme complex• Interaction is due to van der Waals and H-bonding at the

active site• Interaction is complex - studied via Raman Spectroscopy

and X-ray• Enzyme does not affect equilibrium constant or free energy

change


Lowers the activation energy of a reaction - highly specificDownloaded from

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Lock and Key Mechanism

Specific binding site


Enzyme Kinetics

S P r = v = dP/dt = k1(S)E

S + E P + E r = v = dP/dt = k1(S)(E)

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7

Substrate

v

First order

Zero order


Enzyme Kinetics

Michaelis Menten Approach

(Henri)

S + E <------------> ES -----------> E + P

Based on experimental data

SK

Sv

dt

dPvrate

M max

k1

k2

k3


Rapid Equilibrium MM

k1 (E)(S) = k2(ES)

or (1)

Rate Equation (2)

substituting for (ES) from Eq 1 into Eq 2

))(()(2

1 SEk

kES

)(3 ESkdt

dPv


Using the total enzyme balance E0 = E + ES

or

substituting into the rate equation

))((2

13 SEk

kk

dt

dP

))((2

10 SE

k

kEE

Skk

EE

2

1

0

1)(


we obtain

KM = k2/k1 here is a dissociation constant, it characterizes the interaction of an enzyme with a given substrate S= KM when v = ½ vmax

vmax = k3E0 - maximum reaction rate, proportional to the initial enzyme concentration

SS

kk

EkS

Skk

Ekk

k

dt

dPv

1

2

03

2

1

02

13

1

SK

Svv

M max


Vmax and KM

0

1

2

3

4

5

6

0 1 2 3 4 5 6 7

Substrate

v

Vmax

KM

Figure 3.3 in book


In many cases the assumption of rapid equilibrium is not valid although the enzyme-substrate reaction still shows saturation type kinetics.

•In Class Exercise

S + E <----------> ES <-----------> E + P

Assume rapid equilibrium and determine the rate expression for product formationk2

k1

k4

k3


Quasi Steady State Approach

• Briggs and Haldane – another mathematical approach to the observed experimental MM eqn

S + E <--------> ES -----------> E + P

Assume that the change in (ES) with time is very small compared with to changes in S or P

Rate Equation of (ES)

k2

k1k3

))(())(()()())(()(

321321 ESkkSEkESkESkSEkdt

ESd


Assuming quasi steady state d(ES)/dt = 0Solve equation for (ES)

Rate Equation for P is substituting for (ES)

Using the total enzyme balance E0 = E + ES

))(()(32

1 SEkk

kES

))(()(32

133 SE

kk

kkESk

dt

dPv

)(132

10 S

kk

kEE

Hint: Solve SS eqn for whatever quantity set equal to zero


Or

Substituting into the rate equation for E

Rearranging results in the following:

)(132

1

0

Skk

k

EE

)()(

1)(

)())((

32

132

031

32

31

Skk

kkk

SEkkSE

kk

kk

dt

dPv

)(

)(

)(

1

32

03

Sk

kk

SEk

dt

dPv

Hint: Do not multiply through k values


The equation has the same form as the MM eqn

Where vmax = k3E0 and

Or Figure 3.4

• In Class Exercise

S + E <----------> ES <-----------> E + P

Assume rapid quasi steady state and determine the rate expression for product formation

)(

)(

)(

1

32

03

Sk

kk

SEk

dt

dPv

1

32

k

kkKM

SK

Svv

M max

k1

k2k4

k3


enzymes

Documents

enzymesubstrate reaction

e esorsubstituting

e pbased

k1ses e p e r

equilibrium mmk1 es

equilibrium constant

k3e0 maximum reaction

vmax vmax