SBS017 Basic Biochemistry Dr. Jim Sullivan [email protected]

SBS017Basic Biochemistry

Dr. Jim [email protected]

Lecture 9Enzymes: Basic principles

Learning Objectives Lecture 9

• You should be able to able to define the terms Enzyme, Specificity and Co-factor

• You will understand the concept of Gibbs Free Energy and its relation to reaction equilibrium

• You will be able to describe how enzymes effect the rate of biological reactions and be able to define the term Activation energy in the context of Transition state theory

Enzymes

• Biological catalysts

• Almost all enzymes are proteins (but RNA can have enzymic activity too, “ribozymes”)

• Function by stabilizing transition states in reactions

• Enzymes are highly specific

Enzymes accelerate biological reactions

• e.g. Carbonic Anhydrase

• CO2 + H2O H2CO3

• Each molecule of enzyme can hydrate 1,000,000 molecules of C02 per second, 10,000,000 times faster than uncatalysed reaction

Catalase

•2H2O2 2H2O + O2

Enzymes are highly specific

e.g Peptide bond hydrolysis (proteolysis)

A. Trypsin cleaves only after arginine and lysine residues.

B. Thrombin cleaves between arginine and glycine only in particular sequences.

But Papain cleaves all peptide bonds irrespective of sequence.

Specificity is important

e.g. DNA polymerase I

Adds nucleotides in sequence determined by template strand.

Error rate of < 1 in 1,000,000

Due to precise 3D interaction of enzyme with substrate

Many enzymes require cofactors

Cofactors are small molecules essential for enzyme catalysis

Can be:i.Coenzymes (small organic molecules)ii. Metal ions

Holoenzymes

Enzyme without its cofactor “apoenzyme”

With its cofactor “holoenzyme”

Cofactors are essential for activity

e.g. many vitamins are cofactors, many diseases associated with vitamin deficiency due to lack of specific enzyme activity

Energy transformationMany enzymes transform energy into different forms

Adenosine Triphosphate (ATP) is universal currency

Light ATP Photosynthesis

Food ATP Respiration

ATP work

ATP is an energy carriere.g. ATP provides energy to pump Ca2+ across

membranes

Free energyThe free energy of a reaction is the differencebetween its reactants and its products

This called the ΔG

If ΔG is negative, the reaction will occur spontaneously “exergonic”

If ΔG is positive, energy input is required “endogonic”:

Endogonic reactions

Reaction coordinates

Reactants

Products

Energy(G)

ΔGpositive

Exergonic reactions


Reactants

Products

Energy(G)

ΔGnegative

Exergonic reactions


Energy(G)

ΔGnegative

H2O2

2H2O + O2

ΔG is independent of reaction path


Reactants

Products

Energy(G)

ΔGsame

ΔG and reaction equilibria

• A negative ΔG indicates that a reaction can occur spontaneously not that it will

• ΔG tells us nothing about the rate of reaction only the energy of its end points

• The rate of reaction is defined by the equilibrium constant

The equilibrium constant

A + B C + DKeq

€

Keq = [C][D][A][B]

Calculating ΔGThe free energy of a reaction is given by:

€

ΔG = ΔG° + RT ln [C][D][A][B]

⎞ ⎠ ⎟

⎛ ⎝ ⎜

• Where ΔG° is the standard free energy change (i.e. change at 1M concentrations), R is the universal gas constant and T the absolute temperature

Calculating ΔG at pH7 (ΔG’)

€

0 = ΔG°'+RT ln [C][D][A][B]

⎞ ⎠ ⎟

⎛ ⎝ ⎜

At equilibrium and standard pH7:

Calculating ΔG at pH7 (ΔG’)Rearrange and substitute K’eq:

€

ΔG°'= −RT ln K 'eq)(At 25 °C, in log10, rearranged for K’eq:

€

K 'eq =10−ΔG°' / 5.69

This means that a 10-fold change in Keq is equivalent to 5.69 kJ/mol difference in energy

Reaction equilibria


A

B

Energy(G)

5.69kJ/mol

AKeq

B

€

Keq = [B][A]

If [B] = 10 [A] = 1Then ΔG = 5.69kJ/moll

Reaction equilibria


A

B

Energy(G)

11.38kJ/mol

AKeq

B

€

Keq = [B][A]

If [B] = 100 [A] = 1Then ΔG = 11.38 kJ/moll

Catalysis

• Enzymes cannot change equilibrium of reaction

• ΔG is independent of reaction pathway• Enzymes accelerate rate of reaction

AKeq

B AK+1

BK-1

€

Keq = K+1

K−1

Catalysis

AK+1

BK-1

€

Keq = K+1

K−1

= 101

= 1000100

No enzyme + enzyme

Equilibrium is same but rate is 100x greater

Enzymes accelerate reaction rates

Reaction reaches equilibrium much quicker with enzyme catalysis

Why do exergonic reactions (those with –ΔG) not take place spontaneously?

Activation energy

• Reactions go via a high energy intermediate

• This reduces the rate at which equilibrium is reached

• The larger the activation energy the slower the rate

Transition state theory


A

B

Energy(G)

ΔG

Transition state, S‡

ΔG‡


• Enzymes reduces the activation barrier• Transition state energy becomes smaller


• The rate of reaction depends on ΔG‡

• Big ΔG‡ slow reaction• Small ΔG‡ faster reaction• Enzymes works by reducing ΔG‡

• Enzymes stabilise the transition state

Summary

• Enzymes are biological catalysts• Enzymes are highly specific • Enzymes increase reaction rates but do not

alter the equilibrium of reactions• Enzymes stabilise transition states, reducing

activation energy and increasing the rate of reaction

Learning Objectives Lecture 9

• You should be able to able to define the terms Enzyme, Specificity and Co-factor

• You will understand the concept of Gibbs Free Energy and its relation to reaction equilibrium

• You will be able to describe how enzymes effect the rate of biological reactions and be able to define the term Activation energy in the context of Transition state theory



Lecture 10Enzymes Kinetics

Learning objectives Lecture 10

• You should be able to describe the evidence for enzyme/substrate complexes

• You will be able to define the terms Active sites and Active site residues

• You should be able to describe the Michaelis-Menten model for enzyme kinetics and the significance of KM and Vmax

Enzyme-Substrate (ES) complexes

• Most enzymes are highly selective in binding of their substrates

• Substrates bind to specific region of enzyme called Active Site

• Catalytic specificity depends on binding specificity

• Activity of many enzymes regulated at this stage

Evidence for ES complexesi. Saturation effect:

At constant enzyme concentration reaction rate increases with substrateuntil Vmax is reached.

Vmax is rate (or velocity) at which all enzyme active sites are filled.

Evidence for ES complexes

ii. Structural data:

Crystallography (3D structures), shows substrates bound to enzymes

Cytochrome P450 (green) bound to its substrate camphor

Evidence for ES complexes

iii. Spectroscopic data:

Absorption and fluorescence of proteins and cofactors change when mixed together

Changes in fluorescence intensity in the enzyme tryptophan synthetase after addition of substrates

Active sites

• The active site of enzyme is region that binds the substrate

• The catalytic groups are the amino acid side chains in the active site associated with the making and/or breaking of chemical bonds

Active sitesi. The active site is a 3D structure formed by groups that can come from distant residues in the enzyme (tertiary structure)

A. 3D structure of Lysozyme, active site residues coloured

B. Linear schematic of Lysozyme amino acid primary sequence, active site residues coloured

Active sites

ii. The active site takes up only a small volume of the enzyme

iii. Active sites are unique chemical microenvironments, usually formed from cleft or crevice in enzyme.

Active sites

Active sites often exclude water, and non-polar nature of site can enhance binding of substrates and allows polar catalytic groups to acquire special properties required for catalysis

Active sites

iv. Active sites bind substrates with weak interactions

Bonds can be electrostatic, hydrogen bonds, Van der Waals and hydrophobic interactions.

ES complexes have equilibrium constants (Keq) on the 10-2 to 10-8 range, equivalent to -3 to -12 kcalmol-1

Active sites

v. The specificity of an enzyme for its substrate(s) is critically dependent on the arrangement of amino acid residues at the active site

Remember importance of tertiary structure

Lock and key model

• Proposed by Emil Fischer in 1890

Induced fit model

• Substrates and enzymes are not rigid but dynamic and flexible

• Daniel Koshland in 1958 proposed the induced fit model

• In induced fit model the enzyme changes shape in order to optimise its fit to the substrate only after the substrate has bound

Induced fit model

Reaction kinetics

First order reaction (uni-molecular):

Rate of reaction is:

Units of k: s-1

€

V = k[A]

A P

K

Reaction kinetics

Second order reaction (bi-molecular):

Rate of reaction is:

Units of k: M-1s-1

€

V = k[A][B]

A + B P

K

But if [A]>>[B] or [B]>>[A] reaction is Psuedo First Order

Michaelis-Menten Model

• In 1913 Leonor Michaelis and Maud Menten proposed simple model to account for kinetic characteristics of enzymes

• They observed that a maximal reaction velocity with saturating substrate for a fixed amount of enzyme implies a specific ES complex is a necessary intermediate in enzyme catalysis

• Describes the kinetic property of enzymesAt a fixed concentration of enzyme increasing

[substrate] increases reaction rate

Initial velocity V0 for each substrate concentration is determined from the slope of the curve at the beginning of the reaction where any reverse reaction is insignificant


Rate of catalysis V0 varies with substrate concentration

Rate is initially proportional to [S]

Reaches a saturation value Vmax


Michaelis-Menten kinetics

An enzyme E combines with S to from an ES complex with a rate constant of K+1

In initial reaction where reverse reaction is minimal (we can effectively ignore K-2)

ES complex has two possible fates:

i.Dissociate to E + S with rate constant of K-1ii.Form P with rate constant of K+2

E + S ES E + P

k+1

k-1

k+2

k-2

Michaelis-Menten equation

The Michaelis-Menten equation relates the rate of catalysis to the concentration of substrate.

€

V0 =Vmax[S]

[S]+KMFor derivation seeStryer Chapter 8

Where

€

KM = k−1 + k2

k1

Michaelis constant

KM and Vmax are measurable constants for a particular enzymic reaction

Michaelis-Menten kinetics

Plotting initial velocity V0 of a reaction against substrate concentration [S] produces the Michaelis-Menten curve

Understanding KM

E + S ES E + P

k+1

k-1

k+2

At steady-state:

rate of formation of ES = rate of breakdown of ES

€

k+1[E][S] = (k−1 + k2)[ES]

[E][S][ES]

= (k−1 + k2)k+1

=KM

Significance of KM

For many reactions [S] < KM

€

V =Vmax[S]

[S]+KM

€

V = Vmax

KM[S]

X

Significance of KM

At high [S], [S]>>KM, V is effectively independent of [S]

€

V =Vmax[S]

[S]+KM

€

V =Vmax

XX

X

Significance of KM

• KM for any enzyme depends on pH, temp and ionic strength

• KM is equal to the concentration of substrate required for the reaction velocity to be half its maximal value

when [S] = KM then V = Vmax/2

€

Significance of KM

If k2 << k-1 then KM is equal to the dissociationconstant of the ES complex

E + S ES E + P

k+1

k-1

k+2X

€

KM = k−1 + k+2

k+1

= [E][S][ES]

=KESX

Large KM weak binding, small KM strong binding

Under these conditions KM tells us the enzyme-substrate affinity

Vmax

• number of substrate molecules converted into product by an enzyme molecule per unit time when enzyme is fully saturated

i.e [ES] = [ET], ET is total enzyme concentration

€

Vmax = k2[ET ]

Catalytic power

• An enzyme’s turnover is its catalytic power

• The maximum number of substrate molecules converted into product by an enzyme molecule in unit time

• When E is fully saturated, it is equal to the kinetic constant k2 – also known as kcat

We can calculate kcat from Vmax since Vmax= k2[ET]

Catalytic power

E + S ES E + P

k+1

k-1

k+2

Vmax=k2[ET]

€

V =Vmax[S]

[S] +KM

When [S] << KM

€

V = k2

KM[ET ][S]

i.e Rate will increase with [ET] and [S]

The max value of k2/KM is called the diffusion limit

The perfect enzyme

• The perfect enzyme is limited by diffusion

• Max possible value of k2/KM is limited by size of k+1 to between 106 and 109 M-1s-1

• In this case k+1 is effectively a measure of how often the enzyme collides with its substrate

The perfect enzyme

• An enzyme with a k2/KM of between 108 and 109 M-

1s-1 is therefore only limited by the rate of collisions and is called diffusion limited

• The catalytic processes of such enzymes are considered kinetically perfect because they are not slowing the enzyme's rate

• Any lower value of k2/KM suggests that the catalytic processes are slowing the rate

Reactions with multiple substrates

• Most biological reactions start with 2 substrates and yield 2 products

• Multiple substrate reactions can be classified into classes sequential reactions and double displacement (ping-pong) reactions

A + B P + Q

Sequential Reactions

• All substrates bind to enzyme forming ternary complex (can be ordered or random interaction)

A + B P + QE + A EA EA + B EAB

EAB EPQ EPQ EP + Q

EP E + P

Double Displacement Reactions

• Also called “ping-pong” reactions• One or more products released before all

substrates bind to enzyme

A + B P + Q

E A E’ + P E’ + B E’B

E’B E + Q

E + A EA

Limitations of Michealis-Menten

• M-M kinetics is simple model of enzymology

• Many examples of enzymes that do not follow M-M kinetics

e.g. Allosteric enzymes, which consist of multiple subunits and multiple active sites, V0 shows a sigmoidal dependence on [S]

Limitations of Michealis-Menten

Summary

• Enzymes form complexes with their substrates• Substrates bind to the active sites of enzymes• KM is the concentration of substrate required for

the reaction to be half the maximum value• Vmax is the maximum rate of reaction when the

enzyme is fully saturated with substrate


• You should be able to describe the evidence for enzyme/substrate complexes

• You will be able to define the terms Active sites and Active site residues

• You should be able to describe the Michaelis-Menten model for enzyme kinetics and the significance of KM and Vmax



Lecture 11Enzyme Catalysis


• You should be able to use Lineweaver-Burk plots to calculate KM and Vmax

• You will be able to describe competitive, uncompetitive and non-competitive inhibitors and how they alter enzyme kinetics

• You should understand the terms Allosteric and Co-operative regulation in relation to enzymes

Michaelis-Menten Reminder

€

V =Vmax[S]

[S] +KM

V is the velocity (or rate) of reaction[S] is the substrate concentrationVmax is the maximal velocity of the reactionKM is the Michaelis contstant ([S] at Vmax/2)

Michaelis-Menten Reminder

Michaelis-Menton Reminder

E + S ES E + P

k+1

k-1

k+2

At steady state: rate of formation of ES = rate of breakdown

€

k+1[E][S] = (k−1 + k2)[ES]

[E][S][ES]

= k−1 + k2

k+1

=KM

KM

• KM is equal to the concentration of substrate required for the reaction velocity to be half the maximal value

when [S] = KM then V = Vmax/2

When k2<<k-1 then KM is to the dissociation constant of the ES complex

i.e.

€

KM = k−1 + k2

k+1

= [E][S][ES]

=KESX

KM tells us about substrate affinity

Vmax

Is the number of substrate molecules converted per unit time when enzyme is fully saturated

i.e [ES] = [ET]

Vmax=k2[ET]

Where [ET] is total enzyme concentration

Lineweaver-Burk plot

• Method of calculating KM and Vmax

• A plot of 1/V0 against 1/[S]

• Values of KM and Vmax can be obtained from gradient of line and intercept on 1/[S] axis


€

V =Vmax[S]

[S]+KM

i.e. Y = c + mX (formula for straight line graph)

€

1V

= 1Vmax

+ KMVmax

1[S]


1/V

1/S

XX

XX

X

1/Vmax

-1/KM

“Double reciprocal plot”

slope = KM/Vmax

Enzyme Inhibition

• Inhibitors are molecules that prevent enzymes from working

• Many enzymes are regulated through the action of inhibitors

• Enzymes inhibitors can also act as medicinal drugs or toxins

Enzyme Inhibition

• Two main types of enzyme inhibition:

Irreversible, the inhibitor is tightly bound to the enzyme (sometimes covalently)

Reversible, inhibitor can bind and dissociate from the enzyme

Enzyme Inhibitors

• Competitive inhibitors bind to active site of enzyme

• Reduce the effective substrate concentration

Enzyme Inhibitors

• Non-competitive inhibitors stop enzyme from working

• Reduce the effective enzyme concentration

Enzyme Inhibitors

• Uncompetitive inhibitors bind only to ES complex

• Cannot be overcome by adding more substrate

Enzyme Inhibitors

• Some enzyme inhibitors are very important medicinal drugs

• Therefore understanding how inhibitors work is very important

Enzyme inhibitors as drugs

• Methotrexate, structural analogue of substrate for DHFR, prevents nucleotide synthesis, used to treat cancer (reversible inhibitor)

• Penicillin, covalently modifies transpeptidase, inhibiting bacterial cell wall synthesis and killing bacteria (Irreversible inhibitor)

Methotrexate• Substrate analog for Dihydroxyfolate reductase

(DHFR)• DHFR plays important role in biosynthesis of

purines and pyrimidines.

Methotrexate binds to enzyme 1000 times more effciently than dihyrdofolate

Significantly reduces effective substrate concentration

Penicillin

• Inhibits glycopeptide transpeptidase which cross-links bacterial cell walls

• Has similar structure to normal substrate, and binds at active site

Glycopeptide transpeptidase

• Enzyme essential for cell wall synthesis, without it no cross-linking

Penicillin

• Irreversibly inhibits enzyme by reacting with serine residue in active site

• “suicide”-substrate, reaction kills enzyme

Competitive Inhibition

€

K i = [E][I][EI]

Competitive Inhibition

€

K i = [E][I][EI]

Increasing inhibitor increases apparent KM Vmax can still be reached by adding more substrate

€

KMapp =KM 1+ [I]

K i

⎛ ⎝ ⎜

⎞ ⎠ ⎟50

Non-competitive Inhibition

Non-competitive Inhibition

Binding of molecule stops reaction, apparent Vmax is reduced

€

Vmaxapp = Vmax

1+ [I]/K i

No change in KM

Uncompetitive Inhibition

Uncompetitive InhibitionBinding of molecule to ES stops reaction, apparent

Vmax and KM are reduced

Both Vmax and KMlower


1/V

1/S

XX

XX

X

1/Vmax

-1/KM

“Double reciprocal plot”

slope = KM/Vmax

Lineweaver-Burk plots and Inhibitors

• Double reciprocal plots can be used to see effects of inhibitors

KM same, Vmax lower



KM higher, Vmax same



KM and Vmax lower

Transition state analogs

• In 1948 Linus Pauling proposed idea that compounds which mimic the transition state of enzymic reaction should be effective inhibitors


A

B

Energy(G)

ΔG

Transition state, S‡

ΔG‡ “Transition state analogs”

Transition state analogs

• Enzyme work by stabilising transition state

• TS analogs bind tightly to active site

• Very good competitive inhibitors

Proline BiosynthesisConversion of L-Proline to D-Proline requires

proline racemase

Reaction proceeds via TS, Pyrrole 2-carboxylic acid

resembles TS and is potent inhibitor

Catalytic antibodies

• Stabilising transition state should catalyse a reaction

• Antibodies which recognise a transition state should function as catalysts


• Insertion of Fe into porphyrin ring by ferrochelatase proceeds via ‘bent’ porphyrin transition molecule.


• N-Methylmesoporphyrin, used to generate antibodies

• Antibodies can catalyse ferrochelatase reaction• Antibodies catalysed reaction at 2500 times rate

of uncatalysed reaction, only10-fold less efficient than enzyme

Enzyme regulation

• Enzymes are often regulated

• Regulation can be allosteric and co-operative

• In co-operative regulation binding of substrate to one binding site helps binding to other active sites

• Allosteric regulation involves product inhibition, often used to control flux through metabolic pathways

Allosteric regulation

A B C D Product Xa b c d

regulates

Allosteric regulation

• Allosteric enzymes do not show Michaelis-Menton kinetics

• Due to the presence of multiple subunits and multiple active sites

• Multimeric enzymes often have sigmoidal kinetics

Allosteric kinetics

Hyperbolic Sigmoidal

Summary

• Lineweaver-Burk plots can be used to calculate KM and Vmax

• Inhibitors are substances that prevent enzymes from working

• Inhibitors can be competitive, uncompetitive or non-competitive

• Transition state analogues can be used as competitive inhibitors of enzymes

• Enzymes needs to be regulated, regulation can be co-operative and allosteric

Enzyme summary

• Without enzymes biological reactions would not take place

• Without inhibitors and regulation biological reactions would happen that we didn’t want


• You should be able to use Lineweaver-Burk plots to calculate KM and Vmax

• You will be able to describe Competitive, Uncompetitive and Non-competitive inhibitors and how they alter enzyme kinetics

• You should understand the terms Allosteric and Co-operative regulation in relation to enzymes

Documents

SBS017 Basic Biochemistry Dr. Jim Sullivan [email protected]