CP504 Ppt Set 03 DeterminationOfKineticParameters EnzymeReactions OK

Prof. R. Shanthini Updated: 23 Nov 2012

Enzyme kinetics and associated reactor design:

Determination of the kinetic parameters of

enzyme-induced reactions

- learn about the meaning of kinetic parameters- learn to determine the kinetic parameters- learn the effects of pH, temperature and substrate

concentration on enzyme activity (or reaction rates)- learn about inhibited enzyme kinetics- learn about allosteric enzymes and their kinetics

CP504 – ppt_Set 03


E + S ES E + Pk1

k2

k3

which is equivalent to

S

P[E]

S for substrate (reactant)

E for enzyme

ES for enzyme-substrate complex

P for product

Simple Enzyme Kinetics (in summary)


where rmax = k3CE0 = kcatCE0

and KM = f(rate constants)

- rS rmaxCS =

KM + CS rP =

S

P[E]

rmax is proportional to the initial concentration of the enzyme

KM is a constant



- rS rmaxCS =

KM + CS

Cs

rmax

rmax

2

KM

-rs

Catalyzed reactionCatalyzed reaction

uncatalyzed reaction



An exerciseConsider an industrially important enzyme, which catalyzes the conversion of a protein substrate to form a much more valuable product. The enzyme follows the Briggs-Haldane mechanism:

An initial rate analysis for the reaction in solution, with E0 = 0.10 μM

and various substrate concentrations S0, yields the following

Michaelis-Menten parameters: Vmax = 0.60 μM/s; KM = 80 μM.

A different type of experiment indicates that the association rate constant, k1, is k1 = 2.0 x 106 M-1s-1 (2.0 μM-1s-1).

a. Estimate the values of k2 and k-1.

b. On average, what fraction of enzyme-substrate binding events result in product formation?

Source: Jason Haugh, Department of Chemical & Biomolecular Engineering, North Carolina State University


E + S ES E + Pk1

k2

k3

Substrate binding step


Catalytic step

k3 = kcat





How to determine the kinetic parameters rmax and KM ?

Carry out an enzyme catalysed experiment, and measure the substrate concentration (CS) with time.

t Cs - rs

0 given given

10 given given

15 given given

rmaxCS =

KM + CS - rS


How to determine the M-M kinetics rmax and KM ?

Carry out an enzyme catalysed experiment, and measure the substrate concentration (CS) with time.

t Cs - rs

0 given given

10 given given

15 given given

rmaxCS =

KM + CS - rS


rmaxCS =

KM + CS - rS

We could rearrange

to get the following 3 linear forms:

=- rS

CS

rmax

KM

rmax

1+ CS

=- rS

1

rmax

KM

rmax

1+

CS

1

=- rSrmax KM-

CS

- rS

(15)

(14)

(16)


=- rS

CS

rmax

KM

rmax

1+

CS (14)

CS

- rS

CS

1rmax

- KM

The Langmuir Plot


=- rS

CS

rmax

KM

rmax

1+

CS (14)

CS

- rS

CS

1rmax

- KM

The Langmuir Plot

Determine rmax more accurately than the other plots.


(15)

- rS

1

KM

rmax

- KM

The Lineweaver-Burk Plot

=- rS

1

rmax

KM

rmax

1+

CS

1

CS

1

1


(15)

- rS

1

KM

rmax

- KM


=- rS

1

rmax

KM

rmax

1+

CS

1

CS

1

1

- Gives good estimates of rmax, but not necessarily KM

- Data points at low substrate concentrations influence the slope and intercept more than data points at high Cs


(16)

- rS

KM

KM

The Eadie-Hofstee Plot

CS

-rS

rmax

=- rSrmax KM-

CS

- rS


(16)

- rS

KM

KM


CS

-rS

rmax

=- rSrmax KM-

CS

- rS

- Can be subjected to large errors since both coordinates contain (-rS)

- Less bias on point at low Cs than with Lineweaver-Burk plot


CS

(mmol/l)

-rS

-(mmol/l.min)

1 0.20

2 0.22

3 0.30

5 0.45

7 0.41

10 0.50

Data:

Determine the M-M kinetic parameters for all the three methods discussed in the previous slides.


The Langmuir Plot

y = 1.5866x + 4.6417

R2 = 0.94970

5

10

15

20

25

0 2 4 6 8 10CS (mmol/l)

CS/(

-rS)

min

rmax = 1 / slope = 1 / 1.5866 = 0.63 mmol/l.min

KM = rmax x intercept = 0.63 x 4.6417 = 2.93 mmol/l



y = 3.4575x + 1.945

R2 = 0.84630

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 11/CS l/mmol

1/(

-rS)

l.min

/mm

ol

rmax = 1 / intercept = 1 / 1.945 = 0.51 mmol/l.min

KM = rmax x slope = 0.51 x 3.4575 = 1.78 mmol/l



y = -1.8923x + 0.5386

R2 = 0.6618

0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.05 0.1 0.15 0.2 0.25(-rS)/CS per min

(-r S

) m

mol

/l.m

in

rmax = intercept = 0.54 mmol/l.min

KM = - slope = 1.89 mmol/l


The Langmuir

Plot

The Lineweaver-

Burk Plot


rmax

KM

R2

Comparison of the results


The Langmuir

Plot

The Lineweaver-

Burk Plot


rmax 0.63 0.51 0.54

KM 2.93 1.78 1.89

R2 94.9% 84.6% 66.2%



The Langmuir

Plot

The Lineweaver-

Burk Plot


rmax 0.63 0.51 0.54

KM 2.93 1.78 1.89

R2 94.9% 84.6% 66.2%

Determine rmax more

accurately than the other plots

Gives good estimates of rmax, but not

necessarily KM

Can be subjected to large errors





http://www.youtube.com/watch?v=D2j2KGwJXJc


http://academic.brooklyn.cuny.edu/biology/bio4fv/page/enz_act.htm

Effects of temperature on enzyme activity:

Increases in the temperature of a system results from increases in the kinetic energy of the system.

Kinetic energy increase has the following effects on the rates of reactions:

1) More energetic collisions

2) Increase in the number of collisions per unit time

3) Denaturation of the enzyme or substrate




More energetic collisions:

When molecules collide, the kinetic energy of the molecules can be converted into chemical potential energy of the molecules.

If the chemical potential energy of the molecules become great enough, the activation energy of a exergonic reaction can be achieved and a change in chemical state will result.

Thus the greater the kinetic energy of the molecules in a system, the greater is the resulting chemical potential energy when two molecules collide.

As the temperature of a system is increased it is possible that more molecules per unit time will reach the activation energy.

Thus the rate of the reaction may increase.




Increase in the number of collisions per unit time:

In order to convert substrate into product, enzymes must collide with and bind to the substrate at the active site.

Increasing the temperature of a system will increase the number of collisions of enzyme and substrate per unit time.

Thus, within limits, the rate of the reaction will increase.


http://www.woisd.net/moodle/mod/resource/view.php?id=44


Denaturation of the enzyme:

Enzymes are very large proteins whose three dimensional shape is vital for their activity.

When proteins are heated up too much they vibrate.

If the heat gets too intense then the enzymes literally shake themselves out of shape, and the structure breaks down.

The enzyme is said to be denatured.

A denatured enzyme does not have the correct 'lock' structure.

Therefore it cannot function efficiently by accepting the 'key' substrate molecule.







As temperature increases, enzyme activity increases until its optimum temperature is reached. At higher temperatures, the enzyme activity rapidly falls to zero.



Denaturation for most human enzymes:

http://www.woisd.net/moodle/mod/resource/view.php?id=44

Optimal for most human enzymes

The optimum temperature for most human enzymes to work at is around 37ºC which is why this temperature is body temperature.

Enzymes start to denature at about 45°C.


https://wikispaces.psu.edu/display/230/Enzyme+Kinetics+and+Catalysis

Temperature (deg C)

Rea

ctio

n r

ate

Optimal for most human enzymes

Optimal for some thermophillic bacterial enzymes



Effects of pH on enzyme activity:

The structure of the protein enzyme can depends on how acid or alkaline the reaction medium is, that is, it is pH dependent.

If it is too acid or too alkaline, the structure of the protein is changed and it is 'denatured' and becomes less effective.

If the enzyme does not have the correct 'lock' structure, it cannot function efficiently by accepting the 'key' substrate molecule.

In the optimum pH range, the enzyme catalysis is at its most efficient.


https://wikispaces.psu.edu/display/230/Enzyme+Kinetics+and+Catalysis

pH

Rea

ctio

n r

ate

Optimal for pepsin (a stomach enzyme)

Optimal for trypsin (an intestinal enzyme)




Amylase (pancreas) enzyme

Optimum pH: 6.7 - 7.0

Function: A pancreatic enzyme that catalyzes the breakdown/hydrolysis of starch into soluble sugars that can readily be digested and metabolised for energy generation.

www.docbrown.info/page01/ExIndChem/ExIndChema.htm

Amylase (malt) enzyme


Function: Catalyzes the breakdown/hydrolysis of starch into soluble sugars in malt carbohydrate extracts.



Catalase enzyme

Optimum pH: ~7.0

Function: Catalyses the breakdown of potentially harmful hydrogen peroxide to water and oxygen. Important in respiration/metabolism chemistry.

2H2O2(aq) ==> 2H2O(l) + O2(g)




Invertase enzyme

Optimum pH: 4.5

Function: Catalyses the breakdown/hydrolysis of sucrose into fructose + glucose, the resulting mixture is 'inverted sugar syrup'.

C12H22O11 + H2O ==> C6H12O6 + C6H12O6




Lipase (pancreas) enzyme

Optimum pH: ~8.0

Function: Lipases catalyse the breakdown dietary fats, oils, triglycerides etc. into digestible molecules in the human digestion system.


Lipase (stomach) enzyme


Function: As above, but note the significantly different optimum pH in the acid stomach juices, to optimum pH in the alkaline fluids of the pancreas.



Maltase enzyme


Function: Breaks down malt sugars.




Pepsin enzyme


Function: Catalyses the breakdown/hydrolysis of proteins into smaller peptide fragments.




Trypsin enzyme


Function: Catalyses the breakdown/hydrolysis of proteins into amino acids. Note again, the significantly different optimum pH to similarly functioning pepsin.




Urease enzyme

Optimum pH: ~7.0

Function: Catalyzes the breakdown of urea into ammonia and carbon dioxide.

(NH2)2(aq) + H2O(l) ==> 2NH3(aq) + CO2(aq)



Effects of substrate concentration on enzyme activity:



Effect of shear


Complex enzyme kinetics




Inhibited enzyme reactions

Inhibitors are substances that slow down the rate of enzyme catalyzed reactions.

There are two distinct types of inhibitors:

- Irreversible inhibitors form a stable complex with enzymes and reduce enzyme activity (e.g. lead, cadmium,

organophosphorous pesticide)

- Reversible inhibitors interact more loosely with enzymes and can be displaced.


Inhibited enzyme reactions - applications

Many drugs and poisons are inhibitors of enzymes in the nervous system.

Poisons: snake bite, plant alkaloids and nerve gases

Medicines: antibiotics, sulphonamides, sedatives and stimulants


Primary constituents of Snake Venom

EnzymesEnzymes - Spur physiologically disruptive or destructive processes.Proteolysins - Dissolve cells and tissue at the bite site, causing local pain and swelling.Cardiotoxins - Variable effects, some depolarise cardiac muscles and alter heart contraction, causing heart failure.Harmorrhagins - Destroy capillary walls, causing haemorrhages near and distant from the bite.Coagulation - Retarding compounds prevent blood clotting.Thromboses - Coagulate blood and foster clot formation throughout the circulatory system.Haemolysis - Destroy red blood cells.Cytolysins - Destroy white blood cells.Neurotoxins - Block the transmission of nerve impulses to muscles, especially those associated with the diaphragm and breathing.

http://www.writework.com/essay/biochemistry-snake-venom


Inhibited enzyme reactions

Inhibitors are also classified as competitive and non-competitive inhibitors.


Competitive inhibition

- The structure of inhibitor molecule closely resembles the chemical structure and molecular geometry of the substrate.

- The inhibitor competes for the same active site as the substrate molecule.

- It does not alter the structure of the enzyme.

- The inhibitor may interact with the enzyme at the active site, but no reaction takes place.

http://www.elmhurst.edu/~chm/vchembook/573inhibit.html



- The inhibitor is "stuck" on the enzyme and prevents any substrate molecules from reacting with the enzyme.

- However, a competitive inhibition is usually reversible if sufficient substrate molecules are available to ultimately displace the inhibitor.

- Therefore, the amount of enzyme inhibition depends upon the inhibitor concentration, substrate concentration, and the relative affinities of the inhibitor and substrate for the active site.

http://www.elmhurst.edu/~chm/vchembook/573inhibit.html



Competitive inhibitors (denoted by I) compete with substrate to occupy the active site of the enzyme.

E + S ES E + Pk1

k2

k3

E + I EIk4

k5

rP = k3 CES (17)

CE0 = CE + CES + CEI

where

(18)



Assuming rapid equilibrium, we get

k1 CE CS = k2 CES

k4 CE CI = k5 CEI

k2

k1 KM =

CE CS

CES =

k5

k4 KI =

CE CI

CEI =

(19)

(20)



Combining (17) to (20), we get

k3CE0CSrP =

rmaxCS =

KM,app + CS (21)

KM (1 + CI / KI) + CS

where

KM,app = KM (1 + CI / KI) (22)

KM = k2 / k1 (6)

(5)rmax = k3CE0

KM,app > KM



- rS

1

- KM


rmax

1

CS

1

1 - KM, app

1 CI = 0 (no inhibitor)

CI > 0



In the presence of a competitive inhibitor,

the maximal rate of the reaction (rmax) is unchanged,

but the Michaelis constant (KM) is increased.


Competitive inhibition – an example

Ethanol is metabolized in the body by oxidation to acetaldehyde, which is a toxic compound and a known carcinogen.

The enzyme alcohol dehydrogenase (ADH) converts ethanol into acetaldehyde plus two

hydrogen atoms.



Acetaldehyde is generally short-lived; it is quickly broken down to a less toxic compound called acetate in a rapid reaction so that acetaldehyde does not accumulate in the body.

.

The enzyme aldehyde dehydrogenase (ALDH) converts acetaldehyde to acetyl (acetate) radical and a hydrogen atom.



A drug, disulfiram (Antabuse) inhibits the aldehyde dehydrogenase.

Such inhibition results in the accumulation of acetaldehyde in the body.

High levels of acetaldehyde act directly on the heart and blood vessels, causing flushing, a racing heartbeat and a drop in blood pressure that causes dizziness. Other unpleasant symptoms include headache, shortness of breath, palpitations, nausea and vomiting.

This drug is sometimes used to help people overcome the drinking habit.


Non-competitive inhibition

https://ibhumanbiochemistry.wikispaces.com/C.7.5

- The structure of inhibitor molecule is entirely different from that of the substrate molecule.

- The inhibitor forms complex at a point other than the active site (remote from or very close to the active site).

- It does not complete with the substrate.

- It alters the structure of the enzyme in such a way that the substrate can no longer interact with the enzyme to give a reaction.



https://ibhumanbiochemistry.wikispaces.com/C.7.5

- Non competitive inhibitors are usually reversible,

- but are not influenced by concentrations of the substrate as is the case for a reversible competitive inhibitor.



E + S ES E + Pk1

k2

k3

E + I EIk4

k5

EI + S ESIk6

k7

ES + I ESIk8

k9



k2

k1 = KM =

We could drive the rate equation (given on the next page) assuming the following:

k7

k6 = KIM

k5

k4 = KI =

k9

k8 = KMI



rP = rmax,appCS

KM + CS (23)

where

KM = k2 / k1 (6)

(5)rmax = k3CE0

rmax,app < rmax

rmax,app =(1 + CI / KI)

rmax(24)



- rS

1

- KM


rmax

1

CS

1

1

CI = 0 (no inhibitor)

CI > 0

rmax,app

1



In the presence of a non-competitive inhibitor,

the maximal rate of the reaction (rmax) is lower

but the Michaelis constant (KM) is unchanged.


Uncompetitive inhibition

E + S ES E + Pk1

k2

k3

ES + I ESIk4

k5

Inhibitor can only bind to the enzyme-substrate complex, reversibly forming a nonproductive complex.



An uncompetitive inhibitor binds only to the enzyme-substrate complex preventing the formation or release of the enzymatic products.

Unlike with competitive inhibition an uncompetitive inhibitor need not resemble the structure of the enzymes natural substrate.

An uncompetitive inhibitor is most effective at high substrate concentration as there will be more enzyme-substrate complex for it to bind.

Unlike with competitive inhibitors the effects of an uncompetitive inhibitor cannot be overcome by increasing the concentration of substrate.



rP = rmax,appCS

KM + CS (23)

where

KM = k2 / k1 (6)

(5)rmax = k3CE0

rmax,app < rmax


rmax(24)



rP = rmax,appCS

KM,app + CS (25)

KM,app = KM / (1 + CI / KI) (26) KM,app < KM

where


rmax(24)

KM = k2 / k1 (6)

(5)rmax = k3CE0

rmax,app < rmax



KM is reduced

rmax is also reduced

This is because the total ‘pool’ of enzymes available to react has been reduced, effectively our enzyme concentration has reduced.

Can be explained by rmax = k3CE0 = kcatCE0



- rS

1

- KM


rmax

1CS

11

CI = 0 (no inhibitor)

CI > 0

rmax,app

1

- KM, app

1


Competitive versus Uncompetitive inhibition


Mixed inhibition


An exerciseThe kinetic properties of the ATPase enzyme, isolated from yeast, which catalyzes the hydrolysis of ATP to form ADP and P i, are

assessed by measuring initial rates in solution, with various ATP concentrations S0 and a total ATPase concentration E0 = 0.60 μM.

From these experiments, it is determined that

Vmax = 1.20 μM/s; KM = 40 μM.

a. Calculate the values of kcat and the catalytic efficiency for ATPase

under these conditions.

b. An inhibitor molecule is added at a concentration of 0.1 mM, and the experiments are repeated. The apparent Vmax and KM are now

found to be 0.6 μM/s, and 20 μM, respectively. Speculate on how this inhibitor works (i.e., specify which species are engaged by the inhibitor).

Source: Jason Haugh, Department of Chemical & Biomolecular Engineering, North Carolina State University


Substrate / Product inhibition

Either the substrate or product of an enzyme reaction inhibit the enzyme's activity.

This inhibition may follow the competitive, uncompetitive or mixed patterns.

In substrate inhibition there is a progressive decrease in activity at high substrate concentrations.

Product inhibition is often a regulatory feature in metabolism and can be a form of negative feedback.


Substrate / Product inhibition


Assignment

Get the rate equations for substrate and product inhibition


“Food for Thought”

Problem 3.13 from Shuler & Kargi:

The following substrate reaction rate (-rS) data were obtained from enzymatic oxidation of phenol by phenol oxidase at different phenol concentrations (CS).

By plotting (-rS) versus (CS) curve, or otherwise, determine the type of inhibition described by the data provided?

CS

(mg/l)

-rS

(mg/l.h)

10 5

20 7.5

30 10

50 12.5

60 13.7

80 15

90 15

110 21.5

130 9.5

140 7.5

150 5.7


Sigmoid/Hill kinetics

A particular class of enzymes exhibit kinetic properties that cannot be studied using the Michaelis-Menten equation.

The rate equation of these unique enzymes is characterized by Sigmoid/Hill kinetics as follows:

rP = rmaxCS

n

K + CSn

(27)

n = 1 gives Michaelis-Menten kinetics

n > 1 gives positive cooperativity

n < 1 gives negative cooperativity

http://chemwiki.ucdavis.edu/Biological_Chemistry/Catalysts/Enzymatic_Kinetics/Sigmoid_Kinetics

The Hill equation

Hill coefficientHill constant


Sigmoid/Hill kinetics

Examples of the “S-shaped” sigmoidal/Hill curve, which is different from the hyberbolic curve of M-M kinetics.

n = 2n = 4

n = 6


Sigmoid kinetics

1 - θ

CSn

K + CSn

(28)


For an alternative formulation of Hill equation, we could rewrite (25) in a linear form as follows:

θln = n ln(CS) – ln (K)

rmax θ = =

rP


Allosteric enzyme


Find out what it is on your own

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CP504 Ppt Set 03 DeterminationOfKineticParameters EnzymeReactions OK