ENZYMEENZYME

KINETICSKINETICS

Enzyme ActionEnzyme Action

• Each enzyme has a Each enzyme has a unique three-dimensional shape unique three-dimensional shape that binds and recognizes a group of reacting that binds and recognizes a group of reacting molecules called substrates.molecules called substrates.

• The The active siteactive site of the enzyme is a small pocket to of the enzyme is a small pocket to which the substrate directly binds.which the substrate directly binds.

• Some enzymes are Some enzymes are specific only specific only to one substrate; to one substrate; others can bind more than one substrate.others can bind more than one substrate.

Enzyme-Substrate BindingEnzyme-Substrate Binding

Models of Models of Enzyme ActionEnzyme Action

• Early theory: Early theory: lock-and-key modellock-and-key model. Active site (lock) . Active site (lock) had the same shape as the substrate (key). Only the had the same shape as the substrate (key). Only the right shape key could bind.right shape key could bind.

• Current theory: Current theory: induced fit modelinduced fit model. Active site . Active site closely resembles but does not exactly bind the closely resembles but does not exactly bind the substrate. substrate. – Allows for Allows for more flexibilitymore flexibility in type of substrate in type of substrate– Also Also explainsexplains how the reaction itself occurs. As the how the reaction itself occurs. As the

substrate flexes to fit the active site, bonds in the substrate substrate flexes to fit the active site, bonds in the substrate are flexed and stressed -- this causes changes/conversion are flexed and stressed -- this causes changes/conversion to product.to product.

Molecular Molecular RecognitionRecognition

How does an enzyme bind a substrate, reduce the activation barrier, and produce a How does an enzyme bind a substrate, reduce the activation barrier, and produce a product?product?

Lock & Key HypothesisLock & Key Hypothesis Induced Fit HypothesisInduced Fit Hypothesis

vs.

C. C. Factors Affecting Factors Affecting Enzyme ActivityEnzyme Activity

• Enzyme activity is Enzyme activity is defined as defined as how fast an enzyme how fast an enzyme catalyzes its reaction.catalyzes its reaction.

• Many Many factors affect factors affect enzyme activity:enzyme activity:– TemperatureTemperature: most have an optimum temp around 37: most have an optimum temp around 37ooCC– pHpH: most cellular enzymes are optimal around physiological : most cellular enzymes are optimal around physiological

pH, but enzymes in the stomach have a lower optimum pHpH, but enzymes in the stomach have a lower optimum pH– ConcentrationConcentration of enzyme and substrate: have all of the of enzyme and substrate: have all of the

enzyme molecules been used up, even though substrate is enzyme molecules been used up, even though substrate is still available?still available?

Energy of activation: ΔG‡

ΔG‡

ΔGcat‡

ΔGT1‡

ΔGT2‡

(T1 > T2)

Effect of temp

Effect of catalysis

A → B

A → B

Reduced entropy in ES formation.

Destabilization of ES: strain, charge, electrostatics

Stabilization of the transition state: covalent bonds, metals, acid-base, and proximity.

Rate acceleration: mechanisms

Rate acceleration: mechanisms

hydrolysis of a β-glycosidic bond yielding a unit of α-glucose

Major factors: pH, ions, & temp

At pH ~ 7 amino acids exist as zwitterions.

The R group determines pH.

aspartic acid [pKa = 4.0]

arginine [pKa = 12.5]

Major factors: pH, ions, & temp

• pH

• ionic strength

• temperature

barley α-amylase activity plotted as a function of pH

Major factors: pH, ions, & temp

• pH

• ionic strength

• temperature

Having the correct ions is important. Why? barley α-amylase isozyme 1

[crystallized with Ca2+ (green)]

Major factors: pH, ions, & temp

• pH

• ionic strength

• temperature

barley α-amylase with CaCl2

barley α-amylase w/o CaCl2

Michaelis-Menten Kinetics

Assumptions:[1] Steady-state of the intermediate complex ES[2] Neglect back rxn from product (k-2; not shown)[3] Conservation of mass ([ET] = [E] + [ES])

ν = Vmax [S]

Km + [S]

Vmax = k2[ET]where:

Km =(k-1 + k2)

k1

E + S ↔ ES → E + P

k1

k-1

k2

Michaelis-Menten Kinetics

Michaelis-Menten Kinetics

Many types of inhibition can be included in the MM model as well as multiple substrates and steps:

Inhibition:competitive (rev)noncompetitive (rev)mixed (rev)irreversible

Reaction Schemes:single substratemultiple substratesingle displacementdouble disp (ping-pong)

Reaction Rate vs. Enzyme and Substrate Conc.

Control of Enzyme Activity

• We don’t always need high levels of products of enzyme-catalyzed reactions around. What kind of control system is used to regulate amounts of enzyme and products?

• Two main methods: zymogens, and feedback control.

Zymogens

• Many enzymes are active as soon as they’re made.• However, some are made in an inactive form and stored.

This inactive form is called a zymogen or proenzyme. • To become active, the body needs only to cleave off a

small peptide fragment.

Feedback Control• Some enzymes (allosteric enzymes) bind molecules

called regulators (different from the substrate) that can affect the enzyme either positively or negatively– Positive regulator: speeds up the reaction by changing the

shape of the active site -- substrate binds more effectively– Negative regulator: slows down reaction by preventing proper

substrate binding, again, by changing enzyme shape

• Feedback control: the end product acts as a negative regulator. If there is enough of the end product, it will slow down the first enzyme in a pathway.

The kinetics of enzyme catalysis:

Steady state kinetics

A hyperbolic curve between V0 and [S] was revealed by in vitro studies using purified enzymes

• It was the initial velocity (rate), V0, that was measured, so the change of [S] could be ignored.

• The catalysis was assumed to occur as:

• The enzyme will become saturated at high [S]: the V0 will not be affected by [S] at high [S].

Hyperbolic relationship between V0 and [S]

The effect on V0 of varying [S] is measured when the enzyme

concentration is held constant.

Vmax is extrapolated from the plot: V0 approaches but never quite reaches Vmax.

A mathematical relationship between V0 and [S] was established (Michaelis and Menten, 1913; Briggs and Haldane, 1925)

• E + S ES E + P– Formation of ES is fast and reversible.– The reverse reaction from PS (k-2 step) was

assumed to be negligible. – The breakdown of ES to product and free enzyme

is the rate limiting step for the overall reaction.– ES was assumed to be at a steady state: its

concentration remains constant over time.• Thus V0 = k2[ES]

k 1 k 2

( )

• Steady-state assumption:– Rate of ES formation=rate of ES breakdown– k1([Et]-[ES])[S]=k-1[ES] + k2[ES] ([Et] is the total enzyme concentration.)

• Solve the equation for [ES]:

Km is called the Michaelis constant.

k 1

k -1

k 2

V0 = k2[ES]

The maximum velocity is achieved when all the enzyme is saturated by substrate, i.e., when [ES] =[Et]. Thus Vmax =k2[Et]

The Michaelis-Menten Equation

The Michaelis-Menten Equation nicelydescribes the experimental observations.

When [S] << KmWhen [S] >> Km

The substrate concentration at which V0 is half maximal is Km

The Vmax and Km values of a certain enzyme can be measured by the double reciprocal plot (i.e., the Lineweaver-Burk plot).

The double reciprocal plot:

1/V0 vs 1/[S]

The Michaelis-Menten equation, but not their approximated mechanism applies to a

great many enzymes

• Most enzymes (except the regulatory enzymes) have been found to follow the Michaelis-Menten kinetics, but their actual mechanisms are usually more complicated (by having more intermediate steps) than the one assumed by Michaelis and Menten.

• The values of Vmax and Km alone provide little information about the number, rates, or chemical nature of discrete steps in the reaction.

The actual meaning of Km depends on the reaction mechanism

• For• If k2 is rate-limiting, k2<<k-1,

Km = k-1/k 1

– Km equals to the dissociation constant (Kd) of the ES complex;

– Km represent a measure of affinity of the enzyme for its substrate in the ES complex.

k -1

k 1

PLOT EADIE-HOFSTEE DAN HANES - WOOLFPLOT EADIE-HOFSTEE DAN HANES - WOOLF

Plot Lineweaver – Burk mempunyai sedikit kelemahan, yaitu•Sering kali pada saat mengekstrapolasi grafik untuk menentukan harga -1/Km ternyata akan memotong sumbu 1/[S] di luar grafik yang dibuat•Pada konsentrasi substrat yang terlalu rendah, maka akan diperoleh hasil yang kurang akurat•Awal dari kelinearannya sering kurang jelas dibanding dengan plot lain, terutama plot Eadie – Hofstee, padahal hal ini sangat penting pada penentuan mekanisme reaksi

Plot Eadie-Hofstee dan Hanes diturunkan dari persamaan Lineweaver-Burk dengan mengalikan kedua sisi persamaan dengan faktor vo Vmax

sehingga akan diperoleh persamaan garis lurus selanjutnya dipergunakan untuk menghitung Vmax dan Km

Dengan cara penurunan yang mirip, Hanes-Woolf mengalikan perasamaan Lineweaver-Burk dengan [So] maka diperoleh:

Plot Eadie – Hofstee dan Hanes banyak digunakan pada studi kinetik enzim, namun demikian studi enzim secara umum masih menggunakan plot Lineweaver – Burk.

– Rewrite Michaelis-Menten rate expression

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

Lineweaver-Burk (double reciprocal plot)

1v

Km

Vmax

1[S ]

1

Vmax

Graphical Solution

1/ V

1/ [S]

1/ Vmax

-1/ Km

1v

Km

Vmax

1[S ]

1

Vmax

Slope = Km/ Vmax

intercepts

Example: Lineweaver-Burk

[S] x 10-5 M V, M/min x 10-5 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 PlotLineweaver-Burk Plot

y = 0.5686x + 2.8687

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

0.00 2.00 4.00 6.00 8.00 10.00 12.00

1/[S] x 10^(-4)

slope = Km/ Vmax= 0.5686

y intercept = 1/ Vmax= 2.8687

Michaelis-Menten Kinetics

v Vmax[S ]Km [S ]

k 2[E0][S ]Km [S ]

Fit to Data

0.00

0.50

1.00

1.50

2.00

2.50

3.00

0.00 1.00 2.00 3.00 4.00 5.00 6.00

[S] (M) x 10^(-5)

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

• Hanes- Woolf

[S ]v

Km

Vmax

1Vmax

[S ]

v Vmax Km

v[S ]

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!

PERSAMAAN HALDAN UNTUK REAKSI REVERSIBEL

• Reaksi enzimatis dalam sel sering berlangsung secara reversibel.

• Reaksi substrat tunggal, S P, berlangsung melalui pembentukan satu kompleks intermediate, arah ke kanan dianggap sebagai kompleks ES dan sebaliknya kompleks EP

E + S ES/EP P + E

• Persamaan MM arah kekanan pada [Eo] tetap dengan laju awal vf dan Vs

max

• Persamaan MM ke arah sebaliknya pada [Eo] tetap dengan laju awal vb dan VP

max

• Perumusan Haldan hubungan antara konstanta laju dan kesetimbangan reaksi pada reaksi kesetimbangan adalah

• Karena

• Maka:

• Bila konstanta kesetimbangan diketahui, maka persamaan tersebut dapat digunakan untuk memvalidasi konstanta laju yang diperoleh

• Secara umum Km dari arah reaksi metabolisme penting akan sedikit lebih kecil dari arah sebaliknya. Namun arah metabolisme dipengaruhi juga oleh [S] dan [P] dalam sel

KINETIKA REAKSI CEPAT

• Kinetika keadaan steady penting bagi enzimologis yang memungkinkan untuk menentukan Km dan Kcat

• Bilangan peredaran (turn over number), Kcat pada beberapa enzim pada tingkat 100 s-1, yaitu 100 molekul produk dihasilkan setiap detik tiap molekul enzim. Hal ini berarti bahwa tahapan yang paling lambat dari suatu mekanisme reaksi mempunyai waktu hidup hanya beberapa detik

• Untuk reaksi sederhana order pertama A B

• Maka laju reaksi pada t adalah-d[A]/dt = d[B]/dt = k[A]

• Integrasi dari persamaan tersebut memberikan persamaan

ln[Ao] – ln [A] = kt[A] = [Ao]e –kt

konsentrasi

waktu

B

A

Kurva teoritis perubahan reaksi A menjadi B

• Persamaan integrasi dari sebagian besar mekanisme reaksi akan lebih rumit . Untuk reaksi substrat tunggal yang membentuk suatu kompleks intermediate dengan konsentrasi awal dari [S] >>> [E] dinyatakan sebagai:

E + S ES E + P

• Laju penambahan [ES] pada waktu t (pada periode awal dimana pembentukan [P] diabaikan dinyatakan sebagai:d[S]/dt = [E][S] - [ES] - [ES] d[S]/dt = ([Eo] - [ES])([So] – [ES] – [P]) - [ES] - [ES]karena : [So]>>[Eo], maka ([So]-[ES]-[P] ~ [So]Jadi : d[S]/dt = ([Eo] - [ES])([So] ) - [ES] - [ES]

k1k2

k-1 k-2

k-1k1 k2

k1 k-1k2

k1 k-1k2

• Integrasi persamaan tersebut akan menunjukkan perubahan ES terhadap waktu

PS

E ES

E + S ES EP E + P dimana [So] >> [Eo]

Waktu

Konstr.

Induction period

Transient or pre steady state phase

Steady state phase

• Bagian linier dari grafik [P] vs t menunjukkan fase steady state dari reaksi dengan slope = k2[E][So]/([So]/Km) yang didapat dari mensubstitusi (k-1 + k2)/k1 dengan Km dari persamaan integrasinya

• Jika bagian grafik steady state linier, ekstrapolasinya akan memotong sumbu t pada t = 1/(k1[So]+Km) dan disebut sebagai periode induksi.

• Kurva perubahan konsentrasi akan lebih rumit bila reaksinya seperti;

• Dengan tahap laju yang menentukan (rate-limiting-step) adalah EP menjadi E dan P.

E + S ES EP E + P

P

E

ES

Waktu

Konstr.

E + S ES EP E + P dimana EP menjadi E dan P sebagai penentu

EP

Vmax is determined by kcat, the rate constant of the rate-limiting step

• Vmax = kcat[Et]• kcat equals to k2 or k3 or

a complex function of both, depending on which is the rate-limiting step.

• kcat is also called the turnover number: the number of substrate molecules converted to product in a given unit of time per enzyme molecule when the enzyme is saturated with substrate.

40,000,000 molecules of H2O2 are convertedto H2O and O2 by one catalase molecule within one

second!

The kinetic parameters kcat and Km are often studied and compared for

different enzymes

• Km often reflects the normal substrate concentration present in vivo for a certain enzyme.

• The catalytic efficiency of different enzymes is often compared by comparing their kcat/Km ratios (the specificity constant).

•

•

• kcat/Km is an apparent second-order rate constant (with units of M-1S-1), relating the reaction rate to the concentrations of free enzyme and substrate.

The value of kcat/Km has an upper limit (for the perfected enzymes)

• It can be no greater than k1.

– The decomposition of ES to E + P can occur no more frequently that E and S come together to form ES.

• The most efficient enzymes have kcat/Km values near the diffusion-controlled limit of 108 to 109 M-1S-1.

Catalytic perfection (rate of reaction being diffusion-controlled) can be achieved by a

combination of different values of kcat and Km.

Rate enhancement is often used to describe the efficiency of an enzyme

Rate enhancement: ratio of the rates of the catalyzed and the uncatalyzed reactions.

kcat

kcat

catalyzed

uncatalyzed

Nonenzymatic half-life

Uncatalyzed rate

(kun, s-1)

Catalyzed rate

(kcat, s-1)

Rate enhancement

(kcat/kun)Enzyme

Rate enhancement by selected enzymes

Enzyme-catalyzed reactions of two or more substrates can also be analyzed by the Michaelis-Menten approach

• Each substrate will have one characteristic Km value.• Noncovalent ternary complex (with two substrates

bound to the enzyme concurrently) may or may not be formed for the bisubstrate reactions depending on the mechanism.

• Steady-state kinetics can often help distinguish these two mechanisms.

In those enzyme-catalyzed bisubstrate reactions where aternary complex is formed, the two substrates may either bind in a random sequence or in a specific order.

Maintaining the concentrationof one substrate (S2) constant,the double reciprocal plots made by varying the concentrationof the other substrate (S1) willintersect.

For those reactionswhere ternary complexis formed:

No ternary complex is formed in the Ping-Pong (or double displacement)

mechanism: The first substrate is converted to a product that leaves the enzyme active site before the second

substrate enters.

For enzymes having Ping-Pong mechanisms (ternary complex not formed).

Maintaining the concentrationof one substrate (S2) constant,the double reciprocal plots made by varying the concentrationof the other substrate (S1) will notintersect.

As [S2] increases, Vmax increases,as does the Km for S1.

S1

Rates of individual steps for an enzyme-catalyzed reaction may be obtained by pre-steady state kinetics

• The enzyme (of large amount) is used in substrate quantities and the events on the enzyme are directly observed.

• Rates of many reaction steps may be measured independently.

• Very rapid mixing and sampling techniques are required (the enzyme and substrate have to be brought together in milliseconds and measurements also be made within short period of time).

“Rapid kinetics” or “pre-steady-state kinetics”is applied to the observation of rates of systems that occur in very short time intervals (usually ms or sub-ms scale ) and very low product concentrations. This period covers the time from the enzyme encountering its target (either a substrate, inhibitor or some other ligands) to the point of system settling to equilibrium.

The concentration of ES will rise from zero to its steady-state value.

(ms or sub-ms)

• specific activity is the amount of product formed by an enzyme in a given amount of time under given conditions per milligram of enzyme.

• The rate of a reaction is the concentration of substrate disappearing (or product produced) per unit time (mol L − 1s − 1)

• The enzyme activity is the moles converted per unit time (rate × reaction volume). Enzyme activity is a measure of quantity of enzyme present. The SI unit is the katal, 1 katal = 1 mol s-1, but this is an excessively large unit. A more practical value is 1 enzyme unit (EU) = 1 μmol min-1 (μ = micro, x 10-6).

• The specific activity is the moles converted per unit time per unit mass of enzyme (enzyme activity / actual mass of enzyme present). The SI units are katal kg-1, but more practical units are μmol mg-1 min-1. Specific activity is a measure of enzyme efficiency, usually constant for a pure enzyme.

• If the specific activity of 100% pure enzyme is known, then an impure sample will have a lower specific activity, allowing purity to be calculated.

• The % purity is 100% × (specific activity of enzyme sample / specific activity of pure enzyme). The impure sample has lower specific activity because some of the mass is not actually enzyme.

Turnover number

The turnover number of carbonic anhydrase: Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers among known enzymes, it catalyses the reversible reaction of CO2:

H2O + CO2 -> H2CO3

This is an important process in the transport of CO2 from the tissues to the lungs. If 10μg of pure

carbonic anhydrase catalyses the hydration of 0.30g of CO2 in 1min at 37°C at Vmax, and the reaction volume is 1ml. What is the turnover

number (Kcat) of carbonic anhydrase expressed in units of per min and per sec)? Mr of CO2 is 44.