CHM 414

ENZYME CATALYSIS

• Introduction

• Characteristics of enzymes

• Active sites of enzymes

• Models for explaining the enzyme mechanism of reaction

• -Lock and key model

• -Induced-fit model

• Michaeli - Menten method of enzyme catalysis

• Significance of Michael- Menten equation

Enzymes-Introduction

• Enzymes are biological catalysts. • They are special proteins or nucleic acids that contain an active site,

which is responsible for binding the substrates, the reactants, and processing them into products.

• As is true of any catalyst, the active site returns to its original state after the products are released.

• Many enzymes consist primarily of proteins, some featuring organic or inorganic co-factors in their active sites. However, certain RNA molecules can also be biological

• An enzyme can accelerate a reaction because it can provide a proper environment to lower the energy barrier (activation energy) that is required to reach the highest energy transition state of the reaction. Generally, only a few molecules have sufficient energy for reaction

Characteristics of enzyme catalysis

• A single molecule of this catalysis can transform a million molecules of the reactant per second. Hence it is highly efficient.

• These biochemical catalysts are specific in nature i.e. the same catalyst cannot be used in more than one reaction.

• The effectiveness of a catalyst is maximum at its optimum temperature. The activity of the biochemical catalysts declines at either side of the optimum temperature.

• Biochemical catalysis is dependent upon the pH of the solution. A catalyst works best at an optimum pH between 5-7 .

• The activity of the enzymes usually increases in the presence of a coenzyme or an activator such as Na+, Co2+ The rate of the reaction increases due to the presence of a weak bond which exists between the enzyme and a metal ion.

Active sites of enzyme

• Active site of an enzyme is that portion of the enzyme that binds the substrate (reactant) to form enzyme –substrate complex which forms products.

• The active site is a three dimensional entity.

• It can be a cleft or a crevice on the surface of the enzyme on which the substrate is bound by weak interactions such as van der Waal, hydrogen bond and electrostatic interactions.

Models for explaining the mechanism of enzyme catalysis

• Two models proposed to explain how enzyme bind to substrates are

• 1. The lock and key model

• 2. The induced-fit model

Lock and key model

• This was proposed by Emil Fischer in 1894.

• In the lock-and-key model, the active site and substrate have complementary three-dimensional structures and dock perfectly without the need for major atomic rearrangements.

• In other words, the shape of the substrate and the active site of the enzyme are assumed to fit together like a lock and key

Source:chm.bris.ac.uk

Fig.1 Lock and Key model for enzymatic reaction

Induced- fit model

• This model was proposed in 1958 by Daniel E. koshland Jr.

• The substrate binding induces a conformational change on the active site of the enzyme.

• The enzyme may also distort the substrate, forcing it into a conformational change that is similar to the transition state.

• Only after the change does the substrate fit snugly in the active site.

Induced- fit model

Source:khanacademy.org

Fig.2 Induced – fit model for enzymatic reaction

The Michaelis–Menten mechanism of enzyme catalysis

• Experimental investigations of enzyme kinetics are carried out by monitoring the initial rate of product formation in a solution in which the enzyme is present at very low concentration.

• The Michael-Menten mechanism is represented by the following scheme:

• E + S ⇋ ES → E + PWhere E represents the enzyme, S is substrate(reactant), ES is the intermediate complex between the enzyme and the substrate and P is the product. k1, k2 and k3 are the rate constants for the forward reaction, reverse reaction and the formation of products respectively.

The Michaelis–Menten mechanism of enzyme catalysis

• −𝑑 𝑆

𝑑𝑡=

𝑑 𝑃

𝑑𝑡= 𝑘1 𝐸 𝑆 − 𝑘2 𝐸𝑆 = 𝑘3 𝐸𝑆 1

•𝑑[𝐸𝑆]

𝑑𝑡= 𝑘1 𝐸 𝑆 − 𝑘2 𝐸𝑆 − 𝑘3[𝐸𝑆] 2

Eq. 1 gives the disappearance of the substrate or the formation of the product.Eq.2 gives the formation of the complex.Since the concentration of the enzyme is very small compared with that of the substrate, therefore the equation can be written as[E]o = [E] + [ES] 3[E]o is the initial concentration of the enzyme.

The Michaelis–Menten mechanism of enzyme catalysis

• From eq. 3• [E] = [E]o- [ES] 4• Substituting eq. 4 in eq.2

•𝑑[𝐸𝑆]

𝑑𝑡= (𝑘1 𝐸 𝑜 − [𝐸𝑆]) 𝑆 − 𝑘2 𝐸𝑆 − 𝑘3[𝐸𝑆]

= k1[E]o[S]-(k1[S]+k2 + k3) [ES] 5Under steady-state,

𝑑[𝐸𝑆]

𝑑𝑡= 0

Therefore, K1[E]o[S]-(k1[S]+ k2 + k3) [ES] = 0

The Michaelis–Menten mechanism of enzyme catalysis

• [ES] =𝑘1[𝐸]𝑜[𝑆]

𝑘1 𝑆 +𝑘2+𝑘36

• Dividing the numerator and the denominator of the right hand side of eq. 6 by k1[S] gives

• 𝐸𝑆 =[𝐸]𝑜

1+𝐾2+𝐾3𝐾1[𝑆]

7

• Substituting the value of [ES] in eq. 1 gives the

formation of the product i.e. 𝑑 𝑃

𝑑𝑡= 𝑘3 𝐸𝑆

The Michaelis–Menten mechanism of enzyme catalysis

•𝑑[𝑃]

𝑑𝑡=

𝑘3[𝐸]𝑜

1+𝑘2+𝑘3𝑘1[𝑆]

8

• The quantity 𝑘2+𝑘3

𝑘1is the Michaelis constant, characteristic

of a given enzyme acting on a given substrate. Its units are in terms of concentration and it is a combination of rate constants.

• Given that KM= 𝑘2+𝑘3

𝑘1[𝑆]

• because the substrate is typically in large excess relative to the enzyme, the free substrate concentration is approximately equal to the initial substrate concentration and we can write [S] ≈ [S]o. It then follows that

The Michaelis–Menten mechanism of enzyme catalysis

• From eq. 8, the rate of formation V of the product is given by

• V=𝑘3[𝐸]𝑜

1+𝑘2+𝑘3𝑘1[𝑆]

=𝑘3[𝐸]𝑜

1+𝐾𝑀𝑆

𝑜

9

• When [S]o << KM, the rate is proportional to

• V=𝑘3

𝐾𝑀[S]o[E]o 10

• From eq. 9, when [S]o >> KM, the rate gets to its maximum value and is independent of [S]o:

• V = Vmax = k3[E]o 11

• Vmax is the maximum velocity

• The maximal rate, Vmax, is attained when the catalytic sites on the enzyme are saturated with substrate—that is, when [ES] = [E]o

Michael - Mentis Plot

Source:Chemistry LibreTexts

Fig. 3 Diagram of reaction speed and Michaelis-Menten kinetics

• Substituting for Rmax and KM in eq. 9 gives

• V=𝑉𝑚𝑎𝑥

1+𝐾𝑀[𝑆]𝑜

10

• Eq. 10 is Michaelis Menten equation. Rearranging the equation ( by finding the reciprocal of the equation), we have

•1

𝑉=

1

𝑉𝑚𝑎𝑥+

𝐾𝑀

𝑉𝑚𝑎𝑥

1

[𝑆]𝑜11

• A plot of 1

𝑉against

1

[𝑆]𝑜will give a straight line. This is

called the Lineweaver-Burk plot for enzyme catalyzed reaction. The plot should give a straight line.

Lineweaver-Burk plot

Source:Chemistry LibreTexts

Significance of Michael- Mentenequation

• The Michaelis-Menten equation is used to predict the rate of product formation in enzymatic reactions.

• Michaelis constant Km is a measure of the substrate-binding affinity.

• The value of KM is used as a measure of an enzyme's affinity for its substrate. The lower the KM value the higher the enzyme's affinity for the substrate.

• For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

• An enzyme with a high Km has a low affinity (binding) for its substrate, and requires a greater concentration of substrate to achieve Vmax.

Significance of Michael- Menten equation

• Since the unit of KM is same as that of the substrate concentration, it means that there is a relationship between KM and [S]

• What is the implication of KM = [S]?

• V=𝑉𝑚𝑎𝑥[𝑆]𝑜

[𝑆]𝑜+[𝑆]𝑜=𝑉𝑚𝑎𝑥[𝑆]𝑜

2[𝑆]𝑜=𝑉𝑚𝑎𝑥

2

• KM is also the substrate concentration at which the enzyme operates at one half of its maximum velocity i.e.

• KM=[S] at 1

2𝑉𝑚𝑎𝑥

• For certain enzymes under certain conditions, KM can also be a measure of affinity between E and S – approximates the dissociation constant of the ES complex

• If the value of KM is low i.e. a small number, • it means that • 1. the enzyme is held tightly to the Substrate (HIGH

affinity) • 2. It reaches Vmax at a lower [S] • 3. Small number means less than 10-3 M• If the value of KM is high i.e. a large number, • it means that • 1. the enzyme holds weakly to the Substrate (LOW affinity) • 2. It reaches Vmax at a higher [S] • 3. Large number means 10-1 to 10-3 M

NOTE:

Vmax indicates the following:

1. It is the maximum rate that can be observedin the reaction.

2. Substrate is present in excess

3. Enzyme can be saturated (zero order reaction)

Example 1

• For a given enzyme catalyzed reaction, the Michaelis constant is 0.8mM and the substrate concentration is 1.2mM. What is the fractional saturation of the enzyme under these conditions?

• Solution

• Explanation: The fractional saturation of an enzyme isdefined as the amount of enzyme that is bound to substratedivided by the total amount of enzyme. To calculate thefractional saturation, we'll need to use the Michaelis-Menton

equation:

Example 2

• What is the ratio of 𝑉𝑜

𝑉𝑚𝑎𝑥when [S]=4KM?

• From the Michealis Menten equation

• V=𝑉𝑚𝑎𝑥[𝑆]𝑜

[𝑆]𝑜+𝐾𝑀

• Rearranging the equation to find the ratio

•𝑉𝑜

𝑉𝑚𝑎𝑥=

[𝑆]

[𝑆] +𝐾𝑀=

𝑉𝑜

𝑉𝑚𝑎𝑥=

4KM

4𝐾𝑀+𝐾𝑀=

4

5𝐾𝑀

• V=𝑉𝑚𝑎𝑥[𝑆]𝑜

[𝑆]𝑜+𝐾𝑀

• To define the rate and maximum rate in terms of enzyme concentration we have

• 𝑉𝑜 = 𝑘𝑐𝑎𝑡 𝐸𝑆

• 𝑉𝑚𝑎𝑥 = 𝑘𝑐𝑎𝑡[𝐸]𝑇

• Fractional saturation = [𝐸𝑆]

[𝐸]𝑇=

𝑉𝑜

𝑉𝑚𝑎𝑥=

[𝑆]𝑜

[𝑆]𝑜+𝐾𝑀

•[𝑆]𝑜

[𝑆]𝑜+𝐾𝑀=

1.2𝑚𝑀

1.2𝑚𝑀+0.8𝑚𝑀= 0.6 = 60%

References

• 1. Sharma K.K., Sharma. L. K. 2006.Textbook of Physical Chemistry 4th Ed.

• 2. Bahl, B.S., Bahl A, Tuli G.D. 2006. Essential of Physical Chemistry. P. 706-722.

• 3. Atkin P, de Paula J 2006. Physical Chemistry. P840-842.

• 4.