--%>

Describe Enzyme Catalyzed reactions with examples.

Many enzyme catalyzed reactions obeys a complex rate equation that can be written as the total quantity of enzyme and the whole amount of substrate in the reaction system.

Many rate equations that are more complex than first and second order equations and are encountered in chemical rate studies. Such rate equations can be illustrated by considering reactions that occur in biological systems, or at least are affected by enzymes occurring in such systems.

The impact of enzymes on the rate through which chemical reactions move toward their equilibrium position gives one of the most dramatic catalytic effects. Much of the current interest in the subject is centered on the details of the action between the enzyme, which is the catalyst, and the material, known as substrate, whose reaction it effects. It is significant to know that how an enzyme catalyzed reaction proceeds in time and how the catalytic action of the enzyme substrate pair is analysed from the measurement of the development of such reactions.

The experimental data for enzyme catalyzed reactions show a variety of forms that depend on the enzyme, the substrate, the temperature, the presence of interfering substances, and so forth. Many of the behaviors that are found can be looked on as variations from the ideal curve. It is such rate curves for which we now develop a rate equation in a form that is conviently related to the quantities measured in enzymes studies.

Inspection of the curve shows that at high substrate concentrations the rate of the reaction is independent of the substrate concentration. It is, the however, proportional to the total amount of the enzyme. At low substrate concentrations the rate, as shown by the initial straight line section of the curves, is proportional to the substrate concentration. The rate would be found to be proportional to the total enzyme concentration. These features also be found to be proportional by a rate of equation, where R denotes the rate of the reaction, of the form:

R = (const) [Etot ] [ S ] / const' + [S]

To anticipate the notion introduced when the mechanism of enzyme catalyzed reactions is dealt with, we introduce the symbols k2 and KM for the two constants and thus write the equation in the form:

R = k2 [Etot] [S]/const' + [S]

To anticipate the notion when the mechanism of enzyme catalyzed reactions is dealt with, we introduce the symbols k2 and KM for the two constants and thus write the rate equation in the form:

R = k2 [Etot] [S]/KM + [S]

Although the parameters k2 and KM could be determined so that a function corresponding to the experimental more convenient procedure can be found. The initial rate is often obtained by measuring [S] after a time t at which only a small fraction of the substance has been consumed. If[S0] is the initial substrate concentration, we can express the initial rate as [S0] - [S]/t. then it becomes:

[S0] - [S] = k2 [Etot] [S0]/KM + [S0] × t

The "constants" k2 [Etot] and KM can be evaluated from measurements of the initial rate of reaction. This rate, Rinit, is approximately [S0] - [S]/t, where [S] is the concentration after a small time interval t.

Values of Rinit can be obtained for various values of [S0]. A convenient procedure for obtaining the constants is based on the reciprocal of this equation. We write down:

1/Rinit = 1/k2 [Etot] + KM/k2[Etot] × 1/[S0]

Thus, if a plot of 1/Rinit versus 1/S0 gives a straight line, the intercept and slope can be used to obtain k2 [Etot] and KM/k2 [Etot]. From these quantities the value of KM can be calculated. Separation of the factors k2 and [Etot] requires studies of systems with various amounts of enzyme.

   Related Questions in Chemistry

  • Q : Solubility product On passing H 2 S gas

    On passing H2S gas through a particular solution of Cu+ and Zn+2 ions, first CuS is precipitated because : (a)Solubility product of CuS is equal to the ionic product of ZnS (b) Solubility product of CuS is equal to the solubility product

  • Q : Problem on equilibrium constant Ethanol

    Ethanol is manufactured from carbon monoxide and hydrogen at 600 K and 20 bars according to the reaction2 C0(g) + 4 H2(g) ↔ C2H5OH(g) + H2O (g)The feed stream contains 60 mol% H2, 20 m

  • Q : Simulate the column in HYSYS The

    The objective of this work is to separate a binary mixture and to cool down the bottom product for storage. (Check table below to see which mixture you are asked to study). 100 kmol of feed containing 10 mol percent of the lighter component enters a continuous distillation column at the m

  • Q : Explain the process of adsorption in

    The process of adsorption can occurs in solutions also. This implies that the solid surfaces can also adsorb solutes from solutions. Some clarifying examples are listed below: (i) When an aqueous solution of ethano

  • Q : Strength of Nacl in solution To 5.85gm

    To 5.85gm of Nacl one kg of water is added to prepare of solution. What is the strength of Nacl in this solution (mol. wt. of nacl = 58.5)? (a) 0.1 Normal (b) 0.1 Molal (c) 0.1 Molar (d) 0.1 FormalAnswer:

  • Q : Molecular crystals Among the below

    Among the below shown which crystal will be soft and have low melting point: (a) Covalent  (b) Ionic  (c) Metallic  (d) MolecularAnswer: (d) Molecular crystals are soft and have low melting point.

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Number of moles present in water

    Provide solution of this question. How many moles of water are present in 180 of water: (a)1 mole (b)18 mole (c)10 mole (d)100 mole

  • Q : Calculating molarity of a solution

    Select the right answer of the question .The molarity of a 0.2 N N2Co3 solution will be: (a) 0.05 M (b) 0.2 M (c) 0.1 M (d)0.4 M

  • Q : Calculating amount of Sodium hydroxide

    Choose the right answer from following. The amount of NaOH in gms in 250cm3 of a0.100M NaOH solution would be : (a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm