--%>

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 : Problem on decinormal Select the right

    Select the right answer of the question. How much water is required to dilute 10 ml of 10 N hydrochloric acid to make it exactly decinormal (0.1 N): (a) 990 ml (b) 1000 ml (c) 1010 ml (d) 100 ml

  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm

  • Q : Problem on Redlich-Kwong equation i)

    i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be

  • Q : Diffusion Molecular View When the

    When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.To see how the experimental coefficients can be treat

  • Q : Describe Thermodynamics Properties The

    The free energy property leads to convenient expressions for the volume and pressure dependence of internal energy, enthalpy and the heat capacities.All the properties of a chemical system, a sample of a substance, or a mixture of substances have some fixe

  • Q : HCl is polar or non-polar Can you

    Can you please illustrate that HCl is polar or non-polar? Briefly illustrate it.

  • Q : What is chemisorption or chemical

    When the forces of attraction existing between adsorbate particles and adsorbent almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also known as chemisorptions. Since forces of attraction existing b

  • Q : What are Ethers and its types? Ethers

    Ethers are the compounds with general formula or CnH2n+

  • Q : Problem on bubble point The following

    The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery.

    Q : Molarity of acid solution If 20ml of

    If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M  (c) 0.3M (d) 0.4M Choose the right answer fron above.