--%>

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 : Adiabatic compression A lean natural

    A lean natural gas is available at 18oC and 65 bars and must be compressed for economical pipeline transportation. The gas is first adiabatically compressed to 200 bars and then isobarically (i.e. at constant pressure) cooled to 25°C. The gas, which is

  • Q : Number of moles of potassium chloride

    Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5

  • Q : Question based on vapour pressure While

    While a substance is dissolved in a solvent, the vapour pressure of the solvent is decreased. This results in: (a) An increase in the boiling point of the solution (b) A decrease in the boiling point of solvent (c) The solution having a higher freezing point than

  • Q : How reactive is Trimethylindium towards

    Illustrate the reason, how reactive is Trimethylindium towards oxygen and water?

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

  • 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 : Molecular Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Product of HCl Zn Illustrate  the

    Illustrate  the product of HCl Zn?

  • Q : The three facts on the evaporation

    Describe briefly the three facts on the evaporation?

  • Q : Question based on strength of solution

    Help me to go through this problem. On dissolving 1 mole of each of the following acids in 1 litre water, the acid which does not give a solution of strength 1N is: (a) HCl (b) Perchloric acid (c) HNO3 (d) Phosphoric acid