--%>

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 : How molecule-molecule collisions takes

    An extension of the kinetic molecular theory of gases recognizes that molecules have an appreciable size and deals with molecule-molecule collisions. We begin studies of elementary reactions by investigating the collisions b

  • Q : Units of Measurement Unit of

    Unit of measurement- These are also some systems for units:      (1) C.G.S.

  • Q : Vapour pressure of benzene Give me

    Give me answer of this question. The vapour pressure of benzene at a certain temperature is 640mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175g is added to 39.08g of benzene. The vapour pressure of the solution is 600,mm of Hg . What is the mo

  • Q : Molarity what is the molarity of the

    what is the molarity of the solution prepared by dissolving 75.5 g of pure KOH in 540 ml of solution

  • Q : Molality of glucose Help me to go

    Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal

  • Q : Wavelengths which the human eye can see

    Briefly state the wavelengths which the human eye can see?

  • Q : Explain oxygen and its preparation.

    Karl Scheele, the Swedish chemist, was

  • Q : Question based on vapour pressure and

    Give me answer of this question. The vapour pressure of water at 20degreeC is 17.54 mm. When 20g of a non-ionic, substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular weight of the substances: (a) 210.2 (b) 206.88

  • Q : Entropy is entropy on moleculare basis

    is entropy on moleculare basis relates to the tras.,vib.,and rotational motions?

  • Q : Problem on distribution law The

    The distribution law is exerted for the distribution of basic acid among: (i) Water and ethyl alcohol (ii) Water and amyl alcohol (iii) Water and sulphuric acid (iv) Water and liquor ammonia What is the right answer.