--%>

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 : Coordination number of a cation The

    The coordination number of a cation engaging a tetrahedral hole is: (a) 6  (b) 8  (c) 12  (d) 4 Answer: (d) The co-ordination number of a cation occupying a tetrahedral hole is 4.

  • Q : Calculating total number of moles

    Choose the right answer from following. While 90 gm of water is mixed with 300 gm of acetic acid. The total number of moles will be: (a)5 (b)10 (c)15 (d)20

  • Q : Finding Normality Can someone please

    Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N  (b) 19.8 N  (c) 39.6 N  (d) 98

  • Q : Meaning of molality of a solution The

    The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

  • Q : Molarity of solution Help me to go

    Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M

  • Q : Problem on MM equation How to obtain

    How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).

  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

  • Q : Colligative property related question

    Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point

  • Q : Theory of one dimensional motion For

    For motion in one dimension, the distribution of the molecules over quantum states, speeds, and energies can be deduced.Here we show that the energy of a macroscopic gas sample can be described on the basis of our knowledge of the quantum states allowed to

  • 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