--%>
Assignment Help - homework help

Problem on MM equation

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).

E

Expert

Verified

According to M.M Equation,
                                
                           V = Vmax x {S]/Km + {S]
                
Since Vmax = ksec^-1 {Et}

V = Ksec^-1{Et} x {S}/Km + {S}

Substituting the given value of Ksec^-1 in the above equation,we get
 
V = Vmax x mol.wt x 1min/60 sec x {S}/mg of enzyme / Km +{S]

V = Vmax x mol.wt x {S}/mg of enzyme/Km +{S}

Substituting the given value of {S}

V = Vmax x mol.wt x 4.576(log K -10.753-logT+Ea/4.576T)/mg of enzyme/ Km +  4.576(log K -10.753-logT+Ea/4.576T)

   Related Questions in Chemistry

  • Q : Relationship between Pressure and

    The pressure-temperature relation for solid-vapor or liquid vapor equilibrium is expressed by the Clausis-Clapeyron equation.We now obtain an expression for the pressure-temperature dependence of the state of equilibrium between two phases. To be specific,

    		•
  • Q : Organic structure of cetearyl alcohol

    Show the organic structure of cetearyl alcohol and state what the organic family is? Briefly state it.

    		•
  • Q : Freezing point of equimolal aqueous

    The freezing point of equi-molal aqueous solution will be maximum for:            (a) C6H5NH3+Cl-(aniline hydrochloride)  (b) Ca(NO3

    		•
  • Q : Sedimentation and Velocity The first

    The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A

    		•
  • Q : What are lattices and unit cells? The

    The repeating, atomic level structure of a crystal can be represented by a lattice and by the repeating unit of the lattice, the unit cell.It was apparent very early in the study of crystals that the shapes of crystals stem from an ordered array of smaller

    		•
  • Q : Concentration of urea Help me to go

    Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<

    		•
  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

    		•
  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

    		•
  • 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 : Calculating density of water using

    What is the percent error in calculating the density of water using the ideal gas law for the following conditions:  a. 110 oC, 1 bar   b. 210 oC 10 bar  c. 374 o

    		•