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).
Expert
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)
Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules
25 ml of a solution of barium hydroxide on titration with 0.1 molar solution of the hydrochloric acid provide a litre value of 35 ml. The molarity of barium hydroxide solution will be: (i) 0.07 (ii) 0.14 (iii) 0.28 (iv) 0.35
Enzymes are complex proteinous substances, produced by living bodies, such as act as catalysis in the physiological reactions. The enzymes are, also called biochemical catalysts and the phenomenon is known as bio-chemical catalysis because numerous reactions that occur the bodies of animals and p
Choose the right answer from following. When 6gm urea dissolve in180gm H2O . The mole fraction of urea is : (a)10/ 10.1 (b)10.1/10 (c)10.1/ 0.1 (d) 0.1/ 10.1
Provide solution of this question. 1.0 gm of pure calcium carbonate was found to need 50 ml of dilute HCL for complete reaction. The strength of the HCL solution is specified by : (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N
This is a very important aspect of Physical Chemistry in which knowledge of the size, shape, rigidity and electronic structure of molecules deduced from the experimental methods treated here goes hand in hand with the theoretical approaches of chemical reactions. Spec
Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N
During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is
Can someone please help me in getting through this problem. The total number of molecules in 16 gm of methane will be: (i) 3.1 x 1023 (ii) 6.02 x 1023 (iii) 16/6.02 x 1023 (iv) 16/3.0 x 1023
Help me to go through this problem. 1000 gms aqueous solution of CaCO3 contains 10 gms of carbonate. Concentration of the solution is : (a)10 ppm (b)100 ppm (c)1000 ppm (d)10000 ppm
18,76,764
1929487 Asked
3,689
Active Tutors
1439313
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!