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)
Choose the right answer from following. Volume of water needed to mix with 10 ml 10N NHO3 to get 0.1 N HNO3: (a) 1000 ml (b) 990 ml (c) 1010 ml (d) 10 ml
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The catalytic dehydrogenation of 1-butene to 1,3-butadiene, C4H8(g) = C4H6(g)+H2(g) is carried out at 900 K and 1 atm. Discover Q & A Leading Solution Library Avail More Than 1452859 Solved problems, classrooms assignments, textbook's solutions, for quick Downloads No hassle, Instant Access Start Discovering 18,76,764 1927322 Asked 3,689 Active Tutors 1452859 Questions Answered Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!! Submit Assignment
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