Michaelis-menten equation:
Vo=Vmax [S]/Km+[S] where [S] =substrate concentration,Vmax=maximum velocity of enzyme-substrate reaction
Km=michaelis-menten constant,vo=reaction velocity
this can be rearranged to give : (1/Vo)=(Km/Vmax)*1/[S] +1/Vmax
A plot between 1/[S] vs 1/Vo is called lineweaver-Burk plot is linear with slope=Km/Vmax and y-intercept=1/Vmax
Also,Vmax=kcat*[E]
Vo=kcat*[E] [S]/Km+[S]
For high [S], km<<<[S]
So,Vo=kcat*[E] [S]/[S] =kcat [E]
thus kcat =pseudo first order rate constant as substrate is in large excess.
Data:
1/[So] (L/mmol) 1/v(Ls/umol)
1.00 0.222
0.5 0.126
0.25 0.09
0.125 0.066
slope=km/Vmax=0.177 (Ls/umol) *(mmol/L)=0.177 *1000 (s)
intercept=1/Vmax=0.043 Ls/umol
Vmax=1/0.043=23.256 umol/Ls
km=0.177*1000 s *23.256 *10^-6 mol/Ls=4.116 *10^-3 mol/L=4.116 mmol/L
Also [E]=mol/volume
mol of enzyme=2.5ug/ 42300 g/mol=0.0000543 umol [1dalton=1g/mol]
[E]=0.0000543 umol/0.1L=0.000543 umol/L
So kcat=Vmax/[E]=(23.256 umol/Ls)/(0.000543 umol/L)=42828.729 s^-1
kcat=42828.729 s^-1
The following intial rate data were obtained by adding 2.5 μg of a purified enzyme having...
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18.0 6.18 The following data were obtained on the initial rate of isomerization of a compound S catalyzed by an enzyme E: [S]./(mmol dm-) 1.00 2.00 3.00 4.00 vo/(mmol dm-'s-1) (a) 4.5 9.0 15.0 (b) 14.8 25.0 45.0 59.7 (c) 58.9 120.0 180.0 238.0 The enzyme concentrations are (a) 1.00 mmol dm-?, (b) 3.00 mol dm-3, and (c) 10.0 mmol dm-?. Find the orders of reactions with respect to S and E, and the rate constant. Hints...
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6.18 The following data were obtained on the initial rate of isomerization of a compound S catalyzed by an enzyme E: S]./(mmol dm") Va/(mmol dms) 2.00 3.00 1.00 4.00 15.0 9.0 4.5 (a) (b) 18.0 45.0 25.0 14.8 59.7 180.0 120.0 238.0 (c) 58.9 The enzyme concentrations are (a) 1.00 mmol dm, (b) 3.00 mol dm, and (c) 10.0 mmol dm Find the orders of reactions with respect to S and E, and the rate...