Determine the fraction of Vmax that would be obtained at the following substrate concentrations. SHOW ALL WORK: a.) [S] = 0.5Km b.) [S] = 2Km c.) [S] = 10Km
The Correct Answer and Explanation is:
To determine the fraction of Vmax at different substrate concentrations, we use the Michaelis-Menten equation. The Michaelis-Menten equation relates the rate of reaction (v) to the substrate concentration [S] and the maximum velocity (Vmax) of the enzyme: v=Vmax[S]Km+[S]v = \frac{V_{\text{max}} [S]}{K_m + [S]}v=Km+[S]Vmax[S]
Where:
- vvv is the reaction velocity at substrate concentration [S][S][S],
- VmaxV_{\text{max}}Vmax is the maximum reaction velocity,
- KmK_mKm is the Michaelis constant (the substrate concentration at which the reaction rate is half of VmaxV_{\text{max}}Vmax).
The fraction of VmaxV_{\text{max}}Vmax is found by calculating: vVmax=[S]Km+[S]\frac{v}{V_{\text{max}}} = \frac{[S]}{K_m + [S]}Vmaxv=Km+[S][S]
Now, let’s calculate this fraction for the given substrate concentrations:
a.) [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km
Substitute [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km into the formula: vVmax=0.5KmKm+0.5Km=0.5Km1.5Km=13\frac{v}{V_{\text{max}}} = \frac{0.5 K_m}{K_m + 0.5 K_m} = \frac{0.5 K_m}{1.5 K_m} = \frac{1}{3}Vmaxv=Km+0.5Km0.5Km=1.5Km0.5Km=31
Thus, the fraction of VmaxV_{\text{max}}Vmax at [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km is 1/3 or approximately 33.33%.
b.) [S]=2Km[S] = 2 K_m[S]=2Km
Substitute [S]=2Km[S] = 2 K_m[S]=2Km into the formula: vVmax=2KmKm+2Km=2Km3Km=23\frac{v}{V_{\text{max}}} = \frac{2 K_m}{K_m + 2 K_m} = \frac{2 K_m}{3 K_m} = \frac{2}{3}Vmaxv=Km+2Km2Km=3Km2Km=32
Thus, the fraction of VmaxV_{\text{max}}Vmax at [S]=2Km[S] = 2 K_m[S]=2Km is 2/3 or approximately 66.67%.
c.) [S]=10Km[S] = 10 K_m[S]=10Km
Substitute [S]=10Km[S] = 10 K_m[S]=10Km into the formula: vVmax=10KmKm+10Km=10Km11Km=1011\frac{v}{V_{\text{max}}} = \frac{10 K_m}{K_m + 10 K_m} = \frac{10 K_m}{11 K_m} = \frac{10}{11}Vmaxv=Km+10Km10Km=11Km10Km=1110
Thus, the fraction of VmaxV_{\text{max}}Vmax at [S]=10Km[S] = 10 K_m[S]=10Km is 10/11 or approximately 90.91%.
Summary:
- At [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km, the fraction of VmaxV_{\text{max}}Vmax is 1/3 or 33.33%.
- At [S]=2Km[S] = 2 K_m[S]=2Km, the fraction of VmaxV_{\text{max}}Vmax is 2/3 or 66.67%.
- At [S]=10Km[S] = 10 K_m[S]=10Km, the fraction of VmaxV_{\text{max}}Vmax is 10/11 or 90.91%.
These calculations show that as the substrate concentration increases relative to the KmK_mKm, the reaction velocity approaches VmaxV_{\text{max}}Vmax
