Determine the fraction of Vmax that would be obtained at the following substrate concentrations: 0.5 Km, 2 Km, and 10 Km.
The Correct Answer and Explanation is:
To determine the fraction of VmaxV_{\text{max}}Vmax at different substrate concentrations relative to KmK_mKm, we can use the Michaelis-Menten equation: V=Vmax⋅[S]Km+[S]V = \frac{V_{\text{max}} \cdot [S]}{K_m + [S]}V=Km+[S]Vmax⋅[S]
Where:
- VVV is the reaction velocity at a specific substrate concentration [S][S][S],
- VmaxV_{\text{max}}Vmax is the maximum velocity when the enzyme is saturated with substrate,
- KmK_mKm is the Michaelis constant, and
- [S][S][S] is the concentration of the substrate.
To find the fraction of VmaxV_{\text{max}}Vmax, we divide the reaction velocity at a given substrate concentration by VmaxV_{\text{max}}Vmax. This gives us the fraction of the maximum velocity: VVmax=[S]Km+[S]\frac{V}{V_{\text{max}}} = \frac{[S]}{K_m + [S]}VmaxV=Km+[S][S]
1. At [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 equation: 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 0.5Km0.5 K_m0.5Km is 13\frac{1}{3}31 or approximately 0.33.
2. At [S]=2Km[S] = 2 K_m[S]=2Km:
Substitute [S]=2Km[S] = 2 K_m[S]=2Km into the equation: 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 2Km2 K_m2Km is 23\frac{2}{3}32 or approximately 0.67.
3. At [S]=10Km[S] = 10 K_m[S]=10Km:
Substitute [S]=10Km[S] = 10 K_m[S]=10Km into the equation: 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 10Km10 K_m10Km is 1011\frac{10}{11}1110 or approximately 0.91.
Summary of Results:
- At 0.5Km0.5 K_m0.5Km, the fraction of VmaxV_{\text{max}}Vmax is 0.33.
- At 2Km2 K_m2Km, the fraction of VmaxV_{\text{max}}Vmax is 0.67.
- At 10Km10 K_m10Km, the fraction of VmaxV_{\text{max}}Vmax is 0.91.
These results show how the reaction velocity increases as the substrate concentration increases relative to KmK_mKm, but it never exceeds VmaxV_{\text{max}}Vmax, which occurs only when the substrate concentration is very high (typically much greater than KmK_mKm).
