For any enzyme that follows simple Michaelis-Menten kinetics, when the initial Velocity of the reaction is 80% of Vmax what is the substrate concentration?
The answer is: [S] = 4Km
Can you explain why it is?
v = (VmaxS)/(K + S)
when v = 0.8Vmax, then
0.8Vmax = (VmaxS)/(K + S)
0.8VmaxK + 0.8VmaxS = VmaxS
0.8VmaxS - VmaxS = -0.8VmaxK
-0.2VmaxS = -0.8VmaxK
0.2VmaxS = 0.8VmaxK
S = 0.8VmaxK/0.2Vmax
S= 4K
How did you get - 0.2 from?
This entire problem is not about chemistry. It's just an exercise in algebra.
Step 5 is
0.8VmaxS - VmaxS = -0.8VmaxK
So 0.8VmaxS - 1.0VmaxS = -0.8VmaxK and
-0.2VmaxS = -0.8VmaxK
etc
To explain why the substrate concentration ([S]) at which the initial velocity (V0) of a reaction is 80% of the maximum velocity (Vmax) is [S] = 4Km, we need to understand the Michaelis-Menten kinetics equation and how it relates to enzyme-substrate interactions.
The Michaelis-Menten equation is given as:
V0 = (Vmax * [S]) / ([S] + Km)
Where:
- V0 is the initial velocity of the reaction,
- Vmax is the maximum velocity of the reaction,
- [S] is the substrate concentration, and
- Km (Michaelis constant) is a measure of the affinity of the enzyme for the substrate.
In this case, we are told that V0 is 80% of Vmax. We can express this mathematically as:
V0 = 0.8 * Vmax
Now, we can substitute this value into the Michaelis-Menten equation:
0.8 * Vmax = (Vmax * [S]) / ([S] + Km)
To simplify, we can cancel out the Vmax on both sides:
0.8 = [S] / ([S] + Km)
Multiply both sides by ([S] + Km):
0.8 * ([S] + Km) = [S]
Distribute the 0.8:
0.8[S] + 0.8Km = [S]
Rearrange the equation to isolate [S]:
0.8[S] - [S] = -0.8Km
Combine like terms:
-0.2[S] = -0.8Km
Divide both sides by -0.2:
[S] = 4Km
Therefore, the substrate concentration at which the initial velocity of the reaction is 80% of Vmax is [S] = 4Km.
It's important to note that Km is a constant value specific to an enzyme-substrate pair, and it represents the concentration of substrate required for the enzyme to achieve half of its maximum velocity. By knowing the Km value for a specific enzyme, we can determine the corresponding substrate concentration at various velocities of the reaction.