An enzyme has velocities of 0.11 nmol/min at a substrate concentration of 1 μM, 0.34 nmol/min at a substrate concentration of 5 μM, and 0.45 nmol/min at a substrate concentration of 10 μM. What is the Vmax for the enzyme if the KM is 5 × 10–6 M?
A.
0.22 nmol/min
B.
0.68 nmol/min
C.
0.90 nmol/min
D.
1.12 nmol/min
E.
None of the answers is correct.
By definition, 2*.34 nmol/min
So the answer would be none of the above? I had gotten C?
Sorry I just realized what you meant. It's 2 times .34 which equals .68
To determine the Vmax for the enzyme, we need to analyze the relationship between substrate concentration and the enzyme velocity using the Michaelis-Menten equation.
The Michaelis-Menten equation is given by:
V = Vmax * [S] / (KM + [S])
Where:
V is the velocity of the enzyme reaction
Vmax is the maximum velocity of the enzyme-catalyzed reaction
[S] is the substrate concentration
KM is the Michaelis constant, which is equal to the substrate concentration at which the reaction rate is half of Vmax
Given that KM is 5 × 10–6 M, we can now calculate the Vmax.
Let's use the first set of data points with a substrate concentration of 1 μM and a velocity of 0.11 nmol/min:
0.11 nmol/min = Vmax * (1 μM) / (5 × 10–6 M + 1 μM)
To simplify the calculation, we convert the substrate concentration to Molar:
1 μM = 1 × 10–6 M
0.11 nmol/min = Vmax * (1 × 10–6 M) / (5 × 10–6 M + 1 × 10–6 M)
0.11 nmol/min = Vmax * (1 × 10–6 M) / (6 × 10–6 M)
0.11 nmol/min = Vmax * 1/6
Vmax = 0.11 nmol/min * 6 = 0.66 nmol/min
Now, we can check if this Vmax value matches the other data points to verify our calculation:
At a substrate concentration of 5 μM, the observed velocity is 0.34 nmol/min.
0.34 nmol/min = Vmax * (5 × 10–6 M) / (5 × 10–6 M + 5 μM)
0.34 nmol/min = Vmax * (5 × 10–6 M) / (5 × 10–6 M + 5 × 10–6 M)
0.34 nmol/min = Vmax * (5 × 10–6 M) / (10 × 10–6 M)
0.34 nmol/min = Vmax * 1/2
Vmax = 0.34 nmol/min * 2 = 0.68 nmol/min
At a substrate concentration of 10 μM, the observed velocity is 0.45 nmol/min.
0.45 nmol/min = Vmax * (10 × 10–6 M) / (10 × 10–6 M + 5 × 10–6 M)
0.45 nmol/min = Vmax * (10 × 10–6 M) / (15 × 10–6 M)
0.45 nmol/min = Vmax * 2/3
Vmax = 0.45 nmol/min * (3/2) = 0.675 nmol/min
From the above calculations, we can see that the Vmax value is approximately 0.68 nmol/min, which matches the velocity at a substrate concentration of 5 μM.
Therefore, the correct answer is B. 0.68 nmol/min.