The velocity of an enzyme catalyzed reaction was 10% of Vmax at 2.0 μM substrate concentration. What was the Km in μM?
I know Km=1/2Vmax but I do not know if this is the right equation to use or if I am incorporating the known substrate concentration right?
I tried to set up the equation so
2uM= 1/2(1/10*Vmax)
and got v max= 40
then I took the same equation km=1/2vmax and plugged in the vmax of 40 to get a km concentration of 20uM. But the correct answer is 18uM.
Thannk you.
Use
v0= vmax*[S]/(km+[S])
where v0= 1/10*vmax
km= 1/2vmax
plug and chug, solve for vmax and plug into km=1/2vmax equation to get km
To determine the Km in this enzyme-catalyzed reaction, you need to use the correct equation and incorporate the known substrate concentration correctly.
The equation Km = 1/2Vmax is indeed the correct equation to use for determining the Km. In this equation, Km represents the Michaelis constant which indicates the concentration of substrate at which the reaction velocity is half the maximum velocity (Vmax).
To set up the equation correctly:
1. Start with the given information: velocity (V) is 10% of Vmax at 2.0 μM substrate concentration.
2. Find the Vmax by rearranging the equation: V = (Vmax * [S]) / (Km + [S]) where [S] is the substrate concentration. Plug in the known values: 0.1Vmax = (Vmax * 2.0 μM) / (Km + 2.0 μM).
3. Simplify the equation: 0.1 = (2.0 μM) / (Km + 2.0 μM).
4. Solve for Km: Cross-multiply and rearrange the equation to get 0.1(Km + 2.0 μM) = 2.0 μM. This will give you 0.1Km + 0.2 μM = 2.0 μM. Subtract 0.2 μM from both sides to isolate 0.1Km = 1.8 μM, and then divide by 0.1 to calculate Km, which equals 1.8 / 0.1 = 18 μM.
So the correct answer for Km in this case is indeed 18 μM, not 20 μM as you initially calculated.