Asked by RJ
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.
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.
Answers
Answered by
Anonymous
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
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
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