Asked by Anonymous
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.
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.
Answers
Answered by
bobpursley
By definition, 2*.34 nmol/min
Answered by
Anonymous
So the answer would be none of the above? I had gotten C?
Answered by
Anonymous
Sorry I just realized what you meant. It's 2 times .34 which equals .68
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.