问答题
(南京大学2008年考研试题)(1)At what substrate concentration would an enzyme with a K
cat
of 30．OS
-1
and a K
m
of 0．0050 mol/L operate at one—quarter of its maximum rate?
(2)determine the fraction of V
max
that would be obtained at the following substrate concentration：[S]=1／2K
m
，2K
m
，and 10K
m
．
【正确答案】正确答案：(1)当V
0
=1／4V
max
时，求[S]的值。由米氏方程V
0
=V
max
[S]／(K
m
+[S])得： V
0
=V
max
[S]／(K
m
+[S])=1／4V
max
整理得[S]／(K
m
+[S])=1／4，所以：[S]=1／3K
m
=1／3×0．005 mol／L=1．67×10
-3
mol／L 所以，在底物浓度为1．67×10
-3
mol／L时，表现出的反应速度为最大反应速度的1／4。
【答案解析】