c*****t
发帖数: 1879 | 1 kcat = k2
k1 k2
E + S -----> ES -----> E + P
<-----
k(-1)
k2 determintes how fast the ES complex is returned to the
enzyme pool. One can imagine that if an enzyme has a difficulty
to dissociate from ES, soon [E] would be depleted. k2 is usually
the slowest step and therefore determines the rate of enzymatic
reactions. Also, when talking about initial rate, assuming
that there are large amount of [S], then all enzymes are able to
bind | s******o
发帖数: 3 | 2 V=Kcat*(E)eq
(E).*Km
=Kcat* ------
Km+S K_1 + Kat
In R & H equation (not M & M), Km = --------
K1
When Km >> S, We get:
Vmax = Kcat* (E).
Hehe, just finished the midterm.
【在 c*****t 的大作中提到】
: kcat = k2
: k1 k2
: E + S -----> ES -----> E + P
: <-----
: k(-1)
: k2 determintes how fast the ES complex is returned to the
: enzyme pool. One can imagine that if an enzyme has a difficulty
: to dissociate from ES, soon [E] would be depleted. k2 is usually
: the slowest step and therefore determines the rate of enzymatic
: reactions. Also, when talking about initial rate, assuming
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