Trivial enzyme kinetics question

Mavi Gozler mavigozler at
Tue Apr 23 13:00:06 EST 2002

On 21 Apr 2002 08:58:28 GMT, kaj.stenberg at wrote:

>A student asked a question regarding a calculation, the answer must be 
>trivial but I don't get it:
>From a book (not at home so full reference missing):
>E+P --k1,k-1--[ES]--k2-- E+P

Probably you mean  E+S --k1,k-1--[ES]--k2-- E+P

>A) what is the breakdown of ES. Because k2<<k1 k2 can be neglected, Kd
>is k-1/k1. No problem.

Probably you means that k2 << k-1, and so "Kd" (usually called Ks) is
the equilibrium constant for E + S <--> ES

>B) Km is (k-1 + k2)/k1 
>In case B, why can k2 not be neglected? (judging from the answer it
>isn't, a factor of three difference. The ansver is in the 10E5 range)

One of the problems here is that Ks is an equilibrium constant, and
when you have E, S, and ES in the mixture, you don't necessarily have
equilibrium conditions.  Instead there is a steady-state in the
formation and breakdown of ES in these single substrate reactions.
"Breakdown" of ES can be simple dissociation, or instead product
formation (often shown as product formation and instantaneous
release).  It is entirely reasonable for k2 to be much less than k-1,
and for Km to be equivalent to Ks, in a state of pseudo-equilibrium.
But the Km takes into account these competitive "reactions" which the
enzyme-substate complex can undergo.  Anyway, trust the math.  In any
complex mixture (multiple substrates, perhaps with inhibitor,
including effects of product[s]), if your rate equations and
assumptions and equations are sound, your conclusions based on them
will be also.

(One of) the best discussions on this subject is one written by a
member of my qualifying exam committe, Irwin H. Segel, Enzyme
Kinetics, 1975.

More information about the Proteins mailing list