Wolfram - 4-1-2004 at 11:42
In my biochemistry book is written that Km konstant is independent of the koncentration of enzyme how is this possible?
Km konstant is the koncentration of substrate at which half of the active sites are occupied with substrate at a given moment.
Geomancer - 4-1-2004 at 16:22
This is simply a somewhat unconventional way of describing an equilibrium constant. Remember from general chem that for any reaction where the idea of
"concentration" makes sense, the expression
([r<sub>1</sub>]<sup>e<sub>1</sub></sup>*[r<sub>2</sub>]<sup>e<sub>2</sub></sup&g
t;...*[r<sub>n</sub>]<sup>e<sub>n</sub></sup>
/
([p<sub>1</sub>]<sup>e'<sub>1</sub></sup>*[p<sub>2</sub>]<sup>e'<sub>2</sub&g
t;</sup>...*[p<sub>n</sub>]<sup>e'<sub>n</sub></sup>=k
, where (r,p)<sub>i</sub> are reactants and products, respectively, holds for suitable k, e<sub>i</sub>, and
e'<sub>i</sub>.
In this case, it is reasonable to assume that the rate determining step is the docking of the substrate, and involves one molecule of both enzyme
and substrate; therefore, the appropriate equation becomes:
[E][S]/[ES]
Working backwards from your definition, given an enzyme concentration [E], we wish to find [S] s.t. [E]=[ES]. That is, we want to solve:
[E][S]/[E]=k
Clearly the [E] drops out, so the enzyme concentration is irrelavent, and the solution is simply the equilibrium constant of the reaction.
Edit: Since this is a thermodynamic concept, the rate determining step is irrelevant; all that matters is the stoichiometry (1:1 in this case). I got
confused by the related concept of rate constant.
[Edited on 5-1-2004 by Geomancer]
Al Koholic - 6-1-2004 at 20:00
quote: Km konstant is the koncentration of substrate at which half of the active sites are occupied with substrate at a given moment.
You have stated this correctly, I'm not sure why you think this not to be possible.
I'm not sure if I totally follow your post there Geomancer but I do know that the Km of an enzyme/substrate system can be, and is described in
terms of rate determining steps.
In reactions with only 2 steps ie: E+S <--> ES <--> E+P, the forward and backward (association and dissociation w/out reaction) rates of
the first equilibrium are k1 and k-1. The second step in which E+P results cannot go back so it is only forward k2. Since Km = k2+k-1/k1, when the
k2 is rate-limiting, k2 << k-1, and Km reduces to k-1/k1 which is the Kd or dissociation constant of the ES complex. This is an accurate
description of the affinity the substrate and enzyme have for each other. This is the only time when this is applicable though. Some people like to
use the Km as a measure of the affinity between the enzyme and substrate however, this is often incorrect.
Most enzymes do not follow this and have perhaps more steps involving configurational changes before actual product formation. Some enzymes will have
EP (enzyme-product) complexes that can revert back to the starting E+S with some rate. In these cases and in cases where Km approximates k2/k1 or a
complex function of all three parameters in a two step reaction, affinity is not a valid description.
Thermodynamically the products and reactants have set points, as does the equilibrium which is not influenced by catalysts. What happens inbetween
(rate wise) is all up to enzyme efficiency. The Km depends on these rates to some degree but is better thought of with respect to Vo and Vmax; the
initial and max rates of reaction with respect to [S].
When [S] >> [E], you have Vmax where almost all E exists as ES complex for the simple model above. By placing the enzyme into various
concentrations of S, one can measure the initial reaction rate and plot Vo/[S]. Through a series of derivations I don't want to type you end up
with Vo (initial rate) = Vmax[S]/Km+[S], known as Michalis-Menton equation.
By setting Vo = Vmax/2 we obtain
Vmax/2 = Vmax[S]/Km+[S]
1/2 = [S]/Km+[S]
Km + [S] = 2[S]
Km = [S], when Vo = Vmax/2.
Sorry for the little ramble there I'm not even sure if that helped your question but that should be a start...
[Edited on 7-1-2004 by Al Koholic]
Thanx
Wolfram - 7-1-2004 at 07:37
If you think rationaly without abstract formulas. You would find out that if one part of substrate occupies half of the sites of a given amount of
enzyme a a given time it would be strange if the same amount of substrate would also be able to occupy half of the sites of a enzyme concentration
hundred times larger do you understand what I mean?
But i guess its like this that the enzyme is so much larger than the substrate that the enzyme is always rate limiting and not the substrate.
Geomancer - 7-1-2004 at 14:03
You're right, of course, Al Koholic. I had completely ignored product formation. Perhaps I should have mentioned my grasp of biochemistry is
shakey on the best of days 
. But now I'm confused--why doesn't the
concentration of the product affect the concentration of the enzyme/substrate complex? Is Km defined for "low" product concentrations, i.e.
where the reverse reaction rate is negligible?