## kinetics

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zonia30
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Joined: Fri Dec 30, 2005 9:06 pm

### kinetics

can anybody please explain michaelis-menton kinetics in a simplified manner. i understand about the Vmax and Km but thats about it i dont understand the equation that i keep coming across.

sdekivit
King Cobra
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you mean the derivation of the Michaelis-Menten-equation ?

zonia30
Garter
Posts: 19
Joined: Fri Dec 30, 2005 9:06 pm
yeah think thats what i mean !!!!! i confused

kiekyon
Coral
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Location: Malaysia
Km ~ Vmax/2,

k1 k3
E + S <====> ES ----> E + P
k2

k1[E][S] = k2[ES] + k3[ES]

[E][S]
[ES] = ----------
(k2+k3)/k1

Km = (k2 + k3) / k1

[S]
V = Vmax X ---------
[S] + Km

[S]
V = Vmax X --------- (cancel the [S])
[S] + [S]

= 1/2 Vmax

good luck

sdekivit
King Cobra
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Joined: Sat Jul 30, 2005 7:16 pm
Location: holland
Contact:
the above reply is very confusing, so i start from the very beginning here

in order to understand the Michaelis-Menten-kinetics, we have to assume an enzymatic reaction has the following mechanism:

E + S <--> ES --> E + P (with E = enzyme, S = substrate and P = product; ES = enzyme-substrate-intermediate)

here we have 3 reaction rate constants:

E + S<-- --> ES (to the right: k1 to the left k2) and ES --> E + P (k3)

now assume the reaction has a steady-state-phase and that is [ES] = 0. In other words: when ES is formed by the reaction E + S, it is immediately reacting to E + P.

This is important in order to state the Michaeilis-Menten-constant.

Now we have several reaction rate constants, with the follwoing expressions for the formation of the product and the formation of the enzyme-substrate-complex:

formation product: s = k3 * [ES]
formation intermediate: s = k1 * [E] * [S]
breakdown ES: s = (k2 + k3) * [ES]

Now we have the steady-state-assumption that the formationrate of ES equals the breakdown of ES:

k1 * [E] * [S] = (k2 + k3) * [ES]

--> [ES] = k1 * [E] * [S] / (k2 + k3)

Now the fraction (k2 + k3)/k1 is the Michaelis-Menten-constant Km thus we can also state:

--> [ES] = [E] * [S] / Km (k1 / (k2 + k3) = 1/Km)

But now we have an expression with enzyme-concentration, but we want a relation between rate and substrateconcentration. Therefore we need to assume that in the steady-state the enzyme concentration is total concentration of free enzyme minus the intermediate form:

[E] = [E]t - [ES]

this yields:

[ES] = [S]([E]t - [ES]) / Km

This needs to be rearranged, so that [ES] is separated from the rest:

[ES] = [S][E]t - [S][ES] / Km

Km * [ES] = [S][Et] - [S][ES]

Km * [ES] + [S][ES] = [S][E]t

[ES](Km + [S]) = [S][E]t

[ES] = [E]t * [S] / (Km + [S])

Now this looks like it, but we want to know the realtionship between [S] and V so we need to get rid of [E]t. Remember we stated that the formation of product equals k3 * [ES]. This is V.

V = k3 * [ES] = k3 * [E]t * [S] / (Km + [S])

now the maximum rate of the reaction occurs when all the free enzyme is in the form of the intermediate: [E]t = [ES]

And thsu V = K3 * [ES] --> Vmax = k3 * [E]t

and thus this yields the Michaelis-Menten-kinetics:

V = Vmax * [S]/Km + [S])

Now out of this expression we can see the follwoing:

1. V = Vmax when [S] >> Km

2. when [S] = Km, then V = 1/2 * Vmax

3. when [S] << Km V = Vmax * [S]/Km and thus V is directly proportional to [S]

Good luck with it

kiekyon
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Location: Malaysia
yeah, that's much better

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