MSKC052207
Chemically, most single-substrate enzyme-catalyzed reactions are first order with respect to substrate. When the concentration of of substrate becomes saturating, the reaction becomes zeroth order. In 1913, Leonor Michaelis and Maud Menten proposed a mechanism which explains this shift from first order to zeroth order. A reaction between a substrate, substrate, S, and and an enzyme, enzyme, E, takes takes place in two steps. In the first step the substrate and the enzyme combine to produce the ES intermediate which in the second step breaks apart into free enzyme and product. E+S
k1 k2
ES
k3
E+P
If the second step is the rate-determining step, then the rate of the reaction, V = k 3[ES]. Consider the rate of formation of ES, that is rate form = k1[E][S] and the rate of decomposition of ES, that is rate decomposition = k2[ES] + k 3[ES]. It is assumed that [ES] reaches a steady state and changes no more during the course of the reaction in which [S] decreases and [P] increases. Under such a steady state situation rate form = ratedecomposition . K1 [E][S] = (k 2 + k3)[ES] So that: [ES] =
[E][S] . (k2 + k3)/k1
The Michaelis constant, K M is defined as KM = (k2 + k3)/k1 So that: [ES] =
[E][S] KM
The concentration of the uncombined substrate, [S], is approximately equal to the total substrate concentration, provided the concentration of the enzyme is much lower than that of the substrate. The concentration of the uncombined enzyme, [E], equals the total enzyme concentration, [E] T, minus the concentration of the enzyme in the ES complex, [E] = [E] T - [ES]. It follows that: [ES] =
([E]T - [ES])[S] [E][S] = KM KM
[ES] =
[E]T[S] [ES][S] KM KM
So that: (1 +
KM + [S] [E]T[S] [S] )[ES] = ( )[ES] = KM KM KM
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Then: [ES] =
[E]T[S] KM [S] x = [E]T KM KM + [S] KM + [S]
And: V = k3[ES] = k 3[E]T
[S] KM + [S]
V is proportional to [S] until it reaches a maximum value, after which V is no longer dependent upon [S] (first order to zeroth order). This maximum rate is reached when the enzyme sites are saturated with substrate and adding more substrate cannot increase the rate of the reaction. At this maximum maximum rate, [S] is large large and indeed indeed much much larger than than K M So that: [S]/(K M + [S]) = 1 And: Vmax = k3[E]T. Substitution renders the rate law, V = V max
[S] KM + [S]
This expression for the rate explains the first order shift to zeroth order as [S] increases. When [S] << K M, V = [S]V max/KM and V, the rate is proportional to [S]. When [S]>> K M, V = Vmax. At large [S], the rate reaches a maximum value and this value is constant independent of any further increase in [S]. Note that when [S] = K M, then V = V max/2. It follows that a meaning for K M is it is the concentration of the substrate which renders the rate of the reaction equal to half its maximum value. The KM is an index of how easily an enzyme can be saturated by the substrate it is activating. For our experiment, a plot of the Michaelis and Menten equation:
vo
Vmax [S] K m [S]
Vmax 1 Km / S
looks like:
MSKC052207
[S] is the actual substrate concentration present in mM, V 0 is the initial reaction velocity at that particular concentration in moles ONP produced/min, and V max and Km are constants whose values depend upon our solution of -galactosidase and assay conditions employed. From your graph, you will probably not be able to predict exactly what the eventual value of Vmax will be at saturating substrate concentrations without using a nonlinear regression analysis. And, without an accurate determination of V max, you will be unable to determine K M. An alternate way to obtain V max was suggested by Lineweaver and Burke. They suggested that instead of plotting v o vs. [S], we instead plot 1/v o vs 1/[S]. (i.e. a double reciprocal plot). A plot of the Lineweaver-Bur Lineweaver-Burke ke equation: equation:
1 vo
1 Vmax
K m 1 looks like: Vmax [S]
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Since a Lineweaver-Burke plot is linear, it is possible to extrapolate the line to the Y-axis (which is the equivalent of extrapolating S to infinity) and determine the value of 1/V 0 at infinite substrate concentration (i.e. 1/V max). The second constant, K M, can also be determined from a Lineweaver-Burke plot since the slope of the line is K M/VMax. Alternately, you should note that the X-intercept X-interc ept of the Lineweaver-Burke plot is equal to -1/K M. Inhibition
There are three major types of inhibition. Competitive inhibition involves a substrate mimic binding the active site. This does not alter Vmax of the enzyme, because large amounts of substrate will overwhelm the competitive inhibitor and allow the enzyme to function at Vmax. It does increase apparent Km, as more substrate is needed to have the enzyme function at rate Vmax/2 Noncompetitive inhibitors bind an allosteric site of the enzyme, and prevent it from functioning catalytically. This effectively reduces the concentration of enzyme in solution. The observed Vmax will be lower, but Km will not be affected. Uncompetitive inhibitors are rather rare, and do not come up as frequently in problems. They bind allosteric sites on the ES complex and prevent catalysis from proceeding. By Le Chatelier, this increases the enzyme’s apparent affinit y for the substrate (so Km is lower). Since the effective enzyme concentration is lowered, Vmax also decreases. The ratio Km/Vmax remains constant. Know what Michaelis-Menten and Lineweaver-Burke plots look like for each kind of inhibition!
