enzyme kinetics. enzyme kinetics i an enzyme-catalyzed reaction of substrate s to product p, can be...

Download Enzyme Kinetics. Enzyme Kinetics I An enzyme-catalyzed reaction of substrate S to product P, can be written Actually, the enzyme and substrate must combine

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  • Slide 1
  • Enzyme Kinetics
  • Slide 2
  • Enzyme Kinetics I An enzyme-catalyzed reaction of substrate S to product P, can be written Actually, the enzyme and substrate must combine and E recycled after the reaction is finished, just like any catalyst. Because the enzyme actually binds the substrate the reaction can be written as: The simplest reaction is a single substrate going to a single product. E SP E+ S k 1 k 1 ES k 2 P+ E
  • Slide 3
  • Rate or velocity of the reaction depends on the formation of the ES The P -> ES is ignored The equilibrium constant Keq is based on the idea that the reaction is limited to the formation of the ES complex and that only K1 and K-1 are involved because the thermodynamics of the reversal of K2 cause it to be minimal How fast an enzyme catalyzes a reaction is it's rate. The rate of the reaction is in the number of moles of product produced per second Keq= k 1 K -1 rate (v)= d[P] dt = k 2 [ES]
  • Slide 4
  • The relationship between the concentration of a substrate and the rate of an enzymatic reaction is described by looking at the concentration of S and v When the reaction is first order - the rate is dependent on [S] When the reaction is zero order, there is no relationship between v and S A second order is between 1st and 0 order, where the relationship between V and [S] is not proportional to [S] [ Substrate] Initial Velocity (V i or V)
  • Slide 5
  • To study enzymes, first order kinetics must be followed! To study enzymes, first order kinetics must be followed! Think of the graph of [S] vs. v in this way: Think of the graph of [S] vs. v in this way: The velocity increases as the substrate concentration is increased up to a point where the enzyme is "saturated" with substrate. At this point the rate of the reaction (v) reaches a maximal value and is unaffected by further increases in substrate because all of the enzyme active site is bound to substrate
  • Slide 6
  • For the most part enzyme reactions are treated as if there is only one substrate and one product. If there are two substrates, one of them is held at a high concentration (0 order) and the other substrate is studied at a lower concentration so that for that substrate, it is a first order reaction. This leads us to the M and M equation.
  • Slide 7
  • Conditions for Michaelis -Menten Two assumptions must be met for the Michaelis- Menten equation Equilibrium -the association and dissociation of the substrate and enzyme is assumed to be a rapid equilibrium and Ks is the enzyme:substrate dissociation constant.
  • Slide 8
  • Conditions for Michaelis -Menten Two assumptions must be met for the Michaelis- Menten equation Steady state - the enzyme substrate complex ES is at a constant value. That is the ES is formed as fast as the enzyme releases the product. For this to happen the concentration of substrate has to be much higher than the enzyme concentration. That is why we only study the initial velocity. Later in the reaction the substrate concentration is relatively lower and the rate of product starts to be limited by diffusion and not the mechanism of the enzyme. Steady state - the enzyme substrate complex ES is at a constant value. That is the ES is formed as fast as the enzyme releases the product. For this to happen the concentration of substrate has to be much higher than the enzyme concentration. That is why we only study the initial velocity. Later in the reaction the substrate concentration is relatively lower and the rate of product starts to be limited by diffusion and not the mechanism of the enzyme.
  • Slide 9
  • Slide 10
  • Michaelis-Menten Enzyme kinetics Don't for get the two assumptions - They both lead to the same equation, the michaelis-menten equation.Don't for get the two assumptions - They both lead to the same equation, the michaelis-menten equation. What is this awe inspiring equation? The Michaelis- Menten kinetic model explains several aspects of the behavior of many enzymes. Each enzyme has a Km value that is characteristic of that enzyme under certain conditions.What is this awe inspiring equation? The Michaelis- Menten kinetic model explains several aspects of the behavior of many enzymes. Each enzyme has a Km value that is characteristic of that enzyme under certain conditions.
  • Slide 11
  • Graphical model of the representation of the M&M eq. Reaction velocity (V) vs concentration of substrate [S] - as [S] increases, velocity increases and eventually levels off = V max 1st order vs zero order rates of reaction - back to the two assumptions There are two important values for each enzyme that are described by the M&M equation; V max and Km (Michaelis-Menten constant) Graphically, these are shown as 1/2 V max = Km can not reach real V max so....Graphically, these are shown as 1/2 V max = Km can not reach real V max so....
  • Slide 12
  • Mathematical model of the representation of the M&M eq. - For the reaction: 1) The Michaelis constant Km is: Think of what this means in terms of the equilibrium. Large vs. a small Km E+ S k 1 k 1 ES k 2 P+ E K m = K -1 + K 2 K 1
  • Slide 13
  • 2 ) When investigating the initial rate (Vo) the Michaelis- Menten equation is: Graphical representation is a hyperbola. Think of the difference between O 2 binding of myoglobin and hemoglobin. When [S] > Km, the initial velocity is independent of [S] When [S] = Km, then Vo = 1/2 V max Prove this mathematically and graphicaly. Vo = V max [S] [S] + K m
  • Slide 14
  • Km is a measure of the affinity of the enzyme for it's substrate and also informs about the rate of a reaction. The binding constant is appoximated by Km Km is a measure of the affinity of the enzyme for it's substrate and also informs about the rate of a reaction. The binding constant is appoximated by Km Rules for using the M&M equation: Rules for using the M&M equation: The reaction must be first order and [S] >> E (two assumptions) The reaction must be first order and [S] >> E (two assumptions)
  • Slide 15
  • Turnover Number - kcat - the direct measure of the catalytic production of product. The larger the kcat is, the more rapid the catalytic events at the enzyme's active site must be. The number of times a binding and reaction event "turns over" -When the [S]

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