15 - Wednesday Lecture 15 Announcements PyMOL assignment#5...

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Wednesday, September 28, 2011 Lecture 15 Announcements: PyMOL assignment #5, Exploring the Protein Data Bank Friday 9/30 at 2:55PM in Comstock B108: The medical school interview-- what are they looking for? A wallet was lost in the Statler Auditorium right after the Monday lecture. If you have any sort of info about it, please let the prof know! If you are interested in Shoals Marine Lab, SEA semester, or just marine biology at Cornell: Monday 10/3 at 4:45PM in G01 Stimson Hall. Monday's lecture: All the chemical reactions in Life are catalyzed reactions. No exceptions. How do these catalysts work? How catalysts work: bind the TS or take the rxn along a different path that has a lower free energy TS Graph of Vin vs [reactant] is very different for catalyzed vs uncatalyzed rxns Today’s lecture, Top of p. 109 The reason that Vin vs [S] "levels off" at high [S]: the catalyst (the enzyme) becomes saturated with the substrate, and catalytic rate then reaches a maximum. Thus, the observation that V in rises to approach a constant value as [S] becomes large is the hallmark of catalysis in general. The most simple possible reaction scheme that explicitly shows the role of the enzyme postulates that the substrate first binds the enzyme (in one step), then reacts, again in one step, to form product: E + S ES E + P This equation describes binding of free enzyme and substrate to form a complex in which no chemical reaction has yet occurred (“merely” binding has happened), the enzyme-substrate complex ES . This ES complex then undergoes chemical reaction in one step to form product and free enzyme. (Real enzyme-catalyzed reactions can be MUCH more complicated than this simplest mechanism, with the rxn “step” actually having many individual steps). p. 110 Next we seek an equation to express the rxn velocity in terms of easily measured quantities (since it is hard to measure [ES]):
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1. Use the term k 2 to mean the rate constant of the rxn of ES to form E + P. 2. Neglect k -2 , which is the rate constant for the reaction of enzyme + product going back to the ES complex. Neglect this term because we always measure V initial , i.e. at time t = zero when there is no product formed yet. ( We have not assumed that the value of k -2 is nearly zero! ). 3. Assume that [ES] is not changing. The overall rxn is not at equilibrium, but [ES] has built up to a steady value . This is not very obvious, but it turns out to be a really good assumption. Why? Because ES is formed by "mere" collision of S with E, so the energy barrier between E+S and ES is usually but not always small (no very unstable structure must form), and therefore the rate to get over the barrier in either direction is fast. E+S and ES would be in equilibrium, except that some product is forming: some
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15 - Wednesday Lecture 15 Announcements PyMOL assignment#5...

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