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Lecture 16 - Friday Lecture 16 Announcements 1 Quiz 4...

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Friday, September 29, 2006, Lecture 16 Announcements: 1. Quiz 4 results: A 24.9 B 23.9 C 24.4 2. Today in Comstock B106 from 1:25 – 2:15PM: Application to medical schools: writing the personal statement 3. Note! Class moved to Bailey Hall for Monday 10/2 only. 4. Rasmol, Protein Data Bank. Some changes: a. The PDB web site has been modified since July 2006 when this Rasmol exercise was written. b. In the second step of Part I of the exercise, rather than clicking on Other, instead click on the number at the bottom of the Other column. As of 9/28/06, this is 34 . c. Part II . Under Biological Process, some are click-able, some are not! d. Part III . There is no longer a category Biological Unit. Instead, the category is termed Molecular Description. 5. Superb 5-min movie about the cell: http://aimediaserver.com/studiodaily/videoplayer/? src=harvard/harvard.swf&width=640&height=520 Wednesday’s lecture: How do catalysts speed up rxns with no effect on Keq? • Catalysts work by two (and only two!) general types of mechanism: 1. Directly bind the TS and thereby lower its free energy: example of Ni catalyst; 2. Take the rxn along a different pathway than the uncatalyzed rxn-- a pathway with different intermediates and a different, lower free energy TS : example of acid catalysis • The primary data of kinetics are [product] vs time. These data yield Vin Today’s lecture: p. 109 A most useful feature of the Michaelis-Menten equation: It is easy to measure V in and [S]! We can then get some information on how the enzyme works, in particular from the value of K M . When [S] is much greater than K M , then the V in = V max = k 2 [E] total . See this for yourself by making [S] >>K M in the M-M Equation, so that K M + [S] 2245 [S]. Hey, check out this little calculation: when V in = 1/2(V max ) = (V max [S])/(K M + [S]), the V max value cancels out, and 1/2 = [S]/(K M + [S]), so that K M = [S] !! In other words, at the substrate concentration that gives exactly half maximal rxn velocity, K M = [S]. This little calculation informs you that K M has units of concentration! Now let’s consider again the progress of rxn: E + S ES E + P
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** Note that for this simple rxn scheme, each of the elementary rate constants, k 1 , k -1 , k 2 is associated with a barrier on the free energy diagram.
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