Kinetics Review Sheet - Effects on v max Effects on K M No...

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Kinetics Review Sheet E + S E S E + P (where k -2 negligible) Mechaelis-Menten Equation Describes the product formation of a two step catalyzed reaction: v 0 = [ ° ] ±² = k 2 [ES] (because release of product is usually the rate limiting step) @ Steady-state: [E S] = constant AND we are ALWAYS at steady-state!!! (otherwise we’d be sick or even dead) ³ ? = ³ ´µ¶ [ · ] ¸ ¹ + [ · ] (M-M equ.) where v max = k 2 [E T ] (also k 2 [ES] max ) and K M = º ? + º −? º ? Remember: Velocity in the presence of an inhibitor (competitive, reversible) » 0 = » ¼½¾ [ ¿ ] ÀÁ  + [ ¿ ] where α = 1 + [I]/K I , remember: since [E T ] is not affected by I, v max = same [I] = 0 α = 1 note: The higher [I], the smaller v 0 Turnover number UNITS: s -1 k cat = » ¼½¾ [ à ] Ä number of reactions/ units time (for one active site) Catalytic Efficiency UNITS: M -1 s -1 Cat. Efficiency = k cat /K M → frequency of productive encounter of E and S (high = efficient)
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Unformatted text preview: Effects on v max Effects on K M No inhibitor V max = k 2* [E T ] therefore only changes with the change in [E T ] Michaelis-Menten constant Intrinsic property of S-E relationship, therefore never changes Competitive inhibitor Since no effect on [E T ], no change in v max either Constant, but with competitive inhibiter we have K apparent =K M and it may seem like K M changes, but instead we observe a k 1 k 2 k-1 k-2 @ 1/2v max K M = [S] K M = apparent K M (observed K M ) new (apparent) K M (which is based on real K M ) Irreversible inhibitor Decrease in v max since inhibitor inactivates E (resembles decrease in [E T ]) NO!! K M never changes (you should know this by now ) Note: since v max is also effected the relationship @ v = 1/2v max [S]=K M still holds true...
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This note was uploaded on 02/02/2010 for the course BIOCHEM Biochem440 taught by Professor Dr.klevit during the Winter '09 term at University of Washington.

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Kinetics Review Sheet - Effects on v max Effects on K M No...

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