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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 = [ 𝑃 ] 𝑑𝑡 = 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) 𝒗 ? = 𝒗 𝒎𝒂𝒙 [ 𝑺 ] 𝑲 𝑴 + [ 𝑺 ] (M-M equ.) where v max = k 2 [E T ] (also k 2 [ES] max ) and K M = 𝒌 ? + 𝒌 −? 𝒌 ? Remember: Velocity in the presence of an inhibitor (competitive, reversible) 𝑣 0 = 𝑣 𝑚𝑎𝑥 [ ? ] 𝛼𝐾 𝑀 + [ ? ] 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 = 𝑣 𝑚𝑎𝑥 [ 𝐸 ] ? 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 =αK 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] αK 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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