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Lecture10 - Feedback Inhibition In feedback inhibition the...

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Feedback Inhibition In feedback inhibition, the end product of a metabolic pathway shuts down the pathway Active site available Isoleucine used up by cell Feedback inhibition Isoleucine binds to allosteric site Active site of enzyme 1 no longer binds threonine; pathway is switched off Initial substrate (threonine) Threonine in active site Enzyme 1 (threonine deaminase) Intermediate A Intermediate B Intermediate C Intermediate D Enzyme 2 Enzyme 3 Enzyme 4 Enzyme 5 End product (isoleucine) Figure 8.21 Feedback inhibition occurs when isoleucine, the end- product of the pathway, binds to an allosteric site on threonine deaminase, the first enzyme in the pathway.
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Specific Localization of Enzymes Within the Cell Structures within the cell help bring order to metabolic pathways Some enzymes act as structural components of membranes Some enzymes reside in specific organelles, such as enzymes for cellular respiration being located in mitochondria
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Symbols 4 V Reaction rate (velocity) (M/sec) = d[P]/dt; M/sec V o Initial rate - rate of increase in product with time  when  [P] is low V max Maximum rate for catalyzed reaction. V o   approaches V max  at high substrate concentration. V o   then depends on  [E T ] but not [S] [E T ] Total concentration (M) of enzyme (both free and  bound to  substrate) k 1 Rate constant for step #1 in a reaction (1/sec for  first  order reaction; M/sec for second order  reaction) k   Rate constant for reverse of step # 2 in a reaction
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How does amount of product depend on time? Add E to S; S is converted to P over  time. The amount of P increases with time. The rate at which P is produced  decreases with time. So if the rate is changing, what do use  for a "reaction rate?" Use V , initial rate.  Fig. 6-11 time product Fixed ample [E] V o
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How does V depend on [S]? What if you double the  amount of substrate? Then the rate doubles. Fig. 6-11 time product Fixed ample [E] 2xV o It appears that V o  = k[S]   2V o V o
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Continue to increase [S]… V  continues to increase. 
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o 8 But, say you put in more and more S  (lots!)  At some point, amount of product  made doesn't increase because all of  the enzyme molecules are occupied  (saturation) Note curve at right, "x" axis is [S].  Note saturation behavior. V  reaches V  when E becomes  It looked like  V o  = k[S]   Fixed amt. Of E .
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The importance of the ES complex Clearly, we can't describe kinetics of an enzymatically catalyzed  reaction with the simple expression  V o  = k[S] Recall the key to catalysis: the enzyme binds the substrate through  a set of weak interactions … lowering the activation energy  G  of the reaction
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A simple enzymatic reaction
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