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BIOCH499A (12)

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Department
Biochemistry
Course Code
BIOCH499A
Professor
Gail Amort- Larson

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The Duplex by Glenn McCoy Read 17.1, 17.2, 17.4 Practice Solutions to PS #2 are posted online. A sample exam is also posted. Give it a try under simulated conditions: 45 min, closed book, etc. Solutions will be posted on Friday. 1 2 Rate = k [ES] Enzyme mechanism 2 Enzyme highly specific catalyst Can’t measure [ES] See Silberberg 16.26 k 1E][S]  k–1ES] + k [2S] (steady state approx) k1[E][S] [ES] = k–1 k 2 So the rate is k1[E][S] Rate = k 2 k + k –1 2 E = free enzyme You are expected to know how to arrive at this result. S = substrate The next page shows how to obtain the final result. ES = enzyme–substrate complex You are not responsible for the derivation but you Step 1 E + S  ES should know the page after that (p. 5) on how to use  the final equation and the plot on p. 6. Step 2 ES  E + Products Or on one line E + S  ES  E + Products 3 4 Unfortunately, it is not convenient to measure the k + k Define K =M –1 2 concentration of the free enzyme. k 1 K M Michaelis – Menten constant Instead, the total enzyme concentration [Eo] is measured. (it includes the free and bound enzyme). k2[Eo][S] Rate = K + [S] M [Eo] = [E] + [ES] So [E] = [Eo] – [ES] 2 extreme case: And[E= S] k1[E][S] k–1+ k2 1. [S] >> K M (k + k ) [ES] = k [E ][S] – k [ES][S] k2[Eo][S] st –1 2 1 o 1 Rate  [S] = k 2E o 1 order ? (k–1 k 2 [ES] + k 1ES][S] = k 1E ]oS] But [E o] = constant  pseudo zero order! k1[E o[S] [ES] = 2. [S] << K k–1 k +2k [S1 M k1 2[E o[S] k2[Eo][S] k2 st Rate = k2 [ES] = k + k + k [S] Rate  K M = K M [Eo][S]  pseudo 1 order! –1 2 1 5 6 Catalysis A catalyst is a substance that increases the rate of rxn without being consumed. Catalysts occur in all parts of nature, the environment and chemical industry. The human body functions because of enzymes (which are really catalysts). How does the catalyst do it? It provides an alternative path with a lower activation energy for both the forward and reverse rxns. It allows equilibrium to be rea
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