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Unformatted text preview: ↳ the number of substrate molecules that are converted to product by one active site when [S] is high relative to K M . ↳ called the turnover number v = k cat E [ ] total S [ ] total S [ ] total + K M What happens to the equation if [S] total << K M ? v = k cat E [ ] total S [ ] total K M = k cat K M E [ ] total S [ ] total At really low [S], the rate is dependent on [E] total and [S] total k cat /K M is a secondorder rate constant ↳ describes the enzyme’s efFciency . ↳ called the speciFcity constant . k cat (s1 ) K M (mM) k cat /K M (M1 s1 ) Sub. 1 1,000 10 1 x 10 5 Sub. 2 10 0.001 1 x10 7 Example: A hydrolase’s activity is measured with two different substrates. The k cat and K M results are below. Which substrate is the better one for the enzyme?...
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 Fall '11
 T.Baird
 Chemistry, Enzyme, Kinetics, vmax, Kcat

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