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Topic: Michaelis Menten kinetic problem  (Read 8384 times)

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Offline f77126

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Michaelis Menten kinetic problem
« on: February 25, 2009, 10:56:14 PM »
Activity of the mutant enzyme doubled when the substrate concentration was increased from 200 uM to 400 uM. Activity of the wild-type enzyme increased only 10% when the substrate concentration was increased from 200 uM to 400 uM. Which form of the enzyme has higher Km value? Explain your reasoning

Offline Yggdrasil

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Re: Michaelis Menten kinetic problem
« Reply #1 on: February 26, 2009, 12:15:01 AM »
Start with the Michaelis-Menten Equation (see wikipedia or a intro biochem textbook) and see if you can derive anything from that.

Offline mascott

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Re: Michaelis Menten kinetic problem
« Reply #2 on: March 22, 2009, 09:27:42 PM »
Better than using equations is to understand the underlying mechanisms in the MM model. In MM, as I'm sure you know, there are two processes: 1) binding of the substrate to the enzyme - usually modelled as a reversible reaction, and 2) the catalytic reaction forming the product - irreversible.

Km is defined as a mixture of the reaction constants for both these reactions, and when rate of reaction is plotted against substrate concentration, you get a stereotyped saturation curve, with Km being defined as the substrate concentration at half max speed. Crucially, this curve starts with a large slope, and as increases, its slope diminishes. So if the mutant has a rapidly changing speed with changes in substrate concentration, it has clearly not yet approached saturation (ie Km < or ~= ). Conversely, if the wildtype reaction rate is barely changing with , it has approached saturation - and therefore >> Km. Since is the same in both cases, the wild-type must have a smaller Km.

Hope this helps.
:)

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