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Enzyme rate expression

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An enzyme is a large protein molecule (typically with a molar mass greater than 20,000 AMU) that has a structure capable of catalyzing specific biological reactions. In general one or more reactant molecules (called substrates) bind to an enzyme at its "active sites" where the reaction the enzyme is responsible for occurs. The general mechanism for enzyme catalysis is:

Where E stands for the enzyme, S the substrate, ES a short-lived intermidiate enzyme-substrate complex, and P the product of the reaction.

When solving this rate expression it is helpful to replace [E], the total enzyme concentration with [E]0 the initial amount of enzyme present which during the reaction is equal to:
[E]0 = [E] + [ES]
since it is very hard to measure the quantity of enzyme bound to the substrate [ES] during the reaction. This substitution will help you in solving this problem. Use the steady state approximation we learned in class to derive an expression for the rate of formation of product, d[P]/dt, that is only a function of [E]0 (defined above), [S], and the rate constants. Your expression should be independent of [ES] (the quantity of enzyme-substrate present), and [E] (the quantity of unbound enzyme), as these are very hard to measure during a reaction.

The equation you will be deriving was originally derived by Michaelis and Menten and is the standard equation used by biologists to describe most enzyme kinetics.

To help keep the input of your answer simple please make the following letter substitutions (note answers are CASE sensitive, use caps and lower case as defined here):
[E]0 = e k1 = a k2 = c
[S] = S k-1 = b
d[P]/dt =

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The steady-state approximation is based on the fact that we form so much enzyme-substrate complex that its concentration will essentially not change

d[ES]/dt=0

We want to figure out the rate of appearance of the product, d[P]/dt, which can be expressed as this

d[P]/dt = k2[ES]

But we need to change out that [ES] for something more useful. We use the ...

Solution Summary

This solution is a step-by-step guide to deriving the rate law of an enzymatic reaction, especially the steady-state approximation. It is clearly laid out how this approximation works in simple terms.

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Inhibitors: Irreversible and Reversible

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It is possible to distinguish between an irreversible inhibitor and a classical mixed type reversible inhibitor by measuring Vmax of the enzymatic reaction as a function of [E]t in the absence and presence of the inhibitor.

What does the slope represent in each case?

What does the X-axis intercept represent in each case?

Please view attachment to see a diagram for this question.

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