Lecture 11: HOW DOES AN ENZYME CATALYSE A REACTION?2 Flashcards
What does a progress curve do?
Measure the appearance of product or disappearance of substrate with time at a steady state
When is initial reaction velocity measured?
At/near time zero
What happens if there is sufficient excess of substrate?
As the amount of enzyme increases, the rate of reaction increases
What is initial velocity proportional to?
Enzyme concentration when substrate is in excess
How is data usually collected?
With a fixed amount of enzyme and variable amount of substrate
What happens to the initial rate as substrate concentration is increased?
At first when there is a low substrate concentration, it increases linearly and then as all the enzyme active sites become occupied, the rate of reaction stops increasing
What kinetic parameters can be identified on a velocity vs substrate concentration graph?
maximum velocity and Km
What is maximum velocity?
The maximum velocity possibly when substrate concentration is infinity
What is Km?
The substrate concentration at which velocity=maximum velocity/2. The Michaelis constant
What does a better enzyme have?
A smalle Km
What did enzymes evolve?
A Km which matched the substrate concentration in the cell
How I binding and catalyses described?
An enzyme, E, converts a single substrate, S, to a single product P that is instantly released.
What do we assume is irreversible?
Conversion of products back to the enzyme substrate complex
What do the relative speeds of K1 and K-1 do?
Define how tightly substrate binds
What is the rate of catalysis?
K2
What does K2 relate to?
Energy of activation for the transition state
What does the Michaelis-menten equation describe?
The V-obs vs substrate concentration curve
What does [ES]/[E] total describe?
The fraction of vmax the enzyme is running at
What are the assumptions to do with the Michaelis-menten equation?
Product is not converted back to substrate, Haldane’s steady state assumption, measuring initial rate insures [S] does not change significantly, [S] is much greater than [E] and all ES complexes have the same rate of reaction
What is Haldane’s steady state assumption?
The rate of ES formation equals the rate of its breakdown