Lecture 12/13: Enzyme kinectics/Allosteric enzymes Flashcards
a first order reaction
there is only a single reactant
A → B.
rate of first order reaction
A → B
Units for first-order rate constants, K are …
s^-1
second order reaction
there are two reactants
A + B → products
rate of second order reaction
v = k[A][B]
Units for second-order rate constant (K)
M^-1s^-1
What determines the rate of a
reaction?
The height of the energy barrier, ∆Go‡, determines the rate of the reaction.
pseudo first order
a second-order or bimolecular reaction that is made to behave like a first-order reaction
Zero order
Does not depend on concentration of reactants
Initial velocity V0
Moles of products formed per second at very early time points
The simplest enzyme mechanism involves the equation:
Michalis Menten Kinectics assumptions (2)
- ignore E+P going back to ES (ignore K-2)
- Concentration of enzyme substrate complex is constant
Km formula with rate constants
The Michaelis-Menten Model assumes Steady state kinectics. This means
- Measurements of V0 while [ES] is relatively stable
Vmax is directly dependent on
enzyme concentration
Michaelis-Menten, what happens when [S]«Km
When [s]«Km, velocity is directly porportional to [s]
Michaelis-Menten, what happens when [S]»Km
Rate is zero order, independent of [s]
Michaelis-Menten, what happens when [S]=Km
Km is equal to the [s] at which the reaction rate is half its maximal value
Talk about Km value (3)
depends+what it is+ cellular concentration
- Depends on the substrate, PH, temperature, ionic strength
- Km is the concentration of substrate at which half of the enzyme molecules are bound to substrate
- The cellular concentration of a particular substrate is often close to the Km value
Why is it beneficial for the cellular concentration of a particular substrate to be close to the Km value?
- significant catalysis, sensitive to flunctuations in [s]
- If enzyme always operate at vmax (saturated) we will have signifigant velocity but any slight changes in [s] wil not change vmax
slope, x-inyersept and y interscept in lineweaver plot what is that
Turnover (2)
What it is+ condition of
- The number of substrate molecules that the enzyme can turn into products per unit time when fully saturated with substrate
- condition of max velocity
1/ Kcat
Time to convert 1 molecule of substrate into product
Km and Vmax values are always
positive