terms for chem exam 2 Flashcards
the induced- fit model of enzyme actions assumes that the enzyme active site is more flexible pocket whose shape changes to accommodate the substrate molecule leading to a better fit of substrate with enzyme
binding energy is free energy relases n forming multiple weak bonds and interactions between an enzyme and its substrate
binding energy is used to lower the activation energy barrier
an enzyme speeds a reaction by lowering the activation energy and changing the reaction pathway, providing a lower energy route for the conversion of substrate to product
enzymes do decrease the activation energy
how does an enzyme lower activation energy?
stablizing the transition state, puts molecules in close proximity to react(increases the local concentration of reactants)
enzymes put molecules in correct orientation, reactants are not only near each other on enzyme, they’re orientated in optimal position to react, making it possible to always collide in the correct orientation
model of enzyme action
1. initiation: reactants bind to the active site in a specific orientation
2. transition state: shape changes, using less energy
3. termination: products have a lower affinity for the active site
Michaelis-Menten Equation, initial rate measurements [s]»_space; [E], Vo=k2, km=k2+k-1/k1
the theoretical maximal velocity, Vo=Vmax[s]/Km+[s],
vmax is a constant
vmax is the theoretical maximal rate of the reaction-never achieved in reality
to reach Vmax would require all enzyme molecules are bound with the substrate,
Vmax is the maximum rate that can be seen in the reaction
- substrate is present in excess
- enzyme can be saturated (zero order reaction)
Km, the Michaelis constant, a constant for any given enzyme/substrate pair (indep)
- derived from rate constants
Km is an inverse measure of ES binding, a relative measure of the affinity of a substrate for an enzyme (how well it binds)
-small Km means tight binding; large Km means weak binding
at one half of the maximal velocity, the substrate concentration at this velocity will be equal to the Km
higher Km=lower substrate affinity to enzyme=higher [S] required to reach 1/2 Vmax