Chapter 6 - enzymes: kinetics and enzyme inhibition Flashcards
what is Δ G’°
- Biological standard free-energy change
- because H+ concentrations in biological systems are far below 1M
- use Δ G’° for standard free energy at pH7
what are the two components that contribute to Δ G’°
entropy and enthalpy
how does an enzymes affect Δ G’°
enzymes do not change Δ G’°, they only affect the rate
how do enzymes affect equilibria
- they don’t change K’eq
- just help reach equilibrium faster
how do enzymes affect reaction rates
- they increase the rate of reaction by lowering the activation energy
how do enzymes affect activation energy
- they lower the activation energy
what is a transition state
- the activated substrate molecule has undergone a partial chemical reaction
- the point at which decay to product or back to substrate is equally likely
define activation energy
- the difference between the energy levels of the ground state and the transition state
- “energy barrier” between substrate and product
what is binding energy (Δ GB)
- the difference between the activation energy catalyzed and uncatalyzed reactions
what is induced fit
- initial substrate binding causes a conformational change in the enzyme
- conformational change maximizes interactions with the substrate and brings specific functional groups in the active site into position to catalyze the reaction
how is induced fit different from the lock and key model?
- in the lock and key model, the pocket (lock) is already the correct shape for the substrate (key)
- in induced fit, the enzyme doesn’t fit the substrate perfectly until it binds and the enzyme changes conformation to fit
what does a dissociation constant (Kd) represent
- the equilibrium constant for the release of a ligand (substrate) from a protein (enzyme)
- Kd= ([E][S])/[ES]
- lower Kd means higher affinity
what is the relationship between [S], V0 and Vmax?
- low [S] –> linear increase in Vo
- at higher [S] V0 increases by smaller amounts and eventually plateaus to Vmax
describe what the Michaelis-Mentin equation shows
- equation for the curve expressing the relationship between [S] and V0
- applies for one-substrate enzyme-catalyzed reaction
- takes on steady state assumption
how does the Michaelis-Menten equation confirm observations from the plot
- at low concentrations, Km is much bigger than [S], so it has a linear relationship to [S]
- at high concentrations, Km becomes negligible and V0=Vmax
What is the steady state assumption
- the rate of ES formation is proportional to the ES breakdown
- ES can breakdown in two ways, back to E and S or forward to E and P
- ES only forms from E and S
- k2 (formation of product) is rate limiting
define Km in terms of rate constants
Km=[S] at 1/2 Vmax
When does [ES]=[Et] (total enzyme concentration)
- at Vmax all of the enzyme is bound to the substrate so [Et]=[ES]
how would you change the Michaelis-Menten equation into Lineweaver-Burk form
- take the reciprocal of both sides and simplify
- plot 1/V0 vs 1/[S]
On a lineweaver-burk plot, how do you determine Vmax and Km
- the x-intercept is -1/Km
- the y-intercept is 1/Vmax
- slope is Km/Vmax
what are some of the benefits of the LIneweaver-Burk plot
- more accurate Vmax since it is a point rather than an estimate
- more useful to distinguish reaction mechanism
- more useful when looking at enzyme inhibition
what is Kcat
- catalytic constant
- how many substrate molecules one enzyme molecule can convert to product per unit time
- rate constant that describes the limiting rate of an enzyme-catalyzed reaction at saturation
why is Kcat/Km a good measure of enzyme efficiency
- low Km is correlated with high substrate affinity for enzyme
- Kcat tells how fast an enzyme can convert substrate to product when substrate is bound
- reflects how efficiently an enzyme converts a substrate into product relative to its affinity for the substrate
what is a bi-substrate reaction
- enzymes catalyze reactions with two or more substrates
ternary complex
- substrates can bind to enzyme in random order or specific order
ternary mechanism
- both substrates must bind for the reaction to proceed
- increasing [S2] increases Vmax and decreases Km
ping-pong
- S1 binds to get to intermediate state, then S2 can join
- increasing [S2] increases Vmax and Km
how could you experimentally distinguish between the two bisubstrate reaction mechanisms
- hold the concentration of S2 constant while varying S1
- repeat for several values of [S2]
- plot 1/V0 vs 1/[S1]
- ternary - slope will decrease with inc. [S1]
- ping-pong - slope will stay constant, line will shift
what is reversible inhibition and what are the 3 types of reversible inhibition
- reversible means that the enzyme can return to its original state and continue to function
- competitive
- uncompetitive
- noncompetitive
competitive inhibition
- inhibitor binds directly to E at active site
- Km increases becuase more substrate is needed to outcompete the inhibitor
- Vmax stays the same
Uncompetitive inhibition
- inhibitor can only bind to ES
- substrate has to have already bound for inhibitor to bind
- decreases both Vmax and Km
noncompetetive inhibition
- Inhibitor can bind to E or ES
- Vmax decreases but Km stays the same
