Chapter 6 - enzymes: kinetics and enzyme inhibition

Question 1

what is Δ G’°

Answer 1

• Biological standard free-energy change
• because H+ concentrations in biological systems are far below 1M
• use Δ G’° for standard free energy at pH7

Question 2

what are the two components that contribute to Δ G’°

Answer 2

entropy and enthalpy

Question 3

how does an enzymes affect Δ G’°

Answer 3

enzymes do not change Δ G’°, they only affect the rate

Question 4

how do enzymes affect equilibria

Answer 4

• they don’t change K’eq
• just help reach equilibrium faster

Question 5

how do enzymes affect reaction rates

Answer 5

• they increase the rate of reaction by lowering the activation energy

Question 6

how do enzymes affect activation energy

Answer 6

• they lower the activation energy

Question 7

what is a transition state

Answer 7

• the activated substrate molecule has undergone a partial chemical reaction
• the point at which decay to product or back to substrate is equally likely

Question 8

define activation energy

Answer 8

• the difference between the energy levels of the ground state and the transition state
• “energy barrier” between substrate and product

Question 9

what is binding energy (Δ GB)

Answer 9

• the difference between the activation energy catalyzed and uncatalyzed reactions

Question 9

what is induced fit

Answer 9

• 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

Question 9

how is induced fit different from the lock and key model?

Answer 9

• 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

Question 10

what does a dissociation constant (Kd) represent

Answer 10

• the equilibrium constant for the release of a ligand (substrate) from a protein (enzyme)
• Kd= ([E][S])/[ES]
• lower Kd means higher affinity

Question 11

what is the relationship between [S], V0 and Vmax?

Answer 11

• low [S] –> linear increase in Vo
• at higher [S] V0 increases by smaller amounts and eventually plateaus to Vmax

Question 12

describe what the Michaelis-Mentin equation shows

Answer 12

• equation for the curve expressing the relationship between [S] and V0
• applies for one-substrate enzyme-catalyzed reaction
• takes on steady state assumption

Question 13

how does the Michaelis-Menten equation confirm observations from the plot

Answer 13

• 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

Question 14

What is the steady state assumption

Answer 14

• 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

Question 15

define Km in terms of rate constants

Answer 15

Km=[S] at 1/2 Vmax

Question 16

When does [ES]=[Et] (total enzyme concentration)

Answer 16

• at Vmax all of the enzyme is bound to the substrate so [Et]=[ES]

Question 17

how would you change the Michaelis-Menten equation into Lineweaver-Burk form

Answer 17

• take the reciprocal of both sides and simplify
• plot 1/V0 vs 1/[S]

Question 18

On a lineweaver-burk plot, how do you determine Vmax and Km

Answer 18

• the x-intercept is -1/Km
• the y-intercept is 1/Vmax
• slope is Km/Vmax

Question 19

what are some of the benefits of the LIneweaver-Burk plot

Answer 19

• 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

Question 20

what is Kcat

Answer 20

• 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

Question 21

why is Kcat/Km a good measure of enzyme efficiency

Answer 21

• 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

Question 22

what is a bi-substrate reaction

Answer 22

• enzymes catalyze reactions with two or more substrates

Question 23

ternary complex

Answer 23

• substrates can bind to enzyme in random order or specific order

Question 24

ternary mechanism

Answer 24

• both substrates must bind for the reaction to proceed
• increasing [S2] increases Vmax and decreases Km

Question 25

ping-pong

Answer 25

• S1 binds to get to intermediate state, then S2 can join
• increasing [S2] increases Vmax and Km

Question 26

how could you experimentally distinguish between the two bisubstrate reaction mechanisms

Answer 26

• 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

Question 27

what is reversible inhibition and what are the 3 types of reversible inhibition

Answer 27

• reversible means that the enzyme can return to its original state and continue to function
• competitive
• uncompetitive
• noncompetitive

Question 28

competitive inhibition

Answer 28

• inhibitor binds directly to E at active site
• Km increases becuase more substrate is needed to outcompete the inhibitor
• Vmax stays the same

Question 29

Uncompetitive inhibition

Answer 29

• inhibitor can only bind to ES
• substrate has to have already bound for inhibitor to bind
• decreases both Vmax and Km

Question 30

noncompetetive inhibition

Answer 30

• Inhibitor can bind to E or ES
• Vmax decreases but Km stays the same