Quiz 12

Question 1

kinetics

Answer 1

• the branch of science concerned with the rates of reactions

Question 2

enzyme kinetics

Answer 2

seeks to determine the initial and maximal reaction velocity that enzymes can attain and the binding affinities for substrates and inhibitors

Question 3

simple rate equations describe the progress of

Answer 3

first and second order reactions

Question 4

michaelis-menten equation

Answer 4

relates the initial velocity of a reaction to the maximal reaction velocity and the Michaelis constant for a particular enzyme and substrate

Question 5

enzyme catalytic efficiency is expressed as

Answer 5

Kcat/Km

Question 6

bisubstrate reactions can occur by

Answer 6

ordered or random sequential mechanisms or by pingpong mechanism

Question 7

line weaver- Burk plot

Answer 7

can be used to present kinetic data and to calculate values for Km and Vmax. useful for analysis of inhibitors

Question 8

rate law

Answer 8

mathematical relationship between the reaction rate, or velocity, “n” and concentration of reactants
v = Vmax[s]/Km+[s]

Question 9

rate or velocity

Answer 9

the amount of A consumed or P formed per unit time [t]

Question 10

rate constant k

Answer 10

proportionality constant in the rate equation

Question 11

simple first order rxns display

Answer 11

a linear plot of the substrate [a] and product [p]

Question 12

the order for any reactant is given by its

Answer 12

exponent in the rate equation

Question 13

enzyme catalyzed rxns are more complex than

Answer 13

chemically catalyzed rxns

Question 14

Vo [initial velocity] is a number equal to

Answer 14

the slope at the beginning of each time course

Question 15

enzyme catalyzed rxns

Answer 15

• at low [s] the rate is proportional to s as in first-order rxn : n=k [S]
• at high [s], the enzyme reaction approaches zero order kinetics : n= Vmax
  rate of formation of ES: k1
  rate of disassociation: k-1
  rate of product formation : k2

Question 16

initial velocity is plotted as

Answer 16

a function of substrate conc

Question 17

Leonor Michaelis and Maud mentens theory

Answer 17

1. assumes formation of enzyme-substrate complex [ES]
2. assumes the [ES] is constant, “steady-state assumption”
3. assumes that at high [s] the breakdown of ES [k2] to form product P, is the rate limiting step, equal to the catalytic rate constant , Kcat

Question 18

steady-state assumption

Answer 18

• no change in the conc of the ES complex
• rapid flux means the ES complex either dissociates or results in the product formation, as rapidly as new ES complexes are forming

Question 19

lineweaver-burk plots

Answer 19

offer another useful form rate of law
- begin with n= Vmax [s]/ (Km +[s]) and take reciprocal of both sides
- distinguish between types of bisubstrate reactions

Question 20

Km

Answer 20

• substrate conc at 1/2 Vmax
  2. K-1+K2/ K1

small km means little dissociation [tight binding]
large km means lots of dissociation [little binding]

Question 21

Vmax

Answer 21

• constant for each enzyme equal to Kcat [E]t
• the theoretical maximal rare of the reaction [never achieved in reality ]

Question 22

turnover number

Answer 22

Kcat is the turnover number
K2=Kcat= Vmax/ET

Question 23

catalytic efficiency , Kcat/Km

Answer 23

Kcat/Km
- a way to score how well an enzyme catalyzes a reaction
- upper limit is the diffusion limit approaching 1x10^9, the rate at which E and S encounter each other by diffusion

Question 24

enzyme substrate recognition and catalysis are greatly dependent on

Answer 24

pH

Question 25

rates of enzyme-catalyzed reactions generally increase with

Answer 25

increasing temp.
- at temps above 50 to 60c enzymes typically show a decline in activity
2 effects:
1. enzyme rate typically doubles in rate for every 10C as long as the enzyme is stable and active
2. at higher temps, the protein becomes unstable, and denaturation occurs

Question 26

bisubstrate reactions may be sequential or single displacement reactions or double displacement [pingpong] reactions

Answer 26

1. single displacement aka sequential can be two distinct classes
   a. random
   b. ordered
2. double displacement [pingpong] rwactions
   involves a covalent intermediate

Question 27

double displacement [ping-pong] reactions proceed via formation of a

Answer 27

covalently modified enzyme intermediate, E’ product [p] of the reaction with A is released prior to the rxn of the enzyme with the second substrate
- characterized by parallel lines

Question 28

reversible inhibitors may

Answer 28

bind at the active site or at some other site

Question 29

enzymes may also be inhibited in an

Answer 29

irreversible manner
- penicillin is an irreversible, suicide inhibitor

Question 30

competitive

Answer 30

raises Km with no change in Vmax

Question 31

noncompetitive

Answer 31

decreases Vmax with no change in Km

Question 32

mixed noncompetitive

Answer 32

alters both Km and Vmax

Question 33

uncompetitive

Answer 33

alters both Km and Vmax but with same slope, Km/Vmax

Question 34

assessment of kinetic mparamenetes using the michaelis-menten equation and line weaver-burk plots allows for

Answer 34

• characterization of enzymes, identification of bi-substrate reactions and characterization of inhibitors

Question 35

kinetics

Answer 35

• kinetics analysis is a deductive reasoning tool
• kinetics cannot prove a reaction mechanism
• kinetics can only rule out various alternative hypotheses because they dont fit the data