Lecture 12 - Enzyme rate (Inhibition & Activation)

Question 1

The Michaelis-Menten equation

Answer 1

describes The V vs. [S] curve

V = Vmax [S] / Km + [S]

based on model reaction

Question 2

model reaction

Answer 2

E + S ⟷ ES ⟶ E + P

⟷

• K1
• K-1

⟶
- K2

Question 3

Vmax

Answer 3

how fast it can go if all enzyme is in ES complex

Question 4

The more ES complex

Answer 4

faster it will go

Question 5

Lineweaver - Burk Plot

X intercept

Y intercept

Answer 5

Double reciprocal
1 / S concentration vs 1 / V (kinetic rate)
Linear

-1 / Km

1 / Vmax

Question 6

Significance of KM

Answer 6

substrate concentration needed to reach half Vmax

units mmol/ L

Characterises one enzyme-substrate pair (if an enzyme can act on different substrates, it will have different KM values for each).

Question 7

For many enzymes k 2 &laquo_space;k -1 , so approximation neglects k 2 :

Answer 7

Km = K-1 / K1

the ES dissociation constant

Question 8

Low KM

Answer 8

high affinity between E and S;

Question 9

high KM

Answer 9

low affinity between E and S;

Question 10

Physiological significance of KM

in enzyme-substrate interaction,

Answer 10

[S] is below the KM.

rate will rise to accommodate more substrate, tending to maintain steady state.

Question 11

KM: substrate preference and response

KM for each isozyme and substrate is same or different?

Answer 11

different

Question 12

Isozyme glucokinase

Answer 12

stores energy as glycogen in the liver.

Question 13

Turnover number, kcat

Answer 13

number of substrate molecules converted to product, per enzyme, per unit of time, when E is saturated with substrate.

define the activity of one enzyme molecule – a measure of catalytic activity.

Question 14

If the Michaelis-Menten model fits , kcat = k2

kcat describes

Answer 14

the ‘rate limiting’ step.

Vmax = kcat [E]T

Question 15

most effective enzymes should have…

Answer 15

A high kcat

A low KM

Question 16

kcat / KM measure

Answer 16

enzyme efficiency;

the higher the better.

Question 17

A high kcat

Answer 17

(ability to turnover a lot of substrate into product, per second).

Question 18

A low KM

Answer 18

(low substrate concentration required to achieve near Vmax; high affinity for the substrate under the Michaelis-Menten assumptions).

Question 19

kcat, KM and “catalytic perfection”

The upper limit for kcat / KM is the

Answer 19

diffusion-controlled limit; i.e. the rate at which enzyme and substrate diffuse together.

Question 20

Viscosity of water sets an absolute upper limit at

Answer 20

~10^9 s-1 M-1.

Question 21

Enzymes with kcat / KM above 10^8 s-1 M-1 are referred

to as

Answer 21

‘perfect’ catalysts.

Question 22

Enzymes are optimized for

Answer 22

specific roles

Question 23

Inhibitor:

Answer 23

compound that binds to an enzyme and reduces its activity.

Question 24

why are Enzyme inhibitors Important?

Answer 24

o Natural inhibitors regulate metabolism.

o Many drugs, poisons & toxins are enzyme inhibitors.

o Used to study enzyme mechanisms.

o Used to study metabolic pathways.

Question 25

Two classes of inhibitor

Answer 25

Irreversible inhibitor – binds covalently to the enzyme. Reversible inhibitor – not covalently bound to the enzyme.

Question 26

Irreversible inhibitor

Answer 26

binds covalently to the enzyme. reacts with a specific amino acid side chain in the active site, and forms a covalent bond.

Question 27

Reversible inhibitor

Answer 27

binds to the enzyme but can be released, leaving the enzyme in its original condition. not covalently bound to the enzyme. Competitive or Non-competitive (pure or mixed).

Question 28

Covalent inhibitors often react with catalytic residues. eg in chymotrypsin

Answer 28

Addition of the bulky tosyl-L-phenylalanine methylketone to the histidine disables the catalytic triad and fills the active site, blocking substrate binding.

Question 29

Competitive inhibition definition

Answer 29

Inhibitor competes with the substrate for binding in | the active site.

Question 30

Competitive inhibition michaelis-menten plot

Answer 30

No change in Vmax: high Km High [S] than the inhibitor.

Question 31

Competitive inhibition Lineweaver-burk plot

Answer 31

Increases KM: Vmax unchange More substrate is needed to get to V = Vmax / 2.

Question 32

Transition state analogues as drugs

Answer 32

comp inhibitor

Question 33

Enzymes are often targets for

Answer 33

drugs

Question 34

Transition state analogues can make ideal

Answer 34

enzyme inhibitors.

Question 35

Enalapril and Aliskiren lower

Answer 35

blood pressure.

Question 36

Statins lower

Answer 36

serum cholesterol.

Question 37

Protease inhibitors are

Answer 37

AIDS drugs.

Question 38

Juvenile hormone esterase is a

Answer 38

pesticide target.

Question 39

Tamiflu is an inhibitor of

Answer 39

influenza neuraminidase.

Question 40

Transition state analogs make tight binding inhibitors eg in Adenosine deaminase

Answer 40

use tetrahedral intermediate. • A non-reactive analog, 1,6-dihydroinosine, effectively inhibits the enzyme.

Question 41

Substrate analogue inhibitors of HIV protease

Answer 41

•inhibitor fills the active site. • Two catalytic aspartic acid residues.

Question 42

Non-competitive inhibition

Answer 42

Inhibitor binds at a different site than the substrate. Enzyme can bind substrate, or inhibitor (I), or both.

Question 43

In pure non-competitive inhibition,

Answer 43

binding of I has no effect on the binding of S i.e. the substrate binds to E and EI with the same affinity.

Question 44

Competitive vs. pure non-competitive inhibition michaelis menten plor

Answer 44

same Vmax / 2 different Km (comp more than pure)

Question 45

Competitive vs. pure non-competitive inhibition lineweaver burk plot

Answer 45

non - km unchanged - vmax reduce comp - km increase - vmax unchange

Question 46

Pure non-competitive inhibition lineweaver burk plot

Answer 46

Vmax decreases; KM stays the same. Binding I changes the structure of the active site such that S still binds, but transition state stabilisation is no longer optimal.

Question 47

Mixed non-competitive inhibition

Answer 47

Vmax decreases; KM increases. binding of the inhibitor does affect binding of the substrate à mixed non-competitive inhibition.

Question 48

Competitive inhibitors may react

Answer 48

alternate substrate can compete for the active site of an enzyme, as for alcohol dehydrogenase, on the right. Typically, related molecules fail to react., but some do.

Question 49

Michaelis-Menten equation is based on

Answer 49

binding theory and simple chemical reaction rates.

Question 50

Inhibition can be

Answer 50

reversible or irreversible.

Question 51

Inhibition can change either

Answer 51

binding (KM) or | catalysis (kcat) or both.

Question 52

Competitive inhibition gives a change in apparent

Answer 52

KM.

Question 53

Competitive inhibitors often resemble

Answer 53

substrates or transition states.