Topic 6 - Classes of Enzymes, Michaelis-Menten/Lineweaver-Burk/Eadie-Hofstee, Inhibitors Flashcards Preview

Biochem 153A > Topic 6 - Classes of Enzymes, Michaelis-Menten/Lineweaver-Burk/Eadie-Hofstee, Inhibitors > Flashcards

Flashcards in Topic 6 - Classes of Enzymes, Michaelis-Menten/Lineweaver-Burk/Eadie-Hofstee, Inhibitors Deck (29):
1

Oxidoreductases

Oxidation-reduction reactions

Electron Donor: Electron Acceptor + Oxidoreductase

 

Subclasses

Dehydrogenase

Oxidase

Oxygenase

Reductase

2

Transferases

Transfer of functional groups

Reactants + Functional group + Transferase

 

Subclasses:

Kinase

Aminotransferase/Transaminase

3

Hydrolases

Hydrolysis reactions (a single bond cleavage via addition of H2O)

Compound to be cleaved or formed + Hydrolase

 

Subclasses:

Phosphotase

Peptidase/Protease/Proteinase

Glycosidase

4

Lyases

Type 1. Group elimination to change a single bond to a double bond.

Type 2. Breaking a single bond to form two products, one of which has a new double bond.

Reactant + Lyase or Product + Synthase

 

Subclass

Synthase: Reverse of Type 2 lyase above

5

Isomerases

Isomerization (intramolecular rearrangement)

Reactant + Isomerase

 

Subclass:

Mutase: The apparent migration of a functional group from one position on a compound to another position on the same compound

6

Ligases

Bond formation coupled to ATP hydrolysis

Reactants + Ligase (or Synthetase)

 

Subclass

Synthetase: Formation of a new bond at the expense of ATP/GTP hydrolysis

7

Dehydrogenase

Transfer of hydride ion

(Oxidoreductase subclass)

8

Oxidase

O2 is the electron acceptor

(Oxidoreductase subclass)

9

Oxidase

O2 is directly incorporated into the substrate

(Oxidoreductase subclass)

10

Reductase

Electron transfer between any 2 compounds

(Oxidoreductase subclass)

11

Kinase

Transfer of phosphoryl group(s) between one of the adenylates (AMP, ADP, ATP) and another compound

(Transferase subclass)

12

Aminotransferase (Transaminase)

Transfer of an amino group between compounds

(Transferase subclass)

13

Transferase subclass

Hydrolysis of a single bond between a compound and a phosphoryl group, producing inorganic phosphate (Pi)

(Hydrolase subclass)

14

Peptidase/Protease/Proteinase

Hydrolysis of peptide bond

(Hydrolase subclass)

15

Glycosidase

Hydrolysis of a glycosidic bond

(Hydrolase subclass)

16

Synthase

The formation of a single bond from two products, one of which had a double bond (The reverse of Type 2 Lyase above)

(Lyase subclass)

17

Mutase

The apparent migration of a functional group from one position on a compound to another position on the same compound.

(Isomerase subclass)

18

Synthetase

Formation of a new bond at the expense of ATP/GTP hydrolysis 

(Ligase subclass)

19

Key Note About Enzymes

Enzymes will catalyze reactions in either direction depending on thermodynamic favorability.

Therefore, hydrolases (phosphotases, peptidases/proteases/proteinases, glycosidases), lyases (synthases), and ligases (synthetases) will also catalyze the reverse of the applicable reactions.

20

Michaelis-Menten Model - 5 Assumptions

1. For initial velocity [P] = 0. Therefore, velocity of reverse reaction, E + P --> E-S is 0, b/c V = k-2[E][P] = 0.

2. Reaction rate will depend on the rate-limiting step, which is assumed to be E-S ---> E + P. Therefore, the actual equation for the intial reaction rate is: V0 = k2[E-S].

3. [E-S] instantaneously comes to steady state (not equilibrium b/c this would then be a closed system). Therefore [E-S] remains constant throughout the subsequent time course of the reaction.

4. [S] >>>>>>>>>> [E-S]. Therefore, [S] does not change significantly at early time points.

5. [ET] = [EF] + [E-S] --> total enzyme = free enzyme + enzyme-substrate complex

21

Michaelis-Menten Equation

V0 = Vmax[S] / [KM + [S]]

@ Vmax/2, [S] = KM

Vmax = [E] x kcat

Catalytic efficiency = kcat/KM

 

22

Lineweaver-Burk (double reciprocal plot)

Original MM Equ: V0 = Vmax[S] / (KM + [S])

Lineweaver-Burk: 1/V0 = (KM/Vmax)*(1/[S]) + 1/Vmax

(y = mx + b)

X-axis: 1/[S], or 1 over the concentration of the substrate

Y-axis: 1/V0, or 1 over the initial velocity. Y-intercept = 1/Vmax when x = 1/[S] = 0

When x = 1/[S] = 0, then y-intercept = 1/V0 = 1/Vmax 

When y = 1/V0 = 0, then x-intercept = 1/[S] = -1/KM

23

Eadie-Hofstee

Original MM Equ: V0 = Vmax[S] / (KM + [S])

Eadee-Hofstee: V0 = -KM(V0/[S]) + Vmax

(y = -mx + b)

X-axis: V0/[S], or initial velocity over the concentration of substrate (will decrease over time)

Y-axis: V0, velocity, Vmax will occur at the Y-intercept when x = V0/[S] = 0

Y-intercept = Vmax when x = V0/[S] = 0

24

What factors affect Enzyme Activity?

pH

Temperature

Effectors:

-inhibitors (negative effectors)

-positive effectors

25

How does pH affect enzyme activity?

E + S E-S

1. may affect charge of substrate

2. may affect charge of amino acids involved in binding substrates

 

E-S --> E + P

3. may affect charge of amino acids involved in catalysis

4. disrupt enzyme structure and lead to impaired function

26

How does temperature affect enzyme activity?

1. increase the energy of the substrate (increase the activity of the enzyme until...see below)

2. denature the protein

27

How do competitive inhibitors affect enzyme activity?

 

 

 

Q image thumb

The inhibitor competes with the substrate for binding to the enzyme. Both cannot bind simultaneously. The inhibitor will appear to change the enzyme's affinity for the substrate (Alex's presence appears to change Emily's affinity for legos).

E + S E-S 

E + I E-I (enzyme-inhibitor complex --> makes enzyme inactive)

KI = [E]Free*[I] / [E-I] = reactants/product

a = 1 + [I]/KI, then V0 = Vmax[S] / (aKM + [S])

Graph of Lineweaver-Burk & Eadee-Hofstee: 

Lineweaver-Burk: 1/V0 = (KM/Vmax)*(1/[S]) + 1/Vmax

Y-intercept: same, no change in Vmax

X-intercept: change, apparent increase in KM --> will affect the X-intercept by making it less negative (-1/KM --> -1/aKM). Slope will increase for LB (more positive).

Eadee-Hofstee: V0 = -KM(V0/[S]) + Vmax

Y-intercept: same, no change in Vmax

X-intercept: change, apparent increase in KM --> will affect the X-intercept by making it less positive (Vmax/KM --> Vmax/aKM). Slope will decrease for EH (more negative).

Comments: High [S] overcomes inhibition because all of the Enzyme will be bound in the E-S complex. Therefore, greatest inhibition occurs at low substrate concentration. Also: No change in Vmax & apparent increase in KM.

28

How do mixed inhibitors (noncompetitive inhibitors) affect enzyme activity?

Q image thumb

The inhibitor can bind to the free enzyme and also to the E-S complex. The binding constant may be the same for the E-S complex as for free enzyme (KI = K'I, in which it is a noncompetitive inhibitor) or it may be different (KI > or < K'I).

Therefore, will have concentrations of E-S, E-I (same as competitive inhibitor), and E-S-I (same as uncompetitive inhibitor).

KI = [E]Free*[I] / [E-I]

K'I = [E-S][I] / [E-S-I]

a = 1 + [I]/KI & a' = 1 + [I]/K'I --> then V0 = Vmax[S] / (aKM + a'[S])

Graph of Lineweaver-Burk & Eadee-Hofstee: Noncompetitive Inhibitor, where KI = K'I

Lineweaver-Burk: 1/V0 = (KM/Vmax)*(1/[S]) + 1/Vmax

Y-intercept: changes, decrease in Vmax --> will affect the Y-intercept by making it more positive. Slope will increase for LB (more positive).

X-intercept: same, no change in apparent KM

Eadee-Hofstee: V0 = -KM(V0/[S]) + Vmax

Y-intercept: changes, decrease in Vmax --> will affect the Y-intercept by making it less positive. Slope will be the same however.

X-intercept: same, no change in apparent KM

EH lines with & without inhibitor will be parallel.

Comments: High [S] cannot overcome inhibition because inhibitor can also bind to E-S complex. Therefore, proportionate inhibition at all substrate concentrations. Also, decrease in Vmax & no apparent change in KM.

29

How do uncompetitive inhibitors affect enzyme activity?

Q image thumb

The inhibitor can bind only to the E-S complex (will bind to enzyme after substrate has also been bound).

Therefore, will have concentrations of E-S and E-S-I (same as mixed inhibitor).

K'I = [E-S][I] / [E-S-I]

a' = 1 + [I]/K'I --> then V0 = Vmax[S] / (KM + a'[S])

Graph of Lineweaver-Burk & Eadee-Hofstee: 

Lineweaver-Burk: 1/V0 = (KM/Vmax)*(1/[S]) + 1/Vmax

Y-intercept: changes, decrease in Vmax --> will affect the Y-intercept by making it more positive. Slope will be the same however. 

X-intercept: changes, apparent decrease in KM

LB lines with & without inhibitor will be parallel.

Eadee-Hofstee: V0 = -KM(V0/[S]) + Vmax

Y-intercept: changes, decrease in Vmax --> will affect the Y-intercept by making it less positive. 

X-intercept: changes, apparent decrease in KM. Slope will also decrease.

Comments: High [S] cannot overcome inhibition because presence of S is required to provide a site for binding of the inhibitor. Therefore, greatest inhibition occurs at high substrate concentration (opposite of competitve inhibition). Uncompetitive inhibition also overlaps in the beginning. Also, decrease in Vmax & apparent decrease in KM.