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

Question 1

Oxidoreductases

Answer 1

Oxidation-reduction reactions

Electron Donor: Electron Acceptor + Oxidoreductase

Subclasses:

Dehydrogenase

Oxidase

Oxygenase

Reductase

Question 2

Transferases

Answer 2

Transfer of functional groups

Reactants + Functional group + Transferase

Subclasses:

Kinase

Aminotransferase/Transaminase

Question 3

Hydrolases

Answer 3

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

Compound to be cleaved or formed + Hydrolase

Subclasses:

Phosphotase

Peptidase/Protease/Proteinase

Glycosidase

Question 4

Lyases

Answer 4

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

Question 5

Isomerases

Answer 5

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

Question 6

Ligases

Answer 6

Bond formation coupled to ATP hydrolysis

Reactants + Ligase (or Synthetase)

Subclass:

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

Question 7

Dehydrogenase

Answer 7

Transfer of hydride ion

(Oxidoreductase subclass)

Question 8

Oxidase

Answer 8

O2 is the electron acceptor

(Oxidoreductase subclass)

Question 9

Oxidase

Answer 9

O2 is directly incorporated into the substrate

(Oxidoreductase subclass)

Question 10

Reductase

Answer 10

Electron transfer between any 2 compounds

(Oxidoreductase subclass)

Question 11

Kinase

Answer 11

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

(Transferase subclass)

Question 12

Aminotransferase (Transaminase)

Answer 12

Transfer of an amino group between compounds

(Transferase subclass)

Question 13

Transferase subclass

Answer 13

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

(Hydrolase subclass)

Question 14

Peptidase/Protease/Proteinase

Answer 14

Hydrolysis of peptide bond

(Hydrolase subclass)

Question 15

Glycosidase

Answer 15

Hydrolysis of a glycosidic bond

(Hydrolase subclass)

Question 16

Synthase

Answer 16

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)

Question 17

Mutase

Answer 17

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

(Isomerase subclass)

Question 18

Synthetase

Answer 18

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

(Ligase subclass)

Question 19

Key Note About Enzymes

Answer 19

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.

Question 20

Michaelis-Menten Model - 5 Assumptions

Answer 20

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. [E_T] = [E_F] + [E-S] –> total enzyme = free enzyme + enzyme-substrate complex

Question 21

Michaelis-Menten Equation

Answer 21

V₀ = V_max[S] / [K_M + [S]]

@ V_max/2, [S] = K_M

V_max = [E] x k_cat

Catalytic efficiency = k_cat/K_M

Question 22

Lineweaver-Burk (double reciprocal plot)

Answer 22

Original MM Equ: V₀ = V_max[S] / (K_M + [S])

Lineweaver-Burk: 1/V₀ = (K_M/V_max)*(1/[S]) + 1/V_max

(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/V_max when x = 1/[S] = 0

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

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

Question 23

Eadie-Hofstee

Answer 23

Original MM Equ: V₀ = V_max[S] / (K_M + [S])

Eadee-Hofstee: V₀ = -K_M(V₀/[S]) + V_max

(y = -mx + b)

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

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

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

Question 24

What factors affect Enzyme Activity?

Answer 24

pH

Temperature

Effectors:

• inhibitors (negative effectors)
• positive effectors

Question 25

How does pH affect enzyme activity?

Answer 25

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

Question 26

How does temperature affect enzyme activity?

Answer 26

1. increase the energy of the substrate (increase the activity of the enzyme until…see below)
2. denature the protein

Question 27

How do competitive inhibitors affect enzyme activity?

Answer 27

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)

K_I = [E]_Free*[I] / [E-I] = reactants/product

a = 1 + [I]/K_I, then V₀ = V_max[S] / (aK_M + [S])

Graph of Lineweaver-Burk & Eadee-Hofstee:

Lineweaver-Burk: 1/V₀ = (K_M/V_max)*(1/[S]) + 1/V_max

Y-intercept: same, no change in V_max

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

Eadee-Hofstee: V₀ = -K_M(V₀/[S]) + V_max

Y-intercept: same, no change in V_max

X-intercept: change, apparent increase in K_M –> will affect the X-intercept by making it less positive (V_max/K_M –> V_max/aK_M). 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 V_max & apparent increase in K_M.

Question 28

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

Answer 28

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 (K_I = K’_I, in which it is a noncompetitive inhibitor) or it may be different (K_I > or < K’_I).

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

K_I = [E]_Free*[I] / [E-I]

K’_I = [E-S][I] / [E-S-I]

a = 1 + [I]/K_I & a’ = 1 + [I]/K’_I –> then V₀ = V_max[S] / (aK_M + a‘[S])

Graph of Lineweaver-Burk & Eadee-Hofstee: Noncompetitive Inhibitor, where K_I = K’_I

Lineweaver-Burk: 1/V₀ = (K_M/V_max)*(1/[S]) + 1/V_max

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

X-intercept: same, no change in apparent K_M

Eadee-Hofstee: V₀ = -K_M(V₀/[S]) + V_max

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

X-intercept: same, no change in apparent K_M

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 V_max & no apparent change in K_M.

Question 29

How do uncompetitive inhibitors affect enzyme activity?

Answer 29

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 = V_max[S] / (K_M + a’[S])

Graph of Lineweaver-Burk & Eadee-Hofstee:

Lineweaver-Burk: 1/V₀ = (K_M/V_max)*(1/[S]) + 1/V_max

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

X-intercept: changes, apparent decrease in K_M

LB lines with & without inhibitor will be parallel.

Eadee-Hofstee: V₀ = -K_M(V₀/[S]) + V_max

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

X-intercept: changes, apparent decrease in K_M. 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 V_max & apparent decrease in K_M.