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Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

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Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation
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Page 1: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Bio 98 - Lecture 10

Enzymes III:

Competitive Inhibition &

Enzyme Regulation

Page 2: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Note: [E], concentration of free enzyme, is not the same as [E]t

k-1

k1 k2E + S ES E + P

Page 3: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

VI. A better way to plot vo vs [S] data.

[S]

vo

Vmax

vo vs [S] plot

?

Km 1/[S]

1/Vmax

-1/Km

1/vo

Lineweaver-Burk plot

Vmax [S]vo = ––––––––– Km + [S]

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

The Lineweaver-Burk plot reduces uncertainty in estimating Vmax and Km.

Page 4: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Vmax [S]vo = ––––––––– Km + [S]

Take reciprocal of both sides of equation

Expand

= ––––––––vo Vmax [S]

Km + [S]1

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

Thus y = ax + b

Lineweaver-Burk

= ––––––––vo Vmax [S]

Km1 +[S]

Vmax[S]

Page 5: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Vmax [S]vo = ––––––––– Km + [S]

1 Km 1 1–– = –––– ––– + ––––– vo Vmax [S] Vmax

y = a x + b

Solve for y at x=1/[S]=0: y =

Solve for x at y=1/v0=0: x =

1/[S]

1/Vmax

-1/Km

1/vo

Lineweaver-Burk plot

1 1–– = –––– = b vo Vmax

1 1–– = - –––– [S] Vmax

Vmax b –– = - –– Km a

Page 6: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

How do you measure competitive inhibition?

Vmax [S]vo = ––––––––– Km + [S]

[E][I]KI = –––––– [EI]

where [I]

= 1 + ––– KI

Vmax remains unchanged, but apparent Km increases with increasing [I]

-1 Km

-1 Km

K-I

K1

K-1 [I] = KI

[I] = 0

[I] = 2KI

Page 7: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Modes of Enzyme Regulation

1. Allosteric* control/regulation

• homotropic allostery (O2 for hemoglobin)

• heterotropic allostery (H+, CO2, BPG for Hb)

2. Covalent modification

• group addition - often reversible, ie phosphorylation

allosteric* = allo (other); steric (shape, object)

Page 8: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Regulation of Enzyme Activity

It would be wasteful to continue to turn substrate into product if enough is available for proper cellular function. Therefore, enzymes often are highly regulated by binding small molecule regulators that can either decrease or increase activity.

A classic example is in amino acid metabolism. Several enzymes are required to convert simple substrates into more complex amino acids:

In this example, five enzyme-catalyzed steps are required to convert Thr to Ile. When there is sufficient Ile available, Ile will “feedback” inhibit enzyme 1 (E1) that converts Thr to intermediate B. Such inhibition effectively shuts down the entire pathway. There must be a careful balance between the concentration of Ile required for normal function and the concentration of Ile required to inhibit enzyme E1. FEEDBACK INHIBITION!

E1Thr E2

threoninedehydratase

E3 E4 E5 IleB

Page 9: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

A Simple Model of Heterotropic Allosteric Regulation(activation)

Page 10: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

inhibitor activator

Heteroallosteric Regulation 2 subunit enzyme model

T-state R-state substrate

Page 11: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Heteroallostery: Effect of inhibitor or activator on vo vs [S] plot

[S]

vo

Vmax

Vmax

2

Km KmappKm

app

R-state

T-state

Page 12: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Phosphofructokinase-1 is regulated by both activators and inhibitors

+ PPP

F 6-P (mM)

+ ADP

F 6

-bis

F

pyruvate + ATP

When the concentration of phosphenolpyruvate (PPP) reaches a certain level, then P6-F kinase activity is lowered by feedback inhibition from phosphenolpyruvate. On the other hand, ADP is the allosteric activator; this may seem strange but the net objective of glycolysis is generation of ATP. If the ATP levels drop (ie [ADP] goes up), glycolysis needs to be stimulated. Thus ADP binds to and activates phosphofructokinase-1 to increase ATP production.

PEP

deactivates (phosphoenolpyruvate)

activates (ADP)

phosphoenol-pyruvate

Page 13: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Most allosteric enzymes are multi-subunit enzymesPhosphofructokinase has 4 subunits

Inactivated state (T-state)Activated state (R-state)

ADP

F6bisP + ADP

Page 14: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Covalent modifications

Phosphorylation - a reversible modification

protein phosphatase

H20HOPO3

=

activity state 1

ATP ADP

-OPO3

= activity state 2

protein kinase

Enz Enz

Page 15: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Phosphorylation is reversible and is used in many pathways to control activity.

Enzymes that add a phosphate to a hydroxyl side chain are commonly called kinases.

Enzymes that remove a phosphate from a phosphorylated side chain are called phosphatases.

Page 16: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.
Page 17: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

Examples: pancreatic enzymes

Proteolysis – an irreversible modification

inactive enzyme

protease +

active enzyme*

Page 18: Bio 98 - Lecture 10 Enzymes III: Competitive Inhibition & Enzyme Regulation.

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