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Bio 98 - Lecture 10
Enzymes III:
Competitive Inhibition &
Enzyme Regulation
E+S
k1
k-1
ES
k2
E+P
Note: [E], concentration of free enzyme, is not the same as [E]t
VI. A better way to plot vo vs [S] data.
vo vs [S] plot
Vmax
Lineweaver-Burk plot
?
vo
1/vo
1/Vmax
Km
[S]
Vmax [S]
vo = –––––––––
Km + [S]
-1/Km
1/[S]
1
Km
1
1
–– = –––– ––– + –––––
vo
Vmax [S]
Vmax
The Lineweaver-Burk plot reduces uncertainty in estimating Vmax and Km.
Vmax [S]
vo = –––––––––
Km + [S]
Take reciprocal of both sides of equation
Km + [S]
1
= ––––––––
vo
Vmax [S]
Expand
[S]
K
1
m
+
= ––––––––
vo Vmax [S]
Vmax[S]
Thus
1
Km
1
1
–– = –––– ––– + –––––
vo
Vmax [S]
Vmax
Lineweaver-Burk
y = ax + b
Lineweaver-Burk plot
Vmax [S]
vo = –––––––––
Km + [S]
1/vo
1
Km
1
1
–– = –––– ––– + –––––
vo
Vmax [S]
Vmax
y =
a
x
+
1/Vmax
-1/Km
1/[S]
b
Solve for y at x=1/[S]=0:
1
1
y = –– = –––– = b
vo
Vmax
Solve for x at y=1/v0=0:
1
1 Vmax
b
x = –– = - –––– –– = - ––
[S]
Vmax Km
a
How do you measure competitive inhibition?
K1
[I] = 2KI
K-1
[I] = KI
[I] = 0
K-I
-1
Km
Vmax [S]
vo = –––––––––
aKm + [S]
where
-1
aKm
[I]
a = 1 + –––
KI
[E][I]
KI = ––––––
[EI]
Vmax remains unchanged, but apparent Km increases with increasing [I]
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)
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:
Thr
E1
threonine
dehydratase
B
E2
E3
E4
E5
Ile
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!
A Simple Model of Heterotropic Allosteric Regulation
(activation)
Heteroallosteric Regulation
2 subunit enzyme model
T-state
R-state
inhibitor
activator
substrate
Heteroallostery:
Effect of inhibitor or activator on vo vs [S] plot
Vmax
R-state
vo
Vmax
T-state
2
Kmapp
Km
Kmapp
[S]
Phosphofructokinase-1
is regulated by both activators and inhibitors
activates (ADP)
phosphoenolpyruvate
PEP
pyruvate
+ ATP
F 6-bisF
deactivates (phosphoenolpyruvate)
+ ADP
+ PPP
F 6-P (mM)
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.
Most allosteric enzymes are multi-subunit enzymes
Phosphofructokinase has 4 subunits
F6bisP + ADP
ADP
Activated state (R-
Inactivated state (T-state)
Covalent modifications
Phosphorylation - a reversible modification
ATP
ADP
protein kinase
activity
state 1
Enz
Enz
protein
phosphatase
=
HOPO3
H20
=
-OPO3
activity
state 2
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.
Proteolysis – an irreversible modification
protease
inactive enzyme
Examples: pancreatic
enzymes
+
active enzyme*