Lecture16-kinetics -...

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Unformatted text preview: BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 1 Synopsis:
 
 
 Enzyme
 kinetics
 studies
 how
 rates
 of
 enzyme‐catalyzed
 reactions
 depend
 on
 the
 concentrations
 of
 substrates.
 Enzyme
 reactions
 do
 not
 show
 the
 simple
 zero,
 first
 or
 second
 order
 relationships
 of
 chemical
 reactions.
 Instead
 the
 reaction
 reaches
 a
 limiting
 or
 saturation
 rate.
This
 behaviour
is
governed
by
the
Michaelis‐Menten
equation
and
the
two
characteristic
 constants
associated
with
this
equation,
Vmax
and
KM.
 
 Reading:
Lehninger
p.
194‐205
(4th
ed
p.
202‐
212)
 
 
 Enzyme
kinetics
 
 Reaction
rate
and
chemical
kinetics
 
 If
 decrease
 in
 substrate
 concentration
 or
 increase
 in
 product
 concentration
 is
 measured
 over
 a
 period
 of
 time
for
an
enzyme‐catalyzed
reaction,
the
data
can
be
 plotted
as
a
curve
called
the
progress
curve.
 
 Reaction
 rate
 is
 the
 slope
 of
 the
 curve.
 
 The
 initial
 reaction
rate
is
given
the
symbol
 vo,
and
is
the
tangent
 to
the
progress
curve
at
time
zero
(negative
tangent
if
 disappearance
of
reactant
is
measured).
Some
enzyme
 reactions
 remain
 linear
 for
 a
 significant
 time,
 making
 slope
 measurement
 easier;
 some
 reactions
 follow
 curves
as
in
the
examples.
 
 If
 several
 experiments
 are
 run,
 each
 with
 a
 different
 initial
 concentration
 of
 substrate,
 each
 experiment
 produces
 a
 different
 progress
 curve
 with
 one
 value
 of
 initial
rate
vo
for
each
curve
measured.
 
 The
 initial
 rates
 are
 then
 plotted
 as
 a
 function
 of
 initial
 substrate
 concentration
 [S].
 Normal
 chemical
 reaction
 follow
 simple
 rate
 laws,
 which
 describe
 how
 rate
 varies
 as
 a
 function
 of
 reactant
 or
 substrate
concentration.
 Page
1
of
6
 BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 2 The
 enzyme‐catalyzed
 reaction
 does
 not
 follow
 a
 simple
rate
law:
 
 Although
 vo
 is
 directly
 proportional
 to
 [E],
 concentration
 of
 enzyme,
 initial
 rate
 vo
 is
 a
 complex
 function
 of
 [S],
 substrate
 concentration.
 
 Interpretation
of
initial
velocity
dependence
on
[S]
 
 First
 worked
 out
 by
 Henri
 (1905,
 France),
 and
 subsequently
verified
experimentally
by
Michaelis
and
 Menten
 (USA/Canada,
 1913).
 
 Henri,
 Michaelis
 and
 Menten
 all
 assumed
 that
 the
 binding
 step
 was
 at
 equilibrium,
 which
 is
 only
 an
 approximation.
 
 Briggs
 and
Haldane
described
a
more
accurate
derivation
of
 the
equation
in
1926
that
follows
below.
 
 Although
the
overall
reaction
appears
to
be
S
→
P,
the
underlying
process
is
more
complex,
and
 can
be
interpreted
as
a
two‐step
reaction.
 
 1. The
enzyme
is
a
catalyst
and
is
recycled
in
an
unchanged
state
by
the
end
of
the
reaction
 process,
so
the
total
quantity
of
enzyme
[E]total
does
not
change
as
reaction
proceeds.

 
 2. If
we
deal
only
with
initial
reaction
rate
vo,
when
product
P
is
not
yet
present,
there
will
 be
no
reverse
reaction
ES
 →
E
+
P
at
the
catalytic
step
to
complicate
matters.

Omitting
 this
reverse
reaction
will
make
the
algebra
more
manageable.
 
 3. Each
stage
in
the
reaction
has
an
associated
rate
constant,
k1,
 k2,
plus
k–1
for
the
reverse
 of
step
1.

k2
is
sometimes
described
as
kcat,
the
rate
constant
for
catalysis.
 
 
 The
rate
of
appearance
of
product
P
describes
the
overall
rate
of
the
complete
enzyme
reaction,
 and
can
be
 determined
by
taking
the
rate
of
step
2,
which
is
a
 fundamental
chemical
 process,
 for
which
we
can
write
the
first
order
rate
equation:
 
 vo
=
k2
[ES]
 
 where
 [ES]
is
the
concentration
of
enzyme
actually
occupied
by
substrate.

Although
we
know
 [E]total,
 because
 that’s
 how
 much
 enzyme
 we
 put
 into
 the
 reaction,
 we
 don’t
 explicitly
 know
 what
 fraction
 of
 enzyme
 is
 empty
 E
 and
 what
 fraction
 is
 occupied
 ES,
 so
 [E]
 and
 [ES]
 are
 unknown
values,
but
related
by
the
equation

[E]total

=
[E]
+
[ES].
 Page
2
of
6
 BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 3 We
can
determine
[ES]
algebraically
 We
can
write
the
fundamental
rate
equations
for
formation
and
breakdown
of
ES.
 Rate
of
formation
of
ES
=
k1[E][S]
 Rate
of
breakdown
of
ES
=

k2[ES]
+
k–1[ES]
(forward
component
+
reverse
component)
 
 The
steady
state
condition
of
Briggs
and
Haldane
states
that

 rate
of
breakdown
=
rate
of
formation
 If
the
rate
of
breakdown
happens
to
be
greater
than
 the
rate
of
formation,
[ES]
will
decrease,
 causing
the
rate
of
breakdown
to
slow
down.

If
the
rate
of
breakdown
happens
to
be
less
than
 the
rate
of
formation,
[ES]
will
increase,
causing
the
rate
of
breakdown
to
speed
up.

Hence
the
 two
rates
rapidly
tend
to
match
each
other.
 Thus
we
can
write
 (k2
+
k–1)[ES]
=
k1[E][S]
 
 k + k−1 Divide
by
k1
to
put
all
the
constants
on
the
left,
and
let
 K M = 2 k1 

 
 KM[ES]
=
[E][S]
but
both
[ES]
and
[E]
are
unknown.
 
 Now
use
the
relationship
[E]
=
[E]total
–
[ES]
to
eliminate
the
second
unknown
value
[E]
 € KM[ES]
=
([E]total
–
[ES])
[S]
 
 then
take
the
term
[ES][S]
over
to
the
left
side
 (KM
+
[S])
[ES]
=
[E]total[S]
 
 and
divide
by
(KM
+
[S])
 
 [E] total[ S] where
[E]total
and
[S]
are
known
values,
and
KM
is
a
constant
 [ES] = 
 K M + [ S] 

 
 This
is
used
to
replace
the
[ES]
in
the
rate
equation
vo
=
k2[ES]
 k [E] [S] 
 v 0 = 2 total €
 K M + [S ] 

 

 
 The
maximum
observable
rate
Vmax
occurs
when
100%
of
the
enzyme
is
occupied
by
substrate,
 so
[ES]max
=
[E]total
where
[E]total
is
a
measurable
quantity.
 €
 Thus
we
can
write
Vmax
=
k2[E]total
and
replace

k2[E]total
in
the
equation
above
with
Vmax
to
get
 the
Michaelis‐Menten
equation
 
 Fractional
form
 Standard
form
 Page
3
of
6
 BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 4 The
 Michaelis‐Menten
 equation
 shows
 how
 vo
 varies
 as
 a
 function
 of
 substrate
 concentration
 [S],
 in
 terms
 of
 two
 constants,
Vmax
and
KM.
 
 Every
enzyme
has
characteristic
values
of
KM
and
Vmax
that
 describe
its
catalytic
behaviour.
 
 
 Vmax
is
the
upper
limit
for
the
observed
rate,
where
the
kinetic
curve
levels
off
at
high
[S]:

 
 
 When
vo
approaches
Vmax,
almost
all
of
the
 enzyme
 is
 in
 the
 occupied
 ES
 state,
 and
 the
 enzyme
 is
 said
 to
 be
 saturated
 with substrate.
 
 The
 Michaelis
 constant
 KM
 is
 the
 concentration
 of
 substrate
 needed
 to
 give
 a
 rate
 of
 exactly
 0.5
 Vmax.
 [S]
 =
 KM
 when
vo
=
0.5
Vmax
 
 KM
has
units
of
concentration,
and
typical
 KM
values
are
between
10‐6
M
and
10‐2
M.
 
 Low
 KM
 indicates
 that
 the
 enzyme
 binds
 and
utilizes
the
substrate
well;
a
lower
[S]
 is
sufficient
to
occupy
the
enzyme.
 
 A
 high
 KM
 indicates
 that
 the
 enzyme
 binds
 and
 utilizes
 the
 substrate
 poorly;
 a
 higher
 [S]
 is
 needed
to
get
S
to
occupy
the
enzyme.
 
 
 Page
4
of
6
 BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 5 Calculations
based
on
Michaelis‐Menten
equation
 If
 the
 enzyme
 phosphatase
 has
 KM
 =
 2.0
 ×
 10–4
 M
 and
 its
 substrate
 p‐nitrophenyl
 phosphate
is
present
at
5.0
×
10–4
M,
find
the
rate
of
reaction
a)
as
a
fraction
of
Vmax;
and
 b)
if
Vmax
=
5.0

×
10–8
M
.s–1
 
 v0 [S] a)
calculating
vo
as
a
fraction
of
Vmax.

 =
 
 Vmax KM + [S] 
 5.0 × 10 ‐4 
 






 = 
 
 (2.0 + 5.0) × 10 ‐4 
 






 = 
 0.71 

 
 b)
calculating
vo
for
a
given
value
of
Vmax
=
5.0

×
10–8
M
.s–1
 
 v 0 = 
 0.71 × Vmax 
 
 
 



 = 
 0.71 × 5.0 × 10 ‐8 
M ⋅ s ‐1 € ‐8 ‐1 



 

 = 
 3.5 × 10 
M ⋅ s 
 If
glucosidase
has
KM
=
6.0
x
10–5
M,
what
value
of
[S]
is
needed
to
get
a)
vo
=
0.75
Vmax

 and
b)
vo
=
4.0
x
10–9
M
.s–1
given
Vmax
=
2.0
x
10–8
M
.s–1
 
 € a)
 b )
 [S] 0.75 = 
 KM + [S] 0.75(KM + [S]) = [S] 0.75KM = 
0.25[S] [S]
 = 
3KM 0.20KM = 
0.80[S] [S]
 = 
0.25KM ‐4 [ 

S]
 = 
1.8 × 10 
M € 4.0 × 10 ‐9 [S] =
 2.0 × 10 ‐8 KM + [S] 0.20
(KM + [S]) = 
[S] ‐5 [S]
 

 = 
1.5 × 10 
M € Page
5
of
6
 BIOC*2580
Lecture
16:

Enzyme
kinetics
 
 6 If
[S]
=
0.5
KM,
what
is
vo,
expressed
as
a
fraction
of
Vmax?
 
 
 v0 [S] 
 =
 
 Vmax KM + [S] 
 
 0.5KM 
 






 = 
 
 KM + 0.5KM 
 






 = 
 0.33Vmax 
 

 
 
 If
vo/Vmax
=
0.86,
what
is
[S],
expressed
as
a
multiple
of
KM?
 
 
 € [S] 
 0.86 = 
 
 KM + [S] 
 
 0.86(KM + [S]) = [S] 
 
 0.86KM = 
0.14[S] 
 [S]
 
 

 = 
6.1KM 
 
 Note
 that
 vo
 can
 never
 be
 greater
 than
 Vmax,
 but
 [S]
 may
 have
 any
 value,
 and
 may
 be
 much
 greater
than
KM.
 
 € Page
6
of
6
 ...
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This note was uploaded on 09/21/2011 for the course BIOOC 2580 taught by Professor Douger during the Fall '10 term at University of Guelph.

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