1
Session 12
Enzymes:
Kinetics
Tony Carruthers
Room LRB 926
Anthony.Carruthers@umassmed.edu
voice: 508 856 5570
2
Rationale
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3
Goals
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4
WARNING!
Kinetics involves some math and should not be considered a
“Spectator Sport”.
To get the most out of this lecture, try working through the
equations and substituting values for various constants. In this
way you will understand how changing any given constant
affects the behavior of the expression.
You need remember only ONE equation - this will be highlighted
during the class.
!"#"$%#&!'()&*!&($+,
5
*"F'-".4'<"%&"'&:-;'"18&)(0&"%."46'"103;+(0&"01"46'"'&:-;'"NOPQ.85.43+4'"NRP")0;<,'?L"OR=
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13''"O=""@6'.'"+3'T
!=""
"
OR")0;<,'?'."6+2'"5''&"9%3')4,-"2%.8+,%:'9"5-"OU"VWG3+-""
)3-.4+,,073+<6-=
Jack Gri!th developed techniques
that let scientists see the finer
details of DNA. In 1971 he and
Arthur Kornberg published this
photo, the first electron microscope
image of DNA bound to a known
protein - DNA polymerase.
>="""
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Y
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+"<0%&4"I6'3'"<3098)4"103;+(0&"&0",0&7'3"%&)3'+.'.=""@6%.".+483+(0&"%."
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%,,8.43+4'9"5',0I="
v (d[P]/dt), rate of reaction
120
Vm
100
80
M;"%."+"46'03'()+,L";+?%;8;"
2+,8'"103"2=
60 0.5 V
m
$;"%."46+4")0&)'&43+(0&"01"ZR["
40
<3098)%&7"+"2"01"M;Q>=""
20
0
Km
0
10
20
30
[S] mM
40
50
]
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%&.4+&)'"4+F'."46'"7'&'3%)"103;
"
"
"
"
"
"
"
"
"
v=
Vm * [S]
K m + [S]
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_'4"8."'?+;%&'"+&"'&:-;'G;'9%+4'9"3'+)(0&=
k1
E+S
k2
ES
k-1
k3
EP
k-2
E+P
k-3
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k1
E+S
k3
ES
E+P
k2
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!P" f03;+(0&"01"OR
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:7.5""*#8.#AB#C7"$&*.3,#8.#:7.*<58#9#$,*#A(
11
EXAM-TYPE QUESTION
In deriving the Michaelis-Menten equation for enzyme-mediated reactions,
which of the following did we assume?
A.!The concentration of S is reduced by formation of ES.
B.! The rate of the reaction is limited by ES dissociation to form free enzyme and
substrate S.
C.! ES breakdown to form E + S is slower than ES breakdown to form E + P.
D.! An intermediate complex, EP is involved in the reaction.
E.! The reverse reaction is insignificant.
!>
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D4"#.7*"7#.?#$#7"$5-.,#+%#*"8"72+,"*#C1#84"#:.3"7#.?#84"#
5.,5",87$-.,#8"72%#+,#84"#7$8"#"E<$-.,(
In a simple isomerization reaction
A
k1
B
d[B]
= k1 [A]
dt
The units for this first order reaction are derived from molarity of
product formed per second per molarity of reactant or,
molarity per sec = k1.molarity
molarity per sec
= k1 = per sec
molarity
!A
g&"+"3'+)(0&"01"46'"4-<'
k
1
E + S !!
"E#S
@6%."3'+)(0&"%.".')0&9"039'3"+."0&'".%&7,'".4'<"%."%&20,2'9"%&"46'"
3'+)(0&"01"4I0".<')%'.=""
@6'"3+4'"01"46'"3'+)(0&"%."<30<03(0&+,"40"ZO[=ZR[=
d[ES]
= k1[E][S]
dt
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01"<3098)4"103;'9"<'3".')0&9"<'3";0,+3%4->"01"3'+)4+&4."03
;0,+3%4-"<'3".')"h"F!"N;0,+3%4-P>
molarity per sec
= k1 = per molarity per sec
molarity 2
!C
@68."103"083"3'+)(0&L
k1
E+S
k3
ES
E+P
k2
@6'"103;+(0&"01"OR"%."+".')0&9"039'3"<30)'.."+&9"46'"53'+F90I&"01"
OR"40"O"V"R"03"40"O"V"S"+3'"B3.4"039'3"<30)'..'.="
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F!"h"<'3";0,'"<'3".')="i08"I%,,".''",+4'3"I6-"46'.'"8&%4."+3'"%;<034+&4="
!X
+6ABC0D/2E/:@3:0'.6:0@37456/'!')070F:5@D8.G369'.6:0@37456/
@6'"B3.4".4'<"%&"'&:-;'"18&)(0&"%."46'"103;+(0&"01"46'"'&:-;'"NOPQ.85.43+4'"NRP"
)0;<,'?L"OR=
j0&.%9'3"46'"3'+)(0&
k1
E+S
ES
k2
k'B&%&7"O4"+."O"V"OR
D4"#*"7+6$-.,#.?#84+%#
%./<-.,#+%#+,5/<*"*#$%#
F::",*+G#H
[S]
[ES]
=
k
[E t ]
2
+ [S]
k1
I6+4"%."F>QF!J
!Y
1.0
per sec
=M
per M per sec
0.8
0.6
[ES]/[Et]
@6'"3+(0"F>QF!"6+."8&%4."01T
0.4
K6'&"ZR["h"F>QF!L
[S]
[S]
[ES]
=
=
= 0.5
[E t ] k 2 + [S] 2[S]
k1
0.2
k2/k1
0.0
0
25
50
75
[S] µM
@6%.";'+&."46+4"0&'G6+,1"01"ZO4["h"ZOR["I6'&"ZR["h"F>QF!"
100
!]
!4$8#+%#84"#%+),+@5$,5"#.?#&IJ&HK
E+S
k1
ES
k2
&HJ&I#h"$'a"h"'a8%,%53%8;")0&.4+&4"03"+..0)%+(0&")0&.4+&4"103"OGR"
%&4'3+)(0&=
&IJ&H#h"!Q$'a"h"$."03"$9"h"9%..0)%+(0&")0&.4+&4"103"46'"OR")0;<,'?
ES
k2
E+S
k1
!^
g1"$."%."!"lUL"Xem"01"O4"h"OR"I6'&"ZR["h"!"lU"h"!"?"!eGYU=
g1"$."%."!"&UL"Xem"01"O4"h"OR"I6'&"ZR["h"!"&U"h"!"?"!eG`U=
*",0I"2+,8'"103"$R";'+&."46+4"46'"OR")0;<,'?"%.";03'".4+5,'"
N,'.."9%..0)%+4'."40"O"V"RP"468."+4"+&-"ZR[L"46'3'"%."+"6%76'3"
<305+5%,%4-"46+4"OR"%."103;'9="_'.."R"%."3'a8%3'9"40"0))8<-"0&'G
6+,1"01"46'"+2+%,+5,'"5%&9%&7".%4'.="@6'"'&:-;'".60I."6%76"
+b&%4-"103"R=
19
_'4"8."&0I")0&.%9'3"46'"53'+F90I&"01"OR"40"103;"
3'7'&'3+4'9"O"+&9"<3098)4L"S=
@6'"02'3+,,".)6';'"%."2%.8+,%:'9"+."10,,0I.=""
k1
E+S
k3
ES
E+P
k2
20
k'B&%&7"46'"3+4'"01"<3098)4"103;+(0&L"2"+.
"""2"h"FA"ZOR[""""N!
It can be shown that:
V [S]
v= m
K m + [S]
(2
Vm = [Et]k3
(3
where
Km =
k2 + k3
k1
(4
The derivation of
this solution is
included as
Appendix 2
21
Km =
D4"#<,+8%#.?#M2#$7"'##
k2 + k3
k1
<'3".')Q<'3"U"<'3".')"h"U
Vm = [Et]k3
$,*#84"#<,+8%#.?#L2#$7"#'"
U"<'3".')
22
Properties of the Michaelis Menten Equation
At very high [S], !
!
zero-order kinetics
v " Vm
100
!
!
i.e. v does not increase with increasing [S]
at low [S],
v!
i.e. v increases linearly with [S]
v (d[P]/dt)
Vm [S]
Km
80
60
40
20
0
first-order kinetics
0
100
200
[S] !M
However, Vm is rarely measurable because in many instances it is not
possible to add sufficient quantities of substrate to saturate the enzyme.
You may then ask, if Vm is not measurable, how does one determine Vm
and Km for a reaction?
23
You may also ask, “Why would you want to measure Vm and Km?”
You saw on page 21 that Vm = [Et]k3
If Vm is reduced by an inhibitor or by phosphorylation of the enzyme or
by a mutation of the enzyme this could mean either:
1) [Et] is reduced (reduced expression, increased degradation,
relocation to another cellular compartment, etc) or,
2) k3 is reduced.
You also saw on page 21 that Km = (k2+k3)/k1
If Vm is unaffected and Km is changed, the most likely explanation is
that k2 is altered because (as you will see later) k1 is largely
dependent on substrate diffusion rates.
Thus the enzyme displays altered dissociation or affinity for substrate.
Hence, interesting biology can be inferred from changes in Vm or Km
24
*".%;<,'L"+,7'53+%)"43%)F".0,2'."<305,';"01")0;<8(&7"M;"+&9"$;n""
@6'"3')%<30)+,"01"46'"U%)6+',%.GU'&4'&"'a8+(0&"%.
Km
1
[S]
=
+
v Vm [S] Vm [S]
1 Km 1
1
=
+
v Vm [S] Vm
@6%."%."+",%&'+3"'a8+(0&"I%46"46'"1+;%,%+3"103;L"
y = slope * x + intercept
468.";+F%&7T"!Q2"h"-""" +&9L"!QZR["h"?
+"<,04"01"!Q2"6"7%<%"!QZR["I%,,"<3098)'"+".43+%764",%&'"01T
<0.%(2'".,0<'"h"$;QM;""+&9L"-G%&4'3)'<4""h"!QM;
25
K6'&"ZR["h"G$;"N+"46'03'()+,")0&9%(0&"5')+8.'"ZR["o"e"%&"46'"3'+,"I03,9PL"46%."
;'+&."46+4
1 Km 1
1 1
1
1
=
+
= =
+
=0
v Vm ! K m Vm v !Vm Vm
(9
468."I6'&"ZR["h"G$;L""!Q2"h"e="@6'"<,04"NF&0I&"+."+"_%&'I'+2'3Gp83F"<,04P",00F.",%F'
1/v sec per µmol
0.6
y = 0.1 + 2.5 x
0.5
0.4
0.3
slope = Km/Vm
0.2
-1/Km
1/Vm
0.1
0.00 0.05 0.10 0.15 0.20 0.25
1/[S] per µM
26
EXAM-TYPE QUESTION
The Lineweaver–Burk plot is characterized by one of the
following:
A.!! 1/Vm = x-intercept.
B.!! -1/Km = y-intercept
C. ! Vm/Km = 1/slope
D. ! -1/Vm = y-intercept.
E. ! When [S] = Km, 1/v = 0.
27
Significance of Km and Vm
Km is the concentration of [S] at which 1/2 of the active sites are filled
with substrate. The fraction of filled sites [ES]/[Et]!
1
0.8
[ES]/[Et]
!S #
!" ES #$
= " $
!" Et #$ !" S #$ + K m
0.6
0.4
0.2
0
0 20 40 60 80 100
Thus when [S] = Km
[S]
!" ES #$ K m
=
= 0.5
!" Et #$ 2K m
Thus one-half of the enzyme is in the form of ES and the rate of the
reaction, v, is
0.5 [Et] k3 or v = 0.5 Vm
28
We also saw above that
!
!
!
(k + k3 )!
! K =! 2
m
k1
Km =
Thus if k2 >> k3
(ES dissociation is much faster than catalytic steps)
k2
= KS
k1
where Ks is the dissociation constant or Kd for S interaction with E.
E+S
k1
k2
ES
[E][S] k2
=
[ES]
k1
i.e. Km is the dissociation constant (reverse of the equilibrium constant) for ES
formation and is a measure of the stability of the complex.
Thus in some reactions where k3 << k2, Km = KS while in other enzyme
mediated reactions where k3 # k2, Km > KS.
A high Km implies weak binding (it takes more substrate to fill substrate binding sites).
A low Km implies strong binding (it takes less substrate to fill substrate binding sites).
29
Vm reveals the turnover number (kcat) of an enzyme if [Et] is known because
!
Vm = kcat [Et]
memorize
this
if [Et] = 1 µM and Vm = 600 mmols/L/sec,
kcat = 6 x 105 sec-1 - Each round of catalysis is 1/kcat = 1.7 µsec.
30
EXAM-TYPE QUESTION
An enzyme has a Km for substrate (S) of 10 mM and Vm of
5 !mol.L-1.sec-1 at a total enzyme concentration of 1 nM.
At [S] = 10 mM, kcat is:
A.
B.
C.
D.
E.
!
!
!
!
!
2500 per M per sec.
5000 per M per sec.
2500 per sec.
5000 per sec.
1250 per M per sec.
“Kinetic Perfection” and the kcat/Km Criterion
31
K'".+I"'+3,%'3"46+4"I6'&"ZR["qq"$;L"
V [S]
"
"
"
"
"
" v = "m
Km
K'"+,.0"F&0I"46+4"M;"h"F)+4"ZO4["
*&9"103"083"'&:-;'";')6+&%.;
M;""h"FA"ZO4[""""N84<%#?.7#.<7#",012">#&5$8#O#&PQ
.%&)'"M;"h"F)+4"ZO4["h"FA"ZO4[L"46%.";'+&.
v=
k3
[S][Et ]
Km
2"%."46'3'103'"9%3')4,-"<30<03(0&+,"40""
k3
Km
(11
+&9"ZR["+4"B?'9"ZO4[
32
Are there limits upon kcat/Km?
kk
kcat
= 3 1
K m k2 + k3
(12
O?+;%&+(0&"01"'a&"!>"%&9%)+4'."46+4"F!"%.",%;%(&7=
K6'&"F>"qq"FA"N+("(#5$8$/1%+%#+%#2<54#?$%8"7#84$,#AB#*+%%.5+$-.,Q
k2 is the rate of ES dissociation to E + S
k3 is the rate of ES breakdown to E + P
k3 k1
k 2 + k3
!
k3k1
! k1
k3
D4<%#84"#4+)4"%8#6$/<"#84$8#&PJM2#5$,#$54+"6"#+%#&H(
F!"%."46'".')0&9"039'3"3+4'")0&.4+&4"46+4"9'.)3%5'."+..0)%+(0&"01"O"
+&9"R"40"103;"46'"OR")0;<,'?="F!"%&),89'."4'3;."46+4"9'.)3%5'"46'"
)0,,%.%0&+,"13'a8'&)-"01"O"+&9"R=
33
f03" 46'" )+.'" 01" 8&)6+37'9";0,')8,'." %&".0,8(0&L" 46'"'&)08&4'3"
3+4'")0&.4+&4L"F'")+&"5'")+,)8,+4'9"130;
ke =
4!N(DA + DB )(rA + rB ) -1
M .sec-1
1000
note, DA =
kT
6!rA"
I6'3'"k"+&9"3"3'1'3"40"46'"9%D8.%0&")0&.4+&4."+&9"3'+)(0&"3+9%%"
01";0,')8,'."*"+&9"p"+&9"r"%."*2+7+930s."&8;5'3=""f03"+".-.4';"
.8)6"+."+&"'&:-;'";0,')8,'"I%46"+"3+9%8."01"+5084"Ae"t"+&9"+"
.85.43+4'";0,')8,'"I%46"+"3+9%8."01"+5084"X"tL"46'"'&)08&4'3"
3+4'")0&.4+&4")0;'."40"+<<30?%;+4',-"!e`"UG!=.')G!=
34
Thus diffusion tends to limit the rate of encounters of E and S and an upper
limit of k3/Km is therefore 109 M-1.sec-1. Even this requires that the substrate
can encounter the enzyme surface in any orientation.
F)+4Q$;"3+(0."01".0;'"'&:-;'.L"'=7="+)'4-,)60,%&'.4'3+.'"+&9")+350&%)"
+&6-93+.'"+3'"5'4I''&"!e^"G"!e`"UG!".')G!"%&9%)+(&7"46'-"6+2'"+)6%'2'9"F%&'()"
<'31')(0&=
f03"46'.'"'&:-;'.L"5')+8.'"F)+4Q$;"u"!e^"UG!".')G!L"46%.";'+&."46+4"FA"%.".0"1+.4"3',+(2'"
40"F>"46+4"46'"OR")0;<,'?"53'+F."90I&"1+.4'3"40"103;"<3098)4"+&9"O"46+&"%4"9%..0)%+4'."
5+)F"40"R"+&9"O=
@6'"+)(2%4-"01"46'.'"'&:-;'."%.",%;%4'9"0&,-"5-"46'"3+4'"+4"I6%)6"46'-"
'&)08&4'3".85.43+4'"%&".0,8(0&=""*&-"18346'3"7+%&")+&"0&,-"5'"+)6%'2'9"5-"
9')3'+.%&7"9%D8.%0&"(;'.=""@6%.")+&"5'"+)6%'2'9"5-".'a8'.4'3%&7".85.43+4'."
+&9"<3098)4."%&"46'")0&B&'9"20,8;'"01"+";8,(G'&:-;'")0;<,'?L"'=7="
;%40)60&93%+="
35
EXAM-TYPE QUESTION
An enzyme, E, is characterized by Vm = 2 mM/sec and
Km = 100 !M when [Et] = 1 !M. When[S] is 2 mM and
[Et] = 40 !M:
A.
B.
C.
D.
E.
!
!
!
!
!
The enzyme has achieved kinetic perfection.
kcat/Km = 1 x 107 M-1.sec-1.
kcat/Km = 2 x 107 sec-1.
kcat/Km = 8 x 108 M-1.sec-1.
The enzyme has not achieved kinetic perfection.
26
36
kcat
V
mol
= max ;units are
Km [Et ]Km
sec.mol 2
2x10 −3
=
1x10 −6 x1x10 −4
2x10 −3
=
= 2x107 M −1.s −1
−10
1x10
“Perfection” requires that kcat/Km ≥ 1 x108 M-1.s-1
37
ENZYMES ARE INHIBITED BY SPECIFIC MOLECULES
Reversible Inhibitors - May occur naturally or may be
synthesized as drugs.
Competitive Inhibition
S
E+I
+
S
I
King-Altman
Representation
EI
Ki
Ks
enzyme
ES
k3
E+P
A competitive inhibitor, I, competes for binding with S. I is not
transformed into product. The inhibitor, I, resembles the substrate,
S. I reduces the rate of reaction + S by reducing the proportion of
enzyme in the form ES
38
Noncompetitive Inhibition
S
S
E
C
I
C
I
S
S
E
E
I
Ki
EI
+
S
Ks
ES + I
C
k3
E+I
+
S
ESI
C
k3
I
E+P
E+P
S and I are not mutually exclusive but ESI is catalytically inactive.
When S binds, the enzyme undergoes a conformational change
which aligns the catalytic center, C, with the susceptible bonds of S;
I interferes with the conformational change, but has no effect on S
binding.
39
Competitive inhibition
S and I compete for binding to the same site
k3 [S]
KS
v
=
[S]
[I]
[Et ]
1+
+
KS Ki
v=
[Et ]k3 [S]
=
!# [I] %#
K S "1 +
& + [S]
K i '#
$#
Vmapp [S]
K mapp + [S]
What are Vmapp and Kmapp? They are algebraic terms that tell us what
factors affect Vm and Km.
Vmapp = [Et ]k3
!# [I] %#
K mapp = K S "1 +
&
K i '#
$#
i.e. when [I] increases
Kmapp increases because Kmapp includes an [I] term
!
Vmapp is unaffected because = k3[E t] (i.e. includes no [I] terms)
!
40
j0;<'((2'"%&6%5%(0&
1.2
12
V
m
control
+ inhibitor
8
0.8
6
0.6
-1/K
0.4 -1/K
4
K
2
0
Control
+ inhibitor
1
1/V
v, rate of reaction
10
0
20
K
m
40
60
m
0.2
m(app)
80
m(app)
100
0
-0.1
1/V
0
[S] mM
M;"%."8&+D')4'9"584"$;"%."%&)3'+.'9=""
Km(app) = KS(1 + [I]/Ki)
m
0.1
1/[S] per mM
0.2
+H3CF80/'5>'!5CF044I0'&6-.E.:5@/
R$8<7$//1#.55<77+,)#N%.<75"#$,*#8$7)"8%Q
41
k%7%4+,%."N10?"7,02'v"r+L$*@S+.'P
@'43+9040?%&"N<8D'3"B.6v"r+Gj6+&&',P
j-40)6+,+.%&"p"N18&7%v"7,8)0.'"43+&.<034'3.P
*430<%&'"N9'+9,-"&%764.6+9'v"*)'4-,)60,%&'"3')'<403P
B1,84"-5#N84"+7#8$7)"8%#$,*#<%"Q
g58<301'&"N)-),0G0?-7'&+.'"%&6%5%403"G"+&(G%&\+;;+403-P
R8,1+&%,+;%9'"N9%6-930<4'30+4'".-&46'4+.'v"+&(5+)4'3%+,."G"%&6%5%4"
10,+4'".-&46'.%.P
r'0.(7;%&'"N+)'4-,)60,%&'.4'3+.'v"<30,0&7"&'830;8.)8,+3"
43+&.;%..%0&"G"43'+4";-+.46'&%+"73+2%.P
g&9%&+2%3"NHgM"<304'+.'"gg"%&6%5%403v"<3'2'&4"HgM"43+&.;%..%0&"+&9"18,,G
5,0I&"*gkRP
see !
http://www.emdbiosciences.com/docs/docs/LIT/ISB_USD.pdf
42
Noncompetitive inhibition
[S]
KS
v
=
[S] [I] [S][I]
k3 [Et ]
1+
+
+
KS Ki KS Ki
or v =
where Vmapp =
or
[S]
v
=
k3 [Et ]
K S + [S]
!# [I] %#
"1 +
&
K i '#
$#
Vmapp [S]
K mapp + [S]
k3 [Et ]
#! [I] #%
"1 +
&
K i #'
#$
and K mapp = K S
thus the inhibitor I reduces Vmapp but has no effect on Kmapp
43
Noncompetitive inhibition
12
control
+ inhibitor
V
1.4
m
1.2
V
6
m(app)
0.8
0.6
4
1/V
0.4 -1/K
2
0
Control
+ inhibitor
1
8
1/V
v, rate of reaction
10
K
0
20
m
0.2
m
40
m(app)
60
80
1/Vm
0
-0.1
100
0
[S] mM
0.1
1/[S] per mM
0.2
Kmapp is unaffected but Vmapp is decreased.
Vmapp = Vm/(1 + [I]/Ki)
where Ki is the dissociation constant for the EI complex, Ki = [E][I]/[EI]
44
Noncompetitive inhibition is a common theme in feedback
inhibition.
For example, the biosynthesis of isoleucine from threonine in
bacteria involves 4 steps mediated by 4 different enzymes.
The first reaction is catalyzed by threonine deaminase (TD).
This enzyme is noncompetitively inhibited by isoleucine.
Thus as product increases, the first step in the reaction
decreases. As P falls due to shut down of TD, so TD is
released from inhibition due to dissociation of P from the
TD.P complex and the reaction cycle proceeds again.
Threonine
threonine
A
-
deaminase
B
D
Isoleucine
+H3CF80/'5>'*56<!5CF044I0'&6-.E.:5@/
45
R$8<7$//1#.55<77+,)#N%.<75"#$,*#8$7)"8%Q
j+D'%)"*)%9"N40;+40v",%<0?-7'&+.'P
j+D'%&'"N4'+L")0D''v")*US"<60.<609%'.4'3+.'L"7,8)0.'"43+&.<034'3.P
B1,84"-5#N84"+7#8$7)"8%#$,*#<%"Q
H+,0<'3%90,"N53+%&"w"'&9046',%+,")',,"r%43%)"x?%9'"R-&46+.'"%&6%5%403"G"
+&(G<.-)60()P
@3%)60.4+(&"*"NH%.40&'"k'*)'4-,+.'v"+&(G)+&)'3P
;-)0<6'&0,%)"+)%9"Ng&0.%&'";0&0<60.<6+4'"43+&.1'3+.'v"k'&78'"
2%38.P
see !
http://www.emdbiosciences.com/docs/docs/LIT/ISB_USD.pdf
46
EXAM-TYPE QUESTION
A reversible inhibitor was observed to change the Lineweaver–Burk
plot of an enzyme’s activity in two ways. 1. One of the intercepts of
the plot was modified. 2. The slope of the plot was increased.
A.
B.
C.
D.
E.
The inhibitor is noncompetitive if the x-intercept is increased.
The inhibitor is competitive if the y-intercept is increased.
The inhibitor is competitive if the x-intercept is decreased
The inhibitor is competitive if the y-intercept is unchanged.
The inhibitor is noncompetitive when the y-intercept is decreased.
47
EXAM-TYPE QUESTION
In an enzyme mediated reaction, the inhibitor I reduces the rate of
product formation from substrate. If velocity/substrate data are
plotted in Lineweaver-Burk form, the inhibitor is competitive when:
A.! Both the y- and x-intercepts are decreased.
B.! The slope of the plot is decreased.
C.! The x-intercept is decreased, the y-intercept is unchanged and the slope
of the plot increases.
D.! The slope of the plot is increased and the y-intercept decreases
E.! The x-intercept increased and the y-intercept is unchanged.
C^
R8;;+3!="
U+F%&7".%;<,%1-%&7".0,8(0&."I'"9'3%2'9"46'"U%)6+',%.GU'&4'&"'a8+(0&"+&9"
'?<3'..'9"$;"+&9"M;"%&"4'3;."01"3+4'")0&.4+&4."+&9"ZO404+,[
2.!
We discussed the significance of Km, Ks and Vm
Km =
A="
"
"
C="
k 2 + k3
k1
Ks =
k2
k1
Vm = k3[E]t
K'"'?+;%&'9"46'".%7&%B)+&)'"01"F)+4"+&9".+I"46+4"I6'&"F)+4"%.".0"73'+4"46+4"%4"%."
;8)6"1+.4'3"46+&"OR"9%..0)%+(0&"40"O"w"RL"F)+4Q$;"+<<30+)6'."!e^"UG!=.G!"+&9"46'"
'&:-;'"%.".+%9"40"6+2'"+)6%'2'9"F%&'()"<'31')(0&="R8)6"+&"'&:-;'"%."&0I",%;%4'9"
0&,-"5-";0,')8,+3"9%D8.%0&=
K6'&"F)+4"%."2'3-",0IL"$;"+<<30+)6'."$R=
K'"'?+;%&'9"'&:-;'"%&6%5%(0&."Nj0;<'((2'L"&0&)0;<'((2'P"+&9".60I'9"46+4"
46'.'"+3'"9%.(&78%.6+5,'="j0;<'((2'"%&6%5%403."%&)3'+.'"$;+<<"+&9"&0&)0;<'((2'"
%&6%5%403."3'98)'"M;+<<=
Appendix 1
49
Amount or quantity versus concentration
When we write a reaction of the form:
d[B]
= k1 [A]
dt
the rate of product (B) formation is written as d[B]/dt where [B]
is the concentration of B and t is time. The term “d” indicates an
infinitesimal change in [B] with respect to t (in math terms, this
is a derivative).
The concentration of B or MOLARITY is given by
[B] =
moles of B
= molarity = M
liters
Appendix 1
50
Enzyme/substrate interactions
Receptor/ligand interactions
The first step in enzyme function is the formation of the enzyme (E)/substrate (S)
complex, ES.
Consider the reaction
k1
E+S
ES
k2
The rate of ES formation is given by:
+d[ES]
= k1 [E][S]
dt
The rate of ES breakdown is given by:
!d[ES]
= k 2 [ES]
dt
At equilibrium:
Thus:
+d[ES] !d[ES]
=
"k1 [E][S] = k 2 [ES]
dt
dt
[ES] =
k1
[E][S]
k2
Appendix 1
51
If we are interested in how much ES is formed (how much substrate is bound) at any
given [S] and [E], we can express [ES] as as a fraction of total enzyme [Et] as:
[ES]
[ES]
=
[E t ] [E] + [ES]
k1
[E][S]
[ES]
k2
=
[E t ] [E] + k1 [E][S]
k2
and, substituting for [ES],
Canceling [E] thus gives us:
k1
[S]
[ES]
k2
=
[E t ] 1 + k1 [S]
k2
thus
[S]
[ES]
=
[E t ] k 2 + [S]
k1
Appendix 2
52
Defining the rate of product formation, v as
v = k3 [ES]
(1
We must again express [ES] in terms of known quantities!
Rate of [ES] formation
+d[ES]/dt = k1 [E][S] (2
Rate of breakdown of [ES]
-d[ES]/dt = k2[ES] + k3 [ES]
or
!
! !
!
-d[ES]/dt = (k2 + k3) [ES] (3
Appendix 2
53
In the "steady state" the concentrations of intermediates (e.g. ES) are
unchanged, whereas [S] + [P] can change. If we limit measurements
of v to early stages, [ES] does not change (there is no reverse
reaction)
!
!
!
d[ES]/dt = 0 (4
!
!
i.e.
k1 [E] [S] = (k2 + k3) [ES] (5
hence
[ES] =
[E][S]k 1
k2 + k 3
[ES] =
(6
[E][S]
(k 2 + k 3 )
k1
(7
Appendix 2
54
The following steps are algebraic tricks
[Et] = [E] + [ES] (8
where Et is total enzyme
if we divide the velocity equation (1 by Et we obtain
k [ES]
v
= 3
[E t ] [E] + [ES]
!
(9
Appendix 2
55
We can rearrange this to
[ES]
v
=
[E t ]k 3 [E] + [ES]
(10
then substitute for [ES] from equation (6 to give
k1
k1
[E][S]
[S]
v
k2 + k3
k2 + k3
=
=
[E t ]k 3 [E] + k 1 [E][S] 1 + k 1 [S]
k2 + k3
k2 + k3
Appendix 2
(11
56
Define ! Vm as:! !
[Et]k3 = Vm!
Km =
!
(12
k2 + k3
k1
(13
[S]
v
Km
=
Vm 1+ [S]
Km
or,
v=
Vm [S]
K m + [S]
(14
(15