bioc 301 problemset3-2012answers
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Bios. 301 Name: answers
2
2. (5 points) Rieder and Rose (1959, J. Biol. Chem. 234, 1007) performed the following experiment to
examine the stereo-specificities of the three key “middle” enzymes in glycolysis. Dihydroxyacetone phosphate (DHAP) was labeled with tritium using muscle aldolase by incubating DHAP by itself
with the enzyme in T2O. The labeled DHAP was then incubated with muscle phosphotriose
isomerase, glyceraldehyde-3-Pi dehydrogenase, arsenate, and NAD+
(no ADP or ATP was
added). After spectrophotometric measurements showed that the reaction was complete, tritium
labeled NADH was recovered.The isolated NADH was then oxidized with acetaldehyde and yeast alcohol dehydrogenase. Of
the 182,000 cpm (counts per minute) originally present in the labeled dihydroxyacetonephosphate,167,000 cpm were recovered in the oxidized NAD
+, which contains a tritium atom at the C4
position.
a. (1 point) Write out the reactions described above using structural formulas for the glycolytic
metabolites and include cofactors.
(Less detail here is o.k. – I have put some of the answers to the next page in this diagram.)
b. (1 point) Why do you think arsenate was added to the labeled DHAP/enzyme mixture in order to
make the reaction to go to completion?
Arsenate replaces phosphate in the GAPDH reaction. The mixed arsenocarboxylate anhydride
spontaneously hydrolyzes to 3-P i-glycerate, driving the GAPDH reaction to completion.
N
R
T
C
O
NH2
H
CH3CHO
N
R
T
C
O
NH2
C
OH
H H
CH3
++
ADH (re or pro-R or A
side hydrideremoval)
CH2OPO3=
C O
C HHO
H
Dihydroxyacetone-P
Aldolase catalyzesstereospecific removal
of only the pro-S H atom
pro-R
pro-S
aldolase
T2O
CH2OPO3=
C O
C HHO
T
pro-R
pro-S Triose isomeras catalyzes
stereospecific removal
of only the pro-R H atom
TIM
CH2OPO3=
C
CO T
HO H
D-glyceraldehyde-3-P
N
R
H
C
O
NH2
++
NAD+
AsO4H=
instead of Pi
GAPDH ( si or pro-S or
B side hydride addition)
+C
OO
C OHH
CH2OPO3=
AS
O
O
O
1-arseno-3-phospho-D-glycerate
(-)
(-)
C
OO
C OHH
CH2OPO3=
(-)
Spontaneous,
irreversible
HO AS
O
OO
(-)+
3-Pi-D-glycerate
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3
(Problem 2 continued)
c. (1 point) From the observed data, what conclusion can be drawn about the relative stereospecificities
of aldolase and triose isomerase with respect to ! carbanion proton removal and enediol
isomerization?
Aldolase and triose isomerase must remove different hydrogen atoms from the C3 atm of
dihydroxyacetonephosphate or otherwise the radiolabel would have been "lost" from D-
glyceraldehyde phosphate.
d. (1 point) Draw the portion of NAD+
which is involved in oxidation-reduction and indicate the
position to which hydride ion is transferred.
e. (1 point) From the observed data, what conclusion can be made concerning the relative
stereospecificities of glyceraldehyde-3-Pi dehydrogenase and yeast alcohol dehydrogenase with
respect to hydride transfer to and from NAD?
As described above, the two dehydrogenases transfer H -from their substrates to the opposite faces of
the pyridine ring in NAD+. ADH adds and substracts the hydride atom from the A, pro-R, or re-side and
GAPDH adds and substracts from the B, pro-S, or si-side (see lecture 2, p. 8).
O3POH2C
C
CHHO
H N
H OH2O
O3POH2C
C
CHHO
O
H*
DHAP =
+
(-)
Hydrolysis of Schiff's
base and dissociation of DHAP
Aldolase
Proton labeled
with aldolase
=
Note proton isabstracted and added
from the "bottom"
+
pro-S
O C
O
His95
Glu165CH2OPO3
C
C
HO
O
H
H
N N
H
DHAP
=
(-)
abstraction of C1
proton from the "top"
versus
Triose isomerase proton abstractionfrom the "top", Pro-R
N
R
H
C
O
NH2
H
R
N
H
C
O
NH2
H
back
N
R
H
C
O
H2N
H back
Hydride transfer or
removal from the right:
A-type on pro-R side
(ADH, LDH)
Hydride transfer or
removal from the right:
B-type on pro-S side
(GAPDH, glycerolPDH)
pro-R facing outward
pro-S facing outward
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Bios. 301 Name: answers
4
3. (6 points) UDP-glucose pyrophosphorylase catalyzes the formation of UDP-glucose, which is the
first step in the synthesis of glycogen. Glycogen is the main storage polysaccharide in liver andskeletal muscle
In order to determine the reaction mechanism of this enzyme, the following steady state kineticexperiments were carried out.
Initial velocity (vi)
(µmoles/mg-min) for
[UTP]
!M
K M(obs)
!MVmax(obs)
[glucose-1-P]
µ"
[UTP] =
5.0 µ"
[UTP] =
20µ"
[UTP] =
2,000µ" 5.0 10 20
2 3.3 8.3 16.7
5 6.7 16.7 33.3 20.0 10 50
10 10.0 25.0 50.0
30 15.0 37.5 75.0 2000 10 100
100 18.2 45.4 90.91000 19.8 49.5 99.8
V MAX (obs) = 20 50 100
a. (1 points) From these data, fill in the table above for the values of K M(obs) and Vmax(obs) and
indicate below whether or not the mechanism for this synthase involves a ternary complex or two
sets of binary complex reactions. Explain your answer in terms of the general expression for theinitial velocity of a two substrate, enzyme catalyzed reaction.
The pyrophosphorylase or UDPG synthetase catalyzes the reaction by a ternary complex mechanism because the ratio
K M (obs)/V MAX (obs) is decreasing. Because, K M (obs) is unchanging the binding of UTP and glucose-1-P appears to be
random.
b. (2 points) From these data compute the value of the "true" Vmax and the values of the various
Michaelis constants for your mechanism, i.e. K a, K b, and K ab.The general equation for V MAX (obs) is:
V MAX (obs)=V MAX [ B ]
K b + [ B]. You could either solve two
equations in two unknowns for the data pairs in the left-hand
table or assume that V MAX (obs) at 2000 ! M UTP gives the
true V MAX at [glucose-1-P], [UTP] = #. Using the latter
assumption:
50 =100•20
K b +20" K b = 20µ M
The general expression for K M (obs)is:
K M (obs)= K ab + K a [ B ]
K b + [ B ]but since K M (obs) is unchanging
because the mechanism of binding appears random, K M (obs)
= K a = 10 ! M. The random mechanism requires that
K ab=K a•K b , so K ab = 20•10 = 200 ! M 2
c. (1 point) Devise an isotope exchange reaction to test whether or not the enzyme is participating in a
binary complex mechanism with one of the substrates (use either 14C labeled sugar or 32P). Based
on your answer to Part A, will these exchange reactions occur? (hint: Lecture 3, p. 7 and 8) If the enzyme were using a binary complex mechanism, it would be picking up a UMP moiety (~Y) and release PP i (X) and
then transferring the UMP group to glucose-1-P i (Z) to from the final product UDPG (Y~Z). The exchange reactions would
be (see page 8):
NH
O
O N
OOP
O-
O
OP
O-
O
HO OP
O-
O O H
O
CH2OH
P
O
O-
NH
O
O N
OOP
O-
O
O OHP
O-
O
-O OP
O-
O
O H
O
CH2OH
P
O
O-
O-
UDPGUTP
glucose-1-P
++
pyrophosphate (PPi)
UTP* + E E~UMP + PP*i
E~UMP + Glc-1-P* E + UDP*G
X~Y E~Y X
Z~YE~Y Z
K M(obs) is
unchanging
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5
d. (2 points) Another possible mechanism for UDP-glucose pyrophosphorylase is a ternary complexmechanism with random addition of substrates and the rapid equilibrium assumption for the binding
of the two substrates:
K A and K B are equilibrium dissociation constants for the binding of substrates A and B.
(i). (1 point) Derive an expression for the initial velocity of UDP-glucose pyrophosphorylase, vi, in
terms of [A], [B], [E]o, k cat, K A, and K B. First fill in the expressions below and then combine
them.
vi = k cat[EAB]
K A=
[E][A]
[EA]
! [E] =K
A[EA]
[A]
=
K A
K B[EAB]
[B]
[A]
=
K AK
B[EAB]
[A][B]
[E]o = [E]+[EB]+[EA]+[EAB]
[E]0=
K AK
B[EAB]
[A][B]+
K A[EAB]
[A]+
K B[EAB]
[B]+[EAB]
E0=
K AK
B
[A][B]+
K A
[A]+
K B
[B]+1
!
"#
$
%&[EAB]
K B=
[E][B]
[EB] =
[EA][B]
[EAB]! [EB]=
[E][B]
K B
and [EA]=K
B[EAB]
[B]
[EB] =
K AK
B[EAB]
[A][B][B]
K B
= K
A[EAB]
[A]
[EAB]=E
0
K AK
B
[A][B]
+
K A
[A]
+
K B
[B]
+1!
"
#$
%
&
'[A][B]
[A][B]=
E0[A][B]
K AK
B+K
A[B]+K
B[A]+[A][B]( )
=
E0[A][B]
K AK
B+[B]( )+ K
B+[B]( )[A]
'
1
K B+[B]
1
K B+[B]
[EAB]=
E0[A][B]
K B+[B]
K A+[A]
! vi= k
cat[EAB]=
k cat
E0[B]
K B+[B]
[A]
K A+[A]
; vi=
VMAX
[B]
K B+[B]
[A]
K A+[A]
(ii). (1 point) Does this mechanism fit with the data observed on the previous page for UDP-glucose pyrophosphorylase and is the binding of the two substrates ordered or random?
Yes, this expression predicts that K M (obs) should be independent of [B] or [UTP], which is what
is observed on the previous page.
A
B
K A K
B
B
A
K B K
A
Release of products P and Q
kcat
EAB
EB
EA
E
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6
4. (6 points) Citrate synthase catalyzes the condensation of acetyl CoA with oxaloacetate to formcitrate and is the first step in the citric acid cycle for the complete oxidation of fatty acids and sugars.
B A
a. (1 point) The following double reciprocal plots were obtained when oxaloacetate (OAA) was varied
at three different acetyl CoA concentrations.What does this pattern say about the mechanism of
citrate synthase and why?The converging lines imply a ternary complex mechanism
because K M (obs)/V MAX (obs) is decreasing with increasing
[B]. Because K M (obs) is decreasing markedly, an ordered
mechanism is suggested with OAA binding first.
b. Acetonyl CoA is a strong inhibitor of citrate synthase and has the structureshown to the right. Two sets of inhibition studies were carried out to
examine the mode of inhibition exerted by this compound.Acetonyl CoA
i. (1 point) In the first set of experiments, oxaloacetate was fixed
and [acetyl CoA] was varied at three different acetonyl CoAinhibitor concentrations. Based on the pattern to the left, what
kind of inhibition is observed? Explain your answer in terms of effects on K M(obs) and VMAX(obs).
A competitive inhibition pattern is observed with K M (obs) increasing and V MAX (obs) remaining unchanged. This pattern makes sense since
acetonylCoA "looks" line acetylCoA.
ii. (1 point) In the second set of experiments, acetyl CoA wasfixed and [oxaloacetate] was varied at three different acetonyl
CoA inhibitor concentrations. Based on the pattern to the left,what kind of inhibition is observed? Explain your answer in
terms of effects on K M(obs) and VMAX(obs).
This pattern describes uncompetitive inhibition because both K M (obs)
and V MAX (obs) are decreasing by the same extent with increasing
[acetonylCoA] so the slope of the double reciprocal plot is unchanged.
c. (1 point) What do these inhibition patterns say about the mechanism of citrate synthase and theorder of addition of the OAA and acetyl-CoA substrates?
These data show that citrate synthase catalyzes the reaction using a ternary complex, ordered mechanism in which OAA
binds first before AcetylCoA (see LPOB, pp. 622-623).
CH2
C
O
CO2
O2CC
O
S CoACH3 +Citrate
synthaseHSCoA + CH2 C
OH
CO2
O2C CO2CH2
CitrateAcetyl CoA Oxaloacetate(OAA)
1/vi
1/[OAA]
increasing
[Acetyl CoA]]
CH3
C
O
CH2 S CoA
1/vi
increasing
[I]
Fixed OAA
1/[acetyl CoA]
1/vi
increasing
[I]
Fixed Acetyl CoA
1/[OAA]
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d. (2 points) Other workers have reported that citrate synthase Salmonella typhimurium shows anordered ternary complex mechanism, which they wrote as:
K A and K'B are equilibrium dissociation constants for the binding of the substrates A and B.(i). (1 point) Is this mechanism consistent with the kinetic data on the previous page? If so, which
substrate is A and which is B?
Yes, this mechanism specifies an ordered, ternary complex mechanism with A or oxaloacetate binding first
and then AcetylCoA or B.
(ii). (1 point) Derive an expression for the initial velocity of citrate synthase activity, vi, in terms of
[A], [B], [E]o, k cat, K A, and K'B. First fill in the expressions below and then combine them.
vi = k cat[EAB] K A= [E][A]
[EA] ! [E] = K A[EA]
[A]=
K A
"K B[EAB]
[B][A]
= K A"
K B[EAB][A][B]
[E]o = [E]+[EA]+[EAB] !K B=
[EA][B]
[EAB]" [EA]=
!K B[EAB]
[B]
[E]0=
K A
!K B[EAB]
[A][B]+
!K B[EAB]
[B]+[EAB] " E
0=
K A
!K B
[A][B]+
!K B
[B]+1
#
$%
&
'([EAB]"
[EAB]=E
0
K A
!K B
[A][B]+
!K B
[B]+1
)[A][B]
[A][B]=
E0[A][B]
K A
!K B+ !K
B[A]+[A][B]
=E
0[A][B]
K A
!K B+ !K
B+[B]( )[A]
[EAB] =
E0[A][B]
!K B +[B]K
A!K B
!K B+[B]
+ [A]
vi= k
cat[EAB] " v
i=
k cat
E0[B]
!K B + [B]
[A]
K A
!K B
!K B+ [B]
+[A]
or vi=
VMAX
[B]
!K B + [B]
[A]
K A
!K B
!K B+ [B]
+[A]
K'BK A
kcatrelease of productsEABEA + BE + A
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8
5. (4 points) The second step in the catabolism of arginine involves transamination of the $-amino
group on ornithine to !-ketoglutarate (see LPOB-18, Fig. 18-26, p. 698).
The mechanism proposed for this enzyme is:
K A and K B are equilibrium dissociation constants for the binding of orinithine (A) and !-ketoglutarate
(B) to the enzyme and E* is the form of the enzyme containing a nitrogen atom from the $-amino
group of ornithine on the vitamin B6 cofactor (pyridoxamine - see LPOB-18, pp. 677-680).
a. (1 points) Design a set of isotope exchange reactions to test whether or not this mechanism is correct.
Remember:
b. (1 points) What kind of pattern would be observed in double reciprocal plots of 1/vi versus 1/[A] at
different concentrations of B? Explain your answer in one or two sentences.
Parallel lines would be observed because for a binary complex
mechanism K M (obs) and V MAX (obs) both increase to the same
extent, [B]/(K b+[B]). Thus, K M (obs)/V MAX (obs) is unchanging.
ornithine-!-amino-
transferase (E)
ornithine (A)
C
CO2-
H NH3+
CH2
CH2
CH2
NH3+
+
C
CO2-
O
CH2
CH2
CO2-
"-ketoglutarate (B) glutamate (Q)
C
CO2-
NH3+
CH2
CH2
CO2
-
H
+
C
CO2-
H NH3+
CH2
CH2
C
OH
glutamate-5-semialdehyde (P)
k 2k 1Q + EE*BP + E* + BEAE + A
K BK A
Ornithine*-NH2 + E E~NH2 + Glu-semi-aldehyde*
E~NH2 + !-ketoglutarate* E + Glu*
1 / [A]
1/vi
Increasing [B]
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9
c. (2 points) Derive an equation for vi in terms of [A], [B], E0, K A, K B, k 1, and k 2, where E0 is the total
enzyme concentration and the constants are defined in the mechanism above.
(i). (1 point) Write out expressions for the following:
Remember the proposed
mechanism is:
vi=
d[P]
dt= k
1[EA] or k
2[E*B] K
A=
[E][A]
[EA] ! [E] =
K A[EA]
[A]; K
B=
[E*][B]
[E*B]! [E*]=
K B[E*B]
[B]
E0 = [E]+[EA]+[E*]+[E*B]
(ii). (1 points) Using these equations and the following steady state assumption for the modified
enzyme intermediates, derive the equation for vi and find expressions for K M(obs) and Vmax(obs)
in terms of [B], k 1, k 2, K A, K B, and Eo.
d([E*]+[E*B])dt
= k 1[EA]
!k
2[E*B]= 0 ! [E*B]= k 1[EA]
k 2
; then[E*]= [E*]= K B[E*B][B]
=
K B
k 1[EA]
k 2[B]
= K Bk 1[EA]k
2[B]
E0= [E]+ [EA]+ [E*]+[E*B]=
K A
[EA]
[A]+[EA]+
K Bk
1[EA]
k 2[B]
+
k 1[EA]
k 2
=
K A
[A]+1+
K Bk
1
k 2[B]
+
k 1
k 2
!
"##
$
%&&[EA]
[EA]=E
0
K A
[A]+
K Bk
1
k 2[B]
+ 1+k
1
k 2
!
"##
$
%&&
'k
2[A][B]
k 2[A][B]
=
E0k
2[A][B]
k 2K
A[B]+K
Bk
1[A]+ k
1+ k
2( )[A][B]=
E0k
2[A][B]
k 2K
A[B]+ K
Bk
1+ k
1+ k
2( )[B]( )[A]
[EA]=
E0k 2[A][B]K
Bk
1+ k
1+ k
2( )[B]
k 2K
A[B]
K Bk
1+ k
1+ k
2( )[B]+ [A]
=
E0k
2[A][B]
k 1+ k
2
k 1K Bk
1+ k
2
+ [B]
k 2K
A[B]
k 1+ k
2
k 1K
B
k 1+ k
2
+[B]
+[A]
; vi= k
1[EA] =
E0k
1k
2
k 1+ k
2
[B]
k 1K Bk
1+ k
2
+ [B]
[A]
k 2K
A
k 1+ k
2
[B]
k 1K
B
k 1+ k
2
+ [B]
+ [A]
or
VMAX[B]K
b + [B]
[A]
K a[B]
K b+[B]
+ [A]
Where:
VMAX
(obs) =V
MAX[B]
K b
+ [B]and V
MAX=
k 1k
2E
0
k 1+ k
2
; K b=
k 1K
B
k 1+ k
2
K M (obs) = K a[B]K
b+[B]
and K a = k 2K Ak
1+ k
2
; K b = k 1K Bk
1+ k
2
k 2k 1Q + EE*BP + E* + BEAE + A
K BK A