BIOC 301 ProblemSet3-2012Answers

7/30/2019 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




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




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




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




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]




7

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




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]




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

BIOC 301 ProblemSet3-2012Answers

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