PRACTICE SET 2
Glycolysis and glycogen breakdown
1.
In working skeletal muscle under anaerobic conditions, glyceraldehyde-3-phosphate is
converted into pyruvate (the payoff phase of glycolysis), and pyruvate is reduced to
lactate. Write balanced equations for all of the reactions in this process, with the standard
free-energy change for each (obtain from text). Then write the overall or net equation for
the payoff phase of glycolysis (with lactate as the end product), including the net standard
free-energy change.
2.
A “pulse-chase” experiment using 14C -labeled carbon sources is carried out on a yeast
extract maintained under strictly anaerobic conditions to produce ethanol. The
experiment consists of incubating a small amount of 14C -labeled substrate (the pulse)
with the yeast extract just long enough for each intermediate in the pathway to become
labeled. The label is then “chased” through the pathway by the addition of excess
unlabeled glucose. The “chase” effectively prevents any further entry of labeled glucose
into the pathway.
(a) If [1-14C] glucose (glucose labeled at C-1 with 14C) is used as a substrate, what is the
location of 14C in the product ethanol? Explain.
(b) Where would 14C have to be located in the starting glucose molecule in order to
assure that all the 14C activity were liberated as 14CO2 during fermentation to ethanol?
Explain.
14
3.
C-Labeled glyceraldehyde-3-phosphate was added to a yeast extract. After a short time,
fructose-1,6-bisphosphate labeled with 14C at C-3 and C-4 was isolated. What was the
location of the label in the starting glyceraldehyde-3-phosphate? Where did the second
14
C label in fructose-1,6-bisphosphate come from? Explain.
4.
Suppose you discovered a mutant yeast whose glycolytic pathway was shorter because of
the presence of a new enzyme catalyzing the reaction
Glyceraldehyde-3-phosphate + H2O + NAD+ → 3-phosphoglycerate + NADH + H+
Although this mutant enzyme shortens glycolysis by one step, how would it affect
anaerobic ATP production? Aerobic ATP production?
5.
During strenuous activity, muscle tissue demands vast quantities of ATP compared with
resting tissue. In rabbit leg muscle or turkey flight muscle, this ATP is produced almost
exclusively by lactate fermentation. ATP is produced in the payoff phase of glycolysis
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by two enzymatic reactions, promoted by phosphoglycerate kinase and pyruvate kinase.
Suppose skeletal muscle were devoid of lactate dehydrogenase. Could it carry out
strenuous physical activity; that is, could it generate ATP at a high rate by glycolysis?
6.
The oxidation of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate, catalyzed by
glyceraldehyde-3-phosphate dehydrogenase, proceeds with an unfavorable equilibrium
constant (K’eq = 0.08; ∆Go’ = + 6.3 kJ/mol). Despite this unfavorable equilibrium, the
flow through this point in the pathway proceeds smoothly. How does the cell overcome
the unfavorable equilibrium?
7.
Measuring the concentrations of metabolic intermediates in the living cell presents a
difficult experimental problem. Because cellular enzymes rapidly catalyze metabolic
interconversions, a common problem associated with perturbing the cell experimentally
is that the measured concentrations of metabolites reflect not the physiological
concentrations but the equilibrium concentrations. Hence, a reliable experimental
technique requires all enzyme-catalyzed reactions to be instantaneously stopped in the
intact tissue, so that the metabolic intermediates do not undergo change. This objective is
accomplished by rapidly compressing the tissue between large aluminum plates cooled
with liquid nitrogen (-190oC), a process called freeze-clamping. After freezing, which
stops enzyme action instantly, the tissue is powdered and the enzymes are inactivated by
precipitation with perchloric acid. The precipitate is removed by centrifugation, and the
clear supernatant extract is analyzed for metabolites. To calculate the actual intracellular
concentration of the metabolite in the cell, the intracellular volume is determined from
the total water content of the tissue and a measurement of the extracellular volume.
The actual intracellular concentrations of the substrates and products involved in
the phosphorylation of fructose-6-phosphate by the enzyme phosphofructokinase-1 in
isolated rat heart tissue are given in the table below.
Metabolite
Apparent concentration (mM)*
Fructose-6-phosphate
0.087
Fructose-1,6-bisphosphate
0.022
ATP
11.42
ADP
1.32
Source: From Williamson, J.R. (1965) Glycolytic control mechanisms I. Inhibition of
glycolysis by acetate and pyruvate in the isolated, perfused rat heart. J. Biol. Chem. 240, 23082321.
*Calculated as µmol/mL of intracellular water.
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(a) Using the information in the table, calculate the ratio, [fructose-1,6-bisphosphate]
[ADP]/[fructose-6-phosphate] [ATP], for the phosphofructokinase-1 reaction under
physiological conditions.
(b) Given that ∆Go’ for the PFK-1 reaction is –14.2 kJ/mol, calculate the equilibrium
constant for this reaction.
(c) Compare the values of the mass-action ratio and K’eq. Is the physiological reaction at
equilibrium? Explain.
8.
The regulated steps of glycolysis in intact cells are identified by studying the catabolism
of glucose in whole tissues or organs. For example, the consumption of glucose by heart
muscle can be measured by artificially circulating blood through an isolated intact heart
and measuring the concentration of glucose before and after the blood passes through the
heart. If the circulating blood is deoxygenated, heart muscle consumes glucose at a
steady rate. When oxygen is added to the blood, the rate of glucose consumption drops
dramatically, then continues at the new, lower rate. Why?
9.
Glycogen phosphorylase catalyzes the removal of glucose from glycogen. Given that
∆Go’ for this reaction is 3.1 kJ/mol, calculate the ratio of [Pi] to [glucose-1-phosphate]
when this reaction is at equilibrium. (Hint: The removal of glucose units from glycogen
does not change the glucose units from glycogen does not change the glycogen
concentration.) The measured ratio of [Pi] to [glucose-1-phosphate] in muscle cells under
physiological conditions is more than 100 to 1. What does this indicate about the
direction of metabolite flow through the glycogen phosphorylase reaction? Why are the
equilibrium and physiological ratios different? What is the possible significance of this
difference?
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Answer Key
+
1.
G-3-P + Pi + NAD
1,3 Bis PGA + ADP
3PGA
2PGA
PEP + ADP
Pyruvate + NADH + H+
Net reaction:
G-3-P + 2ADP + Pi
a.
∆Go’
6.3
-18.9
4.4
1.8
-31.7
-25.2
→ NADH + H + 1,3bis PGA
→
3PGA + ATP
→ 2PGA
→ PEP + H2O
→ pyruvate + ATP
→ lactate + NAD+
+
→ lactate + 2ATP + H2O ∆Go’= -63.3 kJ/mol
The labeled C is on the C2 of ethanol
14
CHO
OH
H
HO
14CH OPO 22
3
H
H
-O
O
HC
O
C
H
OH
O
CH2 OH
O
14CH
3
OH
CH2 OH
H
CH2OPO3 2-
14
14CH
3
OH
CH2 OH
2.
b. Work backwards from 14CO2:
CH2 OPO32-
O 14C
H
O
CHO
O
14
C
O
O
H
CH2OH
O
HO
C O
CH3
OH
14
+
14
H
or
CH3
H
14
H
O
14
OH
C
H
H
OH
OH
CH2 OH
CH2 OPO32-
Radiolabeled carbons could
be either #3 or #4 of GLC
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3. The label has to be in position C#1 of G-3-P. This carbon derived from C4 of fructose1,6bisP. Label in position C3 comes from the fact that some of the G-3-P has to isomerizes
to DHAP to create fructose-1,6bisP. The DHAP will contribute radiolabel at position C3.
4. Normally, the steps are G-3-P to 1,3bisPGA to 3PGA. 2 ATP per glucose are produced
during 1,3bisPGA to 3PGA conversion. Uncoupling this step from ATP production gives a
net yield of 0 ATP for anaerobic respiration.
Since aerobic respiration produces much more ATP, the loss of 2 from glycolysis is much
less significant.
5. No, without it, there would be no way to regenerate NAD+ and glycolysis would stop.
6. The reaction 1,3bisPGA to 3PGA is very energetically favorable and removes the 1,3bisPGA
so that ∆G is negative.
7. a. [fructose-1,6-bisphosphate] [ADP]/[fructose-6-phosphate] [ATP] =0.0292
b. ∆Go’= -RT ln K’eq ,
Keq = 308
-14,200=-8.315(298) ln Keq
c. not near equilibrium, actual mass-action ratio is much less than Keq. Reaction not at
equilibrium, probably by low activity of enzyme. Good place for regulation.
8. The yield of ATP/per glucose metabolized is greater in the presence of oxygen. The total
amount of glucose consumes drops, because the needed amount of ATP is obtained from less
glucose. Glucose consumption reaches a new steady state based on energy yield from the
oxidation of the carbons all the way to carbon dioxide.
9. Reaction is Glycogen + Pi → glycogen + G-1-P
R=8.315 J/moloK T=298oK
∆Go’= -RT ln K’eq
∆Go’= 3.1 kJ/mol
3100= –8.315 (298) (ln Keq)
Keq=
0.286= [G-1-P]
[Pi]
Problem asks to compare the reciprocal of this:
[Pi] .
[G-1-P]
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or 1/0.286 = 3.5 to the observed ratio of >100. G-1-P is at lower concentration than equilibrium
concentration by conversion to G-6-P. This results in a negative ∆G.
∆G = ∆Go’ + RT ln [products ]
[reactants]
∆G = 3,100 + 8.315 (298) ln .01
∆G = 3,100 + -11,411
∆G = -8,311
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