Selected Solutions to End of Chapter 14 Problems
Class 1
1. This is a good exercise to see if you can do it with the book closed. You need to know the
chemical structure of the molecules, but not the exact ΔG o’ for each reaction:
ΔGo’
Glucose + ATP  Glucose-6-phosphate + ADP
-16.7 kJ/mole
Glucose-6-phosphate  Fructose-6-phosphate
1.7 kJ/mole
Fructose-6-phosphate + ATP  Fructose-1,6-bisphosphate
-14.2 kJ/mole
Fructose-1,6-bisphosphate  DHAP + Glycerald-3-phosphate
23.8 kJ/mole
DHAP  Glycerald-3-phosphate
7.5 kJ/mole
Glucose + 2 ATP  2 Glycerald-3-phosphate + 2 ADP
2.1 kJ/mole
What you can see what is working here is that aldolase has a high thermodynamic tendency to
go backwards (a big positive ΔGo’), but the ΔG in the cell is near zero which means in the cell
instead of having “” aldolase should be “” as it is in the text (page 550).
2. The rest of Glycolysis, question 1 forward, the Pay-off Phase: each reaction….
Glycerald-3-phosphate + Pi + NAD+  1,3 bisphosphoglycerate + NADH
1,3 bisphosphoglycerate + ADP  3-phosphoglycerate + ATP
6.3 kJ/mole
-18.5 kJ/mole
3-phosophoglycerate  2-phosphoglycerate
4.4 kJ/mole
2-phosophoglycerate  Phosphoenolpyruvate + H2O
7.5 kJ/mole
Phosphoenolpyruvate + ADP  Pyruvate + ATP
-31.4 kJ/mole
Glyceraldehyde-3-phosphate + NAD+ + 2 ADP + Pi  Pyruvate + NADH + 2ATP + H2O
-31.7 kJ/mole
Of course, you know that each of the reactions in the 2 nd half of Glycolysis goes twice for one
glucose because one glucose produced two Glyceral-3-phosphates in the first half.
The question goes on to ask about glycolysis and converting the pyruvate to lactic acid (a
fermentation end product but also what muscles do when oxygen is low) and these run twice:
Pyruvate + NADH  Lactic acid + NAD+ ΔGo’ is -25.1 kJ/mole.
then the overall would be:
2Glycerald-3-phosphate + 4ADP + 2Pi  2Lactic acid + 4 ATP + 2H2O ΔGo’ = -113.6 kJ/mole
4. Ethanol production in Yeasts. Pyruvate is converted to ethanol and CO 2. The first enzyme
decarboxylates pyruvate to produce acetaldehyde (2nd carbon of Pyr becomes the first in
acetaldehyde…both forms of a carbonyl. Then the second reaction is ethanol dehydrogenase:
acetaldehyde + NADH  ethanol + NAD+
Calculate the Keq from the standard reduction potentials in Table13-7: Strategy: first get the
ΔEo’ and then calculate the ΔGo’ which can be used then to calculate the Keq
acetaldehyde/ethanol Eo’ = -0.197v
NAD/NADH Eo’ = -0.32v
ΔEo’ = -0.197v – (-0.32v) = 0.123v
ΔGo’ = -nF ΔEo’ = -2(96.5 jK/v mole)( 0.123v) = -23.7 kJ/mole
Remembering that ΔGo’ = - RT ln Keq
So, rearrange this to be 1 / ln Keq = - RT / ΔGo’
which can be inverted to be
ln Keq = ΔGo’ / -RT = -23.7 kJ/mole / 2.48 kJ/mole = 9.56
so Keq = e9.56 = 1.4 x 104
5. Aldolase Thermodynamics. Remember that the aldolase ΔGo’ = 23.8 kJ/mole in the direction of
Glycolysis, which means it is -23.8 kJ/mole going towards glucose! So what is the actual, in the cell,
ΔG taking into consideration the actual concentrations of substrates and products?
Remember: ΔG = ΔGo’ + RT ln Q
So ΔG = 23.8 kJ/mole + (2.58 kJ/mole) ln (
[3x10−6 M][1.6x10−5 M]
[1.4x10−5 M]
)
ΔG = 23.8 kJ/mole + 2.58 kJ/mole (-12.6) = -8.6 kJ/mole
So, you can see the endothermic ΔGo’ gets completely reversed in vivo.
6. Pulse-Chase experiments have been crucial in figuring out metabolic pathways. This takes
advantage of using radioactive molecules in which only one or particular atoms have been made
radioactive and it is rather easy to measure these as they become transformed by metabolism. The
beauty of it is that enzymes react with radioactive atoms as if they were the normal atoms (which they
are in their electron clouds), it is just that the radioactive atoms of C, N, H, S, P have an extra
neutron(s) in the atomic nucleus.
So, in this experiment glucose labeled in carbon 1 (as 14C with all the other carbons being normal 12C
and not radioactive) is used as the pulse. This simply means the cell free extract of yeast, was
inoculated with a small amount of 14C-glucose, then a little while later (can be seconds or minutes)
given a much larger amount of non-radioactive glucose as the “chase”.
a. In this experiment, we are asking which carbon of ethanol (it only has 2 carbons) is the
radioactive carbon? What do you expect knowing how Glycolysis works? Make sure you
know why it is in the methyl carbon of ethanol.
b. It’s carbons 3 or 4. This is where aldolase cuts the hexose-phosphate into the two three
carbon molecules: DHAP and glyceraldehyde-3-phosphate. So both the 3 and 4 carbons on
glucose become the aldehyde carbon (carbon 1) of glyceraldehyde-3-phosphate.
7. Heat from metabolism. Here taking into account the very large industrial fermentation tanks (see
Box 14-3 on page 566) one needs to keep a temperature at which the fermenting microbes do the
optimum fermentation. Metabolism produces heat, here glycolysis while making ATP does not
efficiently convert all the energy available to ATP, the rest, more than 50% is released as heat. This
is what keeps your body warm and makes you regulate heat dissipation through your skin and lungs.
For these large industrial fermenters, they must include pipes in the inside of the tank through which
they can circulate chilled water to remove heat. As usual, this was learned the hard way by making
the first fermenters without ways to get rid of the heat resulting in the microbes producing so much,
they killed themselves.
9. This question concerns the reversibility of aldolase in a cell. Adding to a cell free yeast extract
glyceraldehyde-3-phosphate with one carbon (this is what we will have to figure out) labeled. After
some time, the label could be detected in carbons 3 and 4 of fructose-1,6-bisphosphate. Remember
aldolase can break the bond between carbons 3 and 5 of fructose-1,6,-bisphosphate, and it can make
the that bond bringing DHAP and glycerald-3-phosphate together. It is an equilibrium reaction.
To get the 4th carbon of the fructose-16-bisphosphate labeled, it has to come from the 1st carbon of
glycerald-3-phosphate. And to get the 3rd carbon of fructose-1,6-bisphosphate labeled it has to come
from the 1st carbon of DHAP. Triosephosphate isomerase takes care of interconverting DHAP and
glycerald-3-phosphate.
10. Shortcuts don’t always work. In this case, oxidizing glyceraldehyde-3-phosphate directly to 3phosphoglycerate cuts out the substrate level phosphorylation: putting Pi onto the mixed acid
anhydride 1,3 bisphosphoglycerate, and consequently a loss of an ATP at this points next step, 3phosphoglycerate kinase. If that happened, Glycolysis would not produce any net ATP. Disaster for
organisms living by fermentation. A respiratory organism could still do well because they oxidize all
the NADH produced from glycolysis and the Citric Acid Cycle to produce ATP (we will do that in
Chapter 19).
14. Arsenic poisoning. You can see that arsenate is very much similar to phosphate. It is a poison
because of that: if AsO4= gets into cells, it can react just like inorganic phosphate and form a mixed
acid anhydride in the glyceraldehyde-3-phosphate dehydrogenase reaction.
a. The product is unstable and rapidly breaks down to arsenate and 3-phosphoglycerate.
b. Arsenate lowers the amount of ATP made at this step driving down the energy charge. It is
leathal.
16. The vitamin niacin is required in small amounts. It is converted to NAD + which is constantly
recycling between its reduced and oxidized forms, so very little is needed compared to nutrients that
provide energy and the production of metabolic intermediates used for the synthesis of amino acids,
fats, nucleotides, and sugars used in forming polymers (most of our biomass).
Class 2
21. The Glycolysis part should be immediately easy: for each sugar molecule 2 ATP and 2 NADH
are produced to form 2 pyruvates. Gluconeogenesis (Table 14-4, page 573) from pyruvate to glucose
costs 4 ATP + 2 GTP + 2 NADH. Counting a GTP as an ATP, that would be 6 ATP and 2 NADH. So
every time a cell converted 1 glucose to 2 pyruvate and the converted 2 pyruvates to one glucose
would be -4 ATP. This is the classic example of a futile cycle: if these two pathways were running the
same time in a cell, it would die of every starvation. We will get into how we stop this from happening
in Chapter 15. All metabolic pathways are highly regulated so futile cycles do not occur.
22. Why is important that Gluconeogenesis is not the reversal of Glycolysis. This is because while
several glycolytic enzymes are just about a equilibrium, there are others that are highly exothermic
(exergonic) and can not be reversed. Thereby requiring different gluconeogenic enzymes at those
points to circumvent them: pyruvate kinase, hexokinase and phosphofructokinase.
23. Pyruvate kinase reaction is essentially irreversible. In Glycolysis it produces an ATP, but
Gluconeogenesis requires an ATP and a GTP (essentially 2 ATP’s ) to get pyruvate converted to
PEP.
24. Gluconeogenic substrates. It is all about getting to PEP. Here’s the deal:
a. succinate in the Citric Acid Cycle gets converted to oxaloacetate  PEP. Yes
b. glycerol is converted to DHAP and thereby can be run back up to glucose (2 glycerols 
glucose).
c. acetyl-CoA. No deal, we have no way to get this into PEP.
d. pyruvate, why of course…we did this as the beginning of gluconeogenesis.
e. butyrate. No deal, this can only be converted to acetyl-CoA.
Note: if we had the glyoxylate cycle (coming in Chapter 16) we could convert 2 acetyl-CoA to
succinate  OAA  PEP and then glucose like microbes and plants, but mammals lack the
glyoxylate cycle and can not produce PEP from acetyl-CoA (and thus can not produce glucose
from acetyl-CoA).
26. Blood lactate levels before, during and after a 400 m sprint. This graph does not come from real
data. It came from a cartoonist’s imagination. We will be doing this in class with real data.
a. The fast rise comes from rapid glycolysis in muscles. The contractions in muscles limits
blood flow through these muscles, so they ferment pyruvate to lactic acid to maintain ATP
production for this vigorous exercise.
b. The decline comes from the liver working to remove lactic acid from blood and converting it
to glucose through gluconeogenesis.
c. The equilibrium is in favor of lactate formation (see problem 2), so there is always a small
amount of lactate in blood.
27. The relationship of blood lactate levels of fructose-1,6-bisphosphatase. Some individuals lack
this enzyme, and therefore can not do Gluconeogenesis. Their livers struggle to deal with the high
lactate levels. Some gets slowly oxidized (requiring the unfavorable lactate dehydrogenase reaction).