Selected Solutions to End of Chapter 13 Problems
Class 1
6. The second reaction in Glycolysis (an “isomerase”) is written backwards in this problem (which
makes the real answers different than that in the textbook), it should be:
Glucose-6-phosphate  Fuctose-6-phosphate
Keq = 0.508
This is an aldose  ketose. Think about this reaction: one ring structure going to another.
a. What is the ΔGo’ ?
Well… ΔGo’ = -RT ln Keq
ΔGo’ = -(2.48 kJ/mole) ln 0.51
ΔGo’ = 1.7 kJ/mole
b. If the concentration of glucose-6-phosphate is 0.5M and fructose-6-phosphate is 1.5M, then
what is the ΔG ?
Well… ΔG = ΔGo’ +RT lnQ
So, ΔG = 1.7 kJ/mole + RT ln (1.5 M / 0.5 M )
ΔG = 1.7 kJ/mole + (2.48 kJ/mole x ln 3)
ΔG = 4.4 kJ/mole
Note: this is different from the answer in the text which was calculated from the reverse
of the glycolytic pathway.
c. Why are ΔGo’ and ΔG different? One is under standard conditions, the other, ΔG is the
“actual’ ΔG in the conditions that either are artificial (in this case) or natural in the cell with the
actual concentrations of metabolites in the cell. This is what we will look at to see the actual
thermodynamics of whole metabolic pathways.
9. ΔGo’ for coupled reactions:
glucose-1-phosphate  glucose-6-phosphate
ΔGo’ = -7.3 kJ/mole
glucose-6- phosphate  fructose-6-phosphate ΔGo’ = 1.7 kJ/mole
So, the overall ΔGo’ is -5.6 kJ/mole
now, what’s the Keq?
From: ΔGo’ = -RT ln Keq
ln Keq = - ΔGo’ / RT = - (5.6 kJ/mole) / (2.48 kJ/mole) = 2.25
ln Keq = 2.25 so Keq = 9.5 Note that this answer differs from that in the textbook,
this is because they rounded out too much before getting to the calculated answer.
12. Some more coupled reactions:
a. Phosphocreatine + H2O
ADP + Pi
Creatine + Pi
b. ATP + H2O
Fructose + Pi
ΔGo’ = 30.5 kJ/mole
ATP + H2O
Overall: Phosphocreatine + ADP
ΔGo’ = -43.0 kJ/mole
Creatine + ATP ΔGo’ = -12.5 kJ/mole
ΔGo’ = -30.5 kJ/mole
ADP + Pi
Fructose-6-phosphate + H2O ΔGo’ = 15.9 kJmole
Overall: ATP + Fructose
ADP + Fructose-6-phosphate ΔGo’ = -14.6 kJ/mole.
14. This problem is just like problem 6 above. Do it and get the answers which is for the ΔG in the
cell: -13 kJ/mole. You need to use the RT at 37oC which is 2.58 kJ/mole and not that much different
from that at 25oC, but use the right one anyhow.
19. Daily use of ATP by Human Adults. First of all, the calculation of the ΔG of ATP synthesis from
known concentrations of the substrates and products in a liver cell at 37 oC. Then we can calculate
the weight of ATP used each day in a 150 pound human (68 kg).
a. The ΔG. Data: ΔGo’ of ATP hydrolysis is -30.5 kJ/mole and just the opposite sign for
synthesis. In liver cell: [ATP] = 3.5 mM, [ADP] = 1.5 mM, and [Pi] = 5.0 mM.
So for the synthesis: ADP + Pi  ATP + H2O
ΔG = ΔGo’ + RT lnQ
hints: RT has to be at 37oC (310o K), put concentrations in M.
so, RT = (8.315 J/mole K)(310 K) = 2.58 kJ/mole
ΔG = 30.5 kJ/mole + (2.58 kJ/mole) ln (
[0.0035 M]
)
[0.0015 M][0.005 M]
ΔG = 30.5 kJ/mole + (2.58 kJ/mole)(6.15) = 46.4 kJ/mole
What is the take home message? Thermodynamics in the cell is the real energy changes that
take place or need to be supported. The standard ΔG shows the tendency to produce or
require energy.
b. Data: the average human (68 kg) needs the standard 2,000 diet calories (one diet calorie is
really 1000 cal = kcal) which in kJ is 8,360 kJ / 24 hours. Assume half of food energy is used
to make ATP. What is the weight of ATP your body makes/day.
So the strategy is to get the moles of ATP this represents and use the MW to get the weight.
The MW of ATP = 507 g / mole or 0.507 kg / mole
Half of the energy = 8,360 kJ / 2 = 4,180 kJ.
4,180 kJ / (46.4 kJ / mole) = 90 moles
90 moles x (0.507 kg / mole) = 45.6 kg WOW that is 2/3’s your weight !
c. That is certainly not the mass in your cells, but the metabolic flux is rather rapid (even
though the total amount/day is large) keeping the concentrations low, in the mM range.
20. Rates of turnover of the β and γ phosphates of ATP. Here the data is about having 32P labeled
ATP in either the γ-phosphate or the middle, β-phosphate (the α-phosphate is ester linked to ribose
carbon 5’) and then measuring the radioactivity in the cell’s inorganic fraction as Pi. Data: shows that
rapidly the γ-phosphate occurs in the inorganic fraction, and the β-phosphate is slower entering the
inorganic fraction. This is because there are many reactions that use ATP energy in the β-to-γ acid
anhydride bond (ATP  ADP + Pi). There are reactions that use the other ATP acid anyhydride bond
(ATP  AMP + PPi) between the α-to-β phosphates. See the next question. The pyrophosphate (PPi)
that is formed is then converted to 2 Pi by the enzyme pyrophosphatase.
21. Synthesis of acetyl-CoA from acetate:
Acetate + CoA + ATP  acetyl-CoA + AMP + PPi
a. The ΔGo’ of the hydrolysis of acetyl-CoA to acetate and CoA is -31.4 kJ/mole (in the table
below, -32.5kJ/mole in the EOC problem) which is clearly more than ATP + H2O  ADP + Pi (30 kJ/mole) which means the usually ATP hydrolysis reaction can not drive the synthesis of
acetyl-CoA from acetate + CoA. So, now check out Table 13-6, see below.. The ΔGo’ for ATP
+ H2O  AMP + PPi is -45.6 kJ/mole, now we are talking energy to do the job.
b. Pyrophosphatase even makes this reaction more favorable because it rapidly removes one
of the products, PPi.
Class 2
24. The mitochondrial electron transport pathway from NADH to oxygen, the overall equation is
simple:
NADH + H+ + ½ O2  NAD+ + H2O
Here is Table 13-7 standard reduction potential of some half cells (half reactions)
a. Calculate the ΔEo’ for the net NADH oxidation in the mitochondria.
ΔEo’ = Eo’ oxidized – Eo’ reduced = 0.816 v – (- 0.320v) = 1.14 v
b. Calculate the ΔGo’ for this reaction.
ΔGo’ = -n F ΔEo’ = -2 (96.5 kJ/v mole) (1.14 v) = 220 kJ/mole
c. About how many ATPs can this generate given it costs 52 kJ/mole to make ATP in the
mitochondrion? Easy, 220 kJ/mole / 52 kJ/mole = ~4 (4.2 would be wrong, you can’t have
part of an ATP !
25. What is the E with different concentration of NAD+ and NADH? Basic thermo calculations from:
E = Eo’ +
𝑅𝑇
[𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑎𝑐𝑐𝑒𝑝𝑡𝑜𝑟]
ln (
)
𝑛𝐹
[𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑑𝑜𝑛𝑜𝑟]
Lets first make this simple, and get one number for
𝑅𝑇
𝑛𝐹
=
2.48 𝑘𝐽/𝑚𝑜𝑙𝑒
𝑘𝐽
2 𝑥 96.5𝑣𝑚𝑜𝑙𝑒
= 0.013 v.
Now this is easy:
a. E = 0.32 v + (8.5 x 10-3 v) ln (1mM / 10 mM) = 0.32 v + 0.013 v (-2.3)
= 0. 32v – 0.030 v = 0.29 v
b. E = 0.32 v + (8.5 x 10-3 v) ln (10mM / 10 mM) = 0.32 v + 0.013 v (0)
= 0.32 v + 0 = 0.32v
c. E = 0.32 v + (8.5 x 10-3 v) ln (10mM / 1 mM) = 0.32 v + 0.013 v (2.3)
= 0.32 v + 0.03 v = 0.035 v
26. Making a list of increasing tendency to accept electrons (check Table 13-7 above).
α-ketoglutarate < NADP+ < oxaloacetate < O2
-0.38v
< -0.32v < -0.166v
list goes from most reducing to most oxidizing:
< 0.82v
27. Just for fun! Which would likely go in the direction shown…just reading Table 13-7
Reaction
Malate + NAD+  oxaloacetate + NADH
Acetoacetate + NADH  β-OH-butyrate + NAD+
Pyruvate + NADH  Lactate + NAD+
Pyruvate + β-OH-butyrate  Lactate + acetoacetate
Malate + Pyruvate  oxaloacetate + Lactate
Acetaldehyde + succinate  ethanol + fumarate
? in the Direction Shown
NO
NO
YES
YES
NO
NO
Remember ΔEo’ = Eo’ electron acceptor – Eo’ electron donor….use the Table above.
ΔEo’
-0.154v
-0.026v
0.135v
0.161v
-0.019v
-0.228v