Answer Key
1) False, NAD+ is a suitable oxidation agent, O2 works, but is not required.
2) C. 2 H2O, 2 pyruvate, 2 NADH, and 2 ATP. There are 4 ATP’s made overall, but 2 are used, so the
net is only 2. After 3 cycles of glycolysis, 12 total ATP’s are produced, but the net production is 6
3) A. Eventually you will arrive at 2 molecules of glyceraldehydes-3-phosphate, but initially it is in
the form of 1 DHAP and 1 G3P.
4) C. phosphoglycerate kinase is in the oxidation stage.
5) B. All of these are involved in this step, but pyruvate carboxylase is an enzyme, not a cofactor.
6) A. The reverse reaction is not doable with the same molecules and enzyme, you have to form
an oxaloacetate intermediate, so this is an irreversible step.
7) A and E. You could argue glucose because the glycolysis pathway transforms glucose into
oxaloacetate, but it doesn’t directly feed it.
8) B. Substrate-level phosphorylation is a type of metabolism that results in the formation and
creation of ATP or GTP. In the transformation of Succinyl-CoA to Succinate, GDP picks up an
inorganic phosphate molecule and becomes GTP. This is the only substrate-level
phosphorylation step in TCA cycle.
9) A.
10) False. Acetyl Coenzyme A synthesis is irreversible.
11) False. The pancreas secretes insulin to breakdown carbohydrates, the gallbladder produces bile
to emulsify lipids.
12) D. Biotin is a cofactor for the formation of oxaloacetate from pyruvate. It’s function is only to
carry CO2 to the pyruvate group, adding a carboxylate functional group.
13) A.
14) B. High energy state, or high concentrations of ATP inhibits glycolysis at the
phosphofructokinase step.
15) C. AMP inhibits gluconeogenesis at the fructose-1,6-bisphosphatase step.
16) D. A, B, and C are specific examples talked about in the lecture notes.
17) B. Since pepsin is active in acidic environments, it’s safe to assume that it’s part of the only
acidic amino acid protease on the board, which is aspartate (the carboxylate form of aspartic
acid).
18) B. Cholocystokinin is actually produced by intestinal cells, not in the stomach. Pepsin digestion
in the stomach stimulates the release of CCK.
19) A. Phosphoenolpyruvate release the most energy, because once the phosphate group leaves,
there is an easily stabilized (by resonance forms) enol structure. The alcohol will rapidly convert
to a more stable carbonyl ketone group.
20) A. When GAP  1,3-BPG an electron is trapped as electrical energy as NADH when NAD+
oxidizes GAP. Likewise, chemical energy is trapped as ATP when 1,3-BPG donates one of its
phosphate groups to ADP.
21) C, 22) F, 23) A, 24) G, 25) I, 26) B, 27) E, 28) J, 29) D, 30) H
Honestly, it’s unrealistic to assume we’ll have a matching section on the quiz or test. That being
said, it’s very important to know where and why each enzyme does its thing.
Hexokinase – Glucose  G6P, makes sense. We have a hexose (glucose) and we’re
phosphorylating it, which is the standard function of a kinase.
Phosphoglucoisomerase – G6P  F6P, again just take it for what it is. We have a
phosphoglucose, and we’re rearranging it with the same amount of carbons, and that is what an
isomerase does.
Phosphofructokinase – F6P F-1,6-BP, once more, just take it piece by piece. We have a
phosphofructose compound, and we are adding another phosphate via a kinase.
Aldolase – F-1,6-BP gets cleaved into 1 DHAP and 1 GAP. F-1,6-BP is an aldol (carbonyl group
with an OH at the beta position), and aldolase’s cleave aldols.
Triose phosphate isomerase – DHAP  GAP, this is not a part of the main reaction, but when
aldolase cleaves F-1,6-BP it makes 1 DHAP and 1 GAP. Glycolysis needs GAP to proceed, so this
enzyme creates an ISOMER of DHAP, which is a PHOSphorylated TRIOSE.
GAP dehydrogenase – GAP  1,3-BPG, this one is confusing because it looks like you are just
adding a phosphate group, so you would think this should be a kinase. However you start with
an aldehyde, and it is oxidized into a carboxylic acid group. NAD+ is the oxidizing agent, and gets
reduced to NADH. This step is highly exergonic, so much so that it will drive the next reaction
which is highly endergonic. This reaction is an inorganic phosphate group being attached to the
carboxylate functional group, making 1,3-BPG, a group with very high phosphoryl-transfer
potential. This is an example of phosphorylation coupled to oxidation, and even though the
overall reaction is slightly endergonic, it can proceed because of how exergonic the first part
of the reaction is.
Phosphoglyceratekinase – 1,3-BPG  G3P, you start with a phosphoglycerate, and remove one
of the phosphate groups with a kinase.
Phosphoglyceratemutase – G3P  G2P, start with a phosphoglycerate and produce an isomer,
which is a “mutation” if you will, hence mutase.
Enolase – G2P  PEP, Create an enol group by eliminating H2O from G2P, and forming an alkene
bond next to a phosphate group.
Pyruvate Kinase – PEP  Pyruvate, remove a phosphate group from PEP and you have pyruvate!
31) Glucose  G6P, irreversible because you need a new enzyme to return to glucose, which is
Glucose 6-phosphatase.
32) F6P  F-1,6-BP, irreversible for the same reason, new enzyme is fructose 1,6-biphosphatase
33) PEP  Pyruvate, irreversible, requires an intermediate structure, and 2 new enzymes.
Pyruvate  Oxaloacetate, catalyzed by pyruvate carboxylase.
Oxaloacetate  PEP, catalyzed by PEP carboxykinase.
34) B. The phosphate group actually changes which enzymes will be receptive to the NAD
compound. It segregates the reducing potential that the cell produces for biosynthesis from the
reducing potential that it is dedicating for oxidation and ATP synthesis. Without the phosphate
group there is a preference for catabolic enzymes. With the phosphate group, there is a
preference for biosynthetic enzymes.
35) D. Thioesters have all of these qualities, where as oxygen esters do not.
36) B. The TCA cycle is an excellent source of ATP, much better than glycolysis.
37) C. The purpose of fermentation is to recreate an NAD+ molecule that was used up in an
oxidation during glycolysis. More NAD+ molecules will further drive glycolysis reactions.
38) False. It is actually exactly as efficient because there is still the same Net ATP gain at the
end.
39) B. Lactose is never mentioned as a precursor, lactate is however.
40) D. Gluconeogenesis occurs in the liver.
41) True. After glucose is created in the liver, it is exported to the blood, to build up bloodglucose levels.
42) C. B and D are eliminated because there is no isomerization or mutation occurring. A is
correct for the first reaction, glycerol to glycerol phosphate, but the question asks for the 2nd
reaction, so Glycerol phosphate dehydrogenase is the correct choice. (An ATP is used, and an
NADH is formed)