Explanation:
The preparatory phase converts glucose first to glucose-6-phosphate, then fructose-6-phosphate, and finally fructose-1,6-bisphosphate. Aldolase subsequently cleaves the six-carbon intermediate into two three-carbon triose phosphates.
2. How many ATP molecules are consumed directly during the preparatory phase for each glucose molecule entering glycolysis?
A.
One ATP, used only during conversion of glucose to glucose-6-phosphate
B.
Two ATP, used during phosphorylation of glucose and fructose-6-phosphate
C.
Two ATP, both used during cleavage of fructose-1,6-bisphosphate
D.
Four ATP, because each of the two three-carbon products is phosphorylated twice
Correct Answer:
Two ATP, used during phosphorylation of glucose and fructose-6-phosphate
Explanation:
The preparatory phase requires two ATP investments per glucose: one for glucose phosphorylation to glucose-6-phosphate and one for fructose-6-phosphate phosphorylation to fructose-1,6-bisphosphate.
3. Which statement best explains why the preparatory phase is described as an energy-investment phase?
A.
It produces ATP only after oxidation of the two triose phosphates
B.
It consumes ATP to add phosphate groups and prepare the six-carbon sugar for cleavage
C.
It converts NAD+ into NADH before glucose is split
D.
It generates carbon dioxide while releasing stored chemical energy
Correct Answer:
It consumes ATP to add phosphate groups and prepare the six-carbon sugar for cleavage
Explanation:
The preparatory phase uses two ATP to phosphorylate the substrate. These phosphorylations increase the molecule's reactivity and help prepare fructose-1,6-bisphosphate for cleavage into two three-carbon compounds.
4. If phosphoglucose isomerase were inhibited in the preparatory phase, which immediate conversion would be blocked?
A.
Glucose to glucose-6-phosphate
B.
Glucose-6-phosphate to fructose-6-phosphate
C.
Fructose-6-phosphate to fructose-1,6-bisphosphate
D.
Fructose-1,6-bisphosphate to two triose phosphates
Correct Answer:
Glucose-6-phosphate to fructose-6-phosphate
Explanation:
Phosphoglucose isomerase catalyzes the rearrangement of glucose-6-phosphate into fructose-6-phosphate, which is the second major transformation of the preparatory phase.
5. Which pair correctly identifies the ATP-consuming reactions of the preparatory phase and the enzymes catalyzing them?
A.
Glucose to glucose-6-phosphate by hexokinase; fructose-6-phosphate to fructose-1,6-bisphosphate by phosphofructokinase
B.
Glucose-6-phosphate to fructose-6-phosphate by hexokinase; fructose-1,6-bisphosphate cleavage by aldolase
C.
Glucose to glucose-6-phosphate by aldolase; fructose-6-phosphate to fructose-1,6-bisphosphate by isomerase
D.
Fructose-1,6-bisphosphate cleavage by phosphofructokinase; triose-phosphate conversion by hexokinase
Correct Answer:
Glucose to glucose-6-phosphate by hexokinase; fructose-6-phosphate to fructose-1,6-bisphosphate by phosphofructokinase
Explanation:
Hexokinase catalyzes the first ATP-dependent phosphorylation, while phosphofructokinase catalyzes the second ATP-dependent phosphorylation in the preparatory phase.
6. What is the principal significance of converting glucose-6-phosphate into fructose-6-phosphate before the second ATP-dependent phosphorylation?
A.
It changes the carbonyl position so that subsequent cleavage can yield two three-carbon compounds
B.
It removes the first phosphate group so that no ATP is consumed later
C.
It oxidizes the sugar and produces NADH before cleavage
D.
It directly converts the six-carbon sugar into two glyceraldehyde-3-phosphate molecules
Correct Answer:
It changes the carbonyl position so that subsequent cleavage can yield two three-carbon compounds
Explanation:
Isomerization changes glucose-6-phosphate, an aldose, into fructose-6-phosphate, a ketose. This arrangement permits phosphorylation at carbon 1 and subsequent symmetrical cleavage of fructose-1,6-bisphosphate.
7. A single fructose-1,6-bisphosphate molecule is cleaved during the preparatory phase. Which products result directly from this cleavage?
A.
Two molecules of glucose-6-phosphate
B.
One molecule each of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate
C.
Two molecules of pyruvate
D.
One molecule each of fructose-6-phosphate and glyceraldehyde-3-phosphate
Correct Answer:
One molecule each of glyceraldehyde-3-phosphate and dihydroxyacetone phosphate
Explanation:
Aldolase cleaves fructose-1,6-bisphosphate into the two three-carbon phosphates glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.
8. Why does the preparatory phase ultimately provide two molecules that can enter the subsequent payoff sequence, even though aldolase initially forms two different products?
A.
Dihydroxyacetone phosphate is converted into glyceraldehyde-3-phosphate by triose phosphate isomerase
B.
Glyceraldehyde-3-phosphate is converted into glucose-6-phosphate by hexokinase
C.
Both products are directly converted into fructose-1,6-bisphosphate by phosphofructokinase
D.
Aldolase immediately converts both products into pyruvate
Correct Answer:
Dihydroxyacetone phosphate is converted into glyceraldehyde-3-phosphate by triose phosphate isomerase
Explanation:
Aldolase produces one glyceraldehyde-3-phosphate and one dihydroxyacetone phosphate. Triose phosphate isomerase converts the latter into glyceraldehyde-3-phosphate, yielding two molecules for the subsequent glycolytic sequence.
9. For each glucose molecule, what is the carbon-count change associated with the cleavage step of the preparatory phase?
A.
One six-carbon molecule becomes two three-carbon molecules
B.
One six-carbon molecule becomes three two-carbon molecules
C.
Two three-carbon molecules combine into one six-carbon molecule
D.
One six-carbon molecule loses one carbon and becomes one five-carbon molecule
Correct Answer:
One six-carbon molecule becomes two three-carbon molecules
Explanation:
The six-carbon fructose-1,6-bisphosphate molecule is cleaved into two three-carbon triose phosphates, preserving the total number of carbon atoms while dividing the pathway into two parallel three-carbon branches.
10. Which combined description of the preparatory phase is correct for one glucose molecule?
A.
Two ATP are consumed, fructose-1,6-bisphosphate is formed, and two three-carbon phosphate molecules are produced
B.
Two ATP are produced, glucose-6-phosphate is oxidized, and one three-carbon molecule is formed
C.
One ATP is consumed, pyruvate is formed, and carbon dioxide is released
D.
Four ATP are consumed, fructose-1,6-bisphosphate is reduced, and two-carbon molecules are produced
Correct Answer:
Two ATP are consumed, fructose-1,6-bisphosphate is formed, and two three-carbon phosphate molecules are produced
Explanation:
The preparatory phase invests two ATP in phosphorylation reactions, forms fructose-1,6-bisphosphate, and then generates two three-carbon phosphate molecules through cleavage and isomerization.
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Preparatory Phase
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