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Glycolysis: all 10 enzymes, in order, in 15 seconds.
Glucose becomes 2 pyruvate, with a net gain of 2 ATP and 2 NADH. Knowing where ATP is spent, where it is made and where NADH goes is what the questions test.
A cell extract that contains all the glycolytic enzymes is given fructose 1,6-bisphosphate instead of glucose. Per molecule of fructose 1,6-bisphosphate converted to two pyruvate, what are the NET ATP and the NADH produced?
Pick one. The red pen does the rest.
The deciding lineGlycolysis spends ATP at hexokinase and at phosphofructokinase-1 (PFK-1), which makes fructose 1,6-bisphosphate; aldolase and triose phosphate isomerase then give two glyceraldehyde 3-phosphate, each of which yields one NADH and two ATP on the way to pyruvate.
Why C: Fructose 1,6-bisphosphate is the product of PFK-1, so both ATP-spending steps (hexokinase and PFK-1) lie before the starting point: nothing is invested. It splits into two glyceraldehyde 3-phosphate, and each yields 1 NADH (GAPDH) and 2 ATP (phosphoglycerate kinase and pyruvate kinase): net 4 ATP and 2 NADH.
Why each choice fails
AConditions ignored: net 2 ATP is the yield from glucose, which pays 2 ATP at hexokinase and PFK-1; starting from fructose 1,6-bisphosphate, both payments have already been made, so the net is 4.
BOne step short: this subtracts the PFK-1 ATP as if only hexokinase were skipped, but PFK-1 is the enzyme that makes fructose 1,6-bisphosphate, so its ATP is not spent either: 4 − 0 = 4.
CHolds: no ATP is invested after fructose 1,6-bisphosphate, and its two glyceraldehyde 3-phosphate yield 2 × 2 = 4 ATP and 2 × 1 = 2 NADH.
DScale slip: this counts one triose only; triose phosphate isomerase converts dihydroxyacetone phosphate to a second glyceraldehyde 3-phosphate, so the yields double to 4 ATP and 2 NADH.
In glycolysis, aldolase cleaves fructose 1,6-bisphosphate, whose carbonyl group is at C2, between C3 and C4 by a retro-aldol reaction, giving two three-carbon products. Glucose 6-phosphate, whose carbonyl group is at C1, is first isomerized to fructose 6-phosphate. If the open-chain form of glucose 6-phosphate underwent a retro-aldol cleavage instead, which products would form?
Pick one. The red pen does the rest.
The deciding lineAn aldol reaction joins the α-carbon of one carbonyl compound to the carbonyl carbon of another, giving a β-hydroxy carbonyl compound; a retro-aldol reaction breaks that same bond between the α-carbon and the β-carbon.
Why B: With the carbonyl at C1, C2 is the α-carbon and C3, which carries an OH, is the β-carbon, so a retro-aldol breaks C2–C3: C1–C2 leaves as a two-carbon fragment and C3–C6 as a four-carbon aldehyde whose C3 hydroxyl becomes the new carbonyl. In fructose the carbonyl at C2 puts the α–β bond at C3–C4, which is why the isomerization comes first.
Why each choice fails
ALook-alike terms: C1–C2 joins the carbonyl carbon to the α-carbon; a retro-aldol breaks the bond between the α-carbon and the β-carbon, the one that carries the hydroxyl group.
BHolds: carbonyl at C1 makes C2 the α-carbon and C3 (with its OH) the β-carbon; breaking C2–C3 gives a two-carbon fragment (C1–C2) and a four-carbon aldehyde (C3–C6).
CConditions ignored: C3–C4 is the α–β bond only when the carbonyl is at C2, as in fructose; with the carbonyl at C1 the cleavage moves to C2–C3, which is why glucose 6-phosphate is isomerized first.
DSays more than the data: one ketose example does not show that ketones are required; aldehydes also form aldols and undergo retro-aldol cleavage (two acetaldehyde molecules give 3-hydroxybutanal).
Lactate dehydrogenase catalyzes pyruvate + NADH + H+ → lactate + NAD+, which converts pyruvate (CH₃COCOO−) into lactate (CH₃CH(OH)COO−). Which statement about C2, the carbon of the ketone group of pyruvate, is correct?
Pick one. The red pen does the rest.
The deciding lineA carbon's oxidation state counts +1 for each bond to a more electronegative atom such as O (a double bond counts twice), −1 for each bond to H and 0 for each bond to another carbon.
Why C: In pyruvate, C2 has a C=O double bond (+2) and two C–C bonds (0): +2. In lactate, C2 has one C–O bond (+1), one C–H bond (−1) and two C–C bonds: 0. The oxidation state falls by 2, so C2 gains two electrons (delivered by NADH as a hydride ion) and is reduced.
Why each choice fails
AOne step short: +1 counts the C=O bond becoming C–O but not the new C–H bond, which lowers the count by one more; C2 in lactate is at 0.
BLook-alike terms: the numbers are right, but oxidation and reduction are swapped; a fall from +2 to 0 is a gain of electrons, which is reduction. NADH is the species oxidized.
CHolds: C2 goes from +2 (C=O, two C–C) to 0 (C–O +1, C–H −1, two C–C), a gain of two electrons from NADH, so it is reduced.
DDirection reversed: a fall in oxidation state means C2 gains electrons, so it is reduced, not oxidized; +1 also misses the new C–H bond (lactate's C2 is at 0).
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