MCAT practice · Biological and Biochemical Foundations · 1D Bioenergetics and metabolism
Throttling glycolysis: PFK-1 effectors and NAD+ regeneration in muscle
An original practice passage written for this site, not an AAMC item. Exam pace is about 8 minutes for a passage like this.
Phosphofructokinase-1 (PFK-1) transfers a phosphoryl group from ATP to fructose 6-phosphate (F6P), forming fructose 1,6-bisphosphate in the committed step of glycolysis. Besides serving as a substrate, ATP also binds the enzyme at a separate regulatory site. To characterize how the enzyme responds to several cellular signals, investigators purified PFK-1 from rabbit skeletal muscle and measured its initial velocity across a range of F6P concentrations in a coupled spectrophotometric assay at pH 7.0 and 25 °C. Each condition was run in triplicate. The four conditions differed only in the concentration of ATP and in whether AMP or fructose 2,6-bisphosphate (F2,6BP) was added (Table 1).
In a second experiment, the investigators asked how cultured mouse myotubes sustain glycolysis when oxygen is limited. Replicate dishes (n = 4 per condition) were incubated for 2 h in medium containing 5 mM glucose under 21% O₂ (normoxia) or 1% O₂ (hypoxia). A third group was held in hypoxia with 20 mM oxamate, a pyruvate analog that competitively inhibits lactate dehydrogenase (LDH). Glucose disappearance from the medium and lactate accumulation in the medium were measured enzymatically and normalized to total cell protein (Figure 1). Cell viability, assessed by dye exclusion, exceeded 95% in every group.
Hypoxia raised both glucose consumption and lactate release. Oxamate abolished the hypoxic rise in lactate release and also lowered glucose consumption below the normoxic value. The investigators proposed that oxamate slows glucose use indirectly, by limiting the regeneration of NAD+ in the cytosol of cells whose mitochondria are starved of oxygen. They suggested that the same dependence explains why tissues with little oxidative capacity rely so heavily on lactate production.
| [F6P] (mM) | 0.5 mM ATP | 5 mM ATP | 5 mM ATP + 0.2 mM AMP | 5 mM ATP + 1 µM F2,6BP |
|---|---|---|---|---|
| 0.25 | 85.7 | 0.2 | 10.7 | 51.9 |
| 0.5 | 100 | 1.8 | 33.7 | 82 |
| 1 | 109.1 | 13.3 | 73.2 | 103.1 |
| 2 | 114.3 | 60 | 103.4 | 113.4 |
| 4 | 117.1 | 106.7 | 115.4 | 117.6 |
| 8 | 118.5 | 118.2 | 118.8 | 119.1 |
- Glucose consumed
- Lactate released
Data are illustrative, constructed for practice.
1. According to Table 1, raising the ATP concentration from 0.5 mM to 5 mM changes the kinetic behavior of PFK-1 in which way?
- A. Right: half-maximal velocity (about 60) is reached below 0.25 mM F6P at 0.5 mM ATP but only at about 2 mM F6P at 5 mM ATP, while both curves converge near 118 at saturating F6P. High ATP lowers the enzyme's apparent affinity for F6P, the signature of allosteric inhibition.
- B. Wrong: the half-maximal F6P level rises rather than falls, and the maximal velocity is unchanged, so neither part matches the data.
- C. Wrong: this is the reverse relationship; high ATP shifts the curve to the right, so more F6P, not less, is needed for half-maximal velocity.
- D. Wrong: at 8 mM F6P both ATP conditions reach about 118 µmol/min/mg, so the maximal velocity does not fall; this reads the depressed low-F6P values as a Vmax change.
2. A mutant PFK-1 has a regulatory site that can no longer bind ATP, while its catalytic site is unaffected. Assayed under the conditions of Table 1 at 5 mM ATP and 1 mM F6P with no other effectors, the mutant's velocity (µmol/min/mg) would be closest to:
- A. Wrong: 13.3 is the wild-type velocity at 5 mM ATP, where ATP is bound at the regulatory site; the mutant cannot be inhibited this way.
- B. Wrong: 60.0 is the wild-type value at 5 mM ATP but at 2 mM F6P, a misread row that still reflects inhibition.
- C. Wrong: 73.2 is the value with AMP partly relieving inhibition in the wild type; the mutant has no ATP inhibition to relieve, so it should run faster than this.
- D. Right: with no ATP bound at the regulatory site, the catalytic site sees 5 mM ATP only as saturating substrate, so the enzyme behaves like the uninhibited wild type at low ATP, about 109 µmol/min/mg at 1 mM F6P.
3. Which finding, if obtained, would most weaken the investigators' proposal for how oxamate reduces glucose consumption by hypoxic myotubes?
- A. Wrong: if NAD+ is limiting, the GAPDH step slows and its substrate and the triose phosphates upstream of it should pile up; this supports the proposal.
- B. Wrong: ruling out direct inhibition of the upstream regulated enzymes removes an alternative explanation, which strengthens the proposal.
- C. Wrong: a lower lactate-to-pyruvate ratio is what LDH inhibition predicts, so this finding supports rather than weakens the proposal.
- D. Right: the proposal is that oxamate acts only by starving GAPDH of NAD+. If glucose use still slows when NAD+ is supplied in excess, the effect cannot be explained by NAD+ shortage and must come from some other action of the drug.
4. Based on Figure 1, approximately what percentage of the glucose consumed by normoxic myotubes was released into the medium as lactate?
- A. Wrong: 22% is the share of glucose not recovered as lactate (100% − 78%), the remainder rather than the quantity asked for.
- B. Wrong: 65% is glucose divided by lactate (0.40/0.62), the ratio inverted and without accounting for the two lactates made per glucose.
- C. Right: each glucose can yield two lactates, so 0.40 µmol/h of glucose could give at most 0.80 µmol/h of lactate; 0.62/0.80 ≈ 78%. The rest of the pyruvate is presumably oxidized by the mitochondria, which oxygen allows.
- D. Wrong: 95% is the hypoxic value (2.08/2.20), read from the wrong condition.
5. Assume liver PFK-1 responds to effectors as the muscle enzyme in Table 1 does. In a hepatocyte containing about 5 mM ATP, glucagon signaling would most likely cause:
- A. Right: glucagon lowers F2,6BP, removing the activator that overrides ATP inhibition, so at physiological ATP the enzyme behaves like the strongly inhibited 5 mM ATP curve and glycolytic flux falls.
- B. Wrong: losing F2,6BP does not lower ATP; the 0.5 mM ATP column describes an uninhibited enzyme, the opposite of what losing an activator does.
- C. Wrong: this is the insulin response; glucagon lowers F2,6BP so the liver slows glycolysis and can release glucose.
- D. Wrong: the direction of the F2,6BP change is reversed, and a rise in an activator could not move activity toward the inhibited curve.
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