Driving unfavorable reactions in cells: Reactions with positive ΔG° can still proceed in vivo by which strategy?

Introduction / Context:
Cells routinely drive thermodynamically unfavorable steps by harnessing energy from favorable processes. This is central to metabolism, transport, and biosynthesis, where pathway design and enzyme mechanisms exploit coupling and mass-action effects.

Given Data / Assumptions:

• Standard free energy change ΔG° > 0 for the isolated reaction.
• Cellular systems are open and maintain non-equilibrium concentrations.
• ATP hydrolysis is a common exergonic currency (ΔG°′ ≈ −30.5 kJ/mol).

Concept / Approach:
Actual free energy change is ΔG = ΔG° + RT ln Q. By altering Q (ratio of products to reactants), ΔG can become negative. Enzymes also couple unfavorable reactions to favorable ones (including ATP hydrolysis), often via phosphorylated or otherwise high-energy intermediates, making the overall sum exergonic.

Step-by-Step Solution:
Use mass action: keep products low and reactants high → RT ln Q sufficiently negative. Couple reactions: A ⇌ B (unfavorable) + B ⇌ C (highly favorable) → A ⇌ C overall favorable. Exploit ATP: enzyme forms phosphorylated intermediate, transfers group to drive the endergonic step. Therefore, all listed strategies enable progress of an otherwise unfavorable step.

Verification / Alternative check:
Glycolysis and amino acid activation (aminoacyl-tRNA synthetases) exemplify ATP-coupled steps; metabolic channeling maintains favorable Q values.

Why Other Options Are Wrong:
Choosing only one strategy ignores the complementary biochemical tactics cells actually use.

Common Pitfalls:
Equating ΔG° with inevitability in vivo; forgetting that ΔG depends on concentrations and coupling.

Final Answer:
All of the above.