Biological energy: use coupling, concentrations and carriers

MCAT content library / Biology & biochemistry / Biological energy: use coupling, concentrations and carriers

Calculate the complete coupled process, then separate its driving force from its speed and its carrier chemistry.

About 16 minutes, including the worked examples and checks.

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Ask what is favorable, what is fast, and what is coupled
Standard values are a reference, not the cell’s current state
Worked coupling: add reactions and their free energies
Now change the concentrations
Identify what the carrier actually carries
A steady pool can support continuing work
Check your reasoning
Continue learning
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Ask what is favorable, what is fast, and what is coupled

A cell can use a favorable process to help accomplish an unfavorable one, but the connection must be physical: a shared reaction intermediate, a linked conformational cycle or another defined mechanism. Simply placing ATP beside a substrate does not establish coupling. The useful MCAT task is to identify the complete process, calculate its driving force under the supplied conditions, and distinguish that result from its speed.¹

At fixed temperature and pressure, a negative actual Gibbs free-energy change, ΔG, favors the forward reaction; a positive value favors the reverse direction. At equilibrium, ΔG = 0 for that reaction and there is no net thermodynamic preference. These statements concern the mixture’s present condition. They do not say that an uncatalyzed favorable reaction must occur rapidly.²

Standard values are a reference, not the cell’s current state

The relationship is ΔG = ΔG°′ + RT ln Q when using a biochemical standard reference with the stated buffered conditions. The prime commonly denotes a fixed-pH biochemical convention; use a matching reaction quotient and do not add a separate proton term already absorbed into that convention. Q is dimensionless, constructed from product activities divided by reactant activities, each raised to its stoichiometric coefficient. For dilute idealized solutions, activities can be approximated by concentration divided by the standard concentration.⁷

Products accumulating usually raise Q for the written forward reaction; reactants accumulating usually lower it. Thus, a reaction with positive ΔG°′ can have negative actual ΔG if its reactants are sufficiently abundant relative to products. Conversely, a negative standard value does not guarantee forward favorability under every mixture composition.²

ATP hydrolysis illustrates this distinction. Its driving force depends on ATP relative to ADP and inorganic phosphate, along with the defined solution conditions. Cells maintain these pools away from equilibrium. There is no single actual ATP-hydrolysis ΔG that applies to every cell, compartment and condition.³

Worked coupling: add reactions and their free energies

For an original numerical example, suppose a defined enzyme system couples formation of product AB to one ATP hydrolysis. The following transformed standard values are supplied for the same temperature, pH and reference convention; they are illustrative, not measurements of a named cellular pathway.

Process	Written reaction	Supplied ΔG°′
Uncoupled synthesis	A + B → AB + H₂O	+18 kJ/mol
ATP hydrolysis	ATP + H₂O → ADP + Pi	−30 kJ/mol
Coupled overall process	A + B + ATP → AB + ADP + Pi	−12 kJ/mol

Original reaction accounting, per mole of AB formed

Add the first two reactions and cancel the water appearing on opposite sides. Their free-energy changes add as well: +18 + (−30) = −12 kJ/mol. If the enzyme uses a phosphorylated intermediate, that intermediate also disappears from the overall equation when it is produced in one step and consumed in the next. The summed value belongs to the full coupled stoichiometry.⁴

The calculation does not prove that every proposed intermediate is chemically possible. Nor does ATP hydrolysis in a separate, unconnected enzyme automatically drive AB formation. The example specifies a functioning coupled mechanism; the sum tests whether that complete mechanism has a favorable net direction under the chosen conditions.

Now change the concentrations

For this overall reaction, Q = a_ABa_ADPa_Pi/(a_Aa_Ba_ATP), where each a is a dimensionless activity. At the specified temperature, use the supplied approximation RT ln 10 = 6 kJ/mol, so ΔG = −12 + 6 log₁₀Q.

In mixture 1, Q = 0.01. Then log₁₀Q = −2 and ΔG = −12 − 12 = −24 kJ/mol. In mixture 2, ATP activity is 100-fold lower while the other activities are held fixed. Because ATP is in the denominator, Q rises 100-fold to 1. Actual ΔG is now −12 kJ/mol. The forward process is still favorable, but its driving force is smaller.

To locate the boundary, set 0 = −12 + 6 log₁₀Q. This gives log₁₀Q = 2 and Q = 100. At that composition the overall reaction is at equilibrium. At Q = 1000, ΔG = +6 kJ/mol, and the reverse direction is favored. These hypothetical mixtures hold the listed variables fixed; they are not predictions that a living cell can reduce ATP arbitrarily while leaving every other process unchanged.

Identify what the carrier actually carries

ATP can participate in phosphoryl-group transfer and in coupled chemical, mechanical or transport work. NADH and NADPH instead carry reducing equivalents. When NAD⁺ accepts two electrons and a proton to become NADH, it is reduced; the donating substrate is oxidized. FAD can also accept reducing equivalents, and flavoproteins use bound flavin cofactors in electron-transfer chemistry. The cofactor’s presence does not by itself specify the net direction of the complete redox reaction.⁵

The extra phosphate that distinguishes NADP⁺/NADPH from NAD⁺/NADH helps distinguish their interactions with enzymes; NADPH is not thereby an ATP-like phosphoryl donor. NADH commonly participates in fuel oxidation, while NADPH commonly supplies reducing power for biosynthesis and protective redox processes. The relevant question remains: which species gains electrons, which loses them, and what subsequent process regenerates the carrier?⁴⁵

Consider a supplied reaction R_reduced + NAD⁺ → R_oxidized + NADH. The NADH rise reports reduction of the carrier, not direct ATP production. ATP could be generated later through a coupled pathway, but its amount cannot be calculated from this observation without the pathway and coupling assumptions. Keep electron bookkeeping separate from ATP bookkeeping.

A steady pool can support continuing work

At a dynamic steady state, a metabolite’s concentration can remain roughly constant while its production and consumption rates are both nonzero. A constant ATP concentration therefore does not mean ATP hydrolysis has stopped. It can mean that ATP regeneration balances use. Equilibrium is different: the reaction’s forward and reverse fluxes balance with ΔG = 0. A driven cellular network can hold steady concentrations while individual reactions remain displaced from equilibrium.³

Finally, a catalyst lowers an activation barrier and changes how quickly a system approaches its allowed state; it does not make the same overall reaction’s ΔG more negative at unchanged reactant/product activities. Changing coupling stoichiometry or metabolite concentrations can change the driving force, but that is a different operation.⁶

For a new problem, write the balanced overall process, count ATP or carrier equivalents, add the supplied free-energy changes consistently, and apply any concentration correction once. Then make two separate conclusions: whether the direction is thermodynamically favorable and whether the evidence establishes a rate.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

Under one stated set of conditions, synthesis of one mole of X has actual ΔG = +24 kJ/mol. A defined enzyme mechanism couples each X formed to hydrolysis of one mole of ATP, whose actual ΔG under those same conditions is −40 kJ/mol. What is ΔG for the complete coupled process per mole of X?

A. −64 kJ/mol

Review this choice. Adding two negative magnitudes would ignore that the synthesis contribution is unfavorable and positive.

B. −16 kJ/mol

Correct. The actual values are for the same conditions and the stated 1:1 coupling, so +24 + (−40) = −16 kJ/mol.

C. +16 kJ/mol

Review this choice. This reverses the sign of the summed driving force.

D. +64 kJ/mol

Review this choice. This adds magnitudes while discarding the favorable hydrolysis sign.

The principle to carry forward

Add free-energy changes with the signs and stoichiometric coefficients of the complete coupled reaction.

Check 2

In an assay, substrate R is converted from its reduced form to its oxidized form while NAD+ becomes NADH. No ATP reaction is included in the measured step. Which species is the electron donor in that step?

A. Both R and NAD+

Review this choice. NAD+ gains electrons to form NADH, so it is the acceptor rather than a second donor.

B. NAD+ only

Review this choice. NAD+ is reduced in the written change; a reduced species has accepted electrons.

C. R only

Correct. R loses electrons as it is oxidized and therefore donates them to NAD+. This observation alone does not count ATP production.

D. Neither R nor NAD+

Review this choice. The paired oxidation and reduction identify transfer from R to NAD+.

The principle to carry forward

Name the electron donor from the species oxidized; carrier reduction is not itself an ATP balance.

Check 3

A catalyst increases the initial rate of a specified reaction. Temperature, reactant and product activities, and overall reaction stoichiometry are held unchanged. Which change is consistent with catalysis alone?

A. A lower activation barrier with the same reaction ΔG.

Correct. A catalyst changes the pathway and its activation barrier while the free-energy difference between the same reactant and product states stays the same.

B. A lower activation barrier with a necessarily more negative reaction ΔG.

Review this choice. Lowering the kinetic barrier does not alter the free-energy difference between those fixed states.

C. A more negative reaction ΔG with the same activation barrier.

Review this choice. The stem holds the variables determining reaction ΔG fixed; a rate change does not establish a more favorable reaction.

D. A more positive reaction ΔG with the same equilibrium constant.

Review this choice. Changing reaction ΔG under unchanged conditions is not the effect of catalysis; the statement also confuses actual conditions with equilibrium properties.

The principle to carry forward

A faster reaction and a more favorable reaction are distinct claims; use the measurement and maintained conditions to separate them.