a.Aditi / practice lab

SKILLS · BIOLOGY & BIOCHEMISTRY

Coupled transport energy

A transport cycle can move one substance uphill by moving another downhill. Add every contribution, with its direction and stoichiometry, before judging the cycle.

01 · THE IDEA

Count each transported particle’s chemical and electrical contributions once.

ΔGion = RT ln(Cdestination / Corigin) + zF(ψdestination − ψorigin)

For an ideal dilute model, z is signed charge and voltages are measured in volts. Neutral solutes have z = 0. Sum the terms for a mole of complete carrier cycles. Negative ΔG favors the specified forward cycle; it does not by itself specify transport speed.

Choose a direction first

For inward motion, destination means inside. A lower inside concentration gives a negative chemical term. An inside-negative voltage favors a positive ion moving inward, but opposes a negative ion moving inward.

Multiply the whole ion term

If two identical ions move each cycle, multiply both the chemical and electrical terms by two. Add the neutral solute’s chemical term separately.

Reverse and interpret

Reversing the entire cycle under the same conditions reverses the sign of its ΔG. A favorable cycle still requires a suitable transport mechanism; an ion gradient can supply energy without direct ATP hydrolysis by the carrier.

CHANGE ONE PART OF THE MODEL

Build the energy budget for one cycle

Each inward ion contributes −3 kJ/mol chemically and, when voltage is present, −2 kJ/mol electrically. One neutral nutrient costs +9 kJ/mol inward.

Ions per cycle
Each ion-5 kJ/mol
Neutral nutrient+9 kJ/mol
Full cycle ΔG-1 kJ/mol

2 × (−3 − 2) + 9 = -1 kJ/mol of cycles

The inward cycle is energetically favorable under these conditions.

02 · WORKED EXAMPLES

Follow the setup, then the numbers

1. Add the complete cycle

Two positive ions enter with one neutral nutrient. Each ion has an inward chemical contribution of −3 kJ/mol and electrical contribution of −2 kJ/mol. The nutrient contributes +9 kJ/mol inward.

  1. One ion contributes −3 − 2 = −5 kJ/mol.
  2. Two ions contribute 2 × (−5) = −10 kJ per mole of cycles.
  3. Add nutrient cost: −10 + 9 = −1 kJ/mol of cycles.
  4. The specified inward cycle is energetically favorable.

Carry this forward: The uphill nutrient contribution is permitted because the total cycle is downhill.

2. Remove only one driving force

Keep the concentrations from example 1, but clamp the voltage difference to zero.

  1. Each ion now contributes only its chemical term: −3 kJ/mol.
  2. Cycle total = 2 × (−3) + 9 = +3 kJ/mol of cycles.
  3. The inward cycle is now unfavorable under the stated conditions.

Carry this forward: Removing voltage does not remove the concentration gradient; change only the specified term.

03 · TRY IT IN DIFFERENT FORMS

Practice in different forms

0 of 4 checked

For each question, identify the requested quantity and write the setup before calculating. Use a hint if needed, then choose an answer to see the explanation.

Untimed learning practice. These responses stay on this page and reset when you leave or reload; they do not change your test results.

QUESTION 1 / 4 · B/B

Two ions and one neutral solute

A cycle brings two identical ions inward with one neutral solute. Per ion, chemical ΔG is −4 kJ/mol and electrical ΔG is −1 kJ/mol. The solute contributes +6 kJ/mol. What is ΔG per mole of inward cycles?

Choose one answer
1 of 4

TAKE INTO A FRESH PASSAGE

Your setup checklist

  1. State the transport direction.
  2. Write chemical and electrical terms with signed charge.
  3. Multiply each complete term by the number transported per cycle.
  4. Sum the cycle and interpret the sign separately from the transport rate.

Before calculating, explain what each term represents and what the requested answer would mean.

Concept references

All scenarios and questions here are original learning exercises. These references support the underlying concepts; the simplified assumptions are stated within each example.