Q.82  Active transport involves the movement of a biomolecule against a concentration gradient
across the cell membrane using metabolic energy. If the extracellular concentration of a
biomolecule is 0.005M and its intracellular concentration is 0.5M, the least amount of
energy that the cell would need to spend to transport this biomolecule from the outside to
the inside of the cell is ________ kcal/mol (up to 2 decimal points).
(Temperature T = 298K and universal gas constant R= 1.98 cal/mol.K)

Answer: 2.72 kcal/mol

The calculation uses the formula for the minimum free energy change required for
active transport of an uncharged biomolecule against its concentration gradient:

ΔG = RT ln([inside]/[outside])

---

Calculation Steps

1. The concentration inside the cell is 0.5 M and outside is
   0.005 M.
2. The concentration ratio is:

   [inside]/[outside] = 0.5 / 0.005 = 100

3. The natural logarithm of 100 is:

   ln(100) ≈ 4.605

4. Using R = 1.98 cal/mol·K and T = 298 K:

   RT = 1.98 × 298 ≈ 590 cal/mol

5. Therefore:

   ΔG = 590 × 4.605 ≈ 2,717 cal/mol

6. Converting to kilocalories:

   ΔG ≈ 2.72 kcal/mol

---

Assumptions Explained

• No membrane potential term (zFΔψ) is included because the molecule
  is specified as neutral and no voltage is given.
• This value represents the thermodynamic minimum energy required.
• In real biological systems, ATP hydrolysis
  (ΔG°′ ≈ −7.3 kcal/mol) provides more than sufficient energy.
• Since the question has no options, direct numerical derivation confirms the answer.

---

Introduction to Active Transport Energy Calculation

Active transport energy calculation determines the minimum metabolic energy required
for cells to move biomolecules against a concentration gradient across membranes.
For a system with 0.005 M extracellular and
0.5 M intracellular concentrations at 298 K,
the energy required is 2.72 kcal/mol.
This concept is frequently tested in CSIR NET Life Sciences.

Core Formula and Derivation

For an uncharged solute, the Gibbs free energy change is given by:

ΔG = RT ln(C_in / C_out)

• R (gas constant) = 1.98 cal/mol·K
• T = 298 K
• C_out = 0.005 M
• C_in = 0.5 M

Ratio = 100, ln(100) ≈ 4.605

Result: ΔG = 2,717 cal/mol = 2.72 kcal/mol

Step-by-Step Solution for CSIR NET

1. Compute concentration ratio: 0.5 / 0.005 = 100
2. Calculate ln(100) = 4.60517
3. Calculate RT = 1.98 × 298 ≈ 590 cal/mol
4. ΔG = 590 × 4.605 ≈ 2,717 cal/mol
5. Convert to kcal: 2.72 kcal/mol

Applications in Cell Biology

Cells commonly use ATP-driven pumps (e.g., Na⁺/K⁺-ATPase),
where ATP hydrolysis provides significantly more energy than this minimum requirement.
This principle is important in molecular biology, genetics, and biotechnology.

Common Pitfalls and Tips

• Do not include membrane potential unless the solute is charged.
• Always check units and convert cal to kcal correctly.
• Round final answers as instructed (typically to two decimals).