GATE 2019 BT Q.46: solution and answer

GATE 2019 · BTQ.46 · 2 marks · NAT

The difference in concentrations of an uncharged solute between two compartments is 1.6-fold. The energy required for active transport of the solute across the membrane separating the two compartments is ____ cal mol^-1 (rounded off to the nearest integer). (R = 1.987 cal mol^-1 K^-1, T = 37 °C)

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Answer and solution

ANSWER

The energy required is about 290 cal per mole.

Official answer: 280 to 295Source: Official answer key, GATE 2019 BT (IIT Madras)

For an uncharged solute there is no membrane-potential term, so the free energy needed to move it against a concentration gradient is $Δ G = R T ln \frac{C _{2}}{C _{1}}$ .
The concentration ratio is $C_{2} / C_{1} = 1.6$ , so $ln 1.6 = 0.4700$ .
Convert the temperature to kelvin: $T = 37 + 273 = 310 K$ (using $310.15 K$ gives the same integer).
$R T = 1.987 cal mol^{- 1} K^{- 1} \times 310 K \approx 616 cal mol^{- 1}$ .
$Δ G = 616 \times 0.4700 \approx 289.5 \approx 290 cal mol^{- 1}$ , so active transport must supply about 290 cal per mole.

KEY IDEA

Active transport of an uncharged solute against a concentration gradient costs $Δ G = R T ln (C_{2} / C_{1})$ , with no electrical term, and T must be in kelvin.

WATCH OUT FOR

Using T = 37 in degrees Celsius instead of kelvin, using log base 10 in place of the natural log, or adding a membrane-potential term $z F Δ ψ$ that does not apply to an uncharged solute.

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GATE 2019 BT Q.46: solution and answer

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