GATE 2020 BT Q.53: solution and answer

GATE 2020 · BTQ.53 · 2 marks · NAT

The concentrations of ATP, ADP and inorganic phosphate in a cell are 2.59, 0.73 and 2.72 mM, respectively. Under these conditions, free energy change for the synthesis of ATP at 37 °C is ______ kJ/mol (round off to 2 decimal places). Given: free energy change for ATP hydrolysis under standard conditions is -30.5 kJ/mol and R = 8.315 kJ/mol.K

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Answer and solution

ANSWER

The free energy change for ATP synthesis under these cell conditions is about 48.99 kJ/mol.

Official answer: 48.50 to 49.50, or 18521.20 to 18521.30Source: Official answer key, GATE 2020 BT (IIT Delhi)

Synthesis is the reverse of hydrolysis, $ADP + P_{i} \to ATP + H_{2} O$ . Hydrolysis has $Δ G^{\circ'} = - 30.5 kJ/mol$ , so for synthesis $Δ G^{\circ'} = + 30.5 kJ/mol$ .
Use $Δ G = Δ G^{\circ'} + R T ln Q$ with $Q = \frac{[ ATP ]}{[ ADP ] [ P _{i} ]}$ . Water is left out of $Q$ by the biochemical standard-state convention.
The standard state is 1 M, so convert to molar: $[ATP] = 2.59 \times 1 0^{- 3}$ M, $[ADP] = 0.73 \times 1 0^{- 3}$ M, $[P_{i}] = 2.72 \times 1 0^{- 3}$ M.
$Q = \frac{2.59 \times 1 0 ^{- 3}}{( 0.73 \times 1 0 ^{- 3} ) ( 2.72 \times 1 0 ^{- 3} )} = \frac{2.59 \times 1 0 ^{- 3}}{1.9856 \times 1 0 ^{- 6}} \approx 1304.4 M^{- 1}$ .
$T = 37 + 273 = 310 K$ . The gas constant $8.315 J/mol.K$ equals $8.315 \times 1 0^{- 3} kJ/mol.K$ , so $R T = 8.315 \times 1 0^{- 3} \times 310 \approx 2.5777 kJ/mol$ .
$ln Q = ln 1304.4 \approx 7.1736$ , so $R T ln Q \approx 2.5777 \times 7.1736 \approx 18.49 kJ/mol$ .
$Δ G = 30.5 + 18.49 \approx 48.99 kJ/mol$ . It is positive, so making ATP in the cell needs an input of energy, as expected.

KEY IDEA

Synthesis is the reverse reaction, so the standard value flips to +30.5 kJ/mol. The actual value comes from $Δ G = Δ G^{\circ'} + R T ln Q$ with concentrations in molar and water left out of $Q$ .

WATCH OUT FOR

Using the hydrolysis quotient (ADP times Pi over ATP) or keeping the sign at -30.5 for synthesis. Another slip is leaving concentrations in mM, which gives a quotient of 1.3 instead of 1304 per molar. A third is using R = 8.315 kJ/mol.K literally, which makes RT a thousand times too large.

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GATE 2020 BT Q.53: solution and answer

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