Muscle cells need an input of free energy to be able to contract. One biochemical pathway for energy — Physical Chemistry — Thermodynamics Chemistry Question
Thermodynamics in Biochemistry
Muscle cells need an input of free energy to be able to contract. One biochemical pathway for energy transfer is the breakdown of glucose to pyruvate in a process called glycolysis. In the presence of sufficient oxygen in the cell, pyruvate is oxidized to CO2 and H2O to make further energy available. Under extreme conditions, such as an Olympic 100m sprint, the blood can not provide enough oxygen, so that the muscle cell produces lactate according to the following reaction:
ΔG o ’ = – 25.1 kJ mol–1
In living cells the pH value usually is about pH = 7. The proton concentration is therefore constant and can be included into ΔG o which is then called ΔG o ’, a quantity commonly used in biochemistry.
Calculate ΔG o for the reaction given above.
Model Answer
ΔG 0 = – RT lnK
= – RT ln( c(lactate) c(NAD+) / (c(pyruvate) c(NADH) c(H+)) )
= – RT ln( c(lactate) c(NAD+) / (c(pyruvate) c(NADH)) ) – RT ln( c(H+)–1 )
ΔG 0 ’ = – RT ln( c(lactate) c(NAD+) / (c(pyruvate) c(NADH)) )
ΔG 0 = ΔG 0 ’ – RT ln(c(H+)–1)
= – 25100 J mol–1 – 8.314 J mol–1 K–1 × 298.15 K × ln 10^7
= – 25.1 kJ mol–1 – 40.0 kJ mol–1
= – 65.1 kJ mol–1
Calculate the reaction constant K’ (the proton concentration is included again in the constant, K’ = K · c(H+)) for the reaction above at 25°C and pH = 7.
Model Answer
ΔG 0 ’ = – RT lnK’
K’ = e^(–ΔG°’/(RT))
K’ = e^(25100 / (8.314 × 298.15))
K’ = 2.5×10^4
ΔG o ’ indicates the free enthalpy of the reaction under standard conditions if the concentration of all reactants (except for H+) is 1 mol dm–3. Assume the following cellular concentrations at pH = 7: pyruvate 380 µmol dm–3, NADH 50 µmol dm–3, lactate 3700 µmol dm–3, NAD+ 540 µmol dm–3.
Calculate ΔG’ at the concentrations of the muscle cell at 25 °C.
Model Answer
ΔG’ = ΔG 0 ’ + RT ln( c(prod.) / c(react.) )
= ΔG 0 ’ + RT ln( c(lactate) c(NAD+) / (c(pyruvate) c(NADH)) )
= – 25100 J mol–1 + 8.314 J mol–1 K–1 × 298.15 K × (3700 × 540) / (380 × 50)
= – 25.1 kJ mol–1 + 11.5 kJ mol–1
= – 13.6 kJ mol–1