Question

2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo’) for this reaction is +1.7 kJ/mol. If the cellular concentrations are 2PG = 0.5 mM and PEP = 0.1 mM, what is the free energy change at 37 oC for the reaction 2PG ↔ PEP

5.8 kJ/mol
-5.8 kJ/mol
+2.4 kJ/mol
-2.4 kJ/mol
-4146.4 kJ/mol
+4146.4 kJ/mol

Answer 1

Answer: -2.4 kJ/mol

Step-by-step explanation:

We can relate
`\Delta G` to
`\Delta G\°` according to:

`\Delta G=\Delta G\°+RTln(([products]^n)/([reagents]^m))`

Where [products] is the concentration of products, [reagents] is the concentrarion of reagents, n and m are the corresponding stoichiometric coefficients, T is the temperature in Kelvin and R is the gas constant.

Converting 37°C to Kelvin by adding 273, mM to M dividing by 1000, and substituting we get:

`\Delta G=1.7(kJ)/(mol) +(8.314*10^(-3)(kJ)/(K*mol))*(37+273K)*ln(((0.1)/(1000)M)/((0.5)/(1000)M))`

That gives us a value of -2.4 kJ/mol, the fouth option.