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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

Respuesta :

alepertu

Answer: -2.4 kJ/mol

Explanation:

We can relate [tex]\Delta G[/tex] to [tex]\Delta G\°[/tex] according to:

[tex]\Delta G=\Delta G\°+RTln(\frac{[products]^n}{[reagents]^m})[/tex]

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:

[tex]\Delta G=1.7\frac{kJ}{mol} +(8.314*10^{-3}\frac{kJ}{K*mol})*(37+273K)*ln(\frac{\frac{0.1}{1000}M}{\frac{0.5}{1000}M})[/tex]

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

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