Question & Answer: The Delta G^0 for formation of a phosphodiester bond (shown below) is +25 kJ/mol. (Base)_n…..

22. (1 pts) The ΔGo for formation of a phosphodiester bond (shown below) is +25 kJ/mol. (Base), + Base→ (Base),+1+H20 ΔG0 +25 kJ/mol n+l Figure 4.6 Formation of a polynucleotide by a hypothetical dehydration reaction a) What is the equilibrium constant for this reaction at 37 °C? HO CH Ho CHI b) How might this reaction be made favorable? (Show the math for full credit)

The Delta G^0 for formation of a phosphodiester bond (shown below) is +25 kJ/mol. (Base)_n + Base rightarrow (Base)_n + 1 + H_2O Delta G^0 = +25 kJ/mol a) What is the equilibrium constant for this reaction at 37 degree C? b) How might this reaction be made favorable? (Show the math for full credit)

Expert Answer

Given

$Delta$ G of the reaction = +25KJ/mole = +25 * 10³ J/mole

T = 37⁰C + 273k = 310 K

we know

$Delta$ G = -RTlnK_eq

+25 * 10³ J/mole = – 8.314 J/K.mole * 310K * lnK_eq

on solving equilibrium constant K_eq = 6.2 * 10^-5

here equilibrium constant is less than 1. So it is reactant favourable. i.e, backward reaction occurs mostly.

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