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Question

Staphylococcal nuclease accelerates phosphodiester bond hydrolysis by a factor of 5.6x1014 over the spontaneous rate (1.7x10-13). How much does staphylococcal nuclease decrease the activation energy of its reaction at 25?C?

Answer 1

The relationship between the rate enhancement and the change in activation energy is:

kuncat​kcat​​=eRTΔΔG‡​

Rearranging to solve for the decrease in activation energy (ΔΔG‡):

ΔΔG‡=RTln(kuncat​kcat​​)

Where:

1. R (Ideal Gas Constant) = 8.314 J/mol⋅K (or 1.987×10−3 kcal/mol⋅K)
2. T (Temperature) = 25∘C=298.15 K
3. k uncat​/k cat​​ = 5.6×1014

Using the natural log (ln):

ΔΔG‡=(8.314 J/mol⋅K)×(298.15 K)×ln(5.6×1014)

ΔΔG‡=2478.8 J/mol×33.958

ΔΔG‡≈84,178 J/mol

Staphylococcal nuclease decreases the activation energy of the reaction by:

1. 84.18 kJ/mol

or

1. 20.12 kcal/mol (using 1 kcal=4.184 kJ)