Question

Determine ΔG° of deprotonation for Cys 178 and for the standard Cys (pKa = 8.2). Then use these to determine, ΔΔG°, which is the change in the standard Gibbs energy of deprotonation due to the environment in the enzyme, relative to water.

Answer 1

Step(I)

Using the formula

P^Ka  = -log Ka

Where Ka is dissociation constants of Cys 178

P^Ka = 8.2

Putting the all value in the formula

8.2 = -logKa

Ka = 10^-8.2 = 6.3097×10^-9

^step(ll)

^{Formula of ∆G}

^{∆G° = -RTlnKa = 2.303RTlogKa}

^{where ∆G° = Gibbs free energy}

^{R = gas constant = 8.314 J/(mol-K)}

^{T = Temperature = environment temperature = (25+273)K = 298K}

∆G° = -(2.303)(8.314J/mol-K)(298K)log(6.3097×10^-9)

∆G°= -{5705.84×(-8.2)}

∆G° = 46787.96J/mol

∆G°= 46.78796×10³J/mol

∆G°= 46.788kJ/mol