The solubility of two amino acids in two solvents at 25 degree C are given below

Question

The solubility of two amino acids in two solvents at 25 degree C are given below; the numbers represent the concentrations present in saturated solutions. Calculate the standard free energy change delta mu degree for transfer of 1 mol of glycine from the solid to aqueous solution at 25 degree C. [Recall the standard state of solid is the pure solid and the standard state for solution is 1 M, assuming ideal behavior and using the molarity scale]. Calculate the standard free energy of transfer delta mu degree of 1 mol of glycine from ethanol to the aqueous solution at 25 degree C. Assuming that the effects of the amino acid backbone and side chain are additive (glycine essentially has no side chain) calculate the standard free energy of transfer of the (CH_3)_2CH-side chain of valine from water to ethanol at 25 degree C. Ethanol can be considered to mimic the interior of a protein. Will the mutation of a glycine to a valine in the interior of a protein favor the folding of the protein in water if interaction with the solvent is the dominant effect?

Explanation / Answer

a) Calculate the standard free energy of transfer, µ°

Given-    Water                Ethanol

Glycine 3.09 M   0.000404 M

             Valine   0.60 M      0.00132 M

G =G0 + RT ln Q

G0 is µ0 So equation becomes like this

µ = µ0 +RT ln Q

Here R is the ideal gas constant = 8.314 J/mol-K

T is the temperature in kelvin 25 +273 = 298K

Q is the reaction quotient at that moment in time.

When the reaction is at equilibrium rG = 0

µ0 =- 8.314 J/mol-K x 298K ln [glycine (aq)]/[glycine (s)]

   = -2477.57 ln 3.09/1

2477.57 x 1.128 =- 2.79 KJ/mol

b) ) Calculate the standard free energy of transfer, µ°, of 1 mol of glycine from ethanol to the aqueous solution at 25°c

µ = µ0 +RT ln Q

0 = µ0 =- 8.314 J/mol-K x 298K ln [glycine (aq)]/[glycine (ethanol)]

µ0 =- 8.314 J/mol-K x 298K ln(3.09/ 4.04x10^-4)

       = -22.2 kJ/ mole

c) Assume that the effects of the backbone and side chain are simply additive and calculate the standard free energy of transfer of 1 mol of the (CH3)CH– side chain of valine from water to ethanol at 25°C.

First calculate the energy of transfer of valine from water to ethanol

µ°= -RT ln [Val (ethanol)]/[Val(aq)]

µ0 =- 8.314 J/mol-K x 298K ln(1.32 x 10^-3/ 0.60)

= +15.2 kJ mol-1

The energy of transfer from water to ethanol is -(-22.2 kJ mol-1) = +22.2 kJ mol-1

So µ° for transfer of the Val side chain (only) from water to ethanol is:

15.2 - 22.2 = –7.0 kJ mol-1.

d) The answer to part (c) tells that the transfer of the hydrophobic Val side chain from water to a less polar environment is in fact favorable. Therefore mutation of Gly toVal, where the side chain is in the interior of the protein, should favor folding of the protein.

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