Glu GlA 17. Consider the following reaction: Glutamate+NH4G c+ H20 a. The Req fo

Question

Glu GlA 17. Consider the following reaction: Glutamate+NH4G c+ H20 a. The Req for this reaction as written, from left to right is 3.13 x 10 ,a clearly thermodynamically unfavorable reaction. Calculate dyo for this reaction at 25°C (R-8.3 15 J/mol.K. T = 298k-25°C) Show your work and any--aG equations used. 19,29/noltvea.Ll : murA SrbstDde tant Calculate the minimum concentration of NH4' that would be required to produce a negative G for the reaction described above. Assume the concentration of Glu and Gln are equal. Show your work and any equations used. c. Cells need to make glutamine from glutamate. This is accomplished by coupling the reaction shown above to ATP hydrolysis. Calculate Go for the coupled reaction at 25°C. R = 8.315 J/mol-K. Show your work and any equations used. Hint: the aro for ATP hydrolysis is-30.5 kJ/mol./me-30s063/n_p

Explanation / Answer

Ans. #a. Using the equation dG0’ = - RT lnKeq              - equation 1

Where, dG0’ = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = equilibrium constant under standard condition

R = (0.001987 kcal mol-1K-1 or 0.008314 kJ mol-1 K-1)

Putting the values in equation 1-

            dG0’ = - (0.008314 kJ mol-1K-) x 298.0 K x ln (3.13 x 10-3)

            Or, dG0’ = - 2.477572 kJ mol-1 x 2.303 log (3.13 x 10-3)

            Or, dG0’ = - 2.477572 kJ mol-1 x (- 5.7678)

            Hence, dG0’ = +14.29 kJ mol-1

#b. At dG0’ = 0, the reaction is at equilibrium. dG0’ = 0 when Keq = 1 because ln 1 or (2.303 log 1) is zero.

So, first calculate the [NH4+] that gives Keq = 1.

Given, [Glu] = [Gln]

Now,

            Keq = [Gln] / ([Glu] [NH4+])

            Or, 1 = [Gln] / ([Glu] [NH4+])

            Or, [NH4+] = 1

Therefore, at [NH4+] = 1, the Keq = 1.

# To make the dG0’ value negative, the Keq value must be greater than 1.00 because the product of a negative value (- RT) with a positive value (when Keq in ln Keq is grater than) gives a negative value. (note than lnKeq, where Keq is less than .00 gives a negative value).

To satisfy a positive lnKeq value, the value of Keq must be greater than 1.00.

Therefore, minimum [NH4+] = Just greater than 1.00 M

#c. Glutamate + NH4+ -----------> Glutamine + H2O    ; dG0’ = +14.29 kJ/mol

Since the reaction has positive dG0’ value, it is non-spontaneous.

# The dG0’ value of ATP hydrolysis is -30.5 kJ/mol. Since dG0’ of ATP hydrolysis is negative and numerically greater than that of dG0’ of glutamine synthesis, ATP hydrolysis can drive the reaction. The two reactions can be coupled as follow-

            Glutamate + NH4+ -----> Glutamine + H2O         ; dG0’ = +14.29 kJ/mol

            ATP + H2O -------------> ADP + Pi                       ; dG0 = -30.5 kJ/ mol

            ----------------------------------------------------------------------------

Glutamate + NH4+ + ATP ----> Glutamine + ADP + H2O          ; dG0’net = -16.21 kJ/mol

Therefore, dG0’ of the coupled reaction = -16.21 kJ mol-1

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