Part A Consider the last two steps in the alcoholic fermentation of glucose by b

Question

Part A

Consider the last two steps in the alcoholic fermentation of glucose by brewer's yeast:

pyruvate+NADH+2H+ethanol+NAD++CO2

Calculate the nonequilibrium concentration of ethanol in yeast cells, if G=38.3kJ/mol for this reaction at pH=7.4 and 37C when the reactants and products are at the concentrations given above.

Express your answer with the appropriate units.

Part B

Consider the degradation of glucose to pyruvate by the glycolytic pathway:

glucose+2ADP+2Pi+2NAD+2pyruvate+2ATP+2H2O+2NADH+2H+

Calculate G for this reaction at pH=7.4 and 37C .

Express your answer with the appropriate units.

0.527 M

62.4 kJmol

0.527 M

Explanation / Answer

PART A

pyruvate + NADH + 2H+ ----> ethanol + NAD+ + CO2

DG = DG0 + 2.303RTlog[CO2][NAD+][ethanol/[H+^2][NADH][pyruvate]

[H+] = 10^(-pH) = 10^(-7.4) = 3.981*10^-8 M

(-38*1000) = (-64.4*1000)+(2.303*8.314*310log((0.02)(350*10^(-6))(ethanol)/((3.981*10^(-8))^2(15*10^(-6))(62*10^(-6)))

[Ethanol] = 5.9*10^(-15) M

           = 5.9*10^(-9) micro M

pART B

glucose+2ADP+2Pi+2NAD+ ----> 2pyruvate+2ATP+2H2O+2NADH + 2H+

DG0 RXN = 2*DG0 + 2*DG0'

         = (2*64.4)+(2*(32.2))

     = + 193.3 Kj

DG = DG0 + 2.303RTlog [H+^2][NADH^2][pyruvate^2][ATP^2]/[NAD+^2][pI^2l[ADP^2][Glucose]

   = (193.3*1000)+ (2.303*8.314*310log(((3.981*10^(-8))^2(15*10^(-6)^2)(62*10^(-6)^2)(3.1*10^(-3)^2)/((350*10^(-6)^2)(5.9*10^(-3)^2)(220*10^(-6)^2)(5.1*10^(-3)^2))

    = 123.82 kj

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