homework 2 KEY

# homework 2 KEY - BIBC 102 Fall 2007 Homework 2 KEY 1) In...

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BIBC 102 KEY Fall 2007 Homework 2 1) In glycolysis, glucose-6-phosphate is converted to fructose-6-phosphate. If the reactants and products have the following concentrations at equilibrium , calculate the standard free energy change ( G’°) for the reaction at 25°C. glucose-6-phosphate fructose-6-phosphate (9.6x10 -2 M) (4.8x10 -2 M) Would you expect this reaction to be reversible under cellular substrate and product concentrations? Is it? First calculate K’ eq ; K’ eq = [product] eq /[reactant] eq = .048 M/.096 M = 0.5 G’° can be calculated from K’ eq using the formula G’° = -RT ln K’ eq G’° = -(8.315 J/mol•K)(298 K)( ln 0.5) = -(2477.9 J/mol)(-0.693) = 1717.2 J/mol = 1.72 kJ/mol This is a small free energy change, which makes sense as this reaction is freely reversible, and goes both ways depending on the cytoplasmic concentrations of glucose-6-phosphate and fructose-6-phosphate. 2) From the textbook, chapter 13, answer question 9 on page 519. Answer parts a-c, and look at and think about d and e. a. G’° = -RT ln K’ eq therefore ln K’ eq = G’°/-RT = 13.8 kJ/mol = 13.8 kJ/mol -8.315 J/mol•K(298 K) -2.48 kJ/mol = -5.56 K’ eq = .00385 The equilibrium concentration of glucose 6-phosphate can be calculated by [glucose 6-P] eq = K’ eq therefore, .00385 = [glucose 6-P]/(.0048 M)(.0048 M) [glucose] eq [P i ] eq [glucose 6-P] eq = (.00385)(.0048) 2 = 8.8x10 -8 M

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This reaction does not represent a reasonable metabolic step, because this equilibrium concentration of glucose 6-P is far below the actual physiologic concentration. In other words, you would not be able to make a reasonable amount of glucose 6-P this way. b.
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## This note was uploaded on 12/30/2010 for the course BIOLOGY BIBC 102 taught by Professor Scheffler during the Spring '10 term at CSU Northridge.

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homework 2 KEY - BIBC 102 Fall 2007 Homework 2 KEY 1) In...

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