Exam1key-2 - CH368, Spring 2012 Exam 1 Key 1. D 2. B 3. D...

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1 CH368, Spring 2012 Exam 1 Key 1. D 2. B 3. D 4. C 5. B 6. B 7. C 8. B 9. E 10. A 11. A) It is necessary to graph these data (e.g., a Lineweaver-Burk plot) in order to fully assess the transporter and the impact of fructose on the transporter. The glucose transporter seems to be functioning as a carrier with a Km of about 0.25 mM. The Vmax under these experimental conditions is about 0.10 mM/min. B) Fructose is functioning as a competitive inhibitor. The apparent Km is about 0.50 mM but the Vmax is still about 0.10 mM/min. This is classic competitive inhibition. C) The glucose transporter does not depend upon ion gradients in any way, indicating it is a passive transporter and not a coupled transporter. 12. A) Δ G = RT ln [H + ] out + Z F Δψ Assume Δψ = -0.06 V (inside “-“), T= 310K. [H + ] in Δ G = 8.314 J/mol-K x 310K x ln {0.18 M/1 x 10 -7 M} + (+1) x 96,480 J/V-mol x (0.06 V) = 42.9 kJ/mol This value could have varied from about 41 kJ – 44 kJ if you used 298K or a different value for the membrane potential (between 0.05 V and 0.07 V is reasonable). B) Δ G for ATP hydrolysis under cellular conditions is -50 kJ/mol, so it would take 1 ATP molecule per H + secreted. If you used the standard free energy for ATP hydrolysis (-30.5 kJ/mol), then it would take to 2 molecules of ATP. Since the question says under cellular conditions, then this answer is not as good but would earn partial credit. 13. A) phosphocreatine + H 2 O creatine + Pi Δ G’ ° = -43.0 kJ/mol ADP + Pi ATP + H 2 O Δ G’ ° = +30.5 kJ/mol
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2 Net rxn: Phosphocreatine + ADP ATP + creatine Δ G’ ° = -12.5 kJ/mol Thus, favorable. B)
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Exam1key-2 - CH368, Spring 2012 Exam 1 Key 1. D 2. B 3. D...

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