intro. metabolism problem set key

intro. metabolism problem set key - 2 O -> creatine + Pi...

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BIOC 460 Introduction to Metabolism Problem Set Reading: Chapter 15 End of Chapter Problems (pp. 431-432): 1,2,3,5,8,10, 11, 12, and 15 1. Which yields more free energy per carbon atom during oxidation in a eukaryotic cell, a polymer of three glucose molecules or a saturated fatty acid with 18 carbons? Why? The saturated fatty acid. Glucose molecules are partially oxidized. Saturated fatty acids have more electrons per carbon available. 2. Phosphocreatine is used as a phosphoryl donor for ATP synthesis in muscle (see page 416 of your text). Given that in resting muscle [ATP] = 4 mM, [ADP] = 0.013 mM [creatine phosphate] = 25 mM and [creatine] = 13 mM: Calculate Δ G’ for the creatine kinase-catalyzed reaction creatine phosphate + ADP <-> ATP + creatine in resting muscle at 25 °C. From page 416 ADP + Pi + H+ -> ATP + H 2 O Δ Gº’ = +30.5 kJ/mol Creatine phosphate + H
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Unformatted text preview: 2 O -> creatine + Pi + H+ Δ Gº’ = -43.1 kJ/mol ________________________________________________________________ for the coupled reaction creatine phosphate + ADP -> creatine + ATP Δ Gº’ = -12.6 kJ/mol Δ G’ = Δ Gº’+ RT • ln [B] actual / [A] actual Δ G’= (-12.6 kJ/mol) + R(298 K)ln{([ATP] * [creatine])/([ADP] * [creatine phosphate] Δ G’ = (-12.6 kJ/mol) + (.0083)(298 K)ln{(4 mM)(13 mM)/(0.03 mM)(25 mM)} Δ G’ = -0.1 kJ/mol 3. Explain why low energy charge favors catabolism and high energy charge favors anabolism. When [ATP] is low then then energy charge is low. Under these conditions catabolism is favored because catabolic pathways generate ATP. When [ATP] is high then the energy charge is high. These condition favor anabolism using the ATP to synthesize glycogen and triacylglycerides for long-term energy storage....
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This note was uploaded on 05/06/2010 for the course BIOC 460 taught by Professor Ziegler during the Spring '07 term at Arizona.

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