#1 MCB 102 -Practice Problems_key

#1 MCB 102 -Practice Problems_key - Mcb 102 Discussion...

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Unformatted text preview: Mcb 102 Discussion Sections 101 & 102 Practice Problems: Week I (Answers will be posted next week) Note: Please use the textbook as your guide for the amino acid structures & codes. 1. Draw L and D Tryptophan Please see your text. 2. What is the difference between a pK and a pI? Please see your text. 3. In your own words, how can a peptide composed of D amino acids act as a drug? How are (were) they designed? Lecture Notes 4. Why is the inside of a cell described as jello or gel-like? Densely packed cell environment resembles a gel instead of a dilute solution 5. Why are peptide bonds rigid (cannot rotate freely)? Partial double bond character restricts rotation 6. Relationships between ΔG0’, E0’ and Keq Consider the following the reaction: A Now fill in the gaps: Direction ΔE0’ Left right Right left Formulas: +B C + ΔG0’ >1 D Exergonic/Endergonic Keq negative positive positive negative exergonic endergonic 0<Keq<1 ΔG0’ = -2.3 RT log Keq ΔG0’ = -n F ΔE0’ 7. Aldolase : ΔG0’, ΔG and Keq Consider the aldolase reaction of glycolysis: 2 -O 3PO O H OH OH OPO32CH2 2 -O 3PO O OH CH2 + O HC OH OPO32CH2 H2C OH H H H2C ΔG 0'= + 23.8kJ/mol Dihdroxyacetonephosphate DHAP Glyceraldehydephosphate GAP Fructose-1,6-bisphosphate (FPB) a. Towards which direction does the reaction proceed under standard conditions? Is this with the flow of glycolysis or against it? Under standard conditions, the reaction will go from right to left, because ΔG0’ is positive. This is against the flow of glycolysis. b. Calculate the equilibrium constant Keq for the reaction. ΔG = −2.3RT log K eq 0/ K eq = 10 − ΔG 0 2.3 RT / − 23.8 kJ mol 2.3⋅ 8.314 ⋅10 −3 kJ ( mol ⋅ K )⋅ 298 K = 10 = 6.6 ⋅ 10−5 M c. At body temperature (37ºC, 310 K), what is the actual free-energy change (ΔG) for the reaction? € € Remember: ΔG = ΔG0’ + 2.3 RT log [product]/[reactant] / [ DHAP ][GAP ] = ΔG = ΔG 0 + 2.3RT log [ FBP ] 23.8 kJ mol + 8.314 ⋅ 10−3 kJ ( mol ⋅ K ) ⋅ 310K ⋅ log 3.4 ⋅ 10−6 = −8.6 kJ mol Note, that the reaction will proceed forward inside the cell because ΔG < 0. 8. Coupled Reactions € Consider the following table of standard free energies for the hydrolysis of some phosphorylated compounds: Reactant phosphoenolpyruvate creatine phosphate ATP glucose 6-phosphate glycerol 3-phosphate Product pyruvate creatine ADP glucose glycerol ΔG0’(kJ/mol) -62 -43 -30 -14 -9 ΔG0’(kJ/mol) +13, no -21, yes +32, no -16, yes Now fill out the chart for each of the reactions below: Spontaneous? (a) ATP + creatine (b) ATP + glycerol (c) ATP + pyruvate (d) ATP + glucose creatine phosphate + ADP glycerol 3-phosphate + ADP phosphoenolpyruvate + ADP glucose 6-phosphate + ADP ...
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