Chem 4A - Spring 2006 - - Final

Chem 4A - Spring 2006 - - Final - Name_ GSI_ Chemistry 4B,...

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Name__________________________ GSI_________________ 1 Chemistry 4B, Spring 2006 Final May 15, 2006 (180 min., closed book) N a m e : G S I : S I D : S e c t i o n : Please read this first: Write your name and that of your TA on all 20 pages of the exam Test-taking Strategy In order to maximize your score on the exam: Do the questions you know how to do first. Go back and spend more time on the questions you find more challenging. Budget your time carefully -- don't spend too much time on one problem. Show all work for which you want credit and don't forget to include units. Question Page Points Score Q1 9 24 Q2 10 27 Q3 11 28 Q4 12 44 Q5 13 41 Q6 14 15 Q7 15-16 52 Q8 17-18 25 Q9 19 28 Q10 20 16 Total 300
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Name__________________________ GSI_________________ 2 E=h ν λν = c E kin = ½mv 2 E kin = 3RT/2 v rms = (3 / ) R TM k B = 1.381 × 10 -23 J/K m e- = 9.11 × 10 -31 kg h = 6.626 × 10 -34 J s c = 3.0 × 10 8 m/s 1 nm = 10 -9 m 1 kJ = 1000 J 1 Torr = 1 mmHg 1 atm = 760 mmHg J = (kg m 2 )/s 2 PV = nRT () nRT nb V V n a P = + 2 2 = 6 12 4 V(R) Energy, Potential R R σ ε T(K) = T(C) + 273.15 R = 0.0821 L·atm / (mol·K) R = 8.314 J / (mol K) A = ε lC P a = X a P tot P = F/A N A = 6.022 × 10 23 1 cal = 4.18 J 101.3 J = 1 L atm K a K b = K w K w = 10 -14 pX = -log X pK a + pK b = pK w pH = pKa + log - [] A HA k B = 1.381 × 10 -23 J/K m e- = 9.11 × 10 -31 kg h = 6.626 × 10 -34 J s c = 3.0 × 10 8 m/s 1 kJ = 1000 J J = (kg m 2 )/s 2 PV = nRT T(K) = T(C) + 273.15 R = 0.0821 L·atm / (mol·K) R = 8.314 J / (mol K) P a = X a P tot P = F/A E = q + w H ° = Σ n H ° f (products) - Σ n H ° f (reactants) H ° = Σ n H ° bonds broken - Σ n H ° bonds formed G ° = Σ n G ° f (products) - Σ n G ° f (reactants) S ° = Σ nS ° (products) - Σ nS ° (reactants) for a reversible phase change T H T q S rev = = G ° = H ° - T S ° G = G ° + RT ln Q G ° = -RT ln K q = mC T 101.3 J = 1 L atm anode cathode cell E E E o o o = G o =− nFE cell o Q nF RT E E ln = o K E RT nF ln = o F = 96,485C/mol e- N A = 6.022 × 10 23 1 cal = 4.18 J 1 amp = 1 C/sec 1 Watt = 1 J/sec
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Name__________________________ GSI_________________ 3
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Name__________________________ GSI_________________ 7 Protonated amino acids.
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Name__________________________ GSI_________________ 8 Nucleotides with physiological protonation.
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Name__________________________ GSI_________________ 9 1. Suppose that a histidine residue in a protein is buried in its hydrophobic core. (24 total) a) First, explain what the “hydrophobic effect” is and why protein folding depends on it. Include the intermolecular forces involved in your answer. (8 points) b) Would you expect the pKa of this histidine to be higher or lower than the case where it is exposed to solvent? Indicate the intermolecular forces involved and use the Henderson- Hasselbalch equation in your answer. (8 points) c) To a first approximation, would you expect the protein to be more or less stable to denaturation when compared to the exact same protein containing a phenylalanine instead of the histidine at the same position (see below)? Explain, based on the properties of the two amino acids.
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Chem 4A - Spring 2006 - - Final - Name_ GSI_ Chemistry 4B,...

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