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Buffers and PH, Peptides, Thermodynamics, Protein Isolation | BCHM 4115, Exams of Biochemistry

Material Type: Exam; Professor: Helm; Class: General Biochemistry; Subject: Biochemistry; University: Virginia Polytechnic Institute And State University; Term: Fall 2012;

Typology: Exams

2011/2012

Uploaded on 10/04/2012

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Download Buffers and PH, Peptides, Thermodynamics, Protein Isolation | BCHM 4115 and more Exams Biochemistry in PDF only on Docsity! BCHM  4115—Fall  2012   Exam  One-­‐Answers     Page 1 of 2   BCHM 4115 Exam 1—Answers 1: BUFFERS AND PH. Exam Average: 12.2 out of 20. For Item (a) Parts (i) and (ii) were all or nothing (right or wrong, no partial credit). There were no tricks or hidden items in this problem. (i) Amount (in grams) of NaH2PO4-2H2O required: 15.6 gms Amount (in grams) of NaHPO4 required: 14.2 gms (ii) Ratio of HPO4-2 to H2PO4-1 in the final buffer: 1.995 (iii) Volume of 0.1M NaH2PO4 required: 334 mL Volume of 0.1M NaHPO4 required: 666 mL (iv) 0.83 mL (b) This was similar to several Problem Set Problems. pH = 3.27 2: PEPTIDES AND PROTEIN STRUCTURE Exam Average: 11.1 out of 20. (a) EYMRDGV (b) You were required to mention the fact that the interior of globular proteins are generally hydrophobic so therefore if an a-helix is embedded within this region, the side chains should be hydrophobic as well. Some leeway was given if one wanted to invoke other interactions, but the argument had to be logical. This is what was meant by “justify your selected features.” (c) The ribonuclease experiment demonstrated that the information necessary to properly fold a protein are encoded by the protein. This implies that the protein has limited freedom and cannot adopt all possible spatial orientations. When in the presence of denaturant, the protein cannot form proper intermolecular interactions, with or without the presence of reductant. The denaturant is actually allowing the protein to adopt “unnatural” conformations, and thus when the reductant was removed, incorrect disulfide bonds formed, leading to an inactive protein. (d) You can do that yourself. 3: THERMODYNAMICS AND MORE!   Exam average: 8.5 out of 15. (a) The math was all or nothing. ΔG0’ = -31.7 kJ/mole and Keq = 3.56 x 105. Reaction equilibrium lies very far to the right. (b) +42.3 kJ/mole (c) (i) Points were removed if statements made could also apply to ATP, the product of the reaction. The main difference between 1,3-bisphosphoglycerate and ATP is the bond cleaved is a mixed anhydride and the bond formed is a phosphoanhydride. The functional groups needed to be mentioned. Steric hindrance (NOT hinderance) and resonance stabilization apply to ATP as well. (ii) Needed to see math. ΔG0’ = -18.9 kJ/mole and ΔG = 0.1 kJ/mole. Can let [P]/[R] = x, find x, and then let x=([3PG]10])/[1,3BPG][1]). Ratio was 157:1. 4: PROTEIN ISOLATION AND CHARACTERIZATION Exam average: 11 out of 15. (a) Answers by colum: Protein (mg): 1800, 570, 2.3. Total Activity: 1382, 1100, 950. Specific activity: 0.77, 1.93, 413. Percent Recovery: 100, 890, 69. n-Fold: 1, 2.5, 535. (b) Two ways to do it. Subtract absorbance values first or do each separately. Then had to convert concentration to ”moles. 0.062 ”moles. (c) Calling it a tetramer got you halfway. The salt study was not about pH. Read the question (“a study of the effect of salt). A dimer of dimers is what you have with different forces holding them together. Can’t get too specific about interactions except that the subunits are NOT held together by disulfide bonds as this bond would not be broken by salt. 5. HEMOGLOBIN’S MOVING AROUND Exam average: 8.3 out of 15. (a) Hb Providence curve needed to be above HbA as Hb Providence is less influenced by 2,3-BPG. Should have two curves, sigmoidal in shape, all on the same graph with K0.5 indicated as a dropdown line to the x-axis—it is the pO2 at half fractional saturation. The graph says NOTHING about rate (a source of aggravation to me when stated). An aside: that Hb Providence acts like fetal Hb, it makes me wonder if there are issues with pregnant mothers having this mutation. (b) The Bohr Effect basically states that as H+ increases there will be a concomitant increase in deoxyHb, because the oxy form is a stronger acid. Use of the equilibrium reaction Hb-O2 + H+ = deoxyHb-H+ + O2 to describe the Bohr Effect was
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