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Biochem Exam: Ionizable Groups in Peptides & Proteins, Binding Entropy Change, Exams of Biochemistry

The answers to the first hour examination for biochemistry 462a, focusing on the ionizable groups in peptides and proteins, approximate pka values, and the calculation of isoelectric points. Additionally, it covers the entropy change during the binding of a target peptide to cam-ca2+ and the contribution of conformational entropy, molecularity, and the hydrophobic effect.

Typology: Exams

Pre 2010

Uploaded on 08/31/2009

koofers-user-8id
koofers-user-8id 🇺🇸

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Download Biochem Exam: Ionizable Groups in Peptides & Proteins, Binding Entropy Change and more Exams Biochemistry in PDF only on Docsity! WRITE YOUR NAME LEGIBLY ON EVERY PAGE. PAGES WILL BE SEPARATED FOR GRADING! CHECK TO BE SURE YOU HAVE 7 PAGES, NAME (print): INCLUDING COVER PAGE. Section & T.A. I swear/affirm that I have neither given nor received any assistance with this exam. Signature: Date: BIOCHEMISTRY 462a FIRST HOUR EXAMINATION Form A -- ANSWERS September 17, 2007 A SCIENTIFIC CALCULATOR IS NEEDED FOR THIS EXAM. SHOW YOUR WORK FOR ALL CALCULATIONS, AND BE SURE TO STATE UNITS OF ANY NUMERICAL ANSWERS. If the reasoning, calculations, or answer are shown anywhere other than in the space provided, make a note in the space provided and put answer on back of SAME PAGE for same grader, since pages are separated for grading. USEFUL CONSTANTS: R (gas constant) = 8.315 J•mol–1•Kelvin–1 = 8.315 x 10–3 kJ•mol–1•Kelvin–1 (25 °C) T = 298 K; human physiological temperature (37 °C) T = 310 K. Ionizable group in peptides and proteins Approximate ("generic") pKa in peptides & proteins α-amino 8.0 α-carboxyl 3.0 ε-amino 10.0 guanidino 12.0 thiol 8.5 imidazole 6.5 aromatic hydroxyl 10.0 side chain carboxyl 4.0 p. 2 (31 points) p. 3 (21 points) p. 4 (14 points) p. 5 (14 points) p. 6 (10 points) p. 7 (10 points) TOTAL: (100 points) Biochemistry 462a, Exam #1 Form A ANSWERS NAME (every page!!) September 17, 2007 page 2 1. (15 pts) Consider the peptide whose sequence is A K L Y E N F M. A. (10 pts) At pH 5, what would be the approximate net charge on the peptide? Use the “generic” pKa values on the cover sheet of the exam; show your reasoning in Part A of table below; start by listing all ionizable groups and their approximate pKa values. See lefthand column below. PART A PART B ionizable group pKa approx. charge at pH 5? approx. charge at pH 1 α-carboxyl ~3 –1 0 R-carboxyl (E) ~4 –1 0 <–– pI α-amino ~8 +1 +1 aromatic OH (Y) ~10 0 0 ε-amino (K) ~10 +1 +1 approx. net charge: 0 (pH 5) +2 (pH 1) B. (5 pts) Calculate the isoelectric point (pI) for this peptide, using the “generic” pKa values on the cover sheet of the exam. Start by filling out the Part B column above, listing the charges of all ionizable groups at pH 1. Net charge at pH 1 is +2 , so 2 equivalents of OH– would be needed to get from pH 1 to net charge of zero (pI), so pI is halfway between pKa of side chain carboxyl group of Glu and α- amino group. pI = (4 + 8) / 2 = 6 2. (16 pts) The flat projection structure of a 3-amino acid residue segment of a protein is drawn below. A. (3 pts) What is the sequence shown (1-letter or 3-letter codes)? Lys (K) – Ser (S) – Asn (N) B. (3 pts) Circle 6 atoms that are coplanar (all 6 atoms in the same plane). Circle only ONE group of 6 atoms; you have several correct choices. Do NOT include any side chain atoms. C. (3 pts) With a labeled arrow ↑ , indicate one peptide bond (label only one). D. (2 pts) With 2 labeled arrows ↑ , indicate bonds whose rotations change the φ (phi) and ψ (psi) angles. Label only 1 bond as φ (phi) and 1 bond as ψ (psi). (There are several correct choices.) D. (5 pts) Use dashed lines on the drawing to indicate all the appropriate hydrogen bonds that would be involved in stabilizing a 3-stranded β-sheet, for which this structure is the middle strand; i.e., show hydrogen bonds from this strand to adjacent (imaginary) strands on either side. /31 Biochemistry 462a, Exam #1 Form A ANSWERS NAME (every page!!) September 17, 2007 page 5 6. (8 pts) The overall reaction catalyzed by the enzyme glutamine synthetase is as follows: glutamate + NH3 + ATP glutamine + ADP + Pi (Pi = inorganic phosphate) There are two possible pathways for addition of the amino group to Glu to make Gln, as shown on the scheme below, which has boxes drawn around the reactants and the products. A. (4 pts) What is the overall standard free energy change for formation of glutamine from glutamate? Show your reasoning. ΔGo'overall = ΔGo'1 + ΔGo'2 = 14.2 kJ/mol + –30.5 kJ/mol = –16.3 kJ/mol B. (4 pts) What is the standard free energy change for the reaction of glutamoyl phosphate + NH3 to form glutamine + Pi (ΔG°’4)? Show your reasoning. ΔGo'overall = is independent of pathway, so ΔGo'overall = ΔGo'3 + ΔGo'4 ΔGo'4 = ΔGo'overall – ΔGo'3 = –16.3 kJ/mol – 15.5 kJ/mol = –31.8 kJ/mol 7. (6 pts) List the kinds of forces or bonds that stabilize the quaternary structure of proteins. Hydrophobic interactions resulting from burial of hydrophobic groups, e.g., by packing of secondary structural elements against each other, burying side chains between them Hydrogen bonding (between side chains but also sometimes between backbones, e.g., where there’s a beta sheet that continues across the interface between subunits in a protein) Ionic interactions between buried charged groups in the interface van der Waals interactions (packing of side chains as well as secondary structural elements) (Some people would argue that disulfide bonds could be included here, too.) /14 Biochemistry 462a, Exam #1 Form A ANSWERS NAME (every page!!) September 17, 2007 page 6 8. (10 pts) Examine the three Ramachandran diagrams, each of which corresponds to one of the three proteins shown below. Each dot on the Ramachandran diagram represents the (φ , ψ) coordinates of one residue in that protein. A. (3 pts) Name the predominant (major) secondary structure(s) apparent in each protein below. protein A ____α helix and β sheet (mixed)_________ protein B ________ β sheet _____________________ protein C ________ α helix _____________________ B. (3 pts) In the Ramachandran diagrams (1, 2, and 3 below) for the three proteins, what would you surmise about the majority of the residues whose φ, ψ angles are outside the shaded areas? Residues in “higher energy” parts of the plot (disfavored or “forbidden” areas of φ , ψ space) are likely to be GLYCINE residues, which have much more conformational flexibility. C. (4 pts) Which one of the 3 Ramachandran diagrams corresponds to which one of each of the three proteins shown below? Diagram 1 corresponds to Protein ___C_____ Diagram 2 corresponds to Protein ___A_____ Diagram 3 corresponds to Protein ___B_____ 1) (A) 2) (B) 3) (C) /10 Biochemistry 462a, Exam #1 Form A ANSWERS NAME (every page!!) September 17, 2007 page 7 9. (10 pts) The activity of a certain enzyme requires the participation of a specific glutamate residue. The activity as measured in the laboratory varies with the pH of the solution as shown below at right. A. (3 pts) What is the apparent pKa of the Glu functional group whose state of ionization is affecting activity? How did you choose that value of the pKa? pKa of 5.5 (the pH at which the enzyme is 50% active, so the functional group must be 50% in the state of ionization necessary for activity; arrow on plot indicating pKa on the pH axis, as the pH at which % activity is 50%, would be adequate explanation.) B. (3 pts) Is the active form of the Glu functional group the acid form or the base form? Briefly explain why you gave that answer. The active form of the side chain carboxyl group must be the conjugate acid, because the enzyme LOSES activity as the pH increases, i.e., as the group loses its proton. C. (4 pts) The pKa for the Glu functional group in this enzyme is different from the generic pKa for such a group. What type of structural or environmental factor might account for the unusual pKa for this functional group? •The pKa of 5.5 is higher than a generic side chain carboxyl group pKa, so some environmental factor is making the carboxyl group bind its proton more tightly than a generic carboxyl, i.e., the group is remaining neutral into a higher pH range. •Such an effect could be due to the carboxyl group being in a hydrophobic environment, e.g., the interior of the protein (in the absence of any + charged counter ion). /10
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