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EXAM BIO 1
FALL 2005
NAME:_________________________________________________
ID:_____________________________________________________
1. Write Your name and ID on all pages 2. Write all answers in pen only 3. Limit your answers to the space provided 4. Make sure your exam has seven pages including this cover page
Page 2 Page 3 Page 4 Page 5 Page 6 Page 7 Total
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Maximum of 100 points
NAME:_________________________
ID:_________________________
1. Ionizable side chains. (7 points) (a) List all acidic side chains from strongest to weakest acid. (b) List all basic side chains from strongest to weakest base. (a) Asp, Glu, Cys, Tyr (Asp/Glu, Cys, Tyr also correct) (b) Arg, Lys, His
2. From the list of reagents/techniques choose the one, which is best suited for the task listed in statements (a)-(o). (15 points) 1. Urea 2. NMR spectroscopy 3. MALDI 4. Salt fractionation 5. Dialysis 6. Western blot 7. Chymotrypsin 8. Dabsyl chloride 9. t-Boc 10. Ultracentrifugation 11. Carboxymethyl group 12. DTT 13. Diethylaminoethyl group 14. DCC 15. ELISA (a) Reduces disulfide bonds DTT (12) (b) Determines the three-dimensional structure of proteins in solution NMR (2) (c) Protects the amino group of an amino acid t-Boc (9) (d) Ionizes biological macromolecules MALDI (3) (e) Diagnostic test for HIV infection ELISA (15) (f) Labels the N-terminal residue of a protein Dabsyl chloride (8) (g) Cleaves after aromatic residues Chymotrypsin (7) (h) Binds anions Diethylaminoehtyl group (13) (i) Activates carboxyl group of an amino acid DCC (14) (j) Allows diffusion of small, but not large molecules Dialysis (5) (k) Identifies one protein in a complex mixture Western blo (6) (l) Separates molecules according to their size and shape Ultracentrifugation (10) (m) Unfolds proteins at high concentration Urea (1) (n) Separates proteins according to their solubility Salt fractionation (4) (o) Binds cations Carboxymethyl group (11) 2
NAME:_________________________
ID:_________________________
3. Assume that the average thickness of a biological membrane is 54 . For (a) and (b) show how you derived your answers. (a) An -helix is oriented with its helical axis perpendicular to the membrane. How many residues are required to span the membrane? (6 points) (b) Assume that the angle between the -helix and the membrane is 60. How many residues are now required to traverse the membrane? (Hint: sin (60) = cos (30) ~ 0.9) (4 points) (a) Rise per turn of an -helix is 5.4 , therefore 54 corresponds to 10 turns. There are 3.6 residues per turn and 36 residues per 10 turns. Answer: 36 residues (b) The helix now has to be longer since the helical axis is tilted 30 relative to (a). In the perpendicular triangle formed by the helical axis and the membrane normal (line perpendicular to membrane) the following holds: cos 30 = 54 / x or x = 54 / 0.9 = 60 This corresponds to 36 residues / 0.9 = 40 residues 4. An ion pair is formed between the side chains of an Asp (pKa=5) and a Lys (pKa=9) in a protein. For (a) and (c) show how you derived your answers. (a) Estimate the fraction of the side chains that form the ion pair at pH 5 and pH 7. (6 points) (b) Describe in less than two sentences what you would have expected for the microscopic pKa values and why? (4 points) (c) At pH 6 calculate the fraction where neither side chain is ionized. (5 Bonus Points) (a) pH 5: Asp will be a one to one mixture of ionized and neutral, whereas Lys will be (almost) completely ionized. The fraction of ionized side chains is therefore 0.5 (or 50%). pH 7: Asp and Lys both mostly ionized and fraction will be 1 (or 100%). (b) Expectation would have been that microscopic pKa value of Asp is below 4 (macroscopic value) and microscopic pKa values of Lys is above 10.5 (macroscopic value), since this would have stabilized the ion pair (favorable interaction) over a wider pH range. (c) Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) Asp: log([Asp-]/[H-Asp]) = 1 or ([Asp-]/[H-Asp]) = 10. Fraction of [H-Asp] is 1/11 = 0.0909 (10 Asp are negatively charged and one is uncharged) Lys: log([Lys]/[Lys-H+]) = -3 or ([Lys]/[Lys-H+]) = 0.001. Fraction of [Lys] is 1/1001 = 0.000999 (1000 Lys are positively charged and one is uncharged) Multiplying 0.0909 and 0.000999 yields the probability of both being 9.08 uncharged: x 10-5 (appr. 1/10,000) 3
NAME:_________________________
ID:_________________________
5. Amino acid analysis of a pentapeptide revealed equimolar amounts of Ala, Asp, Lys, Ser, and Trp. Reaction of the pentapeptide with FDNB yields compounds (1) and (2). Digestion with trypsin yields a dipeptide and a tripeptide. The dipeptide strongly absorbs UV light at 280 nm. Reaction of the dipeptide with FDNB yields compound (3). Draw the sequence of the peptide starting at the N terminus. Partial credit will be given if you describe what the reagents do. (16 points)
(1)
NO 2 CH3 O2N H N C H COO -
(2)
NO2 H H N H2 C
O2N
C 4 NH + 3
COO-
(3)
NO 2 H N
OH CH2
FDNB labels amino groups, in this case Nterminal Ala (compound 1) and side chain of internal Lys (Compound 2). Ala is the first residue. Trypsin cleaves after Lys/Arg. Lys is either 2nd or 3rd residue. Aromatic residues absorb at 280 nm, Trp is in the dipeptide FDNB identifies Ser as N-terminal residue of dipeptide (compound 3). Dipeptide sequence: Ser-Trp. Lys has to be third residue and Asp automatically becomes the second residue Sequence: (N) Ala-Asp-Lys-Ser-Trp (C)
O2N
C H
COO -
6. Explain in one sentence why the triphosphate moiety of ATP is often bound to proteins near the N-terminal end of an -helix. (7 points) This generates a favorable interaction between the helix dipole (positive end at Nterminus of the helix) and the negatively charged triphosphate moiety.
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NAME:_________________________
ID:_________________________
7. The diagrams below show a ribbon diagram of the protein carbonic anhydrase. (a) Using Figure 1 describe the quarternary structure and underlying symmetry of this enzyme. (3 points) (b) Using Figure 2 describe the tertiary structure in terms of its secondary structure elements. (4 points) (c) Make a sketch of the Ramachadran diagram and indicate where you expect to find the majority of the residues. (5 points) (d) Take a closer look at the structure, what is unusual about the -sheets? (5 Bonus Points) (a) The enzyme is a homotrimer with a threefold axis of symmetry perpendicular to the paper and passing through the center of the molecule. (b) The monomer contains three parallel -sheets and three -helices. (c)
(d) The -sheets are essentially flat, which means that they do not contain the usual left-handed twist.
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NAME:_________________________
ID:_________________________
8. State the steady state assumption in one sentence AND two equations. (7 points) The steady state assumption postulates that the concentration of the ES-complex is constant. In mathematical terms this means: [ES] = constant or d[ES]/dt = 0 or k1[E][S] = (k-1 + k2)[ES]
9. A ligand binds to its receptor with a dissociation constant of 1 x 10-10 M. (a) Calculate the degree of saturation in the presence of 4 x 10-10 M ligand. (4 points) (b) A substitution of a single amino acid in the receptor alters the dissociation constant in such a way that at the same ligand concentration the degree of saturation is 0.5. What is the dissociation constant now? Does the mutant bind the ligand tighter? (4 points) (a) = [L]/([L] + Kd) = 4 x 10-10 M/(4+1) x 10-10 M = 0.8 (b) Now is equal to 0.5, [L]= 4 x 10-10 M 0.5 = [L]/([L] + Kd) or 0.5 [L] + 0.5 Kd = [L] or [L] = Kd -10 Kd = 4 x 10 M We did not even have to do the calculation since by definition the Kd corresponds to the ligand concentration, which yields 50% saturation. No the mutant binds the ligand more weakly.
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NAME:_________________________
ID:_________________________
10. Histones are proteins present in the nuclei of eukaryotic cells where they tightly interact with DNA. Histones have a pI of around 10.8. Explain each answer in one sentence. (a) Which residues are required to form a tight complex with DNA and what type(s) of interaction(s) is(are) present? (5 points) (b) Do histones bind to anion or cation exchange chromatography columns at pH 7? (3 points) (a) Lys and Arg residues are required to interact with DNA, since they will form ion pairs between the positively charged side chains of Lys and Arg and the negatively charged phosphate groups in the DNA backbone. In addition there will also be hydrogen bonds between the nitrogen donors in Lys/Arg and the oxygen acceptors in the phosphate groups. (b) Histones will bind to cation exchange columns since they are overwhelmingly positively charged at pH 7.
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