361
EXAM BIO 1
FALL 2004
NAME:_________________________________________________
ID:_____________________________________________________
1. Write Your name and ID on all pages 2. Write all answers in pen only 3. Try to limit your answers to the space provided 4. Make sure your exam has six pages including this cover page
Page 2 Page 3 Page 4 Page 5 Page 6 Total
_______________ _______________ _______________ _______________ _______________ _______________
Maximum of 110 points
NAME:_________________________
ID:_________________________
1. Shown below is a modified tripeptide, which resembles the chromophore of red fluorescent protein. The backbone looks unusual because of a cyclization reaction, which is accompanied by a dehydration reaction. (a) Identify (by circling all available atoms) and name the three residues that make up the chromophore. (b) What is the overall charge of this peptide at pH 2 and pH 12? (c) Calculate the pI of this tripeptide. For (c) show how you derived your answer. (12 points)
O
(b) pH 2: +1 and pH 12: -2
Tyr
N HO
N
Gly
COO-
(c) pI = (3+8)/2 = 5.5 The pI is at the midpoint between the pKa of the -carboxy group and the pKa of the -amino group.
NH3+
Gln
H2N
O
2. The sequence Leu-Ile-Thr-Thr-Ala-His-Ala has been found to adopt an -helix in one protein and a -strand in another. (a) Approximately how far apart will be the first and last residue in both conformations? Show how you derived your answer. (b) Label each residue either with or according to its secondary structure preference. (10 points) (a) -helix: rise per residue is 1.5 ; for six residues (from 1 to 7): 6 x 1.5 = 9 -strand: translation per residue is 3.3 ; for six residues: 6 x 3.3 ~ 20 (b) Leu-Ile-Thr-Thr-Ala-His-Ala
2
NAME:_________________________
ID:_________________________
3. The following data were recorded for an enzyme obeying Michaelis-Menten kinetics:
[S] in M 6.25 x 10-6 7.5 x 10-5 1 x 10-4 1 x 10-3 1 x 10-2 V0 in nmoles x liter-1 x min -1 15.0 56.25 60 74.9 74.99
(a) Estimate Vmax (b) Calculate KM (c) (c) What would V0 be at [S] = 1 x 10-4 M if the enzyme concentration were doubled? Show how you derived your answers. (12 points) (a) Vmax will be approached asymptotically at high substrate concentrations, a reasonable estimate is therefore 75 nmoles x liter-1 x min-1. (b) Using the value for Vmax and any pair of values from the table KM can be calculated: V0 = Vmax [S]/([S]+KM] solving for KM: KM = [S] (Vmax-V0)/V0 KM = 1 x 10-4 M (75-60) nmoles x liter-1 x min-1/60 nmoles x liter-1 x min-1= 0.25 x 10-4 M (c) Since Vmax = k2 [E]t doubling the enzyme concentration will double Vmax and this in turn will double V0, which will become equal to 120 nmoles x liter-1 x min-1.
4. You want to study how the composition of the proteome in cultured human skin cells changes in response to an increase in UV radiation. (a) Briefly describe an experiment and its basic principles, which will identify changes in protein expressions levels. (b) Name two techniques, which will allow the identity of these proteins to be determined. (c) After identification of one such protein you discover that an antibody against this protein is commercially available. How can this antibody be used to purify the protein from skin cells? (14 points) (a) 2D gel electrophoresis before and after treatment with UV. In this experiments proteins are first separated by their isoelectric point and then by their size. A comparison of the gels before and after UV treatment will reveal whether the expression levels of a particular protein increased/decreased/did not change. (b) Mass spectrometry and N-terminal sequencing. (c) This antibody can be used to make an affinity column, which will selectively bind only the protein of interest and none of the other proteins.
3
NAME:_________________________
ID:_________________________
5. Based on the known three-dimensional structure of a naturally occurring protein each of the following three variants has been generated by solid phase peptide synthesis: (a) Synthesis with D-amino acids. (b) Circular permutation: The existing N-and C-termini, which are close in the three-dimensional structure, have been linked and new termini have been generated in an existing loop structure. (c) Reversal of the chain direction: The C-terminus became the Nterminus, the second to last residue became the second residue and so on. Predict the effects on the three-dimensional structure in each case? Briefly explain your answer. (12 points) (a) The resulting protein will be the mirror image of the naturally occurring protein. Explanation: The mirror symmetry of the amino acids will translate into mirror symmetrical secondary structure elements and this will translate into a mirror image of the tertiary structure. (b) The structure will be mostly identical to the naturally occurring protein, except for small changes around the old and new termini. Explanation: The primary sequence is the same and has only been shuffled so the information for folding is the same. (c) This is no longer the same protein and certainly almost will not fold; in case of a miracle it will adopt a new fold. Explanation: The primary sequence is altered and this translates into a different fold. Assume you have a protein composed of a single -helix in which each end of the helical dipole is stabilized by corresponding counter charges from side chains at either end of the helix. Assuming that the inverted protein still forms a helix the helical dipole will be destabilized by those charged residues.
6. What is the free energy for the hydrolysis of ADP to AMP and Pi? The standard free energies of the following reactions are given: (1) ATP to ADP and Pi: G'= -30.5 kJ/mol (2) ATP to AMP and PPi: -45.6 kJ/mol. (3) PPi to two Pi: G'= -19.3 kJ/mol (8 points) There are two ways to get to AMP + Pi + Pi: (1) ATP to AMP and PPi followed by PPi to two Pi: The combined G' for these two reactions is: G'= -45.6 kJ/mol + (-19.3 kJ/mol) = -64.9 kJ/mol (2) ATP to ADP and Pi followed by ADP to AMP and Pi: The combined G' for these two reactions will also be -64.9 kJ/mol. Since G' for the first step is -30.5 kJ/mol, the G' for the second step is: G' = -64.9 kJ/mol - (-30.5 kJ/mol) = -34.4 kJ/mol
4
NAME:_________________________
ID:_________________________
7. Amino acid analysis of a peptide yielded equimolar amounts of the following amino acids: Ala, Arg, Asp, Cys, Gly, Lys, Met and Trp. Reaction with dabsyl chloride yields dabsyl-glycine. Cleavage with cyanogen bromide yields a heptapeptide (hepta=7) and Arg. Digestion with trypsin yields a pentapeptide and a tripeptide containing Arg. The tripeptide absorbs UV light at 280 nm. Treatment of the resulting pentapeptide with staphylococcal protease yields a tripeptide containing Gly and a dipeptide, which dimerizes in the presence of oxygen. Give the sequence starting at the N terminus. Partial credit will be given if you explain what the reagents do. (20 points) (1) Dabsyl chloride labels the N-terminal residue: Gly is the first residue (2) Cyanogen bromide cleaves after Met: Met at position 7 and Arg at position 8 (Arg cannot be at position 1, since this has already been assigned to Gly). (3) Trypsin cleaves after Arg/Lys: Lys at position 5 and no cleavage after Arg, since it is the last residue. (4) Aromatic amino acids (especially Trp and Tyr) absorp at 280 nm: Trp at position 6. (5) Staphylococcal protease cleaves after Asp (under special circumstances also Glu): Asp at position 3 because of the N-terminal Gly. (6) Cys will form disulfide linkages in the presence of oxygen: Cys is at position 4. This means Ala is at position 2. N-Gly-Ala-Asp-Cys-Lys-Trp-Met-Arg-C
8. Bonus question: Assume that you have a second peptide with sequence identical to that from question 7, however, this peptide does not react with dabsyl chloride. How could you explain this observation? (5 points) Two possible explanations: (1) We are dealing with a cyclic peptide in which the N- and Ctermini are also linked via a peptide bond. (2) The N-terminus has been covalently modified, for instance through the addition of a formyl or acetyl group.
5
NAME:_________________________
ID:_________________________
9. Shown below is the structure of an -sheet which is another secondary structure element predicted by Pauling and Corey. Alternating residues in each strand have dihedral angles of =87 and =-49 for residue n and =+45 and =+92 for residue n+1. To approximately which region of the Ramachandran diagram belong the dihedral angles of (a) residue n and (b) residue n+1? (c) Are the strands in this sheet parallel or antiparallel? (d) How does the hydrogen bonding pattern in this structure differ from that in a -sheet? (12 points)
R' H N CHR1 C H N CHR2 C H N CHR3 C H N CHR4 C "R O "R O O O H N CHR1 C H N CHR2 C H N CHR3 C H N CHR4 C O "R O O O R' H N CHR1 C H N CHR2 C H N CHR3 C H N CHR4 C O O O O R'
(a) With =-87 and =-49 these residues will be in the region for right-handed -helices. (b) With =+45 and =+92 these residues will be in the region for left-handed -helices. (c) The strands are parallel. (d) Each residue makes one hydrogen bond to the strand on the left and one hydrogen bond to the strand on the right. In both parallel and antiparallel -sheets both hydrogen bonds from one residue go to the same -strand.
10. Bonus question: Amyloid fibrils are found in patients suffering from Alzheimer's disease and are formed from misfolded and aggregated cellular proteins. Recent computer simulations have predicted that -sheets (see question 9) form an intermediate during fibril formation. Which structural feature of the -sheet explains how large aggregates can be formed? (5 points) The carbonyl oxygens all point in the same direction as do the amide hydrogens. Because of this arrangement an existing sheet has a dipole with the negative end on the side of the carbonyl oxygens and the positive end on the side of the amide hydrogens. Addition of the next strand in the correct orientation will be initially favored by long range electrostatic interactions, subsequently the pattern of hydrogen bonds will be established.
6