MCB124_ANSWERS_SET1

MCB124_ANSWERS_SET1 - ANSWER SET 1 MCB124-Baldwin page 1 1....

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MCB124-Baldwin ANSWER SET 1 page 1 1. Amino acid properties M-W P-G L-A H-I R-D A-V E-H W-K Q D-F A-G G-D H-K Y-T C-Y Arrange the above amino acid dipeptides in order of Q, AG, GD, PG, AV, LA, DF, EH~CY, YT, RD, HI, HK, MW, WK Hydrophobcity (I used the Flaucere and Pliska scale) RD, HK, GD, EH, Q, AG, PG, DF, YT, WK, AV, HI, LA, CY, MW Charges at (pKa values from Powerpoint) pH 2 HK WK HI RD EH [PG Q AV LA MW AG] YT~CY DF GD pH 5.5 HK WK HI RD EH [PG Q AV LA MW AG] YT CY DF GD pH 10 HK ~ WK RD HI [PG, Q AV LA MW AG] YT EH CY DF GD 2. POLYPEPTIDES 1 . A) Sketch a tri-peptide Ile-Gly-Pro in the extended conformation . B) draw two boxes around the atoms contained within the two complete peptide planes C) indicate the bonds that correspond to the Gly phi and psi torsion angles which describe the relative positions of the two planes D) Draw a Ramachandran plot on back using the above as a guide. Use it to answer the questions a. Indicate the backbone phi-psi angles for a right-handed helix with an "A" b. Indicate the backbone phi-psi angles for a beta-sheet with a B
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MCB124-Baldwin ANSWER SET 1 page 2 c. Where would you most likely find an alanine residue (mark it with "Ala") d. Where would find only (almost) glycine (mark it with a "Gly) 2 . a. Sketch the bonds and atoms of the following peptide sequence (N to C terminal). Note that this peptide was isolated under oxygen-free conditions. Cys-Trp-Asp-Arg-His b .In the figure, indicate the dominant protonation state for all of the charged groups at pH 5.5. c. What is the one letter code for the peptide? C - W - D – R - H d. Indicate by circling the closest overall charge on the peptide at the following pHs pH net charge
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MCB124-Baldwin ANSWER SET 1 page 3 2 -3 -2 -1 0 +1 +2 +3 5.5 -3 -2 -1 0 +1 +2 +3 8 -3 -2 -1 0 +1 +2 +3 10 -3 -2 -1 0 +1 +2 +3 We have to consider the protonation state for each ionizable group and sum them up. The groups are listed below with their approximate pKas. ionizable group pKa 2 5.5 8 10 C-terminus COOH 4 0 -1 -1 -1 Asp side chain COOH 4 0 -1 -1 -1 N-terminus NH3+ 7 +1 +1 0 0 Cys SH 9 0 0 0 -1 Arg side chain NH2-C=NH2+ 12 +1 +1 +1 +1 His side chain imidazole 6.8 +1 +1 0 0 Peptide overall +3 +1 -1 -2 e. The peptide sat exposed to air for a number of days, and the charges at the above pHs were measured. If any, at which pHs were the charges different? pH 10, since Cys is no longer ionizable What happened ? (One sentence) Cys formed a disulfide link with another peptide from air oxidation. This would, at the very least prevent Cys from ionizing at high pH since the S-S bridge lacks a dissociable proton. It might also shift the pKas of the other groups due to increased charge density in the peptide dimer. f.
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This note was uploaded on 11/10/2011 for the course MCB 124 taught by Professor Baldwin during the Spring '09 term at UC Davis.

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MCB124_ANSWERS_SET1 - ANSWER SET 1 MCB124-Baldwin page 1 1....

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