7_88_answerkey

7_88_answerkey - MIT OpenCourseWare http/ocw.mit.edu 7.88J...

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MIT OpenCourseWare http://ocw.mit.edu 7.88J Protein Folding Problem Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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7.24j/7.88j/5.48j/10.543j Fall 2007 Problem Set Answer Key a) Draw and label the complete structures of the cis and trans prolyl-isomers that may be found at pH 7 for the tripeptide: Ala – Pro – Ser must be drawn correctly, including proper charges on the termini. Your drawing must make clear that the rotation of the peptide bond preceding the proline residue is the difference between the cis and trans isomers, with the alpha carbons of alanine and proline being opposite each other across the peptide bond in the trans form and on the same side in the cis form. b) (6 pts) Given a polypeptide of 100 amino acids in length, how long would the folded conformation be if: i) (2 pts) Folded into an α -helix, as found in globular proteins? As stated in the legend of Figure 2.2 from B&T, there are 3.6 residues per turn and 5.4Å per turn (3.69 res/turn; 5.44Å/turn according to Pauling & Corey paper). Therefore: Rise/residue = (5.4Å/turn) / (3.6 residues/turn) = 1.5Å/res 1.50Å/res * 100 res = 150Å OR Rise/residue = (5.44Å/turn) / (3.69 residues/turn) = 1.47Å/res 1.47Å/res * 100 res = 147Å ii) (2 pts) Folded into a collagen triple helix? As stated on page 285 of B&T, collagen has a rise per residue of 2.9Å along the helix axis. Prof. King gave a more precise value of 2.86Å in class. Therefore: Rise/residue = 2.86Å/res or 2.9Å/res 1
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