Voet Chapter 8 solutions 1-4,12,13,18-22

Voet Chapter 8 solutions 1-4,12,13,18-22 - Chapter 8 Three...

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Chapter 8 Three- Dimensional Structures of Proteins 1. With 3.6 residues per turn and a pitch of 5.4 AI the a-helix is 20 x 5.4/3.6 = 30 A long. The end-to-end distance between two fully extended amino and residues is: .... ""'f------------------------~ "1.33 sin 58°-/- 1.46 sin 64°-j- 1.51 sin 47°+ 1.33 sin 69°+ 1.46 sin 53°+ 1.51 sin 58°"" 7.23. .4 Thus for 20 residues, this is 20 x 7.23 A/2 = 72.3 A. 2. Near ¢ = 0°, there is interference between the carbonyl oxygens of the two residues. Near ¢ = 60 l1 , the H atom substituents of the Nand C u atoms defining ¢
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64 Chilpter 8. Three-DimenSIOnal Structures of Pruteins 3. 4. 5. 6. 7. 8. interfere. Past 90°, the carbonyl 0 interferes with the R group and then, towards 180°, with the H atom substituent of Ca. <j> is relatively free of steric interference between about 240 0 (-120°) and 300 0 (-60°). A 310 helix incorporates two C a atoms in a closed hydrogen bonded ring. Since a "I-amino acid residue also incorporates a ~ and 'Y carbon in its backbone, there must be 10 + 2 x 2 ;:;: 14 atoms in its hydrogen bonded ring cmalogous to a 3 10 helix. With a pitch of 9.9 A and a rise per residuc of 3.2 A, this helix must have 9.9/3.2 == 3.1 residues per turn. Hence, it is designated a 3.1 14 helix. (a) An a-helix (b) An llntiparallel ~ pleated sheet, the two strands of which are connected by a reverse turn (residues 46-48) that resembles a type I ~ bend. (c) Glycine; no other type of residue can have these conformation angles. (d) Coil; these conformation angles have no obvious pattern. (c) Type J ~ bend. Note the Pro-Gly seguence. (f) The torsion angles about the rotatable bonds of the side chains are also reguired to define the three-dimensional structure of a protein. Keratin should most easily split in a direction parallel to its fibrils as there are relatively few covalent bonds perpendicular to this direction that could hold the fibrils together. Thus, the fibrils must run parallel to hair fibers and across the fingers in fingernails. 1\
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This note was uploaded on 09/19/2011 for the course CHEM 43 taught by Professor Therien during the Fall '09 term at Duke.

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Voet Chapter 8 solutions 1-4,12,13,18-22 - Chapter 8 Three...

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