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exam1key - 2 +H 3 N CH C CH 2 OH O CH 2 CH 2 CH 2 NH 3 + +H...

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BC 367 Exam 1 October 8, 2009 Answer Key 1. D 2. E 3. E 4. C 5. B 6. D 7. B 8. B 9. D 10. D 11 . a) A b) B c) Peptide A contains no aromatic amino acids, so it has no A 280 . d) C e) C 12. d, e, g collagen a, c keratin b silk fibroin 13. a) NaH 2 PO 4 (its pKa of 7.21 is the closest to the desired pH), NaOH (need to produce the conjugate base) b) Need 0.10 moles of the acid. 0.1 mol x 138 g/mol = 13.8 g NaH 2 PO 4 Solve Henderson-Hasselbach to determine how much A - , and thus NaOH, is needed. pH = pKa + log [A - ]/[HA] 7.5 = 7.21 + log [A - ]/[HA] [A - ]/[HA] = 1.95 or [A - ] = 1.95[HA] Also know that [A - ] + [HA] =0.1 M 2.95 [HA] = 0.1 M or [HA] = 0.034 M, so [A - ] = 0.066 M Add 66 mL NaOH, H 2 O to 1.0 L 14. Gly-Gly-Phe-Met-Leu-Arg-Phe-Gly-Trp-His
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Unformatted text preview: 2 +H 3 N CH C CH 2 OH O CH 2 CH 2 CH 2 NH 3 + +H 3 N CH C CH 2 O-O CH 2 CH 2 CH 2 NH 3 + 15. gel filtration gel filtration SDS-PAGE SDS-PAGE Normal GDNF Variant GDNF Normal GDNF Variant GDNF No -mercaptoethanol 30,000 30,000 30,000 15,000 With -mercaptoethanol 30,000 30,000 15,000 15,000 b) Wnc interactions keep the two subunits together despite the destruction of the covalent disulfide bond. (This can be inferred from the ultracentrifugation results as well as the data.) c) peptide bonds, 1 structure, net negative charge d) similar 3 structure, homologs, orthologs 16. a) Lysine b) A, C, E c) D d) A, C, E e)...
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exam1key - 2 +H 3 N CH C CH 2 OH O CH 2 CH 2 CH 2 NH 3 + +H...

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