bis102 midterm 2

bis102 midterm 2 - lof2 BIS 102 Name 14 Winter 2010 FirSt...

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Unformatted text preview: lof2 BIS 102 Name 14 Winter, 2010 ' ' FirSt K. Hilt Second Midterm Score (100): Equations: pH = pK. + log {[b]/[a]} K. = xz/(y-x) Kb = x2/ (y-x) pH = (pKal + pKa2)/2 F=(ql q2)/612 AG=AH~TAS Amino acid pK.’s: a—carboxyl group (2.1), a—amino group (9.6) side chains: D (3.9) E (4.2) H (6.0) C (8.3) Y (10.1) Oligopeptide pK.’s: C-terminal carboxyl group (3.6), N-terminal amino group (7.4) K (10.5) R (12.5) 1. Amino acid analysis is done on protein “Z”. The following results are obtained: Number of Residues in War aka I. ' Protein “Z” (20 pts.) a) Calculate the net charge, to the nearest 0.01, of protein “Z” at pH 7. Put your final answer here: 3. H3 . To receive credit, show all of yo work on the back of this page. 30 - 25 = L} Men A 3 A44 cu-Hllow‘nm (3 pts.) b) At pH 7, would you use a cation exchanger or anion exchange resin to bind protein Z, in order to do ion exchange chromatography? Answer: can an flL/bulné er . 2. (25 pts.) Many times in our disc ' n of protein structure, we have used a solid line to represent the covalent backbone of a protein, i.e. /& Use such a line in the space below to indicate your laidwledge of protein Sh‘uCture by drawing a diagram ofa typical gibbular' protein that indicates all levels and M of protein structure. Label each level and sublevel. (A level of protein structure may have one or more sublevels of structure). Then indicate the type(s) of mg; stahilizing each structure. K‘" 50:94 cert saw. as l1i e + r-‘B' A, ‘ wag”? LNreL‘HL) +2. ’/ ~ «H 20 5+vup+w°€v WM Ho |Dina-I‘JA 4" ‘mc.\woL¢, «L reverse -\-v.ka’ +1 H ‘0 \lava AM Waal/A.*‘ J vanilla/Q. P’flm‘i'CéQ 5W, NA‘WW‘“ 'X’Qe H , .w‘-fa,,,,uq @_e\eu\'e& shad?) + *7 '2 \ Ma M Mum. row 2 |0 coqw\%‘\’ MAE, em“ wiw "M5 20 A" 4.] +I at Sky‘» Qucvd- +0 .Hfl‘m, AS H'bml5 in M shew-hurt 0: M H Ma “at uni “LIA u... \1 on '5 ‘032'H. 2 ' z 7 6+5, :- ‘1 P+$. 20f2 BIS 102 Name 3. As polymers, proteins are able to change the pK. of an amino acid R—group by placing the R—group in difierent environments. Sometimes a charged group can move into close proximity, other times a charged group can mave away. In other cases, the pelarity of the environment can change from palm to nOnpolar, or vice versa. (6 pts.) a) Indicate three different changes in a protein that could result in a lysine residue having a lower pKa. r) L.) Mm M M K (kW/>- Nl'lz. \‘5 gmwrex a.) (4—) ¢MM. in 1 +1 ka 3) nowfll’lfid‘ MVKVLMM (6 pts.) b) Indicate three different changes in a protein that could result in an aspartate residue having ahigher \) “mpDqu Mv‘tvonW-l/V‘i— 'Y is gave-NA 25 L_) W ‘9“ +1 8% ‘53 U) Wm“ WM 4. (10 pts.) A protein was completely digested with trypsin and an internal peptide (i.e. not derived from the N— or C-terminals) was purified. Determine its sequence from the following information: a) amino acid analysis yields one mole of E, T, P, A, W, F, K, and R mote of peptide; 5) treatrnent cf the intact peptide with PYl‘C (Edman reagent), followed by acid hydrolysis, gives P'I'H-A; c) treatment of the intact peptide with chymotrypsin yields a tripeptide containing K, T, and E; a tetrapeptide containing P, A, F, and R, and free W; (I) the tripeptide, in part c, yields PTH-T after reaction with PITC and acid hydrolysis. Note: trypsin will not cleave on the carboxyl side of R or K if the next amino acid (donating the amino group) is P. \\ W “W o; Putyouranswerhere: A— K-P— F-W-T—E—K 5. (25 pts.) Using the axes drawn below, on the right, carefully draw in the following curves. Label each curve with its letter, i.e. “a”,“b”, “c”, etc. +3 8) Mb, pH 7.2 N WMW“ 3+ 01- Subum'l's. + 3 b) Hb, pH 7.2 + 3 c) Hb, pH 7.6 44 d) Hb Hiroshima, pH 7.2, where H146 on the B-chain has been mutated to a D 4 4 e) Hb, pH 7.2, where the N-terminus on the a-chain has been mutated from a valine to a leucine + q f) Hb, pH 7.6, where there is 25% sanitation of binding sites with carbon monoxide + H g) Hb, pH 7.2, where 75% of the binding sites 7 . are saturated with carbon dioxide 0 P02 (torr) 100 6- (5 Pm) Explain Why you drew curve “e” in question #5 above the way you did. H+ \OlA&$ +0 ‘HM' oL—Nl‘lz acute am “Wu. ok—swlouni'l'. yw‘l' 'Hq, R-ji‘awP. Rena , M vita/W36 Fl,- \l +1: L. 144/; no ref-eel: ‘—‘—‘>’.3 4— + W MW M-— ...
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bis102 midterm 2 - lof2 BIS 102 Name 14 Winter 2010 FirSt...

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