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Copyright © UC Regents Davis campus, 2005-13. All Rights Reserved. May not be re-distributed without prior written consent of course instructor. 1 of 2 BIS 102 Name Keu Summer, 2013 Last 7 First K, Hilt First Midterm Score (100): Equations: pH =pK,+log {[b)/fal} (Ka) (Ky) = 1x10 Kp=xV (x) Ka=x"Ky-x) pH= (pKa, + pKa,)/2 F=(q'q’)/er AG=AH-TAS pK,’s of amino acid side chains: —_D (3.9), E (4.2), H (6.0), C (8.3), ¥ (10.1), K (10.5), R (12.5) pK, of amino acid a-~COOH (2.1); pK, of amino acid a-NH;* (9.6) pK, of oligopeptide N-terminus (7.4); pK, of oligopeptide C-terminus (3.6) 1. (10 pts.) In class we discussed the photo below. What was this all about? Give a detailed explanation of what was being tested and what the outcome was. What was the overall conclusion? 4 7 - - 2 ayy N tn madae. The 2ffecki 2 Engle ro swryrnmrnd 2 =i 2 aa\ nv deficiency (- 20?) and Mg deficiency © Mg") were being teatod. -motile amd had “te Ws were |e ge ani ey , mtamveliunn ef other mete oe Nar 44 = olds were multi lubed and nam—moh le. pve Q cuncdunion + everything is importers 4o cb, had dung met ella. . 2. In class we worked with the oligopeptide A—H—P. (12 pts.) a) In the space below, write out the complete structure of this tripeptide in the form that would predominate at pH 7. Os See 4 x / me cS A, AN \ c (12 pts.)/6) Calculate the net charge (to the nearest 0.01) of the above tripeptide at pH 7. Show ali work. , coo a) Ne berminua we a] : gh = gka + log oa oe a = + I= 74 + log 2 =| a -p.4 = log B oe Loe 40a oe poe b - ge 7~ + 0.0404 9.24% = % Te " 2At = B = spp = O-T!S” . net charge 0,715 + 0.0907~| = ~ 0.1% ~7o.19) (5 pts.) c) Using your drawing in part “a” above, draw in carefully the planes created by peptide bonds. ite aloe