Spring 2006 Final

Spring 2006 Final - 10f3 Biological Sciences 102 Name K%...

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Unformatted text preview: 10f3 Biological Sciences 102 Name K% Las First Spring, 2006 t, H. Cheng, K. Hilt Final Exam Score (200): PH = PKa + log {[basel/[acid]} AG = AH - TAS F = (‘11 <12)/(8 r2)Vo = {me [3]} / {Km + [3]} Each of the following true/false questions is worth 10 pts. (one point for correctly writing “T” or “F” in the blank; nine points for correctly defending your answer, using an explanation and/or calculation. a? + r‘ 4' ‘°' “50/ vi “3” \ f I! / / 1. F The net charge of the peptide N-Q- - -S is + 2.03 (to the nearest 0.01) at pH 7.5. The pK. ’s of a.a. side chains are: D (3.9), E (4.2), H (6.0), C (8.3), Y (10.1), K (10.5), and R (12.5). The pKa ’s of the N- and C-termini of an oligopeptide are 7.4 and 3.6, respectively. "13:7.q +1.3 {llfl’ 0.1 = .l 3;, +3 ‘ p.z?=a)h _.‘ A: fig: 1+7. ii- wmxwa/ran \ a... 2. _F__ The only negative heterotropic allosteric effector of Hb is H+ and it binds to the heme of each subunit. (Ifthere is more than one, than indigagtes where each binds). 4...; +03 Jr “H H+ ———w C‘*:.OVM\ H 0". fi-gu-VWKH‘S “fl U-+€xwur\i a". 0L~$V"QrWW .5 +1 + -6 .+ 4’( C\- —-) haw K Mg N'W. [UH-3 re, away. 9!. Maw +1 399 ——-7 v.9»..th vow-k)“ H (:02 __, New”..an at ‘11 gm su-bumi-s +1 3. E A student has 100 ml of a 0.10 M solution of histidine at its pl. 100 ml of 0.050 M HCl is added. The final pH will be 7.20. Note: the side chain pKa of histidine is given in question #1. The pKa ’s of an amino acid a—amino group and a—carboxyl group are 9.60 and 2.10, respectively. +7. z.\ ‘-° - D -\ M —> M+\ —) Luv; ‘13—— M alt is» :. «11:6‘0 = (2.0+ has 0,; “I = b. o 4. E When a student measures Km and Vmax for enzyme Z, using a certain concentration of enzyme Z, they determine these values to be 3.2 x 10'6 M and 6.3 umol S —> P/min, respectively. If the student were to re-measure these values again, but this time use ten-times the concentration of enzyme Z, then there would be no change in the reported values. We will assume that the student carries out all manipulations and calculations correctly. +§ Km mi“ no‘l’ okmr.) H’ 16 inMwM 4 “‘W VmM will nm \ag' 63/,mq-Q Sap/M‘ +q 5. F A Lineweaver—Burk plot of an enzyme exposed to an irreversible inhibitor would show a similar plot to that of a competitive reversible inhibitor. In the type of representation shown below, the double arrows would be drawn in the box(es) shown: E+SfiE$—-9E+P + + . *Ll \ \ :‘r: 1 :vcevcfslble. T. halos Mk; nananfdfl'lflvg I. [El M lanes— :uwfl m. E1+sl§§551 *‘ 20f3 BIS 102 Name K% 6. T The pH of an enzyme catalyzed reaction needs to be maintained at pH 6.5. The reaction is: S+ -—> P + H“. A student has the choice of using buffer “X” or bufl‘er “Y”. The pKa of X is 6.0, while the pKa of “Y” is 7.0. The student should use buffer “X”. f q NO «$09» a. CM “Aost m “'4' «veal-Ad _ M k5 ‘\SWMMQK¢¢+*IWM4$wiub—kinm lows-WM. 7. 1’ 0.0020 moles of 02 will be released if a liter of blood containing 45 g of hemoglobin is transferred from 60 torr, pH 7.6 to 20 torr, pH 7.2? Hemoglobin has a molecular weight of 68,000. 15.4—— r. Leis/Ho." M H'b “- M, m M» H a l“ ‘ 2' $ ’0‘: M a; bDWA a; so +mr r * '— % 02 Bound l 75°79 TC‘CG'SeAI b‘AA M M (z - 3 Lo.'15)(2.z.wzz “25M 0,) = M? x 10 m4 9; a, ooze Me . . . . r 5 0 8. F The best column chromatography procedure for protein purification IS a Slzmg column. The 1’ technique is gentle to proteins and provides about a 200 fgld purification. + q - + t W u‘; “\wmn ’ 3600). II— 2000 X waflfiwfim. (Si-Jug column WRA) agm a. Sana,li Qwfi'fcd-im) 9. F The compound drawn below was discussed in our course under the topic of “lipids”. We discussed it because of its importance in the terpenoid class of lipids. + Z a D MM-YMQJ- WOW/9‘ +9 A“ L 2’ a '@ 135*. $5., ow>~a~ws god'an u‘l Yeac’km c _ W W go)- a +1.} + .3 10. T A lysine residue is in a protein. Bringing a glutamate residue into close proximity of the lysine side chain will cause the pKa of the lysine side chain to increase. Another way of causing this lysine pKa to increase would be if a neighboring argimne Slde cham movwanay. (“a R+ / + q J“? K+la / K+ awaan K0 gamut glam K“ \s dm. Hem-e, 9K6. ’r w’ ZauaveA com (2" mm awa-«j. 11. E The structure of phosphatidylcholine, containing fatty acids that would only be present in a plant, is shown below. (If this is not the correct lipid structure, then draw it). a +2 3 u “AA/ ‘5’ a Mia-ngva 0 TH-ro-fi—Wq-w are. I ll _ fl ,_ H - H , ~ C "H W ‘9 f“ 0,-O_<‘“ Q” A: 6101 5 CI“: 0 z'o‘e‘ - z 2"“ 2" ‘J'rzdb QA;0—<:a—cuz:ul’_— Nv (AL; 0 45m. 3 7s, 0 \4H 3 l a, ‘ c, 6‘ WI“; \W/d—‘J '7’ W») «uh M m 6) W + Z 3of3 BIS 102 Name Kg 13 12. F: We are looking at ten amino acid residues that are part of a protein. If these ten residues were present in a form of secondary protein structure, the order of length, from longest to shortest, would be a—helix, reverse turn, and then B—pleated sheet. M p’wew 5M, “PM w vws—e. NW 13. l: A peptide is drawn as shown below. The pH of the solution must be between pH 8 and pH 10 for this to be the predominant form of the peptide. (If this pH range is incorrect, then give the correct pH range). 0 H " rt 3 t n / e z ‘u; " + ' -— c, - N - o - c. - N - ¢ Q H PEN 2» i“ z“? -\ \CH-— <14 1 (0 C 6‘” ‘ 7“ 3 \ / M H 2, Cu \ ‘5” + H + 5 I H" \ u, 14. F The predominant weak bond holding together the various subunits of hemoglobin is the ionic bond. The major form of communication between the subunits is hydrogen bonds. The basis of ionic bonds is electrostatics, as is the case for hydrogen bonds. +2 1. Pvefiwwiwwni’ lama baud-M“ m ‘Su‘oug‘a’s “36% is it)“, Rake-phloem QW_ 1+3 law/s. is We \WMM em‘n’vy‘j fi+t§m, 2 11 ‘ MSW Men—‘4— bma in cmmunicai'im Rs iamlc, \om. “Mob; M1 15. E There are only two histidines that play any role in oxygen delivery by hemoglobin. The two histidines are referred to as the C-terminal histidine of the a—subunits and the distal histidine. Each makes a bond to the heme iron. And of the two, the C-terminal histidine is by far the most important. H +1 _ +1 _ evoflzxjjp‘ ‘4 H. ‘ «Mal (‘_- *MMina—Q H are. 63 —§ubww+§, QW1KM&*Hi$ m°5+ RWNM' 16. F The pH of a 4.50 x 10'12 M HCl solution is approximately 11.3. MAMA? Min-€— H": ht’57N. Ccm;n3 gym H20. 17. T The number of moles of HCl required to lower the pH of 400 m1 of 0.200 M arginine buffer, pH 12.5 toafinal pH of2.l is 0.16 moles. 6+2 ,\ H a... ,, 125 -\ fix ——§ 4.1 a N —-9 (WWW-2M “33’- 0-W M“ “a W "3 4.‘ Jr? | act.“ ‘e“” -:..' 0.11:: wide 18. [—r A Michaelis—Menten enzyme is assayed at a substrate concentration that equals the Km. Th velocity measured is 20 pmoles S —)P/min. 12 pg of enzyme was present in the reaction. The molecular weight of the enzyme is 65,000. The turnover number of this enzyme must be approximately 108,500 min". vac»; t qojmwe sac/M *1 w»): 1.25.1.— = 1.?4sz15w H _‘ ’ Lam filmfi ’ T'U'F 140120 MPtm i ...| ’ , *— .. - L Luz Mom (5 — 2,!Lb x/D mm (o 19. I: When measuring enzyme rates, buffers are used solely to maintain the secondary structure of the enzyme. - + 2’ 7' + L \ mflnw ’5” Max 4° . +00 1 ckénvya, a} “W ant/“WV: sil'e— M8. on subswm; A- 7. 20. F We have studied the 20 amino acids. Threonine reacts easily with oxygen to form another compound. Serine is the only amino acid that has a side chain with an asymmetric carbon in it. And the side chain of arginine has three amino groups! *3 \. rascals wi‘er 0; 1'}. 2. T m& 1‘ impe— am 453MM+WL embw in M Q-yMS 3_ 03,694.“; MM a. jumfélino awe-19) "9+ “wee a.wa ywfilfi. 2. ...
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This note was uploaded on 04/08/2008 for the course BIS 102 taught by Professor Hilt during the Spring '08 term at UC Davis.

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Spring 2006 Final - 10f3 Biological Sciences 102 Name K%...

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