Spring 2007 Final

Spring 2007 Final - lof3 BIS 102 Name Spring, 2007 TuTh...

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Unformatted text preview: lof3 BIS 102 Name Spring, 2007 TuTh Last First K. Hilt Final Exam Score (200): Equations: pH = pKa + log {[b]/[a]} Ka = 2/(y-x) Kb = x2/ (y-x) pH = (pKa1 + pKa2)/2 (Kong) = 1 x 10'“ F = (q‘ qz) / e 12 AG = AH - TAS v0 = {v.m [S]}/ {Km + [3]} Amino acid pKa’s: a-carboxyl group (2.1), a—amino group (9.6) side chains: D (3.9) E (42) H (6.0) C (8.3) Y (10.1) K (10.5) R (12.5) Oligopeptide pK,’s: C-terminal carboxyl group (3.6), N-terminal amino group (7.4) Determine if the following six statements are “true” or “false”. Each statement is worth 10 points —— 2 points for putting in the correct “T” or “F”; the other 8 points for correctly defending your answer. If a statement is “false” for more than one reason, then explain all of the false aspects of the statement. You need to defend each statement, whether it is true or false. 1. 1: Reverse turns are stabilized by a single H-bond between them of the 3M and amino acid residues that comprise the turn. (3 lg” MAL Ll“; aha. raSiA'V-L‘ nfl' R—jrzrw‘m. c MA, 4% + z} + z, \4 2. ’F Buffer A has a pKa of 6.7. Buffer B has a pKa of 7.3. If both buffers were of the same molarity, en buffer A would be the better choice for maintaining the pH at 7.0, given that the buffer has to deal with increasing H+ produced from an enzyme catalyzed reaction. 1' ll \\ ,. A+€H7‘bWAwillAMAWWWOKMbQUIM» “4,5”; m H" me. “t L\ N \o is necessary *1? 3. 'T The peptide sequence CCCCCCCCCCCC would not form an oc-helix at pH 11, but might form . I: anochehxatle. st“ r!» C NM an,“ M wan/3M A+ l m 4' 2' g ‘ ‘ : M H ‘ c u. will ire—(0Q card" 0M. WW?" “3 M 0‘. . 9 " ‘5 no c,th v." A“ 0‘- W “W3 i—WW‘.+L . 4. E In Hb Hiroshima, e C-terminal histidine of the B-subunits is changed to a D. This mutation will result in an increased H+ portion of the Bohr effect, but a decrease in the C02 portion of the Bohr effect. W H +5 l» D rem, “Hut. H+ Bohr M. 1+ vat“ lac-«we no w m 001 Wywh 4.” 5. E The peptide sequence SEGCMALC will always be cut by cyanogen bromide into two separate pieces. ['5""J ‘ +3- Cjamam \wowfifiz but-L an HM. WWW-2Q 471-1 are. M'W‘Wq W X?- “ Azsudfih \auwlL UNA-r7! Wm 04:“ met Le, +0.0 ' Maxi Q: ecu . 4 l, 6' p The PI 0f the Oligopeptide V R is approximately 6.7. (If you disagree with this answer, then calculate the correct pI). __ .. - W“ + + + '0 mi +125 K“ 3.9 6.0 73“ [0.3 “5‘ 0 a; 2 H4. 3—— + Err—a 0 2+. 0 .‘é’—— o .a 0 air + 'P + -\" O = g .r N“! 4. + 9 4" a 0 20f3 Biological Sciences 102 Name Z 7. Enzyme “Z” was assayed in the absence and presence of an inhibitor, “1”. The data are presented in the table below: nmol/min, J“ 7~ 1w % “‘5 no inhibitor “MI-EI-IIHEH-flflfl-Ifll-EE-Iflflllnfll mull-IllllflI-lIfl-IEIIHM7 .me S‘ (30 pts.) a) Using the graph paper below, make a Lineweaver—Burk plot. You should have two lines, one for the three points assayed in the absence of inhibitor, the other line for the three points assayed in the presence of the inhibitor. Plot all of the points. Use the straight edge provided to create. the two lines. 1 Label each line and each axis of the plot. Neatness» counts. Spend some time and make a good graph with both plots. afifia. &‘ fifififl fifli fl flfifi fifi fifl" :- i = ii fififififlflfi I! : Hwflfiflfi Hit; mafiaa aamaa ififia euaaaai I r x . “ rg: I -.: " i -\ l .. ’ p *7, J; H) ‘k L (10 pts.) b) From your Lineweaver-Burk‘ plots above, determine the Km and Vmax for your enzyme Z_. I?“ “‘0” Show your calculations and give the units. a _ 'I Km -A" = —l “Md Vmax JG- : ‘4‘ l0; (“Woe/M) l‘m “‘4 sz+g “EL UM 7. [2'2 anus/Luann (10 pts.) c) Using the expression below, put in arrows ( 2 ) in the appropriate box(es) that describe the action of this inhibitor on enzyme “Z”: E+S ZES -—)E+P + + I ll HM m+slgE$ 3of3 Biological Sciences 102 Name fig 8. (25 pts.) How many moles of H“ are required to change the pH of 100 ml of 0.200 M phosphate buffer from pH 11.7 to pH 2.3? The pKa’s for phosphoric acid are 2.12, 7.21, and 12.32. Put your answer in this blank: 0 . 03 l 8‘ mols. To receive credit, you must show all of your calculations on the back of this page. 9. (25 pts.) Calculate the net charge, to the nearest 0.01 for the oligopeptide AHCEK, at pH 8.0. Put your answer in this blank: ~ I . l 3 . To receive credit, you must show all of your calculations on the back of this page or another page. The rest of this exam is multi le choice. Each question is worth 10 pts. Circle the letter of the best answer. 0 cut: a .— ._-. I‘ ‘ 0 10. The compound drawn to the right is: We... 0 .. C‘-“ a“. cuz—o— g—o—wz— é-» a) phosphatidylcholine with oleic acid and palmitic acid components 0" *r‘au3 b) phosphatidylcholine with a-linolenic acid and stearic acid components c phosphatidylserine with linoleic acid and a-linolenic acid components phosphatidylserine with stearic acid and linoleic acid components e) phosphatidylserine with some other combination of fatty acids 11. To determine that protein “X” has an intact molecular weight of 75,000 g/mol we would run a) an IEF gel b) a native PAGE gel c) a SDS-PAGE gel Q a series of one of the above gels “a” and “c” f) S‘a,’, ‘£b”’ G‘c9, 12. Company A offers 5 ml of trypsin, containing 200 I.U.’s and a specific activity of 56 I.U.lmg, for $100. Company B offers 3 ml of trypsin, containing 300 I.U.’s of trypsin and a specific activity of 44 I.U./mg for the same price. Company C offers 6 ml of trypsin, containing 250 I,U.’s and a specific activity of 33 I.U.lmg for the same price. We plan to use the trypsin for sequencing unknown proteins. We should purchase trypsin from: @companyA .———, 5«‘ 041'. I W b) company B c) company C d) company A or B e) company B or C 13. In enzyme purification, we would rank the following techniques fiom best to worst, in terms of potential fold purification as: @ affinity column, preparative IEF gel, sizing column, ammonium sulfate cuts b) preparative IEF gel, affinity column, sizing column, ammonium sulfate cuts c) sizing column, preparative IEF gel, affinity column, ammonium sulfate cuts (1) affinity column, sizing column, preparative IEF gel, ammonium sulfate cuts e) ammonium sulfate cuts, affinity column, preparative IEF gel, sizing column 4": La.x43}(0,ozywuq 0):.- p. 00391, mat H+ W {of 6% ‘9 5W“; = 0.02 WQV+ h a) Q“- : «\4.‘ +10%; 23. s an. + [0-3, A .. h _ = '1‘: oul?‘ a "' a_ -~ 4rv 2‘51 0 L0,349)€0.az Me P): 0_ 00.176 "are If} a” “Vii; ‘3 TIM 14+ ‘—‘- 0.003% "us—Q + 0.07. McQ- J- o.av'79& we, 5'— 0.031? 1-0 - 7‘ (‘SE—A-‘H-Cia’ E—li‘w‘fl' (H- 8' I s~ _ + 0 (33) «N: (Kh*'”}% 'L efi'zeka «Flaky; 3‘ 7‘” “"313- 7: 3.3 +1 5;. 0b=laj,% ’03—‘Wy-oja’ 3.15”% as , ~75; ' ‘ 1.11, a} bslgs 0.33 a: ’- ‘1— ’ b 5’ 2*02 Nd My: 0,2. +-(-a$$)+(—I)+(Q+(._.\) ...
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Spring 2007 Final - lof3 BIS 102 Name Spring, 2007 TuTh...

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