BIS 102-Hilt-MI

BIS 102-Hilt-MI - ll "--—-I :,-. . I ‘7?é...

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Unformatted text preview: ll "--—-I :,-. . I ‘7?é Biological Sciences 102 Name v: H ’1' 1' __ e M Split-s. 2000 Last, Fm M K. Hilt First Midterm Instructions: You may the back of these pages if you clearly indicate that you are doing so. Some possibly useful equations and pK.’s are given below. " ‘ ' 2.12 7.21 12.32 pH = 403 [H'] pH = 403 3m K. = 2/0 - x) pH = (pKa, + pKaM pH=pKa+los{1base1/1acid1} momb)=1-00x10"‘ Kb=x’I(y-x) v AG=AH-TAS F=(q1q;)/(arz) a=‘rc V IS 102 Name Note: you do not need to uSe the quadratic equation in this exam, even though, in some cases, this equation might yield a slightly better answer. ‘ ' ' ease: I. (15'pts.) Water has several difi'erent properties, physical and chemical, which are conducive to cell function and/or structure. Identify three of these propertiesmeach one being different. For each,- deseribe how the property is important to the cell and how water, at the molecular level, is responsible for that property. ,. It .(fill in the blank). :9: a)’ Why is this important to the cell? H» 275mm ,. H3 i :n/iw ziiks’vwéx‘k‘m Property #2: b) Whyisthisimportant to the cell? (fir-WW floor/d. {flea/U. fxf’WA‘t MAL, QL‘WVMLJ “gt/n, fi _ -..._(fill in the blank). {s3 wsfim, c) Whyisthisirnportanttothecell? (/l/Va’ b) Molecular basis gm w “ff ‘ k; 2. ( 5 pts.) “Water is a good acid/base bufi‘er.” Do you agree or disagree with this statement? Defend your msmon' t H MD A. at, {Mme 5mm“? fiekfifiv ,0 “eifiézfm Q ‘7 ~ pM/fie 1,. l/ 7% a. 0W~fi,.. ,ng ; , i in" _ ,4 Viiwfi’é’ifl/W3'w ’l S 102 Name 3. OneoftheZObanfinoacidsgivesthefoflowingresultswhenyouperformtwodifl‘mentexpefimmts.In experiment #1, the amino acid yieldsthree difl’erent pK.’swhen itds titrated with strong apidor mongjgg. In experiment #2, the sum amino acid is titrated mammal conditio'hs, 'of fa‘raifldehyde. Theremfitsofaperimart#memmepmvnhmofmofthethreepK’smmbstanfiallylowerthanthe-' . conespondinsgroupsinexpuimenfiil. . (lOpts.)a)Drawthestructureoftheaminoacid,inthefonnthatwouldpredominateatthhatwould givetheremmsfistedaboveinacperimemsifl and#2..(Forparfia.lcredit, drnwthesu'ucturejnthefonnthat wouldpredominateatpH'I,ofmaminoaddthathasthreedifi'erentpK.’s). (.6pts.)b)Sketchthefiuafionwvesthatyouwmfldobmeforexpuinmts#land#20nthegraph H papd'bdow.labdyunaxesmdymum.mkethmsketchesfiiflywufimy.(Forparfinlaedigaketchthe mformaltumfivemfimacidthatymseleaedinpmfiflbove). EEW f%" u' i mica-E :fififififlfi%fififi. mm“ gflfifimfififimmflfl%fi%m, fimfififlwflmwfl%fifififlmé ,gfififlfififififlfififlfiflfl: @mgmmmmwfifimmwmmmmfl : amfiammwgmmfiwwmmmm§m m.; gflfilfifiiffiflflfiifilfifiiifi!Hi1?fillfififiiflfilflilEiiii‘él!§i§§lEll§§§ig:_ gldfialihmliahgn _ Jwfifimmammmmmmfifimmmmmwmfifimmwmfi a fl¢%Mflflflfifl%flfl%flfiflmflfififififlfifl%flmmuu W“ "eaifiéiiifi:fitfiflfiififlfifitfiiiiiiflfimmfifi:Elltimiéufiifllnifiiglihififlhiiiufiégfifiifiéafififiii ....... @fififimflmmaammwmmflmmmmmmfi%mmwfimgfifi IS 102 Name 4. One of the 20 L-mfino acids, when you do, a titration curve, gives three different pK.’s. However, if you repeat the experiment after bubbling oxygen into the solution eventight, one of the pK.’s has completely disappeared! ( 10 pts.) 3.) Draw the structure of the amino acid that gives these results, in the form that predominates at pg}, (For partial credit, draw the structure of an amino acid that has three difi‘erent pK.’s, no two of which are the same functional group and the amino acid is not the same one that you drew in question #3.). (10733-4 CGSE‘ , .4 ” L’ii’ t '. "‘ »;,. ; 11 EC} ‘ i l ' M: ,. N54,}. {1} S vs fw fig“: 1 1,53, #Hv“ _¥ ‘/_~7 aw ,, -A h, ,7 ~ ’ s i ' W est: (, 6 pts.) b) Write a balanced chemical reaction that explains what is happening during the bubbling of oxygen. . C5335” goo ' \"x r an,“ x; xv "‘ w my 5. ( 15 pts.) Draw the structure of a tripeptide, in the form that would predominate at pH 7, that satisfies the following conditions: _ a) , one of the residues has a side chain R—group that is an amide N b) one ofthe residues has a side chain R-group that has aggagidygponitm ll 0) one of the residues has a side chain R-group that is aromatic. F: \ ' o . .~ V y : M ' ~ ‘ r3 (/5 Q N; i r V l , V 1‘ t; :1 M ‘3’; :M ’5: r; 6. _ ( 8 pts.) On your structure in question #5 above, clearly indicate all of the atoms that lie in a plane(s) by drawing a box(es) around all of those atoms. Then explain, below, using chemical structures, why those atoms lie in a plane. ,_ ' -/."’ t “ V" m ,6 a "’ Explanation: g 1 a, N , rag ,s ,x.‘ l {g ‘77:», J x if: V" , , *2, wk“; 1“ 4a . m ,J i ’ dedgfly/gvyfire/QM: g:- {7 i. s ' 1.. aw“ r s 102 Name ‘\ if" 7. ( 15 pts.) Calculate, to the nearest 0.1, the net charge of your tripeptide in question #5 0n the PYeViOus page, at pH 10.7. Show all calculations. Label your calculationngo that we can clearly follow what you are doing. " H l/l E“ x x as: w j; A L F 3 w (0:353 my! “ v as,» " NW A), / C ’ l\—" C; l E ‘ ' “ 2“; z 33 §1 ILA twifi ‘: I \ "vuv \ ' ’ . i“ N“: if ‘5 ‘- ~ . , “ix” n3: C: is“ l Nth melts”: A- s, , to W WNW __ _, , “H” s, o s 1 fl, _ I ‘ ‘ I a. P i] g‘ 1‘ V I ‘ ‘2‘” 10"? e: 4.1: “L ~~ fig: < OliOFg-XZW} "‘ ‘21 “*7 ' “ :2: w 2 )- 8. (.10 pts.) What would be the pH of a 0.100 M solution of the protonated arm of your tripeptide in question #5? Show all calculations. : ‘ «o. \OMJ‘é" Pkg?" 1’ ’ i: [4‘1 A, ODD—5&5 : fig" >4 2;; ‘x :Q”"“* as g ’ fi 1"“ 1L)‘1=,_‘v gag—2A x » mull—ts; \ 1.2 t to Q: ., _// / Biological Sciences 102 Name Answer Key Winter, 2001 Last, First K. Hilt First Midterm Instructions: You may use the back of these pages if you clearly indicate that you are doing so. Some possibly useful equations and pKa’s are given below. Page Value Score 2 38 3 34 4 28 Total: (100) Compound pKa’s Phosphate 2.12, 7.21, 12.32 Tris 8 .2 1 Amino acid side chains asp 3.90 glu 4.20 his 6.00 tyr 10.10 lys 10.50 cys 8.30 arg 12.50 Amino acid -carboxyl group 2.10 Amino acid -amino group 9.60 Oligopeptide N-terminal amino group 7.40 Oligopeptide C-terminal carboxyl group 3.60 pH = -log [H+] pH = -log aH+ Ka = x2/(y — X) pH = (pKal + pKa2)/2 pH = pKa + log {[base]/[acid]} (Ka) (Kb) = 1.00 x 10-14 Kb = x2/ (y-x) G=H—TS F=(qlq2)/(r2) a=c BIS 102 Name Key 1. (12 pts.) 15.0 ml of 0.200 M HCl and 0.900 g of alanine in the zwitterionic form were added to water to a final volume of 0.500 liter. What is the pH of the resulting buffer solution? The molecular weight of alanine is 87.0. Show all calculations. Draw out the ionizable forms of alanine and plug in the pKa’s. The zwitterionic form is the intermediate ion. (+3) 0.900 g/87.0 g/mol = 0.01034 mol (+1) (0.015 liter) (0.200 M HCl) = 0.003 mol H+ added (+1) Result is that a moment later we have 0.01034 mol ~ 0.003 mol = 0.00734 mol “b” and 0.003 mol “a” pH = pKa + log {[b]/[a]} = 2.10 + log {(0.00734 mol/0.500 liter)/(0.003 mol/0.500 liter)} (+7 for this whole part) = 2.10 + log 2.4466 = 2.48 2. (8 pts.) What is the net charge on alanine at the pH of the solution in question 1? Show all calculations. 0.003 mol “a”/0.01034 mol = 0.29 (+6) Therefore, (+1 )(0.29) = +0.29 (+2) 3. If some palmitate, H3C—(CH2)14—COO, is added to the buffer in question 1, most of the palmitate will spontaneously form micelles. The pKa of palmitate is 4.6. (6 pts.) a) Which weak bond(s) stabilize the formation of the micelles? Explain. Hydrophobic effect (+2). The increased entropy of water drives the reaction (+4). (6 pts.) b) Which weak bond(s) destabilize the formation of the micelles? Explain. None (+2) . At the pH of 2.48, all of the palmitate carboxyl groups are protonated. Hence, there is no ionic repulsion (+4). (6 pts.) c) Which weak bond(s) neither stabilize nor destabilize the formation of the micelles? Explain. Ionic bonds (+1) , hydrogen bonds (+1) , and van der Waals forces (+1) . There are about the same number of each on either side of the equilibrium an‘ows (+3). BIS 102 Name Key 4. (10 pts.) In class, we discussed some things about this transparency (the transparency was also on our web page): (+6 for first correct answer; (+4) for second correct answer). Indicate two different topics that we discussed, related to this transparency: Topic #1: Cells are chemical engines—-—in a steady state with their environment. (This is required in order to maintain their organized structure). y Topic #2: There are a lot of membranes present in a cell. Hence, we need to pay attention to them in our course (i.e. membrane composition and transport across). Topic #3: Water is involved in the formation of all of these membranes (through the hydrophobic effect). 5. The following questions concern this compound: (3 pts) a) The name of this compound is _glutathione or — glutamylcystylglycine (6 pts.) b) Its importance in our course is: it is the in vivo version of mercaptoethanol or dithiothreitol. (It is helping to keep the interior of a cell chemically reduced). (6 pts.) c) For the above ionized structure to be the predominant form in solution, the pH of the solution must be between _3.6__ and _8.3_. v (+3) (+3) (9 pts.) d) On the structure above, clearly indicate all of the atoms that lie in a ‘v plane(s) by drawing a rectangle(s) around them. Then explain, below, using chemical structures, Why those atoms lie in a plane(s). Two rectangles should be drawn on the above structure, around each of the peptide bonds, with six atoms in each rectangle (+3 for each rectangle). These atoms lie in a plane due to resonance (show structures). (+3) BIS 102 Name Key 6. A mixture of the four amino acids valine, lysine, aspartate, and histidine were dissolved in pH 5.2 buffer and separated using ion exchange chromatography. The column material was a sulfonic acid substituted polystyrene. The amino acid mixture was applied to the column in dilute buffer. The amino acids were then eluted using increasing concentrations of NaCl. The elution pattern is shown below: I II III IV Asp Val His Lys (mM) (10 pts.) a) Predict which amino acid corresponds to which of the peaks 1, II, III, and IV. Give your reasoning. See labeled peaks above. (+1 for each peak = +4) Reasoning (+6): the column is negatively charged. Looking at the amino acid R-groups, at this pH, asp has some negative charge, val has no charge, his is partially positive (pKa = 6.0), and lys has a full positive charge. Amino acids will elute principally due to their net charge---the least positive first. (10 pts.) b) How would you detect the amino acids as they are eluted from the column and determine how much of each amino acid was present? (Give all details). 1. Mix an aliquot with ninhydrin (+3) , heat (+1) , and measure the absorbance in a spectrophotometer (+3) (A570). 2. Determine how much of each is present by comparing the areas beneath each peak to the areas (+1) of the appropriate amino acid standards (+2) chromatographed and detected in the same way. (8 pts.) 0) The above elution profile shows fairly nice separation of all four amino acids. What if, instead, the results of the experiment had been one giant peak coming out in elution volume 10 (i.e. we did not get separation of the four different amino acids). How could you explain that? Give two plausible explanations: Elution volume “10” indicates that the amino acids are not binding to the column. Explanation #1: (+4) the sample that we loaded is too salty. Ionic shielding prevents the binding of the amino acids to the column. Explanation #2: (+4) the pH of the sample is too high or too low. If the pH is too high, the amino acids will not bind since they will have a net negative charge. If the pH is too low, the column will not have a charge. t x “RH/I, 1 . *‘\ l ' 2. ( 23 pts.) Draw the complete structure of the oligopeptide listed in question #1 above, in the :i form that would predominateat pH 5.9.. " ‘imm l 1 of 2 BIS 102 Name Winter, 2003 ' Last First K. Hilt First Midterm Score (100): JL/ Equations: pH = pK. + log {[b]/[a]} (K.) (K) = l x 10"4 K, = xziy-x Kb = x2 ly—x Some possibly useful pK.'s: —m- — _-_ _-E_ . _m— ___ _I_ _ 2-1 -_ ; ‘ ' ' ; u (:3. \ ( 25 pts.) What is the net charge, to the nearest 0.1, of lgs-glu-kéirg-mmhis-cys at pH 5.9? Show all calculations. ,limw .Lwafofiwyi‘ N ‘9thka ‘ \ a”. S.q-*—1\‘~J‘ 91 {79,715 malts {m Rum : E9 :‘.C'5:7Pmehhrt 0? isn‘t l CAjGwM c. J1; A 8 ~ _.-H\I h", ' I ‘- ‘q& [.4 +- 5;?! =t~o + “3% 5+U m. -__ +l-zt’+(.qq)w)+(w]t:) (r1 I143 ;jpwp M h un-gtifrb‘atc E. 01 pg:\—{- .39 7.61% 4—3617 = +16% _ .—.._.____q___ + I ...\ H 0 HQ) H- so H Q ' ++ R I ‘ H t 5? Pt 1 l xx I l o : Hfi — c.~ 2-3-ei'LN—ci;~9 w-cl—vc—m—Cr-f— — N-<f~ C«0‘9 H A Elm. '9'; (CHQS (I‘M; Hafi— ‘g ‘CHL l -. 1:.- l :3 c/ \ S C.—-—b C b (‘11. ( Ht?) % 1 Q v as; 0G) :5, I mu xi: 1 ll - CR1; “1 NH; _ l m 20f2 Biological Sciences 102 Name 4 _ 3. ( 8 pts.) Identify all of the twenty amino acids whose R—groups can act as hydrogen bond 57 donors at pH .14. Use the three-letter code for each amino acid. as“! also lsfll-flnr ljfilat: *l’VE 4. ( 15 pts.) Describe the preparation of 30 liters $9.310 M phosphate buffer, pH 6.8, starting fiom a 3.0 M H3PO4 solution and a 2.0 M NaOH solution. Show all calculations. ‘93; Hai’au ~0H" ill—3 HZ POL“ + Hap—9 HPWZ‘ @ MINUS #——3("1 » ———————————-=5 mfigfiaofll no:me .— 1 .fi 53 .a a ' - .36 [L75 7. -l—— GIPmLilms—W “at” 7 .. {D 5. ( 10 pts.) Fill in the following table (there are 12 boxes; two of them have already been filled in ’5' for you): tem ature is lowered 10 °C I \0 W 6. ( 6 pts.) The organic biomolecules in cells are classified into the following four groups: proteins, lg 9563 , KlULQlC acids ,and Garbo basin/aid . é (Fill in the blanks. Some blanks may have more than one word). 7. Recall our discussion of amino acid analysis in class. We used an ion exchange column. “in An}. -l 0-41“ “714' ‘7' ( 8 pts.) a) Discuss, briefly, how we eluted, detected, and identified the various amino acids. H lova was at)“ RPM! ‘01 l‘wdvofisir, put. cap“ m in h) a», to“,th 04‘? 2J- g in map “it at +60 AA a), F’Wllfl l’b 9* Culuwvi pol-kin Pil‘f SWVCVfi —So3-L WM Jm beats. M wk“ SthGl-dqc-vw) ow PM“ ‘31» Ct‘tfulvc. ?\ 3 m no st... w L. Mu AA—.'§”.‘Sal'l‘ will '3lM'rlA‘tl‘0 3173") (AA,. {up you SW1. «4W cwowhrm'lm it" 1 Cm“ added- “) |SNa Bnabmbamu lair. er‘iwu \u thwfia'mblm Mpwfi‘l‘) F 5 pts.)b) Howmany different peaks do weexpectfromamino acid analysis ofatypical @0133.» '- - - protein? Give a n her and the lain w on chose that number. - I c a -r ufiA a, “1‘13” 0* $309“ ym mtg 13““ g “2+ l 3 Pt"; mum/WW“! . ‘ sad. asnfivnsm hmfiPlel‘lw gwr WM gln'wmmlv 3‘” m H“) [:5 4- + I . . o BIS 102 i Name § Spring, 2005 Last First K. Hflt - z ‘s/ First Midterm Score (100): fig 9 Equations: pH = pr:a + log {{b]/[a]} (1(a) (Kb) = 1 x 10‘” Kb = x2/ (y—X) Ka = XZ/(y — x) pH = (9K01 + pKa2)/2 F = (q1 <12) / 8 r2 I: back ,1. 0 (15,125.) mew ofa sohltionobtainedbymixingSOO ml of 0.400 M NH3 with 200 m1 of 0.250 M HQ. 1? The Kb ofNH3 is 1.78 x 10—5 M. Show all calcaiations. A W “wk .4OOMOLM‘LLv L, 00 X\ ” $03 T \ommt' 200“ ’ ' "‘0‘ NH" 1%" 0.0 S MM \'\0\ “H.180 mgkm. 35‘, .zoom, _» p 22,950,005 Niflfifiq" (mad 3 "' 3% Kb ._ X My; H p0: q. as « \og fi-‘gk 9 m, 3km¥\*\°‘:° TA 1 5‘40 x7/ 14m 50% a or Page, qastr’S 2 (25 pts) An enzyme—catalyzed reaction was carried out. in a s ' 11 buffered with 0.030 M hi ' ' @ éoénésumeg‘b , gbu’ffer2 EH 6.30. As a result of the reaction, 0.005 mole/liter H+ What was the pH at the end of the reaction? The pKa’s ofhistidine are 1.82 (a—COOH), 6.00 (side chain imidazole group), and 9.17 (oz—NHf). Show all calculations. c0 9 e (a 1 “6L fl, , Ahab 4' a 0H I 0 r4\ A N (— N\ “ /° \ ’ k / 7 7, 2 _ - 5/ W _ _ Mm”, xi _ , 2,27 41 6 ® \\”‘p:50 (3 V \ ‘AC + '2. 4— \ V ' . . 2 of 2 .--'Biological Sciences 102 Name 3 I PIER.” Flog ® (10 pts.) What is the pH of a 0.350 M solution of KZHPOH? The pKa’s of phosphoric acid are 2.12, 7.21, 1/ and 12.32. Show all calculations. 1d - J " l1 . 0 Hams-=4 1495304 7‘” H90? ‘13: [3043 - \ :H 'pr-q [2km F‘b _ _ _ ‘ I (4 sons (4 a. x1 Home": 43 W ,4, P w P ° .'° .1 Q. x_:_ 2.5mm =CoH3 Hflfim‘ffl ' 9m; '5 (,1 ._ film 6 mo acrd resrdues 1n the oligopeptide contains a sulfur, but bubbling 02 through a t-‘tionotlhe oligppepfidejfialq creatediwlfidebonds; - H“ (59).. ofigopeptide has a net charge of “f 3” ggfi@m a net charge of “— 0.5” at pH 7.4; _____ _ __ comma, figefi? " m) c) the oligopeptide absorbs UV light. MP, Y .Jo -'"/Ia w.) of- 0H 'i— "1° *5” PKa’s are DM), 13 (4.2), H (6.0),K33), $110.1), MEGS/K (12.5); oligopeptide N—tenninus (7,4), oligopeptide C-terminus (3.6). am a; twat) H“ W \f} Wu W 4M!" 1._ Draw the s c - of the abgve oligopeptide, in the form that \g’Ollld predominate at fill 13. I an. _‘ "F 35B“ + i- i ‘P ’3 R r‘ |;’* ‘ 5 A3,“) I "' *E'fif—‘C-C—N—c-a-UHLflE-‘E b F L cm 5‘,‘ . 1.4 :74 1- ha 1.: g cm i " q a p 0.5 09‘ ‘ l,— .S): “YO-3 °o “‘3 9w hing "‘lf (5 0 O o __ \ _\ . o "' i ': v- 5 ‘1 (3 pts.) a) Which weak bond(s) is (are) weakened by a change in pH? 3 Answer: H Vdmfl We; an; Wong; mo (3 pts.) b) Which weak bond(s) is (are) strengthene H d by an increase in solution temperature? 3 Answer: 6 WI (3 pts.) c) Which weak bond(s) is (are) weakened by a decrease in the WW? Edgar; 1mm ’1 J E } Answer: “Kmth ' - in «mm, id: (3 pts.) (1) Which weak bond(s) has (have) a partial cgyalent flange to their bond? 3 Answer: Huang“ mola- ...
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BIS 102-Hilt-MI - ll &amp;quot;--—-I :,-. . I ‘7?é...

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