Exam 2_Spring 08 - Spring 08 Bisc 220 Exam II Instructions...

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Unformatted text preview: Spring 08 Bisc 220 Exam II Instructions; Use of the Blue Seantron Answer Sheets: l. Enter your name in the NAME field: and fill in the appropriate bubbles. 2. Enter your USC STUDENT ID Nail-{BER (not your Soc. See. No.) in the field labeled lDENTIFICATlON NUMBER, and fill in the bubbles. You need not enter any of the other information (sex, grade, birthday, etc.) 3. Enter the code for your TA’s name in column KI. in the field SPECIAL CODES. Use the following codes: 1 ; . 3 Code for E First name Last name . . . - i Cot-21mm KI.» E Roman Barco E 00 ' t Izlmes Bonaiuto E ()1 E E Prithiviraj Chellanthrhu .-__ 02 E L. Nadine Eiswemtolb —E 03 E V Andrew Fogel 0—": E E Abdoulaye Galhani m E 05 k‘ Jia Hu """""""" "06 E E Maria lntermaggio t 07 J H“ Jeremy lacquer 08 ‘ l -.-__.: Ying Lill- mflffi). E L Jasohw E Mhlrvash 10 E L‘Kimberly Rapp Wm.“ E Joseph So 12- E : Natalia E 'ljoltro __ m_ ml 3 E Michael . E Tringe - l-l E E R0 shanak Varj avzmd 15 E Jagruti \I'edamati —.E.__.. E 16 4. Enter your answers on the front side of the she: Smith the NAB-EB fields, etc. starting #1). Be sure to bubble plainly. lfyou erase, do so Clea-1m; . lnstructions for the Written Portion of the Exam; 1, Write your name and your TAS name in the 2. If you wish to submit the exam for a J'e—gh provided. 5nur answer in PEN only and no wipe-out. 3. When you are done turn in the entire exam with he Blue Scantron Answer Sheet to your TA. 1 xamination BISC 220 Spring 2008 4*M 2008 j; = This examine?” ’ *8 pages and 5% questions. Be sure you have a complete examination. Please put yea? and U80 lD number on the top of each page. Remember to use your time wisely during the.» einalion period. 2%; 1. How do the taro ineiizbefs of a pair of homologous chtomosomes difEer from each other? 2% Eithe idendtv and relative position of the genes present on each of the chromosomes » p a 5: . ,g‘u ‘ . t . e precise sequence of the DNA witlnn each of the ' chromosomes 2. Increasing the substrate concentration in an enzymatic reaction could overcome which of the following? A. denaturation of the enzyme D. satuxation of the enzyme activity )3. allostelic inhibition E. insufficient cofactoxs @ competitive imam: f [j dark to — .1 ution at pH 8. W’hat would be likely to hagpen? % ; 3. Suppose the intexiox of the thylakoids of isolated chloroplasts was made acidic and then transferred in the #334. . e isolated chloroplasts will make ATP. 7-: * cycle will be activated. , Wemphosphoryladon will occur. D. Only A and B will occur. @fl—m C will occur. 4, Zinc, an essential trace element for most organisms, is pzesent in the active site of the enzyme carboxypeptidase (and many others). The zinc most likely functions as 21(11): A. competitive inhibitor of the enzyme. B. noncompetitive iambitoz of the enzyme. C. allostexic activator of the enzyme. cofactox necessary for enzyme activity. 12'. coenzyme derived from a vitamin, Cng—P “)3” “4134;? S. Caffeine is an inhibitor of phosphodiesterase. W’hen you consume a cup of coffee, which of the following effects are most likely to OCCDI in your cells? A. block the response of epinephrine 33¢ decxease the amount of CAMP in the cytoplasm C. block the activation of G proteins in response to epinephrine binding to its receptor @pmlong the effect of epinephtine by ’ mining elevated CAB-[P levels in the cytoplasm E. block the activation of protein kinase A 6. An increase in the levd of cellular ATP is likely to occur under which of the following conditions? 94. increased activity of cambolic pathways 9941 he. Vg. decreased activity of anabolic pathways Kari!» €l (/C. allosterlc inhibition of anabolic pathways $5»be V: D. A and B only @ A, B, and c _.i 7 G proteins and (Er—protein baked ' = A. are found only in animJ » )Biare found oalvm bacterial lls. ’ are thought to have evolved very early,- because of their similar structure and function in a wide variety or modern organisms whose common ancestors diverged billions of years ago. LB? probably evolved from an. adaptation of the citric acid cycle. ii: are not widespread in nature and were unimportant in the evolution of eukaryotes. 8. When electrons move closer to a more electronegative atom what happens? 3A. Energy is released. (£13 The more electronegafise atom is oxidized ,3? Energy is consumed. QA and C are correct. ~E‘EF'C. The more electronegafive atom is reduced. 9. Which of the following statements about NAD+ is fake? VA. NAD" is reduced to NADH during both glycolysis and the citric acid cycle. NAD+ has more chemical energy than NADH NAD+ IS reduced by the action of dehvdrogenases. l/lD. NAD can receive electrons for use in oxidalix-e phosphorvlauon WE. In the absence of NAD , glv coly sis cannot function 10 Substrate— level phosphorylation accounts for approximately what percentage of the ATP formed during glycolysis? Ll W».— A 09’ D, 389/ 17-“ ‘" ’3 81’} o 0 £33 5g Cl z B. 2% E. 100% 3%.»;a @ 4 Acetyl CODA 3—“; WADE ()xnlnncelsle \‘WfrCOA ‘11. An abbreviated form of the citric acid cycle is given at the J ‘ 7. K)? Citrate light. Starting with one molecule of isocitmte and ending with N130: _,‘ [I \\ fumarace, What is the maximum number of ATP molecules that f- ‘1. e could be made through substrate—lav el phosphory lotion? Meme 4%. @- D 12 . '3 Fifi)“ me ' ’NAD’ E. 24 ~41... (30f , g c. 11 I, l NADH l ,FADHz ,3 12. Which of the. following can act as substrates for the manic \I. f RuBP car-bar lase? J K: FAD " T C 01 3‘ C!- Kelumiamle —-—A—.—~¥¥P—arrd BADPH ’ Succinaie 13 C02 and glucose ('0; V/NAnr {C 02 and 02 ATP: ‘-. Succinyl CoAK/ {HAD}! 21105:: P glucose and CO; ADP-r13? E. C02 and ATP 13. You have a friend who lost 7 kg (about 15 pounds) of fat on a "low carb" diet. How did the fat leave her body? @It was released as C02, H20, and heat. B. Chemical energy was converted to heat and then released. W5 converted to ATP, which weighs much less than fat. M broken down to amino acids and eliminated from the body. W converted to molecules in un'ne and eliminated from file body. 14. The old saying "one rotten apple spoils the whole barrel" is due to chemical signaling in plants via A. an increased uptake of carbon dioxide during respiration in target cells. BF—Le‘rzeall) not due to chemical signaling at all. @release of ethvlene gas a plant hormone for opening. D. an a/o: cell signal systemin the rotten apple 1:121 signal transduction pathwav involving glycogen phosplrorylasc. luJ 15. Which of the folloWingis (axejinze a: ligandgaled ion channels? A. They areimpor‘mni in- the nervous system f/ B. They lead to changes in cbnceutra'tion of sodium and calcium (among others. in cells. C. They open or close in response to a chemidal signal. D:\Only A and B are true. (EyIL B, and C are true. 16. If photosynthesizing green algae are provided with C02 synthesized with heavy oxygen (180), later analysis will show that all but one of the following compounds produced by the algae contain the 180 label. That one exception is A. PGA. z/ID. RuBP. B. PGAL. 12.0 5/ C. glucose. (‘9; "T W? “’9 {:5 lie 5 *2 9e 17. After telophase I of meiosis, the chromosomal makeup of each daug'hter ce ls‘x\\‘_ I?" Wand the chromosomes are composed of a single chromatid. ——-B.—di-ploid; and the chromosomes are composed of two chromatids. C. haploid, and the chromosomes are composed of a siagle-skgamatid. " D haploid, and the chromosomes are composed of two chromatids. W and the chromosomes are composed of two chromatids. 18. What would be lI’lIL for The signaling system in an animal cell that lacks the ability to produce GTP? ’/ (ilk would not be able to activate and inactivate the G protein on the cytoplasmic side of the plasma membrane. 23; It could activate only the epinephrine system K It would be able to carry out reception and transduction, but would not be able to respond to a signal. D. Only A and C are true. E. A, B, and C are true. 19. The hydrolvsis of sucrose by the enzyme sucrase results in ( (Aw-baagmg glucose and fructose together to form sucrose. / — ease of Water from sucrose as the bond between glucose and frucnose is broken. if reaking the bond between glucose and fructose and forming new bonds from the atoms of water. Wucfion of water: from the sugar as bonds are broken between the glucose monomers. Mafion of Water as a covalent bond is formed betwe en glucose and fructose to form sucmse. chromosomes? A ernflized egg cell (9:; female somatic cell Baa—sperm cell . both A and D C .Maleeomadc cell 21. Which of the following statements is {We regarding enzymeiom3 m multi—engrme complex contains all the enzymes of a metabolic pathway. .. roduct of a paflnway serves as a competitive inhibitor of an early enzyme in the pathway. What-are molecule bound to an active site affects the active site of several subunits. Meveral substrate molecules can be catalped by the same enzymes Mmbsbrate binds to an active site and inhibits cooperadon between enzymes in a pathway. 22. in g%§m1’ a signal transmitted via phosphorylation of a series of proteins @“ brings a confounadonal change to each protein. X requires binding of a hormone to a qrtosol receptor Xcarmot occur in yeasts because they lack protein phospllatases. D. requires phosphorflase activity. X allows target cells to change their shape and therefore their activity. s_. _. .Maew. ~~ 'ersrawgfl 23. How does the sexual life cycle increase genetic variation in a species? y’A. by allowing independent assortment of chromosomes ME by allowing random fertilization '/C. by allowing crossing over / A and B only L . A, B, and C 24. SW5 benefit cells for all of the following reasons exrejzt l/A. they help c s respond to signal molecules that are too large or too polar to cross the plasma 'membrane. VB. they enable different cells to respond appropriately to the same signal. @ey help cells use up phosphate generated by ATP breakdown. OD- the;r can amplify a signal. IE. variations in the signal transduction pathways can enhance response specificity. 25‘ Which of the following statements is not true? Mitosis produces new nuclei with exactly the same chromosomal endowment as the parent nucleus. I . Mitosis may occur without cytokinesis. i/C. Mitosis and cytokinesis are required for asexual reproduction. CA). All cells come from a preexisting cell. 17E. The mitotic spindles in prokaryotic cells are composed ofmicromhules. 26. All of the following are directly associated with photosystem II way)! % extraction of hydrogen electrons from the splitting of water. ,Bfi release of oxygen. ' j€< harvesting of light energy by chlorophyll. {D‘ENADP‘r mducmse. : ”a KPGW reaction—center chlorophyll. a flflb j V / 27‘ Inside an active mitochondrion, most electrons follow which pathway? Wysis —> NADH -> oxidative phosphorylation —> ATP —> oxygen B. citric acid cycle ~> FADE; —-> electron transport chain w» ATP C. electron transport chain —> citric acid cycle —> ..-\TP —> oxygen pymvate ——> citric acid cycle —r ATP -—> NADH —> oxygen @cltfl'c acid cycle a NADH -+ electron transport chain -> oxygen 28. If the cell whose nuclear material is shown to the right continues toward completion of mitosis, which of the following events would occur next? Areellfitembranc synthesis stpimlle fiber formation Cf‘fificlear envelope breakdown l i/ D. formation of telophase nuclei ’ , sis of chromatids I I 29. A solution of starch at room temperature does not readily decompose to form a solution of simple sugars because A. the starch solution has less free energy than the sugar solution. . the hydrolysis of starch to sugar is endergonic. {_ the activation energy barrier for this reaction cannot be surmounted. D. starch cannot be hydrolyzed in the presence of so much water. E. starch hydrolysis is nonspomaneous. l i 33. Which of the drawings in the figure at the right represents “11 30. You obtain the following data {some stuff; of the length of time spent in each phase of the cell cycle by three eukaryotic oganisms designated bets, aelt'a,’ andpgamma. Minutes Spent in Cell Cycle Phases Cell Type G1 \ S 62 M Beta 18 E g4, 12 16 ' 0 Delta 100 I t o o Gamma 18 L 55$, Q 14 20 “---L The best conclusion you can draw concerning the difference between the 8 phases for beta and gamma is that @ gamma contains more DNA than beta. “M'- m &; beta and gamma contain the same amount of DNA. C. beta contains more RNA than gamma. D. gamma contains 48 times more DNA and RNA than beta. E. beta is a plant cell and gamma is an animal cell. 31. The free energy for the oxidation of glucose to C02 and water is —686 kcal/mole and the free energy for the reduction of REM)" to NADH is +53 heal/mole. Why are only two molecules of NADH formed during glycolysis when it appears that as many as a dozen coulcl be foamed? A. Most of the flea energy available from the oxidation of glucose is useé in the production of ATP in glycolysis. B. Glycolysis is a very inefficient reacfion, with much of the energy of glucose released as heat. @310“ of the free energy available firom the oxidation of glucose remains in pyruvate, one of the ptoducts of glycolysis. :R” There is no C02 or water produced as products of glycolysis. XGlycolysis consists of many enzymatic reactions, each of which extracts some enmgy from the glucose molecule. 32. How do cells at the completion of meiosis compare with cells that have replicated their Dszt and are just about to begin meiosis? 25 They have twice the amount of cytoplasm and half the amount of DNA. 2% ’E’ They have half the number of chromosomes and half the amount of DNA. dream . :8; They have the same number of chromosomes and half the amount of DNA. CK DW®Fhey have half the number of chromosomes and oueiourth the amount of DNA. E. They have half the amount of cytoplasm and twice the amount ofDNA. anaphase of mitosis? é/f , \_\ A. 1 \ , - . — V c. III w ,m D. IV SW E.Vl If [it 34-. W’hich is txue of transcription factors? yfi‘. They regulate the synthesis of DNA in response to a signal. LB? Some transcribe ATP into E‘UVIP. ,€<They initiate the epinephrine response in animal cells. . hey control which genes are turned on to form ntRNi. ~. They are needed to regulate the synthesis of protein in the cytoplasm. u\ 35. If an enzyme has been inhibited W A. the A. G for the reaction it catalyzes-{will always be negative. #3? the active site will be occupied by the inhibitor molecule K raising substrate concentration will increase the inhibition. . more may will be necessary to initiate the reaction. ’ E the inhibitor molecule may be chemically unrelated to the substrate. The following multiple choice questions are worth four points each. 36. You isolate DNA from three different cell types of an organism, determine the relative DNA content for each type, and plot the results on the graph to the right. During meiosis, the level of DNA/ cell would proceed in which of the following sequences? A. Ito II ml to II]. =7 B. l to to C. to D. Ill to I E. always remain ail BMW: in) M W .3 Ila O V! W —L~—-n-|—_...L_~.|._—-L__h..._|_.._l 37. A particular cyclic! called cydin E forms a complex with a cyclin— 9 l i ll dependent kinase calledggkl, This complex is important for the I: progression of the cell from G-. into the S phase of the cell cycle. Which San-p of the following statements is most likely to be mrmi? @7318 amount of cyclin B is greatest during the S phase. )Bil'he amount odek 2 is greater during G1 compared to the S phase. ’ C. The amount of cvclin E is greatest clun'ng G1. =B<The amount of Cdk 2 is greatest dufing GE. fine activity ofthe cyclin E/Cdk 2 complex is greatest during G3. I'll—Ff} \ u f’ - ‘ 38. Theodor W. Engelmann illuminated a filament of algae with light that passed through a prism, thus w E b VG? exposing different segments of algae to different wavelengths of light He overlaid the lament with a suspension of aerobic bacteria and then noted in which amas the hacteda congregated. He noted that the largest groups of bacteria were found in the areas illuminated by [her—ml and hlfilight What did Engelmann conclude about the congregation of bacteria in the red and blue areas? A. Bacteria released excess carbon dioxide in these areas é Bacten'a congregatefi in these areas due to an increase in the temperature of the red and blue light. . Bactcn'a congregatedin these areas because these areas had the most oxygen beingreleased. 94 Bacteria. are attracted to red and blue light and thus these wavelengths are more reactive than other wavelengths. b. bacteria congregated in these areas due to an increase in the temperature caused by an increase in m esi . phows‘ th 5 Certromere 39. The lettered circles (right) show a diEloid nucleus with fan; chromosomes. There are two Hairs of homologous chromosomes, one long and the other short One haploid set is symbolized as black and the other haploid set is gray. The chromosomes in the unlettered circle at the top have not yet replicated. Considenng only normal meiosis or mitosis, which of the five proPosed nuclei contains an impossible configuration of chromosomes? wmg 5 answers is the most correct- ’11 PDGF lg MPF @Tj protein kinase 0' qrclin E Cdk . a protein mamtamed at constant levels throughout the cell cycle that reqmtes cydm to become catalvdcally active r~ L—— 46. triggers the cell 5 passage past the G» checkpoint into De mitosis e each ofthe folio " ' boundaries of worth 3 points each. 4 T. photnres‘pimtion . . ”J .. .2. , .3515?! 5L; 5L1FM‘45'd”J,IV “an; Mal-1 QMZ mafia V ”L185, Maggy umbam‘ci _-_\ “I 49. ligand A smfi wkcuie'fiwr} 9,3735% 'féi'é {aaxéafg fewgajfizxfl‘A‘flLZdefi/ meiiy main? [1 mmszxflzéa Mmieis #5119113; wai'azgr—sxoiubm 50. genome m WM. Ceiiid‘im a? MA @w": 3W9) {w 01 cal? 9r 993de 51. synapsis ‘ lersifi/WPMSLEQQW 33W1mj3§mdzrewwh W353}? £0413=WMW W 551 665m; M V. _, ,,,,,, y...“ A "" " ""‘ 1 .r, _ ' ‘ _ 1 , . , . (I . x . I a Wmfigfigfii iww 1$§qfi~d gab-fl ”Bu. {wié‘sf‘atd dwauae’cdyfmi »% M: ”if? 4m - . ; , - . , awww E ‘Rauk {'0 ...
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