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How many l2-oz (355-mL)bottles of 5% beer could a70-kgpersondrink and remain under the legallimit? A 70-kgperson contains about 40 liters of water. Ignore themetabolism of ethanol, and assumethat the water contentof the person remains constant.D. Ethanol is metabolizedat a constant rate of about 120mg per hour per kg body weight, regardlessof its concentration.
If a 70-kg person were at twice the legal limit (160mg/f 00 mL), how long would it take for their blood alcohollevel to fall below the legal limit?2-12 Specificactivity refers to the amount of radioactivity per unit amount of substance,usually in biologyexpressedon a molar basis,for example,as Ci/mmol. [Onecurie (Ci) corresponds to 2.22 x 1012disintegrations perminute (dpm;.1 As apparent in Table Q2-1, which listsproperties of four isotopes commonly used in biology,there is an inverserelationship between maximum specificactivity and half-life. Do you suppose this is just a coincidence or is there an underlying reason? Explain youranswer.2-13 By a convenientcoincidencethe ion product ofwater,K- = lH+l[OH-],is a nice round number: 1.0x 10-14M2.A.
\AIhyis a solution at pH 7.0 said to be neutral?B. \A/tratis the H+ concentrationand pH of a I mM solutionof NaOH?C. If the pH of a solution is 5.0,what is the concentrationof OH- ions?2-14 Suggesta rank order for the pKvalues (from lowesttohighest)for the carboxylgroup on the aspartateside chain104Chapter2:CellChemistryand Biosynthesisin the following environments in a protein. Explain yourranking.1. An aspartateside chain on the surfaceof a protein withno other ionizable groupsnearby.2. An aspartateside chain buried in a hydrophobic pocketon the surlaceof a protein.3.
An aspartateside chain in a hydrophobic pocket adjacent to a glutamateside chain.4. An aspartateside chain in a hydrophobic pocket adjacent to a lysine side chain.2-15 A histidine side chain is knol,rrnto play an importantrole in the cataly.ticmechanismof an enz).ryne;however,it isnot clear whether histidine is required in its protonated(charged)or unprotonated (uncharged)state.To answerthisquestion you measureenzyrneactivity over a range of pH,with the resultssho\^Trin Figure Q2-1. \Ahich form of histidine is required for enz)ryneactivity?FigureQ2-1 Enzymeactivityasa functionofpH(Problem2-15).cfEoEo,r_ ,_a,,!C- OFigureQ2-2 Threemoleculesthat illustratethesevenmostcommonfunctionalgroupsinbiology(Problem2-17).1,3-Bisphosphoglycerateand pyruvateareintermediatesin glycolysisandcysteineis an aminoacid.HO-CH1 , 3 - b i s p h o s p h o g l y c e r a t e pyruvareSHcyslerneCalculatethe instantaneousvelocity of a water molecule(molecularmass= 1Bdaltons),a glucosemolecule (molecular mass = lB0 daltons),and a myoglobin molecule (molecular mass = 15,000daltons) at 37"C.
Just for fun, convertthesenumbers into kilometers/hour.Beforeyou do any calculations,try to guesswhether the moleculesare moving ata slow crawl (<1km/hr), an easywalk (5 km/hr), or a recordsettingsprint (40km/hr).2-19 Polymerization of tubulin subunits into microtubules occurs with an increasein the orderlinessof thesubunits (Figure Q2-3).
Yet tubulin polymerization occurswith an increasein entropy (decreasein order). How canthat be?o572-16 During an all-out sprint, musclesmetabolizeglucoseanaerobically,producing a high concentrationoflactic acid,which lowers the pH of the blood and of the cytosol andcontributes to the fatigue sprinters experiencewell beforetheir fuel reservesare exhausted.The main blood bufferagainstpH changesis the bicarbonate/CO2system.PKr=PK2=2.338CO2+CO2 +:+H*H2CO3(gas) (dissolved)=PtrGlo"+ HCO3-JsH*POLYMERIZATION.+FigureQ2-3Polymerizationof tubulinsubunitsintoa microtubule(Problem2-19).Thefatesof onesubunit(shoded)anditsassociated(smallspheres)watermoleculesareshown.+ CO32-To improve their performance,would you advisesprinters tohold their breath or to breatherapidly for a minute immediately before the race?Explain your answer.2-17 The three molecules in Figure Q2-2 contain theseven most common reactive groups in biology.
Mostmoleculesin the cell are built from thesefunctional groups.Indicate and name the functional groups in thesemolecules.2-18 "Diffusion" sounds slow-and over everyday distancesit is-but on the scaleof a cell it is very fast.The average instantaneousvelocity of a particle in solution, that is,the velocity betweencollisions,rs2-2O A 70-kg adult human (154lb) could meet his or herentire energyneedsfor one day by eating3 moles of glucose(5a0g). (We don't recommend this.) Each molecule of glucose generates30 AIP when it is oxidizedto CO2.The concentration of AIP is maintained in cellsat about 2 mM, anda 70-kg adult has about 25 liters of intracellularfluid.
Giventhat the ATPconcentrationremains constantin cells,calculate how many times per day,on average,eachAIP moleculein the body is hydrolyzedand reslmthesized.2-21 Assuming that there are 5 x 1013cells in the humanbody and that AIP is turning over at a rate of 10eAIP perminute in each cell, how many watts is the human bodyconsuming?(A watt is a joule per second,and there are 4.18joules/calorie.) Assume that hydrolysis of AIP yields 12kcal/mole.y = (kTlm)hwhere k= 1.38x 10-16g cmzlKsecz,T = temperaturein K(37'Cis 310 K), m = massin g/molecule.2-22 Does a SnickersrMcandy bar (65 g, 325 kcal) provideenough energyto climb from Zermatt (elevation1660m) tothe top of the Matterhorn (4478m, Figure Q2-4), or mightEND-OF-CHAPTERPROBLEMS105FigureQ2-4 TheMatterhorn(Problem2-22).(CourtesyofZermattTourism.)IAI-^Oferta\l-CHr-CH2-CH2-CH2-CH2-CH2-CH2-C.o,,no'Jrnall-lcH2-Cexcretedcompoundtt'work (D = mass (kg)x g (m/sec2;x height gained (m)where gis accelerationdue to gravity (9.8m/sec2).Onejouleis 1 kg m2lsec2and there are 4.lB kf per kcal.\Alhatassumptionsmade here will greatlyunderestimatehow much candy you need?2-23 At first glance, fermentation of pyruvate to lactateappearsto be an optional add-on reaction to glycolysis.Afterall, could cells growing in the absenceof oxygen not simplydiscard pyruvate as a waste product? In the absenceof fermentation, which products derived from glycolysis wouldaccumulate in cells under anaerobic conditions? Could themetabolism of glucosevia the glycoll'tic pathway continue inthe absenceof oxygenin cells that cannot carry out fermentation?\.A/hyor why not?2-24 In the absenceof oxygen,cellsconsumeglucoseat ahigh, steadyrate.When oxygenis added,glucoseconsumption drops precipitouslyand is then maintained at the lowerrate.\.Vhyis glucoseconsumedat a high rate in the absenceof oxygenand at a low rate in its presence?2-25 The liver provides glucose to the rest of the bodybetweenmeals.It doesso by breakingdown glycogen,forming glucose6-phosphatein the penultimate step.Glucose6phosphateis convertedto glucoseby splitting offthe phosphate (AG' = -3.3 kcal/mole).\Mhydo you supposethe liverremovesthe phosphateby hydrolysis,rather than reversingthe reactionby which glucose6-phosphate(G6P)is formedfrom glucose (glucose + AIP -+ G6P + ADB AG' = -4.0kcal/mole)?By reversingthis reactionthe liver could generate both glucoseand AIP//o\o_phenylacetateo*^^n'\-cur-cH2-cH2-cH2-cH2-cH2-ccomoound ll\atyou need to stop at Hdrnli Hut (3260m) to eat another one?Imagine that you and your gear have a mass of 75 kg, andthat all of your work is done against gravity (that is, you arejust climbing straight up).
Rememberfrom your introductory physicscoursethat-|o-seven-carbonchainoexcreted ri\.com-l' o o u n dl l\4.obenzoatefattyacidto analyzelabelingexperimentFigureQ2-5Theoriginal(Problemof anderivatives2-26).( ) Fedandexcretedoxidationderivativesof anfattyacidchain.(B)Fedandexcretedeven-numberfattyacidchain.odd-numberCan you explainthe reasoningthat led him to concludethattwo-carbon fragments, as opposed to any other number,were removed, and that degradation was from the carboxylicacid end, as opposedto the other end?2-27 Pathways for synthesis of amino acids in microorganismswere worked out in part by cross-feedingexperiments among mutant organisms that were defective forindividual steps in the pathway. Results of cross-feedingexperiments for three mutants defective in the trlptophanTrpIl, and TrpE-are sholrryt in Figureparhway-TrpF,were streaked on a Petri dish andThemutantsQ2-6.allowed to grow briefly in the presence of a very smallamount of tryptophan, producing three pale streaks.Asshown, heavier growth was observed at points where somestreakswere close to other streaks.These spots of heaviergrowth indicate that one mutant can cross-feed(supply anintermediate)to the other one.From the pattern of cross-feedingshor,trrin Figure Q2-6,deducethe order ofthe stepscontrolled by the products ofthe TrpB, TrpD, and TrpE genes.Explain your reasoning.C R O S S - F E E D IRNEGS U L TTrpD-TrpE-2-26 In 1904Franz Knoop performed what was probablythe first successfullabeling experiment to study metabolicpathways.
He fed many different fatty acids labeled with aterminal benzene ring to dogs and analyzedtheir urine forexcreted benzene derivatives.\Alheneverthe fatty acid hadan even number of carbon atoms, phenylacetate wasexcreted (Figure Q2-5A). lVhenever the fatty acid had anodd number of carbon atoms, benzoatewas excreted(Figure Q2-58).From theseexperimentsKnoop deducedthat oxidation offatty acidsto CO2and H2Oinvolved the removalof two-carbon fragments from the carboxylic acid end of the chain.o-eight-carbonchain\t'TrDB'usingFigureQ2-6 Definingthe pathwayfor tryptophansynthesis(Problem2-27).Resultsof a crossexperimentscross-feedingfeedingexperimentamongmutantsdefectivefor stepsin thetryptophan biosyntheticpathway.Darkoreason the Petridish showregionsof cellgrowth.CARBONSKELETONSCarbonhasa unique role in the cell becauseof itsabilityto form strongcovalentbondswith othercarbonatoms.Thuscarbonatomscan ioin to formchains.\/,/\/CCCC""\\_ r,,LLLL/\\/\//\^,.."\/\\,,\/-t\./c---\^,r-or branchedtrees^,,./\also written asC-C--c/,21alsowrittenas\a /\Xwrittenarsor, al-)COVALENTBONDSHYDROCARBONSA covalent bond forms when two atoms come very closetogether and shareone or more of their electrons.In a singlebond one electronfrom eachof the two atoms is shared;ina double bond a total of four electronsare shared.Eachatom forms a fixed number of covalentbonds in adefinedspatialarrangement.For example,carbonforms foursinglebondsarrangedtetrahedrally,whereasnitrogenformsthree singlebondsand oxygenforms two singlebondsarrangedas shown below.Carbonand hydrogencombinetogether to make stablecompounds(or chemicalgroups)calledhydrocarbons.Thesearenonpolar,do not formhydrogenbonds,and aregenerallyinsolublein water.Atomsjoined by twoor more covalentbondscannot rotate freelyaround the bond axis.This restriction is amajor influenceon thethree-dimensionalshaoeof many macromolecules.HrC.CH,ALTERNATINGDOUBLEBONDSThe carbonchain can includedoublebonds.lf theseare on alternatecarbonatoms,the bonding electronsmovewithin the molecule,stabilizingthestructureby a phenomenoncalledreS0nance.HrC.Alternatingdouble bonds in a ringcan generate a very stable structure.CHtHrC.CH,HrC.CH,H,CCH,Hzc.the truth is sbmewhereberweenthese two structures9H,H:Coftenwritten.,Opart of the hydrocarbon"tail"of a fatty acid moleculeC_OCHEMICALGROUPSGROUPSC-N CHEMICALMany biologicalcompoundscontaina carbonbondedto an oxygen.For example,Aminesand amidesare two important examplesofcompoundscontaininga carbonlinkedto a nitrogen.Aminesin water combinewith an H+ ion to becomepositivelycharged.lrHt,H_ ct_/ l(/+u* *_q_n_H*r\nr\HAmidesare formed by combiningan acidand anamine.Unlikeamines,amidesare unchargedin water.An exampleisthe peptide bond that joins amino acidsin a orotein.o-c/The -COOHis calledacarboxylgroup.
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