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291:62,2004.With permissionAmerican.)from ScientificAnimalsControlTotalCellMassby UnknownMechanismsThe size of an animal or one of its organs depends largely on the number andsize of the cells it contains-that is, on total cell mass. Remarkably,animals cansomehow assessthe total cell mass in a tissue or organ and regulate it: in manycircumstances,for example, if cell size is experimentally increased or decreasedin an organ, cell numbers adjust to maintain a normal organ size.This has beenmost dramatically illustrated by experiments in salamanders,in which cell sizewas manipulated by altering cell ploidy (in all organisms, the size of a cell is proportional to its ploidy, or genome content).
Salamandersof different ploidies arethe same size but have different numbers of cells. Individual cells in a pentaploid salamander are about five times the volume of those in a haploid salamander, and in each organ the pentaploids have only one-fifth as many cells astheir haploid cousins, so that the organs are about the same size in the two animals (Figure L7-7O andFigure 17-71). Evidently, in this case (and in many others) the size of organs and organisms depends on mechanisms that can somehow measure total cell mass.
How animals measure and adjust total massremains a mystery, however.The development of limbs and organs of specific size and shape depends oncomplex positional controls, as well as on local concentrations of extracellularsignal proteins that stimulate or inhibit cell growth, division, and survival. As wediscuss in Chapter 22, we now know many of the genes that help pattern theseprocessesin the embryo. A great deal remains to be learned, however, about howthese genes regulate cell growth, division, survival, and differentiation to generate a complex organism.The controls that govern these processesin an adult body are also poorlyunderstood.
\A/hen a skin wound heals in a vertebrate, for example, about adozen cell types, ranging from fibroblasts to Schwann cells, must be regeneratedin appropriate numbers, sizes, and positions to reconstruct the lost tissue. TheHAPLOIDDIPLOIDPENTAPLOID11chromosomes22 chromosomes55chromosomesFigure17-70 Sectionsof kidneytubulesfrom salamanderlarvaeof differentploidies,In all organisms,from bacteriatohumans,cellsizeis proportionalto ploidy.for example,havePentaploidsalamanders,cellsthat aremuch largerthan thoseofTheanimalsandhaoloidsalamanders.their individualorgans,however,arethesamesizebecauseeachtissuein thepentaploidanimalcontainsfewercells.Thisindicatesthat the sizeof an organismor organis not controlledsimplybycountingcelldivisionsor cell numbers;totalcell massmustsomehowbe(Adaptedfrom G.
Fankhauser,inregulated.Analysisof Development[8.H.Willier,P.A.Weiss,and V. Hamburger,eds.l,1955.)pp. 126-150.Philadelphia:Saunders,1'112 Chapter17:TheCellCycleFigure17-7'lThe hindbrainin a haploidand in a tetraploidsalamander.(A)Thislight micrographshowsa crosssectionof the hindbrainof a(B)A correspondinghaploidsalamander.crosssectionof the hindbrainof atetraploidsalamander,revealinghow reducedcellnumberscompensatefor increasedcellsize,so that the overallsizeof the hindbrainisthe samein the two animals.(FromG. Fankhauser,lnt.Rev.Cytol.1:165-193,1952.With permissionfrom Elsevier.)mechanisms that control cell growth and proliferation in tissues are likewisecentral to understanding cancer, a disease in which the controls go wrong, asdiscussedin Chapter 20.Su m m a r yIn multicellular animals, cell size,cell diuision,and cell death are carefullycontrolledto ensurethat the organismand its organsachieueand maintain an appropriatesize.Mitogens stimulate the rate of cell diuision by remouing intracellular molecularbrakes that restrain cell-cycleprogressionin Gy Growth factors promote cell growth(an increasein cell mass)by stimulating the synthesisand inhibiting the degradationof macromolecules.
For proliferating cells to maintain a constant cell size, theyemploy multiple mechanismsto ensurethat cell growth is coordinatedwith cetldiuision.Animals maintain the normal sizeof their tissuesand organsby adjusting cellsize to compensatefor changesin cell number, or uice uersa.The mechanismsthatmake this possibleare not known.PROBLEMSWhichstatementsare true?Explainwhy or why not.'17-'lSince there are about 1013cells in an adult human,and about 1010cells die and are replaced each day, webecome new people everythree years.(B)1 0 0p mlike to isolate the wild-type gene that correspondsto thedefective gene in your Cdc mutant.
How might you isolatethe wild-type gene using a plasmid-basedDNA library prepared from wild-t1pe yeastcells?17-7 Many cell-cycle genes from human cells functionperfectly well when expressedin yeast cells. \A/try do yousuppose that is considered remarkable?After all, manyhuman genes encoding enzymes for metabolic reactionsalso function in yeast,and no one thinks that is remarkable.17-9 You have isolateda temperature-sensitivemutant ofbudding yeast.It proliferateswell at25"C, but at 35'C all thecells develop a large bud and then halt their progressionthrough the cell cycle.The characteristicmorphology of thecells at the time they stop cycling is knor,tn as the landmarkmorphology.It is very difficult to obtain s1'nchronouscultures of thisyeast, but you would like to know exactly where in the cellcycle the temperature-sensitivegene product must function-its executionpoint, in the terminology of the field-inorder for the cell to complete the cycle.A cleverfriend, whohas a good microscopewith a heatedstageand a video camera, suggeststhat you take movies of a field of cells as theyexperiencethe temperature increase,and follow the morphology of the cells as they stop cycling.
Sincethe cells donot move much, it is relatively simple to study individualcells.To make senseof what you see,you arrangea circle ofpictures of cellsat the start of the experimentin order of thesizeof their daughterbuds.You then find the correspondingpictures of those same cells6 hours later,when growth anddivision has completely stopped. The results with yourmutant are shor.trnin Figure Qf 7-f .A. Indicate on the diagramin FigureQ17-l where the execution point for your mutant lies.B" Doesthe executionpoint correspondto the time at whichthe cell cycle is arrestedin your mutant? How can you tell?17-8 You have isolated anew Cdcmutant of buddingyeastthat forms coloniesat 25"Cbut not at 37.C.You would now17- 10 The yeastcohesinsubunit Scc1,which is essentialforsister-chromatid pairine, can be artificiallv reeulated for17-2 The regulation of cyclin-Cdk complexes dependsentirely on phosphorylationand dephosphorylation.17-3 In order for proliferatingcellsto maintain a relativelyconstant size, the length of the cell cycle must match thetime it takesfor the cell to double in size.'17-4\A4rileother proteins come and go during the cellcycle,the proteins of the origin recognitioncomplex remainbound to the DNA throughout.17-S Chromosomesare positioned on the metaphaseplate by equal and oppositeforcesthat pull them toward thetwo poles of the spindle.17-6 If we could turn on telomeraseactivity in all our cells,we could prevent aging.Discussthe following problems.1113END-OF-CHAPTERPROBLEMSFigureQ17-1 Time-lapsephotographyof a temperaturesensitivemutant of yeast(Problem17-9).Cellson theinnerring are arrangedin orderof their bud size,whichcorrespondsto their positioninthe cellcycle.After6 hoursat37"C,they havegivenrisetothe cellsshownon the outerrlng.
No further growth ordivisionoccurs.tfbuds i z ea t t i m e o ftemperatureshift/+\expression at any point in the cell cycle. If expression isturned on at the beginning of S phase,all the cells divide satisfactorily and survive. By contrast, if Sccl expression isturned on only after S phase is completed, the cells fail todivide and they die, even though Sccl accumulates in thenucleus and interacts efficiently with chromosomes.'Why doyou supposethat cohesin must be present during S phaseforcells to divide normally?17-1'l If cohesins join sister chromatids all along theirlength, how is it possible for condensins to generatemitoticchromosomes such as that shornmin Figure Ql7-2, whichclearly showsthe two sister chromatids as separatedomains?FigureQ17-2 Ascanningelectronmicrographof afullycondensedmitoticfromchromosomevertebratecells(Problem17-11).(CourtesyofTerryD.Allen.)1*.17-12 High dosesof caffeineinterfere with the DNA replication checkpoint mechanism in mammalian cells.
\A/hythen do you supposethe SurgeonGeneralhas not yet issuedan appropriate warning to heavy coffee and cola drinkers?Atypical cup of coffee (150 mL) contains 100 mg of caffeine(196gi mole). How many cups of coffeewould you have todrink to reach the dose (10 mM) required to interfere withthe DNA replication checkpoint mechanism? (A tlpicaladult contains about 40 liters of water.)stagesofof a singlecellat differentFigureQ17-3Lightmicrographs(Courtesyof ConlyL.Rieder.)17-13).M phase(ProblemCentrosomeswere used to initiate microtubule growth, andthen chromosomeswere added.
The chromosomesboundto the free ends of the microtubules, as illustrated in FigureQl7-4.The complexeswere then diluted to verylowtubulinconcentration (well below the critical concentration formicrotubule assembly)and examined again (Figure Ql7-4).As is evident, only the kinetochore microtubules were stableto dilution.A. Why do you think kinetochore microtubules are stable?B. How would you explain the disappearanceof the astralmicrotubules after dilution? Do they detach from the centrosome, depoll'rnerize from an end, or disintegrate alongtheir length at random?C.
How would a time course after dilution help to distinguish among thesepossiblemechanismsfor disappearanceof the astral microtubules?17-16 'v\hat are the two distinct cytoskeletalmachinesthatare assembledto carry out the mechanical processesofmitosis and cytokinesis in animal cells?17-17 How do mitogens, growth factors, and survival factors differ from one another?17-13 A living cell from the lung epithelium of a newt isshown at different stagesin M phase in Figure Ql7-3.
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