ThisworkbookwasdevelopedforusewithModule2oftheInVitroInsightsCellCultureTrainingProgram,developedbyBectonDickinson.Theproblemthatfollowarepresentedinincreasingdegressofcomplexity.Theyweredesignedtoassistinthedevelopmentofskillsnecessarytodeterminethenumberofviablecellsavailableforsubculturingandtheamountofcellsnecessaryfordifferenttypesofexperiments.

Inordertobetterunderstandtheconceptspresentedinthevideo,itisrecommendedthatviewersusetheHemacytometerWorkbookinconjunctionwiththeHemacytometerReferenceGuide.

Belowisadiagramofthetwohemacytometercountinggrids,withcorrespondinglabelsandterminologyusedthroughoutthisworkbook.

SampleProblem#1:

TwoT-75flaskscontaininganchorage-dependentcultureswretrypsinized.Thecellsfrombothflaskswerepooled,andtheresultingcellsuspensionwastransferredtoaconicaltube.Thetotalvolumeofliquid(cellssuspendedinmedia)was20.5mls.Althoughthesecellsareroutinelysplit1:2,anyonehandlingtheseculturesisrequestedtoassesstheviabilityofthepopulation.

0.5mlofthecellsuspensionwasaddedtoatubecontaininganequalvolumeoftrypanblue,thetubewasaqitated,andahemacytometerwasloadedwiththeresultingsuspension.Bothgridswerecoveredwithappropriateammountofcellsuspension,andthefollowingcountswereobtained:

Whatistheaveragenumberofviablecellspersquare?

Whatisthe%viability?

Whatisthenumberofviablecellspermlinthetubecontainingthecellssuspensionavailbleforsubculturing?

Whatisthetotalnumberofcellsavailableforsubculturing?

Sincethesecellsareroutinelysplit1:2,approximatelyhowmanymillionsofcellswillbeusedtoseedeachoffourflasks?

Ifthesecellsneededtobesplit1:4,approximatelyhowmanymillionsofcellswouldbeusetoseedeachoftheeightflasks?

SampleProblem#2:

Usethedatafromsampleproblem#1todeterminewhatvolumeofthepooledcellsuspensioninthetubeneedstobetransformedtoatissueculturedishrequiringonemillioncells.

SampleProblem#3:

Asampleofcellsuspensionisreadyforcountingandviabilityassessment;however,afterloADIngthehemacytometerandmicroscopicallyobservingthecellsonthecountinggrids,itisdeterminedthatthecountingrangesaretoohigh(i.e.therearetoomanycellstocount.)Whatmustbedonetoobtaingoodcountingranges?

Whatvolumeofmediummustbeaddedto0.1mloftheoriginalcellsuspensiontoobtaina1:10dilution?

WhatisnowtheDilutionFactor?

Ifasampleofthis1:10dilutionisaddedtoanequalvolumeoftrypanbluesolution,whatisthefinalDilutionFactor?

Ifthenewcountsarenowintherangeof3to5cellspercountingsquare,aretheystillvalid?Explain.

Whatmustbedonetoobtainvalidcountingranges?

Whatareoptimalcountingranges?

SampleProblem#4:

AconfluentT-75flaskistrypsinized,andthecellsareresuspendedin15mlofcompletemedium.Thenumberofviablecells/mlhavebeendeterminedtobe1.3x10⁶cells/ml.

Whatisthetotalnumberofviablecellsintheflask?

Whatwasthenumberofviablecellsperunitarea(cm²)inoriginal75cm²flask?

HowmanyT-25flaskscanbeseededwiththiscellsuspension,ifthetargetseedingdensityforeachoftheT-25flasksis2.5x10⁴cellspercm²ofarea?

SampleProblem#5:

YouhavejustbeentoldthatsomeoneinthelabneedseightT-25flasksseededthisafternoon.TheyhavetoldyouthatyoucanuseoneT-175flaskoftheirstockculturetoseedtheeightflasks.Afterdiscussingthetypicalseedingdensitiesforthisparticularcelltype,youdeterminethatyouwillneed4.5x10⁵cellsforeachoftheeightflasksthatyouwillbeseeding.

Whatistheminimumnumberofcellsthatyouwillneedtomeettheseedingdensityrequirementfortheseeightflasks?

SampleProblem#6:

AftercountingthecellsfromtheT-175flaskinthesampleproblem#5,youhavedeterminedtheaveragecellcountpercountingsquaretobe31.

Ifyourdilutionfactorwastwo,whatisthenumberofcellsperml?

Ifyouhave20mlsofthecellsuspensionfromtheoriginalflask,willyouhaveenoughcellstoseedeightT-2flasks?

SampleProblem#7:

Astaticsuspensioncultureofnon-adherentcells(e.g.,leukimiacells)growinginasingleT-7flaskistobesubculturedinto5,60mmdishes(e.g.,BDFalcon™353002).Analiquotoftheoriginal15mlcellsuspensionintheoriginalflaskiscountedwiththehemacytometer,andthecellconcentrationisfoundtobe2.5x10⁶cells/ml.Youwishtoseedthe5dishesatadensityof7x10⁵cells/dish,andinordertopromotegoodgasexchangeinthedishesyouhavedecidedtolimitthedepthofthemediuminthedishesto6mm(0.6cm).Theareaofthedishbottomis28cm².(Ref.InVitroInsightsProgramSupplement#103).

Howdoyoudilutetheoriginalcellsuspensiontoachievethisgoal?

Answers

SampleProblem#1

• Average#ofviablecellspersquare=36.1Grid#1:176-5=171Grid#2:195-5=190171+190=#viablecellsper10squares360/10=#viablecellspersquare
• %Viability361viablecells/371totalcells=0.97=97%
• #Viablecellsperml=72.2x10⁴cells/ml36.1x10⁴x2(DilutionFactor)=72.2x10⁴cells/ml
• Total#cellsavailableforsubculturing=Approx.14millioncells72.2x10⁴cells/mlx20ml=72.2x10⁴x20=14.44x10⁶cells(#viablecells/ml)
• #Cellsperflask(foreachof4flasks)=Approx.3.61millioncells14.44x10⁶cells/4=3.61x10⁶cells
• #Cellsperflask(foreachof8flasks)=Approx.1.8millioncells14.44x10⁶cells/8=1.8x10⁶cells

• #mlsneeded=1.4mlsWehave72.2x10⁴cells/mlor0.722x10⁶cells/mlWeneed1.0x10⁶cellsTherefore;wedivide(1.0x10⁶cells)/(0.722x10⁶)=1.385or1.4mls

• Samplemustbedilutedtoobtaingoodcountingrange
• 0.9mlmediummustbeadded
• TotalDilutionFactor=10
• FinalDilutionFactor=20
• The3-5rangeisnotvalid.Therearetoofewcellstodetermineastatisticallysignificantcount.
• Toobtainvalidcountingranges,tryanintermediatedilution.
• Optimalcountingrangesare20-40cellspersquare(200-400cellsper10squares)

• Total#ofcellsavailable=1.95x10⁷1.3x10⁶cells/mlx15ml=1.3x10⁶x15=1.95x10⁷cells
• #ofviablecells/cm²=1.9x10⁷cells/75cm²=2.6x10⁵cellspercm²
• 31T-25flaskscanbeseededwiththiscellsuspension.2.5x10⁴cells/cm²x25cm²/T-25flask=6.25x10⁵cellsperT-25flask1.95x10⁷cells/0.0625x10⁷cells/flask=31.2=31flasks

• 3.6x10⁶cellsareneededtoseedeightflasks.4.5x10⁵cells/flaskx8flasks=36x10⁵or3.6x10⁶cellstotal

• 6.2x10⁵cellsperml31x10⁴cells/mlx2(dilutionfactor)=62x10⁴=6.2x10⁵cells/ml
• Yes-weneed3.6x10⁶cells(Ref.SampleProblem#5)andhave12.4x10⁶cellsWehave62x10⁴cells/mlx20ml=12.4x10⁶cellstotal

• First,determinethevolumeofmediumperdish:0.6cmx28.0cm²=16.8cm³=16.8ml
• Next,sinceyouneed7x10⁵cellsperdishin16.8mls-youwillneedacellsuspensionconcentrationof4.17x10⁴cells/ml(7x10⁵cell/16.8mls)
• Sincethereare5dishes,youwillneed5x16.8or84mlsofthecellsuspensionattheaboveconcentration.
• Next,youneedtoadd"X"mlsoftheoriginalcellsuspension(2.5x10⁶cells/ml)X=84ml(4.17x10⁴cells/ml)(2.5x10⁶cells/ml)X=3.5028x10⁶cellsX=(3.5028x10⁶/2.5x10⁶)mlsX=1.401mlsTherefore;1.4mlsoforiginalcellsuspensionmustbeaddedto82.6mlsofmediumtomake84mls.Note:Usuallyforpracticalreasons,thevolumeoffinalcellsuspensioncanbeadjustedtoarounded-offnumber(i.e.,90or100mls)

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