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ID:36901931
大小:7.21 MB
页数:42页
时间:2019-05-10
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1、AdvancedBioanalyticalChemistryFall,2016朱志卢嘉锡542zhuzhi@xmu.edu.cn第六节课生物分子的化学标记和探针技术2016年11月3日生物分子的化学标记和探针技术Someofthecommonbioconjugatedesignsoftenusedforlifescienceapplicationsinclude:(A)streptavidin–enzymeconjugate(B)animmobilizedaffinityligandonaparticle(C)anoligomolecularbeaconprobecontainingt
2、wofluorescentlabelsorafluorandaquencherateachend(D)fluorescentlylabeledstreptavidin(E)anaffinityligandattachedtoasurface(F)abiotinylatedenzyme(G)anantibody–enzymeconjugate(H)afluorescentlylabeledantibody(I)abiotinylatedantibody(J)abiotinylatedoligoprobe(K)anantibody–drugconjugate(L)agadoliniumch
3、elate-modifieddendrimercontainingfolatemoleculesfortargeting.蛋白质和核酸的荧光标记LocalizationofMoleculeswithFluorescenceMicroscopyImagefromMolecularProbes(Invitrogen):www.probes.invitrogen.comFluorescenceApplicationsinBiologyFluorescenceTechniquesCanProvidetheFollowingInformation:Simplefluorophores:-Loca
4、lizationofbiomolecules-Assessmentofbinding-CalculationofdiffusioncoefficientsMorecomplexsystems:-Reportingofgeneactivation-Detectionofphysiologicalchanges-MeasurementofdistancesandconformationalchangesExamplesofFluorescentMoleculesFluorescein(Fluorescent)lex=494,lem=518Phenolphthalein(Non-fluore
5、scent)Pyrenelex=345,lem=378Coumarin343lex=454,lem=495BODIPY493lex=500,lem=506TAMRAlex=555,lem=580TexasRedlex=595,lem=6157-Hydroxycoumarinlex=385,lem=445Cy5lex=643,lem=667Fluorescenceofnativeproteinsresidues:Tryptophan:lex=290,lem=330-350,dependingonenvironmentDNAandRNAgenerallyexhibitnofluoresce
6、nce.Partialenergydiagramforaphotoluminescentsystem.荧光的产生Spectralobservablesforfluorescencesensing.Fromlefttoright:Intensity,intensityratio,anisotropy,time-domainlifetime,andphase-modulationlifetime.蛋白质的选择性标记NewFunctionalityChromophoresAffinityLabelsSpinLabelsCatalystsCrosslinkersPolymersEnzymesN
7、anocrystalsMRIContrastAgentsMaterialSurfacesChemicalReactionRadiolabelsCommonExamplesofProteinBioconjugates氨基的反应(赖氨酸侧链与N末端氨基)理想反应条件:pH8~9质子化率的计算:lg([质子受体]/[质子供体])=pH-pKLysε-+NH3pK=10.5求ε-+NH3在pH9.5及pH=11时的质子化%。解:ε-+NH3=ε-NH
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