opg及trail在骨肉瘤中的表达及其意义.docx

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1、结论：l.OPG、TRAIL可能分别在骨肉瘤细胞存活、凋亡机制中起作用；2.在骨肉瘤中OPG的高表达可能抑制了TRAIL的表达，同时可能与大量肿瘤骨的形成有关；3.OPG、TRAIL可能与骨肉瘤侵袭转移有关。关键词：OPG；TRAIL；骨肉瘤；骨软骨瘤；免疫组织化学TheexpressionanditsmeaningofOPGandTRAILinOsteosarcomaAbstractObjective:ThisarticlemainlystudiedtheexpressionofOPGandT

2、RAILinosteosarcomaandosteochondromabyImmunohistochemicalmethods,andpreliminarydiscussionedthepossiblesignificanceandrelevanceofthemintheoccurrenceanddevelopmentofosteosarcoma.Methods:TheexpressionofOPGandTRAILin50casesofosteosarcomaand10casesofosteoc

3、hondromawasdetectedbyImmunohistochemicalSABCmethod,then,SPSS19.0wasusedtoanalyzetheresults;andthus,wecanpreliminarilyexploreitssignificanceandanalyzethecorrelationofthem.Results:LOPGmainlyexpressedinosteosarcomacellcytoplasm,Therewere36caseswereposit

4、ivein50patientswithosteosarcoma,thepositiveratewas72.00%;whileonly2caseswerepositivein10casesofosteochondroma,thepositiveratewas20.00%.Whichhavesignificantdifferencebetweenthetwogroups(P<0.05).TheexpressionofOPGindifferentpathologicaltissuetypingwasn

5、osignificantdifference(P>0.05),However,theexpressionofOPGinthedifferentEnneckingstageswasobviousdifference(P<0.05).2.TRAILmainlyexpressedinthemembraneofosteosarcomacell,Therewere16caseswerepositivein50patientswithosteosarcoma,thepositiveratewas32.00%

6、;while7caseswerepositivein10casesofosteochondroma.thepositiveratewas70.00%.Whichhavesignificantdifferencebetweenthetwogroups(P<0.05).TheexpressionofTRAILwasnosignificantdifferenceindifferentpathologicaltissuetypingandthedifferentEnneckinginstallments

7、inosteosarcoma(P>0.05).3・ThepositiverateofOPGinthetransfergroupofosteosarcomawas93.30%,andinthenon-transfergroupwas58.82%.theexpressionofOPG-positiverateinthetwogroupswassignificantlydifferent(P<0.05).While,thepositiveimmunostainingrateofTRAILinthetr

8、ansfergroupofosteosarcomawas17.39%,andintheon-transfergroupwas31.25%.theexpressionofTRAIL-positiverateinthetwogroupswasdramaticallydifferent(P<0.05).4、OPGandTRAILexpressioninosteosarcomawasnegativelyrelatedandstatisticallysignificant(r=-0.432,P=0.002