Phosphatidylinositol 3-kinase p85{alpha} subunit-dependent interact...
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):8001-8.Phosphatidylinositol 3-kinase p85{alpha} subunit-dependent interaction with BCR/ABL-related fusion tyrosine kinases: molecular mechanisms and biological consequences.1, , , , , .1Molecular Carcinogenesis Section, Center for Biotechnology, College of Science and Technology, Temple University, BLSB 419, 1900 N. 12th Street, Philadelphia, PA 19122, USA.AbstractThe p85alpha subunit of phosphatidylinositol 3-kinase (PI-3k) forms a complex with a protein network associated with oncogenic fusion tyrosine kinases (FTKs) such as BCR/ABL, TEL/ABL, TEL/JAK2, TEL/PDGFbetaR, and NPM/ALK, resulting in constitutive activation of the p110 catalytic subunit of PI-3k. Introduction of point mutations in the N-terminal and C-terminal SH2 domain and SH3 domain of p85alpha, which disrupt their ability to bind phosphotyrosine and proline-rich motifs, respectively, abrogated their interaction with the BCR/ABL protein network. The p85alpha mutant protein (p85mut) bearing these mutations was unable to interact with BCR/ABL and other FTKs, while its binding to the p110alpha catalytic subunit of PI-3k was intact. In addition, binding of Shc, c-Cbl, and Gab2, but not Crk-L, to p85mut was abrogated. p85mut diminished BCR/ABL-dependent activation of PI-3k and Akt kinase, the downstream effector of PI-3k. This effect was associated with the inhibition of BCR/ABL-dependent growth of the hematopoietic cell line and murine bone marrow cells. Interestingly, the addition of interleukin-3 (IL-3) rescued BCR/ABL-transformed cells from the inhibitory effect of p85mut. SCID mice injected with BCR/ABL-positive hematopoietic cells expressing p85mut survived longer than the animals inoculated with BCR/ABL-transformed counterparts. In conclusion, we have identified the domains of p85alpha responsible for the interaction with the FTK protein network and transduction of leukemogenic signaling.PMID:
[PubMed - indexed for MEDLINE] PMCID: PMC1234343 Interaction of the p85α fragments with BCR/ABL. GST fusion proteins containing the indicated p85α fragments (A) were used for the pull-down assay along with the total cell lysates isolated from p210BCR/ABL-positive 32Dcl3 cells (B) or with p210BCR/ABL produced by IVTT (C). The reactions were resolved by SDS-PAGE and analyzed by Western blotting using anti-ABL (upper box) and anti-GST (lower box) primary antibodies to detect BCR/ABL and GST proteins, respectively.Mol Cell Biol. ):.Mutations in the p85α SH2 domains disrupt the interaction with p210BCR/ABL. GST fusion proteins containing the indicated p85α SH2 domains and their mutants (A) were used for the pull-down assay along with the total cell lysates isolated from p210BCR/ABL-positive 32Dcl3 cells (B). The reactions were resolved as described in the Fig. 1 legend.Mol Cell Biol. ):.Mutations in the p85α SH3 domain disrupt the interaction with p210BCR/ABL. GST fusion proteins containing the p85α SH3 domain and the indicated fragments or mutants (A) were used for the pull-down assay along with the total cell lysates isolated from p210BCR/ABL-positive 32Dcl3 cells (B). The reactions were resolved as described in the Fig. 1 legend.Mol Cell Biol. ):.Mutations in the p85α full-length protein affect the interaction with the p210BCR/ABL protein network. GST fusion proteins containing the p85α wild type (wt) or indicated mutants (A) were used for the pull-down assay along with the total cell lysates isolated from p210BCR/ABL-positive 32Dcl3 cells. (B) The reactions were resolved by SDS-PAGE and analyzed by Western blotting using anti-ABL (top), anti-p110α (middle), and anti-GST (bottom) primary antibodies to detect BCR/ABL, p110α, and GST proteins, respectively. (C) p210BCR/ABL and Flag-p85wt, Flag-p85mut, or empty plasmid were expressed in 293T cells and detected in total cell lysates (Lysate). The presence of BCR/ABL and Flag-p85 proteins in anti-Flag immunoprecipitates was also determined (IP:anti-Flag). (D) GST fusion proteins containing the p85α wild type (wt) or the W55F + T72I + R358L + R649L mutant (mut) were used for the pull-down assay along with the total cell lysates isolated from p210BCR/ABL-positive 32Dcl3 cells. The presence of Crk-L, c-Cbl, Gab2, and Shc in the reactions was determined by Western analysis. (E) GST fusion proteins containing p85wt or p85mut were used for the pull-down assay along with the total cell lysates isolated from BaF3 cells transformed by the indicated FTKs. The reactions were resolved by SDS-PAGE and analyzed by Western blotting using the kinase-specific antibody (upper boxes) and anti-GST antibody (bottom box) to detect the indicated FTKs and GST proteins.Mol Cell Biol. ):.cSH2 and SH3 domains of p85α interact with Gab2. GST fusion proteins containing p85α or the indicated p85α fragments were used for the pull-down assay along with the total cell lysates isolated from (A) p210BCR/ABL-positive 32Dcl3 cells, (B) 32Dcl3 cells expressing the p190BCR/ABL wild type (wt) or indicated mutants, and (C) GFP+ p210BCR/ABL-positive Gab2+/+ and Gab2-/- cells expressing similar levels of p210BCR/ABL kinase (Lysates). The reactions were examined by Western analysis to detect Gab2 (A, upper box), BCR/ABL (B, C, upper boxes), and GST (lower boxes) proteins.Mol Cell Biol. ):.The p85α mutant inhibits BCR/ABL-mediated leukemogenesis. (A) The p85α mutant inhibits PI-3k activation and phosphorylation of Akt in vivo. 32Dcl3 parental and BCR/ABL-positive counterparts were infected with p85mutFlag-IRES-GFP (p85mut) or empty IRES-GFP retroviral construct (e). PI-3k enzymatic activity was measured in the kinase assay in vitro using phosphatidylinositol as a substrate. Akt activation was assessed by Western analysis detecting the phosphorylated form of A total Akt protein and actin were detected as loading controls. Anti-Flag antibody was used to confirm the expression of p85mut-Flag protein. (B, C) BCR/ABL-expressing 32Dcl3 cells were infected with p85αmutFlag-IRES-GFP (black bars) or empty IRES-GFP (gray bars) retroviral construct. The proliferation potential of GFP+ cells was analyzed by trypan exclusion in the presence of bovine serum albumin, FBS, or FBS plus IL-3 (B) or clonogenic assay (C). (D) Mouse mononuclear bone marrow cells were coinfected with pSRα-BCR/ABL and p85mutFlag-IRES-GFP retroviral construct or IRES-GFP empty construct. The clonogenic ability of GFP+ cells was evaluated in the absence of growth factors. Results in B, C, and D represent means ± standard deviations from three to four independent experiments (P value calculated by Student's t test). % of clonogenic cells represents the percentage of cells plated that formed colonies. (E) Survival of SCID mice injected with cells transfected with BCR/ABL and empty plasmid (BCR/ABL+E), or BCR/ABL and p85mut (BCR/ABL+p85mut). Survival of the animals was monitored weekly.Mol Cell Biol. ):.Publication TypesMeSH TermsSubstancesGrant SupportFull Text SourcesOther Literature SourcesMedicalMiscellaneous
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External link. Please review our .\_restore{2AA3FABC-CC93-481B-B222-4AA754EAA0BF}\RP19\A0002962.dll - Win32/Delf.NKI 木马的变种_百度知道
\_restore{2AA3FABC-CC93-481B-B222-4AA754EAA0BF}\RP19\A0002962.dll - Win32/Delf.NKI 木马的变种
是不是病毒???
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使用清理专家粉碎下面的文件夹 \_restore{2AA3FABC-CC93-481B-B222-4AA754EAA0BF}\RP19
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出门在外也不愁已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*._百度知道
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Sn = n -5an -85
= n- 5[1-15.(5/6)^(n-1) ] -85
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等比数列的求法吧。设Tn=an-1.然后求的Tn的前n项和再加上n就好了。
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- flag=艾瑞-房产 - ->已知数列{an}的前n项和为Sn，且Sn=n-5an-85，n∈N*．（1）证明：{an-1}是等比数列；（2）求数列{Sn}的通项公式．请指出n为何值时，Sn取得最小值，并说明理由．考点：；．专题：；；；．分析：（1）利用题设中所给的恒等式进行变换，先得到Sn+1=n+1-5an+1-85，n∈N*．与已知中Sn=n-5an-85，n∈N*．作差整理即可得到证明；（2）由（1）Sn+1=n+1-5an+1-85，n∈N*．变形可得Sn+1-（n+1）+90=（Sn-n+90），此是一等比数列，求出它的通项则可以得到数列{Sn}的通项公式，对数列的相邻两项作差，研究其单调性即可得出Sn取得最小值时的n是1解答：解：（1）∵Sn=n-5an-85，n∈N*．∴Sn+1=n+1-5an+1-85，n∈N*．两式作差得an+1=1-5an+1+5an，即6（an+1-1）=5（an-1），即（an+1-1）=（an-1），n∈N*．故{an-1}是等比数列（2）由（1）Sn+1=n+1-5an+1-85，n∈N*．得Sn+1=n+1-5（Sn+1-Sn）-85，n∈N*．得6Sn+1=n+5Sn-84，即6[Sn+1-（n+1）]=5（Sn-n）-90，即Sn+1-（n+1）=（Sn-n）-15整理得Sn+1-（n+1）+90=（Sn-n+90）故{Sn-n+90}是一个等比数列，其公比为，由于a1=1-5a1-85，得a1=-14故{Sn-n+90}的首项为-14-1+90=75故Sn-n+90=75×n-1，即Sn=n-90+75×n-1，由于Sn+1-Sn=1-×n-1，令Sn+1-Sn＞0，对n赋值验证知n＞15时成立，即Sn其最小值是S15点评：本题考查等比关系的确定，求解本题的关键是根据题设中所给的恒成立的方程灵活变形得出形式为公比的形式，故此类题虽然比较抽象，但也有其规律可循，即变形的目标相对确定，解对本题要注意对数列的函数的特性的研究方法即对相邻两项作差确定其单调性，从而求了最小值，此方法不易引起初学者注意，切记．声明：本试题解析著作权属菁优网所有，未经书面同意，不得复制发布。答题：★☆☆☆☆推荐试卷
解析质量好解析质量中解析质量差