【正文】 細(xì)胞通訊信號(hào)有四種，內(nèi)分泌信號(hào)、旁分 泌信號(hào)、 和 。s respiratory epithelium. How is this potential pitfall affected by the presence of gap junctions? Because gap junctions allow ions to diffuse freely between adjacent cells, expression of normal CFTR in one epithelial cell would provide a functional Cl– channel to its neighbors. 3 What is the importance of selectively targeting different glucose transporters to the apical and basolateral domains of the plasma membrane of intestinal epithelial cells? What is the role of tight junctions in this process? The correct localization of transporters mediating active transport and facilitated diffusion is necessary for the polarized function of epithelial cells in transferring glucose from the intestinal lumen to the blood supply. Tight junctions prevent the diffusion of these transporters between domains of the plasma membrane, as well as sealing the spaces between cells of the epithelium. 3 The proliferation of thyroid cells is stimulated by hormones that activate a receptor coupled to Gs. How would inhibitors of cAMP phosphodiesterase affect the proliferation of these cells? Inhibition of cAMP phosphodiesterase would result in elevated levels of cAMP, which would stimulate cell proliferation. 3 The epinephrine receptor is coupled to Gs, whereas the acetylcholine receptor (on heart muscle cells) is coupled to Gi. Suppose you were to construct a rebinant molecule containing the extracellular sequences of the epinephrine receptor joined to the cytosolic sequences of the acetylcholine receptor. What effect would epinephrine have on cAMP levels in cells expressing such a rebinant receptor? What would be the effect of acetylcholine? The rebinant molecule would function as an epinephrine receptor coupled to Gi. Epinephrine would therefore inhibit adenylyl cyclase, lowering intracellular cAMP levels. Acetylcholine would have no effect, since it would not bind to the rebinant receptor. 3 Plateletderived growth factor (PDGF) is a dimer of two polypeptide chains. What would be the predicted effect of PDGF monomers on signaling from the PDGF receptor? PDGF monomers would not induce receptor dimerization. Since this is the first critical step in signaling from receptor proteintyrosine kinases, they would be unable to stimulate the PDGF receptor. 3 How would overexpression of protein phosphatase 1 affect the induction of cAMPinducible genes in response to hormone stimulation of appropriate target cells? Would protein phosphatase 1 affect the function of cAMPgated ion channels involved in odorant reception? Protein phosphatase 1 dephosphorylates serine residues that are phosphorylated by protein kinase A. Cyclic AMPinducible genes are activated by CREB, which is phosphorylated by protein kinase A, so overexpression of protein phosphatase 1 would inhibit their induction. However, protein phosphatase 1 would not affect the activity of cAMPgated ligand channels, since these channels are opened directly by cAMP binding rather than by protein phosphorylation. 3 Protein kinase Cα (PKCα) and protein kinase Cε (PKCε) are two different members of the protein kinase C family, which differ in their regulation. PKCα requires both Ca2+ and diacylglycerol for activation, whereas PKCε requires only diacylglycerol. How would hydrolysis of the phospholipids PIP2 and phosphatidylcholine by phospholipase C affect the activities of these different PKC family members? Hydrolysis of PIP2 by phospholipase C yields both diacylglycerol and IP3, which signals the release of Ca2+ from the endoplasmic reticulum. PIP2 hydrolysis can therefore activate both PKCα and PKCε. Hydrolysis of phosphatidylcholine yields diacylglycerol but not IP3。 2 liulang1979 主題 :請(qǐng)教陳斑竹 . 生物膜在特殊的生理?xiàng)l件下，可能出現(xiàn)什么結(jié)構(gòu)？ RE: 六角相和立方相結(jié)構(gòu) . 參見 :趙南明 周海夢(mèng) 主編 生物物理學(xué) 高教出版社 2022. 2 ebblood 主題 :請(qǐng)教斑竹幾個(gè)問(wèn)題 1. 鞘磷脂和酸性磷酸酶的合成部位是細(xì)胞質(zhì)基質(zhì)中還在在內(nèi)質(zhì)網(wǎng)中？ 2. abo 血型到底由什么決定。這是與 Southern blot 完全不同的一種雜交方法，雜交的受體是完整的Genome DNA， Probe 可以是用 DNA 或 RNA。 2 liuguangming 主題 :跨膜區(qū)域 發(fā)表于： 2022 年 11 月 10 日 10： 45 某跨膜蛋白的氨基酸序列已被測(cè)定 ,發(fā)現(xiàn)它除具有 N端信號(hào)肽之外 ,還存在 14 個(gè)疏水性肽段 ,其中 7 段各含25 個(gè)氨基酸殘基 ,3 段各含 16 個(gè)氨基酸殘基 ,4 段各含 10 個(gè)氨基酸殘基 .經(jīng)過(guò)上述分析可知 ,此肽鏈將來(lái)能形成 __________跨膜區(qū)域 . 我認(rèn)為是四段因?yàn)椋缒^(qū)域 a 螺旋 20 個(gè)氨基酸， B 折疊 1012 個(gè)氨基酸），這地方符合的只有 10 個(gè)氨基酸的肽段，但其余一些肽段又形成 什么結(jié)構(gòu)呢 RE: 14 個(gè)疏水性肽段中 : 跨膜區(qū)域 a 螺旋 2030 個(gè)氨基酸， B 折疊 1014 個(gè)氨基酸 ,數(shù)拒來(lái)自 :趙南明 ,周海夢(mèng) 主編 生物物理學(xué) 高教出版社 2022, 因此可能形成 7 段 aHelix,四段 Bsheet strand. 3 段各含 16 個(gè)氨基酸殘基可能形成 3 段 Bsheet strand 剩下的 2 個(gè) AA REDUES 可留在膜外 ,這樣也可保持EMIN,不過(guò)是諸害取其輕 . 實(shí)際情況可能更為復(fù)雜一些比如暴露在親水環(huán)境下的 AA REDUES 可能接上了其他的側(cè)鏈 ,也可能成為外周PROTEIN 的錨定位 點(diǎn) 疏水相互作用 . 我順便回答另外三個(gè)問(wèn)題 (我找不到原貼了 ): PCR 是否可對(duì) DNA 測(cè)序 ? 當(dāng)然可以 ,原理與 SANGER 加減法一樣 ,不過(guò)是酶 SYSTEM 有區(qū)別 . ? 分子病從其原始定義來(lái)看 ,是指由于 GENES 突變引起的可遺傳的病征 ,但是由于構(gòu)象病的發(fā)現(xiàn)使這一概念發(fā)生動(dòng)搖 ,比如克 雅氏綜合征瘋牛病以及羊騷癢征等 ,皆是由于 PROTEINS 發(fā)生翻譯后加工而后形成了新的構(gòu)象 ,并導(dǎo)致淀粉樣沉淀 ,引發(fā)病理后果 .我請(qǐng)教了北大生科院的三位教授 ,其中兩位是博導(dǎo) ,兩位博導(dǎo)皆認(rèn)為分子病包括 構(gòu)象病并不限于遺傳病 ,另一位教授認(rèn)為分子病可能還是特指遺傳病 ,而構(gòu)象病應(yīng)單歸為一類 .我個(gè)人同意兩位博導(dǎo)的意見 . Blotting 是什么技術(shù) ? 學(xué)術(shù)界有人把 Dot blot 稱為 Eastern blot,這樣?xùn)|南西北四種雜交類型就成了一個(gè)完整的體系，也可能是因?yàn)橛辛四衔鞅倍荒軟](méi)有東的緣故吧，如此說(shuō)來(lái)，將 Dot blot 稱為 Eastern blot 就有牽強(qiáng)之嫌。 1以 Na+K+泵為例，說(shuō)明 ―泵 ‖在物質(zhì)穿膜運(yùn)輸中的作用原理及其在細(xì)胞代謝中的意義。 1 構(gòu)的流動(dòng)鑲嵌模型的要點(diǎn)是什么？ 1細(xì)胞的吞排作用和細(xì)胞內(nèi)的膜泡運(yùn)輸?shù)幕具^(guò)程如何？ 1說(shuō)明籠形蛋白（ clathrin）在膜泡運(yùn)輸?shù)幕具^(guò)程如何？ 1物質(zhì)穿膜運(yùn)輸有哪些方式？比較其異同點(diǎn)。 [提示 ]成帽與成斑現(xiàn)象， P82；（ 2）光脫色恢復(fù)技術(shù)。 花剌酸性會(huì)不會(huì)使纖維素的細(xì)胞壁部分水解 暴露拳擊從而顯色 . 25． .blueki 主題 :體細(xì)胞突變的正負(fù)選擇法 一道考試題 ,大家?guī)兔匆幌?. 什么是體細(xì)胞突變體篩選的正負(fù)選擇法 ,并舉例說(shuō)明 . 26． You are studying an enzyme in which an activesite cysteine residue is encoded by the triplet UGU. How would mutating the UGC also encodes cysteine, so this mutation would have no effect on enzyme function. UGC also encodes cysteine, so this mutation would have no effect on enzyme function. UGA, however, is a stop codon, so this mutation would abolish enzyme activity. 2 tarting with DNA from a single sperm, how many copies of a specific gene sequence will be obtained after 10 cyc