抗体和T细胞受体 The conserved sequences present between the different coding regions within the heavy and light chain loci of immunoglobulin genes that are critical for recombination are called: a.J segments. b.recombination signal sequences (RSS). c.V segments. d.D segments. e.intervening DNA sequences. Somatic recombination of immunoglobulin genes is catalyzed by: a.RAG recombinases. b.terminal deoxynucleotidyl transferase (TdT). c.DNA polymerase. d.exonucleases. e.DNA ligase 4. Terminal deoxynucleotidyl transferase (TdT) catalyzes which of the following reactions? a.cleavage of a single strand of the DNA between the RSS and coding region b.addition of nucleotides in a template-independent manner to the ends of cleaved hairpin sequences c.All of the above d.formation of an asymmetric cleavage in the hairpin structure, leaving a palindromic overhang e.formation of a covalently closed hairpin structure from the cleaved DNA coding regions Without the exonuclease activity that occurs during the final steps of the immunoglobulin gene recombination reaction, what would be expected? a.The unpaired nucleotides between two complementary N region sequences would not be removed. b.The sequence gaps between two paired complementary N regions would not be filled in with nucleotides. c.The N regions would not be added to the cleaved ends of the opened hairpins. d.Cleavage between the coding segment and the RSS would not occur. e.The binding between complementary N region sequences would be prevented. Which factor allows for adaptive immune responses against virtually limitless arrays of antigens? a.genetic diversity in antigens b.genetic diversity in pathogens 免疫球蛋白基因重链和轻链基因座内不同编码区之间的保守序列称为： a、 J段。 b、 重组信号序列（RSS）。 c、 V段。 d、 d段。 e、 干涉DNA序列。 免疫球蛋白基因的体细胞重组由以下催化： a、 RAG重组酶。 b、 末端脱氧核苷酸转移酶（TdT）。 c、 DNA聚合酶。 d、 核酸外切酶。 e、 DNA连接酶4。 末端脱氧核苷酸转移酶(TDT)催化下列哪些反应？选择一项： a.切割rss和编码区之间的单链dna； b.以不依赖于模板的方式将核苷酸添加到切割的发夹序列的末端； c.上述所有 d.在发夹结构中形成不对称切割，留下回文突出 e.从切割的dna编码区形成共价闭合的发夹结构 如果没有在免疫球蛋白基因重组反应的最后步骤中发生的核酸外切酶活性，会发生什么呢？选择一项： a.不会去除两个互补N区序列之间的未配对核苷酸。 b.两对互补N区之间的序列间隙不会被核苷酸填充。 c.N个区域不会添加到打开的发夹的解理端。 d.编码片段和RSS之间不会发生断裂。 e.将防止互补N区序列之间的结合。 哪种因子允许针对几乎无限的抗原阵列进行适应性免疫反应？选择一项： A.抗原的遗传多样性 B.病原体的遗传多样性 c.repeating domains that make up immunoglobulin molecules C.组成免疫球蛋白分子的重复结构域 d.somatic recombination in antigen receptors on B and T D.B和T淋巴细胞上抗原受体的体细胞重组 lymphocytes E.T细胞对宿主抗体的识别 e.T-cell recognition of antibodies in the host Which cell types recognize short snippets of protein antigens presented by MHC? a.B cell b.neutrophils c.T cells d.macrophages e.leukocytes Which term describes any material that can evoke an immune response? a.macrophages b.antibodies c.vaccines d.phagocytes e.antigens What is the significance of the RSS sequences in the mechanism of rearrangement of immunoglobulin gene segments? a.The different lengths of the two types of RSS spacers prevent the joining of two similar segments (for example: D to D). b.All of the answers are correct. c.The boundaries between the folded RSS sequences and the coding sequences are the cleavage sites for the RAG1/RAG2 recombinases. d.RSS sequences are conserved sequences at the 3′ end of V segments and 5′ end of J or D segments that can line up and fold to bring the segments close together for joining. e.None of the answers is correct. Which factor is required for signal transduction by a complete T-cell receptor? a.lack of TCR gene rearrangements b.lack of ITAM domains c.CD3 complex d.V and J segments e.B-cell clonal expansion Which statement is TRUE? a.All of the statements are true. b.Leukocytes circulate throughout the body and take up residence in tissues and lymph nodes. c.Innate immunity provides a second line of defense after mechanical and chemical barriers are crossed. d.Mechanical and chemical boundaries form a first layer of 哪些细胞能识别MHC提供的蛋白质抗原片段？ a、 B细胞 b、 中性粒细胞 c、 T细胞 d、 巨噬细胞 e、 白细胞 哪个术语描述任何能引起免疫反应的物质？ a、 巨噬细胞 b、 抗体 c、 疫苗 d、 吞噬细胞 e、 抗原 RSS序列在免疫球蛋白基因片段重排机制中的意义是什么？ A.两种类型的RSS间隔物的不同长度防止了两个相似片段(例如：D到D)的连接。 B.所有的答案都是正确的。 C.折叠的RSS序列和编码序列之间的边界是RAG1/RAG2重组酶的切割位点。 D.RSS序列是位于V片段的3‘端和J或D片段的5’端的保守序列，其可以对齐和折叠以将片段紧密地结合在一起。 E.答案没有一个是正确的。 一个完整的t细胞受体的信号转导需要哪种因素参与? a、 TCR基因重排的缺乏 b、 ITAM域的缺乏 c、 CD3复合体 d、 V和J段 e、 B细胞克隆扩增 哪个说法是正确的？ a、 所有的陈述都是真的。 b、 白细胞在全身循环，并在组织和淋巴结中定居。 c、 机械和化学屏障被跨越后，先天免疫提供了第二道防线 d、 机械和化学边界构成了抵御病原体的第一层屏障。 e、 病原体通过不同的途径进入人体，并在不同的部位复defense against pathogens. 制。 e.Pathogens enter the body through different routes and replicate at different sites. The immune system can recognize an estimated _____ different molecules. a.10,000 to 20,000 b.1024 or more c.500 x 10 or more d.10 or more e.1 million to 2 million Which is NOT part of the human MHC (HLA complex)? a.B b.A c.DR d.DQ e.2D What is the order of genes in the immunoglobulin heavy-chain constant region? a.μ, δ, γ 3, γ 1, γ 2, ε, α b.α, γ 1, γ 2, γ 3, δ, ε, μ c.γ 1, γ 2, γ 3, δ, ε, μ, α d.α, μ, δ, ε, γ 1, γ 2, γ 3 e.δ, ε, μ, α, γ 1, γ 2, γ _____ destroy virus-infected host cells. a.Helper T cells b.B cells c.Cytotoxic T cells d.Both B cells and cytotoxic T cells e.Both helper T cells and B cells Which statement is FALSE? a.Cytotoxic T cells are guided in the recognition of their targets by MHC molecules. b.B cells do not undergo final differentiation into antibody-secreting plasma cells without assistance from another subset of T cells, helper T cells. c.The constitutive expression of class II MHC molecules is confined to T cells only. d.Cytotoxic T cells recognize and kill the infected targets via the expressed class I MHC molecules that display virus-derived antigen. e.Helper T cells express on their surface CD4 glycoprotein marker and use class II MHC molecules as restriction elements. 69 免疫系统可以识别出 种不同的分子。 a、 10000到20000 b、 1024或更多 c、 500 x 10或以上 d、 10或更多 e、 100万到200万 哪一个不是人类MHC（HLA复合体）的一部分 a.B b.A c.DR d.DQ e.2D 免疫球蛋白重链恒定区的基因顺序是什么？ a.μ, δ, γ 3, γ 1, γ 2, ε, α b.α, γ 1, γ 2, γ 3, δ, ε, μ c.γ 1, γ 2, γ 3, δ, ε, μ, α d.α, μ, δ, ε, γ 1, γ 2, γ 3 e.δ, ε, μ, α, γ 1, γ 2, γ _____破坏病毒感染的宿主细胞。 a、 辅助性T细胞 b、 b细胞 c、 细胞毒性T细胞 d、 B细胞和细胞毒性T细胞 e、 辅助性T细胞和B细胞 哪种说法是错误的？选择一项： A.细胞毒T细胞被MHC分子引导识别其靶细胞。 B.B细胞在没有另一个T细胞亚群--辅助T细胞的帮助下，不会最终分化为分泌抗体的浆细胞。 C.Ⅱ类MHC分子的组成性表达仅限于T细胞。 D.细胞毒性T细胞通过表达的显示病毒衍生抗原的I类MHC分子识别和杀伤受感染的目标。 E.辅助性T细胞表面表达CD4糖蛋白标记物，并以Ⅱ型MHC分子为性元件。 69Within the lymph nodes, soluble antigens are recognized by _____ cells, and antigen-laden dendritic cells present antigens to _____ cells. a.NK; T b.B; T c.T; both B and T d.T; B e.B; both B and T Which cells have a CD4 marker and is class II MHC restricted? a.B cells b.macrophages c.helper T cells d.cytotoxic T cells e.dendritic cells The _____ molecules on the surface of cytotoxic T cells function as co-receptors for binding to conserved portions of _____ MHC molecules. a.CD4; class I b.CD8; class II c.CD4; class II d.All of the answers are correct. e.CD8; class I Immunoglobulin genes encoding intact immunoglobulins do not exist already assembled in the genome, ready for expression. Instead, the required genes segments are: a.brought together and assembled in the course of T-cell development. b.brought together and assembled in the course of B-cell development. c.None of the answers is correct. d.assembled on the pathogen's surface. e.assembled on the antigen's surface. Which statement is TRUE regarding cytotoxic T lymphocytes? a.MHC molecules display protein fragments derived from viral pathogens for cytotoxic T cells to recognize. b.Mice that have recovered from a particular virus infection are a ready source of cytotoxic T cells that can recognize and kill target cells infected with the same virus. c.All of the statements are true. d.Cytotoxic T cells that have receptors of the appropriate specificity unleash a payload of lethal molecules onto the infected target cells, destroying the target cell membrane and killing the infected target. e.MHC molecules play an essential role in the recognition of 在淋巴结内，可溶性抗原被 细胞识别，携带抗原的树突状细胞向 细胞提供抗原。 a、 NK；T b、 B；T c、 T；B和T都是 d、 T；B e、 B和B 哪些细胞有CD4标记物，二类MHC受到？ a、 B细胞 b、 巨噬细胞 c、 辅助性T细胞 d、 细胞毒性T细胞 e、 树突状细胞 细胞毒性T细胞表面的_____分子作为共同受体与_____ MHC分子的保守部分结合。 a、 CD4；I类 b、 CD8；II类 c、 CD4；II级 d、 所有的答案都是正确的。 e、 CD8；I类 编码完整免疫球蛋白的免疫球蛋白基因并不存在于基因组中，可以随时表达。相反，所需的基因片段是： A.在T细胞发育过程中聚集和组装。 B.在B细胞发育过程中聚集和组装。 C.答案没有一个是正确的。 D.在病原体表面组装。 E.在抗原表面组装。 关于细胞毒性T淋巴细胞，下列哪种说法是正确的？选择一项： A.MHC分子展示来自病毒病原体的蛋白质片段，以供细胞毒性T细胞识别。 B.从特定病毒感染中恢复的小鼠是细胞毒性T细胞的现成来源，可以识别和杀死感染相同病毒的靶细胞。 C.所有的陈述都是真实的。 D.具有适当特异性受体的细胞毒性T细胞向受感染的靶细胞释放大量致死分子，破坏靶细胞膜，杀死受感染的靶细胞。 E.在细胞毒T细胞识别病毒感染细胞的过程中，MHC分子起着至关重要的作用。 virus-infected cells by cytotoxic T cells. Alignment of the amino acid sequences of different light chains reveals _____ hypervariable regions that make contact with the antigen in the folded three-dimensional structure of the immunoglobulin. a.four b.three c.two d.five e.six Which statement is TRUE? a.There are two classes of light chains in mammals, κ and λ, and they are characterized by the attributes of their constant regions. b.In mammals, there are five major classes of immunoglobulins. c.Upon recognition of an antigen, only a B lymphocyte that bears a receptor specific for it will be activated and expand clonally. d.All of the statements are true. e.Each individual B lymphocyte expresses an immunoglobulin of unique sequence and is therefore uniquely specific for a particular antigen. _____ proteins are composed of a single-pass transmembrane glycoprotein that interacts noncovalently with a non-MHC subunit called β2-microglobulin. a.Class II MHC b.CD1 c.Class I MHC d.Both Class 1 MHC and CD1 e.All MHC In a classic series of experiments by Emil von Behring and Shibasaburo Kitasato in 1905, exposure of animals to a sublethal dose of diphtheria toxin elicited in their serum a substance that protected against a subsequent challenge with a lethal dose of the toxin. The protective effect of this serum substance could be transferred to a naive (unexposed) animal. Heating of immune serum destroyed its bactericidal activity. However, addition of fresh unheated serum from a naive animal could restore the bactericidal activity of heated immune serum. Why did the fresh unheated serum from the naive animal restore the protection? a.It contained histamine. b.It contained complement. c.It contained anti-diphtheria toxin antibodies. d.It contained NK cells. 不同轻链的氨基酸序列的排列揭示了与免疫球蛋白折叠三维结构中的抗原接触的 个高变区。 a、 四 b、 三 c、 二 d、 五 e、 六 哪个说法是正确的？ a、 哺乳动物中有两类轻链，κ和λ，它们以其恒定区域的属性为特征。 b、 在哺乳动物中，有五大类免疫球蛋白。 c、 一旦识别出一种抗原，只有带有特异性受体的B淋巴细胞才会被激活并克隆性扩张。 d、 所有的陈述都是真的。 e、 每个单独的B淋巴细胞表达一种具有独特序列的免疫球蛋白，因此对特定抗原具有独特的特异性。 _____蛋白质是由一个单程跨膜糖蛋白组成的，它与一个称为β2-微球蛋白的非MHC亚单位非共价相互作用。 a、 二级MHC b、 CD1型 c、 一级MHC d、 1级MHC和CD1 e、 所有MHC 1905年，在埃米尔·冯·贝林（Emil von Behring）和西巴索罗·北泽（Shibasaburo Kitasato）的一系列经典实验中，动物暴露于亚致死剂量的白喉毒素后，其血清中会产生一种物质，这种物质可以保护动物免受随后致命剂量的白喉毒素的攻击。这种血清物质的保护作用可以转移到未暴露的动物身上。免疫血清加热破坏其杀菌活性。然而，添加未经加热的新鲜动物血清可以恢复加热免疫血清的杀菌活性。为什么这只天真动物的新鲜未加热的血清恢复了保护作用？ a、 它含有组胺。 b、 它含有补体。 c、 它含有抗白喉毒素抗体。 d、 它含有NK细胞。 e、 没有一个答案是正确的。 e.None of the answers is correct. 跨膜转运 A Two-Na⁺/One-Glucose Symporter Animation Activity The two-Na⁺/one-glucose symporter moves Na⁺ ions and glucose: a. in the same direction, both against their concentration gradients. b. in the same direction, one with and one against its concentration gradient. c. in opposite directions, one with and one against its concentration gradient. d. in opposite directions, both against their concentration gradients. e. in the same direction, both down their concentration gradients. This symporter takes on how many conformation states in terms of glucose/Na⁺ binding? a. one b. four c. two d. It does not change conformation. e. three Only in the ________ can the glucose and Na⁺ ions dissociate from the transporter to be released into the cytosol. a. absence of glucose inside the cell b. inward-facing conformation of the symporter c. occluded conformation of the symporter d. absence of Na⁺ inside the cell e. outward-facing conformation of the symporter In the occluded conformation of the symporter, the ________ is/are transiently occluded and therefore unable to ________. a. Na⁺ only; move down its concentration gradient b. symporter itself; bind Na⁺ or glucose c. glucose only; move against its concentration gradient d. Na⁺ and glucose; dissociate from the symporter e. Na⁺ and glucose; bind to the symporter for transport The Na+/K+ ATPase acts via _______ to move ions against their concentration gradients, thus requiring _______. a. active transport; Na+ and K+ ions b. passive transport; Na+ and K+ ions c. passive transport; ATP d. active transport; ATP 两个钠离子/一个葡萄糖转运体移动钠离子和葡萄糖： a、 在同一个方向上，都与它们的浓度梯度相反。 b、 在同一个方向上，一个有浓度梯度，一个逆浓度梯度。 c、 在相反的方向，一个有浓度梯度，一个逆浓度梯度。 d、 在相反的方向，两者都与它们的浓度梯度相反。 e、 在同一个方向上，两个都降低了浓度梯度。 这个共转运体在葡萄糖/Na+结合方面有多少构象状态？ a、 一 b、 四 c、 二 d、 它不改变构象。 e、 三 只有在 中，葡萄糖和钠离子才能与转运体分离，释放到细胞质中。 a、 细胞内缺乏葡萄糖 b、 转运体向内构造 c、 共转运蛋白的封闭构象 d、 牢房内没有Na＋ e、 转运体的外向构象 在被遮挡的Symporter构象中， 被暂时遮挡，因此不能 。选择一项： a.仅Na⁺；向下移动其浓度梯度 b.共转运体本身；仅与Na⁺或葡萄糖结合； C仅限葡萄糖；逆其浓度梯度移动 D Na-⁺和葡萄糖；与同向转运蛋白解离 E Na⁺和葡萄糖；结合到转运蛋白上进行转运 Na+/K+ATP酶通过 ，逆着离子浓度梯度移动离子，因此需要 a、 主动输运；Na+和K+离子 b、 被动输运；Na+和K+离子 c、 被动运输，ATP d、 主动运输，ATP The Plasma-Membrane Na⁺/K⁺ ATPase Animation Activity e. diffusion; an established concentration gradient For each molecule of ATP hydrolyzed by the Na+/K+ ATPase: a. 3 Na+ are pumped out of the cell, and 2 K+ are pumped into the cell. b. 3 Na+ and 2 K+ are pumped in or out of the cell depending on their concentration gradients. c. 3 Na+ and 2 K+ are pumped into the cell against their concentration gradients. d. 3 Na+ and 2 K+ are pumped out of the cell against their concentration gradients. e. 3 Na+ are pumped into the cell, and 2 K+ are pumped out of the cell. The binding followed by release of Na+ and K+ ions to the ATPase is achieved by: a. changes in the affinities of the pump for each ion type. b. All of the above c. a change in the extracellular concentration of each ion type. d. a gate within the pump that opens and closes to let ions in and out. e. dimerization of the α and β subunits of the pump. If a mutation were created in the Na+/K+ ATPase that prevents the pump from converting to the E2 from the E1 conformation, what would be expected? a. The pump would bind Na+ only, and not K+. b. All of the above c. Na+ and K+ ions would not be transported against their concentration gradients across the plasma membrane. d. The pump would remain in a high-affinity state for Na+ and a low-affinity state for K+. e. The bound Na+ ions in the pump would not gain access to the extracellular space. ABC transporters: a. function both as exporters and importers. b. All of the answers are correct. c. show a wide range of specificity. d. have ATP-binding domains on the cytoplasmic side. e. have two ATP-binding domains and two transmembrane domains The sarcoplasmic reticulum calcium pump: a. has much higher affinity for calcium ions when open to the cytosol. b. has two calcium ion–binding sites. c. uses ATP to trigger conformational changes. e、 扩散；确定的浓度梯度 每个ATP+被ATP+酶水解： a、 3na+被泵出细胞，2k+被泵入细胞。 b、 根据浓度梯度，3na+和2k+被泵入或排出细胞。 c、 3na+和2k+根据浓度梯度被泵入细胞。 d、 3na+和2k+根据浓度梯度从细胞中泵出。 e、 3na+被泵入细胞，2k+被泵出细胞。 Na+和K+离子与ATP酶的结合和释放是通过以下方式实现的： a、 每种离子类型的泵亲和力的变化。 b、 以上都是 c、 每种离子类型的细胞外浓度的变化。 d、 泵内的一个门，它打开和关闭，让离子进出。 e、 泵的α和β亚基二聚。 如果Na+/K+ATP酶产生突变，阻止泵从E1构象转化为E2构象，会发生什么？选择一项： A.泵只结合Na+，不结合K+。 B.上述所有 C.Na+和K+离子不会逆其浓度梯度跨质膜转运。 D.泵对Na+保持高亲和力状态，对K+保持低亲和力状态。 E 泵中结合的Na+离子不能进入细胞外空间。 跨膜转运quiz ABC转运蛋白： a兼具出口和进口的职能。 b、 所有的答案都是正确的。 c、 表现出广泛的特异性。 d、 在细胞质侧有ATP结合域。 e、 有两个ATP结合域和两个跨膜结构域 肌浆网钙泵：选择一项： A在胞浆中开放时对钙离子有较高的亲和力。 B.有两个钙离子结合位点。 C.使用ATP触发构象变化。 D.所有的答案都是正确的。 d. All of the answers are correct. e. only pumps calcium ions into the sarcoplasmic reticulum. Various GLUT transporters: a. All of the answers are correct. b. comprise a gene family. c. all transport sugars. d. are expressed in different tissues. e. exhibit different kinetics. All of the following are exclusively ion pumps EXCEPT: a. P-class plasma-membrane pumps. b. ABC superfamily pumps. c. V-class vacuolar pumps. d. F-class inner-membrane pumps. e. Actually, these are all exclusively ion pumps. The ability of the organelle to concentrate Ca2+ in skeletal muscle cells depends on which of the following? a. C-class pumps that move Ca2+ in one direction and glucose in the opposite b. calmodulin c. None of the answers is correct. d. F-class pumps producing energy to lower the pH in the organelle's lumen e. P-class pumps driving Ca2+ against its concentration gradient Uniport transport is distinct from simple diffusion in that it: a. None of the answers is correct. b. is somewhat slower. c. moves molecules from low concentration to high. d. requires a protein. e. moves two molecules of different molecular species. The process by which gases cross membranes is known as: a. facilitated diffusion. b. None of the answers is correct. c. osmosis. d. group translocation. e. simple diffusion. Considering the magnitude of the forces that promote Na+ entry into cells, the effect of the: a. electric potential is much larger (one order of magnitude) than the effect of the ion concentration gradient. b. electric potential is larger (two to four fold) than the effect of the ion concentration gradient. c. ion concentration gradient is larger (two to four fold) than the effect of the electric potential. E.只将钙离子泵入肌浆网 各种葡萄糖转运蛋白： a、 所有的答案都是正确的。 b、 组成一个基因家族。 c、 所有都运输糖。 d、 在不同的组织中表达。 e、 表现出不同的动力学。 以下所有设备均为离子泵，除了： a、 P级质膜泵。 b、 ABC超家族泵。 c、 V级真空泵。 d、 F级内膜泵。 e、 实际上，这些都是离子泵。 细胞器在骨骼肌细胞中集中Ca2+的能力取决于以下哪一项？ a、 C级泵，向一个方向移动Ca2+，向相反方向移动葡萄糖 b、 钙调蛋白 c、 没有一个答案是正确的。 d、 F级泵产生能量以降低细胞器内腔的pH值 e、 P级泵驱动Ca2+浓度梯度 单向传输不同于简单扩散，因为它： a、 没有一个答案是正确的。 b、 有点慢。 c、 将分子从低浓度移动到高浓度。 d、 需要蛋白质。 e、 移动两个不同分子种类的分子。 气体通过膜的过程称为： a、 促进扩散。 b、 没有一个答案是正确的。 c、 渗透作用。 d、 群体易位。 e、 简单扩散。 考虑到促进Na+进入细胞的力的大小：选择一项： A电位的影响比离子浓度梯度的影响大得多(一个数量级)。 B电位大于离子浓度梯度的影响(2~4倍)。 C离子浓度梯度大于电位效应(2~4倍)。 D离子浓度梯度比电位的影响大得多(一个数量级)。 E即离子浓度梯度与电位的影响大致相同。 d. ion concentration gradient is much larger (one order of magnitude) than the effect of the electric potential. e. ion concentration gradient is about the same as the effect of the electric potential. Ouabain毒毛旋花苷 is a drug that selectively inhibits the Na+/K+ ATPase P-class ion pump, thereby lowering the cytosolic K+ concentration and at the same time increasing that of cytosolic Na+. If people were given ouabain, what effect would this have on their cardiac muscle contraction? a. Nothing, because the Na+/K+ ATPase P-class ion pump is not present in the heart. b. More Ca2+ ions would be imported into the cell causing stronger contractions. c. None of the answers is correct. d. The Na+/Ca2+ antiporter would be more efficient, causing stronger contractions. e. The Na+/Ca2+ antiporter would be less efficient, causing weaker contractions. What combination of two forces gives rise to the electrochemical gradient? a. diffusion gradient; equilibrium potential b. membrane potential; electric potential across the membrane c. None of the answers is correct. d. electrical gradient across the membrane; concentration potential e. electric potential across the membrane; concentration gradient Studies with bacteria have revealed much about symport function. Which combination uses a symporter to move from the exterior to the interior of the cell? a. K+ and fructose b. ADP and ATP c. Na+ and glucose d. Na+ and Cl– e. K+ and Na+ ABC proteins can export all of the following EXCEPT: a. cholesterol. b. lipophilic drugs. c. bile salts. d. Actually, they can export all of these. e. chloride ions. 哇巴因毒毛旋花苷是一种选择性抑制Na+/K+ATP酶P类离子泵，从而降低细胞内K+浓度，同时升高细胞内Na+浓度的药物。如果人们服用哇巴因，这会对他们的心肌收缩有什么影响？选择一项： A.没有，因为心脏中没有Na+/K+ATP酶P类离子泵。 B.更多的Ca~(2+)离子会被输入到细胞内，导致更强烈的收缩。 C.答案没有一个是正确的。 D.Na+/Ca2+逆向转运蛋白效率更高，导致更强的收缩。 E.Na+/Ca2+逆向转运蛋白效率较低，导致较弱的收缩。 是什么两种力的组合引起了电化学梯度？选择一项： A.扩散梯度；.平衡电位； B.膜电位；跨膜电位。 C没有一个答案是正确的。 D跨膜的电梯度；浓度电位 E跨膜电位；浓度梯度 对细菌的研究揭示了许多有关联合蛋白功能的信息。哪种组合使用symporter从细胞的外部移动到内部？选择一项： a.K+和果糖 b.ADP和ATP c.Na+和葡萄糖 d.Na+和Cl- e.K+和Na+ ABC蛋白可以输出以下所有物质，除了 A.胆固醇。 B.亲脂性药物。 C.胆盐。 D实际上，他们可以把这些都输出 E.氯离子。 F-class pumps belong to the family of ATP-powered pumps, yet F级泵属于ATP驱动的泵家族，但由于其产生的物质，通are often referred to as ATP synthases because of what they 常被称为ATP合成酶。这些泵位于： produce. These pumps are found in: a、 线粒体内膜。 a. inner mitochondrial membranes. b. thylakoid membranes of the chloroplast. c. bacterial plasma membranes. d. All of the answers are correct. e. None of the answers is correct. Secondary active transport is when: a. a pump uses ATP as the energy source for transport. b. None of the answers is correct. c. two molecules move in a cotransport process. d. a molecule is moved from low concentration to high. e. a molecule is moved from high concentration to low. The process by which gases cross membranes is known as: a. simple diffusion. b. None of the answers is correct. c. osmosis. d. group translocation. e. facilitated diffusion. Which ion is typically at HIGHER concentration in the cytoplasm than in extracellular spaces? a. Cl– b. Actually, all of these are typically at lower concentrations in the cytoplasm. c. Mg++ d. Na+ e. K+ The contraction of cardiac muscle is triggered following the rise in cytosolic Ca2+. To bring these elevated levels back to resting levels in preparation for the next contraction, the cell relies on an antiporter where: a. three Na+ ions move into the cell causing one Ca2+ to be exported. b. three Ca2+ ions are exported from the cell for every Na+ ion that enters. c. None of the answers is correct. d. one Ca2+ ion is exported for every Na+ ion that enters. e. two Na+ ions move into the cell causing one Ca2+ to be exported. K+ channels exclude Na+ by: a. All of the answers are correct. b. being too small for the Na+ ion to fit. c. favoring the interaction of the K+ ion with amino groups in the filter. d. None of the answers is correct; they actually move Na+ as well. b、 叶绿体的类囊体膜。 c、 细菌质膜。 d、 所有的答案都是正确的。 e、 没有一个答案是正确的。 次要主动转运是指：选择一项： A.泵使用ATP作为输送的能源。 B.答案没有一个是正确的。 C.两个分子在共运输过程中运动。 D.分子从低浓度移动到高浓度。 E.分子从高浓度移动到低浓度。 气体通过膜的过程称为： a、 简单扩散。 b、 没有一个答案是正确的。 c、 渗透作用。 d、 群体易位。 e、 促进扩散。 哪种离子在细胞质中的浓度比在胞外高？ a. Cl– b. 实际上，所有这些在细胞质中的浓度都很低。 c. Mg++ d. Na+ e. K+ 心肌收缩是随着细胞内钙离子的升高而触发的。为了使这些升高的水平回到静止水平，为下一次收缩做准备，细胞依赖于逆向转运蛋白： A.三个Na+离子进入细胞，导致一个钙离子输出。 B.每进入一个Na+离子，就有三个Ca2+离子从细胞中输出。 C.答案没有一个是正确的。 D.每进入一个Na+离子就输出一个Ca2+离子。 E.两个Na+离子进入细胞，导致一个Ca2+被输出 K+通道不包括Na+的机制是 a.所有答案均正确。 b.对于Na+离子来说太小了，无法容纳。 C.有利于K+离子与过滤器中氨基的相互作用。 d.没有一个答案是正确的，它们实际上也移动了Na+ e不提供能量来除去使Na+离子水合的水分子。 e. not providing energy to remove the water molecules hydrating the Na+ ion. In the Na+/K+ ATPase: a. All of the answers are correct. b. Na+ ions are released and K+ ions are bound while in the same state. c. dephosphorylation induces a conformational change. d. phosphorylation induces a conformational change. Glucose is an energy source and its uptake is essential for maintaining normal cellular physiology. What mechanism(s) is/are needed to allow glucose to be transported across a cellular membrane? a. facilitated transport b. facilitated transport AND cotransport c. simple diffusion d. cotransport e. simple diffusion AND cotransport Na+/K+ATP酶： A.所有答案均正确。 B.在同一状态下，Na+离子被释放，K+离子被结合。 C.去磷酸化诱导构象变化。 D.磷酸化导致构象变化。 葡萄糖是一种能量来源，它的吸收对于维持正常的细胞生理是必不可少的。葡萄糖通过细胞膜转运需要什么机制？ a、 易化扩散 b、 易化扩散和协同运输 c、 简单扩散 d、 协同运输 e、 简单扩散与协同运输 细胞信号转导 Activation of the NF-ĸB Signaling Pathway Animation Activity Which of these is NOT required for activation of NF-ĸB signaling? a. degradation of I-ĸBα b. a stimulus such as viral infection, ionizing radiation, or pro-inflammatory cytokines c. translocation of NF-ĸB into the nucleus d. unmasking of the nuclear localization sequences (NLS) of the NF-ĸB subunits e. deactivation of the IKK complex to relieve inhibition of NF-ĸB Which process is involved in targeting I-ĸBα for degradation to relieve inhibition of NF-ĸB? a. All of the above. b. polyubiquitination of I-ĸBα c. phosphorylation of I-ĸBα d. binding of E3 ubiquitin ligase to I-ĸBα e. proteasomal degradation of I-ĸBα NF-ĸB activates the transcription of all of these EXCEPT: a. I-ĸB. b. toll-like receptors. c. enzymes. d. adhesion proteins. e. chemokines. 以下哪一项不需要激活NF-ĸB信号？ a、 I-ĸBα的降解 b、 一种刺激，如病毒感染、电离辐射或促炎细胞因子 c、 NF-ĸB向细胞核的转运 d、 NF-ĸB亚单位核定位序列（NLS）的揭开 e、 使IKK复合物失活以减轻NF-ĸB的抑制作用 靶向I-ĸBα降解以减轻NF-ĸB的抑制作用涉及哪个过程？ a、 以上都是。 b、 I-ĸbα的多泛素化 c、 I-ĸBα的磷酸化 d、 E3泛素连接酶与I-ĸBα的结合 e、 I-ĸBα的蛋白酶体降解 NF-ĸB激活所有这些的转录，除了： a、 I-ĸB。 b、 toll样受体。 c、 酶。 d、 粘附蛋白。 e、 趋化因子。 If a mutation is introduced into I-ĸBα that inhibits its 如果I-ĸBα中的突变抑制了IKK复合物的磷酸化，这将阻phosphorylation by the IKK complex, this would prevent: 止： a. recruitment of Nemo to the IKK complex. a、 招募尼莫加入IKK综合体。 b. translocation of NF-ĸB into the nucleus. b、 NF-ĸb向细胞核的移位。 c. dimerization of the p50 and p65 subunits. c、 p50和p65亚基的二聚。 d. All of the above. d、 以上都是。 e. TAK1 phosphorylation of IKKβ. e、 IKKβTAK1磷酸化。 GPCR regulation in disease A 10-year-old male complains of exercise intolerance. He describes episodes of breathlessness, chest tightness, and wheezing after moderate exercise, especially in cold weather. Which one of the following pairs of chemical mediators and antagonists is most involved in the pathogenesis of this boy's condition? A. Histamine - clemastine B. Serotonin – sumatriptan C. Prostaglandin E2 - indomethacin D. Leukotriene D4 – montelukast Synthesized by eosinophils and mast cells, leukotriene D4 plays an important role in the pathogenesis of bronchial asthma by inducing bronchospasm and increasing bronchial mucus secretion. An anti-leukotriene medication like montelukast antagonizes leukotriene D4 activity at the cysteinyl leukotriene receptor. E. Thromboxane A2- aspirin A 34-year-old male is brought to the ER with repeated episodes of coffee ground emesis. The patient is frightened and complains of dizziness upon standing. His past medical history is not significant. His blood pressure is 90/60 mmHg supine and 80/55 mmHg sitting. Which of the following substances is most likely to accumulate in this patient's vascular smooth muscle cells? a. 2,3-bisphosphoglycerate (2,3-BPG) b. lnositol triphosphate (IP3) Stimulation of alpha-1 receptors by norepinephrine from the post-ganglionic sympathetic nerves will increase IP3 synthesis c. cGMP d. S-adenosylmethionine (SAM) e. cAMP A 21-year-old male suffers from chronic nasal blockage. He uses over-the-counter nasal decongestants occasionally with minimal symptom relief. He also describes difficulty beathing and nocturnal wheezing when taking aspirin. The latter symptoms are most likely related to an increase in which of the following? 一名10岁男性抱怨运动不耐。他描述了在适度运动后，尤其是在寒冷的天气里，呼吸困难、胸闷和喘息的发作。以下哪一对化学介质和拮抗剂在这个男孩的发病机制中最重要？ A、 组胺-氯马斯汀 B、 血清素-舒马曲坦 C、 前列腺素E2-吲哚美辛 D、 白三烯D4-孟鲁司特 白三烯D4由嗜酸性粒细胞和肥大细胞合成，通过诱导支气管痉挛和增加支气管粘液分泌，在支气管哮喘的发病中起重要作用。像孟鲁司特这样的抗白三烯药物可以拮抗半胱氨酸白三烯受体处的白三烯D4活性。 E、 血栓素A2-阿司匹林 一名34岁男性被带到急诊室，反复发作咖啡地面呕吐。病人害怕，并抱怨站着时头晕。他过去的病史并不重要。他的血压是90/60毫米汞柱仰卧和80/55毫米汞柱坐着。以下哪种物质最有可能积聚在患者的血管平滑肌细胞中？ a、 2,3-二磷酸甘油酯（2,3-BPG） b、 三磷酸肌醇（IP3） 神经节后交感神经的去甲肾上腺素刺激α-1受体可增加IP3的合成 c、 cGMP d、 S-腺苷甲硫氨酸（SAM） e、 营地 21岁男性患有慢性鼻塞。他偶尔会使用非处方的鼻减充血剂，但症状却很少缓解。他还描述了服用阿司匹林时殴打困难和夜间气喘。后一种症状最有可能与以下哪种症状有关？ A、 前列环素 B、 白三烯：阿司匹林敏感性哮喘的发病机制似乎与花生A. Prostacyclin B. Leukotrienes The pathogenesis of aspirin-sensitive asthma appears to involve increased production of the lipoxygenase products of arachidonic acid metabolism (eg, bronchoconstrictive leukotrienes). C. Phospholipase D. Thromboxane A2 E. Prostaglandin E 四烯酸代谢的脂氧合酶产物（如支气管收缩性白三烯）的增加有关。 C、 磷脂酶 D、 血栓素A2 E、 前列腺素E GPCR信号通路 Regulation of Glycogen Metabolism by cAMP and PKA Animation Activity Active PKA plays what role(s) in glycogen metabolism? a. inactivation of glycogen synthase (GS) 活性PKA在糖原代谢中起什么作用？选择一项： A.糖原合成酶(GS)的失活 b. activation of the enzyme GPK c. inhibition of glycogen synthesis d. All of the above e. stimulation of glycogen breakdown The enzyme PP is activated by decreased _______________ levels to promote ______________ glycogen synthesis via activation of _______________. a. GPK; decreased; glycogen synthase (GS) b. cAMP; increased; glycogen synthase (GS) c. glycogen synthase (GS); decreased; PKA d. glycogen synthase (GS); increased; PKA e. cAMP; increased; GPK The enzymes GPK, GP, and GS are all counteracted by: a. PKA. b. glycogen. c. phosphoprotein phosphatase (PP). d. cAMP. e. glucose-1-phosphate. The overall result of increased cAMP levels is: a. decreased glycogen breakdown. b. All of the above. c. elevated glucose 1-phosphate levels in the cell. d. elevated glycogen levels in the cell. e. increased glycogen synthesis. cAMP activation of gene transcription involves all of the following EXCEPT ________. a. binding of cAMP to CREB. b. binding of cAMP to the regulatory subunits of PKA and release of the PKA catalytic subunits. c. movement of the PKA catalytic subunits into the nucleus. d. binding of phosphorylated CREB and a coactivator protein to a CRE sequence in the promoter region of a gene. e. PKA phosphorylation of CREB. Whether cAMP signaling activates transcription of a specific gene depends on ________. a. how much cAMP is present in the cell; high levels of cAMP will turn on all genes. b. the ratio of cAMP molecules to PKA regulatory subunits in the cell. c. whether the gene has a promoter sequence to which the coactivator CBP-P300 binds. Feedback d. the opening of nuclear pores so that cAMP can enter the B.GPK酶的激活 C.抑制糖原合成 d.上述所有 E.刺激糖原分解 酶PP被 水平降低激活，通过激活 促进糖原合成 a、 GPK；降低；糖原合成酶（GS） b、 cAMP；增加；糖原合成酶（GS） c、 糖原合成酶降低 d、 糖原合酶（GS）；增加；PKA e、 cAMP；增加；GPK 酶GPK、GP和GS都被以下因素抵消： a、 PKA b、 糖原。 c、 磷酸蛋白磷酸酶（PP）。 d、 cAMP. e、 葡萄糖-1-磷酸。 cAMP 水平提高的总体结果是： a、 糖原分解减少。 b、 以上都是。 c、 细胞内葡萄糖1-磷酸水平升高。 d、 细胞内糖原水平升高。 e、 增加糖原合成。 cAMP激活基因转录涉及以下所有方面，除了。 a、 cAMP与CREB的绑定。 b、 cAMP与PKA调节亚基的结合及PKA催化亚基的释放。 c、 PKA催化亚单位进入细胞核的运动。 d、 磷酸化CREB和辅活化蛋白与基因启动子区CRE序列的结合。 e、 CREB的PKA磷酸化。 cAMP信号是否激活特定基因的转录取决于。 a、 细胞中存在多少cAMP；高水平的cAMP会激活所有基因。 b、 细胞内cAMP分子与PKA调节亚单位的比率。 c、 该基因是否有与辅激活子CBP-P300结合的启动子序列。 反馈 d、 核孔的打开使cAMP能进入细胞核。 e、 基因启动子中是否含有CRE序列。 Signal Transduction with Adenylyl Cyclase Animation Activity nucleus. e. whether the gene contains a CRE sequence in its promoter. cAMP activation of genes encoding proteins involved in gluconeogenesis ________. a. increases the amount of glucose generated by glycogen breakdown. b. increases the amount of glucose synthesized in the cell. c. decreases the amount of glucose synthesized in the cell. d. decreases the amount of glucose generated by glycogen breakdown. e. decreases the amount of glucose synthesized in the cell and increases the amount of glucose generated by glycogen breakdown. TGF-β signaling pathway receptors RI and RII are: a. tyrosine kinases. b. TGF-β-gated ion channels. c. G protein–coupled receptors (GPCRs). d. serine/threonine kinases. e. phospholipases. Which statement is false regarding the TGF-β signaling pathway when no TGF-β is bound to TGF-β receptor II (RII)? a. RI receptor kinase activity is auto-(self-)inhibited. b. Smad2/3 is unphosphorylated and in the cytoplasm. c. Transcription of the PAI-1 gene is turned off. d. Smad4 is in the nucleus. e. All of the above are true. TGF-β Receptor I (RI) phosphorylation of Smad2/3 does all of the following EXCEPT: a. unmask the Smad2/3 nuclear localization signal (NLS). b. RI phosphorylation of Smad2/3 does all of these things. c. dissociate intramolecular binding of Smad2/3 MH1 and MH2 domains. d. release Smad2/3 from the nucleus into the cytoplasm e. activate Smad2/3 binding to the Co-Smad Smad4 The complex formed by phosphorylated Smad2/3 that moves into the nucleus contains all of the following EXCEPT: a. Smad2/3. b. importin. c. The complex contains all of the listed components. d. Smad4. e. TGF-β receptor I (RI). 糖异生相关蛋白编码基因cAMP的激活。 a、 增加糖原分解产生的葡萄糖量。 b、 增加细胞中葡萄糖的合成量。 c、 减少细胞中葡萄糖的合成量。 d、 减少糖原分解产生的葡萄糖量。 e、 减少细胞中合成的葡萄糖量，增加糖原分解产生的葡萄糖量。 TGF-β Signaling Animation Activity TGF-β信号途径受体RI和RII是： a、 酪氨酸激酶。 b、 TGF-β门控离子通道。 c、 G蛋白偶联受体（GPCR）。 d、 丝氨酸/苏氨酸激酶。 e、 磷脂酶。 当没有TGF-β与TGF-β受体II（RII）结合时，关于TGF-β信号通路的哪个陈述是错误的？ a、 RI受体激酶活性是自抑制的。 b、 Smad2/3不磷酸化，存在于细胞质中。 c、 PAI-1基因的转录被关闭。 d、 Smad4在细胞核内。 e、 以上都是真的。 Smad2/3的TGF-β受体I（RI）磷酸化可实现以下所有功能，除了： a、 解开Smad2/3核定位信号（NLS）的屏蔽。 b、 Smad2/3的RI磷酸化可以做所有这些事情。 c、 解离Smad2/3mh1和MH2结构域的分子内结合。 d、 从细胞核释放Smad2/3进入细胞质 e、 激活Smad2/3与Co Smad Smad4的绑定 磷酸化Smad2/3进入细胞核形成的复合物包含以下所有物质，除了： a、 Smad2/3。 b、 进口蛋白。 c、 复合体包含所有列出的组件。 d、 Smad4。 e、 转化生长因子β受体Ⅰ（RI）。 The catalytic activity of G protein–activated phospholipase C G蛋白活化磷脂酶C（PLC）的催化活性产生了两个第二信(PLC) generates two second messenger molecules: ________ and ________. a. cAMP and Ca2+ b. PIP2 and Ca2+ c. IP3 and Ca2+ d. IP3 and DAG e. IP3 and PIP2 What is the direct target of IP3 and the effect of IP3 binding to its target? a. store-operated Ca2+ channel; influx of extracellular Ca2+ into the cytosol b. ER Ca2+-ATPase; pumping of cytosolic Ca2+ into the ER c. plasma membrane Ca2+-ATPase; pumping of cytosolic Ca2+ out of the cell d. gated Ca2+ channel; release of Ca2+ from the ER into the cytosol e. gated Ca2+ channel; pumping of cytosolic Ca2+ into the ER The second messenger DAG directly ________. a. activates PKC b. activates a plasma membrane Ca2+ ATPase c. activates PLC d. causes a rise in cytosolic Ca2+ concentration e. activates IP3 In order to be active, adenylyl cyclase needs to: a. bind both ATP and an active G protein. b. associate directly with GTP. c. bind ATP. d. both bind ATP and associate directly with GTP. e. bind an active G protein. Activation of a G protein would NOT be considered signal amplification because it: a. is part of a receptor-ligand complex and diffuses to activate a single target, such as adenylyl cyclase. b. triggers the release of intracellular ions. c. is part of a receptor-ligand complex and triggers the release of intracellular ions. d. diffuses to activate a single target, such as adenylyl cyclase. e. is part of a receptor-ligand complex. In a somewhat unusual fashion, rod cells in the dark: a. signal actively to the brain. b. secrete neurotransmitters. c. are relatively depolarized compared with most neurons. 使分子： 和 a. cAMP and Ca2+ b. PIP2 and Ca2+ c. IP3 and Ca2+ d. IP3 and DAG e. IP3 and PIP2 IP3的直接靶点是什么？IP3与靶结合的效果如何？ a、 储存型钙通道；细胞外钙离子流入胞浆 b、 ER-Ca2+-ATPase；胞浆内Ca2+泵入ER c、 质膜Ca2+-ATPase；胞浆Ca2+泵出细胞外 d、 门控钙通道；钙离子从内质网释放到细胞质中 e、 门控Ca2+通道；胞浆Ca2+泵入内质网 第二个信使DAG直接 。 a、 激活PKC b、 激活质膜Ca2+ATPase c、 激活PLC d、 引起细胞内Ca2+浓度升高 e、 激活IP3 信号转导15章quiz1 腺苷酸环化酶需要： a、 结合ATP和活性G蛋白。 b、 直接与GTP联系。 c、 绑定ATP。 d、 两者都与ATP结合并直接与GTP相关联。 e、 结合活性G蛋白。 G蛋白的激活不被视为信号放大，因为它： a、 是受体-配体复合物的一部分，扩散以激活单个靶点，如腺苷酸环化酶。 b、 触发细胞内离子的释放。 c、 是受体-配体复合物的一部分，并触发细胞内离子的释放。 d、 扩散以激活单个靶点，如腺苷酸环化酶。 e、 是受体的一部分。 以一种不寻常的方式，黑暗中的杆状细胞： a、 主动向大脑发出信号。 b、 分泌神经递质。 c、 与大多数神经元相比相对去极化。 d. have open non-selective ion channels. e. All of the answers are correct. All of the following pathways activated by epinephrine result in the breakdown of glycogen EXCEPT: a. inactivation of phosphoprotein phosphatase. b. activation of glycogen phosphorylase. c. activation of glycogen phosphorylase kinase. d. Actually, all of these contribute to glycogen breakdown. e. inactivation of glycogen synthase. In the rod cells of the eye, light triggers: a. phosphorylation of the photoreceptor. b. GTP hydrolysis by the photoreceptor. c. isomerization of the photoreceptor. d. acetylation of the photoreceptor. e. All of the answers are correct. Ligand-activated G protein–coupled receptor signaling involves all of the following EXCEPT: a. the receptor. b. Actually, all of these are involved c. GTPase. d. hormone. e. effector protein. Different G proteins, while associating with different receptors, can: a. lead to activation of the same effector. b. mediate responses to the same extracellular ligand. c. All of the answers are correct. d. use the same second messenger. e. None of the answers is correct. Epidermal growth factor (EGF), as implied, stimulates cell growth. Many cancer cells have elevated levels of the EGF receptor, which means that they will: a. respond to lower levels of EGF. b. None of the answers is correct. c. bind much more EGF. d. bind EGF with higher affinity. e. All of the answers are correct. Adenylyl cyclase acts to: a. produce cyclic AMP. b. use ATP to pump protons across a membrane. c. transfer a phosphate from ATP to second messengers. d. None of the answers is correct. e. function as an ATP/ADP antiporter. Protein kinase A is activated when cyclic AMP: d、 有开放的非选择性离子通道。 e、 所有答案都是正确的。 肾上腺素激活的以下所有途径都会导致糖原分解，除了： a、 磷酸蛋白磷酸酶失活。 b、 糖原磷酸化酶的激活。 c、 糖原磷酸化酶激酶的激活。 d、 实际上，所有这些都会导致糖原分解。 e、 糖原合酶失活。 在眼睛的杆状细胞中，光会触发： a、 光感受器的磷酸化。 b、 光感受器水解GTP。 c、 光感受器异构化。 d、 光感受器的乙酰化。 e、 所有答案都是正确的。 配体激活的G蛋白偶联受体信号涉及以下所有方面，除了： a、 受体。 b、 事实上，所有这些都牵涉其中 c、 GTP酶。 d、 荷尔蒙。 e、 效应蛋白。 不同的G蛋白与不同的受体结合时，可以： a、 导致同一效应器的激活。 b、 介导对同一细胞外配体的反应。 c、 所有的答案都是正确的。 d、 使用相同的第二个信使。 e、 没有一个答案是正确的。 表皮生长因子（EGF）可以促进细胞生长。许多癌细胞EGF受体水平升高，这意味着它们将： a、 对低水平的表皮生长因子有反应。 b、 没有一个答案是正确的。 c、 结合更多的表皮生长因子。 d、 结合EGF具有更高的亲和力。 e、 所有答案都是正确的。 腺苷酸环化酶作用于： a、 产生环状AMP。 b、 用ATP把质子泵过薄膜。 c、 将磷酸盐从ATP转移到第二个信使。 d、 没有一个答案是正确的。 e、 作为ATP/ADP反转运蛋白。 当环腺苷酸： a. binds to the regulatory subunit. b. binds to the catalytic subunit. c. triggers Ca++ entrance to the cytosol. d. None of the answers is correct; protein kinase A is not activated by cAMP. e. All of the answers are correct. Epidermal growth factor is a signaling molecule that travels through the blood and binds to EGF receptor proteins on target cells. This can activate a signaling pathway known as the MAP kinase pathway. This is an example of: a. All of the answers are correct. b. signaling by plasma-membrane–attached proteins. c. paracrine signaling. d. endocrine signaling. e. autocrine signaling. Activation of new gene synthesis by protein kinase A requires all of the following EXCEPT: a. cAMP response elements in the DNA. b. phosphorylation of the co-activator CPB. c. Actually, all of these are required. d. translocation of protein kinase A to the nucleus from the cytoplasm. e. phosphorylation of CREB. Increased protein kinase A activity due to epinephrine leads to: a. increased export of glucose by muscle cells. b. increased glycogen synthase activity. c. decreased glycogen phosphorylase activity. d. None of the answers is correct. e. inhibition of phosphoprotein phosphatase activity. Stimulation of rhodopsin by light causes: a. None of the answers is correct. b. increasing concentration of cGMP. c. decreasing intracellular negativity. d. All of the answers are correct. e. opening of nonspecific cation channels. Protein kinase A activates glycogen synthesis by: a. inhibiting glycogen phosphorylase. b. All of the answers are correct. c. feed-forward stimulation. d. None of the answers is correct; protein kinase A does not activate glycogen synthesis. e. stimulating glycogen synthase. a、 与调节亚单位结合。 b、 与催化亚单位结合。 c、 触发Ca++进入胞浆。 d、 没有一个答案是正确的；蛋白激酶A没有被cAMP激活。 e、 所有答案都是正确的。 表皮生长因子是一种信号分子，在血液中传播并与靶细胞上的EGF受体蛋白结合。这可以激活一种被称为MAP激酶通路的信号通路。这是一个例子： a、 所有的答案都是正确的。 b、 通过质膜附着蛋白的信号传导。 c、 旁分泌信号。 d、 内分泌信号。 e、 自分泌信号。 蛋白激酶A激活新基因合成需要以下所有条件，除了： a、 DNA中的cAMP反应元件。 b、 共激活剂CPB的磷酸化。 c、 实际上，所有这些都是必需的。 d、 蛋白激酶A从细胞质转移到细胞核。 e、 CREB的磷酸化。 肾上腺素引起的蛋白激酶A活性增加导致： a、 增加肌肉细胞输出葡萄糖。 b、 糖原合成酶活性增加。 c、 糖原磷酸化酶活性降低。 d、 没有一个答案是正确的。 e、 抑制磷蛋白磷酸酶活性。 光刺激视紫红质的原因： a、 没有一个答案是正确的。 b、 增加cGMP浓度。 c、 减少细胞内负性。 d、 所有的答案都是正确的。 e、 非特异性阳离子通道的开放。 蛋白激酶A通过以下方式激活糖原合成： a、 抑制糖原磷酸化酶。 b、 所有的答案都是正确的。 c、 前馈刺激。 d、 没有一个答案是正确的；蛋白激酶A不能激活糖原的合成。 e、 刺激糖原合酶。 Muscarinic acetylcholine receptors are found in _____ muscle 毒蕈碱乙酰胆碱受体存在于 肌肉和 肌肉收缩中。 and _____ contraction. a、 骨骼的；慢下来的 a. skeletal; slow down b. None of the answers is correct. c. smooth; speed up d. cardiac; slow down e. skeletal; speed up Termination of signaling from a GPCR by endocytosis involves: a. activation of transcription by CREB. b. arrestin binding the G protein. c. All of the answers are correct. d. hydrolysis of cAMP by PDE. e. arrestin binding the active phosphorylated receptor. Which statement is NOT true regarding the muscarinic acetylcholine receptor? a. It is a GPCR. b. The K+ channel it activates becomes activated when it interacts with the Gα subunit of the receptor. c. Acetylcholine is the ligand. d. Activation of the receptor causes attachment of GTP to the Gα subunit. e. All of the statements are true. Light signals of fewer than 10 photons can be detected by vertebrate eyes due to amplification by all of the following EXCEPT: a. Actually, all of these play parts in the signal amplification. b. activation of multiple G protein subunits by a single rhodopsin. c. closing of tens of thousands of cation channels by cGMP. d. activation of multiple esterases by G protein subunits. e. large-scale hydrolysis of cGMP by esterases Bound Ca++: a. is thought to be much greater in concentration than free Ca++. b. is relatively easy to measure. c. is thought to be about equal in concentration to free Ca++. d. None of the answers is correct. e. can be measured by fura-2. b、 没有一个答案是正确的。 c、 平滑；加速 d、 心脏病；慢下来 e、 骨骼；加速 通过内吞作用终止来自GPCR的信号包括： a、 CREB激活转录。 b、 与G蛋白结合的arrestin。 c、 所有的答案都是正确的。 d、 PDE水解cAMP。 e、 与活性磷酸化受体结合的arrestin。 关于毒蕈碱乙酰胆碱受体，哪种说法不正确？ a、 这是一个GPCR。 b、 它激活的K+通道在与受体的Gα亚单位相互作用时被激活。 c、 乙酰胆碱是配体。 d、 受体的激活导致GTP附着在Gα亚单位上。 e、 所有的陈述都是真的。 脊椎动物的眼睛可以检测到少于10个光子的光信号，这是由于以下所有因素的放大作用，除了： a、 实际上，所有这些都在信号放大中起作用。 b、 单个视紫红质激活多个G蛋白亚基。 c、 cGMP封闭了数以万计的阳离子通道。 d、 G蛋白亚基激活多种酯酶。 e、 酯酶法大规模水解cGMP 绑定Ca++： a、 被认为在浓度上比游离Ca++大得多。 b、 相对容易测量。 c、 被认为在浓度上与游离Ca++相当。 d、 没有一个答案是正确的。 e、 可用fura-2测定。 信号转导16章quiz2 Signaling through the TGF-β receptor includes all of the 通过TGF-β受体发出信号包括以下所有步骤，除了： following steps EXCEPT: a、 三聚体配体与第一受体的结合。 a. binding of the trimeric ligand to the first receptor. b、 受体的异源寡聚。 b. hetero-oligomerization of the receptors. c、 转化生长因子β从一个受体转移到另一个受体。 c. transfer of TGF-β from one receptor to another. d、 实际上，所有这些步骤都包括在内。 d. Actually, all of these steps are included. e、 一种受体被另一种受体磷酸化。 e. phosphorylation of one of the receptors by another. Smad proteins possess all of the following functional regions EXCEPT: a. Actually, they possess all of these. b. nuclear localization sequence. c. activation domains. d. DNA-binding domain. e. kinase domain. What is the CORRECT order the steps of the MAP kinase pathway? a. Raf activation, MEK activation, MAP kinase activation, Ras GTP hydrolysis b. Ras GTP hydrolysis, Raf activation, MEK activation, MAP kinase activation c. Ras GTP hydrolysis, MEK activation, MAP kinase activation, Raf activation d. Ras GTP hydrolysis, MAP kinase activation, Raf activation, MEK activation e. Raf activation, Ras GTP hydrolysis, MEK activation, MAP kinase activation Since there are multiple MAP kinases, each of which associates with a particular signaling pathway, they must associate with the proper proteins, association which is brought about by: a. anchoring them to the membrane with lipids. b. None of the answers is correct. c. All of the answers are correct. d. use of a unique scaffold protein for each pathway. e. expression of the kinases in discrete cell types, so they can only bind the right ligands. In the MAP kinase cascade: a. Ras follows Raf. b. MEK precedes MAPK. c. Ras follows MEK. d. None of the answers is correct. e. All of the answers are correct. Activation of the epidermal growth factor (EGF) receptor: a. includes a conformational shift, exposing the active site of the kinase. b. continues with dimerization, even though the ligand is a monomer. c. None of the answers is correct. d. starts with ligand binding. e. All of the answers are correct. Ligands for receptor tyrosine kinases include: Smad蛋白具有以下所有功能区，除了： a、 实际上，他们拥有所有这些。 b、 核定位序列。 c、 激活域。 d、 DNA结合域。 e、 激酶结构域。 MAP激酶通路的正确顺序是什么？ a、 Raf活化，MEK活化，MAP激酶活化，Ras-GTP水解 b、 Ras-GTP水解、Raf活化、MEK活化、MAP激酶活化 c、 Ras-GTP水解、MEK活化、MAP激酶活化、Raf活化 d、 Ras-GTP水解、MAP激酶活化、Raf活化、MEK活化 e、 Raf活化、Ras-GTP水解、MEK活化、MAP激酶活化 由于存在多个MAP激酶，每一个都与一个特定的信号通路有关，它们必须与适当的蛋白质结合，这种联系是由以下因素引起的： a、 用脂类将它们固定在膜上。 b、 没有一个答案是正确的。 c、 所有的答案都是正确的。 d、 为每个通路使用独特的支架蛋白。 e、 激酶在离散细胞类型中的表达，所以它们只能结合正确的配体。 在MAP激酶级联中： a、 Ras跟随英国皇家空军。 b、 MEK在MAPK之前。 c、 Ras跟着MEK。 d、 没有一个答案是正确的。 e、 所有答案都是正确的。 激活表皮生长因子（EGF）受体： a、 包括构象变化，暴露了激酶的活性部位。 b、 继续二聚，即使配体是单体。 c、 没有一个答案是正确的。 d、 从配体结合开始。 e、 所有答案都是正确的。 受体酪氨酸激酶配体包括： a. nerve growth factor (NGF). b. All of the answers are correct. c. None of the answers is correct. d. fibroblast growth factor (FGF). e. platelet-derived growth factor (PDGF). Which statement is NOT true regarding SH2 domains? a. They only bind to phosphorylated tyrosine residues. b. The SH2 domain binding pocket binds mainly hydrophilic amino acids. c. Different SH2 domains recognize different specific amino acid sequences. d. They are responsible for signal propagation in receptor kinase pathways. e. All of the statements are true. a、 神经生长因子（NGF）。 b、 所有的答案都是正确的。 c、 没有一个答案是正确的。 d、 成纤维细胞生长因子（FGF）。 e、 血小板衍生生长因子（PDGF）。 关于SH2域，哪种说法不正确？ a、 它们只与磷酸化酪氨酸残基结合。 b、 SH2结构域结合袋主要与亲水性氨基酸结合。 c、 不同的SH2结构域识别不同的特定氨基酸序列。 d、 它们负责受体激酶途径中的信号传播。 e、 所有的陈述都是真的。 Signaling pathways that are triggered by protein cleavage 由蛋白质裂解触发的信号通路包括： include: a、 Wnt/连环蛋白 a. Wnt/Catenin b、 所有的答案都是正确的。 b. All of the answers are correct. c、 缺口/三角形。 c. Notch/Delta. d、 阿尔茨海默病患者。 d. Those involved in Alzheimer's disease. e、 没有一个答案是正确的。 e. None of the answers is correct. The Notch/Delta signaling pathway involves all of the following EXCEPT: a. Notch, on the responding cell. b. presenilin-1, a protease, on the signaling cell. c. Actually, all of these are required. d. ADAM 10, a matrix metalloprotease on the responding cell. e. Delta, on the signaling cell. Why does receptor dimerization result in increased activity of the JAK kinases? a. More than one of the above accurately explain the activity. b. Cross-phosphorylation resulting from dimerization allows the recruitment of adaptor proteins that increase kinase activity. c. Phosphorylation of one JAK kinase by the other following dimerization causes a structural change that increases the protein's affinity for ATP. d. One JAK kinase is bound by both dimerized receptors, causing a conformational change that increases activity. e. The kinases bind to ATP which is passed from one kinase domain to the next. In the MAP kinase cascade: a. MAP kinase phosphorylates transcription factors that move to the nucleus. b. Raf is an adaptor protein rather than a kinase Notch/Delta信号通路包括以下所有方面，除了： a、 缺口，在反应室上。 b、 早老素-1，一种蛋白酶，作用于信号细胞。 c、 实际上，所有这些都是必需的。 d、 ADAM 10，反应细胞上的一种基质金属蛋白酶。 e、 在信号细胞上。 为什么受体二聚化导致JAK激酶活性增加？ a、 以上不止一个准确地解释了这个活动。 b、 由二聚作用引起的交叉磷酸化允许增加激酶活性的衔接蛋白。 c、 一个JAK激酶被另一个二聚体磷酸化，引起结构改变，增加蛋白质对ATP的亲和力。 d、 一个JAK激酶被两个二聚体受体结合，引起构象变化，从而增加活性。 e、 激酶与ATP结合，ATP从一个激酶结构域传递到下一个。 在MAP激酶级联中： a、 MAP激酶磷酸化转移到细胞核的转录因子。 b、 它是一种蛋白激酶而不是一种蛋白激酶 c、 没有一个答案是正确的。 c. None of the answers is correct. d、 信号被多个酶步骤放大。 d. the signal is amplified by multiple enzymatic steps. e、 抑制和激活途径结合使用。 e. the pathway uses a combination of inhibitory and activating phosphates. Cytokine receptors can repress transcription by: a. None of the answers is correct. b. activating proteolytic pathways. c. activating the MAP kinase pathway. d. All of the answers are correct. e. activating STAT proteins (signal transducer and activator of transcription). During activation of the epidermal growth factor (EGF) receptor: a. None of the answers is correct. b. the acceptor kinase activation loop gets phosphorylated. c. dimerization is asymmetric. d. All of the answers are correct. e. the donor kinase phosphorylates the acceptor. Ras activation is followed by: a. hormone binding to the receptor. b. None of the answers is correct. c. receptor tyrosine kinase phosphorylation. d. adapter-protein recruitment. e. receptor dimerization. Sos serves as a GEF (guanine exchange factor) for Ras by: a. hydrolyzing ATP. b. inducing a conformational change. c. hydrolyzing GTP. d. phosphorylating it. e. None of the answers is correct. PI-3 kinase functions to do all of the following EXCEPT: a. when mutated, promote tumor transformation. b. phosphorylate tyrosines on adaptor proteins. c. Actually, it does all of these. d. promote cell division. e. block apoptosis. What is required for complete PKB activation? a. recruitment to the plasma membrane by interaction with a phosphotyrosine residue b. phosphorylation by PDK2 c. phosphorylation by PDK1 on its activation site d. recruitment to the plasma membrane by PI 3-phosphates e. All of these are necessary for complete activation. 细胞因子受体可通过以下方式抑制转录： a、 没有一个答案是正确的。 b、 激活蛋白水解途径。 c、 激活MAP激酶通路。 d、 所有的答案都是正确的。 e、 激活STAT蛋白（信号转导和转录激活剂）。 在激活表皮生长因子（EGF）受体期间： a、 没有一个答案是正确的。 b、 受体激酶激活环被磷酸化。 c、 二聚反应是不对称的。 d、 所有的答案都是正确的。 e、 供体激酶磷酸化受体。 Ras激活之后是： a、 与受体结合的激素。 b、 没有一个答案是正确的。 c、 受体酪氨酸激酶磷酸化。 d、 适配蛋白募集。 e、 受体二聚。 Sos作为Ras的GEF（鸟嘌呤交换因子），通过： a、 水解ATP。 b、 引起构象变化。 c、 水解GTP。 d、 磷酸化。 e、 没有一个答案是正确的。 PI-3激酶的功能包括以下所有功能： a、 突变后，促进肿瘤转化。 b、 磷酸化衔接蛋白上的酪氨酸。 c、 事实上，所有这些都可以。 d、 促进细胞。 e、 阻断细胞凋亡。 完整激活PKB需要什么？ a、 通过与磷酸酪氨酸残基的相互作用补充到质膜上 b、 PDK2磷酸化 c、 PDK1在活化位点的磷酸化作用 d、 pi3-磷酸酯对质膜的募集作用 e、 所有这些都是完全激活所必需的。 Activation of protein kinase B requires all of the following 激活蛋白激酶B需要以下所有条件，除了它： EXCEPT its: a、 激活环上的磷酸化。 a. phosphorylation on the activation loop. b. phosphorylation at the C-terminus. c. recruitment to the membrane. d. Actually, all of these are required. e. dimerization. PI-3 kinase functions to do all of the following EXCEPT: a. promote cell division. b. when mutated, promote tumor transformation. c. phosphorylate tyrosines on adaptor proteins. d. Actually, it does all of these. e. block apoptosis. How is ubiquitination involved in Wnt-1 signaling? a. Ubiquitination plays more than one of the roles above. b. Mono-ubiquitination of β-catenin allows for nuclear localization of the protein. c. Wnt-1 prevents the ubiquitination of β-catenin. d. Wnt-1 protein is a target for ubiquitination and degradation. e. Dishevelled is a target for ubiquitination and degradation. Unlike Notch, amyloid β protein: a. undergoes an extracellular cleavage. b. undergoes an intermembrane cleavage. c. None of the answers is correct. d. All of the answers are correct. e. forms insoluble deposits. During the Notch signaling pathway, the: a. extracellular portion of Notch is endocytosed by the signaling cell. b. matrix metalloprotease ADAM 10 cleaves Delta. c. enzyme presenilin 1 cleaves Notch within the membrane. d. extracellular portion of Notch is endocytosed by the signaling cell and the enzyme presenilin 1 cleaves Notch within the membrane. e. enzyme presenilin 1 cleaves Notch within the membrane and the matrix metalloprotease ADAM 10 cleaves Delta. Cytokines include all of the following EXCEPT: a. interferons. b. Actually, these are all cytokines. c. interleukins. d. prolactin. e. defensins. b、 C端磷酸化。 c、 招募到膜上。 d、 实际上，所有这些都是必需的。 e、 二聚作用。 PI-3激酶的功能包括以下所有功能： a、 促进细胞。 b、 突变后，促进肿瘤转化。 c、 磷酸化衔接蛋白上的酪氨酸。 d、 事实上，所有这些都可以。 e、 阻断细胞凋亡。 泛素化是如何参与Wnt-1信号转导的？ a、 泛素化起着以上不止一个的作用。 b、 β-连环蛋白的单泛素化可以使蛋白质的核定位。 c、 Wnt-1可阻止β-catenin泛素化。 d、 Wnt-1蛋白是泛素化和降解的靶点。 e、 蓬乱是泛素化和退化的目标。 与Notch不同，淀粉样β蛋白： a、 细胞外。 b、 经历膜间断裂。 c、 没有一个答案是正确的。 d、 所有的答案都是正确的。 e、 形成不溶性沉积物。 在Notch信号通路中： a、 Notch的胞外部分被信号细胞内吞。 b、 基质金属蛋白酶adam10切割δ。 c、 早老素1酶在膜内切口。 d、 Notch的胞外部分被信号细胞内吞，而prenillin1在膜内裂解Notch。 e、 酶prenillin1在膜内切割缺口，基质金属蛋白酶adam10切割δ。 细胞因子包括以下所有成分，除了： a、 干扰素。 b、 实际上，这些都是细胞因子。 c、白细胞介素 d、 催乳素。 e、 防御素。 When the epidermal growth factor (EGF) receptor is in 当表皮生长因子（EGF）受体为单体形式时： monomeric form: a、 没有一个答案是正确的。 a. None of the answers is correct. b、 它不能与配体结合。 b. it cannot bind ligand. c、 直到信号被触发。 c. it, like rhodopsin, transduces signal until it is triggered. d. its activation loop obscures the kinase domain. e. it has higher than expected kinase activity. Receptor tyrosine kinases possess all of the following domains EXCEPT a: a. membrane-spanning α helix. b. kinase domain. c. ligand-binding domain. d. Actually, they possess all of these. e. dimerization domain. When the activation loop of MAP kinase is phosphorylated, all of the following take place EXCEPT: a. the protein undergoes a conformational change. b. Actually, all of these take place. c. its ability to bind substrate increases. d. its ability to bind ATP is not affected. e. it can translocate to the nucleus. Wnt signaling does NOT involve: a. activation of transcription factors. b. dimerization. c. blockage of degradation. d. Actually, it involves all of these. e. serine/threonine phosphorylation. MAP kinase phosphorylates targets: a. None of the answers is correct. b. All of the answers are correct. c. only when dimerized. d. in the cytoplasm. e. only when phosphorylated. d、 它的激活环掩盖了激酶结构域。 e、 它的激酶活性高于预期。 受体酪氨酸激酶具有除以下哪一项之外的所有结构 :a、 跨膜α螺旋。 b、 激酶结构域。 c、 配体结合域。 d、 实际上，他们拥有所有这些。 e、 二聚结构域。 当MAP激酶的激活环磷酸化时，以下所有情况都会发生，除了： a、 蛋白质发生构象变化。 b、 事实上，所有这些都会发生。 c、 其结合底物的能力增强。 d、 它结合ATP的能力不受影响。 e、 它可以转移到细胞核。 Wnt信令不涉及： a、 转录因子的激活。 b、 二聚作用。 c、 降解受阻。 d、 实际上，它涉及到所有这些。 e、 丝氨酸/苏氨酸磷酸化。 MAP激酶磷酸化靶点： a、 没有一个答案是正确的。 b、 所有的答案都是正确的。 c、 只有当二聚体。 d、 在细胞质中。 e、 只有当磷酸化时。
