分子生物学第八章课后答案教学内容

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杨荣武主编《分子生物学》课后习题答案.doc

第二章1．想想核酸的A260为什么会下降?肯定是形成双螺旋结构引起的。

那为什么相同序列的核酸，RNA 的A260下降，DNA的A260不变?这说明RNA能形成双链，DNA不能。

那么，为什么RNA能形成双链呢?原因肯定就在RNA和DNA序列上不同的碱基U和T上面。

U和T的含义完全一样，差别在于RNA分子上的U可以和G配对，而DNA分子上的T不能和G配对。

如果这时候能想到这一点，题目的答案也就有了。

本题正确的答案是：1个核酸的A260主要是4个碱基的π电子。

当一个核酸是单链的时候，π电子能吸收较大的光；但核酸为双链的时候，碱基对的堆积效应使π电子吸收较少的光。

于是，题目中的数据告诉我们，第一种序列的DNA和RNA在二级结构上没有什么大的差别。

而对于第二种序列的RNA光吸收大幅度减少，意味着RNA形成了某种双链二级结构，RNA二级结构的一个常见的特征是G和U能够配对。

从第二种序列不难看出，它能够自我配对，形成发夹结构，而降低光吸收。

2． RNA的小沟浅而宽，允许接近碱基边缘。

2′－OH位于小沟，提供氢键供体和受体，起稳定作用。

G：U 摇摆碱基对让G的氨基N2 位于小沟，能够与蛋白质相互作用（如在tRNAAla和同源的氨酰－tRNA 合成酶之间）。

3．（1）核酶由RNA组成，所以一定是RNA双螺旋，为A型。

（2）序列交替出现嘌呤和嘧啶，应该是Z型双螺旋。

（3）既然是DNA，在上述湿度条件下，要么是B型，要么是Z型。

由于B型比Z型更紧密（螺距比Z 型短，每个螺旋单位长度具有更多的电荷，相同数目碱基对的总长度要短）。

因此，1号一定是B型，2号为Z型DNA。

4．（1）（2）Arg（3）Asn和Gln5．使用dUTP代替dTTP并不能改变DNA双螺旋的结构。

T和U的差别只是在T嘧啶环上是否有一个甲基，这个甲基位于双螺旋的大沟之中。

如果用2′－OH取代2′－H，则合成出来的是RNA，于是螺旋变成A－型。

多出来的羟基产生空间位阻，致使RNA无法形成B－型双螺旋。

分子生物学第八章课后答案

1，基因家族的分类及其主要表达调控模式。

答：（1），简单多基因家族，真核生物首先是pre rRNA经过特异性甲基化，然后是经RNA 酶的切割便可产生成熟rRNA分子。

原核生物则还要经过核酸酶降解才能产生成熟rRNA分子。

（2），复杂多基因家族，一般由几个相关基因家族构成，基因家族之间由间隔序列隔开，并作为独立的转录单位，可能存在具有不同专一性的组蛋白亚类和发育调控机制。

（3），发育调控的复杂多基因家族，每个基因家族中，基因排列的顺序就是他们在发育阶段的表达顺序。

2，何为外显子和内含子及其结构特点和可变调控？答：大多数真核基因都是由蛋白质编码序列和非蛋白质编码序列组成的，编码序列称为外显子（exon），非编码序列称为内含子（intron）。

结构特点：一个结构基因中编码某一蛋白质不同区域的各个外显子并不连续排列在一起，而是常常被长度不等的内含子所隔离，形成镶嵌排列的断裂方式。

可变调控：不少真核基因的原始转录产物可通过不同剪接方式，产生不同的mRNA，并翻译成不同的蛋白质。

另外，一些核基因由于转录是选择了不同的启动子或者在转录产物上选择了不同的PolyA位点而使转录产物产生不同的二级结构，因而影响剪接过程，最终产生不同的mRNA分子。

3，DNA甲基化对基因表达的调控机制。

答：大量研究表明，DNA甲基化能关闭某些基因达的活性，去甲基化则诱导了基因的重新活化和表达。

三种调控机制：一是DNA甲基化导致了某些区域DNA构象变化，从而影响了蛋白质和DNA的相互作用，抑制了转录因子与启动区DNA的结合效率。

二是促进阻遏蛋白的阻遏作用。

三是DNA的甲基化还提高了该位点的突变频率。

4，真核生物转录元件组成及其分类。

答：启动子，转录模版，RAN聚合酶Ⅱ基础转录所需的蛋白质因子（TFⅡ），RNA聚合酶Ⅱ，增强子，反式作用因子5，增强子的作用机制。

答：增强子是指能使与它连锁的基因转录频率明显增加的DNA序列。

可能有3中作用机制：（1），影响模版附近的DNA双螺旋结构，导致DNA双螺旋弯折或在反式因子的参与下，以蛋白质之间的相互作用为媒介形成增强子鱼启动子之间“成环”连接，活化基因转录。

分子生物学课后答案

分⼦⽣物学课后答案第⼀章绪论1、简述孟德尔、摩尔根与沃森等⼈对分⼦⽣物学发展得主要贡献。

答:孟德尔得对分⼦⽣物学得发展得主要贡献在于她通过豌⾖实验,发现了遗传规律、分离规律及⾃由组合规律;摩尔根得主要贡献在于发现染⾊体得遗传机制,创⽴染⾊体遗传理论,成为现代实验⽣物学奠基⼈;沃森与克⾥克在1953年提出DAN反向双平⾏双螺旋模型。

2、写出DNA与RNA得英⽂全称。

答:脱氧核糖核酸(DNA, Deoxyribonucleic acid), 核糖核酸(RNA, Ribonucleic acid)3、试述“有其⽗必有其⼦”得⽣物学本质。

答:其⽣物学本质就是基因遗传。

⼦代得性质由遗传所得得基因决定,⽽基因由于遗传得作⽤,其基因得⼀半来⾃于⽗⽅,⼀般来⾃于母⽅。

4、早期主要有哪些实验证实DNA就是遗传物质？写出这些实验得主要步骤。

答:⼀,肺炎双球菌感染实验,1,R型菌落粗糙,菌体⽆多糖荚膜,⽆毒,注⼊⼩⿏体内后,⼩⿏不死亡。

2,S型菌落光滑,菌体有多糖荚膜,有毒,注⼊到⼩⿏体内可以使⼩⿏患病死亡。

3,⽤加热得⽅法杀死S型细菌后注⼊到⼩⿏体内,⼩⿏不死亡;⼆,噬菌体侵染细菌得实验:1,噬菌体侵染细菌得实验过程:吸附→侵⼊→复制→组装→释放。

2,DNA中P 得含量多,蛋⽩质中P得含量少;蛋⽩质中有S⽽DNA中没有S,所以⽤放射性同位素35S标记⼀部分噬菌体得蛋⽩质,⽤放射性同位素32P标记另⼀部分噬菌体得DNA。

⽤35P标记蛋⽩质得噬菌体侵染后,细菌体内⽆放射性,即表明噬菌体得蛋⽩质没有进⼊细菌内部;⽽⽤32P标记DNA 得噬菌体侵染细菌后,细菌体内有放射性,即表明噬菌体得DNA进⼊了细菌体内。

三,烟草TMV得重建实验:1957年,Fraenkel-Conrat等⼈,将两个不同得TMV株系(S株系与HR株系)得蛋⽩质与RNA分别提取出来,然后相互对换,将S株系得蛋⽩质与HR株系得RNA,或反过来将HR株系得蛋⽩质与S株系得RNA放在⼀起,重建形成两种杂种病毒,去感染烟草叶⽚。

分子生物学(山东联盟-山东第一医科大学)智慧树知到课后章节答案2023年下山东第一医科大学

分子生物学（山东联盟-山东第一医科大学）智慧树知到课后章节答案2023年下山东第一医科大学山东第一医科大学第一章测试1.真核生物DNA缠绕在组蛋白上构成核小体，核小体含有的蛋白质是答案:H2A、H2B、H3、H4各两分子2.鸟嘌呤和胞嘧啶之间联系是由两对氢键形成答案:错3.核酸中核苷酸之间的连接方式是答案:3',5'磷酸二酯键4.含有稀有碱基比例较多的核酸分子答案:tRNA5.真核细胞m RNA帽样结构中最多见的是答案:m7GpppNmp(Nm)pN6.tRNA的分子结构特征是答案:有反密码环和3'-端C-C-A7.核酸变性后可发生哪种效应答案:增色效应8.Tm值与DNA的分子大小和所含碱基中的G和C所占比例成正比答案:对9.DNA的解链温度指的是答案:A260nm达到最大变化值的50%时的温度第二章测试1.大肠杆菌中主要行使复制功能的酶是答案:DNA聚合酶Ⅲ2.以下哪种酶具有5′-3′外切酶活性答案:DNA聚合酶I3.DNA复制中的引物是答案:由DNA为模板合成的RNA片段4.下列关于DNA复制的叙述，错误的是答案:两条子链均连续合成5.将下列蛋白按其参与DNA复制的顺序排列，正确的是：a = 引物酶b = 解旋酶c = SSBd = DNA 聚合酶 I答案:b,c,a,d6.在原核细胞和真核细胞中，染色体DNA都与组蛋白形成复合体答案:错7.冈崎片段产生的原因是答案:后随链复制与解链方向不同8.原核生物的DNA聚合酶I和DNA聚合酶III都是由一条多肽链组成。

答案:错9.关于大肠杆菌复制起始点的叙述，正确的是答案:其保守序列AT含量较高10.DNA拓扑异构酶的作用是答案:改变DNA分子拓扑构象，理顺DNA链第三章测试1.关于DNA单链断裂的叙述哪一项错误答案:对真核生物来说是致死性的2.关于嘧啶二聚体的形成，下列说法正确的是答案:形式有CC，TC，TT;紫外线照射可引起该损伤;属于DNA链内交联3.所有突变都会引起基因型和表型的改变，因此突变都是有害的答案:错4.常见的DNA损伤修复系统包括答案:切除修复;直接修复;SOS修复;重组修复5.与切除修复过程缺陷有关的疾病是答案:着色性干皮病6.最重要和有效的DNA 修复方式为答案:切除修复7.碱基错配的直接修复是在半甲基化状态下完成的答案:对8.容易产生突变的修复方式是答案:SOS修复第四章测试1.原核生物RNA聚合酶各亚基中与利福平结合的亚基是答案:RNA聚合酶亚基β2.催化真核生物mRNA转录生成的酶是答案:RNA聚合酶Ⅱ3.催化真核生物mRNA生物合成的RNA聚合酶Ⅱ对α-鹅膏蕈碱的反应为答案:高度敏感4.下列关于σ因子的叙述正确的是答案:参与识别DNA模板上转录RNA的特殊起始点5.下列关于DNA复制和转录的描述中哪项是错误的?答案:两过程均需RNA为引物6.原核基因-10区的Pribnow盒是RNA聚合酶对转录起始的辨认位点。

分子生物学智慧树知到课后章节答案2023年下山东农业大学

分子生物学智慧树知到课后章节答案2023年下山东农业大学山东农业大学第一章测试1.格里菲斯转型实验得出了什么结论（）答案:DNA是生命的遗传物质，蛋白质不是遗传物质2.现代遗传工程之父Paul Berg建立了什么技术（）答案:重组DNA技术3.下列哪种技术可以用于测定DNA的序列（）答案:双脱氧终止法4.RNA干扰是指由单链RNA诱发的基因沉默现象，其机制是通过阻碍特定基因的翻译或转录来抑制基因表达。

（）答案:错第二章测试1.比较基因组学是基于基因组图谱和测序基础上，对已知的基因和基因组结构进行比较，来了解基因的功能、表达机理和物种进化的学科。

（）答案:对2.以下哪项是原核生物基因组的结构特点（）答案:操纵子结构3.细菌基因组是（）答案:环状双链DNA4.下列关于基因组表述错误的是（）答案:真核细胞基因组中大部分序列均编码蛋白质产物5.原核生物的结构基因多为单顺反子，真核生物的结构基因多为多顺反子。

( )答案:错6.病毒基因组可以由DNA组成，也可以由RNA组成。

( )答案:对第三章测试1.在原核生物复制子中以下哪种酶除去 RNA 引发体并加入脱氧核糖核苷酸？（）答案:DNA 聚合酶 I2.使 DNA 超螺旋结构松驰的酶是（）。

答案:拓扑异构酶3.从一个复制起点可分出几个复制叉？（）答案:24.所谓半保留复制就是以 DNA 亲本链作为合成新子链 DNA 的模板，这样产生的新的双链 DNA 分子由一条旧链和一条新链组成。

( )答案:对5.DNA 的5′→3′合成意味着当在裸露3′→OH 的基团中添加 dNTP 时，除去无机焦磷酸 DNA链就会伸长。

( )答案:对第四章测试1.对RNA聚合酶的叙述不正确的是（）。

答案:全酶不包括ρ因子2.原核生物RNA聚合酶识别启动子位于（）。

答案:转录起始位点上游3.增强子与启动子的不同在于（）。

答案:增强子与转录启动无直接关系4.启动子总是位于转录起始位点的上游。

基因的分子生物学英文版答案molecular chapter8

Chapter 8 AnswersRNase H is a specific RNase that removes RNA from RNA:DNA hybrids. It leaves the last RNA base pair, which must be removed by a 5’ exonuclease. (This 5’ exonuclease is part of DNA pol I.1. a. See Figures 9.1 and 9.2 for views of the primer:template junction.b. See Figure 9.2. In the reaction, the -OH group at the 3' end of the polynucleotide chain attacks the phosphate of the incoming nucleoside triphosphate, linking the nucleotide to the chain and releasing a pyrophosphate group.The pyrophosphate is rapidly hydrolyzed by an enzyme called pyrophosphatase. The hydrolysis of pyrophosphate releases a substantial amount of free energy (on the order of 10-7 kcal/mol), providing enough free energy to drive the overall reaction forward and to make it effectively irreversible.2. The first mechanism used by DNA polymerase to ensure that the inserted base is correct, called kinetic proofreading, relies on the exclusive ability of correctly paired bases to sit properly within the active site of the enzyme. If the incoming nucleotide is not complementary to the template base at the active site, then the nucleotide cannot properly align with the template and the polymerization reaction becomes much slower. This gives the incorrectly matched nucleotide time to leave the active site and to be replaced by a correctly matched one. Kinetic proofreading reduces polymerase's error rate to about 10-5.The second proofreading mechanism uses the 3' ⌫ 5' exonuclease activity of many DNA polymerases. This activity allows polymerase to backtrack and remove the nucleotide that was most recently incorporated into the primer strand. Although this 3' ⌫ 5' exonuclease activity works on both correctly and incorrectly-matched nucleotides, it is much more active in removing incorrectly matched ones, in part because incorrectly paired bases slow the polymerase down and give the exonuclease more time to work. This second level of proofreading increases the accuracy of replication to about 1 error per 107 bases.3. The "palm," "thumb," and "hand" refer to three different subdomains of the DNA polymerase enzyme.The palm, made up of a β sheet that includes the key components of the polymerase's active site, contributes to replication in several critical ways. First, it binds to metal ions that help remove the hydrogen from the –OH group at the 3' end of the primer, producing a 3' O–that can attack the α phosphate of an incoming nucleoside triphosphate. The metal ions also help to counteract the negative charges of the β and γ phosphates of incoming nucleoside triphosphates as well as the pyrophosphate that is released when a nucleotide is added to a growing primer chain. Finally, the palm contributes to replication accuracy by stabilizing newly synthesized DNA that is correctly base paired; if an incorrect nucleotide is incorporated, the absence of palm-mediated stabilization slows down the polymerase allowing its 3' 5' exonuclease activity to correct the error.The fingers contribute to replication by binding to the incoming nucleoside triphosphates. If the incoming nucleoside triphosphate is complementary to the nucleotide present on the template strand, the fingers move by approximately 40° to "close" around the base pair. This closed state creates a tight pocket that holds the base pair in place and allows catalysis to occur. The fingers also bend the template strand in a way that exposes the nucleotide present at the catalytic site, helping to ensure that the correct base is used for base pairing with the incoming nucleoside triphosphate.The thumb does not contribute directly to catalysis but instead interacts with recently-synthesized DNA to keep the primer and the active site in the correct position and to inhibit the dissociation of the polymerase from the DNA.4. Polymerase can effectively avoid incorporating ribonucleotides into the growing DNA strand because of the precise way in which the enzyme binds deoxyribonucleotides. Specifically, polymerase has two amino acids whose side chains project into the nucleotide binding pocket of the enzyme and occupy the same space that the 2' hydroxyl group of a ribonucleotide would if it were to enter the pocket. Accordingly, ribonucleotides are sterically excluded from the active site of the enzyme.Ribonucleotides are used in vivo to make up the primers that DNA polymerase requires to carry out DNA synthesis. These short RNA primers are synthesized by an enzyme called primase and are required to start both leading and lagging strand synthesis.The ribonucleotides that are incorporated into the strand as a result of primer synthesis (and the extension of the primers by DNA polymerase) are ultimately removed by the 5' ⌫ 3' exonuclease activity of the DNA polymerase I enzyme. DNA polymerase I binds just upstream of the RNA primer, uses its 5' ⌫ 3' exonuclease activity to remove the ribonucleotides, and then replaces them with deoxynucleotides.5. Okazaki observed that the radioactive label ended up in either of two dramatically different size populations. One population, corresponding to the leading strands, was made up of extremely large DNA molecules. The other population, comprising the lagging strands (for example, Okazaki fragments), consisted of very short (1000–2000 nucleotides long) strands of DNA.It was critical for Okazaki to use a short labeling period and then quickly isolate the DNA because Okazaki fragments only exist transiently. As lagging strand synthesis progresses, DNA polymerase I replaces the RNA primers with deoxynucleotides and the strands are linked together by DNA ligase. Accordingly, if Okazaki had waited too long he might have failed to see the short strands because most of the label would have ended up in longer strands of DNA.DNA聚合酶I的具体作用？和RNA primer6. The size of Okazaki fragments in a given organism reflects the specific kinetics governing the association between primase and the other replication components during lagging strand synthesis in the cell. Primase only joins the replication fork periodically, sticking around just long enough to synthesize an RNA primer and thereby trigger lagging strand synthesis. The frequency with which primase joins the fork can differ between organisms, leading to characteristic differences in the length of the Okazaki fragments. In organisms where primase has a high affinity for the other replication factors (such as helicase), for example, the enzyme will associate with the fork more frequently, resulting in shorter Okazaki fragments. Alternatively, when the primase has less affinity for the fork, it will associate less frequently, leading to longer Okazaki fragments.You could experimentally address the role of primase and its association with the replication fork in determining Okazaki fragment length by altering the association and seeing what effect this has on fragment length. For example, you could increase or decrease the concentration of primase in a cell to see if higher concentrations (which should lead tomore frequent primase-fork interactions) result in shorter Okazaki fragments, and if lower concentrations lead to longer ones. You could also isolate mutant forms of primase that have higher or lower affinities for helicase (or other replication components) and ask whether the Okazaki fragment length is altered in the mutant cells.7. DNA upstream of the replication fork is unwound by a class of enzymes called DNA helicases. Helicases, which are typically hexameric protein complexes, encircle one of the two DNA strands upstream of the fork and use the energy provided by ATP hydrolysis to separate the strands. The single-stranded DNA produced by the helicase is rapidly coated by proteins called single-stranded binding protein (SSB), helping to prevent the separated strands from reannealing before the DNA polymerase arrives.The helicase-mediated unwinding of the DNA introduces positive supercoils i nto the double-stranded DNA upstream of the fork. Enzymes called type I topoisomerases remove the supercoils by nicking one of the two strands of the DNA, allowing the strands to rotate until they regain a relaxed state, and then re-sealing the nick. This topoisomerase I activity is critical for replication because, in its absence, the positive supercoils would prevent the replication machinery from advancing.8. Because individual SSB tetramers bind both to DNA and to adjacent tetramers on the DNA, a new tetramer will have more affinity for a spot on the DNA immediately next to an already bound tetramer than it will for DNA with no bound SSB. This difference in affinity provides the basis of the cooperative binding of SSB to single-stranded DNA, and allows SSB to rapidly spread along exposed stretches of single-stranded DNA. In this way, SSB can completely and efficiently coat single-stranded DNA as it ‘from the helicase.9. Sliding clamps are loaded onto and removed from the DNA by specialized, ATP-hydrolyzing protein complexes called sliding clamp loaders. When bound to ATP, the clamp loader opens up the clamp, locates a primer:template junction in the DNA, and places the (still open) clamp around the DNA. The interaction between the clamp loader and the DNA causes the loader to hydrolyze its bound ATP, which causes in turn the clamp loader to release the clamp and dissociate from the DNA. With the clamp loader gone, the clamp closes to form a ring around the DNA.Clamp loaders can also remove clamps from DNA, presumably through a similar ATP-dependent cycle of clamp opening and closing. During replication, however, unloading of the clamp is inhibited by the presence of DNA polymerase, because polymerase and the clamp loader bind to the same face of the clamp. Only when the polymerase has left the clamp (for example, following the completion of the synthesis of an Okazaki fragment), can the clamp loader step in and remove the clamp from the DNA.One major purpose of the clamp's unusual structure is to tether replication components to the DNA, in particular to enhance the processivity of DNA polymerase. Because the clamps are only slowly removed from DNA following replication, the clamps also serve as a marker for recently replicated DNA. Such a marker can be useful for many purposes, for example to target enzymes involved in nucleosome assembly.10. The replisome contains numerous distinct activities, including helicase activity, primase activity, clamp loading activity, and the DNA polymerase and 3' to 5' exonuclease activities provided by DNA Pol III.In a process as complicated as replication, it makes a lot of sense that the enzymes acting to catalyze the various steps of the process work together in a sort of "factory." In this way, for example, sequential steps in the process can occur in a coordinated and efficient way. Having the helicase associated with the polymerase, for instance, allows the DNA to be unwound and then immediately replicated, which is clearly more efficient than if DNA unwinding and replication were uncoupled. Also, the coordinated action of the primase, clamp loader, and polymerase at the replication fork allows rapid synthesis of Okazaki fragments as the fork moves along the chromosome.The association of the various replication components in a single factory also helps them to regulate each other, allowing even greater coordination between the different steps of the process. For example, helicase activity is stimulated by the clamp loader, which actsto keep the helicase from moving too far ahead of the other components of the replication machinery. Another example is provided by the interaction between primase and helicase, the strength of which determines the frequency at which primase joins the replication fork to synthesize RNA primers during lagging strand synthesis, and which therefore also determines Okazaki fragment length.在复制叉(replication fork)处, 由所有参与DNA合成的蛋白质如聚合酶Ⅲ全酶、解旋酶(helicase)、引物酶(primase)组成了一个复制小体(replisome)-----DNA合成工厂.(1)DNA聚合酶Ⅲ全酶与解旋酶间的相互作用: 通过т亚基介导. 相互作用可以使解旋酶活性提高10倍, 因此当解旋酶与DNA聚合酶Ⅲ脱离时,活性的降低使其解链的速率下降,从而保证DNA合成速度与解旋过程高度协调.(2)解旋酶与引物酶间的相互作用, 引物酶与解旋酶及SSB蛋白间的相互作用是一个动态过程,约为1次/秒,而且作用力相对较弱,但与解旋酶的相互作用可使其活性提高1000倍, 一旦引物合成完毕,引物酶从DNA上脱落. 对岗崎片段的长度调节非常重要(原核生物一般为1000-2000bp; 真核生物为100-400bp)11. Initiators discussed in the chapter include DnaA (from E. coli), large T antigen (from the virus SV40), and the ORC complex (found in eukaryotic cells).While these initiators have much in common—they all bind to replicator elements, hydrolyze ATP, and recruit replication factors to the origin, for example—they also differ in important respects. For example, whereas DnaA and large T antigen bind to origins by themselves (or as homomultimers), ORC is a complex made up of six distinct proteins. In addition, both DnaA and T antigen can bind origins even in the absence of ATP, whereas ORC requires ATP binding for any sequence-specific origin binding. Finally, while DnaA and T antigen can both bind to the replicator and directly unwind the DNA, ORC can only bind to the replicator and has to recruit other proteins to do the unwinding.12. The traditional assay used to identify ARS sequences involves inserting random genomic fragments into yeast plasmids containing a selectable marker but lacking a replication origin, and then seeing if any of the plasmids are capable of replicating in yeast. Any genomic fragments that are capable of conferring replication ability on the plasmids must include a functional replication origin. Such fragments are called ARSs, for autonomously replicating sequences.If you identify an ARS using this assay, you can conclude that the sequence is capable of serving as a replication origin in vivo, but you cannot say whether or not the sequence ever actually serves as an origin in vivo. You can imagine a sequence, for example,that is perfectly capable of behaving as an origin (that is, binding the ORC complex, unwinding, and loading replication factors) but which is prevented from ever doing so in vivo because of its genomic location (for example, because of the local chromatin structure, or because it is located next to an even more powerful origin). It is only when the sequence is placed in the relatively simple context of a plasmid during the ARS assay that this potential origin activity is expressed.The best way to assess the in vivo role of a potential origin is to actually observe its behavior in a cell. One standard way to do this involves using a specialized agarose gel assay that can distinguish the "bubble" shaped structure characteristic of replication origins. You could therefore isolate DNA from proliferating cells, digest the DNA using restriction enzymes, run the digested DNA on such a gel, transfer the DNA to a membrane, and probe the transferred DNA using a labeled oligonucleotide specific for your ARS sequence. If your probe hybridizes with the DNA located in the part of the membrane corresponding to the "bubble" structured DNA, you can conclude that your sequence actually does serve as a replication origin in vivo.13. Replication initiation in eukaryotic cells is separated into two distinct steps. In the first step, occurring during the G1 phase of the cell cycle, multiple proteins bind to potential replicators to form a pre-replicative complex (pre-RC) that includes the ORC complex, Cdc6, Cdt1, and the MCM proteins. While pre-RC assembly marks the replicator as a potential site for replication initiation, it does not lead by itself to unwinding of the DNA, nor does it guarantee that the replicator will ultimately serve as a replication origin during S phase. The second step, marked by the appearance of the kinases Ddk and Cdk, occurs at the beginning of S phase. These kinases cause other replication factors to join the origin: first Sld3, Cdc45, and Mcm10, and subsequently the DNA polymerases, sliding clamp loader, primase, and the sliding clamps. Once all of these proteins have joined the origin, replication can begin.The restriction of these two steps to different phases of the cell cycle helps limit replication to once per cell cycle because of the dual role of Cdk in inhibiting pre-RC formation and in promoting origin firing. Specifically, pre-RCs can only form during G1 phase, when Cdk activity is absent, but at the same time the lack of Cdk activity during G1 means that the pre-RCs that form cannot complete origin assembly and initiate replication. Subsequently, when Cdk activity appears at the beginning of S phase, already assembled pre-RCs can proceed with origin assembly and begin replicating, but no new pre-RCs can form because of the continuous presence of Cdk (which persists until the following mitosis, disappearing as cells re-enter G1 phase).14. Bacterial cells rely on the methylation state of a four nucleotide sequence called "Dam" to prevent re-replication of recently-copied DNA. Specifically, OriC contains multiple "Dam" sites that are usually methylated on both strands and fully capable of initiating replication. When OriC is replicated, however, the Dam sites become transiently hemi-methylated, because the newly synthesized strand is initially unmethylated. This hemi-methylated state prevents any further initiation events at OriC, because hemi-methylated Dam sites attract a protein called SeqA that both inhibits DnaA binding to OriC and prevents re-methylation of the site.Mutations in dam methylase should disrupt the transient resistance to re-replication. Without Dam methylase, Dam sites would be permanently unmethylated and therefore never bound by SeqA. Because SeqA normally inhibits DnaA binding, such a loss of SeqA binding would permit DnaA to re-bind OriC immediately after the replication fork has left.15. While leading strand synthesis can proceed until the last nucleotide on the chromosome is copied, the replication fork is incapable of copying the very end of the lagging strand because primase cannot function at the end of a DNA molecule. This is a potentially major problem for the cell because it could result in the loss of several bases from the end of one of the chromosomes with every round of DNA replication.Eukaryotic cells have several strategies for avoiding this potential loss of terminal nucleotides. For example, some bacterial cells and viruses use a specialized protein that can replace the primase at the end of the chromosome by attaching to the chromosome end and priming synthesis of the terminal nucleotides. Another example is seen in eukaryotic cells, where an enzyme called telomerase helps maintain chromosome ends by extending them through the addition of multiple copies of a specific repeated sequence (for example, TTAGGG in humans).。

2024年《分子生物学》全册配套完整教学课件pptx

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运输功能
如载体蛋白，血红蛋白等，在生物体内运输各种物质。
免疫功能
如抗体蛋白，参与生物体的免疫应答。
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蛋白质的功能与调控
调节功能
如激素，生长因子等，调节生物体的生长发育和代谢过程。
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储存功能
如植物种子中的贮藏蛋白，动物体内的肌红蛋白等，储存能量和营养物质。
个性化医疗
根据患者的基因信息，制定个性化的治疗方案。
药物基因组学
预测患者对药物的反应和副作用，指导合理用药。
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基因治疗的原理与应用
基因治疗的原理
通过导入正常基因或修复缺陷基因，从而治疗由基因突变引起的疾病。
遗传性疾病的治疗
如视网膜色素变性、腺苷脱氨酶缺乏症等。
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癌症治疗
利用基因编辑技术，修复或敲除癌症相关基因，抑制肿瘤生长。
基因表达调控的层次
基因表达调控可分为转录前调控、转录水平调控、转录后调控和翻译水平调控等多个层次。
基因表达调控的意义
基因表达调控对于生物体的生长发育、代谢、免疫应答等生理过程具有重要意义，同时也是疾病发生发展的重要因素。
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原核生物的基因表达调控
1 2 3
原核生物基因表达调控的特点
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DNA损伤的修复机制
直接修复
针对某些简单的DNA损伤，如碱基错配，可通过特定的酶直接进行修复。
碱基切除修复
通过识别并切除受损碱基，再合成新的DNA片段进行修复。
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核苷酸切除修复
针对较严重的DNA损伤，如嘧啶二聚体，通过切除一段包含受损部

分子生物学-8_真题(含答案与解析)-交互

分子生物学-8(总分100, 做题时间90分钟)是非题1.基因转录可以发生在DNA分子的任何区域。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] 基因转录只发生在DNA分子上具有转录活性的区域。

2.真核生物RNA pol Ⅲ催化转录的基因启动子都属于基因内启动子。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] RNA pol Ⅲ催化转录的基因启动子并不都属于基因内启动子，例如编码U6-snRNA基因的启动子位于基因上游。

3.大肠杆菌RNA聚合酶全酶由4个亚基(σ2ββ")组成。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] 大肠杆菌RNA聚合酶全酶由6个亚基(α2ββ"ωσ)组成。

4.大肠杆菌RNA聚合酶转录几乎所有蛋白质编码基因。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:正确5.芽孢杆菌基因转录过程中RNA聚合酶结合在oriC位点。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] 芽孢杆菌基因转录过程中RNA聚合酶结合在启动子。

6.细菌RNA聚合酶全酶与非特异性DNA序列具有较高的亲和性。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] 全酶与非特异性DNA序列具有较低的亲和性，而核心酶却具有较高的亲和性。

7.RNA聚合酶催化的反应无需引物，也无校对功能。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:正确8.RNA聚合酶的转录产物大多以腺嘌呤核苷酸(A)开始。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:正确9.** RNA聚合酶的ω亚基是必须亚基，在体外影响酶活性。

SSS_JUDGEMENT正确错误该题您未回答:х该问题分值: 2.5答案:错误[解析] ω亚基在体外不影响酶活性，但在酶的组装过程中起作用。

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分子生物学第八章课
后答案
1，基因家族的分类及其主要表达调控模式。

答：（1），简单多基因家族，真核生物首先是pre rRNA经过特异性甲基化，然后是经RNA酶的切割便可产生成熟rRNA分子。

原核生物则还要经过核酸酶降解才能产生成熟rRNA分子。

（3），发育调控的复杂多基因家族，每个基因家族中，基因排列的顺序就是他们在发育阶段的表达顺序。

2，何为外显子和内含子及其结构特点和可变调控？
答：大多数真核基因都是由蛋白质编码序列和非蛋白质编码序列组成的，编码序列称为外显子（exon），非编码序列称为内含子（intron）。

可变调控：不少真核基因的原始转录产物可通过不同剪接方式，产生不同的mRNA，并翻译成不同的蛋白质。

3，DNA甲基化对基因表达的调控机制。

答：大量研究表明，DNA甲基化能关闭某些基因达的活性，去甲基化则诱导了基因的重新活化和表达。

三种调控机制：一是DNA甲基化导致了某些区域DNA 构象变化，从而影响了蛋白质和DNA的相互作用，抑制了转录因子与启动区DNA的结合效率。

二是促进阻遏蛋白的阻遏作用。

三是DNA的甲基化还提高了该位点的突变频率。

4，真核生物转录元件组成及其分类。

答：启动子，转录模版，RAN聚合酶Ⅱ基础转录所需的蛋白质因子（TFⅡ），RNA聚合酶Ⅱ，增强子，反式作用因子
5，增强子的作用机制。

答：增强子是指能使与它连锁的基因转录频率明显增加的DNA序列。

（2），将模版固定在细胞核内特定位置，如连接在核基质上，有利于DNA拓扑异构酶改变DNA双螺旋结构的张力，促进RNA 聚合酶Ⅱ在DNA链上的结合和滑动。

（3），增强子区可以作为反式作用因子或RNA聚合酶Ⅱ进入染色质结构的“入口”。

6，反式作用因子的结构特点及其对基因表达的调控。

答：反式作用因子是能直接或间接的识别或结合在各类顺式作用元件核心序列上，参与调控靶基因转录效率的蛋白质。

结构特点：
7，举例说明蛋白质磷酸化如何影响基因表达。

答：蛋白质磷酸化主要影响细胞信号转导进而影响基因表达。

举例：在糖原代谢过程中，激素与其受体在肌细胞外表面相结合，诱发细胞质cAMP的合成并活化A激酶，后者再将活化磷酸基团传递给无活性的磷酸化酶激酶，活化糖原磷酸化酶，最终将糖原磷酸化，进入糖酵解途径并提供ATP。

（cAMP介导的蛋白质磷酸化过程）
8，组蛋白乙酰化和去乙酰化影响基因转录的机制。

答：组蛋白乙酰转移酶和去乙酰化酶通过是组蛋白乙酰化和去乙酰化对基因表达产生影响。

组蛋白N端尾部上赖氨酸残基的乙酰化中和了尾部的正电荷，降低了它与组蛋白的亲和性，导致核小体构象发生有利于转录调节蛋白与染色质结合的变化，从而提高了基因转录的活性。

核心组蛋白H2A,H2B,H3,H4通过组蛋白尾部选择性乙酰化影响核小体的浓缩水平和可接近性。

由于乙酰化的组蛋白抑制了核小体的浓缩，使转录因子更容易与基因组的这一部分相接触，有利于提高基因的转录活性。

9，激素影响基因表达的基本模式。

答：许多类固醇激素（如雌激素、孕激素、醛固酮、糖皮质激素和雄激素）以及一些代谢性激素（如胰岛素）的调控作用都是通过起始基因转录而实现的。

靶细胞具有专一的细胞质受体，可与激素形成复合物，导致三维结构甚至化学性质的变化。

经修饰的受体与激素复合物通过核膜进入细胞核与染色质的特定区域结合导致基因转录的起始或关闭。

靶细胞内含有大量激素受体蛋白，而非靶细胞中没有或很少有这类受体，这是激素调节转录组织特异性的根本原因。

10，分子伴侣的分类及其影响基因表达的机制。

答：分子伴侣（molecular chaperone）是一类序列上没有相关性担忧共同功能的保守性蛋白质，它们在细胞内能帮助其他多肽进行正确的折叠、组装、运转和降解。

目前认为分子伴侣至少有两类：热休克蛋白家族和伴侣素。

把能与某个（类）专一蛋白因子结合，从而控制基因特意表达的DNA上游序列称为应答元件（response element），如热休克应答元件，这些应答元件与细胞内转移的转录因子相互作用，协调相关基因的转录。