基因的分子生物学英文版答案molecular chapter14'

Chapter 3 Answers1. Because secondary bonds are significantly weaker than covalent bonds, you could distinguish between these possibilities by assessing the overall stability of the protein or protein complex. Heating the enzyme, for example, would likely be sufficient to separate distinct polypeptide chains, but would not be enough to break apart a single chain.2. Because covalent bonds are stronger, the length of the bonds is much shorter than it is with weak bonds. Also, covalent bonds are more constrained than weak bonds in several respects. For example, covalent bonds involving a double or triple bond have no freedom of rotation, in contrast with weak bonds, which are not limited by their relative orientation. In addition, the angle between any two covalent bonds is fixed, in contrast to the angle between any two weak bonds. Finally, a given atom can only bond with a very small number of other atoms (determined by its valence), whereas a weak bond has more flexibility in this regard.Because secondary bonds are generally too weak to support a stable interaction between molecules by themselves, stability must come from a collection of multiple weak bonds acting together. The flexibility of weak bonds can help in this regard, because it facilitates the formation of these multiple weak interactions. For example, when two proteins bind to each other, numerous secondary bonds are formed between the exposed molecular groups of each of the proteins. If each of these bonds were subject to the same rigid constraints of covalent bonds, the number of possible weak interactions would be limited, and the demands placed on the precise structure of each protein would be that much greater.3. Beyond the obvious importance of strong (covalent) bonds in maintaining the primary structure of polymerase, other enzymes involved in DNA replication, and the DNA strands themselves, strong bonds are also important for providing the energy required for replication. For example, the hydrolysis of ATP provides the energy needed for replication to proceed. Weak bonds also play a critical role in replication. For example, the hydrogen bonds formed between base pairs help ensure the accurate incorporation of nucleotides into the growing polynucleotide chain. Also, the multiple interactions between the various proteins involved in replication, as well as between the proteins and the DNA, all depend on weak bonds and are essential parts of the replication process.4. The hybridization of the two strands of DNA would involve the greatest loss of free energy. Enzyme-substrate interactions involve many fewer weak bonds, which makes sense because this interaction is transitory by nature. Antibody-antigen interactions are typically more extensive, and thus stronger, but still involve only a handful of specific interactions. Hybridization of two DNA molecules yields the most free energy, as energy is provided by at least two hydrogen bonds formed per base pair as well as by the stacking interactions between adjacent bases.5. During a spontaneous reaction, the level of free energy drops. The free energy can be lost as heat or as an increase in entropy. When a collection of atoms reaches its lowest state, it is in a state of equilibrium.6. To calculate the K eq based on a given∆G, we use the formula: K eq = e-∆G/RT. If R = 1.987 cal/deg-mol, and T = 298, then using this formula, we find that the K eq is e-(-2000cal/mol)/(1.987cal/deg-mol)(298 deg), or 29.3.This K eq means that at equilibrium, the ratio of the bound proteins to the unbound proteins is nearly 30:1, meaning that virtually all of the proteins will bind to each other if present at molar concentrations.7. Van der Waals forces are produced from random fluctuations in the distribution of electrons within molecules. Because of these fluctuations, individual atoms experience temporary partial charges, and these partial charges can induce an opposite charge in neighboring atoms. These opposite charges then cause the atoms to be somewhat attracted to each other. Van der Waals attractions are very weak, typically about 1 kcal/mol, which is just slightly greater than the average thermal energy of molecules at room temperature. Van der Waals forces usually involve nonpolar atoms and can only work over a very small range of distances, as the atoms strongly repel each other if too close and the van der Waals attractions rapidly fall to insignificant levels with increasing distance.Hydrogen bonds involve an interaction between a hydrogen with a partial positive charge and a nearby, negatively-charged atom. Such bonds typically arise where a hydrogen is covalently bound to a strongly electronegative atom, such as oxygen or nitrogen, and thus acquires a partial positive charge. Such partially charged hydrogens can then interact with neighboring groups with a negative charge, such as an oxygen or nitrogen that is covalently bound to—and which has obtained a partial negative charge from—a less electronegativeatom such as hydrogen or carbon. Hydrogen bonds are stronger than van der Waals forces, on the order of 3 to 7 kcal/mol.Ionic bonds can form between any two adjacent atoms having opposite charges of at least one unit. Sometimes, such bonds can involve hydrogen bonds, such as an interaction between a negatively charged oxygen within a COO- group and a positively-charged hydrogen within an NH3+ group. Ionic bonds are relatively strong, with energies of about 5 kcal/mol.8. Van der Waals interactions can only occur over a very narrow distance range. The forces are relatively weak to begin with and are inversely proportional to the sixth power of distance, so they rapidly become insignificant as the atoms move apart. Also, if the atoms are too close, strong repressive forces drive them apart, placing a strong limit on their spatial relationship in this sense as well.Because a single van der Waals interaction can only contribute a relatively small amount of energy, molecules relying on such interactions for their binding must use a number of them (and other weak bonds) to achieve a sufficient binding strength. The fact that such interactions can only occur over a relatively limited distance range, then, requires that any two interacting molecules must have complementary surfaces so that the two molecules can fit together in a way that allows multiple van der Waals interactions to take place with appropriate spacing.9. Hydrophobic bonds obtain energy from van der Waals interactions and from the exclusion of water from the site of interaction.Van der Waals bonds result from the fluctuating energy distribution within neighboring (usually nonpolar) atoms, which creates a weak attractive force between them over a narrow distance. Van der Waals forces contribute a small amount of energy to hydrophobic bonds, about 1 kcal/mol.Hydrophobic bonds are often much stronger than what would be predicted by the energetic contribution of van der Waals forces alone. The difference can be explained by entropic effects associated with the exclusion of water from the site of hydrophobic interaction. Specifically, when a nonpolar group is present in an aqueous solution, it disrupts the water matrix, which is very energetically costly. By having nonpolar groups interact only with each other rather than with water, these costly interactions are avoided, greatly contributing to the favorable energetics of the hydrophobic bond.10.a. 50 to 100 kcal/molb. 3 to 7 kcal/molc. 3 to 7 kcal/mold. 1 to 2 kcal/mole. 0.6 kcal/mol11.The energy of weak bonds is higher than that of the average thermal energy of molecules, but not by very much. This means that the random collisions of molecules in solution will often be sufficient to break weak bonds. As a result, molecular interactions depending on weak bonds for their stability must rely on a relatively large number of these bonds, which, together, provide enough energy to withstand the structural threat provided by these random collisions.12. The less symmetrical the distribution of electrons in a covalent bond is (that is, the more polar), the stronger it is. The stronger a bond is, the shorter it is, and the less energy it contains (it already gave up a great amount of energy upon formation).13. In its liquid state, water forms a lattice, with each molecule surrounded by four nearest neighbors (see Figure 3-10). Because water is polar, each oxygen atom has a net negative charge, and each hydrogen has a net positive charge. This allows each oxygen to form a hydrogen bond with two nearby hydrogens and each hydrogen to interact with a single neighboring oxygen. Water is able to do this because it is symmetrical, small, and has a simple dipole spanning the entire molecule.14. Water plays a key role in promoting secondary bonds because of its strong tendency to form hydrogen bonds with itself. This makes any disruption of the lattice formed between water molecules energetically costly. While this is not necessarily so significant for polar regions of a molecule, as these polar regions can themselves form hydrogen bonds with the water molecules (and thereby recover some of the energy lost by the disruption of the water matrix), it is quite costly for nonpolar groups. Nonpolar parts of the molecule cannot form secondary bonds with water, and so the energy lost by the disruption of the water matrix cannot be recovered. On the other hand, nonpolar groups can form secondary bonds withother nonpolar molecules, which drives the nonpolar regions of molecules (such as internal regions of a protein) to interact with each other by forming secondary bonds (van der Waals interactions).Water also promotes the formation of weak bonds because of the fluidity of aqueous solutions. This fluidity, and the heat that constantly alters the relative configurations of molecules in the solution, allows molecules to test their interactions with a large number of other molecules, ultimately finding the energetically most favorable configuration involving the ideal combination of weak bonds.15. Weak bonds are a good choice for maintaining macromolecular structure because they are strong enough to ensure the stability of molecules at physiological temperatures, yet not so strong that they prevent any movement or flexibility within the molecules. This is important, because movement and flexibility within macromolecules can be critical for their function and their regulation. For example, many proteins are subject to allosteric regulation, where a chemical modification or binding event changes the overall shape of the protein, thereby altering its activity. Also, consider the dynamics of DNA within the cell: neither DNA replication nor transcription could occur without locally unwinding the double helix, and this depends entirely on the presence of breakable hydrogen bonds between the strands. If all of the bonds holding macromolecules together in a cell were covalent, then the molecules would be inert, fixed entities, incapable of carrying out the many dynamic functions of the cell.。

分子生物学与基因工程智慧树知到课后章节答案2023年下鲁东大学鲁东大学第一章测试1.分子生物学是研究生物体内分子结构及其相互作用规律的科学答案:错2.分子生物学的准备酝酿阶段人们只针对蛋白质进行了研究答案:错3.分子生物学所研究的生物大分子主要是指答案:核酸和蛋白质4.证明基因就是DNA的Avery是美国的答案:微生物学家5.在DNA分子结构研究中，下列哪位科学家被称为“第三人”答案:威尔金斯6.生物化学家是通过（）来破译遗传密码的答案:三联体-核糖体结合实验7.关于DNA复制机理的认识上，Meselson及Stahl用同位素标记和密度梯度离心技术证实了答案:DNA的半保留复制8.1961年Hall和Spiege-lman用RNA-DNA杂交证明mRNA与DNA序列互补这一研究说明了答案:转录与DNA有关9.中心法则中逆转录的完善是通过哪些科学家的贡献而实现的答案:Howard Temin;David Baltimore10.因遗传工程的奠基性工作而获得诺贝尔奖的是答案:Paul Berg第二章测试1.核酸分为两类,分别为DNA和RNA答案:对2.组成核酸的核糖和脱氧核糖的差别是前者2C位上连接的是一个羟基，后者只连一个H答案:对3.通过Ｔ和Ｕ的存在就可判断是ＤＮＡ还是RNA，组成DNA的碱基--是A、T、U、G，组成RNA的碱基是A、T、C、G 。

答案:错4.核酸是由多个核苷酸聚合而成的多聚核苷酸分子，相邻二个核苷酸之间的连接键是3'，5'-磷酸二酯键答案:对5.在高温下，双链DNA核酸变性，破坏的键是（）答案:氢键6.DNA分子少数是以单链形式存在，绝大多数是双链分子由碱基配对形成的氢键维持，碱基互补配对原则A/T,C/G。

答案:对7.Watson-Crick右手双螺旋结构是（）？答案:Ｂ构型8.DNA碱基对间的氢键也会断裂，双螺旋解开，空间结构破坏，这叫DNA的变异。

答案:错9.变性后的DNA都还具有复性功能，只要消除变性条件，二条互补链就会重新结合，恢复成原来的双螺旋结构。

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》复习指南答案本答案的整理，感谢潘思、殷耒兰、易美玲、包佳文、马娜等一些同学的辛勤付出，但答案还有少数地方不完整，可能也有错误的地方，需进一步完善，先将电子版传到网上，欢迎大家纠错！然后再尽快统一打印或缩印。

《分子生物学》复习指南答案一、名解1、基因：是含有生物信息的DNA片段，根据这些生物信息可以编码具有生物功能的产物，包括RNA和多肽链。

(课件)2、分子伴侣(Molecular Chaperone)：又称为伴侣蛋白，是一类在序列上没有相关性但有共同功能的保守性蛋白质，在细胞内协助其它多肽结构完成正确的折叠、组装、转运和降解，在功能完成后与之分离，不构成这些蛋白质结构执行功能时的组份。

3、RFLP：即限制性片段长度多态性。

高度重复序列中的无间隔反向重复序列很容易形成限制性内切酶识别位点，也很容易由于突变产生或失去一个酶切位点，因而可以造成限制性片段长度多态性。

即用同一种限制性内切酶消化不同个体的同一段DNA时，由于碱基组成的变化而改变限制性内切酶识别位点，从而会产生长度不同的DNA片段，这种方法称为限制性片段长度多态性，简称RFLP技术。

4、DNA的复制（r eplication）：以构成基因组的全套核酸分子为模板，精确合成一套新的核酸分子的过程。

遗传信息通过亲代DNA 分子的复制传递给子代，在保持生物物种遗传的稳定性方面起着重要的作用。

5、反转录：又称逆转录（reverse transcription），是以RNA为模板，在逆转录酶的催化下，合成双链DNA的反应。

6、克隆载体：可携带插入的外源DNA片段并可转入受体细胞中大量扩增的DNA分子。

该分子中含有能够在受体细胞中自主复制的序列和筛选标记，常用于外源基因的克隆，如噬菌体或质粒。

7、功能基因组：细胞内所有具有生物学功能的基因。

表达一定功能的全部基因所组成的DNA序列，包括编码基因和调控基因。

8、核不均一RNA：即hnRNA,即前体mRNA，在真核生物中，最初转录生成的RNA,存在于真核生物细胞核中的不稳定、大小不均的一组高分子RNA之总称。

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).。

2009年分子生物学quizl 参考答案2. What is central dogma? What contents have you learned in each part of the central dogma? (5’)The central dogma is the pathway for the flow of genetic information.We have learnt the maintenance of the genome (the structure of the genetic material and its faithful duplication) and the expression of the genome (the conversion of genetic instructions contained in DNA into proteins).(缺 reverse transcription 扣一分)3. Write out the structure of four bases, label the positions that participate Watson-Crick base pairing and that connect to the ribose? (10’)如果解答中可以 置等，会酌情wg/Q sugar4. What are the features of DNA structure? And how RNA structure differs from DNA? (2 Copyright © 2004 Pearson Education f I nc., publishing as Benjamin Cummings Two antiparallel polynucleotide chains are twisting around each other in the form of a double helix. 6’(polynucleotide(building block) 2’,antiparallel 2’,double helix 2’)Hydrogen Bonding determines the Specificity of base pairing (complimentarity).2’Stacking interaction between bases determines the stability of DNA double helix, hydrogen bond also contributes to stability.2’The double helix has Minor and Major grooves. (A B Z forms)2’Differences ：Primary structure( building block): 2’dNTP vs. rNTP ; T vs. U ; double-stranded vs. single-stranded.Secondary structure: 4’键；AG 是二环，TC 是单环；氢键的位 The features of DNA structure ：间是双键，CGDNA has stable double helical structure. (full complimentarity)RNA chains fold back on themselves to form local regions of double helix similar to A-form DNA. RNA helix are the base-paired segments between short stretches of complementary sequences, which adopt one of the various stem-loop structures, pseudoknots.( inter- and intra-molecular base pairing) Tertiary structures: 2’DNA: no tertiary structure.RNA can fold up into complex tertiary structures, because RNA has enormous rotational freedom in the backbone of its non-base-paired regions.pare the chemistry of DNA synthesis and RNA synthesis. (5’)(clues: compare=differences+the same; chemistry=substrate+direction+ energy)Differences:(1)DNA synthesis requires deoxynucleotide triphosphates while RNA synthesis requires oxynucleotide triphosphates;(1')(2)The base for DNA synthesis are A/T/C/G while that for RNA synthesis are A/U/C/G;(1')(3)DNA synthesis needs a primer:template junction while RNA synthesis do not;(1')The same:(1)The direction of both DNA synthesis and RNA synthesis is 5' to 3';(1')(2)The energy needed for DNA synthesis as well as RNA synthesis is hydrolysis of pyrophosphate (PPi).(1')6.Describe the functions of each domain of the DNA polymerase. (14’+1')DNA polymerase palm domain(1'):(1)Contains two catalytic sites, one for addition of dNTPs (1')and one for removal of the mispaired dNTP.(1')(2)The polymerization site: (1) binds to two metal ions that alter the chemical environment around the catalytic site and lead to the catalysis. (1')(2) Monitors the accuracy of base-pairing for the most recently added nucleotides by forming extensive hydrogen bond contacts with minor groove of the newly synthesized DNA. (1')(3)Exonuclease site/proof reading site . The mechanism of proof reading is kinetic selectivity(1') and the mismatched dNMP is removed by proofreading exonuclease in the direction of 3'-5'(1'). DNA polymerase finger domain:(1')(1)Binds to the incoming dNTP, encloses the correct paired dNTP to the position for catalysis;(1') (2)Bends the template to expose the only nucleotide at the template that ready for forming base pair with the incoming nucleotide;(1')(3)Stabilization of the pyrophosphate;(1')DNA polymerase thumb domain:(1')(1)Not directly involved in catalysis;(2)Interacts with the synthesized DNA to maintain correct position of the primer and the active site, (1')and to maintain a strong association between DNA Pol and its substrate.(1')7.How replication of a DNA molecule is accomplished in bacteria ?Initiation:(1)Recognition and binding of OriC by DnaA(initiator)-ATP.(2)Helicase (DnaB) loading and DnaC(helicase loader) and DNA unwinding.(3)Primase synthesizes RNA primer, and D NA Polymerase III synthesizes the new DNA strand.(The "trombone" model was developed to explain lagging strand and Elongation:leading strand synthesized simultaneously):(1)Leading strand: newly synthesized DNA strand that is continuously copied from the template strand by a DNA polymerase after the first RNA primer was made by a primase. A sliding clamp is usually loaded to the DNA polymerase to increase the polymerase processivity. The3 direction of theing strand is the same as the moving direction of the replication fork.(2)Lagging strand is discontinuously copied from the template strand. The 3 direction of the lagging strand is opposite to the moving direction of the replication fork. Primase makes RNA primers periodically after the template strand is unwound and becomes single-stranded. DNA polymerase extends each primer to synthesis short DNA fragments, called Okazaki fragments. The polymerase dissociates from the template strand when it meets the previous Okazaki fragment. Finally, RNA primers are digested by an RNase H activity, and the gaps are filled by DNA polymerase. At last, the adjacent Okazaki fragments are covalently joined together by a DNA ligase to generate a continuous, intact strand of new DNA.Termination:Type II topoisomerases separate daughter DNA molecules.8.How transcription of a RNA molecule by RNA polymerase II is initiated, elongated and terminated in eukaryotes (25’)Initiation: (11’+附加分)A.1.Promoter recognition: TBP in TFIID binds to the TATA box；(1’) TFIIA and TFIIB are recruited withTFIIB binding to the BRE；(1’)2. RNA Pol II recruitment: RNA Pol II-TFIIF complex is the recruited；(1’)3.TFIIE and TFIIH then bind upstream of Pol II (to form the pre-initiation complex).(1 ()如果提到了pre-initiation complex,有1 分的附加分)B.Promoter melting using energy from ATP hydrolysis by TFIIH .(2’)C.Promoter escapes after the phosphorylation of the CTD tail. (2’)Additional proteins are needed for transcription initiation in vivo:■The mediator complex (1 )■Transcriptional regulatory proteins (1 )■Nucleosome-modifying enzymes (1’)Tips:1、题目已经问了是真核中的情况，所以必须把相关的真核里的factor都回答出来，不然没分；2、B和C两点非常重要，分值也很大，由此可见把握keypoints的重要性；3、很少同学可以回答出in vivo状态下所需要的3种蛋白；4、如果可以答出另外一些细节，比如“TBP binds to and distorts DNA using a p sheet inserted into the minor groove”，会有少量加分。

精选全文完整版（可编辑修改）分子生物学1.插入或缺失碱基对会引起移码突变,下列哪种化合物最容易造成这种突变()。

A. 吖啶衍生物B. 5-溴尿嘧啶C. 咪唑硫嘌呤D. 乙基乙磺酸正确答案: A2.产生移码突变可能是由于碱基对的():A. 转换B. 颠换C. 水解D. 插入正确答案: D3.碱基切除修复中不需要的酶是()A. DNA聚合酶B. 磷酸二酯酶C. 核酸外切酶D. 连接酶正确答案: B4.关于DNA的修复,下列描述中,哪些是不正确的?()A. UV照射可以引起相邻胸腺嘧啶间的交联B. DNA聚合酶III参与修复核苷酸切除修复系统行程的单链缺口C. DNA的修复的过程中需要DNA连接酶D. 哺乳动物细胞可以用不同的糖基化酶来除去特异性的损伤碱基正确答案: B5.镰刀形红细胞贫血病是异常血红蛋白纯合子基因的临床表现。

β-链变异是由下列哪种突变造成的():A. 染色体臂交换B. 单核苷酸插入C. 染色体不分离D. 碱基替换正确答案: D6.在细胞对DNA损伤做出的响应中,哪一种方式可能导致高的变异率?()A. 光复活修复B. 碱基切除修复C. 重组修复D. 跨越合成正确答案: D7.下列哪种修复方式,不能从根本上消除DNA的结构损伤?()A. 核苷酸切除修复B. 错配修复C. 光复活修复D. 重组修复正确答案: D8.紫外线照射对DNA分子的损伤主要是():A. 形成共价连接的嘧啶二聚体B. 碱基替换C. 磷酸酯键的断裂D. 碱基丢失正确答案: A9.紫外线照射引起DNA最常见的损伤形式是生成胸腺嘧啶二聚体。

在下列关于DNA分子结构这种变化的叙述中,哪项是正确的?()A. 是相对的两条互补核苷酸链间胸腺嘧啶之间的共价连接B. 可由核苷酸切除修复系统在内的有关酶系统进行修复C. 是由胸腺嘧啶二聚体酶催化生成的D. 不会影响DNA复制正确答案: B10.光复活修复过程中,以下哪种酶与嘧啶二聚体结合?()A. 光解酶B. 核酸外切酶C. 核酸内切酶D. 连接酶正确答案: A11.在大多数DNA修复中,牵涉到四步序列反应,这四步序列反应的次序是()A. 识别、切除、再合成、再连接B. 再连接、再合成、切除、识别C. 切除、再合成、再连接、识别D. 识别、再合成、再连接、切除正确答案: A12.下列碱基的改变不属于颠换的是():A. A →GB. T →GC. A →TD. C →G正确答案: A13.E. coli中的MutH能识别():A. 扭曲的DNA双链B. 半甲基化的GATCC. 插层剂插入位点D. 冈崎片段间的缺口正确答案: B14.哪一类型的突变最不可逆?()A. 核苷酸的缺失或插入B. 水解脱氨基C. 八氧代鸟嘌呤D. 嘧啶二聚体正确答案: A15.下列何者属于DNA自发性损伤():A. DNA复制时的碱基错配B. 胸腺嘧啶二聚体的形成C. 胞嘧啶脱氧D. DNA交联正确答案: A16.错配修复系统中MutS通过检测子代链序列识别子代链上的错配位点。

基因分子生物学原理复习题及答案好好复习，仅供参考。

——王连庆The control of prokaryotic gene expressionⅠ名词讲明1. Operon 操纵子，指包含结构基因、操纵基因及调剂基因的一些相邻基因组成的DNA片段，其中结构基因的表达受到操纵基因的调控。

它是细菌基因转录调控的差不多模型。

2. Inducible operon 可诱导操纵子，诱导物所诱导的操纵子为可诱导操纵子，如lac操纵子，在诱导物作用下可产生本身代谢所需要的酶。

3. Operator 操纵基因，是操纵子结构基因编码区之前的DNA区段，能够结合阻遏物或活化物。

它位于基因启动子的后面或与启动子重叠，可操纵一个临近基因或基因群的表达。

4. Corepressor 共阻遏物，一类小分子，能阻止产生合成它们的酶，如Trp操纵子结构基因编码的酶所合成的Trp。

5. Repressible operon 可抑制操纵子，共阻遏物所抑制的操纵子为可阻遏操纵子，即共阻遏物的存在能够抑制结构基因所编码酶的表达，如trp操纵子。

6. Attenuator 衰减子，E. coli Trp操纵子上游位于启动子与第一个结构基因之间的序列转录后产物配对形成的终止结构。

该结构有前导肽序列的转录产物3和4互补配对形成。

7. Alarmones 信号素，原核生物应急反应中所产生的信号分子，它们通过与靶蛋白的结合来改变其活性即刻产生调控作用，从而使自身对代谢做出调剂。

8. Prophage 原噬菌体，某些温顺噬菌体侵染细菌后，其DNA整合到宿主染色体上，处于整合状态的噬菌体DNA为原噬菌体。

它是繁育和传递噬菌体本身遗传信息的一种重要方式。

9. Riboswitch 核糖开关，是原核生物中一类调剂RNA元件，它能够直截了当感受小分子的代谢来操纵转录和翻译，是不依靠调控蛋白的一种调剂方式。

10. sRNAs 细菌内小RNA，sRNAs是细菌内长80~100 nt的RNA调控序列。

医学分子生物学复习答案名词解释基因（gene）：基因是细胞和个体世代遗传信息贮存和传递的基本单位；其化学本质是DNA （RNA病毒除外）；一个基因是DNA分子中编码RNA或多肽链的核苷酸排列顺序的一个区段。

Hoogsteen配对（Hoogsteen pair）：DNA双螺旋中的核苷酸除了前述A/T，G/C之间氢键外，还能形成一些附加氢键，如：另一个T与A/T碱基对的A之间，可形成额外的2个氢键，使得这3个碱基形成了T*A/T配对；当PH降低时，胞嘧啶的N-3可以质子化，质子化的胞嘧啶与G/C碱基对的G又可形成2个氢键，3个碱基形成了C*G/C的配对。

这种配对是K.Hoogsteen在1963年发现的，因此称为Hoogsteen配对。

核小体（nucleosome）：真核生物染色质由DNA与蛋白质构成，其基本单位是核小体。

各两分子的H2A、H2B、H3、H4构成八聚体的核心组蛋白，双链DNA缠绕在这一核心上形成核小体的核心颗粒。

颗粒之间再由DNA和组蛋白H1构成的链接区相连形成串珠样结构。

基因表达（gene expression）：指基因所贮存的遗传信息通过RNA 转录和蛋白质生物合成产生具有生物功能的RNA或蛋白质的过程。

复制（replication）：以亲代DNA分子为模板按照碱基配对原则合成子代DNA分子的过程。

广义也指DNA或RNA基因组的扩增过程。

转录（transcription）：以DNA为模板合成RNA的过程称为转录。

翻译（translation）：以RNA为模板合成蛋白质的过程称为翻译。

结构基因（structure gene）：基因中编码RNA或蛋白质的DNA 序列。

调控区域（regulatory region）：与转录有关的，结构基因以外的序列。

断裂基因（split gene）：真核生物编码蛋白质的结构基因最突出的特点是不连续性。

由若干个编码区和非编码区互相间隔开但又连续镶嵌而成。

Chapter 13 Answers1. The discovery that organisms of seemingly different levels of complexity can have similar numbers of genes can be explained in part by the fact that the number of genes an organism has is not necessarily equivalent to the number of different proteins it can make. In particular, alternative splicing can dramatically increase the number of distinct mRNAs that are encoded by a single gene. Accordingly, in the case of humans, because human genes appear to undergo alternative splicing more frequently than do genes in other organisms, it is possible that humans actually produce a substantially greater number of proteins than other organisms, despite the apparent similarity in the number of genes.2. Figures 13-4 and 13-5 show the two reactions involved in splicing, including the reactants and the products of the process. The consensus sequences for the various sites are shown in Figure 13-3.3. See Figure 13-4 for the 5', 3', and 2' linkages of the branch site A.4. The spliceosome and the ribosome are similar in numerous respects, including their size, the fact that both are comprised of a large number of protein and a small number of RNA components, that the key functions of both of the structures are carried out by their RNA components, and that both bind to mRNA to carry out their respective reactions.Ribosomes and spliceosomes differ in several ways, however. For example, the makeup of the spliceosome changes more during its reaction cycle than that of the ribosome, with many of its components coming and going as the reaction proceeds. In contrast, once the large and small ribosomal associate around an mRNA, they generally remain together until protein synthesis is complete. Another difference is that the ribosome scans along the mRNA, acting processively to translate the entire message. A spliceosome, in contrast, binds to a pre-mRNA to carry out a single splicing event, and then dissociates from the mRNA.5.a. U1: U1 acts at the beginning of the splicing pathway to recognize the 5' splice site, using its RNA component to hybridize to the mRNA. Subsequently, once the tri-snRNP particle is formed, U1 leaves the complex and is replaced by U6.b. U2: Following the formation of the Early complex, U2 displaces the BBP at the branch site, base pairing in a way that exposes the branch site A so that it is available to attack the 5' splice site.c. U4: Once the A complex is formed, U4 joins the spliceosome as part of the tri-SNP particle (which also includes U5 and U6), forming the B complex and bringing all three splice sites together. Later, when the assembly pathway is complete, U4 leaves the spliceosome, allowing U6 to interact with U2 to form the active site.d. U5: U5 joins the spliceosome as part of the tri-snRNP particle, bringing the three splice sites together and forming the B complex. U5 also helps bring the two exons together to facilitate the second splicing reaction.e. U6: U6 also joins the spliceosome as part of the tri-snRNP particle. Once the B complex has been formed, U1 leaves the complex and is replaced by U6 at the 5' splice site. Subsequently, U4 also leaves, allowing U6 to interact with U2 to form the active site.f. U2AF: U2AF binds early during the splicing pathway as a dimer, with one subunit binding to the Py tract, and other both interacting with the 3' splice site and helping BBP bind to the branch site. Later, U2AF also helps U2 bind to the branch site, displacing BBP.g. BBP: BBP binds to the branch site at an early stage in the splicing pathway, contributing to the Early complex. Later, following the formation of the A complex, BBP is displaced from the branch site by U2.6. Because splicing is critical for the processing of many mRNAs in eukaryotic cells, mutations that eliminate key splicing factors are usually lethal. Accordingly, one way to screen for splicing mutants would be to isolate lethal mutations and then look to see if any of them show splicing defects. In particular, you could isolate temperature sensitive lethal mutants (i.e., mutations that only eliminate the function of the protein—and thereby kill the cells—at elevated temperature), and then ask whether the mutant cells accumulate unspliced transcripts at high temperatures. This could be done by preparing a northern blot （检测RNA表达水平检测基因表达水平）using mRNA isolated from the mutant cells (at the high temperature), and probing the blot using an intron-specific probe. If the mutation indeed eliminated a splicing factor, then the intron should be present at an abnormally high level.DEAD box-containing proteins are helicases that use ATP to unwind RNA-RNA duplexes. They are involved in many cellular activities, including splicing, where they act in particular to separate RNA-RNA hybrids to facilitate many of the molecular rearrangementsthat take place within the spliceosome. If you isolate a mutation that affects a DEAD-box containing protein, then, it is likely that the phenotype is caused by a block to the splicing pathway resulting from one or more RNA-RNA interactions that cannot be undone.7. One way to test whether the complementary sequences actually hybridize would be to introduce a nucleotide change into one of the sequences—for example, if the first RNA sequence has an A at position 5, and the second sequence has a T, then changing the A in the first strand to a G—and then asking whether the change abolishes the interaction between the splicing factor and the spliceosome subunit. If it does, it is likely that the two sequences actually hybridize, and that this hybridization underlies the interaction between the factor and the subunit. You could then obtain even more convincing proof by altering the second sequence in a way that restores the complementarity between the sequences—i.e., changing the T in the second sequence to a C—and then asking whether this restores the interaction. If it does, you could confidently conclude that the sequences do in fact hybridize.8. Group II introns carry out self-splicing using the same chemistry as that used by the spliceosome during mRNA splicing. Specifically, an A residue at the branch point uses its 2'-OH to attack the phosphoryl group of a G at the 5' splice site. This links the branch site A to the 5' splice site G, cleaving the phosphodiester linkage at the 5' splice site and freeing the 3' end of the (5') exon. This free 3' end then attacks the phosphoryl group at the 3' splice site, joining the two exons together and releasing the intron in the form of a lariat.Group I introns, in contrast, use a free G nucleotide (instead of a branch site A) to attack the 5' splice site. As with group II introns, this cleaves the phosphodiester bond at the 5' splice site, freeing the 3' end of the exon to attack the 3' splice site. In this way, the exons are linked together and the intron is released (although here the intron is linear, not as a lariat as with group II introns).Self-splicing introns are not enzymes because they mediate only a single round of splicing, in contrast to true enzymes which are capable of repeatedly catalyzing a reaction. A group I intron can be readily converted into an enzyme, however, by providing it with a free G and allowing it to react with molecules having a sequence complementary to its internal guide sequence; in this way, the intron can act in trans to cleave an unlimited number of heterologous substrates.9. The proteins that associate with group I introns in vivo help to stabilize the structureof the intron, in part by shielding negatively charged phosphate groups within the polynucleotide backbone so that different regions of the backbone can come into close proximity.Proof that the group I intron-associated proteins play an exclusively structural role was provided by the demonstration that the introns can function even in the absence of the proteins if they are provided with high salt concentrations. Presumably, the positive ions of the salt can substitute for the proteins by interacting with the negatively charged phosphate backbone, thereby allowing the RNA to fold in the proper way.10. The average human gene includes 8 or 9 exons, with the exons averaging about 150 nucleotides in length and the introns about 3,500.The relative size of introns and exons—and in particular the fact that introns are much larger than exons—has important implications for the evolution of genes and of gene function. Because many functional and structural domains are confined to single exons, and because the greater length of introns means that many more recombination events occur within introns than they do within exons, recombination tends to shuffle entire exons to create new combinations of functional and structural domains. Throughout evolution, some of the new combinations created in this way were undoubtedly useful, helping to explain why many individual proteins include repeated domains, and why many unrelated genes share similar domains.11. Accurate splicing requires that the 5' splice site pairs with the correct 3' splice site. While this may be relatively straightforward for genes containing a small number of modestly sized introns, it can present a huge challenge for those that include large numbers of introns (for example, well over a hundred in some cases), and/or introns of vast dimensions (for example, up to 500,000 nucleotides). This challenge is compounded by the relatively weak sequence requirements for splice sites, as this increases the probability that random sequences will be mistakenly used for splicing. The cell has two mechanisms for getting around these difficulties and increasing the accuracy of splice site selection.The first mechanism acts to reduce the possibility that exons are skipped by the splicing machinery. In this case, splicing factors associate with RNA polymerase during transcription, accompanying the polymerase as it progresses along the DNA template.When the polymerase transcribes a 5' splice site, the factors leave the polymerase and bind to the RNA. They then wait until a 3' splice site is transcribed and then interact with factors that bind there, helping to ensure that the 3' site is used.The second mechanism is designed to ensure that true splice sites are used for splicing. Here, a set of proteins, called SR proteins, bind to specific sequences that are located within exons, but close to the intron boundary. Once bound, they help recruit the splicing machinery, thereby ensuring that splicing occurs at sites close to exon-intron boundaries (where it should occur) rather than at cryptic sites located far from any exons. P43112. The observation that introns are widespread in eukaryotic genes but almost completely absent in prokaryotic ones can be explained at an evolutionary level in two ways. The first explanation (called the "intron early" hypothesis) is that the common ancestor of both modern-day bacteria and eukaryotes had introns, but that bacteria have since lost them during the course of evolution. The second possibility (the "intron late" hypothesis) is that the common ancestor did not contain introns, but rather that introns arose later, at some point early in the course of eukaryotic evolution.One major piece of evidence for the intron late hypothesis is the simple observation that introns are widespread in eukaryotes but essentially absent from bacteria. It seems improbable that if bacteria had introns at some point in the past, they could have lost them so completely that it is now virtually impossible to find any bacterial introns. On the other hand, proponents of the intron-early hypothesis have argued that exon shuffling—which requires the presence of introns—is an important evolutionary process that has been occurring since the early stages of evolution.Although it has been very difficult to determine which of these hypotheses is correct, the intron-early explanation would be bolstered by discovering additional introns in bacteria. Alternatively, the exon late hypothesis could be supported by a demonstration that genes common to all eukaryotes (which presumably share a common ancestor dating from early stages of eukaryotic evolution) contain introns that are totally unrelated. This would suggest that the introns arose from separate events that occurred during eukaryotic evolution, rather than representing the descendants of a single ancient intron that was present in an ancestral gene.13. The first type of RNA editing is called site-specific deamination. In this process, which only affects certain genes in specific cell types, an enzyme called a deaminase removes amino groups from particular nucleotides within the mRNA. Deamination can alter the identity of the affected nucleotides, converting, for example, a C to a U, or an A to an inosine. This can have important consequences for the sequence of the encoded protein, causing amino acid substitutions in the protein or replacing amino acid encoding codons with stop codons.The second type of RNA editing, which affects the mRNA sequence more dramatically than deamination, occurs in the mitochondria of trypanosomes. This type of editing involves specialized RNA molecules, called guide RNAs, which are complementary to portions of the mRNA but which contain additional As in their sequence. During editing, the guide RNAs hybridize to their target mRNAs, and an endonuclease cuts the mRNA at a position opposite one of the additional As in the guide sequence. A U is then inserted into the mRNA opposite the additional A. Editing can add a large number of Us to an mRNA, dramatically changing the sequence of the encoded protein either through the addition of multiple amino acids or by causing a frameshift mutation in the mRNA.。