07双语第六章

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《国际贸易务(双语)》三习题库

《国际贸易务(双语)》三习题库

《国际贸易实务(双语)》三习题库第一章国际贸易简介Chapter 1 A Brief Introduction to International Trade预习题1. What is international trade?2. Why do nations trade?练习题3. What are the differences between international trade and domestic trade?4. P17-20 专业术语、课后练习复习题5. What are the characteristics of international trade?6. What are the forms of international trade?7. Useful certificates for your future job-hunting.第二章进出口交易的一般程序General Procedure of International Trade一、国际贸易的基本流程Basic procedure of international trade预习题:1. Suppose you are an exporter, what will you do before you enter into a business with a trade partner?2.我国某出口公司与香港C公司签订了一笔总值25万美元销售猪肉的合同,FOB青岛,目的港韩国釜山,付款方式为D/A远期120天办理托收。

该出口公司按规定的装运期限装运货物后,通过中国银行办理托收手续。

中国银行委托香港南洋商业银行为代收行向香港C公司收款。

单据到香港后,C公司即承兑赎单,又以原提单向韩国收货人收取了货款。

香港南洋商业银行在汇票到期时向C公司催促付款,但此时该公司已宣告破产。

经查核C公司在当地注册资本仅15万港元,其财产远远不够抵偿该公司的欠款。

中级财务会计(双语)第六章

中级财务会计(双语)第六章

Depreciation
Some important concepts:
b. Acquisition cost (or original cost) 取得成本 (原始成本) The initial value of the asset at its acquisition 资 产购买时的初始价值

Time-based methods
Sum-of-the-Years’-Digits (SYD) Method
This method multiplies the depreciable base by a fraction that declines each year and results in depreciation that decreases by the same amount each year. 在这种方法下,折旧 基础乘以一个逐年减小的分数,使得折旧每年以同 样的金额减少。
Time-based methods
Illustration 6-1
Acquisition cost: $250,000 Service life: 5 years Residual value: $40,000
250,000 40,000 Annual Depreciation $42,000 per year 5

Time-based methods
Accelerated method 加速折旧法
A depreciation method which allows faster write-offs than the straight-line method. 比 直线折旧法的折旧速度更快的折旧方法。
Benefits from using the asset

机械工程材料双语课件第六章

机械工程材料双语课件第六章
grain boundaries and subgrain boundaries
①dislocation(位错) (位错) 晶格畸变
• Definition: defects that cause lattice distortion centered around a line. • Formation: created during the solidification of crystalline solids, or formed by the permanent or plastic deformation of crystalline solids and by vacancy condensation and by atomic mismatch in solid solutions.
除细晶强化之外的强化方式都会或 多或少的降低钢的塑性和韧性。 多或少的降低钢的塑性和韧性。
Effects of alloy elements on strengthening and toughening of steels(合金元素对钢强韧化的影响) (合金元素对钢强韧化的影响) •Alloy elements affect the strengthening and toughening of steels mainly through their effect on the phase transformation of steels, and the alloy elements can actively work only if the reasonable heat treatments are carried out.
§6 Strengthening and toughening of steels

英汉翻译教程第六章PPT

英汉翻译教程第六章PPT

广告英语的翻译
• 总之,衡量广告翻译的优劣,主要看译文在多大 程度上体现了原文的宣传效果和表情、说服等功 能,多大程度上保留了原文的语言特色及神韵。 为此,翻译时选词上要力避生涩艰拗,造句上要 尽量简洁明快,修辞上要多保留原文辞格。译文 要注重整体把握,不为一词一句所限,要充分发 挥译者的主观能动性。此外,广告翻译还应充分 考虑中英两种语言间的文化差异。只有这样,才 能使译文体现原文的各项功能,为产品畅销打下 良好的基础。
商贸函电的翻译
• 下面是一份电传的主要部分及其参考译文: TKS F UR OFA BUT RGRTS UR PRICES TOO HIGH F OUR MKT, AS GDS OF JAPANESE MAKE WITH SimILAR DES R ON SALE HR AT MCH LOWER PRICES. IF U RDU UR PRICE BY 5PCT, V WL ODA 1000 SETS. PLS TAKE THIS MATTR INTO CONSIDERATION N RPL ASAP. RGDS.
商贸函电指经济贸易活动中的各类信函、电报、电传等。 其文体特征主要有: 1) 措词简洁明了。为使表达明晰,商贸函电在措词上力求 简明扼要,不太讲究修饰。 2) 用语正式庄重。正式庄重的语言常常显得诚恳、自然、 有礼貌,因此常用于商贸函电中。 3) 专业术语较多。如FOB(离岸价)、CIF (到岸价)、 settlement(理赔)、L/C(信用证)、AC(承兑)等。 4) 具有相对固定的格式。商贸函电格式相对固定,大体上 由以下六个部分组成: a. 信头(the heading) b. 信内地址(the inside address) c. 称呼(the salutation) d. 信文(the body) e. 结束语 (the complimentary close) f. 签名(the signature)

微机原理及应用(双语)知到章节答案智慧树2023年山东科技大学

微机原理及应用(双语)知到章节答案智慧树2023年山东科技大学

微机原理及应用(双语)知到章节测试答案智慧树2023年最新山东科技大学第一章测试1.What kind of device does the MCS-51 microcontroller belong to? ( )参考答案:8-bit device2.The operation of taking a number from memory into the CPU is called ( ).参考答案:READ3.The 8951 microcontroller contains ()参考答案:EEPROM4.What is the highest address in a memory system with 16 data lines? ()参考答案:655355.What are the basic actions that a microprocessor can carry out?参考答案:Taking a number from memory (READ);Arithmeticoperations; Putting a number into memory (WRITE);Making decisions about the program route (jumps and subroutine calls);Logicaloperations (bit manipulation)第二章测试1.In internal RAM, only one of the four register banks is in used at any timewhen the program is running. ()参考答案:对2.When using an external crystal for the 8051 clock, what values of capacitorsshould be connected to it? ()参考答案:30 pF3.On the 8051 what is the storage size (in bits) of accumulator A? ( )参考答案:8 bit4.Which port has alternate function pins? ( )参考答案:Port 35.Which type of memory organization is used by the MCS-51 microcontrollers?参考答案:Harvard第三章测试1.What is the jump range of instruction LJMP? ()参考答案:64KB2.What is the addressing mode of the second operand in the instruction MOVXA, @DPTR ? ()参考答案:Indirect3.If MOV (direct) is used for address 87h which type of memory is accessed?()参考答案:SFRs4.The Stack Pointer will be decremented when a PUSH operation is made? ( )参考答案:错5.Which instruction will clear the low nibble of A if A = C5h?参考答案:ANL A, #0F0h第四章测试1.What is the jump range of instruction CJNE or DJNZ ? ()参考答案:256B2.What is the address of the second operand in the instruction:MOVC A,@A+DPTR ?Assume DPTR=1000h, A=80h. ()参考答案:1080h3.Which programming types does the assembly language belong to? ()参考答案:Linear4.What are the basci program sturcture in assembly language? ( )参考答案:Subroutine structure;Branching structure;Sequential structure;Loop structure5.Which Pseudo-instruction can put bytes in ROM memory? ( )参考答案:DB---------------------------------------------------------------------------------------------------------------------------------第五章测试1.How many interrupt sources are there in 8051 ? ()参考答案:52.The EA bit in IE register does not actually enable all interrupts, but allowsthem to be enabled separately. ()参考答案:对3.For 8051, each interrupt causes the program to jump directly to an addressat which an ISR is located. ()参考答案:对4.In an ISR (Interrupt Service Routine), which of the following instructionsmust be included? ()参考答案:Return from interrupt instruction5.How many interrupts are enabled after running this instruction “MOV IE,#05h”?参考答案:第六章测试1.For timers, which mode produces an 8-bit auto-reload timer? ()参考答案:Mode 22.Which register can we use to set up the mode of operation of Timer 0? ()参考答案:TCON3.For the serial port, which mode is an 8-bit UART? ()参考答案:Mode 14.Where is the parity bit in an RS232 packet? ()参考答案:After data bits5.How many modes can we use in a timer/counter ?参考答案:4第七章测试 the three types of bus which are used when external memory isconnected to the 8051.( )参考答案:Address bus;Control bus;Data bus2.When using external RAM, which port is used for the low address byte? ()参考答案:Port 03.If we need to expand an 8031 by adding 32Kbytes of code memory EPROM,how many address lines should be connected between the 8031 and theEPROM? ()参考答案:154.If an 8031 chip is connected to an external EPROM, Which line can generate asignal to enable reading of data from the EPROM? ()参考答案:/PSEN5.If the program is stored entirely in an external EPROM, what voltage must beapplied to the /EA PIN on the 8051? ()参考答案:0 volts第八章测试1.There are 8 analog input channels that be connected to the ADC0809. ()参考答案:对2.How many bits are required to select which analog channel is active on theADC0809? ()参考答案:33.In order to use the DAC0832, an external operational amplifier must beconnected. ()参考答案:对4.How many port pins would be used to connect a 6*4 matrix keypad to an8051 (excluding the interrupt input)? ()参考答案:105.What are the common forms that the internal wiring of the 7-segmentdisplay can take?( )参考答案:Common anode;Common cathode---------------------------------------------------------------------------------------------------------------------------------第九章测试1.In almost all cases, except completely trivial instances, the design of amicrocontroller-based system must be a top-down activity. This means that one must look first at the broad picture and then refine it down to systemsand sub-systems in both hardware and software. ()参考答案:对2.The lowest level of errors comprises those that result from incorrect use ofthe assembly language, and these are known as syntax errors. ()参考答案:对3.Which errors belong to syntax error of assembly language? ( )参考答案:Symbol not define;Wrong representation of numbers;Misspelling of instructions;Incorrect addressing modes4.The basic steps of microcontroller system design are: Requirement analysis,functional specification, design, implementation, system testing,documentation, and installation. ()参考答案:对5.The function of WDT(Watchdog Timer) is to do the syntax error detection.()参考答案:错。

经济学原理(双语)-教学大纲

经济学原理(双语)-教学大纲

经济学原理(双语)-教学⼤纲《经济学原理》(双语)教学⼤纲Principles of Economics(Class Syllabus)课程编号:032023A课程类型:□通识教育必修课□通识教育选修课□专业必修课□专业选修课学科基础课总学时:48讲课学时:48实验(上机)学时:0学分:3适⽤对象:经济类本科⽣先修课程:⾼等数学⼀、课程的教学⽬标《经济学原理》包括《微观经济学》和《宏观经济学》两部分。

经济学是研究稀缺资源在各种可供选择的⽤途中,如何进⾏最有效的配置,以使得⼈类的⽆限欲望得到最⼤满⾜的⼀门理论经济学。

经济学作为理论经济学,是许多经济学科如管理经济学、公共财政学、市场学、国际经济学、货币银⾏学、财政学、国际贸易学、国际⾦融学等的理论基础。

微观经济学是以个体经济单位例如单个家庭、单个企业、单个市场为研究对象的⼀门理论经济学。

它试图通过对个体经济单位经济⾏为的研究,来说明现代社会中市场机制的运⾏和作⽤以及如何改善这种运⾏的途径。

宏观经济学是以社会总体的经济⾏为即宏观经济运⾏作为研究对象的⼀门理论经济学。

它试图通过对社会总体经济⾏为的研究,来说明市场经济条件下宏观经济的运⾏以及如何改善这种运⾏的途径。

⽬标1:系统理解和掌握经济学基本理论和基础知识⽬标2:熟悉经济学中基本概念和基本原理的英⽂表述⽬标3:能够运⽤所学原理分析实际经济问题和现象⽬标4:为进⼀步学习其他经济类课程打下坚实的理论基础。

⼆、教学基本要求教学内容:第⼆章的消费理论、第三章的⽣产理论、第五章的市场理论、第六章的⽣产要素市场和收⼊分配理论、第七章⼀般均衡、第九章国民收⼊核算、第⼗⼀章产品市场和货币市场的⼀般均衡、第⼗三章财政政策和货币政策、第⼗五章通货膨胀与失业应精讲、细讲。

对各章重点内容,教师应阐述清楚基本原理,并在此基础上以⼀些浅显易懂的案例增进学⽣的理解,启发学⽣的求知欲与好奇⼼。

教学⽅法:为实现教学⽬标,本课程主要采取教师讲授的⽅式授课,任课教师应以多媒体教学为基本授课⽅式,教师也可以所采取的其他教学⼿段如课堂讨论、案例教学等。

太极之从零开始(双语版)知到章节答案智慧树2023年东华理工大学

太极之从零开始(双语版)知到章节答案智慧树2023年东华理工大学

太极之从零开始(双语版)知到章节测试答案智慧树2023年最新东华理工大学第一章测试1.太极拳对腰背部的动作要求是松腰正背。

参考答案:对2.陈有本为适应不同的学习对象,舍弃了某些高难度动作,变发劲为蓄而待发,后人称其架势为老架。

参考答案:错3.目前武术按照源流有序,理法系统,风格独特的要求有拳种()参考答案:1294.吴氏太极拳的特点是参考答案:以柔化为主5.孙氏太极拳的创立人是参考答案:孙禄堂6.24式太极拳的步法有()参考答案:跟步;退步;上步;侧行步7.太极拳的自如阶段在于追求()的完美结合,使动作气势饱满。

神韵自由。

参考答案:气与力;神与形;内与外8.太极拳属于武术参考答案:对9.陈氏太极拳是由()创造的。

参考答案:陈王廷10.杨式太极拳的创始人是()参考答案:杨露禅第二章测试1.不是《中国武术段位制》考评内容的是()参考答案:国学知识2.太极拳运动能加强中枢神经系统功能。

参考答案:对3.参与太极拳运动能加强心血管和呼吸系统功能,改善消化作用。

参考答案:错4.邓小平在()年接见题本太极拳友人三宅正一时写下“太极拳好”的题词。

参考答案:19785.做弓步动作时两脚左右横向距离是()参考答案:30厘米6.向前迈步时脚步()地方先着地。

参考答案:脚跟7.并步云手是两脚相距()厘米参考答案:10-208.太极拳运动有开、合、虚、实与时序结合的要求,那么肢体一开一合,呼吸即参考答案:一呼一吸9.太极拳吸取了当时各家拳法之长,特别是吸收戚继光的()长拳而改编成的。

参考答案:三十二势10.起势开步时两脚的宽度是()参考答案:与肩同宽第三章测试1.下列说法中正确的一项是()参考答案:对练是俩人或俩人以上按照预定的程序进行的攻防格斗套路2.在太极拳演练中讲究"迈步如猫形"要求步法()参考答案:轻灵稳健3.武术的作用是()参考答案:强身健体,娱乐休闲,.修身养性,防身自卫4.太极拳有多种流派,在河南的是杨式。

第六章金融市场与金融机构杰夫马杜拉双语课件资料

第六章金融市场与金融机构杰夫马杜拉双语课件资料

Backing commercial paper 商业汇票的担保
有资产支持的商业汇票的收 益率可以稍微放低,低于无 资产支持的商业汇票
大公司发行的商业汇票风险 相对较低
根据信用额度决定发行情况
Estimating the yield 收益率 的估算
不付利息,折价发行 收益率高于T-bill 风险也相对高 计算例子 交易成本一般包含在买价中
Commercial paper 商业汇票
金额:10万-100万 到期日 20-45 days 个人投资方式:投资于货币市场共同基金,其投资于
商业汇票 一般选择持有到期 Ratings 商业汇票分级 根据公司的财务状况和现金流划分商业汇票的等级 货币市场共同基金一般投资于等级最高和次高的商业
ห้องสมุดไป่ตู้
国库券购买方式:拍卖
竞争性报价:加高者得(优 先购买)
非竞争性报价:平均报价
特点:到期时间越长价格越 低(收益越高)
收益: 到期收益:面值-买价 不到期收益:卖价—买价
Estimating the treasury bill discount国库券贴现的估算
收益率和贴现率的区别
收益率一般高于贴现率
金融机构持有货币市场 工具原因:
收益、流动性
金融机构发行货币市场 工具原因:
为了解决短期现金流紧 张
货币市场工具交易中介:
T-bill 拍卖竞价的方式
其他货币市场工具都有 自己的交易商
货币市场工具的发行方 一般是贷方
Valuation of money market securities 货币市场证券价值的估算(定价)
到期银行会支付给持票人相 应的金额
风险:与NDC 相近,稍高于 T-bill
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Chapter 6 TRANSCRIPTION6.16.1 IntroductionIntroduction b The coding strand The coding strand (Sense strand)(Sense strand) of DNA has the same sequence as of DNA has the same sequence as the mRNA and is related by the genetic code to the protein sequence that it represents.b The antisense strand The antisense strand (Template strand)(Template strand) of DNA is complementary of DNA is complementary to the sense strand, and is the one that acts as the template forsynthesis of mRNA.b RNA polymerases RNA polymerases are enzymes that synthesize RNA using a DNA are enzymes that synthesize RNA using a DNA template (formally described as DNA-dependent RNA polymerases).b A promoter A promoter is a region of DNA where RNA polymerase binds to is a region of DNA where RNA polymerase binds to initiate transcription.bStartpoint (startsite)Startpoint (startsite) (Startsite)(Startsite) refers to the position on DNA refers to the position on DNA corresponding to the first base incorporated into RNA.6.16.1 IntroductionIntroduction bA terminator A terminator is a sequence of DNA that causes RNA polymerase to is a sequence of DNA that causes RNA polymerase to terminate transcription. bA transcription unit A transcription unit is the distance between sites of initiation and is the distance between sites of initiation and termination by RNA polymerase; may include more than one gene. b Upstream Upstream identifies sequences proceeding in the opposite direction identifies sequences proceeding in the opposite direction from expression; for example, the bacterial promoter is upstream of the transcription unit, the initiation codon is upstream of the coding region.b Downstream Downstream identifies sequences proceeding farther in the identifies sequences proceeding farther in the direction of expression; for example, the coding region isdownstream of the initiation codon.bA primary transcript A primary transcript is the original unmodified RNA product is the original unmodified RNA product corresponding to a transcription unit.6.16.1Introduction b Transcription involves synthesis of an RNA chain representing one strand of a DNA duplex. By "representing" we mean that the RNA is identical in sequence with one strand of the DNA, which is called the coding strand. It is complementary to the other strand, which provides the template strand for its synthesis. b Figure 6.1Figure 6.1 recapitulates the recapitulates the relationship between double-strandedDNA and its single-stranded RNAtranscript.6.16.1Introduction bRNA synthesis is catalyzed by the enzyme RNA polymerase.RNA polymerase. Transcription starts Transcription starts when RNA polymerase binds to a special region, the promoter, at the start of the gene. The promoter surrounds the first base pair that is transcribed into RNA, the startpoint. From this point, RNA polymerase moves along the template, synthesizing RNA, until it reaches a terminator sequence. This action defines a transcription unit that extends from the promoter to the terminator. b The critical feature of the transcription unit, depicted in depicted in Figure 6.2Figure 6.2Figure 6.2, is that it constitutes, is that it constitutes a stretch of DNA expressed via theproduction of a single RNA molecule. Atranscription unit may include more thanone gene.6.16.1Introduction b Sequences prior to the startpoint are described as u pstream upstream upstream of it; of it; those after the startpoint (within the transcribed sequence) aredownstream downstream of it. Sequences are conventionally written so thatof it. Sequences are conventionally written so that transcription proceeds from left (upstream) to right(downstream). This corresponds to writing the mRNA in the usual 5` → 3` direction.b Often the DNA sequence is written to show only the coding strand, which has the same sequence as the RNA. Base positions are numbered in both directions away from the startpoint, which is assigned the value +1; numbers are increased goingdownstream. The base before the startpoint is numbered –1, andthe negative numbers increase going upstream. the negative numbers increase going upstream. (There is no base(There is no base assigned the number 0.)6.1 IntroductionbThe immediate product of transcription is called the The immediate product of transcription is called the primary primary transcript transcript. It would consist of an RNA extending from the promoter . It would consist of an RNA extending from the promoter to the terminator, possessing the original 5 ` and 3` ends. However, the primary transcript is almost always unstable. In prokaryotes, it is rapidly degraded (mRNA) or cleaved to give mature products (rRNA and tRNA). In eukaryotes, it is modified at the ends (mRNA) and/or cleaved to give mature products (all RNA).b Transcription is the first stage in gene expression, and the principal step at which it is controlled. Regulatory proteins determine whether a particular gene is available to be transcribed by RNA polymerase. The initial (and often the only) step in regulation is the decision on whether or not to transcribe a gene. Most regulatory events occur at the initiation of transcription, although subsequent stages in transcription (or other stages of gene expression) are sometimes regulated.6.1 Introductionb Within this context, there are two basic questions in gene expression:b How does RNA polymerase find promoters on DNA? This is a particular example of a more general question: how do proteins distinguish their specific binding sites in DNA from other sequences?b How do regulatory proteins interact with RNA polymerase (and with one another) to activate or to repress specific steps in the initiation, elongation, or termination of transcription?6.26.2Transcription occurs by base pairing in a "bubble" of unpaired DNAb RNA polymerase RNA polymerase separates the two strands of DNA in a separates the two strands of DNA in a transient "bubble" and uses one strand as a template to direct synthesis of a complementary sequence of RNA. b The length of the bubble is ~12-14 bp, and the length of RNA-DNA hybrid within it is ~8-9 bp.6.2 Transcription occurs by base pairing in a6.2"bubble" of unpaired DNAb Transcription takes place by the usualprocess of complementary base pairing.illustrates the generalFigure 6.3 illustrates the generalFigure 6.3principle of transcription. RNA synthesistakes place within a "transcriptionbubble," in which DNA is transientlyseparated into its single strands, and thetemplate strand is used to direct synthesisof the RNA strand.6.26.2 Transcription occurs by base pairing in a"bubble" of unpaired DNAb The RNA chain is synthesized from the 5` endtoward the 3` end. The 3` –OH group of the lastnucleotide added to the chain reacts with anincoming nucleoside 5` triphosphate.b The incoming nucleotide loses its terminal twophosphate groups (γ and β); its α group is usedin the phosphodiester bond linking it to the chain.b The overall reaction rate is ~40nucleotides/second at 37°°C (for the bacterial RNAnucleotides/second at 37polymerase); this is about the same as the rate oftranslation (15 amino acids/sec), but much slowerthan the rate of DNA replication (800 bp/sec).6.26.2Transcription occurs by base pairing in a "bubble" of unpaired DNA b RNA polymerase creates the transcription bubble when it binds toa promoter. a promoter. Figure 6.4Figure 6.4Figure 6.4 shows that as shows that asRNA polymerase moves along theDNA, the bubble moves with it, andthe RNA chain grows longer. Theprocess of base pairing and baseaddition within the bubble iscatalyzed and scrutinized by theenzyme.6.26.2Transcription occurs by base pairing in a "bubble" of unpaired DNA b The structure of the bubble within RNA polymerase is shown in theexpanded view of expanded view of Figure 9.5Figure 9.5Figure 9.5. As . AsRNA polymerase moves along theDNA template, it unwinds theduplex at the front of the bubble (theunwinding point), and rewinds theDNA at the back (the rewindingpoint). The length of thetranscription bubble is ~12-14 bp,but the length of the RNA-DNAhybrid region within it is shorter.6.26.2 Transcription occurs by base pairing in a"bubble" of unpaired DNAb There is a major change in thetopology of DNA extending over~1 turn, but it is not clear howmuch of this region is actuallybase paired with RNA at anygiven moment. Certainly theRNA-DNA hybrid is short andtransient. As the enzyme moves on,the DNA duplex reforms, and theRNA is displaced as a freepolynucleotide chain. About thelast 25 ribonucleotides added to6.3 The transcription reaction has three stages b Initiation Initiation describes the stages of transcription up to synthesis of the describes the stages of transcription up to synthesis of the first bond in RNA. This includes binding of RNA polymerase to thepromoter and melting a short region of DNA into single strands.b Elongation Elongation is the stage in a macromolecular synthesis reaction is the stage in a macromolecular synthesis reaction (replication, transcription, or translation) when the nucleotide orpolypeptide chain is being extended by the addition of individual subunits.bTermination Termination is a separate reaction that ends a macromolecular is a separate reaction that ends a macromolecular synthesis reaction (replication, transcription, or translation), bystopping the addition of subunits, and (typically) causingdisassembly of the synthetic apparatus.6.3 The transcription reaction has three stagesb The transcription reaction can be divided into, in which athe stages illustrated in Figure 9.6Figure 9.6, in which abubble is created, RNA synthesis begins, thebubble moves along the DNA, and finally isterminated:b Template recognition begins with the binding ofRNA polymerase to the double-stranded DNAat a promoter to form a "closed complex". Thenthe strands of DNA are separated to form the"open complex" that makes the template strandavailable for base pairing with ribonucleotides.The transcription bubble is created by a localunwinding that begins at the site bound by RNApolymerase.6.3 The transcription reaction has three stagesbInitiation Initiation describes the synthesis of the first nucleotide describes the synthesis of the first nucleotide bonds in RNA. The enzyme remains at the promoter while it synthesizes the first ~9 nucleotide bonds. bThe initiation phase is protracted by the occurrence of abortive events, in which the enzyme makes short transcripts, releases them, and then starts synthesis of RNA again. b The initiation phase ends when the enzyme succeeds in extending the chain and clears the promoter. Thesequence of DNA needed for RNA polymerase to bind to the template and accomplish the initiation reaction defines the promoter .6.3 The transcription reaction has three stagesbDuring During elongation elongation elongation the enzyme moves along the DNA and the enzyme moves along the DNA and extends the growing RNA chain. As the enzyme moves, it unwinds the DNA helix to expose a new segment of the template in single-stranded condition. Nucleotides are covalently added to the 3` end of the growing RNA chain, forming an RNA-DNA hybrid in the unwound region. bBehind the unwound region, the DNA template strand pairs with its original partner to reform the double helix. The RNA emerges as a free single strand.b Elongation Elongation involves the movement of the transcription involves the movement of the transcription bubble by a disruption of DNA structure, in which the template strand of the transiently unwound region is paired with the nascent RNA at the growing point .6.3 The transcription reaction has three stagesb Termination Termination involves recognition of the point at which no involves recognition of the point at which no further bases should be added to the chain. To terminate transcription, the formation of phosphodiester bonds must cease, and the transcription complex must come apart.bWhen the last base is added to the RNA chain, the transcription bubble collapses as the RNA-DNA hybrid is disrupted, the DNA reforms in duplex state, and the enzyme and RNA are both released. The sequence of DNA required for these reactions defines the terminator .6.4 RNA polymerase 6.4.1 Bacterial RNAb6.4.1b6.4.2 Eukaryotic RNA polymerases6.4.1 Bacterial RNA6.4.1b The best characterized RNA polymerases are those of eubacteria, for which E. coli is a typical case.b A single type of RNA polymerase appears to be responsible for almost all synthesis of mRNA, and all rRNA and tRNA, in a eubacterium.b About 7000 RNA polymerase molecules are present in an E. coli cell. Many of them are engaged in transcription; probably 2000-5000 enzymes are synthesizing RNA at any one time, the number depending on the growth conditions.6.4.1 6.4.1Bacterial RNAb The complete enzyme or holoenzyme in E. coli has amolecular weight of ~465 kD.Its subunit composition issummarized in summarized in Figure 9.16Figure 9.16Figure 9.16..6.4.16.4.1 Bacterial RNAb Bacterial RNA polymerase can bedivided into the α2ββ` core enzymethat catalyzes transcription and thesigma subunit that is required only forinitiation.b Sigma factor changes the DNA-bindingproperties of RNA polymerase so thatits affinity for general DNA is reducedand its affinity for promoters isincreased.b Core enzyme does not distinguishbetween promoters and other sequencesof DNA.6.4.16.4.1 Bacterial RNAb Bacterial RNA polymerase can bedivided into the α2ββ` core enzymethat catalyzes transcription and the sigmasubunit that is required only for initiation.b Sigma factor changes the DNA-bindingproperties of RNA polymerase so that itsaffinity for general DNA is reduced andits affinity for promoters is increased.b Core enzyme does not distinguishbetween promoters and other sequencesof DNA.b The inability of free sigma factor torecognize promoter sequences may beimportant.6.4.16.4.1 Bacterial RNAb There are four (perhaps five)conserved features in a bacterialpromoter: the startpoint; thepromoter: the startpoint; the ––10sequence; the ––35 sequence; thesequence; theseparation between the10 and ––35separation between the ––10 andsequences; and (sometimes) the UPelement .6.4.1 Bacterial RNA6.4.1b The startpoint is usually (>90% of the time) a purine. It iscommon for the startpoint to be the central base in the sequence CAT, but the conservation of this triplet is not great enough to regard it as an obligatory signal.b Just upstream of the startpoint, a 6 bp region is recognizable inalmost all promoters. The center of the hexamer generally is close to 10 bp upstream of the startpoint; the distance varies in known promoters from position -18 to -9. Named for its location, thehexamer is often called the -10 sequence. Its consensus isTATAAT, and can be summarized in the formb T80 A95 T45 A60 A50 T966.4.1 6.4.1Bacterial RNA b The distance separating the The distance separating the ––35 and 35 and ––10 sites is between 16-18 bp in 90% of promoters; in the exceptions, it is as little as 15 or as great as 20 bp. Although the actual sequence in the intervening region is unimportant, the distance is critical in holding the two sites at the appropriate separation for the geometry of RNA polymerase.bSome promoters have an A-T-rich sequence located farther upstream. This is called the upstream. This is called the UP element UP element UP element. It interacts with the . It interacts with the α subunit of the RNA polymerase. It is typically found inpromoters that are highly expressed, such as the promoters for rRNA genes.6.4.16.4.1 Bacterial RNAb The optimal promoter is a sequence consisting of thethe ––35 hexamer, separated by 17 bp from the–10 hexamer, lying 7 bp upstream of the startpoint.6.4.2 eukaryotic RNA polymerase b Transcription in eukaryotic cells is divided into three classes. Each class is transcribed by a different RNA polymerase: RNA polymerase I transcribes rRNARNA polymerase II transcribes mRNARNA polymerase III transcribes tRNA and other small RNAs. b For the three eukaryotic enzymes, the f actors factors , rather than the RNA polymerases themselves, are principally responsible for recognizing the promoter. This is different from bacterial RNA polymerase,where it is the enzyme that recognizes the promoter sequences.6.4.2 eukaryotic RNA polymerase b Sequence components of the promoter are defined operationally by the demand that they must be located in the general vicinity of the startpoint and are required forinitiation.b The enhancer is another type of site involved in initiation. It is identified bysequences that stimulate initiation, butthat are located a considerable distance from the startpoint. Enhancer elements are often targets for tissue-specific ortemporal regulation. temporal regulation. Figure 21.1Figure 21.1Figure 21.1 illustratesillustrates the general properties of promoters and enhancers.6.5 There are two types of terminators in E. coli bOnce RNA polymerase has started transcription, the enzyme moves along the template, synthesizing RNA, until it meets a terminator (t ) sequence. At this point, the enzyme stops adding nucleotides to the growing RNA chain, releases the completed product, and dissociates from the DNA template. Termination requires that all hydrogen bonds holding the RNA-DNA hybrid together must be broken, after which the DNA duplex reforms.bCore enzyme can terminate i n vitro in vitro at certain sites in the absence of any other factor. These sites are called intrinsic terminators. b Rho-dependent terminators are defined by the need for addition of rho factor (ρ) in vitro; and mutations show that the factor is involved in termination in vivo.6.5 There are two types of terminators in E. colib Intrinsic terminators have the twostructural features evident in: a hairpin in theFigure 9.46Figure 9.46: a hairpin in thesecondary structure; and a regionthat is rich in U residues at thevery end of the unit. Both featuresare needed for termination. Thehairpin usually contains a G··C-hairpin usually contains a Grich region near the base of thestem. The typical distance betweenthe hairpin and the U-rich regionis 7-9 bases.。

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