07双语第六章
《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
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
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
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
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
第六章 实数(原卷版)
第六章实数【题型一】算术平方根的实际应用典例1.(2022秋·浙江温州·七年级校联考期中)如图,纸上有五个边长为1的小正方形组成的图形纸(图1),我们可以把它剪开拼成一个正方形(图2).(1)图中拼成的正方形的面积是___________;边长是___________;(2)你能把十个小正方形组成的图形纸(图3),剪开并拼成正方形吗?若能,请仿照图的形式把它重新拼成一个正方形.并求出这个正方形的边长是___________.变式1-1.(2022春·福建福州·七年级福建省福州第十九中学校考期中)某小区为了促进全民健身活动的开展,决定在一块面积为21100m的正方形空地上建一个篮球场.已知篮球场的面积为2540m,其中长是宽的5倍,篮球场的四周必须留出1m宽的空地,请你通过计算说明能否按规定在这块空地上建一个篮球场?3变式1-2.(2022春·福建莆田·七年级校考期中)小明要剪一个面积为36cm2的正方形纸片,则边长是多少呢?如果还想剪一个面积为7cm2的正方形纸片,边长又是多少呢?变式1-3.(2022春·陕西渭南·七年级统考期中)如图,从一个大正方形中裁去面积为225cm的两个4cm和2小正方形,求留下的阴影部分的面积.【题型二】已知一个数的平方根求这个数典例2.(2022春·宁夏吴忠·七年级校考期末)已知2a﹣1的平方根为±3,3a+b﹣1的算术平方根为4,求a+2b的平方根.变式2-1(2022春·福建厦门·七年级福建省厦门集美中学校考期中)已知一个数m的两个不相等的平方根分别为a+2和3a-6.(1)求a 的值;(2)求这个数m .变式2-2.(2022春·新疆克拉玛依·七年级校考期末)已知一个正数的两个平方根是m +3和2m ﹣15.(1)求这个正数是多少?的平方根又是多少?变式2-3.(2022秋·山东烟台·七年级统考期末)已知正数x 的平方根是a 和a b+(1)当8b =时,求a 的值.(2)若22()4++=a x a b x ,求x 的值.【题型三】立方根的实际应用典例3.(2022春·新疆乌鲁木齐·七年级新疆农业大学附属中学校考期中)(1)()22125x +=(2)()332240x ++=.变式3-1.(2022春·山西吕梁·七年级统考期中)综合与实践如图是一张面积为2400cm 的正方形纸片.(1)正方形纸片的边长为______;(直接写出答案)(2)若用此正方形纸片制作一个体积为3216cm 的无盖正方体,请在这张正方形纸片上画出无盖正方体的平面展开图的示意图,并求出该正方体所用纸片的面积.变式3-2.(2022春·内蒙古呼伦贝尔·七年级校考期末)小军做了两个正方体纸盒,已知第一个正方体纸盒棱长为3厘米,第二个正方体纸盒比第一个纸盒体积大189立方厘米,试求第二个正方体纸盒的棱长.变式3-3.(2022春·安徽六安·七年级统考期中)把三个半径分别是3,4,5的铅球熔化后做一个更大的铅球,这个大铅球的半径是多少?(球的体积公式是343V R p =,其中R 是球的半径.)【题型四】平方根与立方根综合典例4.(2022春·福建龙岩·七年级统考期中)已知4a +1的平方根是±3,b ﹣1的算术平方根为2.(1)求a 与b 的值;(2)求2a +b ﹣1的立方根.变式4-1.(2022春·云南曲靖·七年级校考期末)已知m A =n -m +3的算术平方根,2m B -=m +2n 的立方根,求B -A 的平方根变式4-2.(2022春·广东中山·七年级统考期中)已知:21a -的算术平方根是3,31b +的立方根是2-,c23a b c +-的值.变式4-3.(2022春·湖北孝感·3=,求x 2+y 2-4的平方根.【题型五】实数与数轴典例5.(2022春·河北承德·七年级统考期末)如图,一只蚂蚁从点A 沿数轴向右爬了2个单位长度到达点B ,点A 表示,设点B 所表示的数为m .(1)求11m m ++-的值;(2)在数轴上还有C 、D 两点分别表示实数c 和d ,且有2c d +与23c d -的平方根.变式5-1.(2022春·湖北黄石·七年级统考期中)实数a 在数轴上的对应点A 的位置如图所示,b =|a |+|2−a |(1)求b 的值;(2)已知b +2的小数部分是m ,8-b 的小数部分是n ,求2m +2n +1的平方根.变式5-2.(2022春·江西宜春·七年级校考期中)(1)表示实数a ,b 的点在数轴上的位置如图所示,化简2b -+的值.(2a b ,求a b +变式5-3.(2022春·江西赣州·七年级统考期末)有理数a 、b 、c 在数轴上的位置如图.(1)用“>”或“<”填空:b ﹣c 0,a +b +3 0b c ++ 0.(2)3b c a b a c +-++--【题型六】实数的混合运算典例6.(2022春·内蒙古呼伦贝尔·七年级校考期末)计算:228)3|--变式6-1.(2022春·内蒙古呼和浩特·七年级呼和浩特市实验中学校考期中)计算:(1)1)2 ||-变式6-2.(2022春·湖南长沙·七年级长沙市长郡双语实验中学校联考期中)计算:()20221-【题型七】实数运算的实际应用典例7.(2022春·上海静安·七年级统考期中)如图,在面积为2平方米的正方形ABCD的木料中,挖去以边BC为直径的半圆,则剩下的木料的面积为多少平方米?( 3.14p»,结果精确到0.1)变式7-1.(2022春·江西上饶·七年级统考期末)如图,将长方形分成四个区域,其中A,B两正方形区域的面积分别是3和9.(1)A,B两正方形的边长各是多少?(2)求图中阴影部分的面积(结果保留两位小数. 1.414 1.732»»).变式7-2.(2022春·湖北宜昌·七年级校考期中)某农场有一块用铁栅栏围墙围成面积为700平方米的长方形空地,长方形长宽之比为7:4.(1)求该长方形的长宽各为多少?(2)农场打算把长方形空地沿边的方向改造出两块不相连的正方形试验田,两个小正方形的边长比为4:3,面积之和为600平方米,并把原来长方形空地的铁栅栏围墙全部用来围两个小正方形试验田,请问能改造出这样的两块不相连的正方形试验田吗,如果能,原来的铁栅栏围墙够用吗?【题型八】新定义下的实数运算典例8.(2022秋·福建厦门·七年级厦门市湖滨中学校考期末)设x 、y 是任意两个有理数,规定x 与y 之间的一种运算“Å”为:345()435()x y x y x y x y x y +-³ìÅ=í+-<î(1)求()11Å-的值;(2)若(2)(3)2m m -Å+=,求m 的值.变式8-1.(2022秋·山东济南·七年级统考期末)用“☆”定义一种新运算:对于任意有理数a 和b ,规定a ☆b =ab 2+2ab +a .如:1☆3=1×32+2×1×3+1=16(1)(﹣2)☆3= ;(2)若(12a +☆3)☆(﹣2)=16,求a 的值;(3)“作差法”是常见的比较代数式大小的一种方法,即要比较代数式M 、N 的大小,只要作出它们的差M ﹣N ,若M ﹣N >0,则M >N ;若M ﹣N =0,则M =N ;若M ﹣N <0,则M <N .若2☆x =m ,(14x )☆3=n (其中x 为有理数),试比较m ,n 的大小.变式8-2.(2022秋·江苏·七年级期末)用“*”定义一种新运算:对于任意有理数a 和b ,规定22a b b ab *=+,如:214421424*=+⨯⨯=.(1)求()25*-的值;(2)若()321x x -*=,求x 的值.变式8-3.(2022秋·湖南长沙·七年级统考期末)对于任意有理数a 、b 、c 、d ,可以组成两个有理数对(),a b 与(),c d .我们规定:()()a,b c,d ac bd Ä=-.例如:()()()2,41,3214314Ä-=⨯--⨯=-.根据上述规定,解决下列问题:(1)有理数对()()2,45,6-Ä-=______;(2)若有理数对()()3,2,418x Ä--=,则x =______;(3)当满足等式()()11229,x x y,y -Ä-=中的x 是整数时,求整数y 的值.。
中国文化概况(汉英双语)第六章 传统服饰与艺术品
第六章
hapter Six
Learning Goals 1. to learn the representative clothing styles and art crafts in different dynasties 2. to learn the cultural codes of traditional clothing and art crafts 3. to compare Chinese clothing and art crafts with those of other countries 4. to know about Chinese bronze ware 5. to know about traditional handicrafts, paper-cutting and embroidery 6. to know about major art works of porcelain decorations
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第一课
传ቤተ መጻሕፍቲ ባይዱ服饰
中文
In the period of Yellow Emperor, there was a saying for clothing called“Shang Yi and Xia Shang”. Although the clothing materials were pretty common and rough, such as linen, natural silk, furs and even tree leaves, Chinese people started to make the Jade accessories on what they called “clothes”. Jade is the symbol of elegance and preciousness in traditional Chinese culture. There is an old saying,“A gentleman is as gentle as a jade.”The pursuit of beauty for Chinese people has never ceased since then.
2024年国际商务合同双语教程
国际商务合同双语教程合同编号:__________第一章:合同双方第二章:股份转让标的2.2转让方同意将其持有的目标公司股份全部转让给受让方。
第三章:股份转让价格及支付方式3.1股份转让价格为人民币__________元/股,总价款为人民币__________元。
3.2.1在签订本合同之日起__________个工作日内,支付人民币__________元作为定金;3.2.2剩余股份转让价款在完成工商变更登记之日起__________个工作日内一次性支付。
第四章:股份转让手续及税费4.1.1签署股权转让协议;4.1.2提交目标公司股东会决议;4.1.3办理工商变更登记;4.1.4其他相关手续。
4.2股份转让过程中产生的税费,按照国家相关法律法规的规定由双方各自承担。
第五章:合同生效及终止5.1本合同自双方签字(或盖章)之日起生效。
5.2.1双方协商一致解除本合同;5.2.2因不可抗力导致本合同无法履行;5.2.3任何一方违反本合同约定,导致合同无法继续履行。
第六章:陈述与保证6.1转让方陈述与保证6.1.1转让方是目标公司股份的合法持有者,拥有完全的权利和授权将其股份转让给受让方。
6.1.2转让方持有的目标公司股份不存在任何质押、抵押或第三方权利限制。
6.1.3转让方保证目标公司的财务报表、经营状况和资产状况真实、准确、完整,不存在虚假记载或误导性陈述。
6.1.4转让方将及时向受让方披露目标公司的所有相关信息,包括但不限于重大合同、诉讼、仲裁、行政处罚等。
6.2受让方陈述与保证6.2.1受让方具有合法的资格和权利购买并持有目标公司的股份。
6.2.2受让方购买目标公司股份的资金来源合法,不存在任何违法或违规行为。
6.2.3受让方同意遵守目标公司的章程、股东协议和其他相关法律文件的规定。
第七章:保密条款7.1双方同意,在合同签订之前和之后,对于与本次股份转让相关的所有商业信息、技术信息、财务信息等,应予以严格保密。
小学数学 第六章 平方和平方根 Squares and Square Roots 英语双语练习题
Maths Chapter 6 Squares and Square Roots with Answers 数学第六章平方和平方根1. The Pythagorean triples whose smallest number is 8:1.毕达哥拉斯的最小数字为8的三元组:a) 8, 16 17 b) 8, 17, 18 c) 8, 15, 17 d) 8, 15, 16 Answer/Explanation 回答/解释Answer: (c) Explanation: The general form of Pythagorean triplets is 2m, m2-1, m2+1答:(c)说明:毕达哥拉斯三胞胎的一般形式是2m,m2-1,m2+1So, m = 4所以,m=4m2–1=42-1=16-1=15m2+1=42+1=16+1=172. Which of the following is a perfect square?2.下列哪一项是完全平方?a) 1057 b) 625 c) 7928 d) 64000Answer/Explanation 回答/解释Answer: (b) Explanation: The natural numbers which end with 0,2,3,7 and 8 are not perfect squares. Hence, 625 is a perfect square.答:(b)说明:以0、2、3、7、8结尾的自然数不是完全平方。
因此,625是一个完全的平方数。
25 x 25 = 6253. Which of the following will have 6 at unit place?3.以下哪一项在个位有6个?a) 192 b) 112 c) 242 d) 132Answer/Explanation 回答/解释Answer: (c) Explanation: 242= 24 x 24 = 576回答:(c)说明:242=24 x 24=5764. If 5278 is squared, then what will be at unit place?4.如果5278平方后,那个位是多少?a) 8 b) 7 c) 6 d) 4Answer/Explanation 回答/解释Answer: (d) Explanation: If we square the unit digit of the number 5278 we get:回答:(d)说明:如果我们把数字5278的个位数字平方,我们得到:82 = 64Hence, at the unit place of 52782 the number is 4.因此,52782的个位数数字是4。
双语自然Manykindsofanimals教案
双语自然Many kinds of animals教案第一章:介绍动物分类教学目标:1. 学习动物的分类词汇,如:哺乳动物、鸟类、爬行动物等。
2. 能够用英语简单描述不同种类的动物。
教学内容:1. 引入动物主题,展示不同种类的动物图片,让学生观察并说出它们的名称。
2. 教授动物分类的词汇,如:哺乳动物、鸟类、爬行动物等,并给出实例。
3. 让学生用英语描述不同种类的动物,如:老虎是哺乳动物,鸟可以飞等。
第二章:学习动物特征教学目标:1. 学习动物的特征词汇,如:有毛发、有翅膀、有爪子等。
2. 能够用英语描述动物的特征。
教学内容:1. 引入动物特征主题,展示不同动物的特征图片,让学生观察并说出它们的特征。
2. 教授动物特征的词汇,如:有毛发、有翅膀、有爪子等,并给出实例。
3. 让学生用英语描述动物的特征,如:猫有毛发,鸟有翅膀等。
第三章:学习动物生活环境教学目标:1. 学习动物的生活环境词汇,如:森林、海洋、草原等。
2. 能够用英语描述动物的生活环境。
教学内容:1. 引入动物生活环境主题,展示不同动物的生活环境图片,让学生观察并说出它们的生活环境。
2. 教授动物生活环境的词汇,如:森林、海洋、草原等,并给出实例。
3. 让学生用英语描述动物的生活环境,如:鱼生活在海洋中,狮子生活在草原上等。
第四章:学习动物行为教学目标:1. 学习动物的行为词汇,如:捕食、迁徙、休息等。
2. 能够用英语描述动物的行为。
教学内容:1. 引入动物行为主题,展示不同动物的行为图片,让学生观察并说出它们的行为。
2. 教授动物行为的词汇,如:捕食、迁徙、休息等,并给出实例。
3. 让学生用英语描述动物的行为,如:鹰捕食小鸟,鲸鱼迁徙等。
第五章:学习动物保护教学目标:1. 学习动物保护的词汇,如:保护、自然保护区、野生动物保护等。
2. 能够用英语描述动物保护的重要性。
教学内容:1. 引入动物保护主题,展示不同动物保护的图片,让学生了解动物保护的重要性。
社会契约论(双语版)
BOOK THREE
CHAPTER 1 Government in General CHAPTER 2 The Principle which Constitutes the Different Forms of Government CHAPTER 3 Classification of Governments CHAPTER 4 Democracy CHAPTER 5 Aristocracy CHAPTER 6 Monarchy CHAPTER 7 Mixed Governments CHAPTER 8 That Every Form of Government Is Not Fit for Every Country
4 第八章论任何
一种政府形式 都不能适用于 所有国家
5
第九章论一个 好政府的标志
第十章论政府 1
的滥用权力和 它衰退的倾向
2
第十一章论政 治体的灭亡
3 第十二章论如
何维持主权权 威
4 第十三章论如
何维持主权权 威(续)
5 第十四章论如
何维持主权权 威(再续)
第十五章论议员或代 表
第十六章论政府的建 立不是一项契约
作者介绍
同名作者介绍
卢梭(Jean-Jacques Rousseau,1712年6月28日-1778年7月2日),出生于日内瓦,法国伟大的哲学家、 教育家、思想家、文学家,18世纪法国大革命的思想先驱,启蒙运动的代表人物之一。 1722年,寄宿在郎拜尔 西埃牧师家,学习古典语文、绘图、数学。1732年,做土地测量工作,自学数学。1737年,因化学实验双眼受伤, 到沙尔麦特村养病。1742年,在巴黎科学院提出《新乐谱记谱法》。1743年,写成歌剧《风雅的缪斯》,以教音 乐、抄乐谱为生。1749年,撰写《百科全书》的音乐部分。1750年,因应征论文《论科学与艺术》获第戎学院奖 金,而声名鹊起。1778年7月2日,在巴黎逝世。卢梭一生颠沛流离,因发表《科学与艺术的进步是否有助敦化风 俗》而闻名,主要著作有《社会契约论》《论人类不平等的起源和基础》等。
细胞生物学(双语)智慧树知到答案章节测试2023年滨州学院
第一章测试1.与高等植物细胞相比,动物细胞特有的结构包括()。
A:核仁B:内质网C:溶酶体D:中心体答案:D2.所有病毒中都含有DNA 和RNA。
()A:对B:错答案:B3.支原体是最小最简单的真核细胞。
()A:对B:错答案:B4.“细胞是生命活动的基本单位”体现在()A:一切有机体都由细胞构成,细胞是构成有机体的基本单位B:细胞是生命起源的标志,是生物演化的起点。
C:细胞是代谢与功能的基本单位,也是有机体生长发育的基础D:细胞是繁殖的基本单位,是遗传的桥梁答案:ABCD5.下列哪种描述不属于细胞的基本共性()A:有相同的遗传装置B:都有线粒体C:具有脂-蛋白体系的生物膜D:一分为二的分裂方式答案:B第二章测试1.为什么透射电镜的照片没有彩色的()A:彩色显微照片的成本太高B:照相的底片接受的是穿过样品的电子C:细胞内部结构是没有颜色的,都是灰白色的暗影D:彩色胶卷还没有发明答案:B2.扫描电子显微镜可用于()A:观察细胞表面的立体形貌B:观察活细胞C:定量分析细胞中的化学成份D:获得细胞不同切面的图象答案:A3.人工培养的细胞中,细胞系是丧失接触抑制的细胞。
()A:对B:错答案:A4.鼠源性单抗在体内应用时,不存在主要组织相容性抗原(MHC)和超敏反应问题。
()A:错B:对答案:A5.细胞培养根据培养方式分为:()A:贴壁细胞培养B:悬浮细胞培养C:半悬浮细胞培养D:静止培养答案:ABC第三章测试1.膜脂成分中含量最多的是()。
A:胆固醇B:脂肪C:磷脂D:糖脂答案:C2.生物膜的液态流动性主要取决于()。
A:糖蛋白B:脂类C:蛋白质D:多糖答案:B3.脂质体是根据磷脂分子可在水相中形成稳定的脂双层膜的趋势而制备的人工膜。
()A:对B:错答案:A4.外周膜蛋白为水不溶性蛋白,形成跨膜螺旋,与膜结合紧密,需用去垢剂使膜崩解后才可分离。
()A:对B:错答案:B5.膜脂分子的运动方式包括()A:脂分子尾部的摆动B:脂分子围绕轴心的自旋运动C:沿膜平面的侧向运动D:双层脂分子之间的翻转运动答案:ABCD第四章测试1.载体蛋白只能进行被动运输。
双语版危险化学品安全管理条例Regulations on the Control over Safety of Dangerous Chemicals dual langu
危险化学品安全管理条例Regulations on the Safety Administration of Dangerous Chemicals危险化学品安全管理条例Regulations on the Safety Administration of Dangerous Chemicals 【法规变迁史】危险化学品安全管理条例(国务院令第344号)2002年1月26日发文(Decree No.344of the State Council of the People's Republic of China was issued on 26th January 2002,and revised by the 144nd Executive Meeting of the State Council on 16th February2011)目录第一章总则 (3)第二章生产、储存安全 (10)第三章使用安全- (21)第四章经营安全 (23)第五章运输安全 (30)第六章危险化学品登记与事故应急救援 (41)第七章法律责任 (45)第八章附则 (68)安监总局总工黄毅对新修订5处重大修改条款解释: (70)ContentsChapter I General Principles (3)ChapterⅡSafety of Manufacture and Storage (10)ChapterⅢSafety of Use (21)ChapterⅣSafety of Operation (23)ChapterⅤSafety of Transportation (30)Chapter VI Registration of Dangerous Chemicals and Emergency Rescue of Accidents (41)Chapter VII Legal Liability (45)Chapter VIII Supplementary Rules (68)State Administration of Work Safety(SAWS)’s chief engineer,Huang Yi explained the5major terms newly revised as below (70)第一章总则Chapter I General Principles第一条为了加强危险化学品的安全管理,预防和减少危险化学品事故,保障人民群众生命财产安全,保护环境,制定本条例。
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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.。