09-10分子生物学期末试卷答案

1、福建农林大学期末考试试卷参考答案2009 2010学年第 学期课程名称： 分子生物学 考试时间 120 分钟 专业 年级 班 学号 姓名 题号一二三四总得分得分评卷人签字复核人签字得分一、填空题（每空0.5分，共10分）1、There are five types of DNA repair, include photo-reactivation repair, _， _，_ and error-prone repair.（切除修复；重组修复；错配修复）2、The holoenzyme of E.coli RNA polymerase is_，the core enzyme is made u

2、p of _，while the_ factor has the ability to recognize specific binding sites.（2；2；）3、The eukaryotes have a well-defined cell cycle.The phase are：G1，_，_ and M.（S；G2）4、Only one strand of a DNA duplex will be transcribed. So the RNA is identical in sequence with another strand of the DNA which is calle

3、d _，and the template strand is _.（有义链；反义链）5、In eukaryotes, mRNA processing includes the following steps: addition of at 5end, addition of at 3end, removal of , joining together the ,And these take place in .（M7G帽子结构；poly(A)尾巴；内含子；外显子；细胞核）6、The structure is a common secondary structural reprensentati

4、on of tRNA molecules. The tertiary structure of tRNA molecules is . （三叶草；L）7、In the large 60S subunit of the eukaryote ribosome, there are rRNA, and rRNA, and rRNA. (5S；5.8S；28S) 得分二、名词解释（每小题2分，共30分）1、Promoter：RNA聚合酶结合并起始转录的一段DNA序列。2、Semi-conservative replication：半保留复制，亲代双链DNA以每条链为模板，按碱基配对原则各合成一条互补链

5、，这样一条亲代DNA双螺旋，形成两条完全相同的子代DNA螺旋，子代DNA分子中都有一条合成的“新”链和一条来自亲代的旧链。3、Intron：真核生物DNA编码序列中间隔的非编码序列，它们也能被转录成RNA，但在mRNA成熟加工中会被切除。4、Transversion：在基因突变中，嘌呤被嘧啶所代替的现象，反之亦然。5、Gene recombination：指由于不同DNA链的断裂和连接而产生的DNA片段的交换和重新组合，形成新的DNA分子的过程。因此，新的DNA分子中含有原来的两个DNA 分子的片段。6、Hybridization：不同来源的核酸单链通过碱基互补配对原则结合成双链的过程。7、T

6、ranscription：以DNA为模板，在RNA聚合酶的催化下，合成RNA的过程。8、Proofreading activity：即DNA聚合酶的35外切酶活性，当复制过程中插入错配碱基时该活性被激活，进而将错配碱基切除。9、RNA processing：RNA加工，前提RNA转变为成熟RNA的过程，发生在细胞核内。10、Triplet code：三联体密码，3个连续的核苷酸编码一个氨基酸。11、Chaperone：分子伴侣，帮助蛋白质分子形成正确空间构象的辅助分子。12、Trans-acting element：反式作用元件，指能直接或间接地识别或结合在各类顺式作用元件核心序列上参与调控靶

7、基因转录效率的蛋白质。13、Open reading frame：特定的核酸序列由一个起始密码子（AUG/ATG）和三个终止密码子组成，核酸序列能翻译为蛋白质序列。14、Ribozyme：核酶，具有催化作用的RNA分子。15、Zinc finger：锌指，一种常出现在DNA结合蛋白中的一种结构基元。是由一个含有大约30个氨基酸的环和一个与环上的4个Cys或2个Cys和2个His配位的Zn2+构成，形成的结构像手指状。得分三、单项选择题（每小题1分，共15分）（ D ）1、What is the percentage of A in a piece of RNA if the percentag

8、e of C is 30%?A. 30% B. 20%C. 60% D. Unable to calculate with the information provided.（ D ）2、Which of the following is NOT a function of RNA?A. It acts a template during translation.B. It is required for RNA splicing.C. It is essential for the structure of ribosomes.D. It attaches the correct amino

9、 acid to tRNA.（ C ）3、Histones are found in close association with DNA. Which of the following statements is incorrect?A. Histone H1 is less evolutionarily conserved than core histones.B. A nucleosome is composed of 146 bp of DNA and a histone octomerC. There are only four types of histonesD. Nucleos

10、omes are evenly distribution within chromatin（ B ）4、The elongation of the lagging strand during DNA synthesis:A. In the 3-5 directionB. Produces Okazaki fragments C. Polymerizes into the direction of fork movementD. Does not require a template strand（ C ）5、Telomeres are found at the end of chromosom

11、es and they are synthesized by telomerase. Which one of the following statements is correct?A. The function of telomerase is to directly elongate the leading strand during DNA replication.B. Telomers code for ribosomal RNA.C. Telomerase is an unusual enzyme made up of RNA template and protein.D. Tel

12、omerase is active in all cell types.（ C ）6、Which of the following is correct concerning transcription and DNA replication?A. Only transcription requires a primer.B. Transcription starts at the origin of replication.C. In DNA replication, both the coding and template strands are copied.D. The entire

13、genome is copied in transcription.（ D ）7、Which of the following statements describes a similarity between the structure of DNA and RNA? In both DNA and RNA, A. the nitrogenous base uracil is found.B. there is a hexose sugar.C. the carbon 2 has an hydroxyl group.D. the nitrogenous base is attached to

14、 carbon 1.（ D ）8、Regulation of the trytophan operon involvesA. repression (at an operator) B. attenuationC. the concentration of free trytophan D. all of the above（ B ）9、Which of the following is not stop codon? A. TAA B. TAT C. TAG D. TGA（ A ）10、The Shine-Dalgarno sequence in mRNAsA. base pairs to

15、the 3'-end of E. coli 16S rRNAB. base pairs to the 5'-end of E. coli 16S rRNAC. has a consensus sequence of YNCURACD. is located just downstream of the initiation codon （ C ）11、Which of the following is not bound to the aminoacyl-tRNA synthetase? A. amino acid B. tRNA C. GTP D. ATP （ D ）12、T

16、he ribosome scanning model for translation was proposed byA. P. Sharp B. R. RobertsC. T. Cech D. M. Kozak（ D ）13、Most nuclear mRNA introns begin with _ and end with _. A. U, G B. GU, NG C. AG, GU D. GU, AG （ A ）14、Which of the following is the covalent linkage of ubiquitin molecule with the protein

17、damaged? A. Gly B.Lys C. Leu D.Thr（ C ）15、In nuclear mRNA splicing, the intron is released as a A. circular RNA B. linear RNA C. lariat RNA D. a “Y” RNA得分四、问答题（45分）1、Please illuminate the main content of central dogma and draw the sketch map.（8 points）DNA是生物遗传的主要物质基础。遗传信息以特定的脱氧核苷酸序列储存在DNA分子中（1分）。DNA

18、能准确地自我复制（1分），并把遗传信息转录到mRNA分子上（1分），然后通过mRNA翻译成为具有特定氨基酸顺序的蛋白质（1分），由蛋白质行使各种生物学功能。Crick把这种遗传信息的流动称为分子遗传的中心法则。后来人们又发现反转录病毒可以其RNA作为模板指导合成DNA（1分）；某些病毒RNA可以进行自我复制，即以自身RNA为模板，合成子代RNA（1分）。这些发现是对中心法则的补充和发展。 DNA RNA 蛋白质 （2分）2、What is the operon? How the lac operon controls gene expression?（8 points）操纵子是原核生物基因表达

19、调控的单位（1分），包括启动子、操纵基因和一系列功能上相关的结构基因（1分）。没有乳糖lacI编码的阻遏蛋白具有活性牢固结合到操纵基因O上结合在启动区的RNA pol不能通过Z、Y、A不能转录（和表达）（3分）加入乳糖乳糖（诱导物）与阻遏蛋白结合阻遏蛋白构象变化阻遏蛋白从操纵基因上脱落RNA pol开始Z、Y、A基因的转录，表达乳糖被利用（3分） 3、Please write out the enzymes and proteins required in prokaryotic DNA replication，and explain their functions.（7 points）拓扑异

20、构酶：消除DNA超螺旋产生的阻力；（1分）解链酶：解开DNA双链，产生单链DNA；（1分）SSB：保护解开的DNA单链，避免复性；（1分）引物酶：合成RNA引物，为DNA聚合酶提供3末端；（1分）DNA聚合酶III：延长DNA片段，是真正起复制作用的酶；（1分）DNA聚合酶I：负责切除引物并填补引物所在的缺口；（1分）DNA连接酶：连接冈崎片段。（1分）4、What are the characters of expression vectors?（5 points）（1）能够进行独立于基因组的自我复制 （1分）（2）容易从细胞中分离提取 （1分）（3）含有多克隆位点，允许外源片段的插入 （1

21、分）（4）含有至少一个选择标记 （1分）（5）具有启动子、终止子等表达元件 （1分）5、Please illuminate the function of the cap structure of mature mRNA in 5emd.（5 points） (1) 对5外切核酸酶形成屏障，保护和稳定转录产物（2分） (2) 有助于前体mRNA的剪接和转运到细胞质（2分） (3) 提高翻译的效率（1分）6、Please illuminate the difference between the initial tRNA molecule and the tRNAMet in elongation in E.coli. （5 points）（1）只有起始tRNA分子的蛋氨酸甲酰化，非甲酰化的fmet-tRNA也能起始，但效率不高，因为甲酰化是与tRNA分子结