武汉大学《分子生物学》考研复习题库

上传人:a****c 文档编号:142543931 上传时间:2020-08-20 格式:PDF 页数:19 大小:158.79KB
返回 下载 相关 举报
武汉大学《分子生物学》考研复习题库_第1页
第1页 / 共19页
武汉大学《分子生物学》考研复习题库_第2页
第2页 / 共19页
武汉大学《分子生物学》考研复习题库_第3页
第3页 / 共19页
武汉大学《分子生物学》考研复习题库_第4页
第4页 / 共19页
武汉大学《分子生物学》考研复习题库_第5页
第5页 / 共19页
点击查看更多>>
资源描述

《武汉大学《分子生物学》考研复习题库》由会员分享,可在线阅读,更多相关《武汉大学《分子生物学》考研复习题库(19页珍藏版)》请在金锄头文库上搜索。

1、考试复习重点资料(最新版)考试复习重点资料(最新版) 封封 面面 第1页 资料见第二页资料见第二页 Quiz Roles Molecular Biology Course (1)桌上只许有笔,答题纸和英文课本。()桌上只许有笔,答题纸和英文课本。(2)不许离 开自己的座位。( )不许离 开自己的座位。(3)不许偷看,不许讲话)不许偷看,不许讲话 (有问题举 手)( 有问题举 手)(4)老师宣布停的时候,必须放下自己的笔。()老师宣布停的时候,必须放下自己的笔。(5) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生

2、违反以上任何 一条规则,第一次违规者,扣除总考试成绩的 ) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违规者,扣除总考试成绩的15%,第 二次违规者必须重修这们课。 ,第 二次违规者必须重修这们课。 Quiz 1: Section A-D 1. Describe the structural difference of the DNA and RNA. How these structural differences interpret the biological functi

3、ons of DNA and RNA (25 points) 2. Describe the major differences of the prokaryotes and eukaryotes. Where do DNA replication, RNA transcription and protein translation occur in prokaryotes and eukaryotes? (25 points) 3. Describe the chromatin structure of eukaryotes, how the different levels of chro

4、mosome packing relate to cell cycle and gene expression? (25 points) 4. Why protein purification is import? What is proteomics? What is the experimental procedure in identifying an interested protein from a protein mix by using the proteomics technique? (25 points) Molecular Biology Course 40 min Di

5、scussions 1: Cells and macromolecules (25 min) 1. Quiz question 1 2. Quiz question 2 3. Quiz question 3 4. Non-covalent interactions: hydrogen bonds, hydrophobic interaction 5. Protein purification 6. DNA purity 7. Open questions Quiz Roles Molecular Biology Course (1)桌上只许有笔,答题纸和英文课本。()桌上只许有笔,答题纸和英文

6、课本。(2)不许离 开自己的座位。( )不许离 开自己的座位。(3)不许偷看,不许讲话)不许偷看,不许讲话 (有问题举 手)( 有问题举 手)(4)老师宣布停的时候,必须放下自己的笔。()老师宣布停的时候,必须放下自己的笔。(5) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违规者,扣除总考试成绩的 ) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违规者,扣除总考试成绩

7、的15%,第 二次违规者必须重修这们课。 ,第 二次违规者必须重修这们课。 Quiz 1: Section E-F 1. What is the semi-conservative replication? Describe the structural and biological reasons for this replication mechanism? (10 points) 2. What is semi-discontinuous replication. Why DNA replication is semi-discontinuous? How the semi-discon

8、tinuous replication is accomplished in E. coli? (20 points) 3. Compare the prokaryotic replication with eukaryotic replication. (20 points) 4. What is the end replication problem in eukaryotic cells? How is the end replication problem resolved (20 points) Molecular Biology Course 50 min 5. What are

9、the mechanisms that keep the high accuracy of DNA replication (10 points) 6. If DNA lesions occur in an organism, cells will attempt to repair the lesions. Describe the known repair mechanisms. How could mutations be generated? (25 points) Quiz Roles Molecular Biology Course (1)桌上只许有笔,答题纸和英文课本。()桌上只

10、许有笔,答题纸和英文课本。(2)不许离 开自己的座位。( )不许离 开自己的座位。(3)不许偷看,不许讲话)不许偷看,不许讲话 (有问题举 手)( 有问题举 手)(4)老师宣布停的时候,必须放下自己的笔。()老师宣布停的时候,必须放下自己的笔。(5) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违规者,扣除总考试成绩的 ) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违

11、规者,扣除总考试成绩的15%,第 二次违规者必须重修这们课。 ,第 二次违规者必须重修这们课。 Quiz 1: Section K-L 1. (20)Compare the processes of DNA replication and transcription. How do they differ in terms of the enzymes involved, the fidelity of the reaction, and the nature of the molecules produced by each reaction? What do the processes h

12、ave in common? 2. (25) What are the major trans-acting factors and cis-acting elements involved in E. coli transcription? How they interact with each other to carry out the transcription process? Describe the roles of the each subunit when you get to the E. coli RNA polymerase. Molecular Biology Cou

13、rse 50 min 3. (10) Regulation of gene expression in prokaryote usually occurs at the transcription _level, and the two major mechanisms are_ and _ . 4. (15) What are the mechanisms that ensure the structure genes of lac operon are only transcribed at a full speed when lactose is the only carbon sour

14、ce. Detail the molecular interactions and events in each mechanism. 5. (15) How the control strategy used by trp operon differs from that of lac operon, why? 6. (15) An operon in E. coli is controlled by a repressor that binds at the two operator sites (O1 and O2) diagrammed in the figure below. In

15、the presence of the appropriate inducer, a transcription rate of 100 is observed, but in the absence of inducer, the transcription rate falls to 5. If either of the two sites is mutated so that the repressor cannot bind, then the transcription rate is observed to be 100. Additionally, if base pairs

16、are inserted at the arrow, the level of transcription is found to vary with the size of the insert as shown in the graph below. Briefly explain this data. Quiz Roles Molecular Biology Course (1)桌上只许有笔,答题纸和英文课本。()桌上只许有笔,答题纸和英文课本。(2)不许离 开自己的座位。( )不许离 开自己的座位。(3)不许偷看,不许讲话)不许偷看,不许讲话 (有问题举 手)( 有问题举 手)(4)老师宣布停的时候,必须放下自己的笔。()老师宣布停的时候,必须放下自己的笔。(5) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同学负责把试卷交给当堂助教。 如果主讲老师或助教发现任何学生违反以上任何 一条规则,第一次违规者,扣除总考试成绩的 ) 老师宣布收试卷时,试卷必须快速由每排的最左边传向 右边,每排最右边的同

展开阅读全文
相关资源
相关搜索

当前位置:首页 > 高等教育 > 工学

电脑版 |金锄头文库版权所有
经营许可证:蜀ICP备13022795号 | 川公网安备 51140202000112号