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1、武汉大学 2003 至 2004 学年第一学期分子生物学期末考试试题 武汉大学生命科学学院 20032004 学年第一学期期末考试 分子生物学试卷 Final exam of Molecular Biology Course (Spring 2004) 年级 _ 专业 _ 学号 _ 姓名 _成绩_ PART I: DESCRIPTION (2 points each) Your answer should describe what each item is and how it functions in the cell. Diagrams, structure and sequence i
2、nformation should be included in your answer, as necessary. 1. Yeast artificial chromosome 2. RNA interference 3. Proteomics 4. Shine-Dalgarno sequence 5. Alternative splicing 6. Ribozyme 7. r-dependent termination 8. RNA editing 9. DNA lesions 10. Protein targeting PART II: MULTIPLE CHOICES (1 poin
3、ts each) Select the one best answer for each question. 1. The catalytic activity for peptide bond formation (the peptidyl transferase activity) is located in the: 1) RNA of the large ribosomal subunit. 2) leader sequence of the messenger RNA. 3) RNA of the small ribosomal subunit. 4) proteins of the
4、 small ribosomal subunit. 5) proteins of the large ribosomal subunit. 2. Bidirectional and semi-conservative are two terms that refer to: 1) transcription. 2) translation. 3) replication. 4) all of the above. 5) none of the above. 3. The fact that most amino acids are specified by multiple codons is
5、 known as: 1) the “wobble ” phenomenon. 2) the universality of the genetic code. 3) codon bias. 4) the anticodon hypothesis. 5) the redundancy of the genetic code. 4. RNA polymerase I is the eukaryotic enzyme responsible for: 1) transcription of ribosomal RNA. 2) transcription of transfer RNA and ot
6、her small RNA species. 3) transcription of messenger RNA. 4) initiation of Okazaki fragment synthesis in DNA replication. 5. Restriction enzymes can cleave DNA that is either single-stranded or double-stranded, as long as it contains the appropriate recognition site. 1) True 2) False 6. Information
7、about the sequence of the coding region of a gene is best obtained from: 1) a YAC clone. 2) a genomic clone. 3) a cDNA clone. 4) the protein. 7. A chromatography method that can be used specifically to purify proteins based on their charge is: 1) gel filtration chromatography. 2) ion-exchange chroma
8、tography. 3) DNA affinity chromatography. 4) antibody affinity chromatography. 8. A nonsense mutation is a change in the DNA sequence that results in: 1) a small deletion or insertion. 2) an amino acid change in the protein encoded by the gene. 3) a premature stop codon. 4) all of the above. 5) none
9、 of the above. 9. A protein complex involved in degradation of proteins within the cell is known as the: 1) ubiquitin/proteasome system. 2) molecular chaperone. 3) chaperonin. 4) ribosome. 5) Krebs/TCA cycle. 10. _binds to the repressor and turn on the transcription of the structural genes in the La
10、c operon. 1) cAMP 2) lactose 3) allolactose 4) CRP 11. Which of the following RNA species is involved in degradation of the mRNA containing complementary sequence 1) miRNA 2) siRNA 3) tRNA 4) 5S RNA 5) U3 snRNA 12. The genome sequencing projects are confirming the theory that genome size is directly
11、 proportional to the number of genes contained within that genome. In other words, a genome that is 10 times as big will contains approximately 10 times as many protein coding genes. 1) True 2) False 13. HeLa cells, derived from a human cervical carcinoma, are able to propagate indefinitely in cultu
12、re and are therefore known as a(n): 1) tissue culture. 2) tumor. 3) transgenic cell line. 4) immortalized cell line. 14. E. coli cells are smaller than yeast cells. 1) True 2) False 15. Which of the following domains is not a DNA binding domain 1) Proline-rich domains 2) Helix-turn helix domains 3)
13、Zinc finger domains 4) Basic domains 16. The aminoacyl-tRNA synthetases distinguish between about 40 different shaped tRNA molecules in the cells. 1) True 2) False PART III: SHORT QUESTIONS (8 points each) 1. How do bacterial replication start and accomplished. Remember to include the proteins/enzym
14、es and important DNA sequence involved in this process. 2. Design experiments to clone a yeast gene and express this gene in yeast. 3. Below is the multiple cloning site (MCS) of the plasmid vector pUC18 and the N-terminal and C-terminal sequence of protein X. Note that the MCS constitutes a part of
15、 the LacZ open reading frame. Suppose that you are going to clone the protein X gene into pUC18, so that your target gene is transcribed under the control of LacZ promoter, and translated with the LacZ gene to produce a fusion protein. You are requested to use BamHI and PstI to the clone X gene, ple
16、ase add these restriction sites on the corresponding position of the Xgene. Remember to maintain the reading frame of the X gene with the LacZgene 4. (1) MCS of pUC18 EcoRI SacI KpnI SmaI BamHI XbaI SalI PstI ACG AAT TCG AGC TCG GTA CCC GGG GAT CCT CTAGAG TCG ACC TGC AGG CAT GCA Thr Asn Ser Ser Ser
17、Val Pro Gly Asp Pro Leu Glu Ser Thr Cys Arg His Ala (2) N-terminal sequence of X gene. ATG ACC CCU CAU AAC GGC GAC , Met Thr Pro His Asn Gly Asp, (3) C-terminal sequence of X gene. , GAU AGU ACA GCU GCC AAG TAA ,Asp Ser Thr Ala Ala Lys PART IV: MAJOR QUESTIONS (20 points each) 1: Please describe how
18、 an mRNA gene is transcribed, processed and translated in human cells. What are the possible mechanisms in regulating the expression of this gene? 2 (20 points): A bacterium is found to metabolize a rare sugar produced by a plant that the bacteria grow on. However, the bacteria prefer glucose as the
19、 energy source. The problem is, if you want to finish this course with a satisfied score, you must figure out the regulatory mechanism that the bacteria used to determine the sugar choice. The gene involving in the rare sugar metabolism has been identified as fun3. You can use northern blot to analy
20、ze the expression of fun3 and use DNA footprinting to analyze the binding of proteins to the control elements of fun3 gene. The following table shows the experimental results Glucose presence Rare sugar presence Levels of fun3RNA Binding of proteins - - - Protein A binds to the promoter region Prote
21、in C binds upstream of the promoter + - - Protein A binds to the promoter region - + + Protein B binds to the promoter region Protein C binds upstream of the promoter + + + Protein B binds to the promoter region Questions: 1. Please propose a mechanism to explain the above results. You should focus
22、on the question “How does the expression of fun3 gene is tightly regulated so that it is only highly expressed when the rare sugar is the only carbon source”. You must answer what proteins A, B and C are. (8 points) 2. How is protein A regulated? (2 points) (1) glucose turns the repressor on (2) glu
23、cose turns the repressor off (3) the rare sugar turns the repressor on (4) the rare sugar turns the repressor off 3. How is protein C regulated? (2 points) (1) glucose turns the activator on (2) glucose turns the activator off (3) the rare sugar turns the activator on (4) the rare sugar turns the ac
24、tivator off 4. How could you make the bacteria always use the rare sugar as the energy source even in the presence of glucose? (8 points) 武汉大学生命科学学院 20032004 学年第一学期期末考试 分子生物学试卷及参考答案 Final exam of Molecular Biology Course (Spring 2004) 写在参考答案前面的话： ?该课程考试目的是考查学生对所学知识掌握的情况，除选择题外，其他题目的答案基本都不是唯一的。 你可以从不同
25、的角度去阐明一个概念。 ?公布参考答案的目的：为该课程画个句号，让学生过个安心暑假 ? Best wishes to all of you PART I: DESCRIPTION (2 points each) Your answer should describe what each item is and how it functions in the cell. Diagrams, structure and sequence information should be included in your answer, as necessary. 1. Yeast artificial
26、chromosome: 酵母人工染色体（0.5 point） contains components required for replication and segregation of the natural yeast chromosome, including two telomeric sequences (TEL), one centromere (CEN) and one autonomously replicating sequences (ARS) (0.75 point ) contains genes act as selective markers in yeast a
27、nd proper restriction sites (0.75 point ) can accommodate genomic DNA fragments of more than 1 Mb (0.5 point) 2. RNA interference: RNA干扰 （0.5 point） a conserved biological response to double-stranded RNA (1 point) regulates the expression of protein-coding genes through siRNA or miRNA (1 point) the
28、dsRNA is restricted by DICER, then RISC mediates the siRNA and miRNA related RNA degradation and translation inhibition, respectively. (1 point) 3. Proteomics 蛋白组学（0.5 point） The study of the proteome using techniques of high resolution protein separation and identification (1.5 point) The best sepa
29、ration method is two dimensional gel electrophoresis, the individual protein spots are then cut from the gel and treated with protease to produce a set of peptides characteristic of that protein. The precise masses of each peptide in the sample are then determined by MALDI mass spectrometry. The res
30、ulted peptide mass fingerprint of that protein is then compared to a database to deduce the function of that protein etc. (1.5 point) 4. Shine-Dalgarno sequence SD 序列（0.5 point） A conserved sequence 8-13 nt upstream of the first codon to be translated (1 point). This sequence was discovered by Shine
31、 and Dalgarno (0.5 point) The sequence is purine-rich and contains all or part of 5-AGGAGGU-3 (0.5 point) Can base-pair with the 3-end of the 16S rRNA (0.5 point) 5. Alternative splicing 可变剪接（0.5 point） The generation of different mature mRNAs from a particular type of gene transcript by choosing di
32、fferent 5- and 3 -splice sites (2 point). 6. Ribozyme 核酶（0.5 point） an RNA molecule capable of catalyzing a chemical reaction (2 point). 7. r-dependent termination r-依赖型转录终止（0.5 point） During bacterial transcription, some terminator sites do not form strong hairpins, thus termination of the transcri
33、ption by bacterial RNA polymerase requires the assistance of an accessory factor called rho (r)protein (2 point). 8. RNA editing RNA编辑（0.5 point） A form of RNA processing in which nucleotide sequence of the primary transcriptis altered by either changing, inserting or deleting residues at the specif
34、ic points along the molecule (2 point). 9. DNA lesions DNA损伤 （0.5 point） An alteration to the normal chemical or physical structure of the DNA (2 point). The lesions can lead to cell death or DNA mutation (1 point). 10. Protein targeting 蛋白定位（0.5 point） Synthesis of eukaryotic proteins is usually oc
35、curred in cytoplasm (0.5 point). However, many proteins need to be transported to specific cellular locations, such as nucleus, mitochondrion or chloroplast, to exert their biological functions. This process is called protein targeting (1 point). The ultimate cellular location of proteins is often d
36、etermined by specific, relative short, amino acid sequences within the proteins themselves. The sequence inside of a protein determining the cellular location of the protein is called signal sequence (1 point). PART II: MULTIPLE CHOICES (1 points each) Select the one best answer for each question. 1
37、. The catalytic activity for peptide bond formation (the peptidyl transferase activity) is located in the: 1)RNA of the large ribosomal subunit. 2) leader sequence of the messenger RNA. 3) RNA of the small ribosomal subunit. 4) proteins of the small ribosomal subunit. 5) proteins of the large riboso
38、mal subunit. 2. Bidirectional and semi-conservative are two terms that refer to: 1) transcription. 2) translation. 3)replication. 4) all of the above. 5) none of the above. 3. The fact that most amino acids are specified by multiple codons is known as: 1) the “wobble ” phenomenon. 2) the universalit
39、y of the genetic code. 3) codon bias. 4) the anticodon hypothesis. 5)the redundancy of the genetic code. 4. RNA polymerase I is the eukaryotic enzyme responsible for: 1)transcription of ribosomal RNA. 2) transcription of transfer RNA and other small RNA species. 3) transcription of messenger RNA. 4)
40、 initiation of Okazaki fragment synthesis in DNA replication. 5. Restriction enzymes can cleave DNA that is either single-stranded or double-stranded, as long as it contains the appropriate recognition site. 1) True 2) False 6. Information about the sequence of the coding region of a gene is best ob
41、tained from: 1) a YAC clone. 2) a genomic clone. 3)a cDNA clone. 4) the protein. 7. A chromatography method that can be used specifically to purify proteins based on their charge is: 1) gel filtration chromatography. 2)ion-exchange chromatography. 3) DNA affinity chromatography. 4) antibody affinity
42、 chromatography. 8. A nonsense mutation is a change in the DNA sequence that results in: 1) a small deletion or insertion. 2) an amino acid change in the protein encoded by the gene. 3)a premature stop codon. 4) all of the above. 5) none of the above. 9. A protein complex involved in degradation of
43、proteins within the cell is known as the: 1)ubiquitin/proteasome system. 2) molecular chaperone. 3) chaperonin. 4) ribosome. 5) Krebs/TCA cycle. 10. _binds to the repressor and turn on the transcription of the structural genes in the Lac operon. 1) cAMP 2) lactose 3)allolactose 4) CRP 11. Which of t
44、he following RNA species is involved in degradation of the mRNA containing complementary sequence 1) miRNA 2)siRNA 3) tRNA 4) 5S RNA 5) U3 snRNA 12. The genome sequencing projects are confirming the theory that genome size is directly proportional to the number of genes contained within that genome.
45、 In other words, a genome that is 10 times as big will contains approximately 10 times as many protein coding genes. 1) True 2) False 13. HeLa cells, derived from a human cervical carcinoma, are able to propagate indefinitely in culture and are therefore known as a(n): 1) tissue culture. 2) tumor. 3
46、) transgenic cell line. 4)immortalized cell line. 14. E. coli cells are smaller than yeast cells. 1) True 2) False 15. Which of the following domains is not a DNA binding domain 1)Proline-rich domains 2) Helix-turn helix domains 3) Zinc finger domains 4) Basic domains 16. The aminoacyl-tRNA syntheta
47、ses distinguish between about 40 different shaped tRNA molecules in the cells. 1) True 2) False PART III: SHORT QUESTIONS (8 points each) 1. How do bacterial replication start and accomplished. Remember to include the proteins/enzymes and important DNA sequence involved in this process. Initiation (3 points): replication of the bacterial chromosome is tightly coupled to the growth cycle. The E. coli origin is within the genetic locus oriC that contains four 9 bp binding sites for the initiator protein DnaA. Synthesis of DnaA is coupled to growth rate so