武汉大学生命科学学院：《分子生物学》2003-2004学年第一学期期末考试试卷及参考答案

武汉大学生命科学学院 2003-2004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査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 chromosome: ①酵母人工染色体_(05 point.) 2 contains components required for replication and segregation of the natural yeast chromosome, includ ing two telomeric sequences(TEL), one centromere(CEN and one autonomously replicating sequences(ARs)(0.75 point contains genes act as selective markers in yeast and proper restriction sites(0.75 4 can accommodate genomic DNA fragments of more than 1 Mb(0.5 point) 2. RNA interference ①RNA干扰_(O5 point) 2 a conserved biological response to double-stranded rNa(I point) 3 regulates the expression of protein-cod ing genes through siRNA or miRNA( DoIn a the dsRNa is restricted by DICER, then RISC med iates the siRNa and miRNA related rna degradation and translation inhibition, respectively. (I point) 3. Proteomics ①蛋白组学(O5 point) The study of the proteome using techniques of high resolution protein separation and identification (1. 5 point) 4 The best separation 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 resulted 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 ①SD序列(O5poin) 2 A conserved sequence 8-13 nt upstream of the first codon to be translated(1 This sequence was discovered by Shine and Dalgarno(0.5 point) 4 The sequence is purine-rich and contains all or part of5'-AGGAGGU-3'(0.5

武汉大学生命科学学院 2003－2004 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 chromosome: ① 酵母人工染色体（0.5 point） ② 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 ) ③ contains genes act as selective markers in yeast and proper restriction sites (0.75 point ) ④ can accommodate genomic DNA fragments of more than 1 Mb (0.5 point) 2. RNA interference: ① RNA 干扰 （0.5 point） ② a conserved biological response to double-stranded RNA (1 point) ③ regulates the expression of protein-coding genes through siRNA or miRNA (1 point) ④ the dsRNA is restricted by DICER, then RISC mediatesthe siRNA and miRNA related RNA degradation and translation inhibition, respectively. (1 point) 3. Proteomics ① 蛋白组学 （0.5 point）The study of the proteome using techniques of high resolution protein separation and identification (1.5 point) ④ The best separation 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 resulted 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 ① SD 序列 （0.5 point） ② A conserved sequence 8-13 nt upstream of the first codon to be translated (1 point). ③ This sequence was discovered by Shine and Dalgarno (0.5 point) ④ The sequence is purine-rich and contains all or part of 5’-AGGAGGU-3’ (0.5

point) 6 Can base-pair with the 3-end of the 16S rRNA (0.5 point) 5. Alternative splicing ①可变剪接_(05 point.) 2 The generation of different mature mRNAs from a particular type of gene transcript by choosing different 5-and 3-splice sites(2 point) 6. Ribozyme ①核酶_(O5 point.) 2 an RNa molecule capable of catalyzing a chemical reaction(2 point) 7. p-dependent termination ①p-依赖型转录终止_(0.5 point.) 2 During bacterial transcription, some terminator sites do not form strong hairpins thus termination of the transcription by bacterial rNa polymerase requires the assistance of an accessory factor called rho(p) protein(2 point) 8. RNA editing ①RNA编辑(O5mmnt 2 A form of RNa processing in which nucleotide sequence of the primary transcript is altered by either changing, inserting or deleting residues at the spec ific points along the molecule(2 point) 9. DNA lesions ①DNA损伤_(O5p0in) 2 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 ①蛋白定位(O5mmnt 2 Synthesis of eukaryotic proteins is usually occurred in cytoplasm(0.5 point) 3 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) 4 The ultimate cellular location of proteins is often determined 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(I point) PART IL: MULTIPLE CHOICES (I points 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 small ribosomal subunit 5) proteins of the large ribosomal subunit 2. Bid irectional and semi-conservative are two terms that refer to transcripti

point) ⑤ Can base-pair with the 3’-end of the 16S rRNA (0.5 point) 5. Alternative splicing ① 可变剪接 （0.5 point） ② The generation of different mature mRNAs from a particular type of gene transcript by choosing different 5’- and 3’-splice sites (2 point). 6. Ribozyme ① 核酶 （0.5 point） ② an RNA molecule capable of catalyzing a chemical reaction (2 point). 7. -dependent termination ① -依赖型转录终止 （0.5 point） ② During bacterial transcription, some terminator sites do not form strong hairpins, thus termination of the transcription by bacterial RNA polymerase requires the assistance of an accessory factor called rho () protein (2 point). 8. RNA editing ① RNA 编辑 （0.5 point） ② A form of RNA processing in which nucleotide sequence of the primary transcript is altered by either changing, inserting or deleting residues at the specific points along the molecule (2 point). 9. DNA lesions ① DNA 损伤 （0.5 point） ② An alteration to the normal chemical or physical structure of the DNA (2 point). ③ The lesions can lead to cell death or DNA mutation (1 point). 10. Protein targeting ① 蛋白定位 （0.5 point） ② Synthesis of eukaryotic proteins is usually occurred in cytoplasm (0.5 point). ③ 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). ④ The ultimate cellular location of proteins is often determined 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. 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 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 known as 2) the universality of the genetic code 3)codon bias 4) the anticodon hypothesis of the genetic code. 4. RNa polymerase I is the eukaryotic enzyme responsible for aI rna 2) transcription of transfer RNa and other 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 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 chromatographv 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) premature stop codon. 4) all of the above 5) none of the above 9. A protein complex involved in degradation of proteins within the cell is known as the 1) ubiquitin/proteasome svstem 2) molecular chaperone 3) cha 5) Krebs/TCA cyo 0. binds to the repressor and turn on the transcription of the structural genes in the Lac operon 1)CAMP

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 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 other 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 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 chromatography. 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 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 Lac operon. 1) cAMP

2)actos 3)allolactose 4) CRP 11. Which of the following RNA species is involved in degradation of the mRNA containing complementary sequence 1)miRNA 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. In other words, a genome that is 10 times as big will contains approximately 10 times as many protein cod ing genes D)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)tu 3) transgenic cell line 4) immortalized cell line 14. E. coli cells are smaller than yeast cell 15. Which of the following domains is not a DNA binding domain 2)Helix-turn helix domains 3)Zinc finger domains 4)Basic domains 16. The aminoacyl-tRNA synthetases distinguish between about 40 different shaped trna molecules in the cells 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): O replication of the bacterial chromosome is tightly coupled to the growth cycle 2 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 that the replication of the bacterial chromosome is coupled to the growth cycle 4 Once the cellular level of DnaA reaches a critical level, DnaA protein forms a complex of 30-40 molecules at the oriC DNA, which facilitates melting of three 13 bp AT-rich repeat sequences to allow binding of DnaB protein

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 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 culture 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) 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/enzymes and important DNA sequence involved in this process. Initiation (3 points): ① replication of the bacterial chromosome is tightly coupled to the growth cycle. ② The E. coli origin is within the genetic locus oriC that contains four 9 bp binding sites for the initiator protein DnaA. ③ Synthesis of DnaA is coupled to growth rate so that the replication of the bacterial chromosome is coupled to the growth cycle. ④ Once the cellular level of DnaA reaches a critical level, DnaA protein forms a complex of 30-40 molecules at the oriC DNA, which facilitates melting of three 13 bp AT-rich repeat sequences to allow binding of DnaB protein

6 Dnab protein is a dn a helicase that utilizes the energy of a tP hydrolysis te melt dsDNA 6 The ssDNA created by DnaB is coated with single-stranded bind ing protein(Ssb) to protect it from breakage and to prevent the dna renat urins O The DNa primase then attaches to the dna and synthesizes a short Rna primer to initiate synthesis of the lead ing strand of the first replication fork Unwinding(1 point) O For replication to proceed away from the origin, DNA helicases must travel along the template strands to open the double helix for copying, which is accomplished by the joint efforts of DnaB and Ssb 2 Unwind ing causes positive supercoiling of the unwound DNA; the positive supercoiling is relaxed continuously by the introduction of further negative supercoils by type ll topoisomerase called DNa gyrase Elongation 3 points) O Initiation of the replication of the lagging strand. As the newly formed replication fork displaces the parental lagging strand, a mobile complex called a primosome which includes the DnaB helicase and DNA primase, synthesizes RNa primer every 1000-2000 nt on the lagging strand 2 Both leading and lagging strand primers are elongated by dna polyi merase IlI holoenzyme. This multisubunit complex is a dimmer, one half synthesizing the lead ing strand and the other the lagging strand Lagging strand synthesis. dNA polymerase III holoenzyme: a subunit-the actual polymerase, s-a 3-5 proofreading exonuclease. DNa polymerase I removes the lagging strand primers and fills in the resulted gaps. DNA ligase makes the final phospho ester bond between fragments Termination and segregation(I points) O Two replication forks meet at the terminator sites(terminus) 2 tus gene product binds to the terminus and acts as an inhibitor of DnaB helicase 3 When replication is completed, the two interlinked daughter circles are unlinked by topoisomerase IV 2. Design experiments to clone a yeast gene and express this gene in yeast Clone (4 points) O PCR amplification of the gene of interest, including two proper restriction sites at the ends of the PCR amplified dna 2 Choose a cloning vector: it could be a regular cloning vector that is capable of propagating in E. coli, easy to extract from E coli and manipulated in test tubes It could also be a shuttle and expression vector that also contains the yeast 2, origin to allow the plasmid replication in yeast, the selective marker to allow the plasmid containing cells grow, the promoter to allow the gene of interest to be

⑤ DnaB protein is a DNA helicase that utilizes the energy of ATP hydrolysis to melt dsDNA. ⑥ The ssDNA created by DnaB is coated with single-stranded binding protein (Ssb) to protect it from breakage and to prevent the DNA renaturing. ⑦ The DNA primase then attaches to the DNA and synthesizes a short RNA primer to initiate synthesis of the leading strand of the first replication fork. Unwinding (1 point) ① For replication to proceed away from the origin, DNA helicases must travel along the template strands to open the double helix for copying, which is accomplished by the joint efforts of DnaB and Ssb. ② Unwinding causes positive supercoiling of the unwound DNA; the positive supercoiling is relaxed continuously by the introduction of further negative supercoils by type II topoisomerase called DNA gyrase. Elongation (3 points) ① Initiation of the replication of the lagging strand. As the newly formed replication fork displaces the parental lagging strand, a mobile complex called a primosome, which includes the DnaB helicase and DNA primase, synthesizes RNA primers every 1000-2000 nt on the lagging strand. ② Both leading and lagging strand primers are elongated by DNA polymerase III holoenzyme. This multisubunit complex is a dimmer, one half synthesizing the leading strand and the other the lagging strand. ③ Lagging strand synthesis. DNA polymerase III holoenzyme:  subunit -the actual polymerase, -a 3’→5’ proofreading exonuclease. DNA polymerase I removes the lagging strand primers and fills in the resulted gaps. DNA ligase makes the final phosphodiester bond between fragments. Termination and segregation (1 points) ① Two replication forks meet at the terminator sites (terminus) ② tus gene product binds to the terminus and acts as an inhibitor of DnaB helicase. ③ When replication is completed, the two interlinked daughter circles are unlinked by topoisomerase IV. 2. Design experiments to clone a yeast gene and express this gene in yeast. Clone (4 points): ① PCR amplification of the gene of interest, including two proper restriction sites at the ends of the PCR amplified DNA ② Choose a cloning vector: it could be a regular cloning vector that is capable of propagating in E. coli, easy to extract from E. coli and manipulated in test tubes. It could also be a shuttle and expression vector that also contains the yeast 2 origin to allow the plasmid replication in yeast, the selective marker to allow the plasmid containing cells grow, the promoter to allow the gene of interest to be