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
transcribed a poly(A) site for add ition of poly (a)tail on the transcript No matter what type of the vector is chosen, plasmid shall be prepared from the e coli cells 8 Restriction digestion of both plasmid and the gene of interest with the same restriction sites Ligation and transformation 6 Selection and identification of the transformants to obtain the clone containing the right recombinant plasmid Expression(4 points) 8 If a shuttle and expression vector is chosen as described in step2, the expression process will include preparation of the recombinant DNA from E. coli cells, transformation of yeast cells with the recombinant DNA, selection of the transformants, growth of the transformants, expression of the gene by induction of the transcription from the promoter and detection of the expressed RNA and/or protein. Induction is not required if a constitutive promoter is used. (4 points) O If a regular cloning vector is chosen as described in step @2, a subclone need to be obtained to allow the gene of interest being cloned into a shuttle and expression vector same as described in step@2. Then expression is performed the same as described in⑥ 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 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 Pstl to the clone X gene, please add these restriction reading frame of the X gene with the Lacz gerte s sites on the corresponding position of the X gene. remember to maintain the (1) MCS of pUC18 EcoRI Sall ACGAAT TC TCG GTA CCC GGG GAT CCTCTA GAG TCG ACCTGC AGG CAT GCA Thr Asn Ser Ser Ser 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..GAUAGUACAGCU GCCAAGTAA … Asp Ser Thr AlaAla Lys ANSWER: GGAT CCN ATG ACC CCU CAU AAC GGC GAC(N-terminus GAU AGUACA GCU GCC AAG TAACIGCAG(C-terminus PART IV: MAJOR QUESTIONS (20 points each) 1: Please describe how an mRNA gene is transcribed, processed and translated in human
transcribed, a poly(A) site for addition of poly(A) tail on the transcript. No matter what type of the vector is chosen, plasmid shall be prepared from the E. coli cells. ③ Restriction digestion of both plasmid and the gene of interest with the same restriction sites. ④ Ligation and transformation ⑤ Selection and identification of the transformants to obtain the clone containing the right recombinant plasmid. Expression (4 points): ⑥ If a shuttle and expression vector is chosen as described in step②, the expression process will include 1preparation of the recombinant DNA from E. coli cells, 2 transformation of yeast cells with the recombinant DNA, 3 selection of the transformants, 4growth of the transformants, 5expression of the gene by induction of the transcription from the promoter and 6detection of the expressed RNA and/or protein. Induction is not required if a constitutive promoter is used. (4 points) ⑦ If a regular cloning vector is chosen as described in step②, a subclone need to be obtained to allow the gene of interest being cloned into a shuttle and expression vector same as described in step②. Then expression is performed the same as described in ⑥. 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 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, please add these restriction sites on the corresponding position of the X gene. Remember to maintain the reading frame of the X gene with the LacZ gene (1) MCS of pUC18 EcoRI SacI KpnI SmaI BamHI XbaI SalI PstI ACGAAT TCGAGC TCG GTA CCC GGG GAT CCT CTA GAG TCG ACC TGC AGG CAT GCA Thr Asn Ser Ser Ser 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 ANSWER: G GAT CC N ATG ACC CCU CAU AAC GGC GAC (N-terminus) GAU AGU ACA GCU GCC AAG TAACTGC AG (C-terminus) PART IV: MAJOR QUESTIONS (20 points each) 1: Please describe how an mRNA gene is transcribed, processed and translated in human Lac Z
cells. What are the possible mechanisms in regulating the expression of this gene? O Transcription: focusing on transcription initiation described on p 222 of the text book. (5 points) 2 Processing: 5,capping(addition of a m'G to the 5'end of the rna pol II transcript catalyzed by mRNA guanyltransferase/capping enzyme )(1.5 point), intron splicing(removal of intron by splicesome that contains 5 snRNAs and many proteins. First step-cleavage at 5 splice site and the conserved A being linked to the 5'-end of the intron, second step-3'cleavage and exon ligation) (2 points), 3'cleavage and polyadenylation(First-assembly of the recognition complex on the polyadenylation site, second-poly (A) polymerase/PAP cleaves the transcript at this site, third -PAP adds up to 250 A residues to the 3'end of the transcript)(1.5 point) Translation: Initiation, elongation and termination(p276, 278)(5 points) 4 Regulation: Firstly, regulation can occur at the transcription level, for example, through interaction of the transcription factors with the promoter or URE to activate or repress the transcription(2.5 point ) Secondly, regulation can occur at the post-transcription level, for example, alternative transcription or alternative poly (A) site(2.5 points) 2 (20 points): A bacterium is found to metabolize a rare sugar produced by a plant that the is, if you want to finish this course with a satisfied score, you must figure out the roblem bacteria grow on. However, the bacteria prefer glucose as the energy source. The regulatory mechanism that the bacteria used to determine the sugar choice use northern blot to analyze the expression of fun3 and use dN A footprinting to analye The gene involving in the rare sugar metabolism has been identified as fun3. You c the binding of proteins to the control elements of fiun3 gene. The following table shows the experimental results Glucose Rare sugar Levels of Binding of proteins presence presence fun3 RNA Protein a binds to the promoter region Protein 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 on the question"How does the expression of fin3 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) Protein A-repressor, protein B-RNA polymerase, protein C-CRP/CAP
cells. What are the possible mechanisms in regulating the expression of this gene? ① Transcription: focusing on transcription initiation described on p222 of the text book. (5 points) ② Processing: 5’ capping (addition of a m7G to the 5’ end of the RNA pol II transcript catalyzed by mRNA guanyltransferase/capping enzyme ) (1.5 point), intron splicing (removal of intron by splicesome that contains 5 snRNAs and many proteins. First step- cleavage at 5’ splice site and the conserved A being linked to the 5’-end of the intron, second step-3’cleavage and exon ligation) (2 points), 3’cleavage and polyadenylation (First-assembly of the recognition complex on the polyadenylation site, second-poly(A) polymerase/PAP cleaves the transcript at this site, third- PAP adds up to 250 A residues to the 3’end of the transcript) (1.5 point). ③ Translation: Initiation, elongation and termination (p276, 278) (5 points) ④ Regulation: Firstly, regulation can occur at the transcription level, for example, through interaction of the transcription factors with the promoter or URE to activate or repress the transcription (2.5 point). Secondly, regulation can occur at the post-transcription level, for example, alternative transcription or alternative poly(A) site (2.5 points). 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 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 analyze 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 fun3 RNA Binding of proteins - - - Protein A binds to the promoter region Protein 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 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) Protein A-repressor, protein B-RNA polymerase, protein C-CRP/CAP
Regulatory mechanism O In the absence of rare sugar, the repressor protein binds to the promoter region preventing the binding of RNA polymerase, transcription of the fun3 gene is not possible. Therefore, no fun3 RNA is observed 2 In the presence of rare sugar, repressor protein cannot bind to the promter, allowing the rna polymerase to bind and initiate transcription. However, the promoter of the fun gene may be a weak promoter, transcription is not efficient in the absence of activator. The activator of the fun3 gene transcription is regulated by glucose level. The absence of glucose activates the activator protein and allows the protein binding upstream of the promoter to activate the transcription of fun3 2. How is protein A regulated? (2 points) (1) glucose turns the repressor on (2)glucose 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 activator off 4. How could you make the bacteria always use the rare sugar as the energy source even In ll the consensus sequence. Thus, fiun3 gene is ahvays highly expresse mnre Ba The simplest way: Change the promoter offun3 gene to a strong promoter contain sugar exists (其他的答案也可以酌情给分
Regulatory mechanism: ① In the absence of rare sugar, the repressor protein binds to the promoter region preventing the binding of RNA polymerase, transcription of the fun3 gene is not possible. Therefore, no fun3 RNA is observed. ② In the presence of rare sugar, repressor protein cannot bind to the promter, allowing the RNA polymerase to bind and initiate transcription. However, the promoter of the fun3 gene may be a weak promoter, transcription is not efficient in the absence of activator. The activator of the fun3 gene transcription is regulated by glucose level. The absence of glucose activates the activator protein and allows the protein binding upstream of the promoter to activate the transcription of fun3. 2. How is protein A regulated? (2 points) (1) glucose turns the repressor on (2) glucose 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 activator 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) The simplest way: Change the promoter of fun3 gene to a strong promoter containing all the consensus sequence. Thus, fun3 gene is always highly expressed when rare sugar exists. (其他的答案也可以酌情给分)
