武漢大學分子生物學_2007期末試卷A

1、武漢大學生命科學學院2007- 2021學年第一學期期末測試?分子生物學?A試卷Final exam of Molecular Biology Course (Fall 2007)年級(Grade)專業(yè)(Major) 姓名 Name學號(Student ID)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 could be i

2、ncluded in your answer, as necessary.1. Trombone model2. TFIID3. Telomerase4. RNA editing5. Tandem mass spectrometry6. tRNA synthetase7. Viral-like retrotransposons8. Transcriptional silencing9. Suppression mutation10. Shine-Dalgarno sequencePART II: MULTIPLE/SINGLE CHOICES (2 points each)1. The fol

3、lowing molecule is1) 2-deoxyadenosine 5-phosphate2) 2-deoxyadenosine3) dAMP4) dATPOII2. The strictness ofthe rules for “Wasckn pairing derives from the complementarities between adenine and thymine and between guanine andcytosine.1) of base stacking2) of shape3) of hydrogen bonding properties4) of s

4、ize3. Which of the following statements correctly describe the difference between DNA and RNA?1) The major groove of the regular DNA double helical structure is rich in chemical information.2) RNA contains deoxyribose and uracil.3) All RNAs are single-stranded while DNA is double-stranded.4) Some RN

5、As can fold up into complex tertiary structures, and function as enzymes.4. Nucleosomes are the building blocks of chromosomes, which of the following statements are CORRECT in describing the structure and function of nucleosome?1) The nucleosome is composed of a core containing eight histone protei

6、ns and the DNA (147 bp) wrapped around them.2) The core histones specifically contact the major groove of the DNA.3) The interaction of DNA with the histone octamer is very stable and cannot be altered unless during DNA replication.4) Nucleosome remodeling by the action of enzymes such as histone ac

7、etylase is very important for gene expression.5. Which of the following statements regarding genomes of living organism is NOT correct?1) Genome size is roughly related to the complexity of the organism.2) The number of genes in a genome cannot explain the complexity of the organism.3) The average n

8、umber of introns per gene increases with the organism complexity.4) The gene density (genes/Mb) increases with the organism complexity.6. 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

9、 anticodon hypothesis.5) The redundancy of the genetic code.7. Which of the following repair mechanisms is involved in repair of the damaged DNA with a double-stranded break?1) Base excision repair2) Nucleotide excision repair3) Translesion repair4) RecBCD pathway repair5) Mismatch repair8. Which of

10、 the following factors/elements regarding translation are CORRECT?1) Ribosome binding site (RBS) is essential for the translation initiation in bacterial and eukaryotic cells.2) The polyA tail in the 3 end of an mRNA promotes the efficient recycling ofribosomes, and therefore the translational effic

11、iency.3) Ribosome is able to discriminate between correctly or incorrectly charged tRNAs.4) The translocation factor EF-G mimics a tRNA molecule so as to displace thetRNA bound to the A site.5) The ribosome is a ribozyme because the large rRNA is responsible for the peptidyl transferase activity.9.

12、RNA polymerase III 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.10 To obtain the sequence of a genome, which of

13、the following steps are required? _1) Obtain a genomic library2) Obtain a cDNA library3) Shotgun sequencing on automated sequencers4) Sequence assembly on computers5) BLAST searchPART III： SHORT QUESTIONS (CHOOSE SIX_QUESTIONS TO ANSWER)(5 points each, total of 30 points)1. Who is your favorite scie

14、ntist among those introduced in this course? Please describe his/her research achievement and contribution to the molecular biology knowledge that you ve learned and how his/her experierfilciences your research attitude.2. How the transcription of an mRNA is terminated in eukaryotic cells?3. Please

15、describe the similarity and difference between group II intron and spliceosome-mediated pre-mRNA splicing.4. What are the general principles of transcription regulation in both prokaryotic and eukaryotic cells?5. Please give an example to demonstrate that the RNA secondary structure can regulate gen

16、e expression in bacteria.Example 1: The attenuation regulation of the tryptophan operon.Example 2: Ribo-switch regulation6. How the activators and repressors regulate gene expression in eukaryotic cells?7. The following DNA sequence contains a small open reading frame (ORF) which encodes only 5 amin

17、o acids. Please list the 5 genetic codons and the stop codon of the ORF. Which strand of the DNA (upper or lower strand) is the template for RNA transcription? The promoter of the gene is in the right or left side of the sequence?5 TCATGCTAGACACGTAATAGCATATGGGA 33 AGTACGATCTGTGCATTATCGTATACCCT 七PART

18、 IV: MAJOR QUESTIONS (10 points each, total of 30 points)1. Please discuss what you have learned from this course, including (1) the general knowledge framework, (2) your most interested knowledge and why this knowledge has impressed you, (3) the value of teamwork, (4) any change of your learning at

19、titude and/or the construction of your interest in science.2. Please discuss the similarity and difference between miRNA and siRNA, and describe the distinct contributions of these two small regulatory RNAs to the fundamental biology and application, respectively.3. It has been recently reported tha

20、t a new protein X functions in repressing the transcription of an oncogene gene Y. Could you design experiments to test if X protein binds to the promoter region of Y gene (1) in vitro and (2) in vivo? If it does bind, could you design an experiment to test if the binding is essential for the transc

21、riptional repression?Notes: You have all DNA sequences, plasmid vectors, cloning enzymes and other reagents that he needs.武漢大學生命科學學院2007- 2021學年第一學期期末測試 ?分子生物學?試卷及參考答案Final exam of Molecular Biology Course (Spring 2021)PART I:1. Trombone modelThis model is proposed to explain the coordinated synthes

22、is of the leading strand and the lagging strand to the direction of the repl ication folk movement at a replication folk (2)2. TFn DA transcription factor composed of TBP and TAFs for RNA polymerase n ;(1 )TBP recognizes TATA box and TBP-DNA complex provides a platform for other transcriptionfactors

23、 and polymerase to the promoter;(1 )Two of TAFs bind the core promoter elements such as Inr and DPE. Several of histone-like TAFsare also associated with some histone modification enzymes.(1)3. Telomerase:Solve the End Replication Problem (1 ) through adding the telomeric sequence tothe 3 end of the

24、 telomere. (1 ) No extra primer nor template needed.4. RNA editing: a way of changing the sequence of RNA after transcription(1 ) by site-specific deamination of insertion.(1 )5. Tandem mass spectrometryAnswer 1: A method that determines the protein sequence based on the accurate mass of protein fra

25、gments obtained by mass spectrometry (2 )Answer 2:(2 )6. Aminoacyl tRNA synthetaseAn enzyme that catalyzes the attachment of an amino acid to a cognate tRNA (2steps: ) through twoadenylylation of amino acid and tRNA charging (+1)7. Viral-like retrotransposons: also called long terminal reapeat (LTR)

26、 retrotransposons. The element includes two long terminal repeat sequences that flank a region encoding two enzymes: integrase and reverse transcriptase. It mediate transposition reaction through a RNA intermediate.8. Transcriptional silencing is a specialized form of repression that can spread alon

27、g chromatin,(1 switching off multiple genes without the need for each to bear binding sites for specific repressors(1 ). The mechanism of this repression is the propagation of certain repressing histone modifications over stretches of chromatin. (1)Insolator elements can block this spreading, thus p

28、rotect some inserted gene from si lencing.(0.5 9. Suppressor Mutation:抑制突變,抑制基因突變,抑制因子突變 Key points: (1) a second mutation(2) the GENOTYPE is mutationally altered but the wild type PHENOTYPE restores(1) maybe on a different gene (intergenetic) or on the same gene (intragenetic)10. Shine-Dalgarno seq

29、uence: 又稱 RBS, ribosome binding siteKey points: (1) in prokaryotic cells(2) a stretch of RNA 3 to 9 nucleosides upstream of the start codon in a mRNA(3) contains conservative sequence: 5 -AGGAGG- 3 , which is complimented to certain region of 16s rRNA(2) the conservation and spacing decides the acti

30、veness of the following OFRPart III: Short questions2 Each mRNA gene contains a poly-A signal sequence near the termination site(1 EUkaryotic transcription termination is highly coupled with polyad enylation:(1 )(1) CstF/CPSF bound at the CTD tail is transferred to poly-A signal sequence after ti is

31、transcripted, resulting in mRNA cleavage and recruitment of enzymes for polyadenylation .(2)(2) Poly-A polynerase(PAP) adds about 200 As to RNA s 3 end.(2 )Two models for polymerase recycle:(3) Transfer of 3 -processing enzymes from CTD tail to RNA triggers comformational change inPol, reduce proces

32、sivity, leading to spontaneous termination.(0.5(4) Absense of 5-cap is sensed by the Pol, recognizes the transcript as improper andterminates.(0.53 .Please describe the similarity and difference between group II intron and spliceosome-mediatedpre-mRNA splicing.(答案有點長,大家可以在精簡一下.)Spliceosome-mediated

33、splicing of mRNA (6 )Chemistry (2 ): splicing occurs as two sequential ester-transfer reactions. Firstly, the 2 OH of the branch point A attacks the phosphoryl group of a conserved 5 G at he 5 splice site, resulting in the free of 5 exon from the intron. Then, the 3 OH of the free 3 end of 5 exon at

34、tacks a phosphoryl group at the 3 splice site, resulting in the ligation of the 5 and 3 exons and release of a lariat form of intron.Mechanisms (4 ):Step 1, formation of the E (early) complex by recognition of the 5 splice site, 3 splice site and A branch point by U1 snRNP, U2AF and BBP respectively

35、.Step 2, U2 snRNP bind to the branch site to replace BBP forms A complex. The base-pairing between U2 snRNA with the branch site make the conserved A residue in the branch site extruded from the paired region, and thus this A is ready to carry the nucleophile attack. The tri-snRNP U4/U6/U5 joins and

36、 A complex is arranged to B complex in which the three splice sites are brought together. In this complex U4/U6 snRNPs are held together tightly by extensive base-pairing between U4 and U6 snRNAs.Step 3, U1 leaves the complex, and U6 occupies the 5 splice site by base-pairing.U4 leaves the complex,

37、allowing the RNA components of U2 and U6 to base pair to produce the active site. The branch site A attacks the 5 splice site, forming the 3-way junction and C complex. The 5 splice site then attacks the 3 splice site, freeing the intron lariat and forming the mRNA product.Group n intron splicing (2

38、 ): the chemistry and the RNA intermediates produced are the same as that of the spliceosome-mediated mRNA splicing, but the splicing is catalyzed by the intronRNA itself, which is also named as self-splicing. The intron itself folds into a specific conformation and catalyzed the chemistry of its ow

39、n release.4 .What are the general principles of transcription of regulation in both prokaryotic and eukaryotic cells?Both regulations are mediated by regulatory proteins: activators and repr essors; (1.5)Activators act using both recruitment and allostery. (1.5)Both regulations are controlled at dif

40、ferent stages: the initiation is the pervasively regulated step, also involves regulation after initiation (2)Both regulations act at a distance and involve DNA looping. (+0.5)Both regulations involve cooperative binding. (+0.5).Example 1.Attenuation of trp operon(1) the 4 regions of the leader RNA

41、could form 3 kinds of possible hairpin loop: region 1 with region 2,region 2 with region 3, region 3 with region 4, ( region 1: two trp codons)(2) transcription & translation in prokaryotic cells are coupled.(3) When the level of tryptophan is moderate, the ribosome “stop6n the region 1&2 when regio

42、n3&4 are transcribed. So, region 3 pair with region 4 to form a hairpin resulting in the termination of the transcription.(4) When the level of tryptophan is low enough, the ribosome stalls at the two adjacent trp codens, leaving region 2 pairs with region 3 , thereby preventing the formation of ter

43、mination hairpin. Thusthe operon could be transcribed and then translated.Riboswitches(1) riboswitches are regulatory RNA elements that sensors the small metabolites to control gene transcription & translation.Two types:I(2) when it binds no metabolites, the anti- terminator is formed in the 5 -UTR

44、of mRNA thereby the transcription is turned on(3) when it binds to certain metabolites, a terminator forms and the transcription is turned off.n(1) when is binds no metabolites, the RBS is exposed to ribosome so that the translation is accomplished.(2) When it binds certain metabolites, the RBS is s

45、equestered by the resulting secondary structure, preventing the translation of the mRNA .6 How the activators and repressors regulate gene expression in eukaryotic cells? (5 ) Activators: a. recruiting the transcriptional machinery to the gene(1)b.recruitingnucleosomemodifiers(1)Repressors: (Competi

46、tion) (Inhibition) (Direct repression) (Indirect repression)a. Competition:competing for the binding site of activators(1)b. Inhibition:inhibiting the function of the activator.(1)c. Direct repression: interacting directly with the transcriptional machinery at the the promoter and inhibiting transcr

47、iption.d. Indirect repression: recruiting nucleosome modifiers (most common)(1)7. Five codons: AUG , CUA, UUA, CGU, GUC stop codon is UAGupper strand; right sidePART IV:1. omitted2. Please discuss the similarity and difference between miRNA and siRNA, and describe the distinct contribution of these

48、two small regulatory RNAs to the fundamental biology and application, respectively.Answer:Similarity: (4)1. Structural: Both are about 20- 23 nt consist of sense and antisense strand (12. Biogenesis: Both are cleaved from precursor by Dicer (1)3. Function: one strand is incorporated into RISC comple

49、x and complement with target mRNA (1 )4. Result: Both could cause gene silencing. (1)Difference (2):Biogenesis: miRNA is endogenously and encoded and siRNA is exogenously and artificial prepared; (1 ) miRNA is produced fromrphRNA to pre-miRNA to mature miRNA, pre-miRNAis stem-loop RNA, but siRNA is

50、directly cleaved from long dsRNA (1)Contribution (4MiRNA: miRNA is phylogenetically conserved regulatory RNAs playing diverse and critical functions in cell proliferation and di fferentiation, and development (2SiRNA: siRNA has significantly contributed to silencing gene function (1 and representing a therapeutic approach to combat disease and viral infection. (13. (1) To test if X binds to th