武汉大学分子生物学_2007期末试卷A

1、武汉大学生命科学学院20072008学年第一学期期末考试分子生物学A 试卷Final exam of Molecular Biology Course (Fall 2007)年级(Grade) _ 专业(Major) _ 姓名 Name _学号(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

2、 be included 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. Th

3、e following molecule is _1) 2-deoxyadenosine 5-phosphate2) 2-deoxyadenosine3) dAMP4) dATP2. The strictness of the rules for “Waston-Crick” pairing derives from the complementarities_ between adenine and thymine and between guanine and cytosine.1) of base stacking2) of shape3) of hydrogen bonding pro

4、perties4) of size3. 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-str

5、anded.4) Some RNAs 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 ei

6、ght histone proteins 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

7、 such as histone acetylase 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 orga

8、nism.3) The average number 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 c

9、ode.3) Codon bias.4) The 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 repair 4) RecBCD pathway repair 5)

10、Mismatch repair8. Which of 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 of ribosomes, and theref

11、ore the translational efficiency. 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 the tRNA bound to the A site.5) The ribosome is a ribozyme because the large rRNA is responsible for the pept

12、idyl transferase activity.9. 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 s

13、equence of a genome, which of 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 poin

14、ts)1. Who is your favorite scientist among those introduced in this course? Please describe his/her research achievement and contribution to the molecular biology knowledge that youve learned and how his/her experience influences your research attitude.2. How the transcription of an mRNA is terminat

15、ed in eukaryotic cells?3. Please 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 sec

16、ondary structure can regulate gene 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 fr

17、ame (ORF) which encodes only 5 amino 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

18、AGTACGATCTGTGCATTATCGTATACCCT 5PART 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,

19、 (4) any change of your learning attitude 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.

20、3. It has been recently reported that 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

21、 binding is essential for the transcriptional repression? Notes: You have all DNA sequences, plasmid vectors, cloning enzymes and other reagents that he needs.武汉大学生命科学学院20072008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 2008)PART I:1. Trombone modelThis model is proposed to

22、 explain the coordinated synthesis of the leading strand and the lagging strand to the direction of the replication folk movement at a replication folk (2)2. TFDA transcription factor composed of TBP and TAFs for RNA polymerase;(1)TBP recognizes TATA box and TBP-DNA complex provides a platform for o

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

24、equence to the 3 end of the 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 accu

25、rate mass of protein fragments 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 (2) through two steps: adenylylation of amino acid and tRNA charging (+1)7. Viral-like retrotransposons: also called long

26、 terminal reapeat (LTR) 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 repressio

27、n that can spread along 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

28、this spreading, thus protect some inserted gene from silencing.(0.5)9. Suppressor Mutation: 抑制突变，抑制基因突变，抑制因子突变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)

29、10. Shine-Dalgarno sequence: 又称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 spacin

30、g decides the activeness 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 polyadenylation:(1)(1) CstF/CPSF bound at the CTD tail is transferred to poly-A signal se

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

32、ol, reduce processivity, leading to spontaneous termination.(0.5)(4) Absense of 5-cap is sensed by the Pol, recognizes the transcript as improper and terminates.(0.5)3.Please describe the similarity and difference between group II intron and spliceosome-mediated pre-mRNA splicing.(答案有点长，大家可以在精简一下。)S

33、pliceosome-mediated splicing of mRNA (6)Chemistry (2): splicing occurs as two sequential ester-transfer reactions. Firstly,the 2OH of the branch point A attacks the phosphoryl group of a conserved 5 G at he5 splice site, resulting in the free of 5 exon from the intron. Then, the 3 OH of thefree 3 en

34、d of 5 exon attacks a phosphoryl group at the 3 splice site, resulting in theligation 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, 3splice site and A branch point by U1 snRNP, U2AF and BBP r

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

36、oins and A complex is arranged to Bcomplex in which the three splice sites are brought together. In this complex U4/U6snRNPs are held together tightly by extensive base-pairing between U4 and U6snRNAs.Step 3, U1 leaves the complex, and U6 occupies the 5 splice site by base-pairing.U4 leaves the comp

37、lex, allowing the RNA components of U2 and U6 to base pair toproduce the active site. The branch site A attacks the 5 splice site, forming the 3-wayjunction and C complex. The 5 splice site then attacks the 3 splice site, freeing theintron lariat and forming the mRNA product.Group 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 own

39、 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 repressors; (1.5) Activators act using both recruitment and allostery. (1.5) Both regulations are controlled at diff

40、erent 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)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 “stop” on the region 1&

42、2 when region 3&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

43、 formation of termination hairpin. Thus the operon could be transcribed and then translated.Example 2 Riboswitches(1) riboswitches are regulatory RNA elements that sensors the small metabolites to control gene transcription & translation.Two types: (1) when it binds no metabolites, the anti-term

44、inator is formed in the 5-UTR of mRNA thereby the transcription is turned on(2) when it binds to certain metabolites, a terminator forms and the transcription is turned off.(1) when is binds no metabolites, the RBS is exposed to ribosome so that the translation is accomplished.(2) When it binds cert

45、ain metabolites, the RBS is sequestered 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. recruiting nucleosome

46、 modifiers (1)Repressors: (Competition) (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

47、the the promoter and inhibiting transcription.d. Indirect repression: recruiting nucleosome modifiers (most common) (1)7Five codons: AUG, CUA, UUA, CGU, GUC stop codon is UAG upper strand; right sidePART IV:1. omitted2. Please discuss the similarity and difference between miRNA and siRNA, and descri

48、be the distinct contribution of these 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 (1)2. Biogenesis: Both are cleaved from precursor by Dicer (1)3. Function: one

49、 strand is incorporated into RISC complex 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 from pri-miRNA to pre-miRNA to mature miRNA

50、, pre-miRNA is stem-loop RNA, but siRNA is directly cleaved from long dsRNA (1)Contribution (4)MiRNA: miRNA is phylogenetically conserved regulatory RNAs playing diverse and critical functions in cell proliferation and differentiation, and development (2)SiRNA: siRNA has significantly contributed to silencing gene function (1) and representing a therapeutic approach to combat disease and viral infection. (1)3. (1) To test if X binds to th