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國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 1 頁,共 14 頁 本試卷共有九個摘要,48 題試題。根據試題之前的摘要內容回答試題。每一試題 2 分 (41-48 題除外,每題 2.5 分)。 一、 1. The purpose of the RT-PCR method is to (A) extract viruses from urban sludge (B) remove PCR inhibitors (C) quantify enterovirus RNA levels (D) quantify poliovirus RNA levels -1- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 2 頁,共 14 頁 2. The amplified PCR products were detected by (A) fluorogenic probes (B) RNeasy mini kit (C) PVP (D) Ribogreen method 3. The Mahoney type I poliovirus RNA served as (A) control samples (B) amplification templates (C) standard samples (D) amplification inhibitors 4. Which of the following is not mentioned as a desired characteristic of the RT-PCR method? (A) high sensitivity (B) high through-put (C) low cost (D) high reproducibility 5. Which journal this article would be most likely to appear in? (A) Journal of Biological Chemistry (B) Trends in Neurosciences (C) Biotechniques (D) Ecology 二、 -2- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 3 頁,共 14 頁 6. Which of the following is an important characteristic for a circadian oscillator (A) unusual neuronal activity (B) spontaneous cycle of neuronal activity (C) being controlled by SCN (D) being connected to other parts of the brains 7. SCN (A) is the only circadian oscillator in mammals (B) is an endocrine tissue (C) dictates our physiological and behavioral rhythms (D) is a universal circadian oscillator for all animals 8. Which of the following statements is true (A) a circadian oscillator is independent of the external world (B) a circadian oscillator is synchronized to the external world (C) the number of SCN neurons is far more than 10000 (D) the molecular mechanism of timekeeping is mostly unraveled 9. Our physiological and behavioral rhythms are (A) unchangable (B) controlled by circadian oscillators (C) independent of any circadian oscillators (D) not synchronized to the external world 10. Which journal this article would be most likely to appear in? (A) Journal of Biological Chemistry (B) Trends in Neurosciences (C) Biotechniques (D) Ecology 三、 Glyceraldehyde-3-phosphate dehydrogenase (GAPDH) and triosephosphate isomerase (TPI) are essential to glycolysis, the major route of carbohydrate breakdown in eukaryotes. In animals and other heterotrophic eukaryotes, both enzymes are localized in the cytosol; in photosynthetic eukaryotes, GAPDH and TPI exist as isoenzymes that function in the glycolytic pathway of the cytosol and in the -3- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 4 頁,共 14 頁 Calvin cycle of chloroplasts. Recent studies showed that diatoms—photosynthetic protists that acquired their plastids through secondary symbiotic engulfment of a eukaryotic rhodophyte—possess an additional isoenzyme each of both GAPDH and TPI. Surprisingly, these new forms are expressed as a TPI-GAPDH fusion protein which is imported into mitochondria prior to its assembly into a tetrameric bifunctional enzyme complex. Homologs of this translational fusion are shown to be conserved and expressed also in nonphotosynthetic, heterokont-flagellated oomycetes. Phylogenetic analyses show that mitochondrial GAPDH and its N-terminal TPI fusion branch deeply within their respective eukaryotic protein phylogenies, suggesting that diatom mitochondria may have retained an ancestral state of glycolytic compartmentation that existed at the onset of mitochondrial symbiosis. These findings strongly support the view that nuclear genes for enzymes of glycolysis in eukaryotes were acquired from mitochondrial genomes and provide new insights into the evolutionary history (host-symbiont relationships) of diatoms and other heterokont-flagellated protists. 11. According to the preceding discussion, how many GAPDH isoenzymes can be found in a diatom cell? (A) (B) (C) (D) None 1 2 3 12. A triosephosphate isomerase (TPI) isoenzyme can not be found in: (A) the cytosol of an animal cell (B) mitochondria of a diatom cell (C) mitochondria of an animal cell (D) chloroplasts of a plant cell 13. According to the preceding paragraph, which of the following statements concerning the relationship between diatoms and oomycetes is most likely to be true? (A) Diatoms and oomycetes are close relatives. (B) Diatoms are plants whereas oomycetes are not. (C) Oomycetes acquire their TPI-GAPDH fusion protein gene from diatoms. (D) Diatoms acquire their TPI-GAPDH fusion protein gene from oomycetes. -4- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 5 頁,共 14 頁 14. The functional conformation of the TPI-GAPDH enzyme complex in diatoms is composed of: (A) one TPI subunit and one GAPDH subunit (B) a single TPI-GAPDH fusion protein (C) two identical TPI-GAPDH fusion proteins (D) four identical TPI-GAPDH fusion proteins 15. Discovery of the mitochondrium-localized TPI-GAPDH enzyme complex in diatoms lends support to the view that: (A) Diatoms depend on their mitochondria for glycolytic enzymes. (B) The diatom mitochondrium represents an evolutionarily primitive stage of this ubiquitous organelle. (C) The diatom mitochondrium is an evolutionary vestige descended from a secondary photosynthetic endosymbiotic protist. (D) Diatoms carry out carbon fixation in their mitochondria using the Calvin cycle 四、 Enzymes that participate in biosynthetic pathways of essential amino acids have been recognized as targets for a number of safe and effective herbicides. The biosynthetic pathway to the branched-chain amino acids valine, leucine, and isoleucine is of special importance in this respect. At least three classes of very potent and extensively used herbicides, the sulfonylureas, the imidazolinones, and the sulfonanilides, are known to inhibit the first common enzyme in this pathway, acetohydroxy acid synthase (AHAS; EC 4.1.3.18). The second common enzyme in the pathway, ketol-acid reductoisomerase (KARI; EC 1.1.1.86), is the target for two additional inhibitors with potential herbicidal activity, 2-dimethylphosphinoyl-2-hydroxyacetate and N-isopropyl oxalylhydroxamate(IpOHA). Sulfometuron methyl (SMM), which belongs to the sulfonylurea class of herbicides and inhibits AHAS activity, is an extremely potent herbicide and bacteriostatic agent. The potency of SMM has been ascribed in part to the toxicity of one of the substrates of AHAS, 2-ketobutyrate, which accumulates upon inhibition of the enzyme. It has been proposed that high 2-ketobutyrate levels interfere with a number of metabolic pathways, including synthesis of coenzyme A, by several mechanisms. IpOHA also causes the accumulation in bacterial growth media of a substrate of the enzyme which it inhibits, acetolactate. However, there are no reports in the literature on the intracellular concentrations of the KARI substrate acetolactate or acetohydroxybutyrate. In the enterobacteria, transcription of ilvC, the gene which encodes KARI, is -5- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 6 頁,共 14 頁 induced by the enzyme’s substrates. Thus, an increase in substrate concentrations due to partial inhibition of the enzyme should lead to a compensating enhancement of the rate of the reaction by increasing enzyme synthesis (as well as by mass action, if KARI normally works well below its Km). 16. In plant and bacterial cells, biosynthesis of which of the following substances can be inhibited by the herbicide 2-dimethylphosphinoyl-2-hydroxyacetate? (A) ketobutyrate (B) sulfometuron methyl (C) isoleucine (D) acetohydroxy acid synthase 17. In enterobacteria, synthesis of the enzyme ketol-acid reductoisomerase increases in direct response to an increase in the concentration of: (A) coenzyme A (B) valine (C) ketobutyrate (D) acetolactate 18. Which of the following manipulations causes the toxic accumulation of 2-ketobutyrate in plant cells? (A) inhibiting the enzyme ketol-acid reductoisomerase (B) stimulating the expression of the gene ilvC (C) administering the compound sulfometuron methyl (D) administering the compound N-isopropyl oxalylhydroxamate 19. Which of the following enzymes or compounds does not participate directly in the synthesis of branched chain amino acids? (A) acetohydroxy acid synthase (B) coenzyme A (C) ketol-acid reductoisomerase (D) acetolactate 20. In keeping with the reasoning in this paragraph, the authors are most likely trying to address which of the following questions? (A)What are the toxic effects of N-isopropyl oxalylhydroxamate on cellular metabolism? (B) How does inhibition of conenzyme A synthesis contribute to the toxic effects of sulfometuron methyl? (C) How are acetohydroxy acid synthase activity and ilvC expression regulated in plants? (D)How do weeds and bacteria develop resistance to herbicides N-isopropyl oxalylhydroxamate and sulfometuron methyl? -6- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 7 頁,共 14 頁 五、 Therapy for patients with advanced thyroid carcinoma is limited. Clinical and in vitro studies suggest that some patients with advanced thyroid cancer may respond to therapy with retinoic acid. mRNA expression of the six retinoic acid (RAR) and retinoid X receptor (RXR) isoforms (RARalpha, -beta, -gamma and RXRalpha, -beta, -gamma) was measured in four human thyroid cell lines, and protein expression was subsequently measured in 10 thyroid cancer cell lines. Two isoforms, RARbeta and RXRgamma, were differentially expressed in the four cell lines. Comparison of 10 thyroid tumors and matched normal thyroid tissue confirmed differential tumor expression of RARbeta and RXRgamma and lack of the RXRgamma isoform in normal thyroid tissue. Cell lines expressing both RARbeta and RXRgamma demonstrated significant growth suppression when treated with retinoids, whereas cell lines lacking these isoforms were unaffected. Expression of RARbeta, the isoform associated with suppression of tumor growth in other cancer types, was not affected by treatment with retinoids in the thyroid cancer cell lines. LG346 increased apoptosis and decreased cells in the S-phase in an anaplastic carcinoma cell line, suggesting that this retinoid causes growth suppression of these cells by multiple mechanisms. 21. How many retinoic acid receptor isoforms in human cells? (A) ten (B) six (C) four (D) three 22. The treatment with retinoids in the thyroid cancer cell may NOT results in ? (A) expression of RARbeta (B) apoptosis (C) growth suppression (D) all of the above 23. What is the reason that advanced thyroid cancer may not respond to therapy with retinoic acid? (A) therapy is limited (B) absence of RARbeta and RXRgamma (C) lacks S-phase (D) LG346 increased apoptosis 24. In normal thyroid tissue, which enzyme does not be expressed? (A) RARbeta (B) RXRgamma (C) RARalpha (D) RXRalpha -7- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 8 頁,共 14 頁 六、 An antiserum, or an antibody, constitutes a useful reagent in protein biochemistry. However, it is a reagent which is specific to the immunogen/antigen couple and often it must be prepared in-house, especially when a novel protein is under investigation. Preparation of an antiserum starts with the immunogen, which is usually a protein isolated by one or more of the methods described in the previous chapters. The more pure the immunogen, the more specific will be the antibodies which it elicits. For most purposes, therefore, it is best to use as pure an immunogen as possible. Alternatively, for the production of so-called peptide antibodies, the immunogen might be a synthetic peptide of ten or more residues. The peptide is chosen from the amino acid sequence of the Ag of interest, i.e. the complete protein which it is hoped the Abs will recognize. For peptide antibodies to recognize the whole protein, it is necessary that the peptide sequence chosen be accessible, i.e. it must be on the surface of the protein. This can be readily determined if the 3-D structure of the protein is known. If the 3-D structure is not known, then the peptide can be chosen by analysis of the amino acid sequence of the Ag of interest for hydrophobicity (since hydrophilic residues will tend to be on the surface) or mobility (since residues on the surface, and especially at the N- or C-terminus are likely to be more mobile). 25. When a novel protein is under investigation, it is better to prepare its antibody by? (A) the investigator (B) the company (C) the competitor (D) anyone 26. How to increase the specificity of the antibodies? (A) more pure antibodies (B) more pure antigen (C) more pure hydrophobicity (D) more pure antiserum 27. The peptide antibodies are made by using? (A) Antiserum (B) synthetic peptide (C) 3-D structure of the protein (D) hydrophobicity -8- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 9 頁,共 14 頁 28. In general, the immunogen is? (A) DNA (B) RNA (C) Protein (D) lipid 29. What information can help to determine whether a specific peptide is on the surface of the protein? (A) 3-D structure of the protein (B) the hydrophobicity of the peptide (C) mobility of the peptide (D) all of the above 30. Why the peptide used for making peptide antibodies is better on the surface of the protein? (A) it is easy to prepare (B) to determine the 3-D structure of the protein (C) for peptide antibodies to recognize the whole protein (D) all of the above 七、 Dosage compensation ensures equal expression of X-linked genes in XX females and XY males. In mammals, this process results in inactivation of one female X-chromosome (XCI) in a random or imprinted manner. In the random form (eutherian), a zygotic counting mechanism initiates dosage compensation and enables a choice mechanism to randomly designate one active (Xa) and one inactive (Xi) X. In the imprinted form, zygotic counting and choice are superseded by parental imprints that direct exclusive paternal X-silencing. Imprinted XCI is found in ancestral marsupials but vestiges remain in the extraembryonic tissues of eutherians such as mice. An epigenetic mark for random and imprinted XCI has long been postulated. The marks are placed at the Xinactivation center (Xic), which includes the cis-acting noncoding gene, Xist, and its antisense counterpart, Tsix. Xist RNA accumulation along the Xi initiates the silencing step, while Tsix represses silencing by blocking Xist RNA accumulation. A cis-acting center for choice and imprinting lies at the 5´ end of Tsix, as its deletion abolishes random choice in epiblast-derived cells to favor inactivation of the mutated X and disrupts maternal Xist imprinting in extraembryonic tissues. Thus, while imprinted XCI is parentally directed and random XCI is zygotically controlled, both work through Tsix to regulate Xist. -9- 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 10 頁,共 14 頁 To date, only X-linked cis-elements have been identified as XCI regulators. Yet, virtually all models invoke transacting factors, which interact with the X-linked sites. In one model for imprinted XCI, a maternal-specific trans-factor confers resistance to XCI. In models for random XCI, an autosomally expressed “blocking factor” protects a single X from silencing. We have proposed that Tsix is the cis-target of both trans-factors. 31. Which of the following animals has imprinted form of X chromosome inactivation? (A) human (B) kangaroo (C) orangutan (D) cow 32. In the imprinted form of X chromosome inactivation, the X chromosomes originated from ______ were inactive. (A) father (B) mother (C) both parents (D) none of above 33. The Xinactivation center does not include (A) Tsix (B) Xist (C) cis-acting center for choice and imprinting (D) blocking factor 34. The synonym of superseded is (A) supervised (B) substituted (C) separated (D) summarized 35. The synonym of abolished is (A) accumulated (B) finished (C) retreated (D) ensured - 10 - 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 11 頁,共 14 頁 36. What is the essence of the cis-acting center for choice and imprinting? (A) DNA (B) RNA (C) protein (D) none of above 37. The X chromosomes are inactivated if (A) Tsix gene is transcribed (B) Xist gene is transcribed (C) both genes are transcribed (D) none of the genes are transcribed 38. Which of the following has not been identified according to the assay? (A) Blocking factor (B) XCI regulators (C) Xist (D) Tsix 39. Which of the following statements is not true? (A) The X chromosome is active if the Xist mRNA is not produced. (B) The X chromosome is active if a maternal-specific transcription factor is present. (C) The X chromosome is not active if a X-linked blocking factor is not present. (D) The X chromosome is not active if the XCI regulator is not bound by transcription factors. 40. Where is the cis-acting center for choice and imprinting is located on the figure in the right? (A) A Tsix A Xist B C D (B) B (C) C (D) D 八、 Transgenic animals have been routinely produced by microinjecting or electroporating naked DNA into 1-cell-stage embryos or unfertilized eggs. However, these techniques are inapplicable to - 11 - 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 12 頁,共 14 頁 live-bearing fish and many crustacean species for which unfertilized or newly fertilized eggs are not readily obtainable. In the present study, replication-defective pantropic retroviral vectors carrying a reporter gene (neo(R) or beta-gal) were used to directly transform the immature ovary or testis of a live-bearing fish (Poeciliopsis lucida) and crayfish (Procambarus clarkii). The fraction of the progeny derived from these treated individuals shown to contain the neo(R) reporter gene by an assay based on polymerase chain reaction (PCR) was significant. The PCR-positive individuals were crossed with nontransgenic individuals, and about 50% of the resulting progeny carried the transgene, suggesting that the F(1) animals are germline transgenic. Integration of the transgenes was confirmed by detecting the junction fragments of the genomic DNA associated with transgene constructs. Expression of reporter genes was detected by a reverse transcription-nested PCR assay. These results showed that transgenic live-bearing fish and crustaceans could be easily produced by directly transforming the immature gonads with replication-defective pantropic retroviral vectors. (Sarmasik A, et al. 2001. Mar. Biotechnol. (NY).3(5):470-7.) 41. How to confirm that progeny of the treated individuals contained the gene neo(R) reporter gene ? (A) by microinjection, (B) by electroporation, (C) by PCR reaction, (D) by replication-defective pantropic retroviral vectors. 42. What is the integration of the transgenes? (A) PCR-positive individuals were crossed with nontransgenic individuals, (B) F(1) animals are germline transgenic, (C) junction fragments of the genomic DNA, (D) the genomic DNA associated with transgene constructs. 43. In this study, what was not detected in the reverse transcription-nested PCR assay? (A) expression of the neo(R) genes, (B) live-bearing fish and crustaceans were transformed successfully, (C) transgenic live-bearing fish and crustaceans were produced, (D) replication-defective pantropic retroviral vectors were transformed. 九、 We estimated a novel phylogeny of tilapiine cichlid fish (an assemblage endemic to Africa and the Near East) within the African cichlid fishes on the basis of complete mitochondrial NADH - 12 - 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 13 頁,共 14 頁 dehydrogenase subunit 2 (ND2) gene sequences. The ND2 (1,047 bp) gene was sequenced in 39 tilapiine cichlids (38 species and 1 subspecies) and in an additional 14 nontilapiine cichlid species in order to evaluate the traditional morphologically based hypothesis of the respective monophyly of the tilapiine and haplochromine cichlid fish assemblages. The analyses included many additional cichlid lineages, not only the so-called tilapiines, but also lineages from Lake Tanganyika, east Africa, the Neotropics and an out-group from Madagascar with a wide range of parental care and mating systems. Our results suggest, in contrast to the historical morphology-based hypotheses from Regan (1920, 1922 ), Trewavas (1983), and Stiassny (1991), that the tilapiines do not form a monophyletic group because there is strong evidence that the genus Tilapia is not monophyletic but divided into at least five distinct groups. In contrast to this finding, an allozyme analysis of Pouyaud and Agnese (1995), largely based on the same samples as used here, found a clustering of the Tilapia species into only two groups. This discrepancy is likely caused by the difference in resolution power of the two marker systems used. Our data suggest that only type species Tilapia sparrmanii Smith (1840) should retain the genus name TILAPIA: One particular group of tilapiines (composed of genera Sarotherodon, Oreochromis, Iranocichla, and Tristramella) is more closely related to an evolutionarily highly successful lineage, the haplochromine cichlids that compose the adaptive radiations of cichlid species flocks of east Africa. It appears that the highly adaptable biology of tilapiines is the ancestral state for all African cichlids and that the more stenotypic lifestyle of the haplochromine cichlids is derived from this condition. We reconstructed the evolution of the highly variable parental care systems on the basis of the most inclusive composite phylogeny to date of the African, Neotropical, and Madagascan cichlids with special emphasis on a group of tilapiines comprising the substrate-spawning genus Tilapia, and the mouthbrooding genera Sarotherodon and OREOCHROMIS: We demonstrate several independent origins of derived mouthbrooding behaviors in the family Cichlidae. (Klett V & Meyer A. 2002. Mol. Biol. Evol. 19(6):865-83.) 44. What was not sequenced to estimate the phylogeny of tilapiine cichlid fish within the African cichlid fishes (A) the complete mitochondrial DNA, (B) mitochondrial NADH dehydrogenase subunit 2, (C) complete ND2 gene, (D) 1,047 bp of the ND2 gene of mtDNA. - 13 - 國立彰化師範大學九十三學年度碩士班招生考試試題 系所:生物技術研究所 科目:專業英文 ☆☆請在答案紙上作答☆☆ 第 14 頁,共 14 頁 45. How many species were used to evaluate the respective monophyly of the tilapiine and haplochromine cichlid fish assemblages? (A) 14 species, (B) 39 species, (C) 52 species, (D) 53 species. 46. What was not the reason to added many other cichlid lineages in this study? (A) They all belong to tilapiines, (B) The all from Lake Tanganyika, (C) To find an out-group from Madagascar, (D) They all have the same parental care and mating systems. 47. What is the reason that results of this study were different from of Regan (1920, 1922 ), Trewavas (1983), and Stiassny (1991)? (A) tilapiines is a monophyletic group, (B) specimens all belong to genus Tilapia, (C) allozyme analysis, (D) different resolution power of the two detected methods. 48. What narration is wrong in this writing? (A) It was suggested that for all African cichlids were highly adaptable in their ancestral state, (B) This study tried to establish the evolutionary history of the African, Neotropical, and Madagascan cichlids, (C) Pouyaud and Agnese (1995) suggested that Tilapia species only divided into two clusters. (D) The mouthbrooding behaviors in the family Cichlidae were derived from the related species. - 14 -