核酸与蛋白质的分析(II).ppt

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1、核酸與蛋白質的分析(II),陽明大學 生物資訊學研究所楊 永 正,上次課程複習,emboss程式簡介Revseq:互補、倒置Transeq:轉譯Backtranseq:反轉譯(可能產生多條序列)密碼表基因體資訊存於GenBank,反轉譯出的核酸序列可能與原序列不同,反轉譯出的核酸序列與原序列不同，是否可在Genbank中找到原序列?,5-ATGGGCGAGAAGGCCCTG-3,提 示,使用資料庫搜尋程式blastn,Submit sequence for blast analysis,Format blast result,Wait for results,Blastn results,Se

2、arch for short,nearly exact matches,反轉譯出的核酸序列，在Genbank中不找到原序列。該怎麼辦?,必須有更長的序列(請回去後做習題組，在週一上午上課前將結果寄到binfoym.edu.tw),上述序列是來自於 Xenopus laevis的TFIIIA 基因，如何才能查到此序列?,關鍵字查詢:Xenoups,TFIIIA,有了整個基因的核酸序列，要怎麼找相似的基因?,利用非洲水生蛙的TFIIIA基因，找其他生物的TFIIIA基因，或與TFIIIA相似的基因,提 示,先找到Xenopus laevis之TFIIIA蛋白質序列再使用資料庫搜尋程式blastp,

3、解答 尋找序列,解 答-blastp,若想知道TFIIIA 蛋白質中是否有蛋白質模組樣式(pattern)，應使用哪一個程式分析?,解 答,http:/bioportal.cgb.indiana.edu/Pise/5.a/patmatmotifs.html,甚麼是 ZINC_FINGER_C2H2_1?,提示:請在網路上尋找Prosite資料庫中，鋅指(zinc finger)模組的說明。,鋅指(zinc finger)模組的發現,Miller,J.,McLachlan,A.D.,and Klug,A.(1985)Repetitive zinc-binding domains in the p

4、rotein transcription factor IIIA from Xenopus oocytes.EMBO 4,1609-1614.,.,contains an unusually large number of Cys and His residues.At first sight these residues appeared to us to form roughly periodic groupings.We therefore made a systematic search for repeats in both amino acid sequence and the c

5、DNA,using the diagonal comparison matrix method and the damped Needleman and Wunsch method(see Materials and methods).,Diagonal MatrixMethod,A T G C G G C G T A C T G A CA 1 1 1T 1 1 1G 1 1 1 1 1C 1 1 1 1G 1 1 1 1G 1 1 1C 1 1 1G 1 1T 1 1A 1 1C 1 1T 1G 1A 1C 1,identical=1different=emptyOff Diagonal L

6、ine Represents Repeat within sequence analyzed,Scoring,Nucleic Acid Identical:1 Different:0Amino Acid blosum62 Matrix,在Emboss套組中是否有繪製dotplot 的程式?,解 答,http:/bioportal.cgb.indiana.edu/Pise/5.a/dotmatcher.html,甚麼是window size與threshold?,Window,Window Size=5,Threshold,88,96,92,74,56,66,85,97,82,77,Thresh

7、old=79,01,01,01,00,00,00,01,01,01,00,三維示意圖,Score,在dotplot 上可以看到多少個鋅指模組?,提示:試調整Window size 與Threshold,討 論,該如何讀取相似的位置?,How to read a dot matrix plot?,189,241,282,230,What is the pattern of repeats?,基因的特徵:分子解剖學,基因的調控:分子生物學,蛋白質可以辨識核酸序列,噬菌體的抑制子與其操作子之間的交互作用。(Taken from for educational purpose),怎樣尋找可能的蛋白質接

8、合位置?,如果一個蛋白質可以調控多個基因，這些基因應有相似的蛋白質接合位置,解 答,由“MGEKAL”反轉譯出的核酸序列在Genbank中會找到多少相似的序列?,EMBOSS:fuzznuc,使用 IUPAC密碼對照表 N代表任何核苷酸 表示，例如 表示此位置，例如.Ambiguities are indicated by listing the acceptable nucleotides for a given position,between square parentheses.For example:ACG stands for A or C or G.Ambiguities are

9、 also indicated by listing between a pair of curly brackets the nucleotides that are not accepted at a given position.For example:AG stands for any nucleotides except A and G.Each element in a pattern is separated from its neighbor by a-.(Optional in fuzznuc).Repetition of an element of the pattern

10、can be indicated by following that element with a numerical value or a numerical range between parenthesis.Examples:N(3)corresponds to N-N-N,N(2,4)corresponds to N-N or N-N-N or N-N-N-N.When a pattern is restricted to either the 5 or 3 end of a sequence,that pattern either starts with a symbol.A per

11、iod ends the pattern.(Optional in fuzznuc).For example,CG(5)TGAN(1,5)C,#Program:fuzznuc#Rundate:Sat May 14 08:45:47 2005#Report_format:seqtable#Report_file:outfile.out#=#Sequence:XELTFIIIA from:1 to:1518#HitCount:2#Pattern:GAATTC#Mismatch:0#Complement:No#=Start End Mismatch Sequence 1 6.GAATTC 1513 1518.GAATTC#-#-,請在網路上尋找Prosite資料庫中，鋅指(zinc finger)模組的說明。,