Chapter eleven RNA transcription
DNAs RNAs Proteins 1 DNA replication: entirety 2 RNA transcription: systematicness 3 protein biosynthesis: individuality 1 3 2
Concept of RNA biosynthesis or Gene transcription The transmission of the genetic infor- mations is from DNA to RNA. Transcription is the synthesis of a single- stranded RNA from DNA template. RNA synthesis occurs in the 5’3’ direction. Transcription is asymmetry.
Section 1 template and enzymes in RNA biosynthesis
coding strand* positive strand Crick strand antisense strand 3’ 5’ new RNA strand template strand* coding strand* GTAC GUAC CATG negative strand sense strand positive strand antisense strand Watson strand Crick strand
transcriptional template 3’ 5’ It is only one of double-strand DNA, therefore, transcription is asymmetric. transcriptional template It is the fragments of gDNA.
types of genes in gDNA mRNA gene rRNA gene spacer tRNA gene 3’ 5’ spacer rRNA gene tRNA gene other RNA gene types of genes in gDNA
RNA polymerase in E.coli molecular weight:480kd consists of 2’
The function of subunits of RNA polymerase in E. coli determines which genes are transcribed catalytic site ’ binds DNA template recognizes transcription initiation site
annotation enzyme in RNA polymerase holoenzyme of RNA polymerase The 2’ complex is core enzyme in RNA polymerase The 2’ complex is holoenzyme of RNA polymerase
RNA polymeras in eukaryote variety transcripts respond to amanitin Ⅰ 45s-rRNA resistant (28S, 18S, 5.8S) Ⅱ* hnRNA(mRNA) very sensitive Ⅲ 5s-rRNA midding tRNA,snRNA sensitive
gene ’ + There is some sequence, which is recognized and bound by RNA polymerase, on 5’ end of transcriptional genes. characters of the sequence There are many A-T pairs There are some consensus sequence 5’ 3’
promoter site gene The binding region transcribed initiation site -50 -40 -30 -20 -10 1 10 5’ gene The binding region termination signal
In prokaryotes promoter 40-80bp gene transcribed initiation site TTGACA AACTGT TATAATPu ATATTAPy transcribed initiation site -50 -40 -30 -20 -10 1 10 5’ Recognition site Binding site Pribnow box promoter 40-80bp In prokaryotes 1 2 3
In eukaryotes promoter TATA box Hogness box CAAT box GC box 5’ gene -50 -40 -30 -20 -10 1 10 5’ CAAT box GC box
5’ Ⅰand Ⅱ Ⅲ + promoter gene RNA pol I or II RNA pol III gene+promoter
G A C C U G A U U U U-OH 3’ 5’ U 5’…GCCGCCAGUUCGGCUGGCGGCAUUUU…3’ RNA 5’…GCCGCCAGTTCGGCTGGCGGCATTTT… 3’ terminator The RNA made from the DNA palindrome is self-complementary and so formed a internal hairpin structure followed by a few U bases. The signals that terminates transcri-ption are localized in the gene 3’ end. A simple termination signal is a GC- rich region that is a palindrome, followed by AT-rich sequence. DNA
Process of RNA biosynthesis Section 2 Process of RNA biosynthesis 1. initiation 2. elongation 3. termination
process of transcription in prokaryote 5’ 2 3 5’ pppG 4 5’ pppGN 5 7 5’ pppG mRNA 6 1 ’
-O P O coding strand template strand 5’ PPi A 3’ P~P~P O G C T H CH2 O OH CH2 P~P~P O A -O P O coding strand template strand
transcription bubble unwinding rewinding 17bp 5’ 17bp RNA polymerase direction of transcription coding strand New RNA strand template strand pppG RNA-DNA hybrid unwinding rewinding
transcription in prokaryote under electromicroscope 5’ 3 1 2 5 4 transcription in prokaryote under electromicroscope
prokaryote 1. Dependent upon factor 5’ c termination of gene transcription in prokaryote 1. Dependent upon factor Promoter Gene RNA polymerase Terminator new RNA
factor function: double strands. It has activity of ATPase and helicase. It can bind with polyC in RNA 3’end. It may denature short RNA-DNA hybrid double strands.
2. Independent to factor: TTGCAGCCTGAGAAATCAGGCTGATGGCTGGTGACTTTTTAGTCACCAGCCTTTTT 2. Independent to factor: The base sequence and transcript RNA secondary structure of rplL gene 3’ end in E. coli. or UUUUU
model about transcription termination for independent to factor RNA polymerase 5’ coding strand template strand new RNA transcript model about transcription termination for independent to factor
post-transcriptional processing of eucaryotes Section 3 post-transcriptional processing of eucaryotes 1. Process after transcription for mRNA 2. Process after transcription for tRNA 3. Process after transcription for rRNA
2. polyadenylation of mRNA 3’-end Process after transcription for mRNA 1. Capping of mRNA 5’-end 2. polyadenylation of mRNA 3’-end 3. splicing of mRNA
Capping of 5’-end hnRNA 5’pppG phosphatase PPi 5’ pG pppG methylase m7G 5’ppp 5’m2’G
Explanation 1. It is carried out in nuclear. 2. It is earlier than the splicing . 3. It relates with protein translation . 4. It can protect mRNA 5’-terminal from nuclease. 5. The mRNA 5’-terminal of prokaryote doesn’t have the capping structure.
modification and termination of transcription about eukaryote gene 5’ RNA polymerase template strand TTATTT CACACA AAUAAA new RNA strand modification point AAA AAA AATAAA GTGTGT coding strand GUGUGU AAA AAA polyA tail
Explanation 1. The polyA tail is independent on DNA template. 2. It is also carried out in nuclear. 3. It is also earlier than the splicing . 4. It can protect mRNA 3’-terminal from nuclease. 5. The mRNA 3’-terminal of prokaryote doesn’t have the polyA tail structure. 1. The polyA tail is independent on DNA template. 6. It relates with protein translation .
P H N H2N O OCH3 CH2 O- OH H2C ……AAAAAAAA-OH 7 CH3 5’
splicing of hnRNA mRNA splicing enzyme DNA double strand intron-1 5’ intron-1 intron-2 intron-3 exon-1 exon-3 exon-4 exon-2 hnRNA strand split gene
5’ hnRNA strand mRNA lariat structure
transcription and past-transcribed modification for albumin gene A B C D E F G L 1 2 3 4 5 6 7 5’ 47 185 51 129 118 143 156 1043 gene-7.7kb hnRNA transcription AAA AAA m7G5’ppp5’ m2’G modification modified mRNA lariat mRNA to form lariat mature mRNA splicing
U1-snRNA U2-snRNA exon-1 exon-2 pG-O-H intron-1 UCCAUACAUAPPPGm AG……. UCCAUACAUAPPPGm AUGAUGU PPPGm exon-1 exon-2 intron-1 U1-snRNA U2-snRNA pG-O-H ….....AGGUAUGU UACUACA
kinds of introns and its way of splicing 1. It is in the rRNA gene mainly. 2. It is in the protein-coding gene of mitochondrion and chloroplast. self-splicing RNA=ribozyme 3. It is in the most protein coding gene. lariat way, by small nuclear ribonucleoprotein requires enzyme and ATP 4. It’s in the tRNA gene.
What is the functions of introns in gene?
Process after transcription for tRNA 1. Splicing . 2. To form rare base. 3. Addition of 3’-terminal CCA-OH.
splicing of primary transcript for tyrosine-tRNA in yeast 3’ OH U primary transcript 5’ 5’ tRNA gene 3’ OH U 5’ spliced tRNA
to form rare base in transcript of tyrosine-tRNA in yeast 1. methylation: catalyzed by tRNA-methyltransferase, A Am, G Gm, C C m spliced tRNA 3’ OH U 5’ 3’ OH U 5’ Gm Cm Am G A C
2. U reduction: U DHU(D) 3’ OH U 5’ Gm Cm Am D to form rare base in transcript of tyrosine-tRNA in yeast
3. translocation in nucleoside:U to form rare base in transcript of tyrosine-tRNA in yeast 3. translocation in nucleoside:U 3’ OH U 5’ Gm Cm Am D 3’ OH U 5’ Gm Cm Am U
Don’t have for tyrosine-tRNA in yeast. 4. deamination Don’t have for tyrosine-tRNA in yeast. For other aminoacyl-tRNA, A I to form rare base in tRNA
nucleotide transferase Addition of 3’-terminal CCA-OH of tyrosine-tRNA in yeast 3’ OH A C U 5’ Gm Cm Am D 3’ OH U 5’ Gm Cm Am D nucleotide transferase 3’ OH U 5’ Gm Cm Am D
cloverleaf pattern: secondary structure mature tyrosine tRNA in yeast 5’ Gm Cm Am D 3’ OH A C cloverleaf pattern: secondary structure mature tyrosine tRNA in yeast T G U DHU loop T loop anticodon loop amino acid arm extra loop
process after transcription for rRNA spacer tandem repeated gene 18S 5.8S 28S 45 S
Explanation 1. In eukaryotic cell the rRNA gene is redundant gene, which arranges on chromosome by tandem. 2. All eukaryote have 18s, 5.8s and 28s rRNA. 3. The rRNA gene is located in nucleolus. 4. The rRNA is spliced by ribozyme. 5. The ribozyme is self-splicing RNA.
The structure of ribozyme 5’ 3’ stem loop catalytic point consensus sequence The structure of ribozyme
What is the significance of ribozyme ?
选择题练习 RNA 转录
1 复制和转录有许多异同点,下列有关这两个过 程的描述哪一项是错误的。 A 转录是以一条DNA链做模板(一个转录本);而复 制是以两条链做模板,并且是同时做模板。 B 复制的产物大于转录的产物。 C 两个过程合成的方向均为 5’至 3’。 D 两个过程均需RNA引物。 E DNA聚合酶和RNA聚合酶均需要Mg2+ 。
2 转录合成RNA的原料是下列哪一类物质。 A NMP B NDP C NTP D dNDP E dNTP ATP CTP GTP UTP TTP ? 合成DNA的原料?
RNA polymerase I RNA polymerase III 3 真核细胞RNA聚合酶 II 所催化合成的RNA是下 列哪一种。 A hnRNA B 45s-rRNA C tRNA D 5s-rRNA E snRNA RNA polymerase I RNA polymerase III
4 大肠杆菌RNA聚合酶中识别转录起始点的亚基是 下列哪一种。 A B ’ C D E F 决定DNA链上哪一基因被转录 与DNA模板结合 催化磷酸二酯键形成 辨认转录起始点 与转录终止相关
A. 5’-agguca-3’ B. 5’-acguca-3’ C. 5’-ucgucu-3’ D. 5’-acgtca-3’ 5. 模板DNA的碱基顺序是 5’-tgacgt-3’,其转录本RNA 的碱基序列是: A. 5’-agguca-3’ B. 5’-acguca-3’ C. 5’-ucgucu-3’ D. 5’-acgtca-3’ E. 5’-acgugt-3’
6. 下列关于promoter的描述,哪一项是正确的? A. 与 起始密码子所在区相对应的那段DNA顺序。 B. 与hnRNA 5’端序列相对应的那段DNA顺序。 C. RNA聚合酶最初与DNA结合的那段 DNA顺序。 D. 阻抑转录蛋白所结合的DNA顺序。 E. 增强转录蛋白所结合的DNA顺序。
7. 下列哪些成分是转录所必需的? A. dNTP。 B. DNA指导下的RNA聚合酶。 C. RNA指导下的DNA聚合酶。 D. NTP。 E. DNA模板。
8. 关于RNA转录,下列哪些是不正确的? A. DNA两条链均有模板功能。 B. 需要引物。 C. 是不对称转录。 E. 因子识别转录起始点。
9. 比较RNA转录与DNA复制,下列哪些描 述是正确的? A. 原料都是 dNTP。 B. 都在细胞核内进行。 C. 链的延长均从5’到3’。 D. 合成产物均需要剪接加工。 E. 在与模板链的碱基配对中均含有G/C或C/G。
10. 关于原核生物RNA聚合酶的叙述,下列 是正确的? A. 全酶由四种,五个亚基( ’ )组成。 B. 核心酶的组成是’ 。 C. 在体内核心酶的任何亚基都不能单独与DNA结合。 D. 亚基也有催化RNA进行转录的功能。 E. 因子协助转录起始。
11. 转录的终止涉及: A. 某些情况,因子与转录本上的终止序列结合。 B. RNA聚合酶识别DNA上的转录终止信号。 C. 在DNA模板上的转录终止区有特殊的碱基序列。 D. 核酸酶参与转录终止。 E. 因子识别转录的终止信号。
12. 转录过程: A. 无校正系统,但具有高度忠实性。 B. 产物RNA与DNA模板之间形成一个短的 RNA-DNA杂交链。 C. 是连续的。 D. 方向是5’到3’。 E. 需要RNA聚合酶 。
13. 真核细胞内基因转录后mRNA的 加工过程包括: A. 在5’端加“帽子”结构。 B. 在3’端加多聚A“尾巴”结构。 C. 去掉内含子,拼接外显子。 D. 在3’端加 CCA 氨基酸臂。 E. 部分碱基甲基化。
14. 关于真核mRNA下列哪些说法是正确的 A. 是由hnRNA生成。 B. 3’端有“帽子”结构。 C. 大部分mRNA的前体中有内含子。 D. 5’端有多聚A“尾巴”。 E. 含少量稀有碱基。
15. tRNA成熟过程包括: A. 在核酸酶作用下切去部分多核苷酸链。 B. 加多聚A 于3’末端。 C. 修饰形成某些稀有碱基。 D. 5’端加“帽子”。 E. 加CCA氨基酸臂。
16. 关于真核RNA分子中“帽子”的叙述下列 哪些说法是正确的 A. 是rRNA的加工过程。 B. 存在于tRNA的3’末端。 C. 是由多聚A组成。 D. 仅存在于真核mRNA上 。 E. 是7-甲基鸟嘌呤核苷三磷酸。
17. 内含子是: A. 不被转录的序列。 B. 不被翻译的序列。 C. 被转录的序列。 D. 编码序列 。 E. 被翻译的序列。