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Section K – Transcription in prokaryotes Contents K1 Basic principles of transcription Transcription: an overview, Initiation, Elongation, Termination K2 Escherichia coli RNA polymerase Escherichia coli RNA polymerase, αSubunit, βSubunit, β’Subunit, Sigma factor K3 E. coli σ70 promoter Promoter sequences, Promoter size, -10 sequence, -35 sequence, Transcription start site, Promoter efficiency K4 Transcription, initiation, elongation and termination Promoter binding, DNA unwinding, RNA chain initiation, RNA chain elongation, RNA chain termination, Rho-dependent termination K1 Basic principles of transcription — Transcription: an overview • The synthesis of a single-stranded RNA from a double-stranded DNA template. • RNA synthesis occurs in the 5’3’direction and its sequence corresponds to that of the DNA strand which is known as the sense strand. • The template of RNA synthesis is the antisense strand. • Necessary components: promoter/template, RNA polymerase, rNTPs, terminator/template (-) strand is antisense strand. (+) strand is sense strand 3‘ 5‘ 5‘ 3‘ • Key: dsDNA 5’ 3’ ssRNA K1 Basic principles of transcription — Initiation (1) Binding of an RNA polymerase to the dsDNA (2) Slide to find the promoter Promoter: The sequence of DNA needed for RNA polymerase to bind to the template and accomplish the initiation reaction; the5’-side (upstream) of the coding region; the short conserved sequence (3) Unwind the DNA helix; For base pairing; Begins at the promoter site (4) Synthesis of the RNA strand at the start site (initiation site), this position called position +1 K1 Basic principles of transcription — Elongation • Add ribonucleotides to the 3’-end • The RNA polymerase extend the growing RNA chain in the direction of 5’ 3’ (E. coli: 40 nt/sec) • The enzyme itself moves in 3’ to 5’ along the antisense DNA strand. The helix is reformed behind the polymerase. K1 Basic principles of transcription — Termination • The dissociation of the transcription complex from the template strand and separation of RNA strand at a specific DNA sequence known as the terminator. These sequences often contain self-complementary regions which can form stem-loop or hairpin structure, some need rho protein as accessory factor. K2 Escherichia coli RNA polymerase — Escherichia coli RNA polymerase RNA polymerase: synthesis of RNA strand from DNA template. • 1. Requires no primer for polymerization • 2. Requires DNA for activity and is most active with a double-stranded DNA as template. • 5’ 3’ synthesis, • Rate: 40 nt per second at 37oC • 3. Require Mg2+ for RNA synthesis activity 4. All RNA polymerases lack 3’ 5’ exonuclease activity, and one error usually occurs when 104 to 105 nucleotides are incorporated. 5. Usually are multisubunit enzyme, but not always. 6. The cylindrical channel in the enzyme complex can bind directly to 16 bp of DNA. The whole polymerase binds over 60 bp. 7. Different from organism to organism The polymerases of bacteriophage T3 and T7 are smaller single polypeptide chains, they synthesize RNA rapidly (200 nt/sec) and recognize their own promoters which are different from E. coli promoters 155 KD 36.5 KD 36.5 KD 11 KD 70 KD 151 KD 465kd Core enzyme(核心酶): 2 ’ for both initiation & elongation Holoenzyme(全 酶): 2 ’ for initiation K2 Escherichia coli RNA polymerase —αSubunit • Two identical subunits in the core enzyme • Encoded by the rpoA gene • Required for core protein assembly • May play a role in promoter recognition K2 Escherichia coli RNA polymerase —βSubunit 1. 2. • • Encoded by rpoB gene. The catalytic center of the RNA polymerase Rifampicin :bind to the β subunit, and inhibit transcription initiation. This class of antibiotic does not inhibit eukaryotic polymerases and has used medically for treatment of Gram-positive bacteria infections and tuberculosis Mutation in rpoB gene can result in rifampicin resistance. Streptolydigins:Inhibits transcription elongation but not initiation. subunit may contain two domains responsible for transcription initiation and elongation K2 Escherichia coli RNA polymerase —β’Subunit 1. Encoded by the rpoC gene . 2. Binds two Zn 2+ ions and may participate in the catalytic function of the polymerase • Heparin:binds to the ’ subunit and inhibits transcription in vitro. • Heparin competes with DNA for binding to the polymerase. ’ subunit may be responsible for binding to the template DNA . K2 Escherichia coli RNA polymerase — Sigma factor 1. Many prokaryotes contain multiple factors to recognize different promoters. The most common factor in E. coli is 70. 2. Binding of the factor converts the core RNA pol into the holoenzyme. 3. s factor is critical in promoter recognition, by decreasing the affinity of the core enzyme for non-specific DNA sites (104) and increasing the affinity for the corresponding promoter • 4. s factor is released from the RNA pol after initiation (RNA chain is 8-9 nt) • 5. Less amount of s factor is required in cells than that of the other subunits of the RNA pol. K3 E. coli σ70 promoter — Promoter sequences • Different factors recognize different promoters • Upstream of the start site of transcription (position +1), thus the promoter sequences are assigned a negative number (编号为负数) • Contains short conserved sequences required for specific binding of RNA polymerase and transcription initiation +1 promoter DNA Transcribed region terminator ATACG TATGC Antisense strand K3 E. coli σ70 promoter — Promoter size • Consists of a sequence of between 40 and 60 bp • -55 to +20: bound by the polymerase • -20 to +20: tightly associated with the polymerase and protected from nuclease digestion by DNaseΙ • Up to position –40: critical for promoter function • -10 and –35 sequence: particularly important for promoter function K3 E. coli σ70 promoter — -10 sequence • A 6 bp sequence is found in the promoters of many different E. coli genes which is centered at around the –10 position (Pribnow, 1975). • A consensus sequence of TATAAT, the first two bases(TA) and the final T are most highly conserved • This hexamer(六聚体) is separated by 5 to 8 bp from position +1, and the distance is critical • The –10 sequence is the unwinding sequence. K3 E. coli σ70 promoter — -35 sequence Enhances recognition and interaction with the polymerase s factor • A conserved hexamer sequence around position –35 • A consensus sequence of TTGACA, the first three positions (TTG) are the most conserved among E. coli promoters. • Separated by 16-18 bp from the –10 box in 90% of all promoters The sequences of five E. coli promoters K3 E. coli σ70 promoter — Transcription start site The sequence around the start site influences initiation • Purine in 90% of all genes • G is more common at position +1 than A • Often, there are C and T bases on either side of the start site nucleotide (i.e. CGT or CAT) K3 E. coli σ70 promoter — Promoter efficiency • There is considerable variation in sequence between different promoters, and the transcription efficiency can vary by up to 1000-fold . • The –35 sequence, -10 sequence, and sequence around the start sites all influence initiation efficiency. • Some promoter require activating factor for initiation. For example, Lac promoter Plac requires cAMP receptor protein (CRP) K4 Transcription, initiation, elongation and termination — Promoter binding • The core enzyme (2 ’) has nonspecific DNA binding (loose binding, 20000 fold less). • The factor enhances the specificity of the core RNA polymerase (2 ’) for promoter binding (100x) • The polymerase finds the promoter –35 and – 10 sequences by sliding along the DNA extremely rapidly and forming a closed complex K4 Transcription, initiation, elongation and termination — DNA unwinding • Necessary to unwind the DNA so that the antisense strand to become accessible for base pairing, carried out by the polymerase. • The initial unwinding of the DNA results in formation of an open complex with the polymerase. K4 Transcription, initiation, elongation and termination — RNA chain initiation • No primer is needed • Start with a GTP (more common) or ATP • Initially incorporates first 2 nucleotides. The first 9 nt are incorporated without polymerase movement along the DNA or σfactor release K4 Transcription, initiation, elongation and termination — RNA chain elongation • σFactor is released to form a ternary complex of the pol-DNA-RNA (newly synthesized), causing the polymerase to progress along the DNA (promoter clearance). • Transcription bubble (unwound DNA region, ~ 17 bp) moves along the DNA with RNA polymerase which unwinds DNA at the front and rewinds it at the rear. K4 Transcription, initiation, elongation and termination — RNA chain termination • Termination: Dissociation of RNA Re-annealing of DNA Release of RNA pol Terminator sequence : • RNA hairpin very common • Accessory rho protein(辅助ρ蛋白) K4 Transcription, initiation, elongation and termination — Rho-dependent termination • Some genes contain terminator sequences requiring an accessory factor, the rho protein (ρ) to mediated transcription termination. • Rho binds to specific sites in the singlestranded RNA. • Rho protein (hexameric protein) binds to certain RNA structure (72bp) • Rho hydrolyses ATP and moves along the nascent RNA towards the transcription complex then enables the polymerase to terminate transcription. Binding Unwinding Initiation Elongation Termination RNA polymerase/transcription and DNA polymerase/replication RNA pol DNA pol Template dsDNA is better ss/dsDNA Require primer No Yes Initiation Promoter Origin Elongation 40 nt/ sec 900 nt /sec Terminator Synthesized RNA Template DNA Multiple choice questions 1. Which two of the following statements about transcription are correct? A RNA synthesis occurs in the 3' to 5' direction. B the RNA polymerase enzyme moves along the sense strand of the DNA in a 5' to 3' direction. C the RNA polymerase enzyme moves along the template strand of the DNA in a 5' to 3' direction. D the transcribed RNA is complementary to the template strand. E the RNA polymerase adds ribonucleotides to the 5' end of the growing RNA chain. F the RNA polymerase adds deoxyribonucleotides to the 3' end of the growing RNA chain. 2. Which one of the following statements about E. coli RNA polymerase is false? A the holoenzyme includes the sigma factor. B the core enzyme includes the sigma factor. C it requires Mg2+ for its activity. D it requires Zn2+ for its activity. 3. A B C D E F 4. A B C D E Which one of the following statements is incorrect? there are two α subunits in the E. coli RNA polymerase. there is one β subunit in the E. coli RNA polymerase. E. coli has one sigma factor. the β subunit of E. coli RNA polymerase is inhibited by rifampicin. the streptolydigins inhibit transcription elongation. heparin is a polyanion, which binds to the β’ subunit. Which one of the following statements about transcription in E. coli is true? the -10 sequence is always exactly 10 bp upstream from the transcription start site. the initiating nucleotide is always a G. the intervening sequence between the -35 and -10 sequences is conserved. the sequence of the DNA after the site of transcription initiation is not important for transcription efficiency. the distance between the -35 and -10 sequences is critical for transcription efficiency. 5. Which one of the following statements about transcription in E. coli is true? A loose binding of the RNA polymerase core enzyme to DNA is non-specific and unstable. B sigma factor dramatically increases the relative affinity of the enzyme for correct promoter sites. C almost all RNA start sites consist of a purine residue, with A being more common than G. D all promoters are inhibited by negative supercoiling. E terminators are often A-U hairpin structures. THANK YOU !