RNA synthesis:transcrition
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Transcript of RNA synthesis:transcrition
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RNA synthesis:transcrition
• Transcription is the process of the synthesis RNA molecule.
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Transcription unit
• Promoter region• Transcribed region• And termination region.
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Promoter region
• Which is certain sequence of base located at the beginning of stretch. It is important for initiation for transcription to occur and can be recognized by RNA polymerase.
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prokaryotes
• 1. The pribno box :stretch of 6 nucleotides (TATAAT) located about 10 bases to the left of the transcription. Initiation site.
• 2. A second nucleotide stretch (TGTTGACA) that located about 35 bases to the left of the transcription initiation site.
• 3. 19 bases (nucleotide) in between two streches.
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Cont.
• Both pribno box and TGTTGACA are region can be recognized by RNA polymerase.
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Eukaryotic promotor
• 1. The Hoges or TATA box: stretch of nucleotide that is almost similar to prokaryotes,located almost 25 nucleotide to the left of the trancrition initiation site.
• 2. CAAT box : located about 70 to 80 nucleotides to the left of the trasncription initiation site.
• 3. 40 bases between the two stretches.
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Cont.
• Transcription region: which is stretch of DNA that is to be transcribed into RNA molecules.
• Termination region: which is a stretch of DNA located at the end of DNA sequence to be trasncribed.
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Requirement for RNA synthesis
• Transcription unit: transcription occurs on one of the two strands of the DNA template and never on both complementary strand
• Four ribonucleotide triphosphate;ATP,GTP,UTP and CTP.
• RNA poymerase ,RNA(=DNA –dependent RNA polymerase)
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Prokaryotes RNA polymerase.
• 1.E.g E.coli ,enzyme synthesize all types of RNA molecules on DNA template.
• 2. the holoenzyme of the RNA polymerase consist of a core molecule and specific protein factor(sigma [σ] factor.
• A. The core enzyme molecule : 4 subunits: 2 of them are identical (alpha subunits.the other 2(β and β’) are similar but not identical .RNAP consist of 2 zinc molecules.
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Sigma factor
• Enables the polymerase to recognize promotor regions on DNA .
• Helps the core enzyme to attach more tightly to the promotor site.
• Some regions on DNA that signal the termination of transcription are recognized by the RNA polymerase itself. Others are recognized by specific termination factor e.g rho factor of E.coli.
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Steps RNA synthesis
• Initiation • Elongation• termination
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Initiation
• 1. RNA polymerase holoenzyme binds with the promotor area:
• A . Sigma subunit ,enables polymerase polymerase to recognize promotor region on DNA.
• B. Β̒subuint to the template.• C. B subunit binds to the nucleotide substrate.
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Cont.
• 2. binding of RNA poymearse DNA template leads to local separation (unwinding)of the DNA double helix into sense and antisense strands.
• 3. first nucleotide of RNA transcript at the initiation site is almost always purine.
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Elongations
• 1.At the sense formation of RNA molecule begins at the end of core enzyme with release of sigma 5 5̒factor.
• 2. Then elongation of the RNA molecule occurs from 5 to end ,antiparallel to its template. 3 ̒ 3̒
• 3. nucleotide building blocks are ribonucleotide 5 5̒triphosphate(ATP ,GTP,CTP,and UTP). They are inserted in the RNA = pairing rule.pyrophosphate is released when each new nucleotide is added to growing chain . GTPGMP + PPi
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Cont.
• 4.RNA polymerase forms a phosphodiester bond between the ‘3” OH of one ribose sugar and 5’ OH of the next ribose.
• 5. the process of elongation of RNA chain continues until a termination region reached.
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Termination:
• Termination region on DNA template can be recognized by :
• A. RNA – polymerase enzyme itself (rho independent termination),.
• B. Rho [ρ] factor which may be required for the release of both RNA strand and RNA polymerase (rho dependant termination).
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Termination result from either
• A. Binding of rho factor polymerase enzyme : when RNA – polymerase enzyme reaches the termination site,rho factor binds with it causing termination.
• Slowing down of RNA – polymerase at the termination site:
• 1. termination site on DNA is charaterized by the presence of palidromes.
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Cont.
• Palidromes is a region of a double stranded DNA in which each of the two strands haas the same sequence when read in the same direction. E.g in the 5’ to 3’direction.
• 2.The RNA transcript of the DNA palidrome can form a stable hairpin structure which is self –complementary structure. This hairpin structure causes slowing of RNA – poylmerase at the termination site.
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antibiotics
• Rifampcin: binds to the beta subunit . Treatment of leprosy and TB patients.
• Actinomycin D: binds to DNA template ,prevent movement of RNA – polymerase along the DNA.
• Alpha –amanitin: toxin from mushroom , inactivates RNA – polymerase II of eukaryotes.
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Post transcriptional modification of RNA
• A. eukaryotes mRNA : primary mRNAs are called heterogeneous nuclear RNA (hnRNA) .hnRNA is modified into mature mRNA in the nucleus by capping additon of nucleotides and splicing.
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5’capping
• The 5’end of the RNA requires a cap which is 7 methyl-guanosine triphosphate. It is attached by a 5’to 5’triphosphate linkage.
• A. reaction needs an enzyme called guanyl transferase.
• B. function of this cap is to facilitate the initiation of translation, and protects the 5’end of mRNA from attack by 5’to 3’exonucleases.
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Addition of poly (A) tail.
• mRNA require almost 40 to 200 adenine nucleotides added at 3’terminus to form a poly adenine (A) tail:
• poly A polymerase . Enzyme for the reaction.• This tail protects the 3’end of mRNA form 3’to
5’ exonuclease attack.
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Splicing (removal of introns)
• hnRNA is formed many pieces of some of them (exons) will be translated into amino acids.
• Others (introns) will be not be translated into amino acids and must be removed before translation takes place.
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spliceosomes
• Responsible for removal of introns from the hnRNA , and splicing (ligation) of both ends of exons to form mature mRNA.
• It consist of primary hnRNA ,4 small nuclear RNA’s (U1,U2,U5, and U4/U6) and an undetermined number of proteins.
• Smaller nuclear RNA’s (snRNA) is to bind each end of the introns by forming base pair with each other.
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cont.
• Spliceosomes facilitate also the transport of mature mRNA from the nucleus to the cytoplasm.
• Defect in the process of spilicing may lead to disease e.g at least one from of beta thalassemia , a disease in which there is absent synthesis of beta chain of hemoglobin appears to result from a nucleotide change at an exon-introns junction.
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Transfer RNA(t RNA)
• 1.It serves as a adapter mol. For the translation of mRNA into protein sequence.
• 2. primary tRNA transcript are subjected to many modifications:
• A. tRNA mol are transcribed as larger precusors. These precursors are reduced in size by specific class of ribonucleases.
• B. Attachment of the characteristics C.C.A terminus at the 3’end of the molecules.
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Cont.
• The tRNAs contain many modification of the standard bases A,U,G and C .Some bases are methylated, alkylated, or attached to carbohydrate residue by glucosidic bonds.
• some tRNA contain near to the anticodon loop a single intron 10-40 nucleotides long. These introns are removed with splicing ofr exons to produce an active tRNA for protein synthesis.
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Ribosomal RNA(rRNA)
• 1.In mammals cells ;rRNA is transcribed as a single larger precursor molecules called 45s.
• 2.In the nucleus ,45s is methylated and cleaved by specific endonucleases and exonucleases to give four kind of r RNA : 5SrRNA,5.8SrRNA,18SrRNA and 28SrRNA.
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Cont.
• 3. The four kind rRNA combine with a number proteins to form a ribosomes.
• 1. ribosomes are cytoplasmic nucleoproteins composed of 4 rRNAs plus a number of proteins.
• 2. These rRNAs and proteins are distributed specifically between the two smaller and larger ribosomal subuints.
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Cont.
• 3. The smaller subunits is called 40S .It contains one 18S rRNA and 33 proteins.
• 4. The larger subunit is called 60S . It contains the remaining r RNAs (28S,5.8S and 5S) and 45 proteins.
• 5. Both smaller (40S) subuint form the whole 80S ribosome.
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Function of ribosomes
• They are site of protein synthesis within cells.