TRANSCRIPTION. Renáta Schipp
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1 TRANSCRIPTION Renáta Schipp
2 Gene expression Gene expression: - is the process by which information from a gene is used for the synthesis of gene products. These products are proteins, but in the case of non-protein coding genes the products are functional RNAs. - has several steps - during gen expression the genotype gives rise to the phenotype (observable trait)
3 Levels of gene expression
4 The central dogma of molecular biology the flow of information in the cells
5 Transcription General features: Replication-Transcription Replication Transcription Symmetry Enzym Proofreading Frequency Direction Starting point Primer
6 Transcription General features: Replication-Transcription Replication symmetrical Transcription asymmetrical Symmetry DNA-polymerase DNA-dependent RNApolymerase Enzym Proofreading 1/ cell cycle /cell cycle Frequency Direction origo promoter Starting point Primer
7 Transcription General features: Prok. vs. Euk. Prokaryotes Eukaryotes chromosome-polysome complex coupled transcription-translation transcription and translation are chronologically and areal separated
8 Transcription General features: Prok. vs. Euk. Prokaryotes coupled transcription-translation no nucleus no maturation same directions: transcription: 5-3 translation: 5-3 Eukaryotes transcription and translation are separated from 1 promoter more genes from 1promotor 1gen 1 RNA polymerase 3 RNA polymerases
9 General features of transcription DNA serves as a template enzyme: RNA polymerase (forms 3-5 phosphodiester bonds) the direction of the synthesis is 5-3 from the point of view from the new RNA molecule!!! (but from the point of view of DNA it is 3 to 5 ) substrate: ribonucleoside-triphosphate β (beta) and γ (gamma) phosphate groups are released during the synthesis bases in RNA: A, U, G, C (no T) 3 steps: initiation, elongation, termination
10 Prokaryotic transcription: promoter A promoter is a region of DNA that initiates transcription of a particular gene. Promoters are located upsteram of the genes they regulated. consensus sequences: are conserved, are located at the -10 and -35 regions upstream of the initiation site, are similar across all promoters
11 Prokaryotic transcription: promoter
12 Prokaryotic transcription: RNA polymerase α: interaction with regulatory proteins β: initiation, elongation β : DNA binding σ: promoter recognation
13 Prokaryotic transcription: initiation
14 Prokaryotic transcription: elongation
15 Prokaryotic transcriptin: termination Rho-factor dependent Rho-factor independent
16 prokaryotische Transkription: Termination Rho-factor dependent Rho-Factor independent Rho-factor
17 Prokaryotic transcription: RNA processing RNA processing characteristic of trna and rrna - nucleolytic cleavages (endonucleases, exonucleases) - nucleotide addition (e.g. trna) - nucleoside modification (e.g. methylation)
18 GENETIC CODE Renáta Schipp
19 The central dogma of molecular biology the flow of information in the cells
20 Definition of the genetic code genetic code is the nucleotide sequence on DNA ( and subsequently on mrna by transcription) which determines the sequence of the amino acids in proteins the code is composed of codons codons are on mrna, anticodons on trna each codon consists of three nucleotides 20 amino acid 4 2 =16 combinations 4 3 =64 combinations
21 Features of the genetic code 61 sense codons amino acids 3 nonsense codons stopcodons the code is redundant (degenerate) more than needed = one amino acid can be coded by more than one codons (Leucin: 6, Glycin: 4, Tryptophan: 1) the code is unambiguous = a codon codes for a single amino acid the code is universal = highly conserved in the living world exceptions (e.g. mitochondrial genetic apparatus)
22 Features of the genetic code the code is continuous = no punctuations, no overlapping
23 wobble = uncertainty at the third codon position usually, the first two nucleotides are enough for the translation of the amino acid in the first two positions the pairing of the codon-anticodon is specific in the 3. position the pairing is less specific
24 Reading frame open reading frame (ORF) monocystronic and polycystronic mrnas polycistronic mrna from prokaryotes (A.); monocistronic mrna from eukaryotes (B.). (I = initiation codon, S = stop codon, ORF = open reading frame)
25 How to translate a codon A table (or dictionary) can be used to translate any codon sequence. Each triplet is read from 5 3 direction on the mrna
26 How to translate a codon
27 Genetic code excersise Given the following DNA sequence, transcribe it into mrna, then translate this sequence into protein. 5 -ATGTTTTATGCACGTACGGAGCTTCGGTAG-3 DNA
28 Results 5 -ATGTTTTATGCACGTACGGAGCTTCGGTAG-3 DNA (coding strand) 3 -TACAAAATACGTGCATGCCTCGAAGCCATC-5 DNA (template strand) 5 -AUGUUUUAUGCACGUACGGAGCUUCGGUAG-3 mrna Met Phe Tyr Ala Arg Thr Glu Leu Arg stop
29 Modification 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAG ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUC UAU GCA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met? Tyr Ala Arg Thr Glu Leu Arg stop 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (template strand)
30 Silent mutation 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAG ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUC UAU GCA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met Phe Tyr Ala Arg Thr Glu Leu Arg stop Met Phe Tyr Ala Arg Thr Glu Leu Arg stop original amino acid sequence no change in amino acid sequence!
31 Modification 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CCT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUU UAU GGA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met Phe Tyr? Arg Thr Glu Leu Arg stop 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (template strand)
32 Missense mutation 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CCT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUU UAU GGA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met Phe Tyr Gly Arg Thr Glu Leu Arg stop Met Phe Tyr Ala Arg Thr Glu Leu Arg stop original amino acid sequence different amino acid sequence!
33 Modification 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATT CGT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUU UAA GCA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met Phe? Ala Arg Thr Glu Leu Arg stop 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (template strand)
34 Nonsense mutation 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATT CGT GCA TGC CTC GAA GCC ATC-5 DNA (mutated template strand) 5 -AUG UUU UAA GCA CGU ACG GAG CUU CGG UAG-3 mutated mrna Met Phe stop Met Phe Tyr Ala Arg Thr Glu Leu Arg stop original amino acid sequence polypeptide synthesis stop!
35 Modification(insertion) 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) Insertion 3 -TAC AAA TAT ACG TGC ATG CCT CGA AGC CAT C-5 DNA (template strand) 5 -AUG UUU AUA UGC ACG UAC GGA GCU UCG GUA G-3 mrna Met Phe?????????????????????????????????????? 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) 3 -TAC AAA ATA CGT GCA TGC CTC GAA GCC ATC-5 DNA (template strand)
36 Frameshift mutation 5 -ATGTTTTATGCACGTACGGAGCTTCGGTAG-3 DNA (coding strand) Insertion 3 -TACAAATATACGTGCATGCCTCGAAGCCATC-5 DNA (template strand) 5 -AUGUUUAUAUGCACGUACGGAGCUUCGGUAG-3 mrna Met Phe Ile Cys Thr Tyr Gly Ala Ser Val Met Phe Tyr Ala Arg Thr Glu Leu Arg stop original amino acid sequence major difference in amino acid sequence!!!!!!!!!!!!!!!!!!!
37 Modification(deletion) 5 -ATG TTT TAT GCA CGT ACG GAG CTT CGG TAG-3 DNA (coding strand) Deletion 3 -TAC AAA A AC GTG CAT GCC TCG AAG CCA TC-5 DNA (template strand) 5 -AUG UUU U-UG CAC GUA CGG AGC UUC GGU AG-3 mrna Met Phe??????????????????????????????????????
38 Frameshift mutation 5 -ATGTTTTATGCACGTACGGAGCTTCGGTAG-3 DNA (coding strand) Deletion 3 -TACAAAA -ACGTGCATGCCTCGAAGCCATC-5 DNA (template strand) 5 -AUGUUUU-UGCACGUACGGAGCUUCGGUAG-3 mrna Met Phe Leu His Val Arg Ser Phe Gly Met Phe Tyr Ala Arg Thr Glu Leu Arg stop original amino acid sequence major difference in amino acid sequence!!!!!!!!!!!!!!!!!!!
39 Summary of mutations
40 Thank you for your attentions!
41 TRANSLATION Renáta Schipp
42 The central dogma of molecular biology the flow of information in the cells
43 Ingredients of translation ribosomes
44 Ingredients of translations transfer RNA: trna
45 Ingredients of translation messenger RNA: mrna Polycistronic mrna from prokaryotes (A.); monocistronic mrna from eukaryotes (B.). (I = initiation codon, S = stop codon, ORF = open reading frame)
46 Ingredients of translation amino acid: 20 essential amino acid synthesis of aminoacyl-trna by aminoacyl-trna synthases step 1 - activation of amino acid amino acid + ATP aminoacyl-amp + pyrophosphate step 2 - formation of aminoacyl-trna aminoacyl-amp + trna aminoacyl-trna + AMP
47 Steps of translation Initiation A: binding of initiation factors to the small subunit B: assembly of 30S initiation complex C: assembly of 70S initiation complex. 30S initiation complex = mrna + 30S subunit + fmet-trna + initiation factors 70S initiation complex = 30S initiation complex + 50S subunit P site: fmet-trna A site: empty
48 Steps of translation Elongation A. Aa-tRNA binding with the help of EF-Tu GTP to A site B. peptide bond formation peptidyl transferase (ribozyme) C. translocation with the help of EF-G GTP
49 Steps of translation Termination stop codon: dose not code for any amino acid releasing factors: release of ribosomal subunites, mrna, polypeptide chain
50 Unique features of eukaryotic translation 5`-cap ribosome binding ribosomes - free ribosomes proteins of cytosol, nucleus, mitochondria - bound ribosomes secretory proteins, proteins of endoplasmic reticulum, Golgi complex, lysosomes, cell membrane
51 General features of translation - direction: 5`end 3`end on mrna N terminus C terminus on polypeptide chain - polysomes are formed - one polypeptide chain/ribosome - requires 4 high energy bonds/peptide bonds (ATP AMP, 2GTP 2GDP) for amino acid activation: ATP AMP=2 high energy bonds for Aa-tRNA binding (EF-Tu) GTP GDP= 1 high energy bond for translocation (EF-G) GTP GDP= 1 high energy bond
52 Inhibitors of protein synthesis chloramphenicol peptidyl transferase erythromycin translocation tetracyclin aa-trna binding streptomycin 30S subunit puromycin early termination Antibiotics are a group of medicines that are used to treat infections caused by some prokaryotes.
53 Thank you for your attentions!
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