IC15 Central Dogma - PDF

IC15: Central Dogma A/Prof Chew Eng Hui Department of Pharmacy Faculty of Science [email protected] Learning Outcomes: To describe the process of DNA replication and compare DNA synthesis on leading and lag...

IC15: Central Dogma A/Prof Chew Eng Hui Department of Pharmacy Faculty of Science [email protected] Learning Outcomes: To describe the process of DNA replication and compare DNA synthesis on leading and lagging strands. (Part 1). To describe the process of transcription (Part 2). To describe the process of translation (Part 2). To gain an overview of the common post-translational modifications that occur (Part 2). 2 What is Central Dogma? The DNA of a gene is transcribed to produce an RNA molecule that is complementary to the DNA. The RNA sequence is then translated into the corresponding sequence of amino acids to form a protein. These transfers of biological information from gene to protein is termed the Central Dogma of Molecular Biology. 3 What is Central Dogma? Replication: DNA replication yields two DNA molecules identical to the original one, ensuring transmission of genetic information with high fidelity. Transcription: Information encoded in the nucleotide sequence of DNA is transcribed through synthesis of mRNA whose sequence is dictated by the DNA sequence (template). Translation: As the sequence of mRNA is read (as groups of three consecutive nucleotides) by the protein synthesis machinery (tRNA, rRNA, other proteins), it is translated into the sequence of amino acids in a protein. 4  Part 1 DNA Replication https://www.youtube.com/watch?v=TNKWgcFPHqw Take note of the enzymes mentioned. They are all needed for DNA replication to take place and complete with high fidelity. 5  Cell Cycle (Mitosis) Prokaryote cell division (by binary fission) Eukaryote cell division Nucleic acids: are hereditary material involved in storage and transmission of genetic information. Before a cell divides, it has to replicate its DNA in high fidelity so that the genetic information is transmitted to the next generation of cells. Cell Cycle: G0, resting stage – nondividing cells; G1 (gap 1) phase – cell enlarges and makes new proteins; S phase – DNA synthesis; G2 (gap 2) phase – cell prepares to divide; M phase (mitosis) – cell divides into 2 new daughter cells. Daughter cells can reenter G1 or stay at G0 stage. 6  DNA Replication – Origin of replication (ori) Is a stretch of DNA sequence in the genome where DNA replication is initiated. In prokaryotes, usually a single origin of replication in a single circular DNA chromosome. In eukaryotes, many origins of replication on each chromosome; activated during S phase of the cell cycle. Eukaryote cell division Cell Cycle: G0, resting stage – nondividing cells; G1 (gap 1) phase – cell enlarges and makes new proteins; S phase – DNA synthesis; G2 (gap 2) phase – cell prepares to divide; M phase (mitosis) – cell divides into 2 new daughter cells. Daughter cells can reenter G1 or stay at G0 stage. 7  The Process of DNA Replication Replication proteins together with DNA polymerase are clustered at origin of replication (ori) to initiate DNA replication. 3’ 5’ 3’ 5’ Replication fork 3’ 5’ 3’ 5’ Replication fork 8  The Process of DNA Replication In DNA replication, primers are made up of RNA (~5 nts 3’ in length, complementary in 5’ sequence to the As helicase continues to unwind DNA). 3’ upstream, another primer synthesized by primase guides Replication fork synthesis of another DNA 5’ fragment. 3’ 5’ 3’ 5’ 3’ 5’ 3’ Polymerase complex acts processively: 5’ once polymerization starts, complex Replication fork; helicase does not dissociate after each act of Earlier primer synthesized by continues to unwind DNA attaching a nucleotide. Chain elongation primerase guides synthesis of continues. double helix 1 DNA fragment 9  The Process of DNA Replication RNase H; to remove RNA primers 5’ 3’ 3’ 5’ DNA ligase; to join fragments of DNA by forming 1 nucleotide bond 10  I have heard before terms like “leading strands” and “lagging strands” formed during DNA replication. Then someone told me about continuous and discontinuous strands… What are they? Are they different? I can clarify! Recall DNA polymerase can read both parental strands as templates and synthesize a daughter strand that is complementary base pairing to the parental template strand? One daughter strand is synthesized as the continuous/leading strand, while another daughter strand is synthesized as the discontinuous/lagging strand. 11  Synthesis of continuous and discontinuous strand Sliding 3’ clamp DNA replication is continuous for one strand 5’ 3’ polymerase and discontinuous for another strand. 3’ 5’ primase 5’ Both Because template DNA strands synthesismove adds through the replication factory in nucleotides only at the 3” end of the the same direction, and DNA elongating DNA strand, DNA polymerase can only synthesize polymerase cannot synthesize the DNA from the 5’ end to the 3’ end. lagging Due strandtwo infactors, to these one one long continuous can polymerase piece as it on remain does for the its DNA leading strand template → the and copy the lagging DNA in strand one is synthesized in small fragments continuous strand while the other called OkazakiDNA complementary fragments (1000- strand must 5000 be basediscontinuously made pairs in length).in short pieces which are later joined together. 12  Synthesis of continuous and discontinuous strand Sliding The Discontinuous Strand 3’ clamp The other polymerase synthesizing the discontinuous strand can only copy a short stretch of DNA before it 5’ 3’ polymerase runs into the primer of the previously sequenced fragment. 3’ 5’ 5’ It is therefore forced to repeatedly release the DNA strand and slide further upstream to begin extension from another RNA primer. The sliding clamp helps hold this DNA polymerase onto the DNA as the DNA moves through the replication machinery. polymerase primase 13  Synthesis of continuous and discontinuous strand 3’ Leading strand, Lagging 5’ 3’ strand, Okazaki fragments & removal of primers 3’ 5’ polymerase 5’ The continuously synthesized strand is known as the leading strand, while the strand that is synthesized in short pieces is known as the lagging strand. The short stretches of DNA that make up the lagging strand are known as Okazaki fragments. Before the lagging-strand DNA exits the replication factory, its RNA primers must be removed and the Okazaki fragments must be joined together to create a continuous DNA strand. polymerase The first step is the removal of the RNA RNAse H primer. RNAse H, which recognizes RNA- DNA hybrid helices, degrades the RNA by hydrolyzing its phosphodiester bonds. 14 Synthesis of continuous and discontinuous strand The replacement of RNA primer with polymerase DNA by DNA polymerase and joining of Okazaki fragments by ligase transform the fragmented lagging strand to a continuous strand. Next, the sequence gap created by RNAse H is then filled in by DNA polymerase which extends the 3’ end of the neighboring Okazaki fragment. Finally, the Okazaki fragments are joined together by DNA ligase that hooks together the 3’ end of one fragment to the 5’ phosphate group of the neighboring fragment in an ATP- dependent reaction. 15 Conclusion: DNA Replication is bidirectional DNA replication is bidirectional with replication forks moving in opposite directions: DNA polymerase moves along each template strand at the replication fork in the 3’ to 5’ direction and “reads” it to determine the correct base to be added to the growing strand. Synthesis of daughter strand is 5’ to 3’. 3’ 5’ 3’ 5’ 5’ 3’ Leading strand Lagging strand synthesized synthesized 5’ 3’ Replication fork Replication fork unwinds  unwinds →→→ direction direction 3’ 5’ Lagging strand synthesized Leading strand Synthesized` 16  DNA Topoisomerases We are not quite done with DNA replication. There is a group of enzymes called topoisomerases that are important for DNA replication. These enzymes are often targets of therapeutic strategies. 17  DNA Topoisomerases Are enzymes that act on DNA to reduce supercoiling stress caused by unwinding during replication and transcription by making transient single-strand (Type I) or double-strand (Type II) breaks. Possess both nuclease and ligase activities. Essential in regulating topological state of genetic material during DNA replication, transcription, DNA repair, chromatin remodelling. A second strand passes through 18  Part 2 Transcription 19  (~15%) (~80%) (~5%) DNA Large Store Genetic Information (chromosome; Transcription is the process of plasmid) synthesizing RNA copies 6 complementary to gene sequences on DNA. Besides mRNA synthesis, it also includes rRNA and tRNA synthesis. 20  Transcription in prokaryotes 1. Initiation RNA polymerase binds to promoter 2. Elongation RNA polymerase unwinds small section of DNA (forms transcription bubble), uses template strand to synthesize RNA copy (nucleotide triphosphate NTP as substrate) RNA copy is synthesized 5’ to 3’ end, antiparallel to template DNA strand (3’ to 5’ end). Same like DNA polymerase (daughter DNA strand synthesized 5’ to 3’). 3. Termination Rho-dependent or Rho-independent. 21  Promoter: starting point of transcription Strong promoter versus weak promoter? -35 region: TTGACA σ factor & RNA sequence located ~35 polymerase bind to σ factor: a single general TATA box to initiate nucleotides upstream of transcription factor in initiation site transcription prokaryotic transcription The nucleotide sequences of representative E. coli promoters. -10 AT-rich The consensus sequences are recognized by RNA Polymerase. region/Pribnow/TATA box: Transcription start point sequence TATAAT of 6 nucleotides centered roughly 10 nucleotides upstream from initiation site +1: initiation site/ transcription start point (1st nucleotide to be transcribed into RNA) In accordance with convention, sequences of the non-template strand where RNA 22  Elongation (prokaryotes) Rapid (~40 nt per second) and simple process RNA polymerase moves along the anti-sense/template strand. If incoming NTP that binds to entry site on the RNA polymerase matches the next base on the DNA template, it is transferred to the polymerase’s active site and a new phosphodiester bond is formed. Meanwhile, double-stranded DNA continues to unwind to expose template strand for elongation to continue. Topoisomerase I and II required in transcription? 23  Termination (prokaryotes) Rho-independent A stem-loop/hairpin structure is formed just upstream of a series of Chain U residues Termination (6-8) located near 3’ (Prokaryote) end ofChain the RNATermination transcript. (Prokaryote) The inverted repeats give rise to a stem-loop, or “hairpin,” structure The invertedending in a give repeats series ofto rise Uaresidues in the stem-loop, or RNA transcript. “hairpin,” Hairpin The hairpin structure structure causes causesofthe RNA RNA polymerase to pause and structure ending in a series U residues in the RNA transcript. polymerase result The hairpin to pause in dissociation structure of the causes and thedislodge transcript from the DNA. RNA polymerase to pause and from result the DNA template. in dissociation of the transcript from the DNA. 5’ Template strand 5’ Template strand 24  Termination (prokaryotes) Rho-dependent RNA transcript contains a binding site for specific protein called rho factor → rho factor slides along RNA chain towards bubble (chases after RNA Pol but is slower than RNA Pol). ATPase-dependent Rho factor has ATP-dependent helicase activity → unwinds RNA transcript from DNA template as it moves along. At rho termination site, Rho factor terminates transcription when it catches the RNA Pol and displaces it from the template. 25  Transcription in eukaryotes How is(takes More complicated than prokaryotes Transcription place in nucleusRegulated vs in cytosol in prokaryotes). in Eukaryotes? How is Transcription Regulat 3 RNA polymerases exist: RNA Pol II for mRNA insynthesis. More complicated than prokaryotes Eukaryotes? Synthetic process (three stages) reorganization Require same as prokaryotes. More complicatedof chromatin for access of than prokaryotes Eukaryotic promoters same as prokaryotic regulatory to promoters: proteinsreorganization Require containfor access promoters.of chromatin TATA box (-26 region) and initiation In addition to site (+1) regulatory sequence. proteins promoters, In addition, to promoters. eukaryotic genes have upstream elements eukaryotic genes have upstream In addition/ to promoters, activation elements/activation eukaryotic sequences. sequences.genes hav upstream elements / activation sequences. Elements Elements of RNA Pol II Promoter of RNA Pol II Promoter RNA Pol II: complex, multiunit regulated enzyme that cannot discriminate between genes, promoters, (Consensus (Consensus and random DNA → (Inr: (Inr: Inr: Initiation Sequences Sequencescontain transcription Inr: Initiation el element contain transcri (Responsive elements) Recognize by interaction bet RNA Pol II (Responsive elements) Recognize by RNA Pol II)start site) start site) RNA Pol II) and promoter mediated by TATAAA other protein factors. What about prokaryotic RNA Pol? 26  Initiation of transcription in eukaryotes At the initiation site (Inr) of the promoter, 1. TATA box-binding protein (TBP) binds to TATA box sequence (5'- TATAAA-3'), unwinds TheDNA helix and assembly of binds DNA dramatically basal (~80). / general (1) TBP: TATA-binding transcription TBP part of transcription factor protein factors complex TFIID → TFIID bindsrequired to TATA box for the initiation of transcription by (2) TAF: TBP associated 2. Transcription factors TFIIB and TFIIA RNA polymerase II. factors next bind. 3. RNA polymerase Watch II (RNA Pol II) (3) recognizes multi-protein complex → ‘Transcription binds Assembly’ on IVLE animation. 4. TFIIE, TFIIF and TFIIH bind → complete assembly of a stable pre- (4) initiation complex Role of transcription factors (have affinity to bind to promoters): help direct RNA Pol II to bind and hold it in proper position to initiate transcription Pre-initiation complex 27  Transcription in eukaryotes DNA looping: permits multiple activator proteins to bind to Formation enhancersof DNA loops (DNA allows specific sequences) transcription located factorsfrom at a distance or activators the that bind to enhancers at a distance from the promoter to interact with regulatory promoter to interact with the regulatory proteins in the pre- proteins in the pre-initiation complex and to maximize transcription. initiation complex → maximize transcription. What happens to transcription if TATA box DNA looping permits multiple activator proteins to bind to multiple is missing? upstream DNA sequences with promoter. Example: Activation of transcription via CREB and CBP through DNA looping 28  Transcription in eukaryotes A complex of 85 proteins or more can be involved in eukaryotic transcription. Eukaryotic cells are more D complex than prokaryotic cells. For ch them to make quick responses to ce changes in the microenvironment they are exposed to, eukaryotic cells rely 1. on not only basal transcription factors, but also additional transcription factors, 2. coactivators, activators and/or repressors to exert regulatory control over expression of genes. 3. Depending on the outcome of such control, we called it either rantscriptional activation T or transcriptional inactivation. p 29 mRNA produced in prokaryotes and eukaryotes mRNA in prokaryotes Prokaryotic mRNAs are polycistronic mRNA in eukaryotes 30  Post-transcriptional processing of mRNA in eukaryotes In the nucleus, the newly synthesized RNA transcript (pre-mRNA) is further modified to mature mRNA for translation. 1. The 5’ end receives a cap of 7- methylguanosine (m7G). Capped mRNA are resistant to 5’-exonucleolytic degradation. 2. A run of 100-200 A residues called a poly(A) tail is added to the 3’ end. This prevents rapid degradation and increases mRNA half-life. 3. Removal of introns and joining of exons (splicing) to form continuous coding sequence for translation. Mature mRNA is exported into cytoplasm for subsequent translation into polypeptide. 31 Translation 55 32 Overview of Translation from mRNA transcript An mRNA transcript is produced from a gene region of DNA. → An mRNA transcript is produced from a gene region of DNA. The nucleotide Nucleotide sequences translated to amino acid sequences by a translation sequences will be translated to amino acid sequences by a translation machinery reading machinery three reading nucleotide three sequences nucleotide sequences as oneas onewhich codon codon which specify specifyone amino one acid following amino a genetic a acid following code (geneticcode genetic dictionary). (genetic dictionary). DNA mRNA mRNA codons amino acids threonine proline glutamate glutamate lysine 4 33  The Genetic Code Each codon made up of three nucleotides is found within the coding region of the mature mRNA; each codon continuous with another and each codes for an amino acid. A total of 64 codons (43) make the entire genetic code. All codons have meaning. Of the 64 codons, 61 specify particular amino acids. Remaining 3: UAA, UAG, UGA specify no amino acids as they are the stop/termination codons (do not encode an amino acid). Genetic code is degenerate. Every amino acid is coded by more than one codon (exception: Met, Trp). Eg: Arg, Leu and Ser are represented by 6 different codons. Genetic code is almost 'universal’; similar in prokaryotes and eukaryotes (with few exceptions). 34  Wobble Hypothesis Discovered by Francis Crick. States that rules of base pairing are relaxed at the third position so that a base can pair with more than one complementary base. → sufficient coding performed by the first 2 bases in the codon (bond strongly to the bases in the anticodon) while the third (wobble) base plays a minor role. 35  The Genetic Code Codons representing same amino acid or chemically similar amino Degeneracy of acids the code tendisto a be similar against ‘buffer’ in sequence. mutational disruption. Third-baseItdegeneracy minimizes the feature (third base damage is less caused relevant for by misreading of coding). the Example: code. Eg. If CUU is misread as CUC,GGX CUAspecifies or CUG Glyduring transcription, UCX specifies Serine it will still be translated GA-py specifies Asp protein as Leu during synthesis. GA-pu specifies Glu (both negatively-charged) CodonThebias: degeneracy tendency of ofgenomes the codetois a ‘buffer’ prefer one codonagainst mutational for an amino acid over disruption. all others. It minimizes G & C rich the genome preferred. damage caused by misreading of the code. Eg. If CUU is misread as CUC, CUA or CUG during transcription, it will still be translated as Leu during protein synthesis. 36 tRNA and Ribosomes are needed in translation Recap A/Prof Wu Wei’s slides in IC6 37  The genetic code First adaptor: Enzyme aminoacyl-tRNA synthetase - couples a particular amino acid to its corresponding tRNA (step also is translated by known as amino-acid activation/amino acid is charged). two adaptors (act Second adaptor: tRNA molecule; its anticodon forms base one after pairs with the appropriate codon on the mRNA. An error in either step would cause the wrong amino acid to another). be incorporated into a protein chain. Active site of enzyme Anticodon binding domain of enzyme anticodon codon on on tRNA mRNA ATP used (energy utilizing) Example: Sequence of events involving amino acid tryptophan (Trp) being selected by the codon UGG on the mRNA. Note: Most organisms have one aminoacyl-tRNA synthetase for each amino acid. 38 Steps involved in Translation Process - Initiation Eukaryotes Small ribosomal subunit forms complex with eukaryotic initiation factors (eIF)(~ 12 eIFs). eIFs specifically recognize initiator tRNA (Met-tRNAMet) → forms mRNA-bound eIF/Met- eIF/Met-tRNA/small subunit complex. tRNA/small ribosomal subunit complex travels to reach AUG eIF in eIF/Met-tRNA/small subunit start codon complex binds to the 5’ cap in mRNA. mRNA-bound eIF/Met-tRNA/small ribosomal subunit complex travels along 5’ untranslated end of mRNA until it reaches the first AUG codon (“start codon”). eIFs are released (ATP-dependent); large ribosomal subunit joins small subunit/Met- tRNA/mRNA complex; Met-tRNA directed Eukaryotic ribosomes to P site. (80S) consist of 60S and 40S subunits 39 Steps involved in Translation Process - Initiation Prokaryotes Generally the same as eukaryotes except: Number of initiation factors involved: 3 The initiation complex: IF/Met-tRNA/small ribosomal subunit complex forms at the Shine-Dalgarno sequence just 5’ to the coding region in the mRNA. Note and recall: prokaryotic mRNAs do not have 5’ m7G cap and 3’ polyA tail Prokaryotic ribosome (70S): Eukaryotic ribosome (80S) 50S + 30S subunits : 60S + 40S subunits 40  Requirement for Ribosomal binding In prokaryotes, ribosomal binding site (RBS) called the Shine-Dalgarno sequence: - purine-rich sequence - sequence lies ~10 nts upstream from the AUG start codon. Prokaryotic (Shine-Dalgarno sequence) +1 5’ – AGGAGGACAGCUAUG – 3’ RBS SPACER INITIATOR In eukaryotes, RBS is either the 5’ cap of mRNA or an internal ribosome entry site (IRES) found along the length of mRNA. In eukaryotes, the Kozak sequence on the mRNA is also recognized by the ribosome as translational start site Eukaryotic (Kozak sequence) +1 5’ – A/GCCACCAUGG – 3’ Eukaryotic RBS/IRES Kozak INITIATOR upstream of Kozak sequence 41 Steps involved in Translation Process - Elongation https://www.youtube.com/watch?v=km 1. On assembled ribosome, rUzDYAmEI Good illustration of tRNA carrying 1st amino acid elongation step in (Met-tRNA) pairs with start translation codon on mRNA in P site. mRNA passes through A 2nd tRNA carrying 2nd amino acid approaches and sits in A ribosomal units. site. tRNAs deliver amino acids to the RBS in the order specified 2. Peptide bond formation by the mRNA. The 1st amino acid is attached to 2nd amino acid by a peptide Peptide bonds form between bond catalyzed by peptidyl the amino acids and the transferase. polypeptide chain elongates. 3. Translocation Elongation process assisted Once peptide bond is formed, by proteins called elongation elongation factor moves ribosome down one codon on factors. the mRNA (5’ to 3’ direction → A site vacant, P site occupied Peptide chain grows in the by 2nd tRNA, E site occupied amino-to-carboxyl direction, by uncharged tRNA (not with ribosome reading the carrying Met anymore), followed by exit of tRNA from mRNA in 5’ to 3’ direction. E site. 42  Steps involved in Translation Process - Termination Stop codon (UGA, UAA, UAG) into place. No tRNA with anticodon to complement with stop codon. Release factors recognize stop codons → bind to the ribosome → promote hydrolysis of peptidyl-tRNA link → polypeptide chain released from mRNA. P site A site 43  Post-translational modification in Eukaryotes Recall: Post-translational Can you recall which modifications for proteins organelle do proteins only occur in eukaryotes, not undergo post-translational prokaryotes. modifications? You can revise slides in A/Prof Wu Wei’s IC2 to recap post-translational modifcations 44  References Principle of Biochemistry by Voet, Voet and Pratt, 4th edition Biochemistry for the Pharmaceutical Sciences by Charles P. Woodbury, Jr. 45