Transcription and Translation 2024 PDF

Central Dogma Transcription and Translation David P. Gardner, Ph.D. M2P Course 2024  Objectives Transcription 1. Summarize the central dogma of molecular biology. 2. Compare and contrast the funct...

Central Dogma Transcription and Translation David P. Gardner, Ph.D. M2P Course 2024  Objectives Transcription 1. Summarize the central dogma of molecular biology. 2. Compare and contrast the function of a DNA dependent RNA polymerase with a DNA dependent DNA polymerase. Associate one of these with de novo synthesis. 3. Describe how prokaryotic RNA polymerase recognizes a promoter region and knows which strand is the template. 4. Describe the steps involved in transcription from recognition of a promoter to initiation of RNA synthesis. 5. Diagram and explain how transcription stops utilizing a termination signal. 6. Describe the organization and function of a 5’ cap and a 3’ polyA tail on an mRNA. 7. Describe the process of intron splicing including the proteins and RNAs involved in each step. 2  Objectives Translation 1. Describe how tRNA aminosynthases link tRNAs and amino acids and why they need to be highly accurate proteins. 2. If given the genetic code and an mRNA sequence, be able to generate a correct amino acid sequence based on the mRNA. 3. Explain how “wobble” allows 4 different codons to specify arginine. 4. Describe the steps and proteins involved in recognition of the first AUG codon in an mRNA molecule. Outline the role of GTP and ATP hydrolysis in the process. 5. Diagram and explain the role of eEF1⍺ and GTP hydrolysis in elongation. 6. List the stop codons and explain how release factor triggers translation termination. 3  Recommended Readings Lippincott Illustrated Reviews: Cell and Molecular Biology, 3e, Chapters 8 and 9 4 Transcription  Central Dogma The central dogma of molecular biology is basically the idea that DNA produces RNA, which produces protein. It was thought to be unidirectional but the discover of reverse transcription (RNA to DNA) changed that thinking. Another component is that DNA can be replicated to yield a copy of the original (DNA to DNA). FIGURE 4.1 From DNA to protein MBoC 6th Edition 6  Central Dogma Today we will focus on the mechanisms of DNA to RNA and RNA to protein. Transcription = DNA to RNA. Translation = RNA to protein. 7  Transcription FIGURE 8.1 E. coli RNA polymerase Much of our basic understanding of The Cell: A Molecular Approach, 8th Edition transcription comes from prokaryotic systems like E. coli. Process is similar in eukaryotes. The key enzyme in the process is a DNA dependent RNA polymerase. Uses DNA as the template for synthesis of _____. DNA polymerase is a DNA dependent The enzyme consists of multiple DNA polymerase. subunits with di erent functions. Uses DNA as the template for synthesis of DNA. 8 ff  Transcription A key question in transcription is how does RNA polymerase know where to start and on which of the two strands. This function requires the sigma ( ) subunit of the polymerase and specific sequences within the promoter region of a gene. Key elements in the promoter include the ____ site and the -10 site. FIGURE 8.2 Sequences of E. coli promoters The Cell: A Molecular Approach, 8th Edition 9 𝜎  Transcription Initiation RNA polymerase first binds non-specifically to the DNA. It moves along the DNA using sigma to look for a -35/-10 sequence. When found, it binds tightly to the DNA forming a closed promoter complex. Figure incorrectly labels the δ factor as ⍺. FIGURE 8.3 Transcription by E. coli RNA polymerase The Cell: A Molecular Approach, 8th Edition RNA polymerase next unwinds DNA in this region forming an open promoter complex. 10  Transcription Initiation RNA polymerase next unwinds DNA in this region forming an open promoter complex. 11  Transcription Initiation How does RNA polymerase know which strand to use as a template? Note that sigma factor must first see a -35 element followed by a -10 element. If it were traveling in the other direction, it would see a -10 first and then a -35. It does not recognize -10/-35 as a promoter region and will keep traveling. 12  Transcription Initiation RNA polymerase next joins two _____ dNTPs together and hydrogen bonds them to the template at the +1 site. +1 is the transcription start. This is a unique function of RNA polymerases that DNA polymerase lacks. This is called de novo synthesis. DNA polymerase can only extend a strand, not initiate a strand of DNA. 13  Transcription Elongation A er addition of about 10 nucleotides, the sigma factor is released, which allows the polymerase to begin moving down the DNA strand. Like all polymerases it can only synthesize in the 5' to 3' direction. It uses the 3' to 5' strand as the template for 5' to 3' synthesis. It reads 3’ to 5’ and creates 5’ to 3’ The 5' to 3' strand is moved out of the way. A er synthesis, the polymerase rewinds the two strands. 14 ft ft  Transcription Termination Downstream of the gene, within the DNA is a termination signal. Most common type in bacteria is a symetrical inverted repeat of a GC rich sequence followed by several A residues on the template strand. FIGURE 8.5 Transcription Termination The Cell: A Molecular Approach, 8th Edition The RNA polymerase reads through the GC rich regions incorporating it into the RNA. 15  Transcription Termination The GC rich sequences form a _____________ structure that causes the RNA polymerase to dissociate. The poly U sequence enhances the release of the RNA from the template. Why? In eukaryotic cells transcription termination is more variable. 16  Eukaryotic Transcription Prokaryotes get by with one RNA polymerase while eukaryotes have four for di erent types of RNAs. 3 nuclear (I, II and III) and one mitochondrial. Mitochondrial RNA polymerase is more similar to bacterial polymerase than any eukaryotic polymerase. Rifampcin inhibits bacterial RNA Rifampcin at high doses can inhibit polymerase. mitochondrial transcription. 17 ff  Regulation of Transcription Initiation Transcriptional control is a critical aspect of cell function in both eukaryotes and prokaryotes and will be consider in detail later in the course. FIGURE 8.8 RNA polymerase II/Mediator complexes and transcription initiation The Cell: A Molecular Approach, 8th Edition 18  RNA Processing Eukaryotic RNAs undergo extensive processing. Prokaryotic tRNA and rRNA also undergo processing but mRNA is rarely processed. We will focus on eukaryotic mRNA processing. Note that transcription and translation are compartmentalized in eukaryotic cells but not prokaryotic cells. This allows for extensive processing before translation begins. 19  mRNA Processing Three critical events in eukaryotic mRNA processing. 5' modification 3' modification Intron splicing We will look at each. 20  mRNA 5' Processing Once exported from the nucleus, translational machinery must e iciently _________ the beginning (5' end) of an mRNA. This end undergoes extensive modification to make it unique. The 5'-methylguanosine cap. FIGURE 8.14 Processing of eukaryotic messenger RNAs The Cell: A Molecular Approach, 8th Edition A unique guanosine (7- methylguanosine) is a ached 7-methylguanosine is a ached via a 5' to 5' bond that maintains the three phosphate to the first base of the mRNA. groups of GTP. This stabilizes the mRNA. The first couple RNA bases are also methylated. 21 ff tt tt  mRNA 3' Processing Eukaryotic mRNA transcripts lack a clearly defined ending and are variable, even from the same gene. They contain an AAUAAA 20-30 bp upstream of what will become the end of the mRNA. The site is recognized by an endonuclease and a polyA polymerase. The endonuclease cleaves at the downstream element. The polymerase adds a sequence of A residues ~200 bp long. PolyA tails regulate both stability FIGURE 8.15 Formation of the 3′ ends of eukaryotic mRNA and translation of mRNAs. The Cell: A Molecular Approach, 8th Edition Longer polyA tails = more stable transcripts. 22  mRNA Splicing Primary RNA transcripts in the nucleus are about 10 times longer than a final transcript in the cytoplasm. ~35% of the human genome is intron sequences. Most eukaryotic mRNAs contain introns and are spliced to remove the intron sequences before exporting from the nucleus. Splicing involves both catalytic ______ and proteins and can be di erentially regulated. 23 ff  Mechanism of Splicing Splicing involves a complex of proteins Small Nuclear RNAs (snRNAs) and RNAs called the spliceosome. U1 U2 U4 U5 U6 U1 snRNP contains the U1 snRNA plus Small Nuclear Ribonuucleoprotein Particles (snRNPs) proteins. U4/U6 snRNP contains both U4 and U6 snRNA. U1 U2 U5 U4/U6 Genes/mRNAs contain sequences that facilitate splicing: GT/GU at the beginning of an intron. AG/AG at the end of an intron. 5' and 3' splice sites. An A nucleotide ~3/4 of the way down the intron. FIGURE 8.17 Splicing of pre-mRNA The Cell: A Molecular Approach, 8th Edition 24  Mechanism of Splicing U1 snRNP binds the 5' SS via base pairing between the U1 snRNA and the mRNA. U2 then binds the same region and holds the intron in a looped conformation, bringing the GU and AG close together. Next a complex of U4/U6 and U5 binds. FIGURE 8.18 Assembly of the spliceosome The Cell: A Molecular Approach, 8th Edition 25  Mechanism of Splicing U1 and U4 leave the complex. Remaining snRNPs then utilize the branch point A to catalyze the excision and ligation of the intron. This releases the now spliced mRNA plus a lariate intron that is degraded. More than 200 human diseases can be caused by aberrant splicing. 26 Translation  Translation While transcription and translation are coupled in prokarytotes, they are sequestered into the nucleus and cytoplasm respectively by the nuclear membrane. Fundamental mechanisms of translation are highly conserved. Key elements include: Charged tRNAs Ribosomes mRNA template 28  Charged tRNAs Aminoacyl tRNA synthetases a ach amino acids to tRNAs. 20 di erent synthetases, one for each amino acid. tRNA synthetases are some of the most __________ enzymes within cells. Translation has no proofreading mechanism. If an incorrect amino acid is a ached to a tRNA, that improper amino acid will get incorporated. First step is to utilize two high energy phosphate bonds of ATP to generate an aminoacyl AMP. This bond between the AMP and the amino acid FIGURE 10.1 Attachment of amino acids to tRNAs holds a lot of energy. The Cell: A Molecular Approach, 8th Edition 29 ff tt tt  Charged tRNAs Next step joins the amino acid to the specific tRNA at the 3' end with the release of the AMP. The high energy bond is now between the tRNA and the amino acid. The tRNA is charged. FIGURE 10.1 Attachment of amino acids to tRNAs The Cell: A Molecular Approach, 8th Edition 30  Genetic Code The 3 bp anticodon of the tRNA hydrogen bonds with a 3 bp codon within the mRNA. 4 bases possible and 3 bases per codon yields 64 possible combinations for 20 amino acids. Thus the genetic code is ___________ in that more than one codon can specify the same amino acid. FIGURE 10.2 Nonstandard codon–anticodon base pairing The Cell: A Molecular Approach, 8th Edition 31  Wobble Further, some tRNAs can recognize more than one codon due to "wobble". Most redundancy is in the 3rd base of the codon due to increased allowance for hydrogen bonding partners. Codon Amino Acid Codon Result GCU Alanine UGA STOP U Go Away GCC Alanine UAA STOP U Are Away GCA Alanine Methionine is specified by UAG STOP U Are Gone GCG Alanine one codon and one tRNA. 32  Translation Initiation Translation initiation is a highly regulated process and involves many proteins. Since mRNA has no grammar, starting at the right nucleotide is critical. At least 12 di erent eIF proteins are involved (eIF = eukaryotic initiation factor). The small subunit of the ribosome also plays a critical role. Process starts with the binding of eIFs to the initiator tRNA, the mRNA and the small subunit of the ribosome. 33 ff  Translation Initiation eIF2 that is bound to a _______ binds to a methionine tRNA. Small subunit is bound to several eIFs. mRNA is bound to several including eIF-4E at the 5' CAP and PABP at the polyA tail. PABP brings the 5' and 3' ends of the mRNA together. 34  Transcription Initiation All three get assembled at the 5' CAP. Complex begins to scan (using ATP) down the mRNA in a 5' to 3' direction. Small subunit looks for the ______ AUG in the mRNA and stops when found. FIGURE 10.8 Initiation of translation in eukaryotic cells The Cell: A Molecular Approach, 8th Edition 35  Translation Initiation Small subunit looks for the first AUG in the mRNA and stops when found. eIF5 triggers the hydrolysis of the GTP bound to eIF2 to a GDP. Heme regulated inhibitor kinase (HRI) phosphorylates eIF2 and prevents release of GDP. It stops further translation. 36  Translation Initiation GTP hydrolysis triggers the release of the various eIFs. The large subunit can now bind and translation can begin. 37  Translation Elongation A key player in elongation is elongation factor 1a (eEF1⍺), which is bound to tRNAs and to GTP. Note that elongation factor 1⍺ is very di erent than eIF1A. Initiation factor vs. elongation factor. 38 ff  Translation Elongation Entrance of the correct tRNA/eEF1⍺ into the ribosome A site triggers GTP _____________ and generation of a peptide bond using the high energy bond of the charged tRNA. The ribosome then moves one codon down the mRNA and the process repeats. FIGURE 10.10 Elongation stage of translation The Cell: A Molecular Approach, 8th Edition 39  Translation Termination When the ribosome encounters a STOP codon, a release factor binds in the A site of the ribosome. Binding of the release factor triggers hydrolysis of the bond between the tRNA in the P site and the polypeptide. Everything then dissociates. FIGURE 10.12 Termination of translation The Cell: A Molecular Approach, 8th Edition 40