BIOL S236F / S236W / 2036SEF Cellular and Molecular Biology Lecture 13 PDF

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Hong Kong Metropolitan University

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This document is a lecture on cellular and molecular biology, detailing nucleic acid structure, DNA functions and packaging, bacterial chromosomes, DNA denaturation and renaturation and related topics. The lecture is from Hong Kong Metropolitan University.

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BIOL S236F / S236W / 2036SEF Cellular and Molecular Biology Lecture 13 Topic 4: Central Dogma and Gene Expression (Part 1-3) Topic 5: Genome Structure and Gene Recombination (Part 1-2) Topic 3: Programmed Cell Death (Part 3) (Revision) Monday, 15 April 2024 Lecture: 09:00-11:00 Tutorial: 11:00-12:00...

BIOL S236F / S236W / 2036SEF Cellular and Molecular Biology Lecture 13 Topic 4: Central Dogma and Gene Expression (Part 1-3) Topic 5: Genome Structure and Gene Recombination (Part 1-2) Topic 3: Programmed Cell Death (Part 3) (Revision) Monday, 15 April 2024 Lecture: 09:00-11:00 Tutorial: 11:00-12:00 2 Topic 4: Central Dogma and Gene Expression (Part 1) Introduction Nucleic acids are to Nucleic Polymeric macromolecules forming the Acids chemical basis for the transmission of genetic trait Two main classes of nucleic acids: Deoxyribonucleic acid (DNA) Ribonucleic acid (RNA). 2 3 Introduction to Nucleic Acids General Structure composed of nucleotides, which are the monomers made of three components 1. 5-carbon sugar (pentose) 2. nitrogenous base Note: Nucleoside = Sugar + Base Nucleotide = Sugar + Base + Phosphate 3. phosphate group 3 4 Introduction to Nucleic Acids Sugar Moiety In DNA: 2-deoxyribose In RNA: ribose Source: http://brussels-scientific.com/?p=5922 4 5 Nucleosides Nucleotides Nitrogenous base + Sugar Moiety Nucleoside with a phosphate group attached at carbon atom 5’ by phosphorylation (Ribonucleoside or Deoxyribonucleoside) (Ribonucleotide or Deoxyribonucleotide) RNA: Ribonucleic acid DNA: Deoxyribonucleic acid 5 Introduction to Nucleic Acids 5. Polynucleotides Two nucleotides covalently boned with the formation of a 3’-5’ phosphodiester bond Bond between the 5’-phosphate of one nucleotide and the 3’-hydroxyl (3’OH) of a second nucleotide Formation of phosphodiester bonds in DNA and RNA = 5’ to 3’ direction i.e. nucleotide being added to the free 3’ end of the preceding one Both DNA and RNA: negatively charged (phosphate group) 6 7 Source: https://www.biologyonline.com/dictionary/phosphodiester-bond General Structure of DNA Two single strands coiled about one another in a spiral → Double helix Both strands are in complementary pairing; hold by non-covalent interactions Lipophilic interaction between adjacent bases Hydrogen bond (H-bond) between bases on opposite strands A = T pairing (2 H-bonds) (lower melting temp) G ≡ C pairing (3 H-bonds) (higher melting temp) Anti-parallel: complementary and opposite direction 5’ → 3’ and 3’ 5’ 5’ end: Phosphate group → 3’ end: Hydroxyl group Formation of phosphodiester bonds in DNA and RNA = 5’ → 3’ direction Addition of a nucleotide: added to the free 3’ end of the preceding one Both DNA and RNA: negatively charged (phosphate group) Note: Mnemonics to remember these bases: "CUT the Py“, CUT: Cytosine, Uracil, Thymine; Py: Pyrimidines "Pure As Gold”, Pur: Purines; A, G: Adenine, Guanine 8 8 9 Functions of DNA 1. Store genetic information 2. Direct RNA transcription → protein synthesis 3. Contains instructions needed to develop, survive, and reproduce Denaturation and Renaturation of DNA 10 Denaturation double strand DNA (dsDNA) 11 Various methods to denature dsDNA e.g. by heat (heating >80-90℃). or alkaline method (pH >11.3) Measurement of dsDNA denaturation: Change in absorbance at 260nm Principle: all nucleotide bases consist of aromatic rings which absorbs light in the UV range dsDNA will have less absorbance at 260 nm than ssDNA due to the stacking interactions between the bases (known as hypochromic effect) When the dsDNA denatures, 1. Absorbance increases 2. Viscosity decreases: reflects the physical change occurred in the DNA structure (and vice versa when the ssDNA renatures) Single stranded DNA (ssDNA) Double stranded DNA (dsDNA) 11 Hyperchromic effect: increasing ability to absorb light Hypochromic effect: decreasing ability to absorb light https://www.biologyexams4u.com/2018/01/ds-or-ssdna-has-more-absorbance-double.html 12 12 13 Bacterial Chromosome (prokaryotes) Bacterial can have single or multiple, circular or linear chromosomes Most common arrangement a) single circular molecular bound to small amounts of protein b) localized to a special region of the bacterial cell called the nucleoid 13 14 DNA Packaging in Prokaryotes Nucleoid area in the cytoplasm which houses the prokaryotes’ genetic material NOT surrounded by nuclear membranes (unlike eukaryotic nucleus) Consists of mostly DNA, together with RNAs and proteins Most of the RNAs are mRNAs Most of the proteins are transcription factor proteins and ribosomes Nucleoid-associated proteins (NAPs) maintain nucleoid structure resemble eukaryotic histone proteins Source: https://commons.wikimedia.org/wiki/File:OSC_Microbio_03_0 3_Nucleoid.jpg 14 The effects of nicks A nick is a discontinuity in a double stranded DNA molecule where there is no phosphodiester bond between adjacent nucleotides of one strand typically through shearing damage or enzymes a) A plasmid is tightly wound into a negative supercoil b) To release the intersecting states, the torsional energy must be released by utilizing nicks c) After introducing a nick in the system, the negative supercoil gradually unwinds d) Until it reaches its final, circular, plasmid state 15 16 DNA Packaging in Prokaryotes Interconversion between relaxed and supercoiled forms (topoisomers) The interconversion between relaxed and supercoiled forms of DNA is catalyzed by enzymes known as topoisomerases 1. Type I : introduce transient single-strand breaks in DNA a) Cut one strand of the double helix b) Allow the DNA to rotate and the uncut strand to be passed through the break before the broken strand is resealed 2. Type II : introduce transient double-strand breaks a) Cut both DNA strands b) Pass a segment of uncut double helix through the break before sealing c) Requires energy derived from the hydrolysis of ATP 16 17 DNA Packaging in Prokaryotes Type I topoisomerase Watson, JD. (2013) 17 18 DNA Packaging in Prokaryotes Type II topoisomerase Watson, JD. (2013) 18 Agarose gel electrophoresis Agarose gel electrophoresis of nicked circular, linear, and supercoiled plasmid DNA Conformation of relaxed circular, linear, and supercoiled plasmid DNA Electrophoretic migration of the same DNA in various conformations Nicked plasmids assume a relaxed, open circular conformation and take up the most volume, migrating most slowly through the gel; Linearized plasmids move through the gel at a slightly higher rate; intact, Supercoiled plasmids, being the most compact, migrate the fastest https://www.thermofisher.com/hk/en/home/life-science/cloning/cloning-learning-center/invitrogen-school-of-molecular-biology/na-electrophoresis-education/na-electrophoresis-considerations.html 19 19 DNA Packaging in Eukaryotes 20 Level of DNA packaging 1. Winding of DNA around histone octamers to create a nucleosome structure. 2. Nucleosomes are connected by strands of linker DNA like beads on a string. 3. Pack nucleosomes into 30-nm chromatin fibers When a cell undergoes mitosis, the chromosomes condense even further. Chromosomes need to be less condensed to allow for gene expression 4. Form looped domains 5. Compress and fold 300-nm fiber to produce a 250-nm wide-fiber 6. Tight coiling of the 250-nm wide-fiber to produce the chromatid of a chromosome 20 21 DNA Packaging in Eukaryotes Histone protein and Nucleosome Two H2A-H2B dimers and two H3-H4 dimers form the histone octamer ~147 bp of DNA wrapping around the core histone octamer, becoming a nucleosome The histone cores are linked with a stretch (~50 bp in length) of DNA, called linker DNA Source: https://www.researchgate.net/figure/Schematic-representation-of-thenucleosome-The-nucleosome-core-is-composed-of-a-histone_fig1_320671837 21 22 DNA Packaging in Eukaryotes Packing of nucleosome to form chromatin fibers and chromosome 1. 30-nm chromatin fiber DNA makes ~2 complete turns wrapping around the nucleosome H1 is not involved in the formation of histone octamer H1: bind to the nucleosomal core particle around the DNA entry and exit sites Stabilize both nucleosome structure and higher-order chromatin architecture (i.e. 30-nm chromatin fiber) 22 23 DNA Packaging in Eukaryotes Packing of nucleosome to form chromatin fibers and chromosome 3. Further condensation to 250-nm-wide chromatin fiber then chromosome Segment of chromatin is highly compact that show up as dark spots in G-banding, called heterochromatin Contains DNA that is transcriptionally inactive (gene silencing) Consists of two forms: a) Facultative heterochromatin: can transition to and from relaxed state (reversible) a) Constitutive heterochromatin: always in a condensed state (irreversible) (e.g. centromere) Another form, which is more loosely packed and diffuse chromatin is called euchromatin ◼Associated with DNA that is being actively transcribed. ◼Much of the chromatin in metabolically active cell is loosely packed as euchromatin. 23 Structure of Chromosome in Eukaryotes 1. Heterochromatin Tightly wound, highly compacted sequence Also known as heterochromatic region telomere centromere heterochromatic region euchromatic region telomere stain more darkly in G-banding (incorporates more Giemsa stain) AT-rich and relatively gene poor; rich in repetitive sequences and sequences that are not to be transcribed in that cell, thus those genes are silenced 2. Euchromatin Less condensed chromatin Also known as euchromatic region Source: https://www.differencebetween.com/difference-betweenconstitutive-and-vs-facultative-heterochromatin/ Appear as a light band in G-banding (incorporates less Giemsa stain) GC-rich, more transcriptionally active (DNA are more accessible to protein binding) as it harbors 24 protein-coding genes 24 25 Chromosomes as seen at metaphase during cell division Light bands GC-rich, early replicating, relatively gene rich Telomere Satellites Nucleolar organizer regions (NORs) Short arm p (petit) Centromere Long arm q Dark (G) bands Telomere AT-rich, late replicating, gene poor Source: http://ib.bioninja.com.au/standard-level/topic-3genetics/32-chromosomes/chromosome-types.html Chr 1 Chr 14 26 Topic 4: Central Dogma and Gene Expression (Part 2) Content to be covered ▪ What is gene? ▪ Prokaryotic gene transcription ▪ Eukaryotic gene transcription Topic 4: Central Dogma and Gene Expression (Part 2) What is gene? 27 ▪ The term gene was coined by a Danish botanist, pharmacist, plant physiologist, and geneticist, named Wilhelm Johannsen ▫ He also coined the terms of genotype and phenotype ▪ ▪ Gene is the basic functional unit of hereditary Genes will have various coding and non-coding regions ▪ The genomes of eukaryotes and prokaryotes are distinguished by the amount of coding and noncoding regions present ▫ Human genome: 3.3G bp ▫ E. coli bacteria genome: 4.6M bp ▫ Mycoplasma genitalium genome: 0.58M bp Structure of a gene 28 ▪ ▪ Genes and DNAs are packed in the chromosome. The location of a gene on a chromosome is called the locus. ▪ Genes can exist in different forms known as alleles e.g. Human blood group type: The ABO gene has 3 types of alleles: A, B, and O. ▫ ▫ ▫ Every normal individual inherits two alleles of the gene, one from each parent. The combination of your two alleles determines your blood type. The different combinations among these three alleles generate the different blood groups Maternal allele (mother) Paternal allele (father) Allele A B O A AA AB AO B BA BB BO O OA OB OO Alleles Genotypes Phenotype (blood type) A+A AA A A+O AO A A+B AB AB B+B BB B B+O BO B O+O OO O Transcription (DNA → RNA) 30 Transcription In transcription, the information stored in the DNA is passed (transcribed) to the RNA through the complementarity of DNA nucleotides, where the RNA will be subsequently translated into a gene protein product. To transcribe a gene accurately, RNA polymerase must recognize where on the genome to start and where to finish. RNA polymerases perform these tasks differs somewhat between prokaryotes and eukaryotes The processes in bacteria (prokaryotes) are simpler: a basal transcription complex comprised of the large and small subunits of RNA polymerase (RNAP) and additional sigma () factors assembles at the start site In eukaryotes, a transcription complex requires RNA polymerase (RNAP) and up to 20 additional factors for accurate initiation DNA sequences / regions involving in transcription Downstream of TSS Upstream of TSS Non-template, coding or sense strand Template, non-coding or antisense strand Promoter Recognition sequences or Transcriptional factor binding sites Transcribed region Transcription start site (TSS) 31 Prokaryotes -35 sequence Transcripti on start +1 site Pribnow Box Eukaryotes Enhancers and/or Silencers Transcription start site 32 33 Machinery for Prokaryotic Transcription Molecular Biology of the Cell 6th Edition ▪ RNA polymerase ▪ Promoter ▪ Elongation ▪ Termination 34 Prokaryotic Promoter ▪ Two 6-base sequences upstream of the start site ▫ Core promoter (-35nt) (TTGACA) ▫ TATA box (-10nt) (TATAAT) Consensus sequence Sequence logo Molecular Biology of the Cell 6th Edition 35 Prokaryotic RNA Polymerase ▪ Structure ▫ Consists of 5 subunits ▫ Cover 75bp ▪ Function ▫ Recognize promoter sequence ▫ Unwind the double helix of DNA in correct direction ▫ Catalyze the formation of phosphodiester bonds that links ribonucleotide together (50nt / sec) ▫ Proofreading of transcripts (error rate 1 / 104 nt) Molecular Biology of the Cell 6th Edition 36 General Process of Prokaryotic Transcription ▪ Initiation – Step 1 - 4 RNAP synthesizes only a few bp at first and then releases ▪ Elongation – Step 5 ▪ Termination – Step 6 - 8 RNAP breaks free from the promoter (promoter clearance) and undergoes a conformational change that makes its association with the DNA very stable Molecular Biology of the Cell 6th Edition RNA polymerase (RNAP) in prokaryotes + 37 http://faculty.samford.edu/~djohnso2/44962w/405/_05transcription.html 38 RNA polymerase (RNAP) in prokaryotes Image source: https://www.ioc.kit.edu/pianowski/downloads/Handout%201v%20SL%20WS17.pdf In order to bind promoters, RNAP core associates with the transcription initiation factor sigma (σ) to form RNA polymerase holoenzyme. Sigma reduces the affinity of RNAP for nonspecific DNA while increasing specificity for promoters, allowing transcription to initiate at correct sites RNAP holoenzyme Transcription initiation in prokaryotes Binding of 𝜎 subunit of RNAP to the promoter on the non-template (or coding) strand Open complex RNAP holoenzyme detects this -35 sequence on the DNA strand and forms the closed complex, subsequently the RNAP holoenzyme moves down to the DNA strand and recognize and bind to the Pribnow box This will create an open complex by unwinding the dsDNA helix → resulting in open complex (transcription bubble) at the Pribnow box That means RNAP holoenzyme performs the helicase activity to unwind the DNA and break the two H-bonding between T:A (requires less energy to melt down this TA region) 39 Transcription Elongation Unlike DNA synthesis, RNA synthesis does not require priming The first ribonucleoside triphosphate retains all of its phosphate groups as the RNA is polymerized in the 5' to 3' direction Subsequent ribonucleoside triphosphates retain only the alpha phosphate, the one closest to the ribose sugar The other two phosphate groups are released as orthophosphate during the synthesis reaction Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 40 Transcription termination in prokaryotes RNA synthesis terminates differently in prokaryotes and eukaryotes Transcription termination in prokaryotes High levels of a gene product induce termination of its own synthesis Rho-independent termination Rho-independent termination occurs at G:C-rich regions in the DNA, followed by A:Trich regions. The G:C bases are transcribed into RNA and fold into a short double-stranded hairpin, which slows the elongation complex. The elongation complex then dissociates as it reaches the A:T-rich area. Rho-dependent termination (https://www.youtube.com/watch?v=cxhXCsvHVuI) Rho (a helicase enzyme) associates with RNA polymerase and inactivates the elongation complex at a cytosine-rich termination site in the DNA. Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 41 Rho-independent transcription termination in prokaryotes RNA-DNA hybrid region In this mechanism, transcription is terminated due to a specific sequence in the mRNA. The specific sequence contains inverted repeats which cause complimentary pairing as transcript RNA form hairpin loop structure. Inverted repeats This inverted repeat is followed by TTTTTTTT (~8 bp) on template DNA (appearing as UUUUUUUU in RNA) The large hairpin loop displaces A:U RNA-DNA hybrid as A:T of double strand DNA form is more stable Therefore, the RNA get separated from RNA-DNA hybrid region 42 43 Promoter Promoter Promoter Promoter Promoter Promoter Promoter Some genes are transcribed using one DNA strand as a template, while others are transcribed using the other strand The direction of transcription is determined by the orientation of the promoter at the beginning of each gene DNA regions that are not transcribed are indicated in gray 44 Overview of Eukaryotic Transcription Molecular Biology of the Cell 6th Edition Pre-mRNA (“primary RNA transcript”) Overview of Eukaryotic Transcription ▪ Initiation – Step 1 - 4 ▪ Elongation – Step 5 ▪ Termination – Step 6 - 8 Molecular Biology of the Cell 6th Edition 45 General Process of Eukaryotic Transcription ▪ Initiation ▫ Transcription factor binds on promoter ▫ Recruitment of RNA polymerase (II) forming initiation complex ▪ Elongation ▫ Recruitment of elongation factor to form elongation complex ▫ Addition of multiple methyl-guanine nucleotides at the 5’ end of the transcript (5’ cap) after the transcription of first 20bp ▪ Termination ▫ Transcript cleaved at specific site (AAUAAA) ▫ Addition of 3’ poly-A tail ▪ Co-transcriptional and post-transcriptional process Molecular Biology of the Cell 6th Edition 46 Prokaryotes -35 sequence Transcripti on start +1 site Pribnow Box Eukaryotes Enhancers and/or Silencers Transcription start site 47 48 Eukaryotic transcription initiation ▪ Equivalent to the RNAP holoenzyme used in prokaryotes, transcription factors binding to the promoter and associating with RNA polymerase II are used in initiation transcription in eukaryotic transcription RNA Polymerases In contrast to prokaryotes (1 type of RNA polymerase), eukaryotic nuclei have 3: (a) RNA polymerase I, (b) RNA polymerase II, and (c) RNA polymerase III They are structurally similar to one another and share some common subunits, but they transcribe different categories of genes RNA polymerases I and III: transcribe genes encoding transfer RNA (tRNA), ribosomal RNA (rRNA), and various small RNAs RNA polymerase II: transcribes protein-coding genes and non-coding RNA genes Molecular Biology of the Cell 6th Edition 49 50 Transcription factors needed for transcription initiation TATA-binding protein (TBP) Molecular Biology of the Cell 6th Edition Transcription initiation in eukaryotes Initiation complex 51 Transcription Elongation Unlike DNA synthesis, RNA synthesis does not require priming The first ribonucleoside triphosphate retains all of its phosphate groups as the RNA is polymerized in the 5' to 3' direction Subsequent ribonucleoside triphosphates retain only the alpha phosphate, the one closest to the ribose sugar The other two phosphate groups are released as orthophosphate during the synthesis reaction Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 52 mRNA processing in eukaryotes The newly synthesized RNA transcript is known as pre-mRNA mRNA processing in eukaryotes involves: 1. Addition of 5’ cap 2. Addition of a poly(A) tail 3. Splicing (removal of introns) (Prokaryotic mRNA does not require processing) 53 mRNA Processing- Capping Eukaryotic mRNA is blocked at the 5' terminus by an unusual 5'-5' pyrophosphate bridge to a methylated guanosine The structure is called a cap The cap confers a protective function (resistant to 5′ exonucleases) as well as serves as a recognition signal for the translational apparatus One terminal phosphate groups on the 5’ mRNA is removed by RNA triphosphatase, leaving a bisphosphate group GTP is added to the terminal bisphosphate by mRNA Guanylyl-transferase, losing a pyrophosphate from the GTP substrate The 7-nitrogen of guanine is methylated by mRNA (guanine-N7-)-methyltransferase Cap-adjacent modifications can occur, normally to the first and second nucleotides Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 54 Transcription Termination in Eukaryotes In transcription of pre-mRNA, RNA polymerase continues elongated along the DNA template. Until RNA polymerase encounters the polyadenylation signal sequence (or cleavage signal), AAUAAA, the endonuclease enzymes starts to cleave the pre-mRNA strand ~ 10-35 bp downstream of the polyadenylation signal sequence Once the pre-mRNA strand is cleaved, Poly(A) polymerase enzyme, add poly(A) tail to the 3’ end of the pre-mRNA strand in a template-independent manner Poly(A) tail is added after pre-mRNA is cleaved; therefore, it does not use DNA strand as a template to add this poly(A) tail with size up to 200 nucleotides of poly(A) in mammalian cells Watson, J. D., Baker, T. A., Gann, A., Bell, S. P., Levine, M., & Losick, R. M. (2004). Molecular biology of the gene. 東京電機大学出版局 Rosonina, E., Kaneko, S., & Manley, J. L. (2006). Terminating the transcript: breaking up is hard to do. Genes & development, 20(9), 1050-1056. 55 Modification to RNA molecules 56 General Structure of RNA 1. 2. Ribose + Nitrogenous base + Phosphate group Single stranded structure 3. Can form secondary structure by singlestranded RNA molecular folding over Form hairpin loops and stabilized by intramolecular H-bond between complementary bases Critical for many RNA functions e.g. ability of tRNA to bind to the correct sequence of mRNA during translation 4. Tertiary structure: related to structural and catalytic functions Source: https://www.researchgate.net/publication/344325555/figure/fig1/AS:938041292099585 @1600657880380/RNA-primary-secondary-and-tertiary-structures.png Messenger RNA (mRNA) Post-transcriptional process: Pre-mature RNA (or Pre-mRNA) → Mature mRNA Newly made mRNA (pre-mRNA) will be further processed before it is functional 3 steps 1.Capping at the 5' end (co-transcriptional) 2.Addition of a poly(A) tail at the 3' end (co-transcriptional) 3.Splicing to remove introns (post-transcriptional) Pre-mature RNA (pre-mRNA or a.k.a. heteronuclear RNA (hnRNA)) Capped and tailed → splicing Source: https://bio.libretexts.org/@api/deki/files/ 6083/Figure_5.5.2.png?revision=1 58 Messenger RNA (mRNA) Composes only about 1-5% of total cellular RNA with the shortest half- life The amount of particular mRNA in a cell is usually related to the level of its final protein product Found in the nucleus and cytoplasm Sequence complementary to the non-coding strand of DNA i.e. sequence same as the coding strand/non-template strand (except Thymine substituted by Uracil in mRNA) 59 Messenger RNA (mRNA) Splicing Remove introns from the pre-mRNA by spliceosome Spliceosome: proteins + small nuclear RNA (snRNA) → protein-RNA enzymes called small nuclear ribonucleoproteins or snRNPs Recognize sequences that are specific to splice junction a) sequence at the 5'-end (also known as the splice donor or donor site) of an intron is GU b) sequence at the 3'-end (also known as the splice acceptor or acceptor site) of the intron is AG. c) branch point: interact with snRNA Correctly cut out the introns and join the exons together to make the spliced and mature mRNA 60 Splicing sites: Donor site is at the 3’ end of an exon Acceptor site is at the 5’end of the next exon Splicing process: Ribonucleoprotein complexes (snRNPs) bind exon-intron junctions at splice donor and splice acceptor sites The branch point nucleotide attacks the 5’ splice site cutting the mRNA at that position and forming a loop. The spliceosome joins the end of one exon to the start of the next exon, hence the intron is removed. Source: https://en.wikipedia.org/wiki/Exon 61 Alternative splicing A process in which exons from the same gene are joined in different combinations, leading to different, but related, mRNA transcripts These different mRNAs transcripts of the same gene can be translated to different proteins with distinct structures and functions A source gives rising phenotypic diversity from a single gene Chen et al. Oncogene (2015) PMID: 24441040 62 Topic 4: Central Dogma and Gene Expression (Part 3) Outlines Part 1: Regulation of gene expression by epigenetics (by microRNA and siRNA; not to be covered) Part 2: Synthesis of polypeptides and proteins Describe the structure and chemical nature of the 20 amino acids Show how the chemistry of the amino acids affects the chemical characteristics and functions of proteins Define primary, secondary, tertiary, and quaternary structure of protein organization Describe how amino acids are polymerized into a polypeptide, using RNA as a guide (translation) Translation process 63 Regulation of gene expression by epigenetics Gene expression can be altered in ways other than by cis elements and transcription factors Epigenetics: Histone modification and DNA methylation, Gene silencing by double-stranded or antisense RNA regulate gene expression Epigenetic regulation are mechanisms controlling gene expression, but without altering the underlying DNA sequence The term epigenetics reflects the idea that the phenotypic effects of these heritable (imprinting) and reversible epigenetic factors which are not encoded in the DNA Genomic Imprinting: - inheritable during somatic cell division and gametogenesis - One copy of the gene/allele is active and the other copy of the gene/allele is silenced (inactive) in a diploid cell according to the parent of origin Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 64 Regulation of gene expression by epigenetics Histone Modification regulate access of trans factors and RNA polymerase to the DNA helix affects the activity of chromatin associated proteins and transcription factors that activate or silence gene expression through these protein interactions, chromatin can switch between transcriptionally active (euchromatin) and transcriptionally silent (heterochromatin) states (refer to Topic 4 – Part 2) Details of modification of histone regulating gene expression would not be covered in this course. Please refer to “Ch.4.3 The Effect of Chromatin Structure on DNA Function” in Albert B. et al. Molecular Biology of the Cell. 7th ed. Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 65 Regulation of gene expression by epigenetics DNA methylation DNA methylation is another type of epigenetic regulation of gene expression, particularly in vertebrates and plants In vertebrates, methylation occurs in cytosine-guanine (CG) rich sequences in the DNA (CpG islands) → 5-methylcytosine (mC) CpG islands are initially defined as regions >200 bp in length with an observed/expected ratio of the occurrence of CpG >0.6 (5-methylcytosine) This definition may be modified to a more selective GC content to exclude unrelated regions of naturally high GC content CpG islands are found around the first exons, promoter regions, and sometimes toward the 3' ends of genes Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 66 Imprinting In imprinted genes, the parent of origin is often marked, or “stamped,” on the gene during the formation of egg and sperm. This stamping process mostly involves DNA methylation, which turns the gene off, and demethylation which makes the gene active. Most imprinted genes are developmental genes which affect fetal growth, nutrient transfer through placenta, cell proliferation, and brain development. Usually growth-related genes are maternally imprinted. Inheritance of imprinted genes across generations Ritu, Giri, P., Mohapatra, B. (2018). Genomic Imprinting. In: Vonk, J., Shackelford, T. (eds) https://doi.org/10.1007/978-3-319-47829-6_23-1 67 DNA methylation and gene silencing Post implantation Early embryogenesis See the description in the next slide methyl-binding proteins trigger a silencing cascade DNMT1 scans DNA and restores methylation newly synthesized DNA (during replication process) is unmethylated 68 DNA methylation and gene silencing In early embryogenesis, DNA is largely lacking of methylation (top left) Post implantation, de novo methylation begins (red lollipops), mediated primarily by DNA (cytosine-5-)-methyltransferase-3alpha (DNMT3A) and -3beta (DNMT3B) (top). When methylation happens in CpG islands, methyl-binding proteins trigger a silencing cascade (illustrated by green stars) whereby histone H3K9 is sequentially deaceylated and then methylated, allowing heterochromatin protein 1 (HP1) to bind; eventually resulting in closed chromatin. After DNA replication, newly synthesized DNA (green letters) is unmethylated. However, DNMT1 rapidly scans DNA and deposits methyl groups on newly synthesized DNA, opposite methyl groups present on the old DNA strand. This results in faithful replication of methylation patterns (Restored methylation) and the maintenance of gene silencing. Adult patterns of methylation are erased by epigenetic reprogramming in early embryogenesis (top left). 69 Regulation of gene expression by DNA methylation DNA methylation Proteins that bind to methylated DNA also form complexes with proteins involved in deacetylation of histones Therefore, when the DNA is in a methylated state, nearby histones are deacetylated, resulting in compact and silent chromatin Change of DNA methylation at these sites is a source of dysregulation of genes in disease states For example, methylated C residues in the promoter regions of tumor suppressor genes is a mechanism of inactivation of these genes in cancer Imprinting maintains the balanced expression of genes in growth and embryonic development by selective methylation of homologous genes This controlled methylation occurs during gametogenesis Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 70 Translation of RNA: Synthesis of polypeptides and proteins Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 71 Amino Acids & Polypeptide Polypeptide is a polymers of amino acids (linked via peptide bond) A polypeptide folds into a complex 3-dimensional function shape which can be referred to a protein molecule Each amino acid has characteristic biochemical properties determined by the nature of its amino acid side chain 20 amino acids can be grouped according to their polarity (tendency to interact with water at pH 7) as follows: Nonpolar Polar uncharged polar negatively or positively charged polar The properties of amino acids that make up a protein determine the shape and biochemical nature of the protein Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 72 Amino acids & Polypeptides Peptides with additional units are tri-, tetra-, pentapeptides, and so forth, depending on how many units are attached to each other At one end of the peptide will be an amino group (the amino terminal, or N-end), and at the opposite terminus of the peptide will be a carboxyl group (the carboxyl terminal, or C-end) Like the 5' to 3' direction of nucleic acids, peptide chains grow from the amino to the carboxyl terminus (N → C) Forming a polypeptide backbone Two amino acids joined together by a peptide bond Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 73 Peptide Bond Formation Condensation reaction between carboxyl (C) and amino group (N) gives a peptide bond; a water molecule is released Image source: https://www.drawittoknowit.com/course/biochemistry/glossar y/biochemical-pathway/peptide-bond-formation 74 Amino Acids Interactions between amino acid side chains fold a polypeptide/protein into predictable configurations These include ordered beta sheets and less-ordered alpha helices, or random coils The alpha helix and beta sheet structures in proteins were first described by Linus Pauling and Robert Corey in 1951 This level of organization is the secondary structure of the polypeptide Some polypeptides, especially structural proteins, consist almost entirely of alpha helices or beta sheets Globular proteins have varying amounts of alpha helix and beta sheets The secondary structure of polypeptide are further folded and arranged into a tertiary structure Tertiary structure is also important for protein function If a protein loses its tertiary structure, it is denatured Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 75 Secondary structure of polypeptides or protein Image source: https://www.khanacademy.org/science/biology/macromolecules/proteins-and-amino-acids/a/orders-of-protein-structure 76 Protein structure Proteins that work together are called oligomers, each component protein being a monomer (2, 3, 4 proteins → dimer, trimer, tetramer) Quaternary structure of proteins The combinatorial nature of protein function may account for the genetic complexity of higher organisms without a concurrent increase in gene number Proteins are classified according to function as enzymes and as transport, storage, motility, structural, defense, or regulatory proteins Enzymes and transport, defense, and regulatory proteins are usually globular in nature, making them soluble and allowing them to diffuse freely across membranes Structural and motility proteins are fibrous and insoluble 77 Haemoglobin Image source: https://alevelbiology.co.uk/notes/hemoglobin/ One of the most familiar examples of a conjugated protein Hemoglobin is a tetramer, having four Fe2+-containing heme groups, one covalently attached to each monomer A protein's primary structure is defined as the amino acid sequence of its polypeptide chain; A secondary structure is the local spatial arrangement of a polypeptide's backbone (main chain) atoms; A tertiary structure refers to the three-dimensional structure of an entire polypeptide chain; A quaternary structure is the three-dimensional arrangement of the subunits in a multi-subunit protein Image source: https://www.researchgate.net/figure/Th-e-primary-secondary-tertiary-andquaternary-structures-of-hemoglobin-Source_fig1_279784579 78 Aminoacyl tRNA Polypeptide synthesis starts with activation of the amino acids by covalent attachment to tRNA (or tRNA charging reaction), a reaction catalyzed by 20 aminoacyl tRNA synthetases: The aminoacyl tRNA synthetase reaction requires energy (ATP) and takes place in two steps 1st: the amino acid is activated by addition of AMP: amino acid + ATP → aminoacyl-AMP + PPi 2nd: the activated amino acid is joined to the tRNA: aminoacyl-AMP + tRNA → aminoacyl-tRNAAA + AMP The product of the reaction is an ester bond between the 3' hydroxyl of the terminal adenine of the tRNA and the carboxyl group of the amino acid Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA ARS: aminoacyl tRNA synthetase tRNA charging reaction Source: //doi.org/10.1016/j.ijbiomac.2017.12.157 79 80 Transfer RNA (tRNA) The amino acid attaches to the 3’ hydroxyl group of the acceptor end (CCA) Anticodon loop: complementary codon (3 nucleotides sequence) present on mRNA Each codon is identified by a specific tRNA. Source: https://www.researchgate.net/publication/335155516/figure/fig1/AS:791603652358144@1565744426214/tRNA-structure-a-tRNAs-are-represented-as8 cloverleaf-structures-in-two-dimensions-b.png 1 Transfer RNA (tRNA) Relatively short, single-stranded polynucleotides of 73-93 bases in length, 24 – 31,000 in molecular weight Composed about 10% of total cellular RNA Predominantly located in the cell cytoplasm General characteristics of tRNA 1. Secondary structure resembles a cloverleaf 2. Amino acid attach to the 3’-CCA terminus 3. Anticodon loop recognize the triplet codon of the template mRNA 8 2 Codons table Due to degeneracy in the genetic code, base substitution does not necessary change an amino acid 83 Ribosomes and tRNA in translation process Translating mRNA from 5’ to 3’ direction →the initiator tRNA carrying the Met amino acid is bound to the P-site (peptidyl) on the ribosome →the A-site (aminoacyl) is aligned with the next codon which will be bound with the anticodon of the next tRNA → the amino acid or peptide on the tRNA at P-site will be translocated (peptide bond formation) to the amino acid of the tRNA at the A-site → the uncharged tRNA at the P-site is moved to the E-site (exit), and it will be released from the ribosome, while the tRNA with the polypeptide at the A-site will be moved to the P-site -> the empty A-site will be available for the next tRNA 84 85 86 Translation process Polypeptide synthesis in the ribosome almost always starts with a “start codon” of Methionine in eukaryotes N-formylmethionine in bacteria, mitochondria, and chloroplasts Initiating factors that participate in the formation of the ribosome complex differentiate the initiating methionyl tRNAs from those that add methionine internally to the protein Termination of the amino acid chain is signaled by one of the three nonsense, or termination (UAA, UAG, or UGA), of which tRNA are not charged with an amino acid When the ribosome encounters a termination codon, termination or release factors (R1, R2, and S in E. coli) causes hydrolysis of the finished polypeptide from the final tRNA, release of that tRNA from the ribosome, and dissociation of the large and small ribosomal subunits In eukaryotes, termination codon–mediated binding of polypeptide chain release factors (eRF1 and eRF3) triggers hydrolysis of peptidyl-tRNA at the ribosomal peptidyl transferase center Buckingham, L. (2019). Molecular diagnostics: fundamentals, methods and clinical applications. FA Davis. 87 Ribosomal RNA (rRNA) See Figure 1.27 in Ref 1. Largest component of cellular RNA: 80-90% of total cellular RNA Half life: 7 days Various types of rRNA are named for their sedimentation coefficient (S, Svedberg unit) S: Correspond to the related rate of sedimentation during ultra high-speed centrifugation Greater the S value, denser the particle r r r r rRN A r r 88 Ribosomal RNA (rRNA) & Ribosomes Ribosomal RNAs (rRNAs) Structural and functional part of ribosome Ribosomal RNAs are assembled with ribosomal protein to make the subunits that will form the complete ribosome Synthesis of rRNA: Single precursor RNA (pre rRNA) ➔ processed into 5.8S, 28S and 18S 5S is synthesized separately Ribosome: nucleoprotein serves as the primary component of cell protein synthesis; present in the cytoplasm 2 subunits (e.g. 60S and 40S in eukaryotes) 60S ribosomal large subunit: 5S, 5.8S, 28S 40S ribosomal small subunit: 18S 89 Prokaryotic transcription is coupled to translation In prokaryotes, the mRNA begins to be translated before transcription is finished. ie. Transcription and translation are coupled Would this happen in eukaryotes? (Hints: location of transcription and translation occurs) Prokaryotic genes are often clustered together. This kind of grouping functionally related genes is referred to as an operon A single transcription unit that usually encodes multiple genes or enzymes necessary for a biochemical pathway 90 Messenger RNA (mRNA) Structure of mRNA Open reading frame (ORF) Source: https://ars.els-cdn.com/content/image/3-s2.0B9780123944474100409-f10040-01-9780123944474.jpg 91 Translation in eukaryotes Translation occur either in the cytoplasm or on the rough endoplasmic reticulum (RER) Some proteins are targeted to be translated on the RER They will have a short series of polypeptide called a signal sequence, which will be specifically recognized and bound by a signal recognition particle (SRP) The complex of signal sequence and SRP will dock to a receptor protein on ER membrane The newly synthesized protein in ER can be modified, e.g. addition of sugars (glycosylation), and transported to the Golgi apparatus. 92 Post-translational modification Critical molecular events can alter protein conformation after their synthesis and diversity protein properties by modulating their stability, localization, interacting partners or the activity of their substrates, Exerting essential roles in regulating the functions of many important eukaryotic proteins Involve the formation of a covalent bond linking a specific functional/chemical group to specific amino acid side chains on the protein a phosphate group (by phosphorylation): alter cellular signaling and protein’s function an acetate group (by acetylation) attach lipid/hydrophobic groups (lipid modification) add carbohydrates (by glycosylation): affect protein’s solubility add ubiquitin (by ubiquitination): for protein degradation Such post-translational modifications are generally reversible (ie. one enzyme adds the modifying group and another can remove it) e.g. phosphorylation by enzymes known as protein kinases, while phosphotases remove such phosphate groups 93 Codons to amino acids A single base substitutional change from A to T, resulting the substitution of Val for Glu 94 Types of single nucleotide variants (SNV) b. Silent variant: change of a nucleotide does not change an amino acid c. Missense variant: change of a nucleotide changes an amino acid d. Nonsense variant: change of a nucleotide introduces a stop codon 95 A template strand of DNA has the following sequence: 5′ – GCAATGGGCTCGGCATAAGTTTAA – 3′ a. Determine the sequence of the RNA from this DNA template b. What is the possible open reading frame (ORF) sequence? c. Predict the amino acid sequence 96 A template strand of DNA has the following sequence: DNA 5′–GCA ATG GGC TCG GCA TAA GTT TAA–3′ mRNA 5’–GCA AUG GGC UCG GCA UAA GUU UAA–3’ ORF AUG GGC UCG GCA UAA GUU Peptide: Met Gly Ser Ala Ter Val DNA 5′–GCA ATG GGC TCG GCA TAA GTT TAA–3′ Substitutional change Codon Amino acid Type of change or variant A UAG Ter (stop) Nonsense variant C CCA Pro Missense variant 97 Topic 5: Genome structure and gene recombination (Part 1) Size of variants Genomic/Genetic Variants ⚫ Chromosomal variants ⚫ XXY, Monosomy X, Trisomy 21, etc. ⚫ Structural variants ⚫ insertions / deletions / duplication / translocations Chromosomal level >1-5 Mb ⚫ Several Mb – >50 bp 1-10 bp ⚫ Copy number variants (CNVs) / Sequence repeats ⚫ Short/Long interspersed nuclear elements (SINEs / LINEs) ⚫ Microsatellites / small tandem repeats (STRs) (e.g. TAn, AATAn) Single nucleotide variants (SNVs) / Indels Feuk, Lars;, et al. Structural variation in the human genome. Nature Reviews Genetics 2006, 7 (2): 85–97 98 99 Components of Human Genome Long interspersed nuclear elements (LINEs) Short interspersed nuclear elements (SINEs) Retroviral-like element DNA-only transposon Mobile genetic element Multiplied by replicating and inserting new copies in different position Microsatellites; 13b long; near the heterochromatin around centromere Minisatellite; repeats of 14 – 500 repeats; scattered around genome 99 Microsatellite / Short tandem repeats (STR) markers e.g. Repeating 2-6 nucleotides in DNA sequence Consider as double stranded DNA Maternal allele 5’ 3’ 3’ 5’ Paternal allele 5’ 3’ 3’ 5’ 2-nt: dinucleotides (e.g. TA, CG) 3-nt: trinucleotides (e.g. CAG, CTA) 4-nt: tetranucleotides (e.g. GCAA) 5-nt: pentanucleotides (e.g. ATTGT) 6-nt: hexanucleotides (e.g. GGGGCC) The repeat region is variable between individuals: Heterozygous = number of repeats between two alleles differ Homozygous = both alleles contain the same number of repeats 100 Slippage effect Misalignment with mismatch repairing PMID: 17572359 Fan, H., & Chu, J. Y. Genomics Proteomics Bioinformatics. 2007 Feb;5(1):7-14. 101 Microsatellite / Short tandem repeats (STR) markers PCR Allele 1 prime r Consider as double stranded DNACTA CTA CTA 4 repeats Allele 2 prime r 8 repeats The repeat region is variable between individuals while the flanking regions where primers bind are constant Heterozygous = number of repeats between two alleles differ; they can be resolved by electrophoretic separation (with different PCR product size) Homozygous = both alleles contain the same number of repeat (with same PCR product size) 102 Human Genome Nuclear Genome Mitochondrial Genome Gene and generelated sequence Coding DNA (10%) Pseudogene Non-coding DNA (90%) Gene fragments Extragenic DNA Unique or low copy number (~60%) Introns, Untranslated sequence, etc. 2 mt-rRNA gene Moderate to highly repetitive (~40%) Tandemly repeated or clustered repeats Interspersed repeats 22 mt-tRNA genes 37 genes 103 13 polypeptide encoding genes Mitochondrial Genome (mtDNA) Structure and gene organization of mtDNA is highly conserved among mammals Maternally inherited Representing the female ancestry of an individuals Circular double-stranded DNA of 16,569 bp with 44% (G+C) Contains 37 genes coding for: 2 mt-rRNAs, 22 mt-tRNAs and 13 polypeptides, which are the components of the oxidative phosphorylation system 104 Source: en.wikipedia.org/wiki/Mitochondrial_DNA Mitochondrial Genome (mtDNA) Structure 1. Heavy strand (H-strand): outer circle Base composition: rich in G Coding gene: two rRNAs, 14 tRNAs and 12 polypeptides 2. Light strand (L-strand): inner circle Base composition: rich in C Coding gene: 8 tRNAs and a single polypeptide 3. Displacement loop (D-loop): Large proportion of the mtDNA duplexes 105 Complementary to the L-strand, displaces the H-strand. Major control site for mtDNA expression Initiation of mtDNA replication and the major promoters for transcription Contain two hypervariable regions (HV1 and HV2) ➔ majority of polymorphic site Source:https://link.springer. com/article/10.1007/s00467 -019-04404-6/figures/1 Mitochondrial Genome (mtDNA) Characteristics of the genome Circular DNA; genes lack introns Intergenetic sequences are absent, except for a 1118 bp regulatory region (D-loop region) Both mt-rRNA and mt-tRNA molecules are small Some of the protein genes are overlapping and, in many cases, part of the termination codons are not encoded but are generated posttranscriptionally by polyadenylation of the mRNAs. All RNA components required for mitochondrial translation are 106 supplied by mitochondria themselves But protein components, including ribosomal proteins, translational factors, aminoacyl-tRNA synthetases, and various factors required for processing and modification of RNAs, as well as biogenesis of mito-ribosome, are encoded in the nuclear genome. They are translated in cytoplasm and imported to mitochondria. Nuclear Genome VS Mitochondrial Genome of Human Nuclear Genome 2 × 3.3 Gb (diploid) Mitochondrial Genome 16.6 kb 23 pairs of chromosomes, all linear One circular DNA molecular Several Class of histone and nonhistone protein Largely free of protein 6,500 - 80,000 genes 37 genes Large fraction of repetitive DNA Very little 107 repetitive DNA Intron found in most genes No intron About 3% are coding DNA About 93% are coding DNA Mendelian inheritance (ie. Maternal & Paternal) Maternal Inheritance DNA Replication 1. Initiator protein binds to double-stranded DNA at the origin of replication Use ATP to slightly unwind the DNA 108 DNA Replication 2. Unwinding of DNA DNA helicase binds to the unwound DNA and continue the unwinding DNA gyrase in prokaryote (Topoisomerase in eukaryote) promotes strand separation by inducing negative supercoiling in front of the DNA helicase Strand separation is maintained by molecules of single stranded DNA binding protein (SSB) Stabilize the unwound region Allow the separated strands to serve as template Replication fork is formed (Topoisomerase in eukaryote) 109 DNA Replication 3. DNA Primase (in prokaryote) DNA primse forms a complex with DNA polymerase α in eukaryote) binds to the first initiating sequence on the leading strand template Synthesize a short RNA primer that is complementary to the DNA template; no self-correction in the primer synthesis 4. DNA polymerase III (in prokaryote) DNA polymerase δ (lagging strand) / ε (leading strand) with proofreading in eukaryote uses the primer to initiate DNA extension/replication by adding nucleotide to the 3’ end of RNA primer or nucleotides The leading strand requires only one priming events because DNA extension is carried out in the 5’ to 3’ direction 110 Self-Correction Mechanism of DNA Polymerase Correct base pairing before polymerization (Proofreading) Exonucleolytic proofreading 3’ to 5’ exonuclease activity of the DNA polymerase 111 DNA Replication 5. Primase intersperses RNA primers in the lagging strand DNA polymerase extends the RNA primers, forming discontinuous short fragments, known as Okazaki fragments, in the lagging strand DNA 6. The RNA primer is then removed by the 5’ to 3’ exonuclease (exonuclease activity of DNA polymerase I in prokaryote) (enzymes of FEN1 and RNase H in eukaryotes) and replaced with DNA nucleotides by the DNA polymerase (Note: The mechanism of RNA primers removal differs between eukaryotes and prokaryotes. In generic term, exonuclease removes RNA primers) 112 DNA Replication 7. DNA ligase links together adjacent Okazaki fragments in the lagging strand DNA DNA ligase, polymerase 113 I, polymerase III, primase, DNA helicase and DNA gyrase will be working simultaneously in the vicinity of the replication fork Key Points (3 main stages: initiation, elongation, and termination) During initiation, proteins bind to the origin of replication while helicase unwinds the DNA helix and two replication forks are formed at the origin of replication. During elongation, a primer sequence is added with complementary RNA nucleotides, which are then replaced by DNA nucleotides. During elongation the leading strand is made continuously, while the lagging strand is made in pieces called Okazaki fragments. During termination, primers are removed and replaced with new DNA nucleotides and the backbone is sealed by DNA ligase. Key Terms origin of replication: a particular sequence in a genome at which replication is initiated leading strand: the template strand of the DNA double helix that is oriented so that the replication fork moves along it in the 3′ to 5′ direction lagging strand: the strand of the template DNA double helix that is oriented so that the replication fork moves along it in a 5′ to 3′ manner 114 A replication fork is formed by the opening of the origin of replication; helicase separates the DNA strands. An RNA primer is synthesized by primase and is elongated by the DNA polymerase. On the leading strand, only a single RNA primer is needed, and DNA is synthesized continuously On the lagging strand, DNA is synthesized in short stretches, each of which must start with its own RNA primer. 115 Replication of Mitochondrial DNA Falkenberg, M. (2018) Essays Biochem 62(3):287-296 mtDNA H-strand and L-strand Noncoding region (NCR) Harbors promoters for transcription Also harbors the origin for H-strand DNA replication (OH) The second origin for L-strand DNA replication (OL), locating outside the NCR, within a tRNA cluster The genome encodes for 13 mRNA (green), 22 tRNA (violet), and 2 rRNA (pale blue) molecules. The major NCR contains the heavy strand promoter (HSP), the light strand promoter (LSP), three conserved sequence boxes (CSB1-3, orange), the H-strand origin of replication (OH), and the termination-associated sequence (TAS, yellow). The triple-stranded displacement-loop (D-loop) structure is formed by premature termination of nascent H-strand DNA synthesis at TAS. The short H-strand replication product formed in this manner is termed 7S DNA. A minor NCR, located approximately 11,000 bp downstream of OH, contains the L-strand origin of replication (OL). 116 Replication of Mitochondrial DNA mitochondrial DNA-directed RNA polymerase (POLRMT) initiates primer synthesis Falkenberg, M. (2018) Essays Biochem 62(3):287-296 The most accepted model of replication of mtDNA: Strand Displacement Model 117 Replication is initiated at OH and proceeds unidirectionally to produce the full-length nascent H-strand (broken line). mtSSB binds and protects the exposed, parental Hstrand. DNA helicase TWINKLE travels in front of DNA polymerase γ (POLγ) to unwind the double-stranded DNA template. Replication of Mitochondrial DNA mitochondrial DNA-directed RNA polymerase (POLRMT) initiates primer synthesis Falkenberg, M. (2018) Essays Biochem 62(3):287-296 When the replication passes OL , a stem–loop structure is formed 118 that blocks mtSSB binding A single-stranded loop-region is formed, where mitochondrial DNA-directed RNA polymerase (POLRMT) initiates primer synthesis. The L-strand DNA synthesis takes place after about 25 nt, where POLγ replaces POLRMT at the 3′-end of the primer. Similar in the replication of nuclear DNA, Topoisomerase I relieves torsional strain while unwinding mtDNA Topic 5: Genome structure and gene recombination (Part 2) Outline Role of homologous DNA recombination Exchange of DNA strands between a pair of homologous duplex DNA sequences Allow one stretch of duplex DNA to acts as a template to restore lost or damaged information of a second stretch of duplex DNA Allow repairment of double DNA breakage Create new combination of gene in meiosis 119 Homologous DNA recombination An exchange of DNA strands between a pair of homologous chromosomes or duplex DNA sequences Such a strand exchange between two regions of double helix that are very similar or identical in nucleotide sequence allows one stretch of duplex DNA to restore lost or damaged information on a second stretch of duplex DNA During meiosis, homologous chromosomes line up and recombination occurs between them Also known as crossing-over happens in meiosis (prophase I) This results in gametes with unique combinations of alleles on each chromosome and therefore unique individuals. Lecture 6: Topic 3 (Part 2) 120 Homologous recombination can repair many types of DNA damage Cells can repair double-strand breaks in one of two ways. A. Nonhomologous end joining (NHEJ) The break is first “cleaned” by a nuclease that chews back the broken ends to produce flush ends. The flush ends are then stitched together by a DNA ligase. Some nucleotides are usually lost in the repair process, as indicated by the black lines in the repaired DNA. B. Homologous recombination If a double-strand break occurs in one of two duplicated dsDNA after DNA replication has occurred, but before the chromosome copies have been separated, the undamaged dsDNA can be used as a template to repair the damaged dsDNA through homologous recombination 121 Nonhomologous end joining (NHEJ) NHEJ joins broken ends together – no template is required NHEJ is the only readily available pathway in G1 and is the main mode of repair in most organisms NHEJ can occur by simple ligation of ends (a) but end processing is usual (b) Path (a): Nuclease digestion can remove a few nucleotides before ligation, or Path (b): Resection can expose singlestrand regions This sequence loss means NHEJ is mutagenic 122 ▪ Homology-directed repair using homologous chromosome is much less error-prone than NHEJ ▪ Single-stranded DNA (at least 50 nucleotides) is generated at break sites by helicases and exonucleases (1) ▪ One of the ends invades the intact duplex and base pairs to make a heteroduplex and form a Displacement-loop (D-loop) (2) ▪ Pairing in the D-loop is specific – ~90-100% accuracy over a 20-30 bp stretch ▪ Once the D-loop forms, repair can proceed in several different ways, including synthesis-dependent 123 strand annealing (SDSA) ▪ In synthesis-dependent strand annealing, after a D-loop forms, new DNA is synthesized (green; 3) ▪ DNA synthesis continues until the new DNA can pair with the other 3′ overhang (4) ▪ Replication and ligation finishes the process (5, 6) ▪ The new DNA spans the area of the break and both strands are the result of new DNA synthesis 124 ▪ Homology-directed repair (a) and homologous recombination (b) have different outcomes ▪ Homology-directed repair can result in gene conversions, whereas homologous recombination exchanges DNA and can result in new genetic combinations ▪ Homologous recombination is a defined feature of meiosis 125 Difference between Crossover and Gene Conversion During crossover the two DNAs reciprocally exchange part of their genetic information (c-c' and C-C’). Gene conversion is a nonreciprocal transfer of genetic information the red DNA "donates" part of its genetic information (c-c' region) to the black DNA. 126 Chromosome crossing-over occurs in meiosis Maternal (red) and paternal (blue) genetic information has been reassorted through chromosome crossovers. In addition, many short regions of gene conversion occur. Homologous recombination in meiosis starts with breaking both strands of the DNA in one of the recombining chromosomes 127 Sequence of Homologous Recombination (resulting in new genetic combinations) DNA double-stand break Spo11 protein breaks both strands of a DNA double helix and covalently bound to the broken DNA Degradation to expose 3’ singlestranded overhangs Mre11 Nuclease degrade the ends bound by Spo11 to create 3’ overhangs Spo11 dissociates from the break 3’ overhangs (single-stranded DNA) 128 128 Sequence of Homologous Recombination (resulting in new genetic combinations) 3’ overhang 3’ 5’ 5’ 3’ overhang 3’ RecA binds to the 3’ overhangs to form a protein-DNA filament The protein-DNA filament pairs with the corresponding sequence in sister chromatid 3’ overhang 3’ 5’ 5’ 3’ overhang 3’ 129 Sequence of Homologous Recombination (resulting in new genetic combinations) RecA dissociate from the DNA double helix at the expense of an ATP Second strands interwoven with the non-paired strand in the sister chromatid DNA polymerase extends the DNA sequences of the 3’ overhang of the two strands forming double Holliday junction The extended 3’ overhang dissociate with the sister chromatid The extended 3’ overhang ligate with the original DNA 3’ overhang 3’ overhang 3’ overhang 5’ 3’ 5’ 3’ overhang 3’ Extended 3’ overhang Extended 3’ overhang 130 Sequence of Homologous Recombination (resulting in new genetic combinations) Holliday junction Holliday junction The double Holliday junction is cleaved (cut or resolved) (at the blue arrows) by enzymes, e.g. resolvase The two original portions of each chromosome upstream and downstream from the two Holliday junctions are thereby swapped The chromosomes are crossed-over 131 BIOL S236F / S236W / 2036SEF Cellular and Molecular Biology Lecture 11 Topic 3 (Part 3): Programmed Cell Death 132 Cause of Cell Death Necrosis Loss of ion gradient Tissue damage across membrane Depletion of energy Many non-apoptotic ways for severely damaged or stressed animal cells to die. These are usually lumped together under the term cell necrosis. Apoptosis Extracellular signal (extrinsic pathway) Intracellular signal (mitochondrial / intrinsic pathway) Damaged or infected cells can also die by apoptosis, ensuring that they are eliminated before they can threaten the health of the animal. Our focus: apoptosis in vertebrates, its molecular mechanism and regulation 133 Functions of Apoptosis At least a million cells die this way each second in a healthy adult human (and are replaced by cell division) Control cell number Embryo development Killing damaged cells Removal of unnecessary cells Selection of cells 134 Control cell number 135 Apoptotic cells Embryo development Apoptosis helps sculpt our hands and feet during embryonic development: start out as spade-like structures, and the individual digits separate only as the cells between them die, as illustrated for a mouse paw (A) The paw in this mouse fetus is stained with the fluorescent dye (acridine orange which give green under apoptosis ), which enters apoptotic cells and thereby brightly labels them in the normal developing paw. The apoptotic cells appear as bright green dots concentrated between the developing digits. (B) (B) The interdigital cell death has eliminated much of the tissue between the developing digits, as seen one day later, when there are very few apoptotic cells. 136 Key morphological features of the cell during the events of apoptosis Cell shrink and condense Flipping of plasma membrane Collapse of cytoskeleton Disintegration of organelle Disassembly of nuclear envelop Condensation and break up of chromatin Formation of apoptotic bodies https://commons.wikimedia.org/wiki/File:Apoptosis_DU145_cells_mosaic.jpg by Egelberg Irreversible 137 Cell circle checkpoints If cells can’t pass the checkpoints (G1 and G2 checkpoint) in cell circle, the cell may be signaled to undergo apoptosis. G1 checkpoint G2 checkpoint In order for cells to pass the G1 checkpoint, several conditions must be met: Before cells can proceed out of G2 and into mitosis, several conditions are checked: The DNA isn’t damaged. The cell copied all the chromosomes. Signals tell the cell to proceed into mitosis. Signals tell the cells to divide. Cells must have plenty of nutrients. The DNA must be in good condition. Cells must be large enough to divide. 138 139 Apoptotic Pathway 139 Apoptotic signal triggers the assembly of large protein platforms that bring multiple initiator caspases together into large complexes Normally exist as inactive, soluble monomers in the cytoplasm In the cytoplasm The caspase cascade is not only destructive and selfamplifying but also irreversible, so that once a cell starts out along the path to destruction, it cannot turn back. Activation of multiple executioner caspases 140 When the executioner caspases are cleaved by the activated initiator caspase at a site in the protease domain, rearrange to an active conformation. 140 Caspase activation during apoptosis An initiator caspase contains a protease domain in its C-terminal region and a small protein interaction domain near its N-terminus. It is initially an inactive monomeric form (called procaspase) Apoptotic signals trigger the assembly of adaptor proteins carrying multiple binding sites for the caspase N-terminal domain. Upon binding to the adaptor proteins, the initiator caspases dimerize and are thereby activated, leading to cleavage of a specific site in their protease domains. Each protease domain is then rearranged into a large and small subunit. Executioner caspases are initially formed as inactive dimers. Upon cleavage at a site in the protease domain by an initiator caspase, the executioner caspase dimer undergoes an activating conformational change. The executioner caspases then cleave a variety of key proteins, leading to the controlled death of the cell. 141 How is the initiator caspase first activated in response to an apoptotic signal? The two best-understood activation mechanisms in mammalian cells are called the extrinsic pathway and the intrinsic (or mitochondrial) pathway Each uses its own initiator caspase and activation system Extracellular signal proteins binding to cell-surface death receptors activate the extrinsic pathway of apoptosis Fas ligand and Fas death receptors binding to Initiator caspases: forming a deathinducing signaling complex (DISC) Intrinsic pathway of apoptosis depends on mitochondria Cytochrome c from mitochondria and Apaf1 oligomerize into a wheel-like heptamer, called an apoptosome Note: Cytochrome c is a water-soluble component of the mitochondrial electron-transport chain 142 Extrinsic pathway of apoptosis Example: Death receptor triggering the extrinsic pathway of apoptosis by binding Fas ligand on the surface of a killer (cytotoxic) lymphocyte. When activated by the binding of Fas ligand, the death domains on the cytosolic tails of the Fas death receptors bind intracellular adaptor proteins, which in turn bind initiator caspases (primarily caspase-8), forming a deathinducing signaling complex (DISC). 143 Intrinsic Pathway of apoptosis (from mitochondria) Cell can activate their apoptosis program from inside the cell, often in response to stresses, such as DNA damage, or in response to developmental signals. Intrinsic pathway of apoptosis depends on the release of cytochrome c (a cytosol of mitochondrial proteins) Cytochrome c binds to a Apaf1, forming a wheel-like heptamer, called an apoptosome, which further recruit initiator caspase-9 protein → the activated caspase-9 activate downstream executioner caspases to 144 induce apoptosis End of Revision 145 Final examination (2 hours) Section A (40%) : 20 multiple choice questions (2% each) To each question, one best answer out of 4 choices Section B (30%): 5 short-answer questions (6 % each) Attempt all questions Section C (30%): Long-answer questions (10% each) Attempt any 3 out of 4 questions 146 Written Assignment #1 Upload all these files to OLE 1) Written Assignment (.docx) (do not accept other formats) 2) Powerpoint file (.pptx) 3) Script file (.docx) 4) Presentation audio & video recording (.mp4) If your recording file is too large to be uploaded to OLE, please upload a word file (.docx) containing a URL link for downloading Please refer to the details in the assignment instruction document 147

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