Coding and Non Coding Strand in Transcription
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Resistance to HIV (Exercise)
1 Evolution in fast motion – Resistance to HIV (Exercise) Resistance to HIV In the early 1990s, a number of studies revealed that some people – although they had repeatedly contact with HIV – did not become carriers of HIV or, in the case of a confirmed HIV-infection, showed a delayed onset of the disease (AIDS) (a delay of several years was reported). First attempts to explain these phenomena were observed a few years later, when scientists identified important co-receptor molecules on the surface of the host cells, which are essential for HIV to infect the host cell. Scientists assumed that resistant persons may carry an aberrant version of the co-receptor molecule, which makes it impossible for the virus to enter the host cell. Such a co-receptor is the chemokine co-receptor CCR5, which is normally involved in the host’s immune answer (Dean & O’Brien, 1998). In order to test their hypothesis, the scientists sequenced the genes which code for the co- receptor CCR5. They investigated more than 700 samples from HIV-infected patients and compared them with the CCR5-sequences from more than 700 healthy persons. The results of the DNA-sequencing revealed mutations in the CCR5-gene in HIV-infected persons with a delayed onset of AIDS, as well as in some samples of healthy persons (but not in HIV- patients with typical onset of AIDS) (Samson et al., 1996). Exercises 1. In material 1, you can find two CCR5-gene-sequences selected from the data set of the scientists. Compare the sequences and a. find out where the mutation is (identify the position and label it in both! sequences). -
Gene Therapy Glossary of Terms
GENE THERAPY GLOSSARY OF TERMS A • Phase 3: A phase of research to describe clinical trials • Allele: one of two or more alternative forms of a gene that that gather more information about a drug’s safety and arise by mutation and are found at the same place on a effectiveness by studying different populations and chromosome. different dosages and by using the drug in combination • Adeno-Associated Virus: A single stranded DNA virus that has with other drugs. These studies typically involve more not been found to cause disease in humans. This type of virus participants.7 is the most frequently used in gene therapy.1 • Phase 4: A phase of research to describe clinical trials • Adenovirus: A member of a family of viruses that can cause occurring after FDA has approved a drug for marketing. infections in the respiratory tract, eye, and gastrointestinal They include post market requirement and commitment tract. studies that are required of or agreed to by the study • Adeno-Associated Virus Vector: Adeno viruses used as sponsor. These trials gather additional information about a vehicles for genes, whose core genetic material has been drug’s safety, efficacy, or optimal use.8 removed and replaced by the FVIII- or FIX-gene • Codon: a sequence of three nucleotides in DNA or RNA • Amino Acids: building block of a protein that gives instructions to add a specific amino acid to an • Antibody: a protein produced by immune cells called B-cells elongating protein in response to a foreign molecule; acts by binding to the • CRISPR: a family of DNA sequences that can be cleaved by molecule and often making it inactive or targeting it for specific enzymes, and therefore serve as a guide to cut out destruction and insert genes. -
COMMENTARY Does Transcription by RNA Polymerase Play a Direct Role in the Initiation of Replication?
Journal of Cell Science 107, 1381-1387 (1994) 1381 Printed in Great Britain © The Company of Biologists Limited 1994 COMMENTARY Does transcription by RNA polymerase play a direct role in the initiation of replication? A. Bassim Hassan* and Peter R. Cook† CRC Nuclear Structure and Function Research Group, Sir William Dunn School of Pathology, University of Oxford, South Parks Road, Oxford, UK *Present address: Addenbrooke’s NHS Trust, Hills Rd, Cambridge CB2 2QQ, UK †Author for correspondence SUMMARY RNA polymerases have been implicated in the initiation of replication in bacteria. The conflicting evidence for a role in initiation in eukaryotes is reviewed. Key words: cell cycle, initiation, origin, replication, transcription PRIMERS AND PRIMASES complex becomes exclusively involved in the synthesis of Okazaki fragments on the lagging strand. A critical step in this DNA polymerases cannot initiate the synthesis of new DNA process is the unwinding of the duplex. chains, they can only elongate pre-existing primers. The opposite polarities of the two strands of the double helix coupled with the 5′r3′ polarity of the polymerase means that RNA POLYMERASES AND THE INITIATION OF replication occurs relatively continuously on one (leading) REPLICATION IN BACTERIA strand and discontinuously on the other (lagging) strand. The continuous strand probably needs to be primed once, usually The first evidence of a role for RNA polymerase came from at an origin. Nature has found many different ways of doing the demonstration that the initiation of replication in this, including the use of RNA primers made by an RNA poly- Escherichia coli was sensitive to rifampicin, an inhibitor of merase (e.g. -
Non-Coding Transcription Influences the Replication Initiation Program Through Chromatin Regulation
Downloaded from genome.cshlp.org on October 9, 2021 - Published by Cold Spring Harbor Laboratory Press Non-coding transcription influences the replication initiation program through chromatin regulation Julien Soudet1*, Jatinder Kaur Gill1 and Françoise Stutz1*. 1Dept. of Cell Biology, University of Geneva, Switzerland. *Correspondence: [email protected], [email protected] Keywords (10 words): non-coding RNA, non-coding transcription, histone modifications, nucleosomes, replication timing, replication initiation, yeast 1 Downloaded from genome.cshlp.org on October 9, 2021 - Published by Cold Spring Harbor Laboratory Press ABSTRACT In Eukaryotic organisms, replication initiation follows a temporal program. Among the parameters that regulate this program in Saccharomyces cerevisiae, chromatin structure has been at the center of attention without considering the contribution of transcription. Here, we revisit the replication initiation program in the light of widespread genomic non-coding transcription. We find that non-coding RNA transcription termination in the vicinity of ARS (Autonomously Replicating Sequences) shields replication initiation from transcriptional readthrough. Consistently, high natural nascent transcription correlates with low ARS efficiency and late replication timing. High readthrough transcription is also linked to increased nucleosome occupancy and high levels of H3K36me3. Moreover, forcing ARS readthrough transcription promotes these chromatin features. Finally, replication initiation defects induced by increased transcriptional readthrough are partially rescued in the absence of H3K36 methylation. Altogether, these observations indicate that natural non-coding transcription into ARS influences replication initiation through chromatin regulation. 2 Downloaded from genome.cshlp.org on October 9, 2021 - Published by Cold Spring Harbor Laboratory Press INTRODUCTION DNA replication is a fundamental process occurring in all living organisms and ensuring accurate duplication of the genome. -
Ch 7 -Brock Information Flow in Biological Systems
Systems Microbiology Monday Oct 2 - Ch 7 -Brock Information flow in biological systems •• DNADNA replicationreplication •• TranscriptionTranscription •• TranslationTranslation Central Dogma DNA Replication Transcription Images removed due to copyright restrictions. RNA Reverse Transcription Translation Protein Flow of information replication DNA → DNA transcription ↓ RNA translation ↓ protein 5' end ring numbering system for -P-O-C deoxyribose 5’ -C O O 4’ 1’ P O- O 3’ 2’ C ssDNA 3’ end HO In a nucleotide, e.g., adenosine monophosphate (AMP), the base is bonded to a ribose sugar, which has a phosphate in ester linkage to the 5' hydroxyl. NH NH NH2 2 2 adenine N N N N N N N N N N N N H −2 HO O3P O CH CH 5' 2 O 2 O 4' H H 1' H H ribose H 3' 2' H H H OH OH OH OH adenine adenosine adenosine monophosphate (AMP) Nucleic acids have a NH2 backbone of adenine N N alternating Pi & ribose moieties. N NH − N 2 Phosphodiester 5' end O cytosine − 5' O P O CH N linkages form as the 2 O 4' 1' O H H ribose 5' phosphate of one N O H 3' 2' H nucleotide forms an O OH − 5' ester link with the 3' O P O CH 2 O OH of the adjacent O H H ribose nucleotide. H 3' H O OH − O P O (etc) nucleic acid 3' end O H N H O N Guanine Cytosine N H N N N N O H N Backbone Backbone Hydrogen H bond H O H N CH3 N Thymine N H N N Adenine N N Hydrogen O bond Backbone Backbone Figure by MIT OCW. -
DNA : TACGCGTATACCGACATT Transcription Will Make Mrna From
Transcription and Translation Practice: Name _____________________________________ Background: • DNA controls our traits • DNA is found in the nucleus of our cells • Our traits are controlled by proteins • DNA is the instructions to make proteins • Proteins are made in ribosomes (outside the nucleus) • Proteins are made of amino acids Transcription makes RNA from DNA • RNA is complementary to DNA • RNA can leave the nucleus (DNA cannot) Translation makes proteins using RNA • Takes place at the ribosome • mRNA is “read” to put together a protein from amino acids Example: Beyonce has brown eyes. Her eyes look brown because her DNA codes for a brown pigment in the cells of her eyes. This is the gene that codes for brown eyes. DNA : T A C G C G T A T A C C G A C A T T Transcription will make mRNA from DNA mRNA: A U G C G C _________________________ Transcription will join amino acids to make the protein Rules for Transcription: Methionine - Arginine - ____________________________________ Base of DNA → Base in mRNA A → U Rules of Translation: C → G Three letters of mRNA = a codon G → C A codon “codes” for an amino acid We use the “Genetic Code” to determine the amino acids T → A RNA contains Uracil instead of Thymine The complete chain of amino acids will complete the protein that will give Beyonce her brown eyes. If you have brown eyes you have the same protein. If you have blue or green eyes your DNA sequence is a little different which will make the amino acid sequence (protein) a little different. -
Lecture 1 1. to Understand the Flow of Biochemical Information from DNA to RNA to Proteins and Ultimately to Biochemical And
Lecture 1 1. To understand the flow of biochemical information from DNA to RNA to proteins and ultimately to biochemical and cellular function and dysfunction (disease) Central Dogma: DNARNAProteinPhenotype NucleotidegeneDNAchromosomegenome - In reality it is much more complex DNA sequence RNA sequence Protein sequence Protein structure Protein function RNA structure RNA function Know: - Structure of DNA/RNA and proteins - Intermolecular interactions: o Hydrogen bonds o Disulfur bonds o Ion bonds o Polar bonds - Main aspects of the central dogma o Translation (at the ribosome) o Transcription o Genetic code DNA structure: - Sugar-phosphate backbone - 4 bases o A/G = purines o T/C = pyrimidines o A-T and G-C - 5’-3’ antiparallel (always read 5’3’) o Addition occurs at the 3’ end o 5’ position to commence reading depends on promotor o Reading frame depends on promotor - Coding and template strand: o Depends on the position of the promotor RNA structure: - U replaces T - Self-complementarity = annealing of strand to itself - tRNA/mRNA/pre-RNA - Spliced (exons remain) to form mature RNA Therefore 6: - 3 from the codon from the top 5’ end, 3 from the codon from the bottom 5’ end. Protein Structure: Chemical Properties Depends on N or C-terminus, peptide bonds and side chains - Non-polar aliphatic - Polar but uncharged - Aromatic - Positively charged - Negatively charged pKa = pH at which the protein has a charge of zero. Alpha-helices: side chains point sideways Beta-helices: - Parallel and anti-parallel to produce alternate the direction the side chains point - In reality, there is a combination of parallel and anti-parallel side chains - Different bonds between NH and CO groups in each direction of side chain Sample Question 1. -
Transcription Study Guide This Study Guide Is a Written Version of the Material You Have Seen Presented in the Transcription Unit
Transcription Study Guide This study guide is a written version of the material you have seen presented in the transcription unit. The cell’s DNA contains the instructions for carrying out the work of the cell. These instructions are used by the cell’s protein-making machinery to create proteins. If the cell’s DNA were directly read by the protein-making machinery, however, it could be damaged and the process would be slow and cumbersome. The cell avoids this problem by copying genetic information from its DNA into an intermediate called messenger RNA (mRNA). It is this mRNA that is read by the cell’s protein-making machinery. This process is called transcription. Components In this section you will be introduced to the components involved in the process of RNA synthesis, called transcription. This process requires an enzyme that uses many nucleotide bases to copy the instructions present in DNA into an intermediate messenger RNA molecule. RNA What is RNA? · Like DNA, RNA is a polymer made up of nucleotides. · Unlike DNA, which is composed of two strands of nucleotides twisted together, RNA is single-stranded. It can also sometimes fold into complex three-dimensional structures. · RNA contains the same nucleotides as DNA, with the substitution of uraciluridine (U) for thymidine (T). · RNA is chemically different from DNA so that the cell can easily tell the two apart. · In this animation, you will see one type of RNA, messenger RNA, being put together. · There are three types of RNA: mRNA, which you will read more about; tRNA, which is used in the translation process, and rRNA, which acts as a structural element in the ribosome (a translation component). -
IGA 8/E Chapter 8
8 RNA: Transcription and Processing WORKING WITH THE FIGURES 1. In Figure 8-3, why are the arrows for genes 1 and 2 pointing in opposite directions? Answer: The arrows for genes 1 and 2 indicate the direction of transcription, which is always 5 to 3. The two genes are transcribed from opposite DNA strands, which are antiparallel, so the genes must be transcribed in opposite directions to maintain the 5 to 3 direction of transcription. 2. In Figure 8-5, draw the “one gene” at much higher resolution with the following components: DNA, RNA polymerase(s), RNA(s). Answer: At the higher resolution, the feathery structures become RNA transcripts, with the longer transcripts occurring nearer the termination of the gene. The RNA in this drawing has been straightened out to illustrate the progressively longer transcripts. 3. In Figure 8-6, describe where the gene promoter is located. Chapter Eight 271 Answer: The promoter is located to the left (upstream) of the 3 end of the template strand. From this sequence it cannot be determined how far the promoter would be from the 5 end of the mRNA. 4. In Figure 8-9b, write a sequence that could form the hairpin loop structure. Answer: Any sequence that contains inverted complementary regions separated by a noncomplementary one would form a hairpin. One sequence would be: ACGCAAGCUUACCGAUUAUUGUAAGCUUGAAG The two bold-faced sequences would pair and form a hairpin. The intervening non-bold sequence would be the loop. 5. How do you know that the events in Figure 8-13 are occurring in the nucleus? Answer: The figure shows a double-stranded DNA molecule from which RNA is being transcribed. -
Consequences and Resolution of Transcription–Replication Conflicts
life Review Consequences and Resolution of Transcription–Replication Conflicts Maxime Lalonde †, Manuel Trauner †, Marcel Werner † and Stephan Hamperl * Institute of Epigenetics and Stem Cells (IES), Helmholtz Zentrum München, 81377 Munich, Germany; [email protected] (M.L.); [email protected] (M.T.); [email protected] (M.W.) * Correspondence: [email protected] † These authors contributed equally. Abstract: Transcription–replication conflicts occur when the two critical cellular machineries respon- sible for gene expression and genome duplication collide with each other on the same genomic location. Although both prokaryotic and eukaryotic cells have evolved multiple mechanisms to coordinate these processes on individual chromosomes, it is now clear that conflicts can arise due to aberrant transcription regulation and premature proliferation, leading to DNA replication stress and genomic instability. As both are considered hallmarks of aging and human diseases such as cancer, understanding the cellular consequences of conflicts is of paramount importance. In this article, we summarize our current knowledge on where and when collisions occur and how these en- counters affect the genome and chromatin landscape of cells. Finally, we conclude with the different cellular pathways and multiple mechanisms that cells have put in place at conflict sites to ensure the resolution of conflicts and accurate genome duplication. Citation: Lalonde, M.; Trauner, M.; Keywords: transcription–replication conflicts; genomic instability; R-loops; torsional stress; common Werner, M.; Hamperl, S. fragile sites; early replicating fragile sites; replication stress; chromatin; fork reversal; MIDAS; G-MiDS Consequences and Resolution of Transcription–Replication Conflicts. Life 2021, 11, 637. https://doi.org/ 10.3390/life11070637 1. -
How Genes Work
Help Me Understand Genetics How Genes Work Reprinted from MedlinePlus Genetics U.S. National Library of Medicine National Institutes of Health Department of Health & Human Services Table of Contents 1 What are proteins and what do they do? 1 2 How do genes direct the production of proteins? 5 3 Can genes be turned on and off in cells? 7 4 What is epigenetics? 8 5 How do cells divide? 10 6 How do genes control the growth and division of cells? 12 7 How do geneticists indicate the location of a gene? 16 Reprinted from MedlinePlus Genetics (https://medlineplus.gov/genetics/) i How Genes Work 1 What are proteins and what do they do? Proteins are large, complex molecules that play many critical roles in the body. They do most of the work in cells and are required for the structure, function, and regulation of thebody’s tissues and organs. Proteins are made up of hundreds or thousands of smaller units called amino acids, which are attached to one another in long chains. There are 20 different types of amino acids that can be combined to make a protein. The sequence of amino acids determineseach protein’s unique 3-dimensional structure and its specific function. Aminoacids are coded by combinations of three DNA building blocks (nucleotides), determined by the sequence of genes. Proteins can be described according to their large range of functions in the body, listed inalphabetical order: Antibody. Antibodies bind to specific foreign particles, such as viruses and bacteria, to help protect the body. Example: Immunoglobulin G (IgG) (Figure 1) Enzyme. -
LESSON 4 Using Bioinformatics to Analyze Protein Sequences
LESSON 4 Using Bioinformatics to 4 Analyze Protein Sequences Introduction In this lesson, students perform a paper exercise designed to reinforce the student understanding of the complementary nature of DNA and how that complementarity leads to six potential protein reading frames in any given DNA sequence. They also gain familiarity with the circular format codon table. Students then use the bioinformatics tool ORF Finder to identify the reading frames in their DNA sequence from Lesson Two and Lesson Three, and to select Class Time the proper open reading frame to use in a multiple sequence alignment with 2 class periods (approximately 50 their protein sequences. In Lesson Four, students also learn how biological minutes each). anthropologists might use bioinformatics tools in their career. Prior Knowledge Needed • DNA contains the genetic information Learning Objectives that encodes traits. • DNA is double stranded and At the end of this lesson, students will know that: anti-parallel. • Each DNA molecule is composed of two complementary strands, which are • The beginning of a DNA strand is arranged anti-parallel to one another. called the 5’ (“five prime”) region and • There are three potential reading frames on each strand of DNA, and a total the end of a DNA strand is called the of six potential reading frames for protein translation in any given region of 3’ (“three prime”) region. the DNA molecule (three on each strand). • Proteins are produced through the processes of transcription and At the end of this lesson, students will be able to: translation. • Amino acids are encoded by • Identify the best open reading frame among the six possible reading frames nucleotide triplets called codons.