GEP Glossary
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Biomolecules
CHAPTER 3 Biomolecules 3.1 Carbohydrates In the previous chapter you have learnt about the cell and 3.2 Fatty Acids and its organelles. Each organelle has distinct structure and Lipids therefore performs different function. For example, cell membrane is made up of lipids and proteins. Cell wall is 3.3 Amino Acids made up of carbohydrates. Chromosomes are made up of 3.4 Protein Structure protein and nucleic acid, i.e., DNA and ribosomes are made 3.5 Nucleic Acids up of protein and nucleic acids, i.e., RNA. These ingredients of cellular organelles are also called macromolecules or biomolecules. There are four major types of biomolecules— carbohydrates, proteins, lipids and nucleic acids. Apart from being structural entities of the cell, these biomolecules play important functions in cellular processes. In this chapter you will study the structure and functions of these biomolecules. 3.1 CARBOHYDRATES Carbohydrates are one of the most abundant classes of biomolecules in nature and found widely distributed in all life forms. Chemically, they are aldehyde and ketone derivatives of the polyhydric alcohols. Major role of carbohydrates in living organisms is to function as a primary source of energy. These molecules also serve as energy stores, 2021-22 Chapter 3 Carbohydrade Final 30.018.2018.indd 50 11/14/2019 10:11:16 AM 51 BIOMOLECULES metabolic intermediates, and one of the major components of bacterial and plant cell wall. Also, these are part of DNA and RNA, which you will study later in this chapter. The cell walls of bacteria and plants are made up of polymers of carbohydrates. -
Site-Selective Artificial Ribonucleases: Renaissance of Oligonucleotide Conjugates for Irreversible Cleavage of RNA Sequences
molecules Review Site-Selective Artificial Ribonucleases: Renaissance of Oligonucleotide Conjugates for Irreversible Cleavage of RNA Sequences Yaroslav Staroseletz 1,†, Svetlana Gaponova 1,†, Olga Patutina 1, Elena Bichenkova 2 , Bahareh Amirloo 2, Thomas Heyman 2, Daria Chiglintseva 1 and Marina Zenkova 1,* 1 Laboratory of Nucleic Acids Biochemistry, Institute of Chemical Biology and Fundamental Medicine SB RAS, Lavrentiev’s Ave. 8, 630090 Novosibirsk, Russia; [email protected] (Y.S.); [email protected] (S.G.); [email protected] (O.P.); [email protected] (D.C.) 2 School of Health Sciences, Faculty of Biology, Medicine and Health, University of Manchester, Oxford Rd., Manchester M13 9PT, UK; [email protected] (E.B.); [email protected] (B.A.); [email protected] (T.H.) * Correspondence: [email protected]; Tel.: +7-383-363-51-60 † These authors contributed equally to this work. Abstract: RNA-targeting therapeutics require highly efficient sequence-specific devices capable of RNA irreversible degradation in vivo. The most developed methods of sequence-specific RNA cleav- age, such as siRNA or antisense oligonucleotides (ASO), are currently based on recruitment of either intracellular multi-protein complexes or enzymes, leaving alternative approaches (e.g., ribozymes Citation: Staroseletz, Y.; Gaponova, and DNAzymes) far behind. Recently, site-selective artificial ribonucleases combining the oligonu- S.; Patutina, O.; Bichenkova, E.; cleotide recognition motifs (or their structural -
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). -
Molecular Biology and Applied Genetics
MOLECULAR BIOLOGY AND APPLIED GENETICS FOR Medical Laboratory Technology Students Upgraded Lecture Note Series Mohammed Awole Adem Jimma University MOLECULAR BIOLOGY AND APPLIED GENETICS For Medical Laboratory Technician Students Lecture Note Series Mohammed Awole Adem Upgraded - 2006 In collaboration with The Carter Center (EPHTI) and The Federal Democratic Republic of Ethiopia Ministry of Education and Ministry of Health Jimma University PREFACE The problem faced today in the learning and teaching of Applied Genetics and Molecular Biology for laboratory technologists in universities, colleges andhealth institutions primarily from the unavailability of textbooks that focus on the needs of Ethiopian students. This lecture note has been prepared with the primary aim of alleviating the problems encountered in the teaching of Medical Applied Genetics and Molecular Biology course and in minimizing discrepancies prevailing among the different teaching and training health institutions. It can also be used in teaching any introductory course on medical Applied Genetics and Molecular Biology and as a reference material. This lecture note is specifically designed for medical laboratory technologists, and includes only those areas of molecular cell biology and Applied Genetics relevant to degree-level understanding of modern laboratory technology. Since genetics is prerequisite course to molecular biology, the lecture note starts with Genetics i followed by Molecular Biology. It provides students with molecular background to enable them to understand and critically analyze recent advances in laboratory sciences. Finally, it contains a glossary, which summarizes important terminologies used in the text. Each chapter begins by specific learning objectives and at the end of each chapter review questions are also included. -
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. -
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. -
Nucleosides & Nucleotides
Nucleosides & Nucleotides Biochemistry Fundamentals > Genetic Information > Genetic Information NUCLEOSIDE AND NUCLEOTIDES SUMMARY NUCLEOSIDES  • Comprise a sugar and a base NUCLEOTIDES  • Phosphorylated nucleosides (at least one phosphorus group) • Link in chains to form polymers called nucleic acids (i.e. DNA and RNA) N-BETA-GLYCOSIDIC BOND  • Links nitrogenous base to sugar in nucleotides and nucleosides • Purines: C1 of sugar bonds with N9 of base • Pyrimidines: C1 of sugar bonds with N1 of base PHOSPHOESTER BOND • Links C3 or C5 hydroxyl group of sugar to phosphate NITROGENOUS BASES  • Adenine • Guanine • Cytosine • Thymine (DNA) 1 / 8 • Uracil (RNA) NUCLEOSIDES • =sugar + base • Adenosine • Guanosine • Cytidine • Thymidine • Uridine NUCLEOTIDE MONOPHOSPHATES – ADD SUFFIX 'SYLATE' • = nucleoside + 1 phosphate group • Adenylate • Guanylate • Cytidylate • Thymidylate • Uridylate Add prefix 'deoxy' when the ribose is a deoxyribose: lacks a hydroxyl group at C2. • Thymine only exists in DNA (deoxy prefix unnecessary for this reason) • Uracil only exists in RNA NUCLEIC ACIDS (DNA AND RNA)  • Phosphodiester bonds: a phosphate group attached to C5 of one sugar bonds with - OH group on C3 of next sugar • Nucleotide monomers of nucleic acids exist as triphosphates • Nucleotide polymers (i.e. nucleic acids) are monophosphates • 5' end is free phosphate group attached to C5 • 3' end is free -OH group attached to C3 2 / 8 FULL-LENGTH TEXT • Here we will learn about learn about nucleoside and nucleotide structure, and how they create the backbones of nucleic acids (DNA and RNA). • Start a table, so we can address key features of nucleosides and nucleotides. • Denote that nucleosides comprise a sugar and a base. -
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. -
Base Paring Rules for Transcription
Base Paring Rules For Transcription Jerry window-shop his Jon traversings unduly or shortly after Giff vacillated and nerves turbulently, Silvanintoxicated labialised and camouflaged. unofficially if attentionalSynchronistic Martin Moore compel twigging, or puzzling. his clapboard curses leasing grindingly. Since the template strand will contain base pairs that bond precisely with the. Dna Base Pair Worksheet Teachers Pay Teachers. Most images show 17 base pairs The coding strand turns gray page then disappears leaving the template strand see strands above Anti-codons in the template. DNA Base Pairing Worksheet 1 CGTAAGCGCTAATTA 2. One strand whereas the DNA duplex the template strand is transcribed into a segment of mRNA shown according to the prior base-pairing rules used in DNA. As surgery the damage of DNA replication base-pairing rules apply However. Base pairs refer them the sets of hydrogen-linked nucleobases that window up nucleic acids DNA and RNA. Students begin by replicating a DNA strand and transcribing the DNA strand into RNA Then they create base pairing rules and search the. Topic 27 DNA Replication Transcription and Translation. Transcribe the DNA strand until the complementary base pairs. Therefore translation of the messenger RNA transcribed from this mutant. DNA RNA Transcription and Translation Protein Synthesis. The base-pairing rules summarize which The template strand despite the DNA contains the gene and is being transcribed pairs of nucleotides are complementary. Transcription Translation & Protein Synthesis Gene. The cell's DNA contains the instructions for carrying out the work of border cell. Rules of Base Pairing A with T the purine adenine A always pairs with the pyrimidine thymine T C with G the pyrimidine cytosine C always. -
DNA Wrap: Packaging Matters an Introduction to Epigenetics UNC-Chapel Hill’S Superfund Research Program
DNA Wrap: Packaging Matters An Introduction to Epigenetics UNC-Chapel Hill’s Superfund Research Program What causes the physical appearance and health status of identical twins to diverge with age? In this lesson, students learn that the environment can alter the way our genes are expressed, making even identical twins different. After watching a PBS video, A Tale of Two Mice, and reviewing data presented in the Environmental Health Perspectives article Maternal Genistein Alters Coat Color and Protects Avy Mouse Offspring from Obesity by Modifying the Fetal Epigenome, students learn about epigenetics and its role in regulating gene expression. Author Dana Haine, MS University of North Carolina at Chapel Hill Superfund Research Program Reviewers Dana Dolinoy, PhD University of Michigan Rebecca Fry, PhD University of North Carolina at Chapel Hill Superfund Research Program Banalata Sen, PhD, Audrey Pinto, PhD, Susan Booker, Dorothy Ritter Environmental Health Perspectives The funding for development of this lesson was provided by the National Institute of Environmental Health Sciences and the UNC Superfund Program. Learning Objectives By the end of this lesson students should be able to: define the term “epigenetics” describe DNA methylation as a mechanism for inhibiting gene transcription describe how gene expression can vary among genetically identical offspring Alignment to NC Essential Science Standards for Biology Bio.3.1 Explain how traits are determined by the structure and function of DNA. Bio.3.2.3 Explain how the environment can influence the expression of genetic traits. Bio.4.1 Understand how biological molecules are essential to the survival of living organisms. Bio.4.1.2 Summarize the relationship among DNA, proteins and amino acids in carrying out the work of cells and how this is similar in all organisms. -
Yeast Ribonuclease H(70) Cleaves RNA-DNA Junctions
View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Volume 206, number 2 FEBS 4062 October 1986 Yeast ribonuclease H(70) cleaves RNA-DNA junctions Robert Karwan and Ulrike Wintersberger Department of Molecular GEnetics, Institute for Tumorbiology and Cancer Research, University of Vienna, Borschkegasse 8a, A-1090 Wien, Austria Received 7 August 1986 A specific substrate, Ml3 DNA:RNA-[3zP]DNA, was synthesized to investigate the mode of cleavage of enzymes with RNase H activity. RNase H(70) from Saccharomyces cerevisiae hydrolyzes the phosphodiester bond at the RNA-DNA junction of this substrate, thereby producing a 5’-monophosphate-terminated poly- deoxyribonucleotide and 3’-hydroxyl-terminated oligoribonucleotides. RNase H DNA-RNA hybrid DNA replication RNA primer removal (Saccharomyces cerevisiae) 1. INTRODUCTION ing short RNA primers. The phosphorus atom of the phosphate bridging the RNA and DNA parts Recently we have purified a ribonuclease H of this polynucleotide is radiolabeled (see fig.1). (RNase H, i.e. an enzyme which specifically Here we show that, in vitro the yeast RNase H(70) hydrolyzes the RNA strand of a DNA-RNA preferentially hydrolyzes the phosphodiester bond hybrid) from the yeast, Saccharomyces cerevisiae, between the ribonucleotide and the deox- which stimulates the in vitro DNA synthesis by yribonucleotide at the RNA-DNA junction. DNA polymerase A from the same organism [ 1,2]. Under certain conditions this protein also exhibits 2. MATERIALS AND METHODS reverse transcriptase activity [3]. As with RNases H from other organisms [4] no definite RNase H(70) and DNA polymerase A from S. -
212 Chapter 28 Biomolecules: Heterocycles and Nucleic Acids
Chapter 28 Biomolecules: Heterocycles and Nucleic Acids Heterocycles: cyclic organic compounds that contain rings atoms other than carbon (N,S,O are the most common). 28.1 Five-Membered Unsaturated Heterocycles (please read) 28.2 Structures of Pyrrole, Furan, and Thiophene (please read) 3 3 3 2 2 2 N 1 O 1 S 1 H Cyclopentadienyl Pyrole Furan Thiophene anion 28.3 Electrophilic Substitution Reactions of Pyrrole, Furan, and Thiophene (please read) 28.4 Pyridine, a Six-Membered Heterocycle (please read) 28.5 Electrophilic Substitution of Pyridine (please read) 28.6 Nucleophilic Substitution of Pyridine (please read) 28.7 Fused-Ring Heterocycles (please read) These sections contain some important concepts that were covered previously. 418 28.8 Nucleic Acids and Nucleotides Nucleic acids are the third class of biopolymers (polysaccharides and proteins being the others) Two major classes of nucleic acids deoxyribonucleic acid (DNA): carrier of genetic information ribonucleic acid (RNA): an intermediate in the expression of genetic information and other diverse roles The Central Dogma (F. Crick): DNA mRNA Protein (genome) (proteome) The monomeric units for nucleic acids are nucleotides Nucleotides are made up of three structural subunits 1. Sugar: ribose in RNA, 2-deoxyribose in DNA 2. Heterocyclic base 3. Phosphate 419 212 Nucleoside, nucleotides and nucleic acids phosphate sugar base phosphate phosphate sugar base sugar base sugar base phosphate nucleoside nucleotides sugar base nucleic acids The chemical linkage between monomer units in nucleic