Module 2 DNA Structure PDF

Principles of Biochemistry Fourth Edition Donald Voet Judith G. Voet Charlotte W. Pratt Module 2 Chapter 3 Nucleotides, Nucleic Acids, and Genetic Information  Specific outcomes Describe in detail each component of a nucleotide Identify...

Principles of Biochemistry Fourth Edition Donald Voet Judith G. Voet Charlotte W. Pratt Module 2 Chapter 3 Nucleotides, Nucleic Acids, and Genetic Information  Specific outcomes Describe in detail each component of a nucleotide Identify and draw the components of nucleotides Discuss the metabolic function of ATP and its derivatives Give an overview of the structure of different nucleic acids Explain Chargaff’s rules of DNA base composition Discuss in detail the DNA double helix structure  Specific outcomes Discuss the function of DNA and RNA in the central dogma of life Explain the principle and function of the following in laboratory techniques: Restriction endonucleases Electrophoresis and restriction mapping Genome sequencing Recombinant DNA technology  Introduction Nucleotides are the building blocks of nucleic acids Nucleotides participate in redox reactions, energy transfer, cell signalling and biosynthetic reactions Nucleic acids (DNA & RNA) store and decode genetic information The structure and chemistry of nucleotides and nucleic acids determine how biological information is expressed in relation to evolution, metabolism and disease Biological information is expressed through the central dogma Transcription Translation DNA RNA Protein  Nucleotides Nucleotides consist of: I. A sugar (ribose or deoxyribose) II. A nitrogenous base attached to the sugar (purine or pyrimidine) III. One to three phosphate groups attached to the sugar (C3’ or C5’ atom) There are 8 nucleotides DNA contains adenine, guanine, cytosine and thymine deoxyribonucleotides RNA contains, adenine, guanine, cytosine and uracil ribonucleotides  Nucleotides: Sugar The sugar molecule is a pentose Ribonucleotides (RNA) – ribose Deoxyribonucleotides (DNA) - deoxyribose C2’ - C2’ - H OH  Nucleotides: Base The nitrogenous base is a purine or a pyrimidine Purines – Adenine (A) and Guanine (G) Pyrimidines – Thymine (T), Uracil (U) and Cytosine (C) The nitrogenous base attaches to C1’ of the pentose Purines attach via N9 atom Pyrimidines attach via N1 atom  Nucleotides: Phosphate One or more phosphates at C3’ or C5’ of the pentose 0 – nucleoside 1 – nucleoside monophosphate Sugar? 2 – nucleoside diphosphate 3 – nucleoside triphosphate  Nucleotides: Overview A, G and C are found in both DNA and RNA T is only found in DNA U is only found in RNA Learn Table 3.1 Nucleotides: Metabolic reactions Free nucleotides and their derivatives can perform metabolic functions Best example is the energy carrier ATP Breakdown of carbohydrates and fatty acids ADP + Pi + Energy → ATP + H2O Cellular work – phosphate groups are transferred ATP + H2O → ADP + Pi + Energy → AMP + Pi + Energy  Nucleotides → Nucleic Many (poly) nucleotidesacids – nucleic acids Two nucleotides are joined by a DNA or phosphodiester bond between the RNA? C3’ of one sugar and the C5’ of the Why? other sugar At physiological pH, nucleic acids are anionic due to the acidic nature of the phosphate groups Each nucleotide (in DNA/RNA) is called a nucleotide residue 5’ end – terminal residue C5’ unlinked 3’ end – terminal residue C3’ unlinked Nucleotide sequence usually written from 5’ – 3’  Structure of DNA Watson and Crick (1953) derived the double helix structure of DNA using Chargraff’s rules (in DNA, A = T and G = C) The keto tautomeric forms of nucleic acid bases are dominant Rosalind Franklin’s X-ray diffraction photograph of a DNA fiber  Structure of DNA – Watson & Crick Model Two polynucleotide chains wind around a common axis to form a double helix The two strands are antiparallel but each form a right-handed helix Bases occupy the core while sugar-phosphate chain runs along the periphery. The surface contains a major groove and minor groove of unequal width A base pair between the two strands occurs by H-bonding between A-T and G-C. This is known as complementary base pairing How many H-bonds between A-T and G-C?  DNA is described in base pairs (bp) or thousands of bps (kb)  Structure of RNA RNA occurs primarily as single strands arranged into compact structures Certain viruses can carry ss-DNA or ds-RNA An RNA strand can base pair with a complementary RNA or DNA strand A base pairs with U (RNA) or T (DNA) Stem-loop structures can occur due to intramolecular base  DNA carries genetic information 1866 – Gregor Mendel postulates that an individual plant contains a pair of genes, inherited from each parent Assumption – genes were made of protein 1944 - Oswald Avery, Colin McLeod and Maclyn McCarthy showed that virulent DNA transformed a nonpathogenic form of Diplococcus pneumoniae into the pathogenic strain. DNA carries genetic info in all cells and many viruses  DNA carries genetic information Double stranded (ds) nature facilitates replication Cell division - each strand is a template for the complementary strand Progeny – complete set of DNA molecules made of 1 parent strand and 1 daughter strand Polymerisation of nucleotides that pair with the bases on the parental (template) strand Meselson & Stahl experiment  Genes direct protein synthesis George Beadle and Edward Tatum theory based on Neurospora crassa nutrient- deficient mutants: 1 gene = 1 enzyme (protein) Crick’s central dogma Beadle & Tatum experiment  Genes direct protein synthesis The 3’-5’ DNA strand is the template strand for messenger RNA (mRNA) synthesis mRNA corresponds to a protein coding gene (it is the same sequence as the 5’-3’ DNA stand) mRNA moves to the ribosome - made of ribosomal RNA (rRNA) Three nucleotides on mRNA pair with 3 nucleotides on transfer RNA (tRNA) Each tRNA has a corresponding amino acid The ribosome catalyses the polymerisation of amino acids to form a protein How many types of RNA are there? Standard genetic code Standard genetic code  Genes direct protein synthesis DNA 5’- ACCGTGCTTACGGGGTACATA – 3’ Complementary 3’- TGGCACGAATGCCCCATGTAT – 5’ mRNA 5’- ACCGUGCUUACGGGGUACAUA -3’ t-RNA 3’- UGGCACGAAUGCCCCAUGUAU -5’ Protein Thr-Val-Leu-Thr-Gly- A change Tyr-Ile in DNA sequence (mutation) = change in protein sequence = change in structure and  What is the difference between deoyribose and ribose? 1. Deoxyribose has a 3’ OH and ribose has a 3’ H 2. Deoxyribose has a 3’H and ribose has a 3’ OH 3. Deoxyribose has a 2’ OH and ribose has a 2’ H 4. Deoxyribose has a 2’ H and ribose has a 2’ OH  When there are no phosphate groups attached to a cytidine-deoxyribose, it is called a 1. Nucleoside 2. Nucleotide  Which of the following bases are purines? 1. T, C, U 2. A, G 3. A, G, U 4. A, T  Which of the following is not a difference between DNA and RNA 1. RNA – ribonucleotides; DNA – deoxyribonucleotides 2. RNA – U; DNA – T 3. RNA – single stranded; DNA – double stranded 4. RNA – only found in viruses; DNA – only found in animals  Which of the following is a complementary base pair? 1. A-G 2. T-U 3. A-U 4. G-T  Restriction endonucleases Bacteria can resist bacteriophage infection (bacteria specific viruses) by methylating (CH3) specific nucleotides in its DNA using a methylase Restriction endonucleases recognise the same sequence as methylase and cleave any DNA that has not been methylated on at least one of its strands Why is the host daughter DNA not cleaved immediately after replication? Type II restriction endonucleases cleave within a 4 – 8 base sequence recognised by the methylase Useful in the laboratory  Type II Restriction endonucleases Naming of restriction endonucleases 1st letter – Genus 2nd two letters – Species 4th letter – serotype or strain Roman numeral –number of the restriction enzyme if more than one type E.g. HindIII – Haemophilus influenzae Rd Cleave a palindromic (mirror) sequence 5’- GAATTC - 5’- AAGCTT - 5’- TCGA - 3’ 3’ 3’ 3’- CTTAAG - 3’- TTCGAA - 3’- AGCT- 3’ 3’ 3’  Type II Restriction endonucleases Cleavage at staggered positions = sticky ends (can base pair with other restriction fragments generated by the same restriction enzyme) Cleavage at symmetry axis – blunt ends Sticky or blunt? EcoRI HindIII AluI 5’- G*AATTC 5’- A*AGCTT 5’- AG*CT -3’ -3’ -3’ 3’- CTTAA*G 3’- TTCGA*A 3’- TC*GA- EcoRI -5’ -5’ HindIII 5’ AluI 5’-G AATTC 5’- A 5’- AG -3’ AGCTT -3’ CT -3’  Type II Restriction endonucleases The recognition site for TaqI is T↓CGA. Indicate the products for the following DNA sequence 5’-ACGTCGAATC-3’ 3’-TGCAGCTTAG-5’ Did TaqI produce sticky or blunt ends?  Nucleic acid gel electrophoresis Nucleic acid separated according to size by gel electrophoresis In an electric field, the velocity of a charged molecules is proportional to overall charge density, size and shape Nucleic acids have constant shape and charge density; velocity depends on size  Nucleic acid gel electrophoresis Gel-like matrix (agarose or polyacrylamide) sandwiched between glass plates Molecules applied at one end of an electric field Move to the other end according to size Smaller molecules move through the pores faster and migrate farther What is the charge found on DNA or RNA?  Nucleic acid gel electrophoresis Separated fragments are visualised as bands after DNA Undigest NdeI + staining: marker ed DNA NdeI SspI SspI Dye that binds to DNA Radioactive labelling Fluorescence Can add a molecular weight ladder to estimate size Set of standards of known sizes Derived from cleavage of a recognised DNA sequence by restriction enzymes or PCR or DNA ligation DNA fingerprinting Which suspect’s profile matches the crime scene? DNA fingerprinting Who is the baby’s daddy?  Restriction enzyme Recognition sequence (5’→3’) EcoRI G↓AATTC HaeIII GG↓CC BamHI G↓GATCC PvuII CAG↓CTG 3’-ATCCGTGTCGACCGTACCATGTC-5’ 1. Which restriction enzyme(s) cuts this 5’-TAGGCACAGCTGGCATGGTACAG-3’ fragment? 2. How many fragments are generated? 3’-ATCCGTGTC-5’ 3’-GACCGTACCATGTC-5’ 3. Show the fragments generated. 5’-TAGGCACAG-3’ 5’-CTGGCATGGTACAG-3’ M 1 2 3 4 5 4. Blunt or sticky? 5. What are the sizes of the fragments? 20 bp 6. How would they separate on an agarose gel? 10 bp 5 bp  Restriction enzyme Recognition sequence (5’→3’) EcoRI G↓AATTC HaeIII GG↓CC BamHI G↓GATCC PvuII CAG↓CTG 1. Which restriction enzyme(s) cuts 5’-TAGGCACGGATCCGAGAATTCGAGGGCT-3’ this fragment? 3’-ATCCGTGCCTAGGCTCTTAAGCTCCCGA-5’ 2. How many fragments are 5’-AATTCGAGGGCT-3’ 5’-TAGGCACG-3’ 5’-GATCCGAG-3’ generated? 3’-GCTCCCGA-5’ 3’-GCTCTTAA - 3. Show the fragments generated. 3’-ATCCGTGCCTAG -5’ 5’ M 1 2 3 4 5 4. Blunt or sticky? 5. What are the sizes of the 20 bp fragments? 6. How would they separate on an 10 bp agarose gel? 5 bp  Restriction enzyme Recognition sequence (5’→3’) AluI AG↓CT EcoRV GAT↓ATC BamHI G↓GATCC XhoI C↓TCGAG 1. Which restriction enzyme(s) cuts 3’-TACGACGAGCTCTCAAGCGTCGACCGG-5’ this fragment? 5’-ATGCTGCTCGAGAGTTCGCAGCTGGCC-3’ 2. How many fragments are generated? 3’-TACGACGAGCT-5’ 3’- CTCAAGCGTC -5’ 3’-GACCGG-5’ 5’- ATGCTGC -3’ 5’-TCGAGAGTTCGCAG -3’ 5’-CTGGCC-3’ 3. Show the fragments generated. 4. Blunt or sticky? M 1 2 3 4 5 5. What are the sizes of the fragments? 20 bp 6. How would they separate on an agarose gel? 10 bp 5 bp  Polymerase Chain Reaction PCR is a way of amplifying specific DNA (≤ 6 kb) Mix target DNA with Heat stable DNA polymerase I (e.g. Taq polymerase) – sequentially adds nucleotides in the 5’→3’ direction Deoxynucleoside triphosphates (dNTPs) - list them Two oligonucleotide primers – flank the DNA segment of interest dsDNA is separated into single strands by heating to break H-bonds (between ?) - https://www.youtube.com/watch?v=c07_5BfIDTw denaturation  Polymerase Chain Reaction DNA polymerase requires a free 3’-OH group to add a nucleotide to Primers complementary to the 3’ end of the template strand can base-pair with the template and allow DNA polymerase to add nucleotides NB: The amplified strand is complementary to the template strand and is synthesised in the 5’→3’ direction  Polymerase Chain Reaction Multiple cycles of PCR results in the doubling of the amount of the target DNA i.e. amplification from as little as a single gene copy At the end of 1 cycle there are 22 strands of DNA Number of DNA molecules at the end of a cycle = 2n+1 where n is the number of cycles PCR can be used for: Sequencing e.g. forensics, infectious disease diagnosis, detection of mutations Cloning e.g. study models, therapeutics, What is PCR used for? 1. To sequence DNA 2. To amplify DNA 3. To digest DNA 4. To separate DNA Which of the following is not part of the standard PCR reaction? 1. dNTPs 2. Primer 3. DNA polymerase 4. Restriction enzyme Primers are used to enable extension of the DNA strand in PCR because 1. DNA polymerase can add nucleotides to both the 5’ and 3’ ends 2. DNA polymerase can only add nucleotides to a free 5’ OH group 3. DNA polymerase can only add nucleotides to a free 3’ OH group 4. They are short  Recombinant DNA technology The isolation, amplification and modification of specific DNA sequences Also called molecular cloning or genetic engineering Cloning refers to the production of many identical organisms from a single ancestor A clone is a collection of cells containing a vector carrying the DNA of interest Aim of cloning: To produce lots of DNA To express lots of protein  Overview of cloning 1. Generate the fragment of DNA of interest by restriction enzyme, PCR or chemical synthesis 2. Ligation - insert the fragment into another DNA molecule called a vector that carries the sequences necessary to direct replication (why?). 3. Transformation - introduce the vector (with DNA of interest) into the host cells and allow them to replicate 4. Selection/screening – cells with the desired DNA (positive transformants) are identified and selected  Cloning: vectors Small, autonomously replicating DNA molecules Plasmids are circular DNA molecules (upto 200 kb) found in bacteria or yeast Present in hundreds of copies within the cell Small and replicate easily Carry genes encoding antibiotic resistance Carry restriction endonuclease sites for foreign DNA insertion Can clone ≤ 10 kb Example is pUC18  Cloning: ligation A restriction fragment is inserted into a cut made in a cloning vector by the same restriction enzyme (sticky ends) The ends of the DNA fragment and vector are complementary and base pair (anneal) The sugar-phosphate backbone are covalently ligated by the enzyme DNA ligase The inserted fragment of foreign DNA can again be excised using the same restriction enzyme Cloning: ligation  Cloning: transformation A chimeric plasmid is taken up by a host bacterium The plasmid becomes permanently established in the host (transformation) and can multiply indefinitely Large amounts of recombinant DNA are produced Typically bacteria are cultured/grown on semisolid growth medium in a low density to enable the formation of single colonies Assumption: one colony arises from a single cell Only host organisms that have been transformed and contain a properly constructed vector must be selected Cloning: transformation Bacterial Transformation  Cloning: selection Selection of positive transformants that carry the constructed vector can be done by: Antibiotic resistance Chromogenic substances Sequencing Antibiotic resistance genes such as ampR, tetR, kanR Grow on growth medium containing antibiotic Untransformed – no growth, transformed – growth  Cloning: selection Chromogenic substances e.g. blue/white screening β-galactosidase is an enzyme encoded by the lacZ gene present in some plasmids like pUC18 It is able to cleave X-gal (colourless) to a blue product When cells of E. coli are transformed with an unmodified pUC18, they form blue colonies because of the enzyme When cells of E. coli are transformed with a pUC18 carrying an insert DNA, the lacZ gene is disrupted and the enzyme is not expressed. The X-gal remains colourless making the colonies appear white Some cells will not take up any pUC18 but they can be excluded using antibiotic selection SELF STUDY SECTION 5: D - Recombinant DNA technology has numerous practical applications