3.a.1- DNA and RNA 10/19/2014. Big Idea 3: Living systems store, retrieve, transmit and respond to information essential to life processes.

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1 3.a.1- DNA and RNA Big Idea 3: Living systems store, retrieve, transmit and respond to information essential to life processes. EU 3.A: Heritable information provides for continuity of life. EU 3.B: Expression of genetic information involves cellular and molecular mechanisms. EU 3.C: The processing of genetic information is imperfect and is a source of genetic variation. EU 3.D: Cells communicate by generating, transmitting and receiving chemical signals. EU 3.E: Transmission of information results in changes within and between biological systems. A. Genetic information is transmitted from one generation to the next through DNA or RNA. 1. Noneukaryotic organisms have circular chromosomes, while eukaryotic organisms have multiple linear chromosomes, although in biology there are exceptions to this rule. 2. Prokaryotes, viruses and eukaryotes can contain plasmids, which are small extra-chromosomal, double-stranded circular DNA molecules. 3. The proof that DNA is the carrier of genetic information involved a number of important historical experiments. These include: a. Frederick Griffith b. Avery-MacLeod-McCarty c. Erwin Chargaff d. Hershey-Chase e. Franklin-Wilkins-Watson-Crick 4. DNA replication ensures continuity of hereditary information. a. Replication is a semiconservative process; that is, one strand serves as the template for a new, complementary strand. b. Replication requires DNA polymerase plus many other essential cellular enzymes, occurs bidirectionally, and differs in the production of the leading and lagging strands. 1

2 5. Genetic information in retroviruses is a special case and has an alternate flow of information: from RNA to DNA, made possible by reverse transcriptase, an enzyme that copies the viral RNA genome into DNA. This DNA integrates into the host genome and becomes transcribed and translated for the assembly of new viral progeny. [See also 3.C.3] B. DNA and RNA molecules have structural similarities and differences that define function. [See also 4.A.1] 1. Both have three components 2. The basic structural differences include: 3. Both DNA and RNA exhibit specific nucleotide base pairing that is conserved through evolution 4. The sequence of the RNA bases, together with the structure of the RNA molecule, determines RNA function C. Genetic information flows from a sequence of nucleotides in a gene to a sequence of amino acids in a protein. 1. Transcription 2. In eukaryotic cells the mrna transcript undergoes a series of enzyme-regulated modifications. 3. Translation and genetic code 4. In prokaryotic organisms, transcription is coupled to translation of the message. Translation involves energy and many steps, including initiation, elongation and termination. 5. Polyribosomes D. Phenotypes are determined through protein activities. 1. Enzymatic reactions (Bubble Boy- adenosine deaminase doesn t break down purines) 2. Transport by proteins (High cholesterol defective LDL receptors) 3. Synthesis (Alzheimers- excess protein plaque) 4. Degradation (PKU) E. Genetic engineering techniques can manipulate the heritable information of DNA and, in special cases, RNA. 1. Polymerase Chain Reaction (PCR) 1. Restriction enzyme analysis of DNA 2. Electrophoresis 3. Plasmid-based transformation 2

3 F. Examples of products of genetic engineering 1. Genetically modified foods 2. Transgenic animals 3. Cloned animals 4. Pharmaceuticals, such as human insulin or factor X (coagulant) Frederick Griffith (1928) A. DNA Can Transform Bacteria B. Conducted experiments with Streptococcus pneumoniae 1. Injected mice with two strains: (S) strain and a (R) strain. 2. The S strain is virulent (mice died); it has a mucous capsule 3. The R strain is not virulent (mice lived); it has no capsule. Frederick Griffith (1928) 4. He injected mice with heat-killed S strain bacteria; the mice lived. 5. He injected mice with a mixture of heat-killed S strain and live R strain bacteria; the mice died and living S strain pneumococcus were recovered from their bodies. 6. Griffith concluded, some substance transformed the R strain. Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) A. Used deoxyribonucleodepolymerase to make the Griffith extract non transformable B. Shows the transforming substance was DNA. MacLeod Back Avery McCarty A. Performed detailed analysis of base content of DNA. B. Purine bases 1. double-ring structure 2. adenine (A) and guanine (G). C. Pyrimidine bases 1. single-ring structure 2. Thymine (T) and cytosine (C) D. Chargaff's Rules: 1. The amount of A, T, G, and C in DNA varies from species to species. 2. In each species, the amount of A=T and G=C. Erwin Chargaff (1947) Alfred Hershey and Martha Chase(1952) A. Used bacteriophage T2 in their experiments. B. See if protein coat or DNA directed reproduction of virus. C. In two separate experiments, they labeled the protein coat with radioactive 35 S and the DNA with radioactive 32 P. 3

4 Alfred Hershey and Martha Chase(1952) D. Viral coats are sheared away from bacterial cells and are separated by centrifugation. E. Results: radioactive 32 P alone is taken up by bacterial host and incorporated in virus reproduction. F. Their results reinforced the notion that DNA is the genetic material. Rosalind Franklin and Maurice Wilkins (1953) A. Franklin produced X-ray diffraction photograph of DNA. B. Wilkins gave Watson/Crick photo C. Photo provided evidence that DNA had the following features: 1. DNA is a helix. 2. One part of the helix is repeated. James Watson and Francis Crick(1953) A. Used information generated by Franklin B. Built a model of DNA as double helix C. Sugar-phosphate molecules on outside D. Paired bases on inside. E. Sugar-phosphate backbones are antiparallel. James Watson and Francis Crick(1953) F. Using information generated by Chargaff. 1. Width is 2 nm. 2. Width of DNA due to purines paired to pyrimidines. 3. Chargaff s rules are consistent a) A hydrogen-bonded to T b) G hydrogen-bonded to C. G. Watson and Crick received the Nobel Prize in 1954 for their model of DNA. How is DNA Replicated A. Is Replication Conservative, Semi-conservative, or Dispersive? How is DNA Replicated B. Matthew Meselson and Franklin Stahl (1958) 1. Confirmed a model of DNA replication. 2. They grew bacteria with heavy nitrogen ( 15 N), then switched to light nitrogen ( 14 N) 4

5 How is DNA Replicated 4. After one division, only hybrid DNA molecules were in the cells. 5. After two divisions, half the DNA molecules were light and half were hybrid. 6. These were exactly the results to be expected if DNA replication is semiconservative. Steps in DNA Replication A. Begins at a specific sequence of nucleotides (origins of replication) B. Unwinding 1. Hydrogen bonds between paired bases are broken. 2. Parent DNA strand is unwound. 3. Both catalyzed by the enzyme helicase. 4. Single-strand binding proteins keep the strands apart. Steps in DNA Replication C. Priming DNA Synthesis 1. Short segments of RNA (10 nucleotides long) formed by primase. 2. RNA primer must be bound to DNA segment to begin replication. D. Complementary base pairing 1. Free nucleotides (Nucleoside triphosphates) bind with complementary bases on unzipped portions of DNA. 2. New strands grow in the 5 3 direction. Steps in DNA Replication 3. 3 end contains hydroxyl end contains phosphate. 5. Process is catalyzed by DNA polymerase. 6. Polymerase can only elongate at 3 end. a) Continuous synthesis creates a leading strand. b) Discontinuous synthesis of the complementary strand creates a lagging strand. c) Lagging strand produces Okazaki fragments. d) Fragments are linked by DNA ligase. A. Sugar, phosphate and a nitrogenous base (nucleotide) B. Connected by covalent bonds to form a linear molecule with 3 and 5' ends. DNA and RNA A. DNA contains deoxyribose (RNA contains ribose). B. RNA contains uracil instead of thymine in DNA. C. DNA is usually double stranded, RNA is usually single stranded. D. The two DNA strands in double-stranded DNA are antiparallel in directionality. Differences 5

6 Nucleotide Base Pairing A. Adenine pairs with thymine or uracil (A-T or A-U) B. Cytosine pairs with guanine (C-G). C. Purines (G and A) have a double ring structure. D. Pyrimidines (C, T and U) have a single ring structure. Functions of RNA A. Messenger RNA (mrna) takes a message from DNA in nucleus to ribosomes in cytoplasm. B. Ribosomal RNA (rrna) and proteins make up ribosomes where proteins are synthesized. C. Transfer RNA (trna) transfers a particular amino acid to a ribosome. D. RNAi (interference) 1. Regulation of gene expression at the level of mrna transcription. 2. Includes mirna (micro) and sirna (small interferring) Transcription A. RNA polymerase attaches to a promoter on DNA. B. Promoter defines start of gene, direction of transcription, and strand copied. (TATA Box in Eukaryotes) C. Transcription factor is needed for polymerase to bind to DNA. D. Complementary RNA nucleotides pair with DNA nucleotides of the strand. Transcription E. RNA polymerase covalently bonds RNA nucleotides. F. Terminator sequence causes RNA polymerase to stop transcribing DNA, and to release mrna G. Many copies of mrna are made from the DNA molecule at the same time. Post Transcription Modification A. Addition of a GTP cap (modified guanine) B. Addition of a poly-a tail C. Excision of introns (by spliceosomes) 1. Exon is portion of mrna transcript eventually expressed. 2. The simpler the eukaryote, the less likely that introns will be present. Translation A. Takes place in cytoplasm B. One language (nucleic acids) is translated into another language (protein) C. Transfer RNA 1. Transfers amino acids to the ribosomes. 2. One end binds to amino acid; other end has an anticodon that binds to mrna codon 6

7 How Genes Code for Amino Acids A. mrna contains the genetic code. B. Codes for an amino acid or stops translation. C. Code is comprised of 64 codons. D. Codon consists of 3 sequential nucleotide bases of mrna. (4 3 =64) How Genes Code for Amino Acids E. 64 mrna triplets code for 20 amino acids. F. Each codon has only one meaning. G. There is one start codon and three stop codons. H. The Code Is Universal 1. Suggests the code dates to very first organisms. 2. Once established, changes would be very disruptive. 1. Small ribosomal subunit attaches to mrna at the start codon (AUG). 2. First or initiator trna pairs with this codon; then large ribosomal subunit joins to small subunit 3. Initiation factor proteins are required to bring necessary components together Chain Initiation Chain Elongation 1. New trna arrives at A site 2. Amino acid at P site is attached by a peptide bond to the newly arrived amino acid. 3. Reaction is catalyzed by a ribozyme of the larger subunit. Chain Elongation 4. The trna molecule in the P site leaves. 5. Translocation occurs when trna, moves from site A to P. 6. As ribosome has moved forward three nucleotides, there is new codon located at empty A site. 1. Occurs at stop codon that does not code for amino acid. 2. Release factor protein binds to stop codon. 3. The polypeptide is enzymatically cleaved from the last trna. 4. trna and polypeptide leave the ribosome, which dissociates into its two subunits. Chain Termination 7

8 Polyribosomes Remember- Proteins are not synthesized one at a time as each protein finishes production. Polyribosomes are clusters of several ribosomes synthesizing the same protein EK 3.C.3: Viral replication results in genetic variation, and viral infection can introduce genetic variation into the hosts. B. The reproductive cycles of viruses facilitate transfer of genetic information. 8