History & DNA Structure PDF

Course: General Genetics AGN 101 NUCLEIC ACID (DNA) STRUCTURE BY DR. Shereen Abu El-Maaty HISTORY OF GENETICS In 1989- The National Center for Human Genome Research is created. In 2000: A rough draft of the human genome is completed and published by the H...

Course: General Genetics AGN 101 NUCLEIC ACID (DNA) STRUCTURE BY DR. Shereen Abu El-Maaty HISTORY OF GENETICS In 1989- The National Center for Human Genome Research is created. In 2000: A rough draft of the human genome is completed and published by the Human Genome Project and Celera. The project was planned to last 15 years, but rapid technological advances accelerated the expected completion date. Project goals are to discover all 30,000 to 40,000 human genes (the human genome) and make them accessible for further study and to determine the complete sequence of the 3 billion DNA bases in the human genome.  Scientists for a long time were trying to know what exactly which of the cell four macromolecules is genetic materials:  Lipids  Carbohydrates  Proteins  Nucleic acid (DNA-RNA)??? Evidence that proves the DNA is genetic material 1- Griffith experiment In 1928, Frederick Griffith while working with Streptococcus pneumoniae (the bacterium that causes pneumonia) observed a miraculous transformation in this bacterium. When you grow this bacterium on a culture plate, some produce shiny colonies (a deadly virulent strain S) and some produce rough colonies (non-virulent strain R). The S strain bacteria have a polysaccharide coat which gives rise to smooth, shiny colonies. The R strain lacks this coat and hence, it gives rough colonies.  He performed the following experiment with these strains and saw different observations. o S strain → Inject into mice → Mice develop pneumonia and die. o R strain → Inject into mice → Mice live. o Heat-killed S strain → Inject into mice → Mice live. (Griffith found that heating kills the bacteria). o Heat-killed S strain + R strain → Inject into mice → Mice die. Observations : Not only did the mice injected with the heat- killed S strain + R strain die, but Griffith also recovered live S strain bacteria from these dead mice. Conclusions : He concluded that this was because the R strain had somehow been ‘transformed’ by the heat-killed S strain. This he argued was due to the transfer of a ‘transforming principle‘ from the S strain to the R strain, which made the R strain virulent. Although significant, his observations did not identify the biochemical nature of the transforming principle.  2-Avery, Macleod and Mccarty experiment (1944) Avery, MacLeod, and McCarty, together set out to determine the biochemical nature of the ‘transforming principle’ identified by Griffith. These people purified DNA, RNA, and proteins from the heat-killed S strain and determined which macromolecule converted the R strain into the S strain. Experiment: They first treated the heat-killed S strain with proteases to break down proteins. Subsequently, they treated it with RNAses and then DNAses to break down RNA and DNA, respectively. Observations: Both protease and RNAse treatments did not affect the transformation of the R strain into the virulent one. Finally, treatment with DNAses inhibited the transformation of the R strain. Conclusions: They concluded that the genetic material is not protein or RNA, but it is DNA. However, this discovery was not accepted by all biologists.  Alfred Hershey and Martha Chase In 1952, Hershey and Chase were the ones to conclusively prove that DNA is the genetic material. They worked with bacteriophages – viruses that infect bacteria. A bacteriophage attaches and delivers its genetic material into a bacterial cell, where it generates more virus particles. Hersey & Chase used bacteriophages to experiment as follows: Labelling: Some viruses were grown on a medium containing radioactive phosphorus (32P) and some on a medium with radioactive sulfur. Viruses – grown on radioactive phosphorus have radioactive DNA but not protein since DNA contains phosphorus but protein does not. Contrarily, viruses grown on radioactive sulfur (35S) have radioactive protein but not DNA since DNA does not contain sulfur. Infection : The radioactive phages were then allowed to infect the bacteria – E. coli. Blending and Centrifugation : As the infection progressed, the viral coats were removed from the bacteria by blending. Then, centrifugation was used to separate the viral particles from the bacteria. Observations : Bacteria infected with viruses that have radioactive DNA, were radioactive, while bacteria infected with viruses that have radioactive protein, were not radioactive. Conclusions : This experiment conclusively showed that DNA is the genetic material transferred from virus to bacteria, and not protein. Features of the genetic material  It must possess four important characteristics. 1-Genetic material must contain complex information. the genetic material must be capable of storing large amounts of information—instructions for all the traits and functions of an organism. 2-the genetic material must be stable, because most alterations to the genetic instructions (mutations) are likely to be detrimental. 3- Genetic material must replicate independent. that genetic material must have the capacity to be copied accurately. 4- Genetic material must encode phenotype. The genetic material (the genotype) must have the capacity to “code for” (determine) traits (the phenotype). The product of a gene is often a protein; so there must be a mechanism for genetic instructions to be translated into the amino acid sequence of a protein. The genetic material must be capable of carrying large amounts information, replicating faithfully, and translating its coding instructions into phenotypes.  The structure of DNA The primary structure of DNA consists of a string of nucleotides (a polymer) joined together by phosphodiester linkages. DNA Stands for: Deoxyribonucleic acid RNA Stands for: Ribonucleic acid each nucleotides, comprising three parts: 1. Phosphate group 2. 5-carbon sugar 3. Nitrogenous base  1-Pentose Sugar The pentose sugars have five carbon atoms, numbered 1, 2, 3, and so forth. Four of the carbon atoms are joined by an oxygen atom to form a five-sided ring; the fifth (5) carbon atom projects upward from the ring. Hydrogen atoms or hydroxyl groups (OH) are attached to each carbon atom  2-Nitrogenous Bases The second component of a nucleotide is its nitrogenous base, which may be of two types a purine or a pyrimidine.Each purine consists of a six-sided ring attached to a five- sided ring, whereas each pyrimidine consists of a six-sided ring only. DNA and RNA both contain two purines, adenine and guanine (A and G), which differ in the positions of their double bonds and in the groups attached to the six-sided ring. There are three pyrimidines found in nucleic acids: cytosine (C), thymine (T), and uracil (U). Cytosine is present in both DNA and RNA; however, thymine is restricted to DNA, and uracil is found only in RNA. In a nucleotide, the nitrogenous base always forms a covalent bond with the 1-carbon atom of the sugar. A deoxyribose (or ribose) sugar and a base together are referred to as a nucleoside. Double ring PURINES Adenine (A) Guanine (G) Single ring PYRIMIDINES Thymine (T) Cytosine (C)  BASE-PAIRINGS Purines only pair with Pyrimidines Three hydrogen bonds required to bond Guanine & Cytosine  Two hydrogen bonds are required to bond adenine & thymine  3-phosphate group Consists of a phosphorus atom bonded to four oxygen atoms.Phosphate groups are found in every nucleotide and frequently carry a negative charge, which makes DNA acidic. The phosphate is always bonded to the 5-carbon atom of the sugar in a nucleotide. sugar+ nitrogen Nucleoside+ base =Nucleoside phosphate group = Nucleotide The nucleotides of DNA are joined in polynucleotide strands by phosphodiester bonds that connect the 3 carbon atom of one nucleotide to the 5 phosphate group of the next. Each polynucleotide strand has polarity, with a 5 end and a 3 end.  DNA BACKBONE The DNA backbone is a polymer with an alternating sugar- phosphate sequence. The deoxyribose sugars are joined at both the 3'-hydroxyl and 5'- hydroxyl groups to phosphate groups in ester links, also known as "phosphodiester" bonds.  Chargaff's rules  In 1950, Chargaff developed the principle of base-pairing. He determined the Amount of DNA and overall compositions vary from organism to organism but are constant within the individuals of a species and within the various tissues of a single organism.  Quantity of Adenine [A] = quantity of Thymine [T]  Guanine [G] =Cytosine [C]  Adenine must pair with Thymine  Guanine must pair with Cytosine  The bases form weak hydrogen bonds T A G C  Wilkins and Franklin  Wilkins and Franklin studied the structure of DNA crystals using X-rays  The X pattern suggested the structure of DNA was a helix.  Distance between the two “backbones” of DNA is constant along the length of the molecule. By collecting existing information about the chemistry of DNA and building molecular models, Watson and Crick were able to discover the three-dimensional structure of the DNA molecule (DNA , it as a double helix of two anti parallel strands). Note: The secondary structure of DNA refers to its three dimensional configuration its fundamental helical structure. DNA’s secondary structure can assume a variety of configurations, depending on its base sequence and the conditions in which it is placed. The double helix A fundamental characteristic of DNA’s secondary structure is that it consists of: 1) two polynucleotide strands wound around each other—it’s a double helix. 2) The sugar–phosphate linkages are on the outside of the helix, and the bases are stacked in the interior of the molecule. 3) The two polynucleotide strands run in opposite directions they are antiparallel, which means that the 5- end of one strand is opposite the 3- end of the second. 4) The strands are held together by two types of molecular forces. Hydrogen bonds link the bases on opposite strands. 5) These bonds are relatively weak compared with the covalent phosphodiester bonds that connect the sugar and phosphate groups of adjoining nucleotides.  Watson-Crick Model of Secondary Structure  DNA has two strands, each strand coiled about the other like strand in a two- stranded rope or a spiral staircase--called a right-handed helix.  20 A in diameter  Internucleotide distance of 3.4 A.  Therefore 34 A repeat would be one complete turn and thus 10 nucleotides/turn. Also 36 degrees rotation/pair.  Pairing relationship giving rise to the specificity is A hydrogen bonds to T with two H-bonds,G hydrogen bonds to C with three H-bonds  Major and minor grooves.  Opposite polarity of strands  Antiparallel strands  One strand of DNA goes from 5’ to 3’ (sugars)  The other strand is opposite in direction going 3’ to 5’ (sugars)  Types of DNA DNA can assume different secondary structures, depending on the conditions in which it is placed and on its base sequence. B-DNA is thought to be the most common configuration in the cell. Local variation in DNA arises as a result of environmental factors and base sequence. A, B, C = right-handed helix Z = left-handed helix (found in vitro under high salt)  Z-DNA B-DNA A-DNA 66  RNA  Unlike DNA, RNA is synthesized as a single strand.  There are double stranded RNA structure:  RNA can fold back on itself depends on bases sequence gives stem (double strand) and loop (single strand structures) RNA SECONDARY STRUCTURE  Forces that stabilize nucleic acid structures  Sugar-phosphate chain conformation  Base pairing  Ionic interactions  Types of RNA mRNA tRNA rRNA Question: If there is 30% Adenine, how much Cytosine is present?  ANSWER: There would be 20% Cytosine Adenine (30%) = Thymine (30%) Guanine (20%) = Cytosine (20%) Therefore, 60% A-T and 40% C-G PROBLEM1: the following drawings are the outlines of two dna base pairs, with the bases identified as a, b, c, and d. what are the real identities of these bases? Problem 2 What are the complimentary base pairs to a DNA strand that has he following order ? 5′ATGCGTGACTAATTCG3′