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Transcript
Objective # 25 Module 3E – DNA Structure and Replication Summarize the evidence from the 1920s through the early 1950s that convinced scientists that DNA is the genetic material. In this module, we will examine: the molecular structure of the genetic material how the genetic material replicates how damage to the genetic material is repaired 1 2 Objective 25 the late 19th and early 20th centuries, scientists studying patterns of inheritance concluded that inherited traits are controlled by “hereditary factors” or alleles which are located on chromosomes. However, a key question remained unanswered: what are these “hereditary factors” made of? Objective 25 Scientists During 3 knew the genetic material must carry out 2 basic functions: code information replicate itself They also knew coding could be done by varying the sequence of monomers that make up a polymer. [Similar to the way we code information in books by varying the sequence of 26 letters.] Objective 25 lipids are not true polymers, and since most polysaccharides are made of repeating glucose units, this left 2 main candidates: proteins and nucleic acids. Proteins were considered the more likely candidate because coding information would be more efficient using molecules made of 20 different monomers rather than just 4. 4 Objective 25 Since An important breakthrough came in 1928, when Frederick Griffith determined that something can pass from one cell to another and alter the characteristics of the recipient cell. Griffith called this process transformation:: transformation 5 6 1 Griffths Experiment Live Virulent Strain of S. pneumoniae Polysaccharide coat Mice die Live Nonvirulent Strain of S. pneumoniae Objective 25 Mixture of Heat-Killed Virulent and Live Heat-killed Virulent Strain Nonvirulent Strains of S. pneumoniae of S. pneumoniae + No Polysaccharide coat Mice live Mice die Mice live Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Their Lungs contain live pathogenic strain of S. pneumoniae Since the agent that passed from one cell to another in Griffith’s experiment actually alters the characteristics of the recipient cell, many scientists believed it must be the genetic material. In 1944, Avery, MacLeod, and McCarty determined that the transforming agent was DNA. 7 8 9 10 Objective 25 In spite of this evidence, many scientists continued to believe that proteins, rather than DNA, function as the hereditary material. The question was finally settled in 1952 by Hershey and Chase who carried out a classic experiment with bacteriophages: Objective # 26 Objective 26 Once scientists determined that the genetic material was composed of DNA, there was intense competition to determine the structure of this molecule molecule. Scientists hoped that the structure of DNA would provide important clues to understanding how it works to replicate itself and control genetic traits. Name the scientists who originally discovered the structure off DNA. DNA 11 12 2 Objective 26 Chargaff’s Rules The structure of DNA was finally discovered at Cambridge University in 1953 by James Watson and Francis Crick. Their discovery was based primarily on x-ray crystallography data collected by Maurice Wilkins and Rosalind Franklin as well as on the chemical analysis of the base composition of DNA carried out by Irwin Chargaff. 13 Erwin Chargaff determined that Amount of adenine = amount of thymine Amount of cytosine = amount of guanine Always an equal proportion of purines (A and G) and pyrimidines (C and T) 14 James Watson and Francis Crick – 1953 Rosalind Franklin Watson and Crick deduced the structure of DNA using evidence from Chargaff, Franklin, and others Did not p perform a single g experiment p themselves related to DNA Proposed a double helix structure Performed XX-ray diffraction studies to identify the 33--D structure Discovered that DNA is helical Us Usingg Maurice au ce Wilkins’ W s DNA DN fibers, discovered that the molecule has a diameter of 2 nm and makes a complete turn of the helix every 3.4 nm 15 16 Objective # 27 Objective 27 All Describe the structure of a DNA nucleotide and explain h iit diff how differs from f the h structure of an RNA nucleotide. nucleotides have 3 parts: A pentose (5(5-carbon) sugar A nitrogenous g base attached to the 1 prime carbon of the sugar A phosphate group (PO4) attached to the 5 prime carbon of the sugar: 17 18 3 P O– NH2 7N 5 1 8 Phosphate group O 4 N 9 –O O CH2 5’ P 2 N 3 2´ OH in RNA OH Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. N C C H N H C N C C N H N C C NH2 H Adenine N Guanine 1’ H in DNA Sugar OH in RNA H C H C C N H3C N C O O NH2 3´ 2’ N H 1´ O OH N C C O 4´ 3’ C 5´ O– 4’ H CH2 O O NH2 6 Purines O Nitrogenous Base Nitrogenous base Pyrimidines Phosphate group –O There are 5 possible nitrogenous bases and 2 possible sugars: Nitrogenous base NH2 6 7N 5 N1 8 2 9 N 4 N3 Structure of a Nucleotide O H Cytosine (both DNA and RNA) H C C C N N H H C C H C H Thymine (DNA only) O C N N H C O H Uracil (RNA only) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. R Sugar H in DNA 19 20 Objective 27 Objective # 28 DNA nucleotides contain the sugar deoxyribose and the nitrogenous bases A, G, C, T. Describe the structure of a DNA molecule and explain how it differs from the structure of a RNA molecule. RNA nucleotides contain the sugar ribose and the nitrogenous bases A, G, C, U. 21 22 Objective 28 Double helix DNA: consists of 2 unbranched chains of DNA nucleotides twisted into a double helix.. helix the 2 chains are held together by hydrogen bonds between the nitogenous bases. A always pairs with T, and G with C. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5´ Phosphate group P 5´ O 1´ 4´ 3´ Phosphodiester bond 2´ P 5´ O 4´ 1´ 3´ P 2´ 5´ O 1´ 4´ 3´ 5-carbon sugar 2´ Nitrogenous base P 5´ O 1´ 4´ 3´ OH • each strand is a polymer of nucleotides • Phosphodiester backbone – repeating sugar and phosphate sugar and phosphate units joined by phosphodiester bonds • Nitrogenous bases project from the sugar‐ phosphate backbone 2´ 3´ 23 24 4 Complementarity of bases: Hydrogen bond N H • G forms 3 hydrogen bonds with C • A forms 2 hydrogen bonds with T bonds with T • Gives consistent diameter O N N G H H N N H H H C N N Sugar N H Sugar H Hydrogen bond H N H N N Sugar A N CH3 O H H N H T N N H Sugar Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 25 26 Objective 28 RNA is a polynucleotide composed of a single, unbranched chain of RNA nucleotides. 27 5’ end RNA P P Objective # 29 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. In detail, describe the process of DNA replication including the names and functions of the enzymes involved. involved Be able to explain why DNA replication is discontinuous along one strand and continuous along the other. Phosphate group O Phosphodiester bonds O P O 5-carbon sugar (ribose) P O OH 3’ end 28 Nitrogenous base (A,C,G or U) 29 30 5 Objective 29 In 1958, Meselson and Stahl carried out an experiment at Cal Tech supporting the hypothesis that the replication of DNA is semiconservative semiconservative.. According to this hypothesis, during replication, the 2 strands of DNA separate and each one serves as a template for a new strand. 31 32 33 34 Objective 29 The following animation provides a brief overview of the process of semiconservative DNA replication. Note that the ability of one strand of DNA to act as a template for the matching strand is based on the fact that A always pairs with T and G always pairs with C. Objective 29 The next animation provides a more detailed description of the process of DNA replication including the names and functions of the major enzymes involved. Note that the main enzyme involved in strand elongation, DNA polymerase III, cannot initiate synthesis of a new nucleotide strand. It can only add nucleotides to the 3’ end of an existing strand. 35 36 6 Objective # 30 Objective 30 DNA replication in prokaryotes and eukaryotes is fundamentally the same. On the next slide we can see how the single circular DNA molecule found in prokaryotic organisms like E. coli is replicated. (Note that the circular DNA molecule is drawn as linear in order to simplify the animation.) Explain how DNA replication in prokaryotes differs from DNA replication in eukaryotes. 37 38 39 40 Objective 30 Objective 30 In order to speed up the replication of such a large amount of DNA, the linear DNA molecule found in each eukaryotic chromosome may have hundreds or even thousands of replication origins, while the circular DNA molecule in a prokaryotic cell has just one. Although DNA replication in prokaryotes and eukaryotes is very similar, there are a few differences. One difference is due to the fact that eukaryotes have so much more DNA than prokaryotes. 41 42 7 Objective 30 Objective 30 Because DNA polymerase III can only add new nucleotides to the 3’ end of an existing strand, there is no way to complete the final segment of the lagging strand located at each end of a linear DNA molecule. As a result, linear DNA molecules gets shorter and shorter with each round of replication: Another difference in the replication process is related to the fact that eukaryotes y have linear rather than circular DNA. This creates a unique problem for eukaryotes. 43 44 Objective 30 Original strands shown in green and newly synthesized strands in blue: Eukaryotes have 2 ways to deal with this problem: 1) The ends of eukaryotic chromosomes have extra DNA in the form of telomeres.. Telomeres consist of a short telomeres nucleotide sequence that is repeated over and over again. As a result, when eukaryotic chromosomes replicate the telomeres shorten rather than the actual protein--coding genes. protein Replication first round 3 5 5 3 Leading strand (no problem) Lagging strand (problem at the end) 3 5 5 3 Last primer Origin Leading strand Pi Primer removall 3 5 5 Lagging strand 3 Replication second round Removed primer cannot be replaced 5 3 3 5 3 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 46 5 3 Shortened template Objective 30 Objective 30 Scientists 2) Some cells, such as germ cells, have an enzyme called telomerase. telomerase. This enzyme has an internal RNA template that is used to lengthen the telomeres so that chromosomes can continue to replicate without shortening the actual protein protein--coding genes. 47 believe that normal shortening of telomeres during DNA replication may help protect against cancer by limiting the number of divisions that cells can undergo. undergo Interestingly, telomerase activity has been detected in cancer cells. By lengthening the telomeres, this may allow the cells to continue to divide indefinitely. 48 8 Objective # 31 Describe the importance and the mechanisms of DNA repair. 49 50 51 52 Objective 31 Many DNA polymerases can “proofread” added bases and correct mistakes as DNA is being replicated. This increases the accuracy of replication but some errors still occur replication, occur. These mistakes or mutations are a mixed blessing. They provide the genetic variation that is essential for evolution but, unfortunately, most are harmful. In Objective 31 addition to mistakes that occur during replication, DNA is constantly exposed to damaging agents such as UV light, XX-rays, and chemicals. Mechanisms to repair this damage fall into 2 categories: specific and non non--specific. Specific repair mechanism target a particular type of damage. Some examples are shown in the next animation. 53 54 9 Objective 31 Excision repair is a nonnon-specific repair mechanism that can be used if only one strand of the DNA is damaged. It involves 3 steps: Recognition of the damage Removal of the damaged strand Synthesis of a new strand using the undamaged strand as a template 55 56 57 10