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1 Module 3E DA Structure and Replication 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 bjective # 25 Summarize the evidence from the 1920s through the early 1950s that convinced scientists that DA is the genetic material. 1 2 bjective 25 During the late 19 th and early 20 th centuries, scientists studying patterns of inheritance concluded that inherited traits are controlled by hereditary factors or alleles which are located on chromosomes. owever, a key question remained unanswered: what are these hereditary factors made of? 3 bjective 25 Scientists 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.] 4 bjective 25 bjective 25 Since lipids are not true polymers, and since most polysaccharides are made of repeating glucose units, this left 2 main candidates: proteins and nucleic acids. roteins were considered the more likely candidate because coding information would be more efficient using molecules made of 20 different monomers rather than just 4. 5 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: 6 1

2 Live Virulent Strain of S. pneumoniae olysaccharide coat Griffths Experiment Live onvirulent Strain of S. pneumoniae o olysaccharide coat eat-killed Virulent Strain of S. pneumoniae Mixture of eat-killed Virulent and Live onvirulent Strains of S. pneumoniae Mice die Mice live Mice live Mice die Their Lungs contain live pathogenic strain of S. pneumoniae 7 opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. + bjective 25 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 Mcarty determined that the transforming agent was DA. 8 bjective 25 In spite of this evidence, many scientists continued to believe that proteins, rather than DA, function as the hereditary material. The question was finally settled in 1952 by ershey and hase who carried out a classic experiment with bacteriophages: 9 10 bjective # 26 ame the scientists who originally discovered the structure of DA. 11 bjective 26 nce scientists determined that the genetic material was composed of DA, there was intense competition to determine the structure of this molecule. Scientists hoped that the structure of DA would provide important clues to understanding how it works to replicate itself and control genetic traits. 12 2

3 bjective 26 The structure of DA was finally discovered at ambridge University in 1953 by James Watson and Francis rick. 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 DA carried out by Irwin hargaff. 13 hargaff s Rules Erwin hargaff 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 ( and T) 14 Rosalind Franklin erformed X-ray diffraction studies to identify the 3-D structure Discovered that DA is helical Using Maurice Wilkins DA fibers, discovered that the molecule has a diameter of 2 nm and makes a complete turn of the helix every 3.4 nm James Watson and Francis rick 1953 Watson and rick deduced the structure of DA using evidence from hargaff, Franklin, and others Did not perform a single experiment themselves related to DA roposed a double helix structure bjective # 27 Describe the structure of a DA nucleotide and explain how it differs from the structure of an RA nucleotide. bjective 27 All nucleotides have 3 parts: A A pentose (5-carbon) sugar A A nitrogenous base attached to the 1 prime carbon of the sugar A A phosphate group ( 4 ) attached to the 5 prime carbon of the sugar:

4 Structure of a ucleotide hosphate group itrogenous base There are 5 possible nitrogenous bases and 2 possible sugars: hosphate group 2 8 itrogenous base Sugar 4 3 in RA in DA 2 urines yrimidines 2 Adenine 2 ytosine (both DA and RA) itrogenous Base 3 Guanine Thymine (DA only) 2 Uracil (RA only) opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. 3 2 R Sugar in RA in DA 19 opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. 20 bjective 27 bjective # 28 DA nucleotides contain the sugar deoxyribose and the nitrogenous bases A, G,, T. RA nucleotides contain the sugar ribose and the nitrogenous bases A, G,, U. Describe the structure of a DA molecule and explain how it differs from the structure of a RA molecule DA: bjective 28 consists of 2 unbranched chains of DA nucleotides twisted into a double helix. the 2 chains are held together by hydrogen bonds between the nitogenous bases. A A always pairs with T, and G with. 23 opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display hosphate group hosphodiester bond Double helix 5-carbon sugar itrogenous base 1 2 each strand is a polymer of nucleotides hosphodiester backbone repeating sugar and phosphate units joined by phosphodiester bonds itrogenous bases project from the sugarphosphate backbone 24 4

5 omplementarity of bases: G forms 3 hydrogen bonds with A forms 2 hydrogen bonds with T Gives consistent diameter Sugar G ydrogen bond ydrogen bond Sugar 3 A Sugar opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. T Sugar bjective 28 RA is a polynucleotide composed of a single, unbranched chain of RA nucleotides end 3 end RA opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. hosphate group hosphodiester bonds 5-carbon sugar (ribose) itrogenous base (A,,G or U) 29 bjective # 29 In detail, describe the process of DA replication including the names and functions of the enzymes involved. Be able to explain why DA replication is discontinuous along one strand and continuous along the other. 30 5

6 bjective 29 In 1958, Meselson and Stahl carried out an experiment at al Tech supporting the hypothesis that the replication of DA is semiconservative. According to this hypothesis, during replication, the 2 strands of DA separate and each one serves as a template for a new strand bjective 29 The following animation provides a brief overview of the process of semiconservative DA replication. ote that the ability of one strand of DA 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 bjective 29 The next animation provides a more detailed description of the process of DA replication including the names and functions of the major enzymes involved. ote that the main enzyme involved in strand elongation, DA polymerase III, cannot initiate synthesis of a new nucleotide strand. It can only add nucleotides to the 3 end of an existing strand

7 bjective # 30 Explain how DA replication in prokaryotes differs from DA replication in eukaryotes. bjective 30 DA replication in prokaryotes and eukaryotes is fundamentally the same. n the next slide we can see how the single circular DA molecule found in prokaryotic organisms like E. coli is replicated. (ote that the circular DA molecule is drawn as linear in order to simplify the animation.) bjective 30 Although DA replication in prokaryotes and eukaryotes is very similar, there are a few differences. ne difference is due to the fact that eukaryotes have so much more DA than prokaryotes. bjective 30 In order to speed up the replication of such a large amount of DA, the linear DA molecule found in each eukaryotic chromosome may have hundreds or even thousands of replication origins, while the circular DA molecule in a prokaryotic cell has just one

8 bjective 30 Another difference in the replication process is related to the fact that eukaryotes have linear rather than circular DA. This creates a unique problem for eukaryotes. bjective 30 Because DA 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 DA molecule. As a result, linear DA molecules gets shorter and shorter with each round of replication: riginal strands shown in green and newly synthesized strands in blue: rigin Replication first round Leading strand Lagging strand Leading strand (no problem) Lagging strand (problem at the end) opyright The McGraw-ill ompanies, Inc. ermission required for reproduction or display. i rimer removal Replication second round Shortened template Last primer Removed primer cannot be replaced bjective 30 Eukaryotes have 2 ways to deal with this problem: 1) The ends of eukaryotic chromosomes have extra DA in the form of telomeres. Telomeres consist of a short 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. 46 bjective 30 2) Some cells, such as germ cells, have an enzyme called telomerase. This enzyme has an internal RA template that is used to lengthen the telomeres so that chromosomes can continue to replicate without shortening the actual protein-coding genes. 47 bjective 30 Scientists believe that normal shortening of telomeres during DA replication may help protect against cancer by limiting the number of divisions that cells can 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

9 bjective # 31 Describe the importance and the mechanisms of DA repair bjective 31 Many DA polymerases can proofread added bases and correct mistakes as DA is being replicated. This increases the accuracy of replication, but some errors still occur. These mistakes or mutations are a mixed blessing. They provide the genetic variation that is essential for evolution but, unfortunately, most are harmful bjective 31 In addition to mistakes that occur during replication, DA is constantly exposed to damaging agents such as UV light, X-rays, and chemicals. Mechanisms to repair this damage fall into 2 categories: specific and non-specific. Specific repair mechanism target a particular type of damage. Some examples are shown in the next animation. 54 9

10 bjective 31 Excision repair is a non-specific repair mechanism that can be used if only one strand of the DA 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