DNA The Genetic Material Chapter 14 Learning Objectives

DNA: The Genetic Material Chapter 14

Learning Objectives 14. 1 The Nature of the Genetic Material • Understand experiments of – Griffith & Avery – Avery, Mac. Leod, and Mc. Carty – Hershey and Chase • For both experiments, know/understand major findings 2

Frederick Griffith – 1928 • Studied Streptococcus pneumoniae, a pathogenic bacterium causing pneumonia • 2 strains of Streptococcus – S strain is virulent – R strain is nonvirulent 3 http: //o. quizlet. com/i/GEJK 81 o. Hl. TEYut. Tzm. SQK 6 Q. jpg

Griffith’s Experiment • Griffith infected mice with these strains hoping to understand the difference between the strains 4

Griffith’s Experiment Live Nonvirulent Strain of S. pneumoniae Live Virulent Strain of S. pneumoniae Polysaccharide coat Mice die a. Mice live b. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5

Griffith’s Experiment Mixture of Heat-killed Virulent and Live Nonvirulent Strains of S. pneumoniae Heat-killed Virulent Strain of S. pneumoniae + Mice die Their lungs contain live pathogenic strain of S. pneumoniae Mice live c. d. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 6

• Griffith’s results – Live S strain cells killed the mice – Live R strain cells did not kill the mice – Heat-killed S strain + live R strain cells killed the mice 7

• Transformation – Information specifying virulence passed from the dead S strain cells into the live R strain cells • Our modern interpretation is that genetic material was actually transferred between the cells 8

Avery, Mac. Leod, & Mc. Carty – 1944 • Repeated Griffith’s experiment using purified cell extracts 9 http: //biology. kenyon. edu/courses/biol 114/KH_lecture_images/How_DNA_works/FG 11_02. JPG

Avery, Mac. Leod, & Mc. Carty – 1944 • Removal of all protein from the transforming material did not destroy its ability to transform R strain cells • DNA-digesting enzymes destroyed all transforming ability • Supported DNA as the genetic material 10

Hershey & Chase – 1952 • Investigated bacteriophages – Viruses that infect bacteria • Bacteriophage was composed of only DNA and protein • Wanted to determine which of these molecules is the genetic material that is injected into the bacteria 11

Hershey & Chase – 1952 • Bacteriophage DNA was labeled with radioactive phosphorus (32 P) • Bacteriophage protein was labeled with radioactive sulfur (35 S) • Radioactive molecules were tracked 12

Hershey & Chase Experiment 35 S-Labeled Bacteriophages + Phage grown in radioactive 35 S, which is incorporated into phage coat Virus infect bacteria 32 P-Labeled Blender separates phage coat from bacteria Centrifuge forms bacterial pellet 35 S in supernatant Bacteriophages + Phage grown in radioactive 32 P. which is incorporated into phage DNA Virus infect bacteria Blender separates Centrifuge forms phage coat from bacterial pellet 32 P in bacteria pellet Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 13

Hershey & Chase – 1952 • Only the bacteriophage DNA (as indicated by the 32 P) entered the bacteria and was used to produce more bacteriophage • Conclusion: DNA is the genetic material! 14

Question 16 Mixing a killed virulent strain of bacteria and a living strain of benign bacteria together produces virulent bacteria. What does this demonstrate? a. Genes are inactivated when a cell dies b. DNA can only be passed on during reproduction c. Cells can pick up genes from the environment d. All bacteria are virulent e. None of the above

Learning Objectives 14. 2 DNA Structure • Understand the contributions of the following people in elucidating DNA’s structure – Chargaff – Wilkins & Franklin – Watson & Crick 16

DNA Structure • DNA is a nucleic acid • Composed of nucleotides – 5 -carbon sugar called deoxyribose – Phosphate group (PO 43 -) • Attached to 5′ carbon of sugar – Nitrogenous base • Adenine, thymine, cytosine, guanine – Free hydroxyl group (—OH) • Attached at the 3′ carbon of sugar 17

Nucleotide Subunits of DNA and RNA Nitrogenous Base Nitrogenous base 1 8 Phosphate group O N 9 P O 4 N 2 H C N C N H C N N C C H H 3 H CH 2 C N N H C NH 2 Guanine Adenine 5′ O– O 1′ 4′ 3′ 2′ OH in RNA OH Sugar O NH 2 H in DNA Pyrimidines –O 6 5 O NH 2 Purines 7 N NH 2 H C C N N C O H Cytosine (both DNA and RNA) H 3 C C H C C N O N H H C C O H C H Thymine (DNA only) C N N H C O H Uracil (RNA only) Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 18

• Phosphodiester bond – Bond between adjacent nucleotides – Formed between the phosphate group of one nucleotide and the 3′ —OH of the next nucleotide • The chain of nucleotides has a 5′to-3′ orientation Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5′ PO 4 Base CH 2 O C O Phosphodiester bond –O hydroxyl group O P Phosphate group O Base CH 2 O OH 3′ 19

Chargaff’s Rules • 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) 20

Representation of Chargaff’s Data Table (1952) Organis m %A %G %C %T A/T G/C %GC %AT φX 174 24. 0 23. 3 21. 5 31. 2 0. 77 1. 08 44. 8 55. 2 Maize 26. 8 22. 8 23. 2 27. 2 0. 99 0. 98 46. 1 54. 0 Octopus 33. 2 17. 6 31. 6 1. 05 1. 00 35. 2 64. 8 Chicken 28. 0 22. 0 21. 6 28. 4 0. 99 1. 02 43. 7 56. 4 Rat 28. 6 21. 4 20. 5 28. 4 1. 01 1. 00 42. 9 57. 0 Human 29. 3 20. 7 20. 0 30. 0 0. 98 1. 04 40. 7 59. 3 Grassho pper 29. 3 20. 5 20. 7 29. 3 1. 00 0. 99 41. 2 58. 6 Sea Urchin 32. 8 17. 7 17. 3 32. 1 1. 02 35. 0 64. 9 Wheat 27. 3 22. 7 22. 8 27. 1 1. 00 45. 5 54. 4 Yeast 31. 3 18. 7 17. 1 32. 9 0. 95 1. 09 35. 8 64. 4 E. coli 24. 7 26. 0 25. 7 23. 6 1. 05 1. 01 51. 7 48. 3 http: //en. wikipedia. org/wiki/Chargaff%27 s_rules

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Rosalind Franklin • Performed X-ray diffraction studies to identify the 3 -D structure – Discovered that DNA is helical – Using Maurice Wilkins’ DNA fibers, discovered that the molecule has a diameter of 2 nm and makes a complete turn of the helix every 3. 4 nm a. b. Courtesy of Cold Spring Harbor Laboratory Archives 22

James Watson and Francis Crick – 1953 • Deduced the structure of DNA using evidence from Chargaff, Franklin, and others • Did not perform a single experiment themselves related to DNA • Proposed a double helix structure 23

Double helix Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5´� Phosphate group P 5� O 1� 4� 3� P Phosphodiester bond 2� 5� O 4� 3� P 1� 2� 5� O 1� 4� 3� 5 -carbon sugar 2� Nitrogenous base P 5� 4� 3� OH 3� • 2 strands are polymers of nucleotides • Phosphodiester backbone – repeating sugar and phosphate units joined by phosphodiester bonds • Wrap around 1 axis • Antiparallel O 1� 2� 24

Antiparallel Nature of DNA http: //academic. brooklyn. cuny. edu/biology/bio 4 fv/page/molecular%20 biology/ds. DNA. jpg 25

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 2 nm 5′ A T G Minor groove 3′ 3. 4 nm C T G A T 0. 34 nm C G Major groove G A G T C Major groove Minor groove 26 3′ 5′

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Hydrogen bond • Complementarity of bases • A forms 2 hydrogen bonds with T • G forms 3 hydrogen bonds with C • Gives consistent diameter O N H N G N H H H N N H N Sugar H C N N H Sugar H Hydrogen bond H N N Sugar A N O H H N CH 3 H T N N H Sugar 27

Question 7 Chargaff’s rule states that a. DNA strands are in antiparallel alignment b. G matches with C, and T matches with A c. The DNA molecule is a double helix d. DNA transformation occurs when an organism incorporates DNA from the environment e. The nuclei of cells are totipotent

Question 11 If a strand of DNA had the sequence 5’- AGTCCA 3’, which of the following would be the complementary DNA strand? a. b. c. d. e. 3’- CATGGT- 5’ 3’- TCAAAU- 5’ 3’- AGTCCA- 5’ 3’- GGTTCA- 5’

Question 12 If a DNA molecule contains 40% thymine, how much guanine will it contain? a. 10% b. 20% c. 30% d. 40%

Learning Objectives 14. 3 Basic Characteristics of DNA Replication • How did Meselson and Stahl figure out DNA replication? • What is required for DNA replication? 31

DNA Replication 3 possible models 1. Conservative model 2. Semiconservative model 3. Dispersive model 32

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Conservative

Meselson and Stahl – 1958 • Bacterial cells were grown in a heavy isotope of nitrogen, 15 N • All the DNA incorporated 15 N • Cells were switched to media containing lighter 14 N • DNA was extracted from the cells at various time intervals 36

Meselson and Stahl – 1958 • Bacterial cells were grown in a heavy isotope of nitrogen, 15 N • All the DNA incorporated 15 N • Cells were switched to media containing lighter 14 N • DNA was extracted from the cells at various time intervals 37

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. DNA E. coli 15 N medium 14 N medium 0 min 0 rounds E. coli cells grown in 15 N medium Cells shifted to 14 N medium and allowed to grow 20 min 1 round 40 min 2 rounds Samples taken at three time points and suspended in cesium chloride solution Samples are centrifuged 0 rounds 1 round 2 rounds 0 1 2 Top Bottom Rounds of replication From M. Meselson and F. W. Stahl/PNAS 44(1958): 671 38

Meselson and Stahl’s Results • Conservative model = rejected – 2 densities were not observed after round 1 • Semiconservative model = supported – Consistent with all observations – 1 band after round 1 – 2 bands after round 2 • Dispersive model = rejected – 1 st round results consistent – 2 nd round – did not observe 1 band 39

DNA Replication • Requires 3 things – Something to copy • Parental DNA molecule – Something to do the copying • Enzymes – Building blocks to make copy • Nucleotide triphosphates 40

• DNA replication includes – Initiation – replication begins – Elongation – new strands of DNA are synthesized by DNA polymerase – Termination – replication is terminated 41

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Template Strand HO New Strand 3′ HO 5′ C New Strand 3′ 5′ P G P C G O O Sugar– phosphate backbone Template Strand P P P T A O O P P A DNA polymerase III O T O P C O G O P O G P P 3′ P OH A O A P T P P O T A OH P O P P Pyrophosphate P O 3′ 5′ O T O P P O O O OH A 5′ 42

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5� 5� 3� 3� 5� RNA polymerase makes primer DNA polymerase extends primer • DNA polymerase – Matches existing DNA bases with complementary nucleotides and links them – All have several common features • Add new bases to 3′ end of existing strands • Synthesize in 5′-to-3′ direction • Requires a primer of RNA 43

Question 5 The Meselson-Stahl experiment demonstrated that DNA replication is a. b. c. d. Conservative Semi-conservative Disruptive Differentiated

Learning Objectives 14. 4 Prokaryotic Replication • How many DNA pol’s does E. coli have and what are their functions? • What other enzymes are needed for DNA replication (in E. coli) • What occurs at the replication fork? • How is DNA replication semidiscontinous? • What are the leading and lagging strands? 45

Prokaryotic Replication • E. coli model • Single circular molecule of DNA • Replication begins at one origin of replication • Proceeds in both directions around the chromosome • Replicon – DNA controlled by an origin 46

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Replisome

• E. coli has 3 DNA polymerases – DNA polymerase I (pol I) • Acts on lagging strand to remove RNA primers and replace them with DNA – DNA polymerase II (pol II) • Involved in DNA repair processes – DNA polymerase III (pol III) • Main replication enzyme – All 3 have 3′-to-5′ exonuclease activity – proofreading – DNA pol I has 5′-to-3′ exonuclase activity 48

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Supercoiling Replisomes No Supercoiling Replisomes DNA gyrase • Unwinding DNA causes torsional strain – Helicases – use energy from ATP to unwind DNA – Single-strand-binding proteins (SSBs) coat strands to keep them apart – Topoisomerase prevent supercoiling • DNA gyrase is used in replication 49

Semidiscontinous • DNA polymerase can synthesize only in 1 direction!!!!! • Leading strand synthesized continuously from an initial primer • Lagging strand synthesized discontinuously with multiple priming events – Okazaki fragments 50

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5′ 3′ First RNA primer Lagging strand (discontinuous) 5′ 3′ Open helix and replicate further Open helix and replicate Second RNA primer 5′ 3′ 3′ 3′ 5′ 5′ RNA primer 5′ 3′ Leading strand (continuous) RNA primer 5′ 3′ 51

• Partial opening of helix forms replication fork • DNA primase – RNA polymerase that makes RNA primer – RNA will be removed and replaced with DNA 52

Leading-strand synthesis – Single priming event – Strand extended by DNA pol III • Processivity – subunit forms “sliding clamp” to keep it attached Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. a. b. a-b: From Biochemistry by Stryer. © 1975, 1981, 1988, 1995 by Lupert Stryer. Used with permission of W. H. Freeman and Company 53

Lagging-strand synthesis – Discontinuous synthesis • DNA pol III – RNA primer made by primase for each Okazaki fragment – All RNA primers removed and replaced by DNA • DNA pol I – Backbone sealed • DNA ligase • Termination occurs at specific site – DNA gyrase unlinks 2 copies 54

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5′ DNA ligase Lagging strand (discontinuous) RNA primer DNA polymerase I Okazaki fragment made by DNA polymerase III Primase Leading strand (continuous) 3′ 55

Replication fork Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. New bases β clamp (sliding clamp) Leading strand 3� 5� Single-strand binding proteins (SSB) Clamp loader DNA gyrase Open β clamp 5� 3� Parent DNA Helicase Primase DNA polymerase III Lagging strand Okazaki fragment New bases 3� 5� DNA polymerase I DNA ligase RNA primer 56

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Question 3 Where are the Okazaki fragments found? a. b. c. d. e. On the lagging strand On the leading strand At the replication origin In the cytoplasm On both strands

Question 4 Name the enzyme that links Okazaki fragments a. DNA polymerase I b. RNA primase c. DNA ligase d. Helicase e. ATP synthase

Question 6 During DNA replication, what enzyme is responsible for untwisting the DNA helix? a. DNA polymerase I b. RNA primase c. DNA ligase d. DNA polymerase III e. Helicase

Question 9 Why does replication proceed in opposite directions on the leading and lagging strands? a. The polymerase enzyme needs a primer b. DNA polymerase III can only add to the 3´ end of a strand c. The Okazaki fragments are only on the leading strands d. The parent strands are oriented in the same direction e. Helicase only allows for replication of one strand at a time

Question 15 What would be the immediate consequence of a non -functional primase enzyme? a. The strands would break due to the torsional strain from rapid untwisting b. The helix could be opened c. The DNA polymerase III enzyme would have nothing to bind to d. The Okazaki fragments would not be linked together e. The single DNA strands could not be held open

Learning Objectives 14. 5 Eukaryotic Replication • What are differences between Prok and Euk replication? • What are telomeres and how are they replicated? 14. 6 DNA Repair • What are three forms of DNA repair? • Why is DNA repair important for the cell? 63

Eukaryotic Replication • Complicated by – Larger amount of DNA in multiple chromosomes – Linear structure • Basic enzymology is similar – Requires new enzymatic activity for dealing with ends only 64

Telomeres • Specialized structures found on the ends of eukaryotic chromosomes • Protect ends of chromosomes from nucleases and maintain the integrity of linear chromosomes • Gradual shortening of chromosomes with each round of cell division 65 http: //www. scientificamerican. com/media/inline/telomeres-telomerase-and_1. jpg

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Replication first round 5´ 3´ 3´ 5´ Leading strand (no problem) Lagging strand (problem at the end) 5´ 3´ 3´ 5´ Last primer Origin Leading strand Primer removal 3´ 5´ 5´ Lagging strand 3´ Replication second round Removed primer cannot be replaced 5´ 3´ 3´ 5´ Shortened template 66

• Telomeres composed of short repeated sequences of DNA • Telomerase – enzyme makes telomere section of lagging strand using an internal RNA template (not the DNA itself) – Leading strand can be replicated to the end • Telomerase developmentally regulated – Relationship between senescence and telomere length • Cancer cells generally show activation of telomerase 67

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. 5 T T G 3 Synthesis by telomerase Telomerase 5 3 T A G C G G C C G T C A A Telomere extended by telomerase G T C Template RNA is part of enzyme Telomerase moves and continues to extend telomere 5 3 T T G G G Now ready to synthesize next repeat G T T G A A C T G C C C A A T C 68

DNA Repair • Errors due to replication – DNA polymerases have proofreading ability • Mutagens – any agent that increases the number of mutations above background level – Radiation and chemicals • Importance of DNA repair is indicated by the multiplicity of repair systems that have been discovered 69

DNA Repair Falls into 2 general categories 1. Specific repair – Targets a single kind of lesion in DNA and repairs only that damage 2. Nonspecific – Use a single mechanism to repair multiple kinds of lesions in DNA 70

Photorepair • Specific repair mechanism • For one particular form of damage caused by UV light • Thymine dimers – Covalent link of adjacent thymine bases in DNA • Photolyase – Absorbs light in visible range – Uses this energy to cleave thymine dimer 71

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. DNA with adjacent thymines T A UV light Helix distorted by thymine dimer T T A A Photolyase binds to damaged DNA Photolyase T T A A Visible light Thymine dimer cleaved T A 72

Excision repair • Nonspecific repair • Damaged region is removed and replaced by DNA synthesis • 3 steps 1. Recognition of damage 2. Removal of the damaged region 3. Resynthesis using the information on the undamaged strand as a template 73

Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Damaged or incorrect base Excision repair enzymes recognize damaged DNA Uvr A, B, C complex binds damaged DNA Excision of damaged strand Resynthesis by DNA polymerase 74

Question 13 The enzyme telomerase attaches the last few bases on the lagging strand. As cells age, telomerase activity drops. What would happen to the chromosomes in the absence of telomerase activity? a. Chromosome replication would be terminated b. Okazaki fragments would not be linked together c. Chromosomes would shorten during each division d. The leading strand would become the lagging strand e. The cells would become cancerous

Question 10 Mutations can be caused by copying mistakes, and by exposure to chemicals or electromagnetic radiation. a. This is True b. This is False