DNA: The Genetic Material Chapter 14 1

Learning Objectives 14.1 The Nature of the Genetic Material Understand experiments of Griffith & Avery Avery, MacLeod, and McCarty Hershey and Chase For both experiments, know/understand major findings

Frederick Griffith – 1928 Studied Streptococcus pneumoniae, a pathogenic bacterium causing pneumonia 2 strains of Streptococcus S strain is virulent R strain is nonvirulent http://o.quizlet.com/i/GEJK81oHlTEYutTzmSQK6Q.jpg

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

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

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 McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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

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

Avery, MacLeod, & McCarty – 1944 Repeated Griffith’s experiment using purified cell extracts http://biology.kenyon.edu/courses/biol114/KH_lecture_images/How_DNA_works/FG11_02.JPG

Avery, MacLeod, & McCarty – 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

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

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

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

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

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

Learning Objectives 14.2 DNA Structure Understand the contributions of the following people in elucidating DNA’s structure Chargaff Wilkins & Franklin Watson & Crick

DNA Structure DNA is a nucleic acid Composed of nucleotides 5-carbon sugar called deoxyribose Phosphate group (PO43-) Attached to 5′ carbon of sugar Nitrogenous base Adenine, thymine, cytosine, guanine Free hydroxyl group (—OH) Attached at the 3′ carbon of sugar

Nucleotide Subunits of DNA and RNA Nitrogenous Base Nitrogenous base NH2 NH2 O 7 N 6 5 N C N C C N C N H 1 H C H C Phosphate group 8 Purines C N C H C N C NH2 2 N N O N 4 N H H 9 3 Adenine Guanine –O P O CH2 5′ O– O NH2 O O 1′ C C C 4′ H C N H3C C N H H C N H H C C O H C C O H C C O 3′ 2′ OH in RNA Pyrimidines N N N OH H H H Sugar H in DNA Cytosine (both DNA and RNA) Thymine (DNA only) Uracil (RNA only) Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

The chain of nucleotides has a 5′-to-3′ orientation 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 McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5′ PO4 Base CH2 O C O hydroxyl group Phosphodiester bond –O P O Phosphate group O Base CH2 O OH 3′

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)

Representation of Chargaff’s Data Table (1952) Organism %A %G %C %T A/T G/C %GC %AT φX174 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 1.02 43.7 56.4 Rat 28.6 21.4 20.5 1.01 42.9 57.0 Human 29.3 20.7 20.0 30.0 1.04 40.7 59.3 Grasshopper 41.2 58.6 Sea Urchin 32.8 17.7 17.3 32.1 35.0 64.9 Wheat 27.3 22.7 27.1 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 51.7 48.3 http://en.wikipedia.org/wiki/Chargaff%27s_rules

Courtesy of Cold Spring Harbor Laboratory Archives Rosalind Franklin Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 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

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

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

Antiparallel Nature of DNA http://academic.brooklyn.cuny.edu/biology/bio4fv/page/molecular%20biology/dsDNA.jpg

2nm 5′ 3′ T A G C 3.4nm Minor groove T G A T 0.34nm G C Major groove G Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2nm 5′ 3′ T A G C 3.4nm Minor groove T G A T 0.34nm G C Major groove G A T G C Major groove Minor groove 3′ 5′

Complementarity of bases A forms 2 hydrogen bonds with T Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Hydrogen bond H Complementarity of bases A forms 2 hydrogen bonds with T G forms 3 hydrogen bonds with C Gives consistent diameter H N O H N H N G N H N C H Sugar N N N H Sugar H Hydrogen bond H H N N H O CH3 N A N H N T H Sugar N N H Sugar

Question 7 Chargaff’s rule states that DNA strands are in antiparallel alignment G matches with C, and T matches with A The DNA molecule is a double helix DNA transformation occurs when an organism incorporates DNA from the environment The nuclei of cells are totipotent Bloom’s level: Knowledge/Understanding,

Question 11 3’- CATGGT- 5’ 3’- TCAGGT- 5’ 3’- TCAAAU- 5’ If a strand of DNA had the sequence 5’- AGTCCA- 3’, which of the following would be the complementary DNA strand? 3’- CATGGT- 5’ 3’- TCAGGT- 5’ 3’- TCAAAU- 5’ 3’- AGTCCA- 5’ 3’- GGTTCA- 5’ Bloom’s level: Application,

Question 12 If a DNA molecule contains 40% thymine, how much guanine will it contain? 10% 20% 30% 40% Bloom’s level: Application

Learning Objectives 14.3 Basic Characteristics of DNA Replication How did Meselson and Stahl figure out DNA replication? What is required for DNA replication?

DNA Replication 3 possible models Conservative model Semiconservative model Dispersive model

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

Conservative Semiconservative 34 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 34 Conservative Semiconservative

Conservative Semiconservative Dispersive 35 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 35 Conservative Semiconservative Dispersive

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

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

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

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 1st round results consistent 2nd round – did not observe 1 band

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

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

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

All have several common features Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 5  5  3  3  5  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

Question 5 Conservative Semi-conservative Disruptive Differentiated The Meselson-Stahl experiment demonstrated that DNA replication is Conservative Semi-conservative Disruptive Differentiated Bloom’s level: Knowledge/Understanding

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?

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

Replisome Termination Origin Replisome Origin Termination Termination Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Replisome Origin Termination Termination Replisome Origin Termination Termination Termination Origin Origin Origin

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

Unwinding DNA causes torsional strain Copyright © The McGraw-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

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

5′ 3′ First RNA primer Lagging strand (discontinuous) 5′ 3′ Copyright © The McGraw-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′

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

Leading-strand synthesis Single priming event Strand extended by DNA pol III Processivity –  subunit forms “sliding clamp” to keep it attached Copyright © The McGraw-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

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

Leading strand (continuous) Copyright © The McGraw-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′

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

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Question 3 On the lagging strand On the leading strand Where are the Okazaki fragments found? On the lagging strand On the leading strand At the replication origin In the cytoplasm On both strands Bloom’s level: Knowledge/Understanding

Question 4 DNA polymerase I RNA primase DNA ligase Helicase Name the enzyme that links Okazaki fragments DNA polymerase I RNA primase DNA ligase Helicase ATP synthase Bloom’s level: Knowledge/Understanding

Question 6 DNA polymerase I RNA primase DNA ligase DNA polymerase III During DNA replication, what enzyme is responsible for untwisting the DNA helix? DNA polymerase I RNA primase DNA ligase DNA polymerase III Helicase Bloom’s level: Knowledge/Understanding

Question 9 Why does replication proceed in opposite directions on the leading and lagging strands? The polymerase enzyme needs a primer DNA polymerase III can only add to the 3´ end of a strand The Okazaki fragments are only on the leading strands The parent strands are oriented in the same direction Helicase only allows for replication of one strand at a time Bloom’s level: Knowledge/Understanding

Question 15 What would be the immediate consequence of a non-functional primase enzyme? The strands would break due to the torsional strain from rapid untwisting The helix could be opened The DNA polymerase III enzyme would have nothing to bind to The Okazaki fragments would not be linked together The single DNA strands could not be held open Bloom’s level: Application

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 the three forms of DNA repair? Why is DNA repair important for the cell?

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

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 http://www.scientificamerican.com/media/inline/telomeres-telomerase-and_1.jpg

Leading strand Lagging strand Removed primer cannot be replaced Copyright © The McGraw-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) 3´ 5´ 3´ 5´ Last primer Origin Primer removal Leading strand 3´ 5´ 5´ Lagging strand 3´ Removed primer cannot be replaced Replication second round 5´ 3´ 3´ 5´ 5´ 3´ 3´ 5´ Shortened template

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

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

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

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

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

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

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

Damaged or incorrect base Copyright © The McGraw-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 DNA polymerase

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? Chromosome replication would be terminated Okazaki fragments would not be linked together Chromosomes would shorten during each division The leading strand would become the lagging strand The cells would become cancerous Bloom’s level: Application

Question 10 This is True This is False Mutations can be caused by copying mistakes, and by exposure to chemicals or electromagnetic radiation. This is True This is False Bloom’s level: Knowledge/Understanding