Objective: You will be able to construct explanations that use

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1 Objective: You will be able to construct explanations that use the structures of DN to support the claim that DN is the genetic material Do Now: Read through the Enduring Understanding section Heritable information provides for continuity of life

2 Essential knowledge: DN, and in some cases RN, is the primary source of heritable information. Genetic information is stored and transmitted from one generation to the next through DN or RN.

3 Information is stored within the DN molecule due to the sequence of bases.

4 Noneukaryotic organisms have circular chromosomes, while eukaryotic organisms have multiple linear chromosomes, although in biology there are exceptions to this rule.

5 Prokaryotes, viruses and eukaryotes can contain plasmids, which are small extra-chromosomal, double-stranded circular DN molecules.

6 Viral Genomes Viral genomes may consist of either: Double- or single-stranded DN, or Double- or single-stranded RN Depending on its type of nucleic acid, a virus is called a DN virus or an RN virus (retrovirus)

7 Figure 17.2 RN Capsomere DN Membranous envelope RN Capsid Head DN Capsomere of capsid Tail sheath Tail fiber Glycoprotein Glycoproteins nm nm (diameter) nm (diameter) nm 20 nm (a) Tobacco mosaic virus 50 nm (b) denoviruses 50 nm (c) Influenza viruses 50 nm (d) Bacteriophage T4

8 DN Replication ensures the continuity of hereditary information.

9 Objective: You will be able to justify the selection of data from historical investigations that support the claim that DN is the source of heritable information. Do Now:

10 What is the Genetic Material? Scientists recognized that chromosomes contained the genetic material because they were passed on to new cells in mitosis and meiosis. Chromosomes were made up of DN and protein So which is the genetic material?

11 Show Me the Evidence! The proof that DN is the carrier of genetic information involved a number of important historical experiments. These include: Griffith and the very-macleod-mccarty experiments Hershey-Chase experiment

12 Figure 13.2 Experiment Living S cells (control) Living R cells (control) Heat-killed S cells (control) Mixture of heat-killed S cells and living R cells Results Mouse dies Mouse healthy Mouse healthy Mouse dies Living S cells

13 Transformation is the change in genotype and phenotype due to assimilation of foreign DN is called transformation

14 Figure nm Phage head Tail sheath Tail fiber DN Bacterial cell

15 Protein DN

16 Figure 13.4a Experiment Batch 1: Radioactive sulfur ( 35 S) in phage protein 1 Labeled phages 2 gitation frees 3 infect cells. outside phage parts from cells. Radioactive protein Centrifuged cells form a pellet. 4 Radioactivity (phage protein) found in liquid Centrifuge Pellet

17 Figure 13.4b Experiment Batch 2: Radioactive phosphorus ( 32 P) in phage DN 1 Labeled phages 2 gitation frees 3 infect cells. outside phage parts from cells. Radioactive DN Centrifuged cells form a pellet. Centrifuge Pellet 4 Radioactivity (phage DN) found in pellet

18 Figure 13.4 Experiment Batch 1: Radioactive sulfur ( 35 S) in phage protein 1 Labeled phages 2 gitation frees outside 3 infect cells. phage parts from cells. Radioactive protein Centrifuged cells form a pellet. 4 Radioactivity (phage protein) found in liquid Centrifuge Batch 2: Radioactive phosphorus ( 32 P) in phage DN Radioactive DN Pellet Centrifuge Pellet 4 Radioactivity (phage DN) found in pellet

19 Objective: You will be able to explain how work on DN s structure helped to support the claim that DN is the source of heritable information. Do Now: Complete SSE on p. 257

20 Chargaff s Rules 1. DN base composition varies between species 2. For each species, the % of and T bases are roughly equal as is the % of G and C bases. How do these rules support the claim that DN is the source of heritable information?

21 Watson and Crick

22 Where oh Where did Watson and Crick get their ideas? 2011 Pearson Education, Inc.

23 Figure 13.7 C G G C C G C G end Notice the antiparallel nature of DN Hydrogen bond T end T 3.4 nm G C C G G C T 1 nm C G C G G C T C G T T T T 0.34 nm end end (a) Key features of DN structure (b) Partial chemical structure (c) Space-filling model Why would it take longer to denature DN that contained many Guanine bases as opposed to denine bases?

24 Objective: You will be able to describe the similarities and differences between DN and RN. Do Now: Make a list of what you can remember as the differences between DN and RN

25 DN and RN molecules have structural similarities and differences that define function. Both have: sugar, phosphate group and a nitrogenous base These form nucleotide units that are connected by covalent bonds Forms a linear molecule with 3' and 5' ends DN/RN Nucleotide

26 Figure 13.5 Sugar phosphate backbone end Nitrogenous bases end DN or RN nucleotide

27 What differences between DN and RN can you identify?

28 The basic structural differences include: DN contains deoxyribose (RN contains ribose). RN contains uracil in lieu of thymine in DN. DN is usually double stranded, RN is usually single stranded. The two DN strands in double-stranded DN are antiparallel in directionality.

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30 Both DN and RN exhibit specific nucleotide base pairing that is conserved through evolution: (In RN U) Purines (G and ) have a double ring structure. Pyrimidines (C, T and U) have a single ring structure.

31 Roles of RN

32 Functions of RN The sequence of the RN bases, together with the structure of the RN molecule, determines RN function. mrn carries information from the DN to the ribosome. trn molecules bind specific amino acids and allow information in the mrn to be translated to a linear peptide sequence. rrn molecules are functional building blocks of ribosomes. The role of RNi includes regulation of gene expression at the level of mrn transcription.

33 Objective: You will be able to describe representations and models illustrating how genetic information is copied for transmission between generations. Do Now:

34 DN replication ensures continuity of hereditary information. Replication is a semiconservative process; that is, one strand serves as the template for a new, complementary strand. Replication requires: DN polymerase plus many other essential cellular enzymes Occurs bidirectionally

35 Figure Semiconservative Replication T C T G G T C (a) Parental molecule

36 Figure Semiconservative Replication T T C G C G T T T T G C G C (a) Parental molecule (b) Separation of parental strands into templates

37 Figure (a) Parental molecule (b) Separation of parental strands into templates (c) Formation of new strands complementary to template strands T C G C G T T T T C G C G T T C G C G T T T C G C G T T Semiconservative Replication

38 Figure New strand Template strand Sugar Phosphate Base T T C G C G G C DN polymerase G T C Nucleotide C P P i Pyrophosphate C 2 P i

39 Figure Leading strand template Overview Origin of replication Leading strand Lagging strand Single-strand binding proteins Helicase Parental DN Lagging strand template DN pol III Primer Primase Leading strand DN pol III Lagging strand Leading strand Overall directions of replication Lagging strand DN pol I DN ligase

40 Figure Nuclease DN polymerase DN ligase

41 Genetic information in retroviruses is a special case and has an alternate flow of information: From RN to DN Made possible by reverse transcriptase, an enzyme that copies the viral RN genome into DN. This DN integrates into the host genome and becomes transcribed and translated for the assembly of new viral progeny.

42 Figure 17.7 Glycoprotein Viral envelope HIV Membrane of white blood cell Capsid Reverse HIV transcriptase RN (two identical strands) Viral RN RN-DN hybrid DN HOST CELL Reverse transcriptase 0.25 m HIV entering a cell Chromosomal DN RN genome for the next viral generation mrn NUCLEUS Provirus New virus New HIV leaving a cell

43 Objective: You will be able to design a representation that illustrates how genetic information is transferred from DN to RN Do Now:

44 Genetic information flows from a sequence of nucleotides in a gene to a sequence of amino acids in a protein. How would you define transcription? How would you define translation?

45 Transcription is the synthesis of RN using information in DN Transcription produces messenger RN (mrn) Translation is the synthesis of a polypeptide, using information in the mrn Ribosomes are the sites of translation

46 Transcription The enzyme RN-polymerase reads the DN molecule in the 3' to 5' direction and synthesizes complementary mrn molecules that determine the order of amino acids in the polypeptide.

47 Figure Promoter Transcription unit 1 Initiation Start point RN polymerase Unwound DN RN Template strand of DN transcript

48 Figure Promoter Transcription unit 1 Initiation Start point RN polymerase 2 Elongation Unwound DN Rewound DN RN transcript RN Template strand of DN transcript Direction of transcription ( downstream )

49 Figure Promoter Transcription unit 1 Initiation Start point RN polymerase 2 3 Elongation Termination Unwound DN Rewound DN RN transcript RN Template strand of DN transcript Completed RN transcript Direction of transcription ( downstream )

50 Figure RN polymerase Nontemplate strand of DN RN nucleotides T C C end U C C T G G T T Newly made RN Direction of transcription Template strand of DN

51 Objective: You will be able to design a representation that illustrates how genetic information of RN is modified to form mrn. Do Now:

52 RN Processing In eukaryotic cells the mrn transcript undergoes a series of enzyme-regulated modifications. ddition of a poly- tail ddition of a GTP cap Excision of introns

53 Figure modified guanine nucleotide added to the end G P GTP Cap Protein-coding segment P P U Start codon Stop codon adenine nucleotides added to the end Polyadenylation signal Poly- tail

54 Figure Pre-mRN Cap Intron mrn Intron Introns cut out and exons spliced together Poly- tail Cap UTR Coding segment Poly- tail UTR U

55 Objective: You will be able to design a representation that illustrates how genetic information of mrn is translated into a sequence of amino acids Do Now:

56 Translation Translation of the mrn occurs in the cytoplasm on the ribosome. Translation involves energy and many steps, including initiation, elongation and termination.

57 Salient Features of Protein Synthesis: Part 1 The mrn interacts with the rrn of the ribosome to initiate translation at the (start) codon. The sequence of nucleotides on the mrn is read in triplets called codons. Each codon encodes a specific amino acid, which can be deduced by using a genetic code chart. Many amino acids have more than one codon.

58 Figure 14.5 DN template strand C C C C G G T T G G T T T G G C T C TRNSCRIPTION mrn U G G U U U G G C U C TRNSLTION Codon Protein Trp Phe Gly Ser mino acid

59 First mrn base ( end of codon) Third mrn base ( end of codon) Figure 14.6 Second mrn base U C G U UUU UUC UU UUG Phe Leu UCU UCC UC UCG Ser UU UC U UG Tyr Stop Stop UGU UGC UG UGG Cys Stop Trp U C G C CUU CUC CU CUG Leu CCU CCC CC CCG Pro CU CC C CG His Gln CGU CGC CG CGG rg U C G UU UC U UG IIe Met or start CU CC C CG Thr U C G sn Lys GU GC G GG Ser rg U C G G GUU GUC GU GUG Val GCU GCC GC GCG la GU GC G GG sp Glu GGU GGC GG GGG Gly U C G

60 Salient Features of Protein Synthesis: Part 2 trn brings the correct amino acid to the correct place on the mrn. The amino acid is transferred to the growing peptide chain. The process continues along the mrn until a "stop" codon is reached. The process terminates by release of the newly synthesized peptide/protein.

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63 Figure Initiation U C U G P site Large ribosomal subunit Initiator trn mrn GTP Start codon Small mrn binding site ribosomal subunit 1 Small ribosomal subunit binds 2 to mrn. P i GDP E Translation initiation complex Large ribosomal subunit completes the initiation complex.

64 Figure mino end of polypeptide 1 Codon recognition mrn E P site site GTP GDP P i Elongation E P

65 Figure mino end of polypeptide 1 Codon recognition mrn E P site site GTP GDP P i E P 2 Peptide bond formation E P

66 Figure mino end of polypeptide 1 Codon recognition Ribosome ready for next aminoacyl trn mrn E P site site GTP GDP P i E E P P 3 Translocation GDP P i GTP 2 Peptide bond formation E P

67 Figure Termination Release factor Free polypeptide 1 Stop codon (UG, U, or UG) Ribosome reaches a stop codon on mrn. 2 GTP 2 Release factor 3 promotes hydrolysis. 2 GDP P i Ribosomal subunits and other components dissociate.

68 Figure TRNSCRIPTION DN RN transcript RN PROCESSING CYTOPLSM Exon NUCLEUS RN polymerase RN transcript (pre-mrn) Intron mino acid trn minoacyl-trn synthetase MINO CID CTIVTION mrn E P Ribosomal subunits minoacyl (charged) trn TRNSLTION E U G G U U U U G nticodon Ribosome Codon

69 Objective: You will be able to compare and contrast protein synthesis in eukaryotic and prokaryotic cells. Do Now:

70 Prokaryotic vs. Eukaryotic Protein Synthesis Compile a lost of similarities from your notes and t diagram above Compile a list of differences.

71 Differences In prokaryotic organisms, transcription is coupled to translation of the message. In eukaryotic cells the mrn transcript undergoes a series of enzyme-regulated modifications. ddition of a poly- tail ddition of a GTP cap Excision of introns

72 Phenotypes are determined through protein activities. Synthesis Transport by proteins Enzymatic reactions Degradation Using your knowledge of biology, how would you explain the difference between eye color in humans? Sickle cell RBC versus normal

73 Objective: You will be able to explain how heritable information can be manipulated by using restriction enzymes and gel electrophoresis. Do Now:

74 Genetic engineering techniques can manipulate the heritable information of DN and, in special cases, RN. Restriction enzyme analysis of DN Electrophoresis Plasmid-based transformation Polymerase Chain Reaction (PCR)

75 Using Restriction Enzymes to Make Recombinant DN Bacterial restriction enzymes cut DN molecules at specific DN sequences called restriction sites restriction enzyme usually makes many cuts, yielding restriction fragments

76 Figure DN Restriction enzyme cuts the sugar-phosphate backbones. DN fragment added from another molecule cut by same enzyme. Base pairing occurs. Restriction site G C G T T C C T T G G Sticky end G 3 DN ligase seals the strands. G T T C C T T G G T T C C T T G One possible combination Recombinant DN molecule

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78 Enzyme Enzyme Enzyme Enzyme X Y X X ,200 1,300 1,500 () Explain how the principles of gel electrophoresis allow for the separation of DN fragments. (B) Describe the results you would expect from the electrophoretic separation of fragments from the following treatments of the DN segment above. ssume that the digestion occurred under appropriate conditions and went to completion. I. DN digested with only enzyme X II. DN digested with only enzyme Y III. DN digested with enzyme X and enzyme Y combined IV. Undigested DN (C) Explain both of the following. (1) The mechanism of action of restriction enzymes. (2) The different results you would expect if a mutation occurred at the recognition site for enzyme Y.

79 Researchers use gel electrophoresis to separate restriction fragments Fragments separate based on length Occurs because DN has a negative charge

80 Figure Mixture of DN molecules of different sizes Cathode Wells Power source node Gel (a) Negatively charged DN molecules will move toward the positive electrode. Restriction fragments (b) Shorter molecules are impeded less than longer ones, so they move faster through the gel.

81 Figure 20.9 Using restriction fragment patterns to distinguish DN from different alleles

82 Figure DN fingerprints from a murder case

83 Objective: You will be able to explain how heritable information can be manipulated by using the polymerase chain reaction. Do Now:

84 The Polymerase Chain Reaction (PCR) The polymerase chain reaction, PCR, can produce many identical copies of a fragment of DN

85 Figure Technique Genomic DN Target sequence 1 Denaturation 2 nnealing Cycle 1 yields 2 molecules Primers 3 Extension New nucleotides Cycle 2 yields 4 molecules Cycle 3 yields 8 molecules; 2 molecules (in white boxes) match target sequence

86 Objective: You will be able to explain how heritable information can be manipulated by plasmid-based transformation. Do Now:

87 Figure Bacterium 1 Gene inserted into plasmid Cell containing gene of interest Bacterial chromosome Plasmid Recombinant DN (plasmid) 2 Gene of interest Plasmid put into bacterial cell DN of chromosome ( foreign DN) Recombinant DN Recombinant bacterium 3 Host cell grown in culture to form a clone of cells containing the cloned gene of interest Gene of interest Copies of gene Protein expressed from gene of interest Protein harvested Gene for pest resistance inserted into plants 4 Basic research and various applications Human growth hormone treats stunted growth Gene used to alter bacteria for cleaning up toxic waste Protein dissolves blood clots in heart attack therapy

88 Examples of products of genetic engineering include: Pharmaceuticals, such as human insulin Genetically modified foods Transgenic animals Cloned animals

89 Figure 14.7 Transgenic Organisms (a) Tobacco plant expressing a firefly gene (b) Pig expressing a jellyfish gene

90 Reproductive Cloning of a Mammal by Nuclear Transplantation