1. © 2012 Pearson Education, Inc. Lecture by Edward J. Zalisko PowerPoint Lectures for Campbell Biology: Concepts & Connections, Seventh Edition Reece, Taylor, Simon, and Dickey Chapter 10 Molecular Biology of the Gene

2. Figure 10.0_1 Chapter 10: Big Ideas The Structure of the Genetic Material DNA Replication The Genetics of Viruses and Bacteria The Flow of Genetic Information from DNA to RNA to Protein

3. THE STRUCTURE OF THE GENETIC MATERIAL DNA & RNA © 2012 Pearson Education, Inc.

4. 10.1 SCIENTIFIC DISCOVERY: Experiments showed that DNA is the genetic material  Griffith’s experiment on bacterial transformation  Hershey-Chase blender experiment  Skip the experimental details © 2012 Pearson Education, Inc.

5. 10.2 DNA and RNA are polymers of nucleotides  DNA and RNA are nucleic acids.  Both are polymers made up of nucleotides.  DNA is made from 4 DNA nucleotides.  Nucleotides contain –Nitrogenous base –5-carbon sugar –Phosphate group © 2012 Pearson Education, Inc.

6. Figure 10.2A_3 Phosphate group Sugar (deoxyribose) DNA nucleotide Thymine (T) Nitrogenous base (can be A, G, C, or T)

7.  DNA is made up of four DNA nucleotides.  Each nucleotide has a different nitrogen-containing base: –adenine (A), –cytosine (C), –thymine (T), and –guanine (G). 10.2 DNA and RNA are polymers of nucleotides © 2012 Pearson Education, Inc.

8. Figure 10.2B Thymine (T) Cytosine (C) PyrimidinesPurines Adenine (A) Guanine (G)

9. Figure 10.2A_2 DNA nucleotide Covalent bond joining nucleotides A Two representations of a DNA polynucleotide G G C T Sugar Nitrogenous base Phosphate group Sugar-phosphate backbone A G G C T

10. Figure 10.5B 5 end 3 end 5 4 3 2 1 1 2 3 4 5 P P P P P HO A T C G GC P P P A T OH 5 end 3 end

11.  The nucleotides are joined to one another by a sugar- phosphate backbone.  The two DNA strands are anti-parallel. This means that one strand runs in 5’ to 3’ direction and the other in 3’ to 5’ direction. 10.2 DNA and RNA are polymers of nucleotides © 2012 Pearson Education, Inc.

12. Figure 10.2A_1 A A A A A A T T T T T C C C C G G G G G CG A T A DNA double helix T

13. 10.2 DNA and RNA are Polymers of Nucleotides  RNA (ribonucleic acid) is made up of 4 RNA nucleotides  RNA –uses the sugar ribose (instead of deoxyribose in DNA) and –RNA has the nitrogenous base uracil (U) instead of thymine. © 2012 Pearson Education, Inc.

14. Figure 10.2C Phosphate group Sugar (ribose) Uracil (U) Nitrogenous base (can be A, G, C, or U) An RNA nucleotide

15. Figure 10.2D Uracil Adenine Cytosine Guanine Ribose Phosphate

16. © 2012 Pearson Education, Inc. Animation: DNA and RNA Structure Right click on animation / Click play

17. 10.3 SCIENTIFIC DISCOVERY: DNA is a double-stranded helix  In 1953, James D. Watson and Francis Crick deduced the secondary structure of DNA, using –X-ray crystallography data of DNA from the work of Rosalind Franklin and Maurice Wilkins and –Chargaff’s observation that in DNA, –the amount of adenine was equal to the amount of thymine and –the amount of guanine was equal to that of cytosine. © 2012 Pearson Education, Inc.

18. Figure 10.3B

19. Figure 10.3A Rosalind Franklin and her X-ray image

20.  Watson and Crick reported that DNA consisted of two polynucleotide strands wrapped into a double helix. –The sugar-phosphate backbone is on the outside. –The nitrogenous bases are perpendicular to the backbone in the interior. –Specific pairs of bases give the helix a uniform shape. –A pairs with T, forming two hydrogen bonds, and –G pairs with C, forming three hydrogen bonds. 10.3 SCIENTIFIC DISCOVERY: DNA is a double-stranded helix © 2012 Pearson Education, Inc.

21. 10.3 SCIENTIFIC DISCOVERY: DNA is a double-stranded helix  In 1962, the Nobel Prize was awarded to –James D. Watson, Francis Crick, and Maurice Wilkins. –Rosalind Franklin probably would have received the prize as well but for her death from cancer in 1958. Nobel Prizes are never awarded posthumously.  The genetic information in a chromosome is encoded in the nucleotide sequence of DNA. © 2012 Pearson Education, Inc.

22. Figure 10.3C Twist

23. Figure 10.3D_1 Base pair Ribbon model C G G C T A A T C G A T T A C G G C A T T A

24. Figure 10.3D_2 Hydrogen bond Partial chemical structure G C T A A T C G

25. Write the base sequence in the complementary DNA strand and label the ends 5’ C T A T G C A A G T C A T T __’ __’ ______________________ __’ DNA Problem:

26. DNA REPLICATION © 2012 Pearson Education, Inc.

27. 10.4 DNA replication depends on specific base pairing  DNA replication refers to making of two strands of DNA starting with one parental strand  Takes place in the synthesis (S) phase of the cell cycle © 2012 Pearson Education, Inc.

28. 10.4 DNA replication depends on specific base pairing  DNA replication follows a semiconservative model. –The two DNA strands separate. –Each strand is used as a pattern to produce a complementary strand, using specific base pairing. –Each new DNA helix has one old strand with one new strand. © 2012 Pearson Education, Inc.

29. Figure 10.4A_s1 A parental molecule of DNA G C A T T A A T C G

30. Figure 10.4A_s2 A parental molecule of DNA A C G C A T T A The parental strands separate and serve as templates Free nucleotides T A T T A A T A G G G C C A T C G C

31. Figure 10.4A_s3 A parental molecule of DNA A C G C A T T A The parental strands separate and serve as templates Free nucleotides T A T T A A T A G G G C C A T C G C Two identical daughter molecules of DNA are formed A T T A C G G C

32. Figure 10.4B Parental DNA molecule Daughter strand Parental strand Daughter DNA molecules A T G C A T T A T C G A T G C T C G T A C G C A T G C A T G A A

33.  DNA replication begins at the origins of replication where –DNA unwinds at the origin to produce a “bubble,” –replication proceeds in both directions from the origin, and –replication ends when products from the bubbles merge with each other. 10.5 DNA replication proceeds in two directions at many sites simultaneously © 2012 Pearson Education, Inc.

34.  DNA replication occurs in the 5 to 3 direction. –Replication is continuous on the 3 to 5 template. This strand is called the leading strand –Replication is discontinuous on the 5 to 3 template, forming short segments. This DNA strand is called the lagging strand 10.5 DNA replication proceeds in two directions at many sites simultaneously © 2012 Pearson Education, Inc.

35. 10.5 DNA replication proceeds in two directions at many sites simultaneously  Key enzymes involved in DNA replication. 1.DNA helicase unwinds the double helix and breaks hydrogen bonds 2.DNA polymerase –adds nucleotides to a growing chain and –proofreads and corrects improper base pairings. 3.DNA ligase joins small fragments into a continuous chain. © 2012 Pearson Education, Inc.

36. Figure 10.5A Parental DNA molecule Origin of replication “Bubble” Parental strand Daughter strand Two daughter DNA molecules

37. Figure 10.5C Overall direction of replication DNA ligase Replication fork Parental DNA DNA polymerase molecule This daughter strand is synthesized continuously This daughter strand is synthesized in pieces 3 5 3 5 3 5 3 5

38. THE FLOW OF GENETIC INFORMATION FROM DNA TO RNA TO PROTEIN © 2012 Pearson Education, Inc.

39. 10.6 The DNA genotype is expressed as proteins, which provide the molecular basis for phenotypic traits  DNA specifies traits by dictating protein synthesis.  The molecular chain of command is from –DNA in the nucleus is transcribed to RNA and –RNA in the cytoplasm is translated to protein.  Transcription is the synthesis of RNA under the direction of DNA.  Translation is the synthesis of proteins under the direction of RNA. © 2012 Pearson Education, Inc.

40. Figure 10.6A_s1 DNA NUCLEUS CYTOPLASM

41. Figure 10.6A_s2 DNA NUCLEUS CYTOPLASM RNA Transcription

42. Figure 10.6A_s3 DNA NUCLEUS CYTOPLASM RNA Transcription Translation Protein

43. 10.7 Genetic information written in codons is translated into amino acid sequences  The sequence of nucleotides in DNA provides a code for constructing a protein. –Protein construction requires a conversion of a nucleotide sequence to an amino acid sequence. –Transcription rewrites the DNA code into RNA, using the same nucleotide “language.” © 2012 Pearson Education, Inc.

44. 10.7 Genetic information written in codons is translated into amino acid sequences –The flow of information from gene to protein is based on a triplet code: the genetic instructions for the amino acid sequence of a polypeptide chain are written in DNA and RNA as a series of nonoverlapping three- base “words” called codons. –Translation involves switching from the nucleotide “language” to the amino acid “language.” –Each amino acid is specified by a codon. –64 codons are possible. –Some amino acids have more than one possible codon. © 2012 Pearson Education, Inc.

45. Figure 10.7 DNA molecule Gene 1 Gene 2 Gene 3 A Transcription RNA TranslationCodon Polypeptide Amino acid AAC C GG C AAAA UU GGCCG UUUU DNA U

46. Figure 10.7_1 A Transcription RNA Translation Codon Polypeptide Amino acid AAC C GG C AAAA U U GGCCG U U U U DNA U

47. 10.8 The genetic code dictates how codons are translated into amino acids  Characteristics of the genetic code –Three nucleotides specify one amino acid. –61 codons correspond to amino acids. –AUG codes for methionine and signals the start of transcription. –3 “stop” codons signal the end of translation. © 2012 Pearson Education, Inc.

48. 10.8 The genetic code dictates how codons are translated into amino acids  The genetic code is –redundant, with more than one codon for some amino acids, –unambiguous in that any codon for one amino acid does not code for any other amino acid, –nearly universal—the genetic code is shared by organisms from the simplest bacteria to the most complex plants and animals, and –without punctuation in that codons are adjacent to each other with no gaps in between. © 2012 Pearson Education, Inc.

49. Figure 10.8A Second base Third base First base

50. Figure 10.8B_s1 T Strand to be transcribed A C T TC A A A A A T DNA AA T C T T T T GAG G

51. Figure 10.8B_s2 T Strand to be transcribed A C T TC A A A A A T DNA AA T C T T T T GAG G RNA Transcription AAAA U U U U U G G G

52. Figure 10.8B_s3 T Strand to be transcribed A C T TC A A A A A T DNA AA T C T T T T GAG G RNA Transcription AAAA U U U U U G G G Translation Polypeptide MetLysPhe Stop codon Start codon

53.  Write the mRNA sequence and the tRNA sequence. Using the genetic code, write the amino acid sequence DNATAC GCT TAA TCA GGC GTA ATT mRNAAUG UAA tRNAUAC Amino acid Met (stop) DNA Problem:

54. 10.9 Transcription produces genetic messages in the form of RNA  Overview of transcription –An RNA molecule is transcribed from a DNA template by a process that resembles the synthesis of a DNA strand during DNA replication. –RNA nucleotides are linked by the transcription enzyme RNA polymerase. –Specific sequences of nucleotides along the DNA mark where transcription begins and ends. –The “start transcribing” signal is a nucleotide sequence called a promoter. © 2012 Pearson Education, Inc.

55. 10.9 Transcription produces genetic messages in the form of RNA –Transcription begins with initiation, as the RNA polymerase attaches to the promoter. –During the second phase, elongation, the RNA grows longer. –As the RNA peels away, the DNA strands rejoin. –Finally, in the third phase, termination, the RNA polymerase reaches a sequence of bases in the DNA template called a terminator, which signals the end of the gene. –The polymerase molecule now detaches from the RNA molecule and the gene. © 2012 Pearson Education, Inc.

56. Figure 10.9A RNA polymerase Free RNA nucleotides Template strand of DNA Newly made RNA Direction of transcription T G A G G A A U C CA C T T A A C C G G U T U T A ACC T A T C

57. Figure 10.9B RNA polymerase DNA of gene Promoter DNA Initiation 1 2 Terminator DNA 3 Elongation Area shown in Figure 10.9A Termination Growing RNA RNA polymerase Completed RNA

58. Figure 10.9B_1 RNA polymerase DNA of gene Promoter DNA Initiation 1 Terminator DNA

59. Figure 10.9B_2 2 Elongation Area shown in Figure 10.9A Growing RNA

60. Figure 10.9B_3 Termination RNA polymerase Completed RNA 3 Growing RNA

61. 10.10 Eukaryotic RNA is processed before leaving the nucleus as mRNA  Messenger RNA (mRNA) –Has codons that encodes amino acid sequences –Conveys genetic messages from DNA to the translation machinery of the cell, which in –prokaryotes, occurs in the same place that mRNA is made, but in –eukaryotes, mRNA must exit the nucleus via nuclear pores to enter the cytoplasm. –Eukaryotic mRNA has –introns, interrupting sequences that separate –exons, the coding regions. © 2012 Pearson Education, Inc.

62. 10.10 Eukaryotic RNA is processed before leaving the nucleus as mRNA  Three ways Eukaryotic RNA is processed –RNA splicing removes introns and joins exons to produce a continuous coding sequence. –Addition of a 5’ cap –Addition of a 3’ tail  The cap and tail protects mRNA –facilitate the export of the mRNA from the nucleus, –protect the mRNA from attack by cellular enzymes, and –help ribosomes bind to the mRNA. © 2012 Pearson Education, Inc.

63. Figure 10.10 DNA Cap Exon IntronExon RNA transcript with cap and tail ExonIntron Transcription Addition of cap and tail Introns removed Tail Exons spliced together Coding sequence NUCLEUS CYTOPLASM mRNA

64. Translation  Four requirements for translation to occur –mRNA –tRNA –Ribosome –Amino acids © 2012 Pearson Education, Inc.

65. 10.11 Transfer RNA molecules serve as interpreters during translation  Transfer RNA (tRNA) molecules function as a language interpreter, –converting the genetic message of mRNA –into the language of proteins.  Transfer RNA molecules perform this interpreter task by –picking up the appropriate amino acid and –using a special triplet of bases, called an anticodon, to recognize the appropriate codons in the mRNA. © 2012 Pearson Education, Inc.

66. Figure 10.11A Amino acid attachment site Hydrogen bond RNA polynucleotide chain Anticodon A simplified schematic of a tRNA A tRNA molecule, showing its polynucleotide strand and hydrogen bonding

67. 10.12 Ribosomes build polypeptides  Ribsome: Translation occurs on the surface of the ribosome. –Ribosomes coordinate the functioning of mRNA and tRNA and, ultimately, the synthesis of polypeptides. –Ribosomes have two subunits: small and large. –Each subunit is composed of ribosomal RNAs and proteins. –Ribosomal subunits come together during translation. –Ribosomes have binding sites for mRNA and tRNAs. © 2012 Pearson Education, Inc.

68. Figure 10.12B tRNA binding sites mRNA binding site Large subunit Small subunit P site A site

69. Figure 10.12C mRNA Codons tRNA Growing polypeptide The next amino acid to be added to the polypeptide

70. 10.13 An initiation codon marks the start of an mRNA message  Translation can be divided into the same three phases as transcription: 1.initiation, 2.elongation, and 3.termination.  Initiation brings together –mRNA, –a tRNA bearing the first amino acid, and –the two subunits of a ribosome. © 2012 Pearson Education, Inc.

71. 10.13 An initiation codon marks the start of an mRNA message  Initiation establishes where translation will begin.  Initiation occurs in two steps. 1.An mRNA molecule binds to a small ribosomal subunit and the first tRNA binds to mRNA at the start codon. –The start codon reads AUG and codes for methionine. –The first tRNA has the anticodon UAC. 2.A large ribosomal subunit joins the small subunit, allowing the ribosome to function. –The first tRNA occupies the P site, which will hold the growing peptide chain. –The A site is available to receive the next tRNA. © 2012 Pearson Education, Inc.

72. Figure 10.13B Initiator tRNA mRNA Start codon Small ribosomal subunit Large ribosomal subunit P site A site Met AUG U A C 2 AUG U A C 1 The initiation of translation

73. 10.14 Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation  Once initiation is complete, amino acids are added one by one to the first amino acid.  Elongation is the addition of amino acids to the polypeptide chain. © 2012 Pearson Education, Inc.

74.  Elongation: Each cycle of elongation has three steps. 1.Codon recognition: The anticodon of an incoming tRNA molecule, carrying its amino acid, pairs with the mRNA codon in the A site of the ribosome. 2.Peptide bond formation: The new amino acid is joined to the chain. 3.Translocation: tRNA is released from the P site and the ribosome moves tRNA from the A site into the P site. 10.14 Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation © 2012 Pearson Education, Inc.

75.  Termination: –When the ribosome reaches a stop codon, –the completed polypeptide is freed from the last tRNA, and –the ribosome splits back into its separate subunits. 10.14 Elongation adds amino acids to the polypeptide chain until a stop codon terminates translation © 2012 Pearson Education, Inc.

76. Figure 10.14_s1 Polypeptide mRNA Codon recognition Anticodon Amino acid Codons P site A site 1

77. Figure 10.14_s2 Polypeptide mRNA Codon recognition Anticodon Amino acid Codons P site A site 1 Peptide bond 2 formation

78. Figure 10.14_s3 Polypeptide mRNA Codon recognition Anticodon Amino acid Codons P site A site 1 Peptide bond 2 formation Translocation 3 New peptide bond

79. Figure 10.14_s4 Polypeptide mRNA Codon recognition Anticodon Amino acid Codons P site A site 1 Peptide bond 2 formation Translocation 3 New peptide bond Stop codon mRNA movement

80. 10.15 Review: The flow of genetic information in the cell is DNA  RNA  protein  Transcription is the synthesis of RNA from a DNA template. In eukaryotic cells, –transcription occurs in the nucleus and –the mRNA must travel from the nucleus to the cytoplasm. © 2012 Pearson Education, Inc.

81. 10.16 Mutations can change the meaning of genes  A mutation is any change in the nucleotide sequence of DNA.  Types of Mutations –Single base substitutions –Frame-Shift mutations (through deletions and insertions) © 2012 Pearson Education, Inc.

82. 10.16 Types of Mutations 1.Base substitutions involve the replacement of one nucleotide with another. Base substitutions may –have no effect at all, producing a silent mutation, –change the amino acid coding, producing a missense mutation, which produces a different amino acid, –lead to a base substitution that produces an improved protein that enhances the success of the mutant organism and its descendant, or –change an amino acid into a stop codon, producing a nonsense mutation. © 2012 Pearson Education, Inc.

83. Figure 10.16A Normal hemoglobin DNAMutant hemoglobin DNA mRNA Sickle-cell hemoglobin Normal hemoglobin Glu Val C T T G A A CT GA A U The molecular basis of sickle-cell disease

84. Figure 10.16B Normal gene Nucleotide substitution Nucleotide deletion Nucleotide insertion Inserted Deleted mRNA Protein Met Lys Phe Lys Phe Ala Gly Ser AUGAAGUUU GGC G CA GC G CA A G UUU AUGAA Met Lys Ala His Leu GUU AUGAA GGC G CA U U Met Lys Ala His Leu GUU AUGAA G G C UG G C

85. Figure 10.8A Second base Third base First base

86. 10.16 Types of Mutations 2.Frame-Shift mutations –Mutations that result in deletions or insertions that may alter the reading frame (triplet grouping) of the mRNA, so that nucleotides are grouped into different codons, –lead to significant changes in amino acid sequence downstream of the mutation, and –produce a nonfunctional polypeptide. © 2012 Pearson Education, Inc.

87.  Write the mRNA sequence and the tRNA sequence. Using the genetic code, write the amino acid sequence DNATAC GCT TAA TCA GGC GTA ATT mRNAAUG …… ….. AGU ……. …… UAA tRNAUAC Amino acid Met ……. ……. Ser ……. …….(stop) DNA Problem: Normal DNA

88.  Write the mRNA sequence and the tRNA sequence. Using the genetic code, write the amino acid sequence for the mutated DNA DNATAC GCT TAA GCA GGC GTA ATT mRNAAUG …… …….CGU……. …… UAA tRNAUAC Amino acid Met …… ……. Ala …….. ……. (stop) DNA Problem: Mutation #1 What kind of mutation is this?

89.  Write the mRNA sequence and the tRNA sequence. Using the genetic code, write the amino acid sequence for this mutated DNA. DNATAC ACT TAA TCA GGC GTA ATT mRNAAUG …... tRNAUAC Amino acid Met DNA Problem: Mutation #2 What kind of mutation is this?

90.  Write the mRNA sequence and the tRNA sequence. Using the genetic code, write the amino acid sequence, if the T base is deleted DNATAC GCT TAA TCA GGC GTA ATT mRNAAUG CGA tRNAUAC Amino acid Met DNA Problem: Mutation #3

91. 10.16 Mutations can change the meaning of genes  Mutagenesis is the production of mutations.  Mutations can be caused by –spontaneous errors that occur during DNA replication or recombination or –mutagens, which include –high-energy radiation such as X-rays and ultraviolet light and –chemicals. © 2012 Pearson Education, Inc.