1. DNA & Protein Synthesis

2. The Major Players & Their Work James Watson and Francis Crick (1953) ◦ double-helical model for DNA structure ◦ substance of inheritance  Hereditary information  reproduced in all cells of the body  directs the development of:  biochemical  Anatomical  Physiological  behavioral traits (to some extent)

3. Frederick Griffith (1928) ◦ discovery of the genetic role of DNA (the beginning) ◦ worked with two strains of a bacterium  pathogenic “S” strain  harmless “R” strain ◦ mixed heat-killed remains of the pathogenic strain with living cells of the harmless strain, some living cells became pathogenic  Transformation  a change in genotype and phenotype due to assimilation of foreign DNA

4. LE 16-2 Living S cells (control) Living R cells (control) Heat-killed S cells (control) Mixture of heat-killed S cells and living R cells Mouse dies Living S cells are found in blood sample Mouse healthy Mouse dies RESULTS

5. Oswald Avery, Maclyn McCarty, and Colin MacLeod (1944) ◦ announced that the transforming substance was DNA ◦ conclusion was based on experimental evidence that only DNA worked in transforming harmless bacteria into pathogenic bacteria

6. Viral DNA can program cells ◦ Evidence from studies of a virus that infects bacteria  bacteriophages (or phages) Animation: Phage T2 Reproductive Cycle Animation: Phage T2 Reproductive Cycle

7. LE 16-3 Bacterial cell Phage head Tail Tail fiber DNA 100 nm

8. Alfred Hershey and Martha Chase (1952) ◦ Found that DNA is the genetic material of a phage known as T2 ◦ they designed an experiment showing that only one of the two components of T2 (DNA or protein) enters an E. coli cell during infection Animation: Hershey-Chase Experiment Animation: Hershey-Chase Experiment

9. LE 16-4 Bacterial cell Phage DNA Radioactive protein Empty protein shell Phage DNA Radioactivity (phage protein) in liquid Batch 1: Sulfur ( 35 S) Radioactive DNA Centrifuge Pellet (bacterial cells and contents) Pellet Radioactivity (phage DNA) in pellet Centrifuge Batch 2: Phosphorus ( 32 P)

10. Erwin Chargaff (1947) reported that DNA composition varies from one species to the next made DNA a more credible candidate for the genetic material Animation: DNA and RNA Structure Animation: DNA and RNA Structure

11. Maurice Wilkins and Rosalind Franklin ◦ X-ray crystallography to study molecular structure Franklin ◦ produced a picture of the DNA molecule using this technique  Watson to deduce that:  DNA was helical  the width of the helix  suggested that the DNA molecule was made up of two strands, forming a double helix  the spacing of the nitrogenous bases Animation: DNA Double Helix Animation: DNA Double Helix

12. LE 16-6 Franklin’s X-ray diffraction photograph of DNA Rosalind Franklin

13. LE 16-7 5 end 3 end 5 end 3 end Space-filling modelPartial chemical structure Hydrogen bond Key features of DNA structure 0.34 nm 3.4 nm 1 nm

14. Franklin ◦ two antiparallel sugar-phosphate backbones, with the nitrogenous bases paired in the molecule’s interior Watson and Crick ◦ built models of a double helix to conform to the X-rays and chemistry of DNA ◦ purine - pyrimidine paring  results in uniform width consistent with the X-ray  Paring of like bases would cause an inconsistent width of the molecule  adenine - thymine and guanine - cytosine

15. LE 16-UN298 Purine + purine: too wide Pyrimidine + pyrimidine: too narrow Purine + pyrimidine: width consistent with X-ray data

16. LE 16-8 Adenine (A) Thymine (T) Guanine (G) Cytosine (C) Sugar

17. DNA Replication two strands of DNA are complementary each strand acts as a template for building a new strand in replication DNA replication ◦ parent molecule unwinds ◦ two new daughter strands are built based on base-pairing rules Animation: DNA Replication Overview Animation: DNA Replication Overview

18. LE 16-9_1 The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.

19. LE 16-9_2 The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C. The first step in replication is separation of the two DNA strands.

20. LE 16-9_3 The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C. The first step in replication is separation of the two DNA strands. Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand.

21. LE 16-9_4 The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C. The first step in replication is separation of the two DNA strands. Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand. The nucleotides are connected to form the sugar-phosphate back- bones of the new strands. Each “daughter” DNA molecule consists of one parental strand and one new strand.

22. semiconservative model of replication ◦ Watson and Crick ◦ each daughter molecule will have:  one old strand (derived or “conserved” from the parent molecule)  one newly made strand

23. LE 16-10 Conservative model. The two parental strands reassociate after acting as templates for new strands, thus restoring the parental double helix. Semiconservative model. The two strands of the parental molecule separate, and each functions as a template for synthesis of a new, comple- mentary strand. Dispersive model. Each strand of both daughter molecules contains a mixture of old and newly synthesized DNA. Parent cell First replication Second replication

24. LE 16-12 In eukaryotes, DNA replication begins at may sites along the giant DNA molecule of each chromosome. Two daughter DNA molecules Parental (template) strand Daughter (new) strand 0.25 µm Replication fork Origin of replication Bubble In this micrograph, three replication bubbles are visible along the DNA of a cultured Chinese hamster cell (TEM). Animation: Origins of Replication Animation: Origins of Replication

25. DNA polymerases ◦ catalyze the elongation of new DNA at a replication fork ◦ add nucleotides to the free 3  end of a growing strand  a new DNA strand can elongate only in the 5  to  3  direction

26. DNA Replication template strand = leading strand of DNA, ◦ complementary strand syhthesized continuously, moving toward the replication fork other new strand = lagging strand, ◦ DNA polymerase must work in the direction away from the replication fork ◦ synthesized as a series of segments (Okazaki fragments)  joined together by DNA ligase Animation: Leading Strand Animation: Leading Strand Animation: Lagging Strand Animation: Lagging Strand

27. LE 16-14 Parental DNA 5 3 Leading strand 3 5 3 5 Okazaki fragments Lagging strand DNA pol III Template strand Leading strand Lagging strand DNA ligase Template strand Overall direction of replication

28. Primer ◦ DNA polymerases cannot initiate synthesis ◦ initial nucleotide strand is a short one called a primer ◦ Only one primer is needed to synthesize the leading strand ◦ for lagging strand, each Okazaki fragment must be primed separately

29. Proteins Assisting with DNA Replication Helicase ◦ untwists the double helix ◦ separates the template DNA strands at the replication fork Single-strand binding protein ◦ binds to and stabilizes single-stranded DNA until it can be used as a template Topoisomerase ◦ corrects “overwinding” ahead of replication forks by breaking, swiveling, and rejoining DNA strands

30. Proteins Assisting With DNA Replication Primase ◦ synthesizes an RNA primer at the 5 ends of the leading strand and the Okazaki fragments DNA pol III ◦ continuously synthesizes the leading strand and elongates Okazaki fragments DNA pol I ◦ removes primer from the 5 ends of the leading strand and Okazaki fragments ◦ replaces primer with DNA and adding to adjacent 3 ends DNA ligase ◦ joins the 3 end of the DNA that replaces the primer to the rest of the leading strand ◦ joins the lagging strand fragments

31. Animation: DNA Replication Review Animation: DNA Replication Review LE 16-16 5 3 Parental DNA 3 5 Overall direction of replication DNA pol III Replication fork Leading strand DNA ligase Primase OVERVIEW Primer DNA pol III DNA pol I Lagging strand Lagging strand Leading strand Leading strand Lagging strand Origin of replication

32. DNA polymerases ◦ proofread newly made DNA ◦ replace any incorrect nucleotides mismatch repair of DNA ◦ repair enzymes correct errors in base pairing nucleotide excision repair ◦ enzymes cut out and replace damaged stretches of DNA

33. Transcription ◦ synthesis of mRNA under the direction of DNA Translation ◦ synthesis of a polypeptide ◦ occurs under the direction of mRNA ◦ Ribosomes

34. LE 17-3-1 TRANSCRIPTION DNA Prokaryotic cell

35. LE 17-3-2 TRANSCRIPTION DNA Prokaryotic cell Ribosome Polypeptide mRNA Prokaryotic cell

36. LE 17-3-3 TRANSCRIPTION TRANSLATION DNA mRNA Ribosome Polypeptide DNA Prokaryotic cell Nuclear envelope TRANSCRIPTION Eukaryotic cell

37. LE 17-3-4 TRANSCRIPTION TRANSLATION DNA mRNA Ribosome Polypeptide DNA Pre-mRNA Prokaryotic cell Nuclear envelope mRNA TRANSCRIPTION RNA PROCESSING Eukaryotic cell

38. LE 17-3-5 TRANSCRIPTION TRANSLATION DNA mRNA Ribosome Polypeptide DNA Pre-mRNA Prokaryotic cell Nuclear envelope mRNA TRANSLATION TRANSCRIPTION RNA PROCESSING Ribosome Polypeptide Eukaryotic cell

39. LE 17-4 DNA molecule Gene 1 Gene 2 Gene 3 DNA strand (template) 3 TRANSCRIPTION Codon mRNA TRANSLATION Protein Amino acid 3 5 5

40. codons ◦ no codon specifies more than one amino acid ◦ must be read in the correct reading frame (correct groupings) in order for the specified polypeptide to be produced

41. LE 17-5 Second mRNA base First mRNA base (5 end) Third mRNA base (3 end)

42. RNA synthesis ◦ catalyzed by RNA polymerase  pries the DNA strands apart  hooks together the RNA nucleotides ◦ uracil substitutes for thymine promoter ◦ The DNA sequence where RNA polymerase attaches  in prokaryotes, the sequence signaling the end of transcription is called the terminator Part of DNA transcribed = transcription unit Animation: Transcription Animation: Transcription

43. Initiation ◦ Promoters (TATA box in eukaryotes) ◦ Transcription factors  Mediate the binding of RNA polymerase  Mediate initiation of transcription ◦ transcription initiation complex  The completed assembly of transcription factors and RNA polymerase II bound to a promoter

44. The three stages of transcription: Elongation ◦ RNA polymerase moves along the DNA,  untwists the double helix, 10 to 20 bases at a time ◦ Transcription progresses at a rate of 60 nucleotides per second in eukaryotes A gene can be transcribed simultaneously by several RNA polymerases

45. The three stages of transcription: Termination ◦ In prokaryotes  the polymerase stops transcription at the end of the terminator ◦ In eukaryotes  the polymerase continues transcription after the pre-mRNA is cleaved from the growing RNA chain  the polymerase eventually falls off the DNA

46. Each end of a pre-mRNA molecule is modified in a particular way: ◦ The 5 end receives a modified nucleotide cap ◦ The 3 end gets a poly-A tail Functions of modifications: ◦ facilitate the export of mRNA ◦ protect mRNA from hydrolytic enzymes ◦ help ribosomes attach to the 5’ end

47. Introns noncoding regions exons eventually expressed usually translated into amino acid sequences RNA splicing removes introns and joins exons creates an mRNA molecule with a continuous coding sequence  carried out by spliceosomes  consist of a variety of proteins and several small nuclear ribonucleoproteins (snRNPs) that recognize the splice sites  Ribozymes  catalytic RNA molecules that function as enzymes and can splice RNA

48. LE 17-11 Exon 1 5 IntronExon 2 Other proteins Protein snRNA snRNPs RNA transcript (pre-mRNA) Spliceosome 5 components Cut-out intron mRNA Exon 1Exon 2 5

49. Translation Molecules of tRNA are not identical: ◦ Each carries a specific amino acid on one end ◦ Each has an anticodon on the other end  the anticodon base-pairs with a complementary codon on mRNA ◦ consists of a single RNA strand that is only about 80 nucleotides long

50. LE 17-13 Polypeptide tRNA with amino acid attached Ribosome tRNA Anticodon 3 5 mRNA Amino acids Codons

51. LE 17-14a Amino acid attachment site Hydrogen bonds 3 5 Two-dimensional structure Anticodon Amino acid attachment site 3 5 Hydrogen bonds Anticodon Symbol used in this book Three-dimensional structure 35

52. Accurate translation requires two steps: ◦ First step:  a correct match between a tRNA and an amino acid  done by the enzyme aminoacyl-tRNA synthetase ◦ Second step:  a correct match between the tRNA anticodon and an mRNA codon

53. LE 17-15 Amino acid Aminoacyl-tRNA synthetase (enzyme) Pyrophosphate Phosphates tRNA AMP Aminoacyl tRNA (an “activated amino acid”)

54. Ribosomes ◦ facilitate specific coupling of tRNA anticodons with mRNA codons ◦ two ribosomal subunits  large and small  made of proteins and ribosomal RNA (rRNA)

55. LE 17-16a tRNA molecules Exit tunnel Growing polypeptide Large subunit mRNA 3 Computer model of functioning ribosome Small subunit 5 E P A

56. A ribosome has three binding sites for tRNA: ◦ The P site  holds the tRNA that carries the growing polypeptide chain ◦ The A site  holds the tRNA that carries the next amino acid to be added to the chain ◦ The E site  is the exit site  discharged tRNAs leave the ribosome

57. LE 17-16b P site (Peptidyl-tRNA binding site) E site (Exit site) mRNA binding site A site (Aminoacyl- tRNA binding site) Large subunit Small subunit Schematic model showing binding sites EPA

58. LE 17-16c Amino end mRNA 5 3 Growing polypeptide Next amino acid to be added to polypeptide chain tRNA Codons Schematic model with mRNA and tRNA E Animation: Translation Animation: Translation

59. LE 17-18 Ribosome ready for next aminoacyl tRNA mRNA 5 Amino end of polypeptide E P site A site 3 2 2 GDP E PA GTP GDP E PA E PA

60. Termination of translation ◦ stop codon reaches the A site  A site accepts a protein called a release factor  causes the addition of a water molecule instead of an amino acid

61. LE 17-19 3 The release factor hydrolyzes the bond between the tRNA in the P site and the last amino acid of the polypeptide chain. The polypeptide is thus freed from the ribosome. The two ribosomal subunits and the other components of the assembly dissociate. Release factor Stop codon (UAG, UAA, or UGA) 5 3 5 3 5 Free polypeptide When a ribosome reaches a stop codon on mRNA, the A site of the ribosome accepts a protein called a release factor instead of tRNA.

62. LE 17-20 Ribosomes mRNA 0.1  m This micrograph shows a large polyribosome in a prokaryotic cell (TEM). An mRNA molecule is generally translated simultaneously by several ribosomes in clusters called polyribosomes. Incoming ribosomal subunits Growing polypeptides End of mRNA (3 end) Start of mRNA (5 end) Polyribosome Completed polypeptides

63. Free ribosomes ◦ mostly synthesize proteins that function in the cytosol Bound ribosomes ◦ make proteins of the endomembrane system ◦ make proteins that are secreted from the cell

64. ProkaryoteEukaryote lack a nuclear envelope, translation to begin while transcription progresses nuclear envelope separates transcription from translation

65. Mutations changes in the genetic material of a cell or virus Point mutations ◦ chemical changes in just one base pair of a gene ◦ leads to production of an abnormal protein

66. LE 17-23 Wild-type hemoglobin DNA mRNA 3553 5335 Mutant hemoglobin DNA mRNA Normal hemoglobinSickle-cell hemoglobin

67. Point Mutations Base-pair substitutions ◦ replaces one nucleotide and its partner with another pair of nucleotides ◦ can cause:  missense mutations  code for an amino acid, but not necessarily the right amino acid  More common type  nonsense mutations  change an amino acid codon into a stop codon  nearly always leading to a nonfunctional protein

68. LE 17-24 Base-pair substitution No effect on amino acid sequence U instead of C Missense A instead of G Nonsense U instead of A Stop Amino end Protein 53 Carboxyl end Stop mRNA Wild type

69. Point Mutations Base-pair insertions or deletions ◦ additions or losses of nucleotide pairs in a gene ◦ disastrous effect on the resulting protein more often than substitutions do ◦ May alter reading frame = a frameshift mutation

70. LE 17-25 Base-pair insertion or deletion Frameshift causing immediate nonsense Extra U Missing Frameshift causing extensive missense Insertion or deletion of 3 nucleotides: no frameshift but extra or missing amino acid Missing Stop Amino end Carboxyl end Stop Wild type mRNA Protein 53

71. Spontaneous mutations ◦ occur during DNA replication, recombination, or repair Mutagens ◦ physical or chemical agents that can cause mutations

72. 2. Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. What percent of the nucleotides in this sample will be thymine? a)12 b)24 c)31 d)38 e)It cannot be determined from the information provided.

73. 2. Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. What percent of the nucleotides in this sample will be thymine? a)12 b)24 c)31 d)38 e)It cannot be determined from the information provided.

74. 3. In an analysis of the nucleotide composition of DNA, which of the following is true? a)A = C b)A = G and C = T c)A + C = G + T d)A + T = G + C e)Both B and C are true

75. 3. In an analysis of the nucleotide composition of DNA, which of the following is true? a)A = C b)A = G and C = T c)A + C = G + T d)A + T = G + C e)Both B and C are true

76. A mutation results in a defective enzyme a. In the following simple metabolic pathway, what would be a consequence of that mutation? a)an accumulation of A and no production of B and C b)an accumulation of A and B and no production of C c)an accumulation of B and no production of A and C d)an accumulation of B and C and no production of A e)an accumulation of C and no production of A and B enzyme aenzyme b a b c

77. A mutation results in a defective enzyme a. In the following simple metabolic pathway, what would be a consequence of that mutation? a)an accumulation of A and no production of B and C b)an accumulation of A and B and no production of C c)an accumulation of B and no production of A and C d)an accumulation of B and C and no production of A e)an accumulation of C and no production of A and B enzyme aenzyme b a b c

78. 3. A portion of the genetic code is UUU = phenylalanine, GCC = alanine, AAA = lysine, and CCC = proline. Assume the correct code places the amino acids phenylalanine, alanine, and lysine in a protein (in that order). Which of the following DNA sequences would substitute proline for alanine? a)AAA-CGG-TTA b)AAT-CGG-TTT c)AAA-CCG-TTT d)AAA-GGG-TTT e)AAA-CCC-TTT

79. 3. A portion of the genetic code is UUU = phenylalanine, GCC = alanine, AAA = lysine, and CCC = proline. Assume the correct code places the amino acids phenylalanine, alanine, and lysine in a protein (in that order). Which of the following DNA sequences would substitute proline for alanine? a)AAA-CGG-TTA b)AAT-CGG-TTT c)AAA-CCG-TTT d)AAA-GGG-TTT e)AAA-CCC-TTT

80. 4. A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is a)TTT. b)UUA. c)UUU. d)AAA. e)either UAA or TAA, depending on wobble in the first base.

81. 4. A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is a)TTT. b)UUA. c)UUU. d)AAA. e)either UAA or TAA, depending on wobble in the first base.

82. 5. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following activated transfer RNA molecules are available. Two of them can correctly match the mRNA so that a dipeptide can form. The dipeptide that will form will be a)cysteine-alanine. b)proline-threonine. c)glycine-cysteine. d)alanine-alanine. e)threonine-glycine. tRNA AnticodonAmino Acid GGCProline CGUAlanine UGCThreonine CCGGlycine ACGCysteine CGGAlanine

83. 5. A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The following activated transfer RNA molecules are available. Two of them can correctly match the mRNA so that a dipeptide can form. The dipeptide that will form will be a)cysteine-alanine. b)proline-threonine. c)glycine-cysteine. d)alanine-alanine. e)threonine-glycine. tRNA AnticodonAmino Acid GGCProline CGUAlanine UGCThreonine CCGGlycine ACGCysteine CGGAlanine

84. 7. Each of the following is a modification of the sentence THECATATETHERAT. A.THERATATETHECAT B.THETACATETHERAT C.THECATARETHERAT D.THECATATTHERAT E.CATATETHERAT Which of the above is analogous to a frameshift mutation? a)A b)B c)C d)D e)E

85. 7. Each of the following is a modification of the sentence THECATATETHERAT. A.THERATATETHECAT B.THETACATETHERAT C.THECATARETHERAT D.THECATATTHERAT E.CATATETHERAT Which of the above is analogous to a frameshift mutation? a)A b)B c)C d)D e)E

86. 8. Each of the following is a modification of the sentence THECATATETHERAT. A.THERATATETHECAT B.THETACATETHERAT C.THECATARETHERAT D.THECATATTHERAT E.CATATETHERAT Which of the above is analogous to a single substitution mutation? a)A b)B c)C d)D e)E

87. 8. Each of the following is a modification of the sentence THECATATETHERAT. A.THERATATETHECAT B.THETACATETHERAT C.THECATARETHERAT D.THECATATTHERAT E.CATATETHERAT Which of the above is analogous to a single substitution mutation? a)A b)B c)C d)D e)E

88. 9. What is the relationship among DNA, a gene, and a chromosome? a)A chromosome contains hundreds of genes, which are composed of protein. b)A chromosome contains hundreds of genes, which are composed of DNA. c)A gene contains hundreds of chromosomes, which are composed of protein. d)A gene is composed of DNA, but there is no relationship to a chromosome. e)A gene contains hundreds of chromosomes, which are composed of DNA.

89. 9. What is the relationship among DNA, a gene, and a chromosome? a)A chromosome contains hundreds of genes, which are composed of protein. b)A chromosome contains hundreds of genes, which are composed of DNA. c)A gene contains hundreds of chromosomes, which are composed of protein. d)A gene is composed of DNA, but there is no relationship to a chromosome. e)A gene contains hundreds of chromosomes, which are composed of DNA.