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Genetics Topic 3:  Dna & Protien Synthesis
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  • 06_05_replic.origin.jpg
  • 06_09_Replic.forks.jpg
  • 06_12_asymmetrical.jpg
  • Living organisms store information in their genetic material. For an organism to survive, it must be able to read the encoded information and use it. This process is called gene expression . (BSCS, p. 233)

Transcript

  • 1. Genetics Topic 3: DNA & Protein Synthesis
  • 2. History of the DNA Molecule
  • 3. Late 1800’s: What is really controlling the cell?
    • Fred Meischer of Bern Switzerland discovered DNA
      • Studies pus cells(while changing bandages)
      • Salmon sperm cell
    • What is common between them?
      • Both have enormous nucleii
  • 4. Transformation
    • 1928--Fred Griffith worked with 2 strains of Pneumonia causing bacteria
    • Smooth strain (Virulent S) slime capsule ( not seen by immune system and kills mice) and
    • Rough strain ( Nonvirulent R) no capsule (easily killed)
    • He found that R strain could become VIRULENT when it took in DNA from heat-killed S strain
    Pneumoccocus bacteria Study suggested that DNA was probably the genetic material
  • 5. Fred Griffith Bacterial Transformation .
  • 6. Griffith Experiment
  • 7. Avery-MacLeod-McCarty Experiment
    • Repeated Griffith’s experiment adding enzymes to destroy 1. lipids, 2. carbohydrates 3. proteins 4. RNA
    • Transformation still occurred only DNA was left
  • 8.
    • The Hershey-Chase experiment showed that certain viruses reprogram host cells to produce more viruses by injecting their DNA
    Head Tail DNA So what was the substance that was causing the transformation?
  • 9.
    • Phage reproductive cycle
    Figure 10.1C Phage attaches to bacterial cell. Phage injects DNA. Phage DNA directs host cell to make more phage DNA and protein parts. New phages assemble. Cell lyses and releases new phages.
  • 10.
    • Two groups of viruses-one with radioactively labeled protein (has S but no P) and another with radioactively labeled NA( has P but no S)
    • The first group produced no radioactivity in the bacteria while the second group showed radioactivity
    • Conclusion : DNA was the transforming factor. Results matched with Avery.
  • 11.
    • The Hershey-Chase Experiment
    Figure 10.1B Mix radioactively labeled phages with bacteria. The phages infect the bacterial cells. Phage Bacterium Radioactive protein DNA Empty protein shell 1 2 Agitate in a blender to separate phages outside the bacteria from the cells and their contents. 3 Centrifuge the mixture so bacteria form a pellet at the bottom of the test tube. 4 Measure the radioactivity in the pellet and liquid. Batch 1 Radioactive protein Batch 2 Radioactive DNA Radioactive DNA Phage DNA Centrifuge Pellet Radioactivity in liquid Radioactivity in pellet Pellet Centrifuge
  • 12.
    • James Watson and Francis Crick worked out the three-dimensional structure of DNA, based on work of Rosalind Franklin
    Who discovered the structure of DNA ?
  • 13. DNA Structure Review
  • 14. DNA
    • Two strands coiled called a double helix
    • Sides made of a pentose sugar Deoxyribose bonded to phosphate (PO 4 ) groups by phosphodiester bonds
    • Center made of nitrogen bases bonded together by weak hydrogen bonds
  • 15. DNA Double Helix Nitrogenous Base (A,T,G or C) “ Rungs of ladder” “ Legs of ladder” Phosphate & Sugar Backbone
  • 16. Helix
    • Most DNA has a right-hand twist with 10 base pairs in a complete turn
    • Left twisted DNA is called Z-DNA or southpaw DNA
    • Hot spots occur where right and left twisted DNA meet producing mutations
  • 17. DNA
    • Stands for Deoxyribonucleic acid
    • Made up of subunits called nucleotides
    • Nucleotide made of:
    • 1. Phosphate group
    • 2. 5-carbon sugar
    • 3. Nitrogenous base
  • 18. DNA Nucleotide O O=P-O O Phosphate Group N Nitrogenous base (A, G, C, or T) CH2 O C 1 C 4 C 3 C 2 5 Sugar (deoxyribose)
  • 19. Pentose Sugar
    • Carbons are numbered clockwise 1’ to 5’
    C H2 O C 1 C 4 C 3 C 2 5 Sugar (deoxyribose)
  • 20. DNA P P P O O O 1 2 3 4 5 5 3 3 5 P P P O O O 1 2 3 4 5 5 3 5 3 G C T A
  • 21. Antiparallel Strands
    • One strand of DNA goes from 5’ to 3’ (sugars)
    • The other strand is opposite in direction going 3’ to 5’ (sugars)
  • 22. Nitrogenous Bases
    • Double ring PURINES
    • Adenine (A)
    • Guanine (G)
    • Single ring PYRIMIDINES
    • Thymine (T)
    • Cytosine (C)
    T or C A or G
  • 23. Base-Pairings
    • Purines only pair with Pyrimidines
    • Three hydrogen bonds required to bond Guanine & Cytosine
    C G 3 H-bonds
  • 24.
    • Two hydrogen bonds are required to bond Adenine & Thymine
    T A
  • 25. Question:
    • If there is 30% Adenine , how much Cytosine is present?
  • 26. Answer:
    • There would be 20% Cytosine
    • Adenine (30%) = Thymine (30%)
    • Guanine (20%) = Cytosine (20%)
    • Therefore, 60% A-T and 40% C-G
  • 27. Nucleosome Structure:
    •  
    • Nucleosome Structure: (Eukaryotes only)
    • DNA (double loop around the histones)
    • Histone Core of eight proteins
    • H1 histone securing the structure
  • 28. Chromosome :
    • Note the supercoiling of the nucleosomes
    • There is actually another protein(not shown) around which the nucleosomes condense
    • Note that in Prokaryotes there are no nucleosomes just 'naked DNA'
  • 29.
    • RNA is also a nucleic acid
      • RNA has a slightly different sugar (ribose) from DNA( deoxyribose)
      • RNA has U instead of T
    Phosphate group Nitrogenous base (A, G, C, or U) Uracil (U) Sugar (ribose)
  • 30. DNA Replication
  • 31. Synthesis Phase (S phase)
    • S phase during interphase of the cell cycle
    • Nucleus of eukaryotes
    Mitosis -prophase -metaphase -anaphase -telophase G 1 G 2 S phase interphase DNA replication takes place in the S phase .
  • 32.
    • DNA replication is very specific to the arrangements of base pairs
    • In DNA replication, the strands separate
      • Enzymes use each strand as a template to assemble the new strands
    DNA REPLICATION Parental molecule of DNA Figure 10.4A Both parental strands serve as templates Two identical daughter molecules of DNA Nucleotides A A
  • 33. 06_05_replic.origin.jpg
  • 34. 06_09_Replic.forks.jpg
  • 35. 06_12_asymmetrical.jpg
  • 36. DNA Replication
    • Begins at Origins of Replication
    • Two strands open forming Replication Forks (Y-shaped region)
    • New strands grow at the forks
    Replication Fork Parental DNA Molecule 3’ 5’ 3’ 5’
  • 37. DNA Replication
    • As the 2 DNA strands open at the origin, Replication Bubbles form
    • Prokaryotes (bacteria) have a single bubble
    • Eukaryotic chromosomes have MANY bubbles
    Bubbles Bubbles
  • 38. DNA Replication
    • Enzyme Helicase unwinds and separates the 2 DNA strands by breaking the weak hydrogen bonds
    • Single-Strand Binding Proteins attach and keep the 2 DNA strands separated and untwisted
  • 39. DNA Replication
    • Before new DNA strands can form, there must be RNA primers present to start the addition of new nucleotides
    • Primase is the enzyme that synthesizes the RNA Primer
    • DNA polymerase can then add the new nucleotides
  • 40. DNA Replication
    • DNA polymerase can only add nucleotides to the 3’ end of the DNA
    • This causes the NEW strand to be built in a 5’ to 3’ direction
    Direction of Replication RNA Primer DNA Polymerase Nucleotide 5’ 5’ 3’
  • 41. Remember HOW the Carbons Are Numbered! O O=P-O O Phosphate Group N Nitrogenous base (A, G, C, or T) CH2 O C 1 C 4 C 3 C 2 5 Sugar (deoxyribose)
  • 42. Remember the Strands are Antiparallel P P P O O O 1 2 3 4 5 5 3 3 5 P P P O O O 1 2 3 4 5 5 3 5 3 G C T A
  • 43. Synthesis of the New DNA Strands
    • The Leading Strand is synthesized as a single strand from the point of origin toward the opening replication fork
    RNA Primer DNA Polymerase Nucleotides 3’ 5’ 5’
  • 44. Synthesis of the New DNA Strands
    • The Lagging Strand is synthesized discontinuously against overall direction of replication
    • This strand is made in MANY short segments It is replicated from the replication fork toward the origin
    RNA Primer Leading Strand DNA Polymerase 5’ 5’ 3’ 3’ Lagging Strand 5’ 5’ 3’ 3’
  • 45. Lagging Strand Segments
    • Okazaki Fragments - series of short segments on the lagging strand
    • Must be joined together by an enzyme
    Lagging Strand RNA Primer DNA Polymerase 3’ 3’ 5’ 5’ Okazaki Fragment
  • 46. Joining of Okazaki Fragments
    • The enzyme Ligase joins the Okazaki fragments together to make one strand
    Lagging Strand Okazaki Fragment 2 DNA ligase Okazaki Fragment 1 5’ 5’ 3’ 3’
  • 47. Replication of Strands Replication Fork Point of Origin
  • 48. Conservation of base sequence
    • The DNA base sequence is usually read down one side of the molecule or the other. The sequence is usually read with reference to the bases and their corresponding identifying letter.
    • e.g. (1) ATG CTC ATT  TTA  GGG CCC ATA CTC
    • = 24 bases thus we can write the complementary sequence of the other helix as:
    • (2) TAC GAG TAA AAT CCC GGG TAT GAG
    • In DNA replication
    • will act as the template for a new complementary sequence of bases
    • copy 1:   (1) ATG  CTC ATT TTA GGG CCC ATA CTC
    •           
    • (2)TAC GAG TAA AAT CCC GGG TAT GAG
  • 49.
    • and (2) will act as a template for the other new complementary sequence (1)
    • Copy 2:             
    • (2) TAC GAG TAA AAT CCC GGG TAT GAG             
    • (1)ATG CTC  ATT TTA  GGG CCC ATA CTC
    • therefore after replication the base sequence of copies 1
    • (1) (2) and copy 2 (1) (2) are identical to each other.
    • Importantly they are also identical in base sequence to the
    • original base sequence of (1)(2) .
    • • The DNA base sequence of the double helix is conserved
    • from one replication to another.
    • • As cells divide the DNA is copied so that each new cell
    • possess a copy of each of the original DNA molecules
  • 50. 6.2.3 Replication points in Eukaryotes
    • Eukaryotic DNA
          • Forms Nucleosomes
          • Linear Chromosomes
          • Many Replication Forks
    • Prokaryotic DNA
          • No Nucleosomes/ no histones
          • Loop DNA
          • No Replication Fork/ Single point
  • 51. How do organisms use the information stored in their genetic material? . . . through a process called gene expression .
  • 52. A gene provides the information for making a specific protein.
  • 53. What is a gene?
    • A gene is a sequence of nucleotide bases in DNA.
    • Genes provide the instructions for making proteins.
  • 54. What are proteins?
    • Proteins are the keys to almost everything that living cells are and do.
    • Proteins are polymers constructed from monomers called amino acids.
  • 55. What are amino acids?
    • Each amino acid monomer consists of a central carbon bonded to four partners:
      • an amino group,
      • a carboxyl group (acid group), and
      • a hydrogen atom.
      • The fourth bond is with a unique side group.
  • 56. Why are proteins important?
    • Proteins are important to the structures of cells and organisms and perform most functions in cells.
    • Proteins can be . . .
      • structural,
      • contractile,
      • used for storage,
      • defensive,
      • used for transport,
      • hormones or
      • enzymes.
  • 57. Two main steps from gene expression to protein 1. transcription 2. translation 1. 2.
  • 58. 1. Transcription: The synthesis of a strand of mRNA
    • Uses an enzyme RNA polymerase.
    • Proceeds in the same direction as replication ( 5’ to 3’ )
    • Forms a complementary strand of mRNA
    • It begins at a promotor site which signals the beginning of gene is not much further down the molecule (about 20 to 30 nucleotides).
    • After the end of the gene is reached there is a terminator sequence that tells RNA polymerase to stop transcribing.
  • 59. 2. Translation synthesis of a polypeptide chain using the genetic code on the mRNA.
    • Location: The ribosomes in the cytoplasm that provide the environment for translation.
    • The genetic code is brought by the mRNA molecule.
    • What is the genetic code?
    • The genetic code consists of the sequence of bases found along the mRNA molecule.
    • There are only four letters to this code (A, G, C and U).
    • The code needs to be complex enough to represent 20 different amino acids used to build proteins
  • 60. THE GENETIC CODE
  • 61.
    • Transcription the expression of a gene
  • 62. 6.3.1 The Direction of Transcription
    • Transcription is carried out in the 5' to 3' direction.
    • mRNA is made by adding 5' free end of the nucleotide to the 3' end of the newly transcribed mRNA sequence.
    • Gene Expression : When a gene is switched on to transcription and translation
    • Gene Regulation : The process that controls which and how particular genes are activated within a cell. It should be noted that there are differences between Prokaryotes and Eukaryotes.
  • 63. Prokaryotic vs. Eukaryotic Protein Synthesis
  • 64. Prokaryotic Gene Expression (transcription) : Operon theory
    • Operons:
    • Groups of genes that are expressed together
    • Only found in Prokaryotes
          •   e.g. Lac Operon : Lactobacillus bacteria require two enzymes and a transport protein when metabolizing lactose sugar.
  • 65.  
  • 66.
    • R egulator gene produces a repressor protein.
    • Promotor gene is where RNA polymerase binds
    • Operator gene is where the repressor binds
    Lac Z , Lac Y , and Lac A are the genes for the three proteins that are required to metabolize lactose. Prokaryotic Gene Expression (transcription)
  • 67.
    • In the absence of lactose the Regulator gene produces a protein that blocks RNA polymerase from the Operator gene
    Lac Z, Lac Y and Lac A are not transcribed. Repressor Protein
  • 68.
    • The Bacteria is now in the presence of Lactose sugar.
    • This binds to the repressor protein which cause molecular shape changes that stop means the repressor cannot bind to the operator
    • RNA polymerase now binds to the operator and begins the transcription of all three genes.
  • 69.  
  • 70.  
  • 71. Eukaryotic gene Expression:
    • The genes of Eukaryotes are expressed separately from each other and not in the form of operons.
    • The genes of eukaryotes contain regions called introns.
    • Introns are transcribed but not translated.
    • Gene Expression in Eukaryotes has a number of stages:
    • Transcription of the gene.
    • Processing of the mRNA to remove introns (post-transcriptional modification).
    • Translation of the mRNA into a protein.
    • Post- Translational modification of the protein which take place in the rough endoplasmic reticulum or the golgi apparatus.
  • 72. Regulation of Transcription:
  • 73. Regulation of Transcription:
    •   a) The gene has a promotor region and a terminator region
    • b) Transcription requires the presence of a regulator protein from another gene (possible from another chromosome).
    • c) The RNA polymerase can now bind to the promotor and begin the transcription of the gene.
    • d) The mRNA is transcribed including introns
    • e)The completed mRNA which will require post transcriptional modification to remove the introns.
  • 74. Sense and Anti-Sense Regions of DNA
    • Excepting the change of bases U for T in RNA. Then
    • we can say that:
    • The sense strand is identical to the mRNA strand. This is the codon sequence.
    • The anit-sense sequence is the same as the anti-codon sequence
    • RNA polymerase moves along the DNA strand in the 3' to 5' direction, adding nucleotides to the 3' end of the RNA chain.
  • 75. Post Transcriptional Modification
  • 76. Post Transcriptional Modification
    •   1. Promotor region
    • 2. Free Nucleotide Phosphates
    • 3. Addition of Nucleotides to the new mRNA
    • 4. Early mRNA
    • 5. Early mRNA showing introns (non-coding)
    • 6.Introns removed allowing exons to combine
    • 7. Mature mRNA ready for translation
    • 8. mRNA going to cytoplasm.
  • 77. 2. Translation construction of a protein
  • 78. The role of RNA in Protein Synthesis
    • 3 Types of RNA molecules in the steps from gene to protein:
      • Messenger RNA (mRNA), Complementary copy of DNA
      • Transfer RNA (tRNA) carries amino acid to the site of synthesis.
      • Ribosomal RNA (rRNA), stabilizes the site of synthesis
  • 79. Translation stages (creation of m RNA) : Initiation, Elongation and Termination
    • Translation occurs in the 5' to 3' direction along the mRNA
    • Initiation begins with the attachment of the ribosome to the mRNA
    • Elongation begins at the mRNA start codon AUG and continues with the addition of amino acids to the polypeptide.
    • Termination occurs at the STOP codon (UGA, UAG or UAA).
    Step 1
  • 80.  
  • 81. Activation : attaching specific amino acids to specific tRNA molecules.
    • Amino acid which is specific to each tRNA.
    • (b) CCA base sequence to
    • which the amino acid is
    • attached by the
    • 'Activating Enzyme'.
    • (c) Complementary base
    • pairing sequence. Helical
    • in shape.
    • (d) 8 free bases
    • non-pairing giving one
    • loop of RNA.
    Step 2
  • 82.
    • (e) 7 free bases non-pairing giving second loop of RNA.
    • (f) Small open loop of RNA which is variable in shape between different tRNA.
    • (g) Anti-codon (3 bases) which binds to the mRNA codon (3 bases) this is specific to the amino acids being carried. The anti-codon is complementary to the sense DNA.
  • 83. Ribosome Structure
    • Proteins and Ribosomal RNA combine in the structure
    • Large sub-unit and a small sub-unit
    • Large sub-unit has two binding sites for tRNA molecules ( P and A site)
    • Small sub-unit has a binding site for mRNA
    P P A Step 2
  • 84. Ribosome Structure
  • 85. Building a Protein
  • 86.
    •   (a) tRNA and amino acids are 'activated‘
    • (b) mRNA produced in transcription and now located in the cytoplasm. The translation will occur in the 5' to 3' direction along the mRNA.
  • 87.
    • (c) Initiation begins with the attachment of the small sub-unit to the mRNA. This is already carrying the tRNA with the Anti-codon, UAC.
    • (d) The large sub-unit binds to the small at the mRNA START codon AUG. The process of elongation of the polypeptide can begin.
  • 88.  
  • 89.  
  • 90.  
  • 91.  
  • 92.  
  • 93.  
  • 94.  
  • 95. Ribosomes effect in translation
    • Ribosome are found in Prokaryotes (70's) and Eukaryotes (80's). Including P and A sites. START codons and STOP codons begin and termination translation.
    • Polyribosome= Polysomes
        • Multiple ribosomes on the same mRNA at the same time.
        • All ribosome move 5' to 3' in sequence.
        • In protein synthesis polyribosomes increase the quantity of polypeptide synthesized.
  • 96.
    • Information in DNA is used to synthesize a particular protein
    • Proteins provide the molecular basis for phenotypic traits
    • The information constituting an organism’s genotype is carried in its sequence of bases
    What does DNA have to do with proteins?
  • 97.
    • A specific gene specifies a polypeptide
      • The DNA is transcribed into RNA, which is translated into the polypeptide
    Figure 10.6A DNA DNA Protein TRANSCRIPTION TRANSLATION
  • 98.
    • The “words” of the DNA “language” are triplets of bases called codons
      • The codons in a gene specify the amino acid sequence of a polypeptide
      • 64 (4 3 ) possible triplets code for 20 amino acids, so more than one triplet codes for an amino acid.
    Genetic information written in codons is translated into amino acid sequences
  • 99. Figure 10.7 DNA molecule Gene 1 Gene 2 Gene 3 DNA strand TRANSCRIPTION RNA Polypeptide TRANSLATION Codon Amino acid
  • 100.
    • Virtually all organisms from the blue whale to the smallest bacteria share the same genetic code
    The genetic code helps us decode the formula for life Figure 10.8A