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Transcription,translation and genetic code(cell biology)by welfredo yu


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Welfredo Lubrico Yu,Jr.

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Transcription,translation and genetic code(cell biology)by welfredo yu

  1. 1.  Cells are governed by a cellular chain of command  DNA RNA protein Transcription  Is the synthesis of RNA under the direction of DNA  Produces messenger RNA (mRNA) Translation  Is the actual synthesis of a polypeptide, which occurs under the direction of mRNA  Occurs on ribosomes The Genetic Code  a non-overlapping sequence, with each amino acid plus polypeptide initiation and termination specified by RNA codons composed of three nucleotides.
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  4. 4.   Genes can be expressed at different efficiencies  • Gene A is transcribed much more efficiently than gene B  • This allows the amount of protein A in the cell to be  greater than protein B  • The lower expression of gene B is a reason behind  incomplete dominance  BIOL211
  5. 5.  RNA is the bridge between genes and the proteins for which they code. what is RNA?
  6. 6.  Monomers of proteins are amino acids what are proteins made of?
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  8. 8.  Codons • A codon is three nucleotides in a row on an RNA molecule that codes for a single amino acid • A specific three-nucleotide sequence encodes for each amino acid
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  11. 11.  Template Strand • During transcription, one of the two DNA strands, called the template strand, provides a template for ordering the sequence of complementary nucleotides in an RNA transcript – The template strand is always the same strand for a given gene – However, different genes may be on opposite strands
  12. 12.  • The genetic code is nearly universal, shared by the simplest bacteria to the most complex animals – Some species prefer certain codons (codon bias) • Genes can be transcribed and translated after being transplanted from one species to another EVOLUTION OF THE CODE
  13. 13.  History: linking genes and proteins  1900’s Archibald Garrod  Inborn errors of metabolism: inherited human metabolic diseases (more information)  Genes are the inherited factors  Enzymes are the biological molecules that drive metabolic reactions  Enzymes are proteins  Question:  How do the inherited factors, the genes, control the structure and activity of enzymes (proteins)?
  14. 14.  History: linking genes and proteins  Beadle and Tatum (1941) PNAS USA 27, 499–506.  Hypothesis:  If genes control structure and activity of metabolic enzymes, then mutations in genes should disrupt production of required nutrients, and that disruption should be heritable.  Method:  Isolated ~2,000 strains from single irradiate spores (Neurospora) that grew on rich but not minimal medium. Examples: defects in B1, B6 synthesis.  Conclusion:  Genes govern the ability to synthesize amino acids, purines and vitamins.
  15. 15.  History: linking genes and proteins  1950s: sickle-cell anemia  Glu to Val change in hemoglobin  Sequence of nucleotides in gene determines sequence of amino acids in protein  Single amino acid change can alter the function of the protein  Tryptophan synthase gene in E. coli  Mutations resulted in single amino acid change  Order of mutations in gene same as order of affected amino acids
  16. 16.  Ribosomal structure E P A Large subunit Peptidyl-tRNA binding site Aminoacyl-tRNA binding site mRNA 5’ Exit site Small subunit 3’
  17. 17.  From gene to protein: transcription  Gene sequence (DNA) recopied or transcribed to RNA sequence  Product of transcription is a messenger molecule that delivers the genetic instructions to the protein synthesis machinery: messenger RNA (mRNA)
  18. 18.  Transcription: evidence for mRNA  Brenner, S., Jacob, F. and Meselson, M. (1961) Nature 190, 576–81.  Question: How do genes work?  Does each one encode a different type of ribosome which in turn synthesizes a different protein, OR  Are all ribosomes alike, receiving the genetic information to create each different protein via some kind of messenger molecule?
  19. 19.  Transcription: evidence for mRNA  E. coli cells switch from making bacterial proteins to phage proteins when infected with bacteriophage T4.  Grow bacteria on medium containing “heavy” nitrogen (15N) and carbon (13C).  Infect with phage T4.  Immediately transfer to “light” medium containing radioactive uracil.
  20. 20.  Transcription: evidence for mRNA  If genes encode different ribosomes, the newly synthesized phage ribosomes will be “light”.  If genes direct new RNA synthesis, the RNA will contain radiolabeled uracil.  Results:  Ribosomes from phage-infected cells were “heavy”, banding at the same density on a CsCl gradient as the original ribosomes.  Newly synthesized RNA was associated with the heavy ribosomes.  New RNA hybridized with viral ssDNA, not bacterial ssDNA.
  21. 21.  Transcription: evidence for mRNA  Conclusion  Expression of phage DNA results in new phage- specific RNA molecules (mRNA)  These mRNA molecules are temporarily associated with ribosomes  Ribosomes do not themselves contain the genetic directions for assembling individual proteins
  22. 22.  Transcription: overview  Transcription requires:  ribonucleoside 5´ triphosphates:  ATP, GTP, CTP and UTP  bases are adenine, guanine, cytosine and uracil  sugar is ribose (not deoxyribose)  DNA-dependent RNA polymerase  Template (sense) DNA strand  Animation of transcription
  23. 23.  Transcription: overview  Features of transcription:  RNA polymerase catalyzes sugar-phosphate bond between 3´-OH of ribose and the 5´-PO4.  Order of bases in DNA template strand determines order of bases in transcript.  Nucleotides are added to the 3´-OH of the growing chain.  RNA synthesis does not require a primer.
  24. 24.  Transcription: overview  In prokaryotes transcription and translation are coupled. Proteins are synthesized directly from the primary transcript as it is made.  In eukaryotes transcription and translation are separated. Transcription occurs in the nucleus, and translation occurs in the cytoplasm on ribosomes.  Figure comparing eukaryotic and prokaryotic transcription and translation.
  25. 25.  Transcription: RNA Polymerase  DNA-dependent  DNA template, ribonucleoside 5´ triphosphates, and Mg2+  Synthesizes RNA in 5´ to 3´ direction  E. coli RNA polymerase consists of 5 subunits  Eukaryotes have three RNA polymerases  RNA polymerase II is responsible for transcription of protein-coding genes and some snRNA molecules  RNA polymerase II has 12 subunits  Requires accessory proteins (transcription factors)  Does not require a primer
  26. 26.  Stages of Transcription
  27. 27.  Transcription: promoter recognition  Transcription factors bind to promoter sequences and recruit RNA polymerase.  DNA is bound first in a closed complex. Then, RNA polymerase denatures a 12–15 bp segment of the DNA (open complex).  The site where the first base is incorporated into the transcription is numbered “+1” and is called the transcription start site.  Transcription factors that are required at every promoter site for RNA polymerase interaction are called basal transcription factors.
  28. 28.  Promoter recognition: promoter sequences  Promoter sequences vary considerably.  RNA polymerase binds to different promoters with different strengths; binding strength relates to the level of gene expression  There are some common consensus sequences for promoters:  Example: E. coli –35 sequence (found 35 bases 5´ to the start of transcription)  Example: E. coli TATA box (found 10 bases 5´ to the start of transcription)
  29. 29.  Promoter recognition: enhancers  Eukaryotic genes may also have enhancers.  Enhancers can be located at great distances from the gene they regulate, either 5´ or 3´ of the transcription start, in introns or even on the noncoding strand.  One of the most common ways to identify promoters and enhancers is to use a reporter gene.
  30. 30.  Promoter recognition: other players  Many proteins can regulate gene expression by modulating the strength of interaction between the promoter and RNA polymerase.  Some proteins can activate transcription (upregulate gene expression).  Some proteins can inhibit transcription by blocking polymerase activity.  Some proteins can act both as repressors and activators of transcription.
  31. 31.  Transcription: chain initiation  Chain initiation:  RNA polymerase locally denatures the DNA.  The first base of the new RNA strand is placed complementary to the +1 site.  RNA polymerase does not require a primer.  The first 8 or 9 bases of the transcript are linked. Transcription factors are released, and the polymerase leaves the promoter region.  Figure of bacterial transcription initiation.
  32. 32.  Transcription: chain elongation  Chain elongation:  RNA polymerase moves along the transcribed or template DNA strand.  The new RNA molecule (primary transcript) forms a short RNA-DNA hybrid molecule with the DNA template.
  33. 33.  Transcription: chain termination  Most known about bacterial chain termination  Termination is signaled by a sequence that can form a hairpin loop.  The polymerase and the new RNA molecule are released upon formation of the loop.  Review the transcription animation.
  34. 34.  Transcription: mRNA synthesis/processing  Prokaryotes: mRNA transcribed directly from DNA template and used immediately in protein synthesis  Eukaryotes: primary transcript must be processed to produce the mRNA  Noncoding sequences (introns) are removed  Coding sequences (exons) spliced together  5´-methylguanosine cap added  3´-polyadenosine tail added
  35. 35.  Transcription: mRNA synthesis/processing  Removal of introns and splicing of exons can occur several ways  For introns within a nuclear transcript, a spliceosome is required.  Splicesomes protein and small nuclear RNA (snRNA)  Specificity of splicing comes from the snRNA, some of which contain sequences complementary to the splice junctions between introns and exons  Alternative splicing can produce different forms of a protein from the same gene  Mutations at the splice sites can cause disease  Thalassemia • Breast cancer (BRCA 1)
  36. 36.  Transcription: mRNA synthesis/processing  RNA splicing inside the nucleus on particles called spliceosomes.  Splicesomes are composed of proteins and small RNA molecules (100–200 bp; snRNA).  Both proteins and RNA are required, but some suggesting that RNA can catalyze the splicing reaction.  Self-splicing in Tetrahymena: the RNA catalyzes its own splicing  Catalytic RNA: ribozymes
  37. 37.  From gene to protein: genetic code  Central Dogma  Information travels from DNA to RNA to Protein  Is there a one-to-one correspondence between DNA, RNA and Protein?  DNA and RNA each have four nucleotides that can form them; so yes, there is a one-to-one correspondence between DNA and RNA.  Proteins can be composed of a potential 20 amino acids; only four RNA nucleotides: no one-to-one correspondence.  How then does RNA direct the order and number of amino acids in a protein?
  38. 38.  From gene to protein: genetic code  How many bases are required for each amino acid?  (4 bases)2bases/aa = 16 amino acids—not enough  (4 bases)3bases/aa = 64 amino acid possibilities  Minimum of 3 bases/aa required  What is the nature of the code?  Does it have punctuation? Is it overlapping?  Crick, F.H. et al. (1961) Nature 192, 1227–32. ( )  3-base, nonoverlapping code that is read from a fixed point.
  39. 39.  From gene to protein: genetic code  Nirenberg and Matthaei: in vitro protein translation  Found that adding rRNA prolonged cell-free protein synthesis  Adding artificial RNA synthesized by polynucleotide phosphorylase (no template, UUUUUUUUU) stimulated protein synthesis more  The protein that came out of this reaction was polyphenylalanine (UUU = Phe)  Other artificial RNAs: AAA = Lys; CCC =Pro
  40. 40.  From gene to protein: genetic code  Nirenberg:  Triplet binding assay: add triplet RNA, ribosomes, binding factors, GTP, and radiolabeled charged tRNA (figure)  UUU trinucleotide binds to Phe-tRNA  UGU trinucleotide binds to CYS-tRNA  By fits and starts the triplet genetic code was worked out.  Each three-letter “word” (codon) specifies an amino acid or directions to stop translation.  The code is redundant or degenerate: more than one way to encode an amino acid
  41. 41.  From gene to protein: Translation  Components required for translation:  mRNA  Ribosomes  tRNA  Aminoacyl tRNA synthetases  Initiation, elongation and termination factors  Animation of translation
  42. 42.  Translation: initiation  Ribosome small subunit binds to mRNA  Charged tRNA anticodon forms base pairs with the mRNA codon  Small subunit interacts with initiation factors and special initiator tRNA that is charged with methionine  mRNA-small subunit-tRNA complex recruits the large subunit  Eukaryotic and prokaryotic initiation differ slightly
  43. 43.  Translation: initiation  The large subunit of the ribosome contains three binding sites  Amino acyl (A site)  Peptidyl (P site)  Exit (E site)  At initiation,  The tRNAfMet occupies the P site  A second, charged tRNA complementary to the next codon binds the A site.
  44. 44.  Translation: elongation  Elongation  Ribosome translocates by three bases after peptide bond formed  New charged tRNA aligns in the A site  Peptide bond between amino acids in A and P sites is formed  Ribosome translocates by three more bases  The uncharged tRNA in the A site is moved to the E site.
  45. 45.  Translation: elongation  EF-Tu recruits charged tRNA to A site. Requires hydrolysis of GTP  Peptidyl transferase catalyzes peptide bond formation (bond between aa and tRNA in the P site converted to peptide bond between the two amino acids)  Peptide bond formation requires RNA and may be a ribozyme-catalyzed reaction
  46. 46.  Translation: termination  Termination  Elongation proceeds until STOP codon reached  UAA, UAG, UGA  No tRNA normally exists that can form base pairing with a STOP codon; recognized by a release factor  tRNA charged with last amino acid will remain at P site  Release factors cleave the amino acid from the tRNA  Ribosome subunits dissociate from each other  Review the animation of translation
  47. 47. 48 Genetic code: Def. Genetic code is the nucleotide base sequence on DNA ( and subsequently on mRNA by transcription) which will be translated into a sequence of amino acids of the protein to be synthesized. The code is composed of codons Codon is composed of 3 bases ( e.g. ACG or UAG). Each codon is translated into one amino acid. The 4 nucleotide bases (A,G,C and U) in mRNA are used to produce the three base codons. There are therefore, 64 codons code for the 20 amino acids, and since each codon code for only one amino acids this means that, there are more than one cone for the same amino acid. How to translate a codon (see table): This table or dictionary can be used to translate any codon sequence. Each triplet is read from 5′ → 3′ direction so the first base is 5′ base, followed by the middle base then the last base which is 3′ base.
  48. 48. 49 Examples: 5′- A UG- 3′ codes for methionine 5′- UCU- 3′ codes for serine 5′ - CCA- 3′ codes for proline Termination (stop or nonsense) codons: Three of the 64 codons; UAA, UAG, UGA do not code for any amino acid. They are termination codes which when one of them appear in mRNA sequence, it indicates finishing of protein synthesis. Characters of the genetic code: 1- Specificity: the genetic code is specific, that is a specific codon always code for the same amino acid. 2- Universality: the genetic code is universal, that is, the same codon is used in all living organisms, procaryotics and eucaryotics. 3- Degeneracy: the genetic code is degenerate i.e. although each codon corresponds to a single amino acid,one amino acid may have more than one codons. e.g arginine has 6 different codons (give more examples from the table).
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  50. 50. 51 Gene mutation (altering the nucleotide sequence): 1- Point mutation: changing in a single nucleotide base on the mRNA can lead to any of the following 3 results: i- Silent mutation: i.e. the codon containg the changed base may code for the same amino acid. For example, in serine codon UCA, if A is changed to U giving the codon UCU, it still code for serine. See table. ii- Missense mutation: the codon containing the changed base may code for a different amino acid. For example, if the serine codon UCA is changed to be CCA ( U is replaced by C), it will code for proline not serine leading to insertion of incorrect amino acid into polypeptide chain. iii- Non sense mutation: the codon containing the changed base may become a termination codon. For example, serine codon UCA becomes UAA if C is changed to A. UAA is a stop codon leading to termination of translation at that point.
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  52. 52. 53 Types of point mutation: U A A (termination codon) Nonsense mutation ↑ U C A → U C U Silent mutation (codon for serine) (codon for serine) ↓ C C A ( codon for proline) Missense mutation: Give other examples on missense mutation which leads to some Hb disease.
  53. 53. 54 2- Frame- shift mutation: deletion or addition of one or two base to message sequence, leading to change in reading frame (reading sequence) and the resulting amino acid seuence may become completely different from this point.
  54. 54. 55 Translation Components required for protein synthesis: 1- Amino acids: all amino acids involved in the finished protein must be present at the time of protein synthesis. 2- Ribosomes: the site of protein synthesis. They are large complexes of protein and rRNA. In human, they consist of two subunits, one large (60S) and one small (40S). 3- tRNA: at least one specific type of tRNA is required to transfer one amino acid. There about 50 tRNA in human for the 20 amino acids, this means some amino acids have more than one specific tRNA. The role of tRNA in protein synthesis is discussed before. (amino acid attachment and anticodon loop). 4- aminoacyl-tRNA synthetase: This is the enzyme that catalyzes the attachment of amino acid with its corresponding tRNA forming aminoacyl tRNA
  55. 55. 56 5- mRNA: that carry code for the protein to be synthesized 6- protein factors: Initiation, elongation and termination (or release) factors are required for peptide synthesis 7- ATP and GTP : are required as source of energy. Steps: (movie) 1- Initiation: Initiation (start) codon is usually AUG which is the codon of methionine, so the initiator tRNA is methionnyl tRNA (Met. tRNA). a- The initiation factors (IF-1, IF-2 and IF-3) binds the Met. tRNA with small ribosomal subunit then to mRNA containing the code of the protein to be synthesized. IFs recognizes mRNA from its 5' cap
  56. 56. 57 b-This complex binds to large ribosomal subunit forming initiation complex in which Met. tRNA is present in P- site of 60 ribosomal subunit. NB:- tRNA bind with mRNA by base pairing between codon on mRNA and anticodon on tRNA. - mRNA is read from 5′ → 3′ direction P-site: is the peptidyl site of the ribosome to which methionyl tRNA is placed (enter). 2- Elongation: elongation factors (EFs) stimulate the stepwise elongation of polypeptide chain as follow: a- The next aminoacyl tRNA (tRNA which carry the next amino acid specified by recognition of the next codon on mRNA) will enter A site of ribosome A site or acceptor site or aminoacyl tRNA site: Is the site of ribosome to which each new incoming aminoacyl tRNA will enter.
  57. 57. b) ribosomal peptidyl transferase enzyme will transfer methionine from methionyl tRNA into A site to form a peptide bond between methionine and the new incoming amino acid to form dipeptidyl tRNA. c) Elongation factor-2 (EF-2), (called also, translocase): moves mRNA and dipeptidyl tRNA from A site to P site leaving A site free to allow entrance of another new aminoacyl tRNA. The figure shows the repetitive cycle of elongation of chain. Each cycle is consisting of 1) codon recognition and the entrance of the new aminoacyl tRNA acid ( amino acid carried on tRNA) into A site, 2) The growing chain in P site will moved to A site with peptide bond formation with the new amino acid 3) Translocation of growing chain to P site allowing A site free for enterance of new amino acid an so on………………….. Resulting in elongation of poly peptide chain.
  58. 58. 59 repetitive cycle of elongation
  59. 59. 60 3- Termination: occurs when one of the three stop codons (UAA, UAG or UGA) enters A site of the ribosome. These codons are recognized by release factors (RFs) which are RF-1, RF-2, RF-3. RFs cause the newly synthesized protein to be released from the ribosomal complex and dissociation of ribosomes from mRNA (i.e. cause dissolution of the complex)