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Chapter 17 From Gene to Protein

Chapter 17 From Gene to Protein. Question?. How does DNA control a cell? By controlling Protein Synthesis. Proteins are the link between genotype and phenotype. For tests:. Name(s) of experimenters Outline of the experiment Result of the experiment and its importance.

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Chapter 17 From Gene to Protein

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  1. Chapter 17From Gene to Protein

  2. Question? • How does DNA control a cell? • By controlling Protein Synthesis. • Proteins are the link between genotype and phenotype.

  3. For tests: • Name(s) of experimenters • Outline of the experiment • Result of the experiment and its importance

  4. 1909 - Archibald Garrod • Suggested genes control enzymes that catalyze chemical processes in cells. • Inherited Diseases - “inborn errors of metabolism” where a person can’t make an enzyme.

  5. Example • Alkaptonuria - where urine turns black after exposure to air. • Lacks - an enzyme to metabolize alkapton.

  6. George Beadle and Edward Tatum • Worked with Neurospora and proved the link between genes and enzymes. Neurospora Pink bread mold

  7. Experiment • Grew Neurospora on agar. • Varied the nutrients. • Looked for mutants that failed to grow on minimum agar.

  8. Results • Three classes of mutants for Arginine Synthesis. • Each mutant had a different block in the Arginine Synthesis pathway.

  9. Conclusion • Mutations were abnormal genes. • Each gene dictated the synthesis of one enzyme. • One Gene - One Enzyme Hypothesis.

  10. Current Hypothesis • One Gene - One Polypeptide Hypothesis (because of 4th degree structure).

  11. Central Dogma DNA Transcription RNA Translation Polypeptide

  12. Explanation • DNA - the Genetic code or genotype. • RNA - the message or instructions. • Polypeptide - the product for the phenotype.

  13. Genetic Code • Sequence of DNA bases that describe which Amino Acid to place in what order in a polypeptide. • The genetic code gives the primary protein structure.

  14. Code Basis If you use: • 1 base = 1 amino acid • 4 bases = 4 amino acids • 41 = 4 combinations, which are not enough for 20 AAs.

  15. If you use: • 2 bases = 1 amino acid • Ex – AT, TA, CA, GC • 42 = 16 amino acids • Still not enough combinations.

  16. If you use: • 3 bases = 1AA • Ex – CAT, AGC, TTT • 43 = 64 combinations • More than enough for 20 amino acids.

  17. Genetic Code • Is based on triplets of bases. • Has redundancy; some AA's have more than 1 code. • Proof - make artificial RNA and see what AAs are used in protein synthesis (early 1960’s).

  18. Codon • A 3-nucleotide “word” in the Genetic Code. • 64 possible codons known.

  19. DNA vs RNA DNARNA Sugar – deoxyribose ribose Bases – ATGC AUGC Backbones – 2 1 Size – very large small Use – genetic code varied

  20. Codon Dictionary • Start- AUG (Met) • Stop- UAA UAG UGA • 60 codons for the other 19 AAs.

  21. For Testing: • Be able to “read” a DNA or RNA message and give the AA sequence. • RNA Genetic Code Table will be provided.

  22. Code Redundancy • Third base in a codon shows "wobble”. • First two bases are the most important in reading the code and giving the correct AA. The third base often doesn’t matter.

  23. Code Evolution • The genetic code is nearly universal. • Ex: CCG = proline (all life) • Reason - The code must have evolved very early. Life on earth must share a common ancestor.

  24. Reading Frame and Frame Shift • The “reading” of the code is every three bases (Reading Frame) • Ex: the red cat ate the rat • Frame shift – improper groupings of the bases • Ex: thr edc ata tat her at • The “words” only make sense if “read” in this grouping of three.

  25. Transcription • Process of making RNA from a DNA template.

  26. Transcription Steps 1. RNA Polymerase Binding 2. Initiation 3. Elongation 4. Termination

  27. RNA Polymerase • Enzyme for building RNA from RNA nucleotides.

  28. Binding • Requires that the enzyme find the “proper” place on the DNA to attach and start transcription.

  29. Binding • Is a complicated process • Uses Promoter Regions on the DNA (upstream from the information for the protein) • Requires proteins called Transcription Factors.

  30. TATA Box • Short segment of T,A,T,A • Located 25 nucleotides upstream for the initiation site. • Recognition site for transcription factors to bind to the DNA.

  31. Transcription Factors • Proteins that bind to DNA before RNA Polymerase. • Recognizes TATA box, attaches, and “flags” the spot for RNA Polymerase.

  32. Transcription Initiation Complex • The complete assembly of transcription factors and RNA Polymerase bound to the promoter area of the DNA to be transcribed.

  33. Initiation • Actual unwinding of DNA to start RNA synthesis. • Requires Initiation Factors.

  34. Elongation • RNA Polymerase untwists DNA 1 turn at a time. • Exposes 10 DNA bases for pairing with RNA nucleotides.

  35. Elongation • Enzyme moves 5’ 3’. • Rate is about 60 nucleotides per second.

  36. Comment • Each gene can be read by sequential RNA Polymerases giving several copies of RNA. • Result - several copies of the protein can be made.

  37. Termination • DNA sequence that tells RNA Polymerase to stop. • Ex: AATAAA • RNA Polymerase detaches from DNA after closing the helix.

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