Bio 230 Exam 1

  1. Describe the relationship between the LECA and FECA in reference to the nucleus, mitochondria and chloroplasts.
    • So endomembrane system came first and then FECA, which produced the nucleus, and then the cell developed phagocytosis, where the mitochondria was engulfed, and then LECA came last, which produced the chloroplast.
    • Before LECA, there was no or limited internal compartments. After LECA, there was the formation of the nucleus and internal compartments that were MORE SOPHISTICATED THAN PROKARYOTES 
    • Adaptive radiation of LECA led to diversity of cells and AR took place due to increased sophistication and diversity
  2. Adaptive radiation
    organisms diversify rapidly from an ancestral species into a multitude of new forms, particularly when a change in the environment makes new resources available

    • AR led to diversity of cells 
    • AR took place bc due to increased sophistication and diversity
  3. FECA
    • First eukaryotic common ancestor (FECA) - nucleus
    • Major selective advantage
    • Creates an environment favorable for the reactions that take place in the nucleus (transcription, replication)
    • Genes that code for the proteins in the cell continue to mutate
  4. Explain how adaptive radiation played a role in the evolution of the endomembrane system via protein function (from Worksheet 2)
    • Organelles of the endomembrane system evolved through alterations of the plasma membrane
    • Membranes in a cell are dynamic and can move in multiple pathways
    • Transport of membranes resulted in formation of organelles
    • Alterations required specific proteins
    • Adaptive radiation promoted specialized internalization of membranes
  5. Endomembrane system
    • membrane bound
    • nuclear membrane, golgi, er, plasma membrane, vesicles
  6. Endosymbiotic
    • membrane bound
    • mitochondria, plastids
  7. Protein-based
    • non membrane
    • cytoskeleton, proteasome, ribosome
  8. Magnification
    makes an image larger
  9. Resolution
    • ability to tell two points apart
    • (shorter WL = more energy = better solution)
  10. Reflective light
    • light bounces off specimen to eye
    • Can view living specimen
    • Ex: dissecting scope
  11. Scanning electron
    • Electrons bounce off specimen to detector
    • Specimen coated in metal
    • Image produced by a computer
    • Used for high resolution or magnification
  12. Transmission light
    • Light travels thru specimen to eye
    • Specimen stained
    • Used for routine visualization
  13. Transmission electron
    • Electrons travel thru specimen to detector
    • Can see inside of specimen
    • Very thin sections required
    • Metals are used
    • Used for high resolution or magnification
  14. Emitted light
    • Uses light originated from specimen
    • Fluorescent
    • visualizing specific cellular structures, HAS MOST ENERGY BC ENERGY IS LOST THRU HEATING A SPECIMEN
  15. Describe why DNA is better as an information storage molecule compared to RNA.
    DNA is more stable bc it is DS, RNA is unstable
  16. Explain the benefits of a nucleus.
    Without a nucleus, transcription and translation would occur at the same time because there is limited regulation w/o a nucleus. A nucleus, however, allows transcription and translation to occur in different regions because there is increased regulation and processing.
  17. Monomeric form of DNA
    • Monomeric form: nucleotides
    • Contains base, sugar, phosphate
    • Bases are purines or pyrimidines
    • Purines: adenine, guanine, double ring
    • Pyrimidines: cytosine, uracil, thymine, single ring
    • Sugar is deoxyribose
    • DNA nucleotides are monophosphate
  18. Polymeric form of DNA
    DNA itself
  19. Describe how monomers form the DNA polymers including the role of base pairing.
    • Monomers are nucleotides and nucleotides are connected by 3’-5’ phosphodiester bonds
    • Phosphate attached to the 5’ carbon of the incoming nucleotide completes the phosphodiester bond with the oxygen on the 3’ carbon of the ribose. Then the next nucleotide will come in and do the same thing. In this way the growing polymer forms a sugar/phosphate backbone which grows in the 3’ to 5’ direction.
    • Template strand read: 3’ to 5’
    • mRNA synthesized : 5 to 3
  20. Describe the role of hydrogen bonding between DNA strand and how it affects DNA strand separation
    • Hold strands of DNA together
    • A/T 2 bonds, G/C 3 bonds
    • G/C more stable, more bonds = more energy required to break
  21. Evaluate how changes to the structure of DNA may impact function.
    The most important feature of the structure of DNA is that it forms an "anti-parallel" structure. This double helix allows the cell to make two new, identical copies, by reading and copying each strand separately. This is how the information is permanently retained.
  22. Describe the grooves of DNA and how proteins interact with the groove.
    Helix shape results in a major and minor groove

    • Proteins bind to the major groove bc
    • 1.) more room for the protein to access the sides of the bases
    • 2.) pattern of H-bond donors and acceptors on the bases are more distinct on the major groove of DNA than on the minor groove
  23. Exonuclease
    Cuts DNA at the end of a DNA polymer
  24. Endonuclease
    cleave the phosphodiester bond
  25. Ligase
    Joins 3’-OH with 5’ PO4
  26. Topoisomerases
    Cut one strand to release rotational tension
  27. Helicase
    Use ATP to break hydrogen bonds between bases
  28. Methylase
    Adds new functional group to DNA
  29. Explain the overall structure of a nucleosome.
    • Nucleosome: basic unit of chromatin in eukaryotes
    • Packing is dynamic bc it changes with the needs of the cell
  30. Euchromatin
    • a lightly packed form of chromatin, occurs during interphase
    • exists for transcription
  31. Heterochromatin
    • a tightly packed form of chromatin, occurs during mitosis
    • inactive for transcription
  32. Describe the general mechanism of epigenetics from Sci Lit reading.
    • Methylation:
    • Histone post-translational modifications:
    • Non-coding RNAs
  33. Methylation
    • Methyl groups are added to DNA
    • Blocks access to promoters
    • Inhibits gene transcription by altering chromatin structure
  34. Histone post-translational modifications:
    Affect gene expression by altering chromatin structure
  35. Genome
    complete genetic material of an organism
  36. Prokaryotes
    • Circular chromosomes
    • Contains double stranded DNA
    • Contain plasmids
    • Contains limited, non-coding dna which results in less regulation
  37. Eukaryotes
    • DNA is contained in nucleus
    • DNA is linear, double stranded
    • Large amount of non-coding DNA
  38. Viruses/Bacteriophage
    • Chromosome can be linear or circular, SS or DS, RNA or DNA
    • Limited noncoding DNA
  39. Prokaryotic regulatory
    • Promoters/operators
    • Terminators
    • UTRs
    • Some archae have introns
  40. Prokaryotic structural
    Protein coding sequence
  41. Promoter
    Transcription factors help RNA polymerase bind to them to initiate transcription.
  42. Operator-
    stretch of DNA that a repressor binds to. Typically, between the promoter and the coding region.
  43. UTRs-
    Regions that get transcribed but not translated. Contain Ribosome Binding Sites and transcription termination sites.
  44. Introns
    stretches of DNA that get transcribed but then excised before translation
  45. Eukaryotic regulatory
    • Enhancers/silencers
    • Promoters
    • Introns
    • UTRs
  46. Eukaryotic structural
    Exons (coding region)
  47. Explain connection between multicellularity and genome size.
    Larger genome size = more multicellularity this is bc there is more regulatory DNA (non-coding)
  48. differential gene expression
    Certain genes are on and certain genes are off in different cell types
  49. Basal transcription factors
    • bind to DNA to initiate transcription process
    • have to be at start of coding region
  50. Co-activators
    Bridge between activators and basal transcription factors
  51. Repressor
    bind to silencer and slow transcription
  52. Activators
    Speed the rate of transcr
  53. Explain the relationship of the proteins to DNA (enhances, silencer, TATA box, ect.) (Worksheet 5). DNA
    promoter, enhancer, silencer, intron, UTR

    All regulatory regions
  54. Explain the relationship of the proteins to DNA (enhances, silencer, TATA box, ect.) (Worksheet 5). Proteins
    basal TF, activator, repressor
  55. Analyze how the PIC forms
    BTF found at promoter region, recruit RNA polymerase at the beginning of the code region
  56. what results would occur if any protein or DNA sequence is altered
    • If a mutation occurred in TFIH transcription would not occur - helps position helicase for transcription to occur
    • If a mutation occurred in CRSP/MED, transcription would be able to occur but would be slower and less stable
  57. Explain how the transcription rate can be altered through the PIC.
    • If activators are present - increase mRNA production
    • If repressors are present - decrease mRNA production
  58. Identify the DNA ends (5' and 3') involved in transcription and how those ends are used to describe the direction of transcription (Worksheet 6).
    Polymerase adds nucleotides to the 3' end of the previous nucleotide and then mRNA is transcribed in the 5'-3' direction but must read the template in the 3'-5' direction
  59. Sense
    • Coding
    • Non template
  60. Anti sense
    • Non coding
    • Template
Author
user011
ID
350781
Card Set
Bio 230 Exam 1
Description
Bio 230
Updated