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BIO (Molec) - Vocab and Smart Pages.docx

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Biology
Course
BIOL 2021
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Winter

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BIOB11 Molecule STUDY NOTES Lecture 1 Vocab words: Diploid – contains both members of each pair of homologous chromosomes, as shown by most somatic cells. Diploid cells are produced from diploid parental cells during mitosis. (10.5) Haploid – Containing one member of each pair of homologous chromosomes. Haploid cells are produced during meiosis, as shown by sperm. (10.4) Allele – Alternate forms of the same gene. (10.1) Dominant/Recessive – Dominant allele masks over the recessive allele Homologous Chromosome – Paired chromosomes of diploid cells, each carrying one of the two copies of the genetic material carried by that chromosome Bivalent = Tetrad – The complex formed during meiosis by a pair of synapsed homologous chromosomes that includes four chromatids Recombination/Reciprocal genetic exchange – Reshuffling of the genes on chromosomes that occurs as a result of breakage and reunion of segments of homologous chromosomes Homozygous – The two alleles are identical to one another (Dominant and Dominant or Recessive and Recessive) Heterozygous – The two alleles are non-identical (Dominant and Recessive) Synaptonemal Complex (SC) – A ladder-like structure composed of three parallel bars with many cross fibres. The SC holds each pair of homologous chromosomes in the proper position to allow the continuation of genetic recombination between strands of DNA Non-disjunction: 1. Primary - Failure of segregation at first division (all haploids affected) 2. Secondary – Failure of segregation at second division (2 of 4 haploids) Trisomy 21 – Three chromosomes (21) … (Monosomy – one chromosome) Down`s syndrome – Presence of three chromosomes at chromosome 21. Exhibit mental impairment, alteration in body features, circulatory problems, increased infectious diseases, great risk of leukemia, and early Alzheimer`s disease Autosomes – A chromosome that is not a sex chromosome (allosome). Come in pairs, in humans 22 pairs of autosomes. Sex chromosomes come in pairs, are letters X or Y BIOB11 Molecule STUDY NOTES Smart pages: Figure 10.1 (Pg.380) – Overview of most important early discoveries on nature of the gene.  1865 – Discovery of discrete units of inheritance  1880 – Discovery of chromosomes  1903 – Discovery of homologous chromosomes  1909-1991 – Discovery of crossing over  1911-1913 – Discovery that genes could be mapped in order along length of chromosomes  1944-1952 – Discovery of DNA as genetic material  1953 – Discovery of DNA structure Figure 10.3 (Pg. 382) – Homologous chromosomes:  Sutton`s drawing of the homologous chromosomes of the male grasshopper that have associated during meiotic prophase to form bivalents. Eleven pairs of homologous chromosomes (a-k) and an unpaired X chromosomes Figure 10.7 (Pg.384) – Crossing over: Mechanism for reshuffling alleles between chromosomes a) Single crossover in Fruit Fly (Drosophila) heterozygote (BbWw) at chromosome 2 and the resulting gametes. If either of the cross-over gametes undergoes fertilization, the offspring will have a chromosome with a combination of alleles that was not present on a single chromosome in the cells of either parent. b) Bivalent (tetrad) formation during meiosis, showing three crossover intersections (chiasmata, indicated by red arrows) Figure 14.39 (Pg.590) – The stages of meiosis 1. Interphase 2. Prophase 1 3. Metaphase 1 4. Anaphase 1 5. Telophase 1 6. Prophase 2 7. Metaphase 2 8. Anaphase 2 9. Telophase 2 10. Haploid cells (4 cells) BIOB11 Molecule STUDY NOTES Figure 14.41 (Pg.592) – The stages of gametogenesis in vertebrates - Compare egg and sperm:  Both sexes, small population of primordial germ cells present in the embryo proliferates by mitosis to form a population of gonial cells (spermatogonia or oogonia) from which the gametes differentiate  Male – Sperm : o Meiosis (Spermatogenesis) occurs before differentiation o Each spermatocyte generally gives rise to four viable gametes  Female - Egg: o Both meiotic divisions occur after differentiation. o Meiosis (Oogenesis) o Each primary oocyte forms only one fertilizable egg  With 2-3 polar bodies Figure 14.44 (Pg.593) – The Synaptonemal Complex (SC – train tracks): a) Electron micrograph of human pachytene bivalent showing a pair of homologous chromosomes held in a tightly ordered parallel array. K (kinetochore). b) Schematic diagram of SC a. Recombination nodules (dense granules) i. Enzymatic machinery to complete genetic recombination ii. Though to begin at Prophase 1 b. Closely paired loops of DNA from two sister chromatids of each chromosome i. Loops maintained in paired configuration by cohesion c. Genetic recombination occurs between the DNA loops from non-sister chromatids Figure 14.46 (Pg.595) – Separation of homologous chromosomes during meiosis 1 and 2: a) Pair of homologous chromosomes at metaphase 1. The chromatids are held together along both arms and centromeres by cohesion. The pair of homologues are maintained bivalent by chiasmata. The kinetochores of sister chromatids are on one side of the chromosome, facing the same pole b) Anaphase 1 – the cohesion holding the arms of the chromatids is cleaved, allowing the homologues to separate. Cohesion remains at centromere, holding chromatids together. c) Metaphase 2 – chromatids held together at centromere, with microtubules from opposite poles attached to the two kinetochores. Kinetochores of the BIOB11 Molecule STUDY NOTES sister chromatids are now on opposite sides of the chromosomes, facing opposite poles. d) Anaphase 2 – Cohesion holding the chromatids together is cleaved, allowing chromosomes to move to opposite poles HP Figure 1 (Pg.596) – Meiotic nondisjunction:  Nondisjunction occurs when chromosomes fail to separate from each other during meiosis.  Primary nondisjunction – failure to separate occurs during the first meiotic division, all of the haploid cells have abnormal number of chromosomes  Secondary nondisjunction – failure to separate occurs during second meiotic division, only two of the four haploid cells are affected. BIOB11 Molecule STUDY NOTES Lecture 2 Vocab words: Absorption spectrum – A plot of the intensity of light absorbed relative to its wavelength Denaturation – Separation of the DNA double helix into its two component strands Nucleic Acid Hybridization – Complementary strands of nucleic acids from different sources can be mixed to form double-stranded (hybrid) molecules. Renaturation/Reannealing – Re-association of complementary single strands of a DNA double helix that had been previously denatured In situ hybridization – Technique to localize a particular DNA or RNA sequence in a cell or on a culture plate or electrophoretic gel Centromere – Marked indentation on a mitotic chromosome that serves as the site of kinetochore formation Telomere – An unusual stretch of repeated DNA sequences, which forms a “cap” at each end of a chromosome Mutation – A spontaneous change in a gene that alters it in a permanent fashion so that it causes heritable change Unequal Crossing Over – A pair of homologous chromosomes comes together during meiosis in such a way that they are not perfectly aligned and because of misalignment, genetic exchange between the homologues causes one chromosome to acquire an extra segment of DNA (a duplication) and the other chromosome to lose a DNA segment ( a deletion) Duplication – A chromosomal aberration that occurs when one chromosome acquires an extra segment of DNA Deletion – A chromosomal aberration that occurs when a portion of a chromosome is missing, lost a DNA segment Transposition – Movement of DNA segments from one place on a chromosome to an entirely different site, often affecting gene. BIOB11 Molecule STUDY NOTES Smart Pages: Figure 10.10 (Pg.388-389) – The double helix: a) Schematic representation of the DNA double helix a. Two chains in opposite directions (Antiparallel) – 5’ to 3’ and 3’ to 5’ b. –sugar-phosphate-sugar-phosphate backbone, phosphate gives negative charge c. Bases occupy perpendicular planes, to the long axis of molecule, stacked one on top of another. Hydrophobic interactions and van der Waals forces for stability of DNA. d. Strands held by hydrogen bonds between bases of one strand and other strand e. Distance from phosphorus atom of backbone to center is 1 nm ( width of double helix is 2nm) f. Pyrimidine always paired with purine, produces 2nm wide length g. Only possible pairs of adenine-thymine and cytosine-guanine h. Space between adjacent turns of helix forms two grooves, Major and Minor i. The two chains of double helix are complementary, 5’-AGC-3’ & 3’-TCG-5’ b) Space-filling model of the B form of DNA c) The Watson-Crick base pairs a. Pyrimidines – single ring (thymine (T) and cytosine (C)) b. Purines – double ring (adenine (A) and guanine (G)) c. Adenine-thymine – Have 2 hydrogen bonds d. Cytosine-guanine – Have 3 hydrogen bonds d) Electron micrograph of DNA a. DNA released from head of T2 bacteriophage i. Linear DNA molecule is 68 µm in length Figure 10.15 (Pg.393) – Thermal denaturation of DNA:  Thermal denaturation curve for bacteriophage T6 DNA in 0.3 M sodium citrate  Melting of DNA (strand separation) occurs over a narrow range of temperature  The temperature at which the shift in absorbance is half completed is termed the melting temperature (T m o Higher the Guanine-Cytosine content, higher the T .m Figure 10.17 (Pg.394) – An idealized plot showing the kinetics of renaturation of eukaryotic DNA  When single-stranded DNA is allowed to reanneal, three classes of fragments are distinguished by their frequency of repetition within the genome: 1. Highly repeated DNA fraction 2. Moderately repeated DNA fraction 3. Non-repeated (single-copy) DNA fraction BIOB11 Molecule STUDY NOTES Figure 10.19 (Pg.398) – Fluorescence in situ hybridization and the localization of satellite DNA: a) Steps to carry out fluorescence in situ hybridization a. Double-stranded DNA treated with hot salt solution to denature DNA b. Single-stranded DNA incubated with biotinylated DNA probe, making a double- stranded DNA hybrid (Hybridization). i. Washed so that un-hybridized DNA is removed c. Incubated with fluorescently labeled avidin to reveal location of bound, labeled DNA probe. i. Avidin is a protein that binds to biotin with very high affinity d. Chromosomes counterstained with propidium iodide so they appear red b) Localization of alpha-satellite DNA at centromere of human chromosome a. Location of bound, biotin-labeled, satellite DNA is revealed by yellow fluorescence. i. Yellow fluorescence stands out against background of red b. Fluorescence appears only at the site where each chromosome is constricted, marks the location of the centromere Figure 10.22 (Pg.401) – Unequal crossing over between duplicated genes: a) Initial state shown had two related genes (1 and 2) a. In a diploid individual, gene 1 on one homologue can align with gene 2 on the other homologue during meiosis. b. If crossover occurs during this misalignment, half the gametes will be missing gene 2 and half will have an extra gene 2. b) As unequal crossing over continues during meiotic divisions in subsequent generations, a cycle repeated array of DNA sequences will gradually evolve Figure 10.23 (Pg.401) A pathway for the evolution of globin genes:  Seven evolutionary steps 1) Exon Fusion: a. Modern globin polypeptide arose from the ancestral form as the result of the fusion of two of the globin exons 800 million years ago. 2) A number of primitive fish are known that have only one globin gene 3) Duplication: a. The first duplication of globin gene 4) Divergence by mutation: a. Approximately 500 million years ago, two copies diverged by mutation to form two distinct globin types, and alpha and beta type, located on a single chromosomes. BIOB11 Molecule STUDY NOTES 5) Separation of genes: a. In subsequent steps, the alpha and beta forms are thought to have become separated from one another by a process of rearrangement that moved them to separate chromosomes 6) Duplications and divergence: a. Each gene underwent subsequent duplications and divergence 7) Generating the arrangement of globin genes that exists today in humans BIOB11 Molecule STUDY NOTES Lecture 3 Vocab words: Transposition – Movement of DNA segments from one place on a chromosome to an entirely different site, often affecting gene expression Transposon – DNA segments capable of moving from one place in the genome to another Retrotransposon – Transposable elements that require reverse transcriptase for their movements within the genome Reverse transcriptase – An RNA-dependent DNA polymerase. An enzyme that uses RNA as a template to synthesize a complementary strand of DNA. [An enzyme that is found in RNA- containing viruses and used in the laboratory to synthesize cDNAs] Centimorgan - Genetic map – Assignment of genetic markers to relative positions on a chromosome based on crossover frequency Physical map – Restriction map: A type of physical map of the chromosome based on the identification and ordering of sets of fragments generated by restriction enzymes Polymorphism – Sites in the genome that very with relatively high frequency among different individuals in a species population RFLP – Restriction fragment length polymorphism: SNP – Single Nucleotide Polymorphisms: sites in the genome where alternate bases are found with high frequency in the population. SNPs are excellent genetic markers for genome mapping studies Haplotype – A block of the genome that tends to be inherited intact from generation to generation. Haplotypes are generally defined by the presence of a consistent combination of a single nucleotide polymorphisms (SNPs) BIOB11 Molecule STUDY NOTES Smart Pages: Figure 10.26 (Pg.403) – Schematic pathways in the movement of transposable elements: a) DNA transposons move by a cut-and-paste pathway, whose mechanism is depicted in (Figure10.25). a. Cut and paste mechanism utilized i. DNA transposons are excised from DNA at the donor site and inserted at a distant target sit. b. Approx. 3% of human genome consists of DNA transposons, none of which are capable of transposition (all are relics left in genome as result of ancestry) b) Retrotransposons move by a copy-and-paste pathway. a. Steps involved in retroptransposition take place both in nucleus and cytoplasm and require numerous proteins, including those of the host b. Copy and paste mechanism utilized 1. Involves an RNA intermediate 2. The DNA of the transposable element is transcribed by an enzyme called ‘reverse transcriptase’ producing a complementary DNA 3. The DNA copy is made double-stranded and integrated into a target DNA site a. Many retrotransposons itself contain sequence for reverse transcriptase, such as AIDS c. More than 40 percent of human genome consists of retrotransposons, only few (40-100) are thought to be capable of transposition Figure 10.27 (Pg.405) – Genome comparisons  The number of protein coding genes (blue bars) - estimations o Varies from 6,000 – 37,000 protein-coding genes  Biologists estimate around 20,000 protein-coding genes in humans  The amount of DNA in a genome: base pairs (red bars) o Wide variety of base pairs, estimated values of 90 billion base pairs (salamander) Figure 10.28 (Pg.407) – Small segments of DNA conserved between humans and related species  Top line represents the nucleotide sequence of a segment of human DNA  Each line below represents nucleotide sequence from another primate  Block on each line represent segments whose nucleotide sequence varies from human  Bottom line shows those parts of DNA segment that are highly conserved in all species shown o Sequence that does not vary in any of the eight species represented BIOB11 Molecule STUDY NOTES Figure 10.30 (Pg.409) – Structural variants a) Schematic representation of major types of genomic polymorphisms a. Normal gene compared to genes with variants - ABCDEFGHI b. Gene duplication – A copy of a gene is made ( B gene duplicated) - ABBCDEFGH c. Deletion – Removal of a gene (C gene is deleted) - ABDEFGHIJ d. Inversion – Part of the gene is reversed (CDE becomes EDC) - ABEDCFGHI e. Insertion – Insert new genes (XYZ inserted) - ABXYZCDEF b) Human chromosome #9 – Normal copy and large inversion copy (arrows) SNP Figure (Blackboard) – Chromosome region that has polymorphism  Some individuals might have A at one point and other may have C (or another base) at the same point. o Sum of all SNPs that make everyone unique (millions of SNPs)  Hypothetical SNP map of patients response to drug test o A Positive response – Second base A and fourth base T o Negative response – Second base T and fourth base G  Some people SNPs mean their protein does not bind or transport drug, or metabolize properly Figure 11.1 (Pg.420) – Beadle-Tatum experiment: isolation of genetic mutants in Neurospora  Several steps taken in the experiment to come to conclusion “One gene-One enzyme” 1. Spores (Neurospora) were irradiated to induce mutations 2. Mutated spores allowed to grow into colonies in tubes containing supplements 3. Individual spores produced by the colonies were then tested for ability to grow on minimal medium  Those that failed to grow were mutants and the task was to find mutant gene 4. Sample of mutant cells were found to grow in minimal medium supplemented with vitamins but not in medium supplemented with amino acids  Observation led to indication of enzyme deficiency leading to formation of vitamin 5. Growth of these same cells in minimal medium supplemented with one or another of the vitamins indicates the deficiency resides in gene involved in formation of pantothenic acid (part of Co-enzyme A) BIOB11 Molecule STUDY NOTES Figure 11.2 (Pg.421) – Overview of the flow of information in Eukaryotic cell  DNA of chromosomes located in nucleus contains entire store of genetic information 1. Selected sites on DNA are transcribe into pre-mRNAs 2. Which are processed into messenger RNAs 3. Messenger RNAs are transported out of the nucleus into cytoplasm 4. Translated into polypeptides by ribosomes that move along mRNA 5. Polypeptide folds to assume native conformation BIOB11 Molecule STUDY NOTES Lecture 4 Vocab words: Core enzyme – A single type of RNA polymerase composed of five subunits tightly associated Sigma factor – Purified accessory polypeptide, attaches to -35/-10 specific sequence Holoenzyme – Complete enzyme: Core enzyme + Sigma factor TATA box – Second conserved sequence found at 10 bases upstream (-10) from initiation site, also now as Pribnow box, responsible for identifying the precise nucleotide at which transcription begins Consensus sequence – The most common version of a conserved sequence. The TTGACA sequence of a bacterial promoter (-35 element) and TATAAT (-10 element) are examples α-amanitin – Used to determine which type of RNA polymerase are present. RNA polymerase 1 is insensitive, RNA polymerase 2 is highly sensitive (inhibited at 1µg/ml) and RNA polymerase 3 is moderately sensitive (inhibited at 10µg/ml) Ribosome – Consist of several molecules of rRNA together with dozens of ribosomal proteins. More than 80 percent of the RNA in most cells, rRNA. Nucleolus – clusters of rDNA (DNA sequence encoding rRNA), gathered together as part of one or more irregularly shaped nuclear structures that function in producing ribosomes. Bulk of nucleolus composed of nascent ribosomal subunits that give the granular appearance. Nucleolar organizer region (NOR) – rRNA gene clusters .Nucleolar organizers are sites where ribosomes begin to be synthesized. Nucleolar organizers are sites where rRNA genes on different chromosomes are found in close association. Pulse-Chase experiment – Two parts of an experiment to tag molecules and determine their fates over a period of time: 1. Pulse: add radioactive precursor to cells for short period. Get incorporation of label into macromolecule pool. 2. Chase: wash cells to remove radioactive precursor, then examine radioactive macromolecules after an incubation period. BIOB11 Molecule STUDY NOTES Smart Pages: Figure 11.4 (Pg.424) – Chain Elongation during transcription a) Elongation of a newly synthesized RNA molecule during transcription a. Polymerase covers 35 base pairs of DNA b. Transcription bubble composed of single-stranded (melted) DNA contains 15 base pairs c. DNA-RNA hybrid has 9 base pairs d. Enzyme generates overwound (positively supercoiled) DNA ahead of itself e. Under-wound (negatively supercoiled) DNA behind itself b) RNA polymerase in transcription elongation a. Downstream DNA lies inside polymerase, clamped by two subunits of enzymes b. DNA makes sharp turn in region of active site, so upstream DNA extends upward c. Nascent RNA exits from enzyme’s active site through separate channel c) Chain elongation results following attack by 3’ OH of nucleotide at end of growing strand on the 5’ alpha-phosphate of incoming nucleoside triphosphate a. Pyrophosphate released is cleaved, which drives reaction in polymerization b. Geometry of base-paring b/w nucleotide of template and incoming nucleotide determines four possible nucleoside triphosphates is incorporated into growing RNA chain d) Electron-micrograph of several RNA polymerase molecules bound to DNA template Figure 11.6 (Pg.425) – Schematic representation of initiation of transcription in bacteria a) Absence of sigma factor, the core enzyme does not interact with DNA at specific initiation sites b) – d) a. When core enzyme is associated with sigma factor it becomes the complete enzyme (haloenzyme) b. It is able to recognize and bind to promoter regions of DNA c. Separate the strands of the DNA double helix and initiate transcription at proper start sites d. Sigma factor dissociates from the core enzyme, which is capable of transcription elongation i. Some cases sigma factor may stay with polymerase Figure 11.7 (Pg.426) – The basic elements of promoter region in DNA of E.coli BIOB11 Molecule STUDY NOTES  Ket regulatory sequences required for initiation of transcription are found in regions located at -35 and -35 base pairs from site at which transcription is initiated  Initiation site marks boundary between the + and – sides of the gene Figure 11.9 (Pg.428) – Macromolecular composition of mammalian ribosomes  Schematic drawing shows components present in each subunits of mammalian ribosomes Figure 11.12 (Pg.430) – The rRNA transcription unit  Top drawing is portion of DNA from a nucleolus as it’s transcribed into rRNA  Bottom drawing is transcription units that codes for rRNA in Xenopus and mouse  Parts of DNA that encode mature rRNA products are blue  Regions of transcribed spacer, portions of DNA that are transcribed, but corresponding RNAs are degraded during processing are yellw  Non-transcribed spacer, which lies between transcription unites, contains promoter region at 5’ side of gene Figure 11.13 (Pg.431) – Kinetic analysis of rRNA synthesis and processing 14  Mammalian cells incubated for 10 minutes in [ C]methionine and then chased in unlabeled medium for various times (10 min, 40 min, 80 min and 150 min)  After chase, cells were washed free of isotope and homogenized and nucleolar and cytoplasmic fractions were prepared  RNA extracted from each fraction and analyzed by sucrose density-gradient sedimentation o Sedimentation: this technique separates RNAs according to size (larger the size, the closer to the bottom of the tube)  Continuous line represents the UV absorbance of each cellular fraction, which provides a measure of the amount of RNA of each size class (Does not change with time)  Dotted line shows radioactivity at vrious times during the chase  Graphs of nucleolar RNA show synthesis of 45S rRNA precursor and subsequent conversion to 32S molecule, which is precursor to 28S and 5.8S rRNAs  Other major product of 45S leaves nucleus rapidly, therefore does not appear prominently in nucleolar RNA  18S rRNA appears in cytoplasm before larger 28S, which correlates with rapid exodus of former from nucleolus Figure 11.14 (Pg.432) – Proposed scheme for processing of mammalian ribosomal RNA BIOB11 Molecule STUDY NOTES  Primary transcript for rRNAs is 45S molecule of about 13,000 bases  Principal cleavage events during processing of this pre-rRNA are indicated by circled number  Cleavage of primary transcript at sites 1 and 5 removes 5’ and 3’ external transcribed sequences and produces a 41S intermediate  Second cleavage occurs at either site 2 or 3 depending on cell type  Site 3 cleavage generates 32S intermediate seen in curves of previous figure  During final processing steps, the 28S and 5.8S sections are separated from one another and the ends of various intermediates are trimmed down to final mature size BIOB11 Molecule STUDY NOTES Lecture 5 Vocab words: HnRNA – Heterogeneous Nuclear RNAs: A large group of RNA molecules that share the following properties: 1) Large molecular weights (up to 80S, or 50,000 nucleotides) 2) They represent many different nucleotide sequences 3) They are found only in nucleus. Includes pre-mRNAs Pre-initiation complex – The assembled association of general transcription factors and RNA polymerase, required before transcription of gene can be initiated Protein kinase – An enzyme that transfers phosphate groups to other proteins, often having the effect of regulating the activity of other proteins Phosphorylation – mRNA cap – 5’ cap containing, 5’methylguanosine cap Poly A tail A string of adenosine residues at the 3’ end of an mRNA added post-transcriptionally Un-translated region (UTRs) – Noncoding segments contained at both 5’ and 4’ ends of mRNAs Complementary DNA (cDNA) – DNA made in vitro by reverse transcriptase using mRNA template. Used because restriction enzymes don’t cleave RNAs and is complimentary sequence to mRNA Reverse transcriptase – An RNA-dependent DNA polymerase. An enzyme that uses RNA as a template to synthesize a complementary strand of DNA. [An enzyme that is found in RNA- containing viruses and used in the laboratory to synthesize cDNAs.] Intron – Those parts of a split gene that correspond to the intervening sequences Exon – Those parts of a split gene that contribute to a mature RNA product Splice junction – Breaks at 5’ and 3’ ends of introns, which are removed BIOB11 Molecule STUDY NOTES Smart Pages: Figure 11.17 (Pg.434) – Formation of hnRNAs and its conversion into smaller mRNAs a) Curves showing sedimentation pattern of total RNA extracted from duck blood cells after exposure to [ P] for 30 minutes – pulse labelled a. Larger RNA, farther it travels during centrifugation and closer to bottom of tube b. Absorbance indicates total amount of RNA in different regions of tube c. Most new synthesized RNA are very large, larger than 18S and 28S rRNAs i. The large RNAs are hnRNAs ii. Much more rRNA than hnRNAS 1. mRNAs half-life in minutes to days 2. rRNAs half-life days to weeks to months b) Absorbance and radioactivity of RNA extracted from cells a. pulse-labelled for 30 minutes b. Chased for 3 hours with actinomycin D (prevents synthesis of more RNA) c. Large hnRNAs processed into smaller RNA products Figure 11.18 (Pg.435) – Initiation of transcription from eukaryotic polymerase II promoter a) Nucleotide sequence of region just upstream from site where transcription is initiated a. TATA box (-10 bases from initiation) b. Second conserved core promoter called initiator (Inr) includes site where transcription is initiated i. Most eukaryotic promoters lack recognizable TATA box ii. Numerous promoter elements (ex. DPE) lie down-stream of transcription start site c. Different genes contain different combinations of promoter elements, only subset required to nucleate (pre-initiation complex) PIC assembly b) Highly schematic model of steps in assembly of pre-initiation complex for RNA polymerase II a. Polymerase denoted RNAPII bind with GTFs to form PIC i. General transcription factors (GTFs) – required for accurate initiation of transcription of diverse array of genes in wide variety of different organisms ii. PIC – pre-initiation complex assembly lie to the 5’ side of each transcription unit BIOB11 Molecule STUDY NOTES b. TFIID includes the TBP subunit, binds to TATA box and other subunits called TBP- associated factors (TAFs) i. TFIID – transcription factor for polymerase II, fraction D ii. TBP – TATA-binding protein, recognized TATA box promoters 1. Bacterial cells identify TATA box on their own c. TFIIB provides a binding site for RNA polymerase d. TFIIF contains subunit homologous to bacterial sigma factor and binds to entering polymerase e. TFIIH contains 10 subunits, 3 possess enzymatic activities i. Attaches to complex and transforms polymerase into active transcribing machine ii. Phosphorylate RNA polymerase, and unwinding enzymes Figure 11.20 (Pg.437) – Initiation of transcription by RNA polymerase II is associated with phosphorylation of C-terminal domain (CTD)  Initiation of transcription associated with TFIIH-catalyzed phosphorylation of serine residues at 5 position of each heptad repeat of CTD o CTD – carboxyl-terminal domain, 52 repeats of heptapeptide
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