Objectives for Exam 3 Chromatin and DNA Sequence Organization – Chapter 11•Be able to define “chromatin” and describe its structure (nucleosomes, histone proteins plus associated DNA). •chromatin- complex of DNA, RNA, histones, and nonhistone proteins that make up uncoiled chromosomes, characteristic of the eukaryotic interphase nucleus•as the cell cycle progresses, most cells reenter mitosis, where the chromatin coils into visible chromosomes once again. •the associated proteins are divided in basic, positively charged histones and less positively charged histones•histones contain large amounts of the positively charged AAʼs lysine and arginine, making it possible for them to bind to the phosphate groups of the nucleotides (which are negatively charged)•nucleosome- (beads on a string) a nuclear complex consisting of 4 pairs of histone molecules (an octamer of H2a, H2b, H3, and H4) wrapped by 2 turns of a DNA molecule. The major structure associated with the organization of chromatin in the nucleus. •each repeating nucleosome is associated with about 200bp DNA•Know the difference between euchromatin and heterochromatin in terms of its structure, composition (cytosine methylation of DNA, histone tail modification) and transcriptional activity. •euchromatin- uncoiled; true chromatin, or chromosomal regions that are relatively uncoiled during the interphase portion of the cell cycle •capable of transcription; uncondensed•heterochromatin- condensed; the heavily staining, late-replicating regions of chromosomes that are prematurely condensed in interphase. Thought to be devoid of structural genes.• condensed DNA; transcription is repressed•Heterochromatic areas are genetically inactive because they either lack genes or contain genes that are repressed.•heterochromatin also replicates later than euchromatin does.•constitutive heterchromatin- heterochromatic DNA in all cells (centromeres and telomeres)•facultative heterochromatin- heterochromatic in some cells (X-inactivation)•Modification of histone tails regulates chromatin structure and function•histone acetylation is associated with euchromatin and transcription•cytosine is methylated in many inactive genes•ex: X inactivation involves formation of facultative heterochromatin•inactive X has deacetylated histones and methylated cytosines•Know the difference between constitutive and facultative heterochomatin and be able to name two places on chromosomes where constitutive heterochromatin is found. •see above•Be able to interpret a “Cot curve” for the renaturation of eukaryotic DNA and know which part of the curve is due to the renaturation of repeated DNA sequences in the genome. •the renaturation kinetics of DNA showed the existence of repeated DNA sequences in eukaryotes•the curves with multiple slopes have repeated DNA sequences.•remember that the x axis is C0t and the y axis is the fraction remaining single stranded•Differences between simple sequence DNAs (satellite DNA sequences) and interspersed repeats and the percentage of the human genome that is composed each of these classes of repeated DNA sequences. •satellite DNAs or Simple Sequences- not transcribed, long tandem arrays of long repeats, short sequences (5-100bp), often in large #ʼs of copies, localized sites on chromosomes; basically short sequences repeated many times in the genome•5-10% of the genome•Interspersed Repeats- longer sequences (300-5000), variable # of copies, not in tandem array, repeats are not identical, dispersed over the chromosome, sometimes transcribed•about 40% of the human genome•40-60% of the genome of eukaryotes is repetitive sequences•most repetitive DNA has no known function (except telomeres and centromeres)•Know the size of the human genome and the approximate percentage of the genome that codes for proteins •genes that code for proteins only account for 5-10% of the genome•we have about 25,000 genes and 3 x 109 bpGenetic Code, Transcription - Chapter 12 •Transcription- process by which base sequence in DNA is converted into RNA. The enzyme responsible is RNA polymerase •the linear sequence of deoxyribosenucleotides making up DNA ultimately dictates the components constituting proteins, the end product of most genes•the question is how much information stored as a nucleic acid is decoded into a protein?• in the first step of of gene expression, info on one of the strands (template strand) is transferred into an RNA complement through transcription•once synthesized, this molecule acts as a ʻmessengerʼ molecule bearing the coded information--hence its name messenger RNA (mRNA)•the mRNAs then associate with ribosomes where decoding into proteins takes place•Understand how the genetic code was deciphered using synthetic mRNAs and in vitro translation. •Nirednberg and Matthaei used in vitro (cell-free) protein synthesizing system and an enzyme called polynucleotide phosphorylase to decipher the code•the in vitro mixture contained all the necessary things for protein synthesis (ribosomes, tRNAs, AAʼs, etc.)•one or more of the AAʼs was radioactively labeled in order to follow the synthesis•finally an mRNA is added to serve as a template•the polynucleotide phosphorylase enzyme catalyzed the reaction•each addition of a ribonucleotide is random, based on the concentrations added to the mixture•the probability of insertion is proportional to its availability relative to the other ribonucleotides•Using Homopolymers•RNA homopolymers- UUU, AAA, CCC; only one type of ribonucleotide•they had added a mRNA molecule only like UUUUUU...., AAAAAA......, etc. to get these and then would see what the RNA homopolymers coded for•Be able to determine possible codon assignments in experiments involving repeating copolymers as the mRNA. •pg. 245-246: this was another technique use to decipher the genetic code•di-, tri-, and tetra-nucleotides that are replicated many times and then enzymatically joined together to form long polynucleotides•depending upon the point of initiation, you can get several different repeating triplets from the copolymers•Know what base transition and base transversions are. Know that adenine and guanine are purines and that cytosine of thymine are pyrimidines. •base transitions- mutations where a purine is switched with another purine (A with G) or a pyrimidine is switched with another pyrimidine (U
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