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Genetics Midterm 2 Notes.docx

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Department
Biology
Course
BIOL 2030
Professor
Dr.Bentzen
Semester
Winter

Description
LECTURE 1 – MOLECULAR BASIS OF HEREDITY For a molecule to serve as the genetic material, it must be able to: 1. Replicate accurately (progeny has same info as parents) 2. Stable and store large amounts of info (structure, function, development, reproduction) 3.Allow for phenotypic variation and change Evidence for DNAas genetic material emerged in the early 1900s from bacteria & viruses (rather than proteins). Johann Meischer (1869): Discovered nuclein (DNA), couldn’t be protein because it lacked sulfur but had phosphorous and nitrogen. Frederick Griffith (1928): Demonstrated cells could be transformed (uptake genetic material from external source, phenotype change) Used two types of pneumonia causing bacteria, Rough (R, non-virulent) and Smooth (S, virulent: bacteriophage that reproduces only through lytic cycle and kills its cell). Mice injected w/ pure R or heat killed S lived, but mice injected w/ pure S and pure R w/ heat killed S died (heat combats S pneumonia) Avery, MacLeod & McCarty: Demonstrated that DNAwas the genetic material He used heat killed virulent S strain (same as Griffith) and extracted cell contents. Treated with enzymes (RNase, protease and DNase). Treated cell extract was mixed with the nonvirulent R strain. Transformation occurred in ALL BUT DNase sample! Hershey & Chase: DNA, not protein, is transmitted or passed on to progeny Used bacteriophage genetic material injection process to determine whether genetic material was protein or DNAusing radioactive markers. When the protein was marked, the fluid was “hot”. When the DNAwas marked, the infected bacteria were “hot”. Subsequently, in the phage progeny, only the labeled DNAwas found, but none of the labeled protein. RNAas genetic material: Some viruses contain RNAinstead of DNAand yet can still reproduce (Heinz Frankel-Conrat). Frankel-Conrat & Singer (1956): Demonstrated that RNAcan serve as genetic material (single or double stranded). Took two strains of TMV isolated the RNAand proteins to produce hybrids (RNAAand Protein B) but the progeny were identical to the parental (ie. Protein was not a factor) Albrecht Kossel: Nucleic acid contains 4 nitrogenous bases:Adenine (A), Cytosine (C), Guanine (G) and Thymine (T). Nucleotides are composed of a base (A,G,C or T) attached to a sugar with a phosphate (sugar + phosphate = nucleoside) Purines: Two rings (A+ G) Pyrimidines: One ring (C + T) • RNAis different from DNAin that it has an OH at the 2’carbon instead of an H.All RNAnucleotides use a RIBOSE sugar. • DNAis essentially deoxy-blank. (ie. Deoxyadenosine, deoxyribonucleotide) • RNAis the same as DNAbut without the deoxy (ie.Adenosine, ribonucletide) Aaron Levene: DNAis made of repeating units called nucleotides. Tetranucleotide hypothesis (WRONG): Thought DNAwas made up of equal amounts of all nucleotides. Structure was too simple to carry much information. Edwin Chargaff: Analyzed the composition of DNA;A=T and G=C. (A+G/T+C = 1) LECTURE 2 – STRUCTURE OF DNAAND RNA • The sequence of bases of each strand of any given double helix are complementary. • Watson & Crick proposed the 3D structure of DNAusing the findings of all the other DNAdiscoveries. o 2 forms of double stranded DNA,Aform in low humidity and B form in high humidity. Purine + purine was too wide, pyrimidine + pyrimide was too narrow, but purine + pyrimidine was consistent with the xray data. 4 key concepts needed to solve DNAstructure: 1. Phosphates on outside, bases on inside 2. Structure of molecule was a double helix 3. The strands were antiparallel 4. Specific base pairing Key characteristics of the double helix: - Constant diameter - Two strands run antiparallel - Bases on the inside - Sugar-phosphate backbone on the outside - Specific base pairing • DNA3’oxygens are linked to 5’oxygens by phosphodiester links (5’phosphate to 3’ carbon) • Strands are held together by hydrogen bonds (G-C 3 bonds, A-T 2 bonds). • Single stranded nucleic acids are also called polynucleotides. Each chain has polarity: a 5’phosphate end and 3’hydroxyl end. • Aform: Right hand turns, 11 bases/turn, narrower major groove, found in DNA-protein complexes. • B form: Right hand turns, 10 bases/turn, usually found in cells • Z form: Left hand turns, 12 bases/turn, only minor grooves (NO major grooves), least common. Hairpin & Stem: Occur in ssDNAwith inverted sequence (ie. TGCGA-TCGCA). Hairpin has non-inverted sequence in the middle, stem has nothing separating the inverted sequence. These can promote mutations (gain or loss of repeating units). In dsDNA, these can produce cruciform (like a plus sign). RNAcan also fold into hairpins and stems. Melting: The separation of two DNAstrands (also known as denaturation). Must occur for replication, transcription, recombination and repair. Allows hybrids to form. Occurs by temperature increase, reduced salt concentration, increased pH or solvents. • As DNAseparates, absorbance increases. Tm is an indication of the stability of the hybridized DNAmolecule (higher is more stable). • The higher the GC content, the higher the melting temperature (each G:C pair contributes more to the stability than A:T pairs). • Higher salt concentration protects the helix, meaning higher melting temperature. Tm = 81.5 + 16.6log[M] + 0.41(%GC) – 675/L • On the other hand, higher pH (basicity) and higher organic solvent concentrations lower DNAstability and therefore lower Tm. LECTURE 3 – DNAReplication DNAreplication is semiconservative. Semiconservative replication: 1. Two strands of parental molecule separate 2. Each parental strand serves as a template for replication 3. Rules of base-pairing ensure genetic info is correctly copied 4. Each “daughter” DNAmolecule consists of one parental strand and one newly synthesized strand Minimum requirements for DNAsynthesis: 1. Template of ssDNA 2. All four dNTPs (one for eachA, G, C and T) 3. DNApolymerase and other enzymes and proteins 4. Free 3’OH group • DNAPolymerase uses dNTP to catalyze phosphodiester bonds (nucleotide-nucleotide). • MustABSOLUTELY have a preexisting 3’OH. Cannot create a new chain, can only extend a chain.Always moves 5’-3’, reading the template in the 3’-5’direction. • Synthesis occurs at the replication fork (beings at the sequence for origin of replication), takes place in the replication bubble, where both strands are simultaneously synthesized at the replication fork and is bi-directional. Replicon isANY DNAmolecule or region of DNAthat replicated from a single origin of replication. • One stand is synthesized continuously (leading strand) and the other in okazaki fragments (lagging stand). Replication takes place at both forks in the bubble.At each fork, replication is semi-discontinuous; the leading strand follows the fork and the lagging strand synthesizes in the opposite direction, away from the fork. Modes of Replication: 1. Theta replication (bacteria): Single replicon (entire chromosome), bidirectional at both replication forks, replication terminates on the other side of the circular DNA. 2. Rolling circle replication (viruses): No replication bubble, cut at origin site and replicate only uncut strand multiple times 3. Linear replication (eukaryotes): Multiple replicons, origins of replication and replication bubbles How does the double helix get unwound? 1) Initiation: Several proteins are involved: • Initiator proteins bind to the origin of replication (oriC) to unwind a short section of DNA. • Helicase binds to ssDNAlagging strand template to break hydrogen bonds • Single-strand binding proteins keep DNAstrands separated 2) Unwinding: • DNAhelicase separates the two DNAstrands by breaking the hydrogen bonds • DNAgyrase (a topoisomerase) travels ahead of the rep fork and alleviates supercoiling caused by the unwinding by breaking and then resealing the strand 3) Elongation: • Short stretch of RNAprimer is synthesized by primase. It provides a free 3’OH for the DNApolymerase to use, and is later removed (Why RNAprimer? It doesn’t require a 3’end to be produced) • Proof reading lowers error rate from 1/1000000 to 1/10mill. Pol III is a proof reading polymerase, which stalls when incorrect base pairing occurs. Uses 3’-5’ exonuclease activity to remove the incorrect nucleotide • Replacing RNAnucleotides: Pol I 5’-3’exonuclease removes RNAprimers, 5’-3’ polymerase activity fills in gaps with DNAnucl. • DNA ligase seals the nick in the sugar phosphate backbone. Eukaryotes • Initiation of Replication: Activated in clusters (20-80 at a time), each cluster is called a replication unit. Controlled initiation: 1. An origin must be selected or “licensed” by replication licensing factor that binds 2. Origin activation – replication begins 3. Once activated/replicated an origin is deactivated using geminin • Fork speed is much slower in eukaryotes than in prokaryotes, with much more genome, which is why there are multiple origins. • Eukaryotic DNAis packaged into chromatin. Need to disassemble, produce histones and reassemble nucleosomes. • Telomeres: ends of linear chromosomes made up of G-rich short repeated sequences that stabilize chromosomes. Each round of replication leaves up to 200b DNAunreplicated (shortening the telomeres each time) from the removal of primer. o Telomerase is a reverse transcriptase that adds telomere repeat sequences to lengthen parental DNAby RNA-templated synthesis. LECTURE 4 – MOLECULAR BASIS OF RECOMBINATION Homologous Recombination: Exchange of genetic material between two homologous chromosomes (crossing over during prophase 1 of meiosis or mitosis). Generates diversity among the progeny. • Holliday Model: Single strand break during alignment of homologous chromosomes, strand invasion to form Holliday junction (chiasma) and then branch migration until the end of each chromosome. • Noncrossover recombinants are those that have the same original ends at both ends of the chromosome (ie. Requires a Double crossover) • Double Strand Break Model: 5’ends at the break are degraded to produce 3’overhangs; strand invasion into homologous sequence and then extension of the 3’end of the invading strand. The displaced strand serves as a template to form newly synthesized DNA. DSBM causes the formation of two Holliday junctions. Resolution of Holliday junctions can result in either non-crossover (-H plane) or crossover (-V plane) products. Study of recombination: Tetrad Analysis Mostly done in simple haploid organisms.All haploid genes are expressed; easy to track. MI: First division, alleles segregate (separation of homologous chromosomes), no crossing over MII: Second division, alleles segregate (separation of sister chromatids), crossing over Gene conversion: related to recombination, produces abnormal ratios of gametes following meiosis and results from heteroduplex formation during recombination. Duplexes with mismatched base pairs will be repaired using one strand or the other as the template, resulting in the conversion of that gene to the alternate allele. This changes the ratio of gametes from 2:2 to 3:1. Failure to repair the mismatched bases can lead to post-meiotic segregation which yields 5:3 ratios. In short: Recombination leads to heteroduplexes which can lead to base pair mismatches. RecBCD Pathway (recombination in E.coli): Complex coded by 3 genes (RecB, RecC and RecD) which binds to a free duplex at the double strand break, which occurs following conjugation. It combines both helicase and nuclease activity, but requiresATP. RecBCD serves in recombinational repair. Generation of a 3’single stranded terminus: 1. RecBCD binds at a double strand break 2. Initiates unwinding of DNAduplex 3. Continues unwinding, cleaving 3’strand much more frequently than 5’strand 4. Encounters chi sequence; stops digesting 3’and increases cleavage of 5’ 5. Loads RecAonto (now protruding) 3’strand 6. RecBCD unbinds DNA, leaving RecAfilament on 3’ RecAassembles on the single stranded DNAand promotes strand invasion and pairing with homologous DNA. RuvAB complex promotes branch migration and heteroduplex formation during crossover. It drives DNAunwinding and rewinding that is necessary for branch migration. RuvC resolvase is the ENDOnuclease that resolves the Holliday juctions. Nicks strands for either horizontal or vertical resolution. LECTURE 5 – TRANSCRIPTION Transcription + translation = expression Prokaryotic gene expression: Transcription & translation both occur in cytoplasm. Transcription:All RNAare transcribed from DNA, eukaryotes and prokaryotes both have mRNA, rRNAand tRNA. Transcription is the selective synthesis of RNAfrom DNA, and it is complementary and antiparallel to the template DNAstrand. Template is read 3’-5’; chain synthesis begins de novo with ribonucleotides added to the 3’OH group of the chain. RNAis synthesized in the 5’-3’direction. Numerous RNAmolecules are transcribed simultaneously from each DNAstrand. Requirements for transcription: 1. RNAnucleoside triphosphates 2. DNAtemplate 3. RNApolymerase and other proteins mRNA= non-template strand (where T=U for RNA) The gene (transcription unit): Promoter: upstream of start site, adjacent to gene RNAcoding region: downstream of start site (only part transcribed) Termination site: downstream of start site Procaryotic Transcription: 2 consensus sequences upstream of start site (-10 and -35) and come in different strengths (variable amino acids). Mutations can be up or down (up strengthens by becoming more similar to the consensus sequence, down weakens by becoming less similar) RNAPolymerase: Control center for transcription, reads DNA3’-5’and produces RNA5’-3’ using rNTP to catalyze the formation of phosphodiester bonds. RNAPol has sigma factors which help recognize promoters to initiate transcription. Initiation: 1. Recognition sigma factor guides core enzyme to promoter, orients polymerase to the strand to be transcribed. 2. Polymerase binds to specific regions of promoter (consensus sequences) 3. Polymerase unwinds the duplex and pairs ribonucleotide with the DNAtemplate, no primer required. Elongation: 1. Escape of the polymerase from the promoter 2. Conformational change in polymerase releases it from the consensus sequence 3. Sigma factor is released 4. Unwinds and rewinds the DNA as it moves, maintains a bubble of ~20nt 5. RNAis dislodged from the DNA, only 8nt remain paired. Termination: Transcription ends after a terminator sequence is transcribed Rho-dependent: 1. Rho binds a stretch of RNAupstream of the terminator 2. RNApolymerase pauses when it reaches the terminator sequence and Rho catches up 3. Rho unwinds DNA-RNAhybrid using helicase activity Rho-independent: Located directly downstream from stop codon. Creates hairpin to destabilize DNA-RNAhybrid. RNAtranscript dissociates from RNApol, DNAreanneals. Eukaryotic Transcription: RNApolymerase I, II and III (unique to eukaryotes), plus IV and V (unique to plants) RNApol II (of subsequent interest) is pre-mRNA RNApol II promoter = core promoter (binding) + regulatory promoter (influence polymerase function) Core promoter: Extends upstream of transcription start site into transcribed region, minimal sequence for transcription initiation and includes a number of consensus sequences (TATAbox) for factor binding. Regulatory promoter: Located upstream of core promoter. Transcriptional activator proteins bind to consensus sequences to change rate of transcription, when and what cell type transcription occurs, etc. Basal Transcription Apparatus: RNApol II, general transcription factors & mediator protein - Binds to core promoter near transcription start site - Necessary for transcription at minimal or basal levels - Extra regulatory proteins bind to the promoter and enhance sequences to affect transcription rate Termination of RNApol II transcription: Does not end at specific sequence. Requires cleavage of mRNAat a specific site.A5’-3’exonuclease degrades the remaining mRNAterminating transcription. LECTURE 6 – RNAMOLECULESAND MRNAPROCESSING Crick (1958): Continuous sequence of nucleotides encodes continuous sequence of amino acids. Number of nucleotides in the gene is proportional to the number of amino acids in the protein. Suggests one amino acid for every 3 nucleotides. Structure of mRNA: Shine-dalgarno sequence = consensus sequence for ribosome binding (prokaryotes only) mRNAin eukaryotes is shorter than DNAtemplate, is not encoded as an equal collinear segment of DNA, derived from segments of DNAcalled exons which are separated by blocks of noncoding sequences called introns. Introns are transcribed, but must be SPLICED from pre-mRNAto produce mRNA. The eukaryotic gene includes DNAsequence that codes for all exons, introns and the untranslated regions of RNA. Prokaryotic: Protein coding genes usually found in a continuous array (operon) which operate as a unit using a single start site for multiple genes Eukaryotic: Each gene is transcribed from its own start site to yield premRNAthat is processed into functional mRNAfor 1 protein Eukaryotic premRNAprocessing: modified after transcription 1. Capping of the 5’end a. Amethylated guanine nucleotide is added to the 5’end: 1. Phosphate group removed from 5’end of premRNA 2.
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