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Genetics Notes Ch 6,7,8,9,11,12,14,17,18

by: Madeline Kaufman

Genetics Notes Ch 6,7,8,9,11,12,14,17,18 BIL 250

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Madeline Kaufman
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These notes cover what is in chapters 6, 7, 8, 9, 10, 11, 12, 14, 17 and 18 of Introduction to Genetic Analysis (10th Edition) by Anthony Griffith et Al.
Dr. Wilson
expressivity, replication, dominance, Codominance, complementation, epistasis, penetrance, Enzymes, Proteins, telomerase, transcription, ribosomes, regulation, Gene Expression, lacoperon
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Date Created: 09/20/16
GENETICS EXAM 2 Chapter 6: Gene Interaction 1. Introduction a. Interactions between a set of genes can be responsible for phenotypic properties b. To deduce interactions underlying a property i. Use protein as bait in vitro and see which others attach ii. Analyze mRNA transcripts present when a process is going on c. Two types of gene interactions i. Between alleles of one locus 1. Variations on dominance ii. Between two or more loci 1. Reveal number and types of genes in program underlying a biological function 2. Interactions Between Alleles of Single Gene: Variations on Dominance a. Multiple Alleles/Allelic Series: the known mutant alleles of a gene and its wild-type allele b. Dominance is a manifestation of how the alleles of a single gene interact in a heterozygote i. Interacting alleles can be wild and mutant alleles (+/m) or different mutant alleles (m1/m2) c. Several types of dominance for different allelic interaction i. Full/Complete Dominance: a fully dominant allele will be expressed when only one copy is present 1. At phenotypic level, heterozygotes and homozygous dominant cannot be distinguished 2. PKU; fully recessive mutation single gene disease (wild type dominant) a. Recessive because one copy of wild type allele P produces enough phenylalanine hydroxylase to break down phenylalanine- haplosufficient 3. Fully Dominant Mutation; several molecular mechanisms a. When wild-type allele is haploinsufficient: one dose is not enough to achieve normal levels of function b. Null mutation: produces nonfunctional protein c. Example: deletion on chromosome 22, DiGeorge syndrome 4. Dominant Negative: a. Gene product can be unit of homodimeric protein i. Homodimeric: protein composed of two units of the same type ii. In heterozygote (+/M), spoiler polypeptide mutant binds to wildtype polypeptide and interferes with it (spoiler to wildtype) b. Same type of spoiling can hinder heterodimer function composed of polypeptides from different genes (spoiler to polypeptide of different gene) c. Gene product can be monodimer; mutant binds the substrate leaving little available on which the wild type protein can act (spoiler to substrate blocking wild type) i. Collagen protein ii. Incomplete Dominance: the phenotype of a heterozygote is intermediate between those of the two homozygotes on some quantitative scale of measurement ;; intermediate phenotype 1. How it happens on molecular level a. Each wild type allele produces set dose of protein product, b. The number of doses determines the concentration of a chemical made by protein (such as pigment) [quantitative differences in gene dose] 2. Example: four o’clock plants a. Two doses produces a lot of pigment b. One dose produces less pigment, flower is pink c. No dose, no pigment 3. Example: human hypercholesterolemia a. HH; homozygous making a lot of LDL receptors, Hh makes half # of receptors, hh doesn’t make receptors 4. 1:2:1 F 2ratio iii. Codominance: expression of both alleles in heterozygote 1. Example: human ABO blood types and codominance of antigen alleles a. Blood groups determined by 3 alleles of one gene interacting in different ways to produce 4 blood types b. Refer to two carbohydrates (A and B) found on the surface of red blood cells i. Surfaces can be coated with A, B, both, or neither ii. A and B codominant, O is recessive 1. Phenotype A: genotype AA or AO 2. Phenotype B: genotype BB or BO 3. Phenotype AB: genotype AB 4. Phenotype O: genotype OO c. Compatible blood type is critical for transfusions i. If donor’s blood cells have foreign carbohydrates, recipients body will produce antibodies causing donor blood cells to clump together (fatal) A B d. 3 major alleles- i, I , I i. Person only has 2 out of the 3 alleles (or 2 copies of one of them) e. Alleles A and B are codominant; both are expressed in the phenotype 2. Example: chevron patterns on white clover a. Determined by alleles at single gene b. Examples of dominance, codominance, and recessiveness all at one locus iv. Recessive Lethals 1. Lethal allele: capable of causing death of organisms 2. Recessive lethals masked by wild type allele a. Two copies of recessive lethal causes death 3. Example- diagnostic test: coat color allele in mice a. Normal wildtype mice have dark coats b. Mutation; yellow- weird inheritance pattern c. Yellow mouse x homozygous wild type mouse i. 1:1 ratio of yellow to wild type d. Yellow mouse x yellow mouse i. 2/3 yellow, 1/3 wild type ii. ¼ A /A : lethal and die iii. ½ A /A: yellow iv. ¼ A/A wild type e. 1:2:1 ratio in zygotes, 2:1 ratio in progeny because of death 4. Example of pleiotropy: one gene affecting multiple characters (in this case, coat color and survival) 5. About lethal mutations a. Change in DNA sequence causing change in amino acid sequence b. Often result in lack of activity or abnormal activity of encoded protein c. Lethality affected by genetic and physical environment d. Detected by abnormal genetic ratios e. Genetic load: total of deleterious and lethal genes in a population d. Summary i. Alleles of single gene can show various degrees of dominance ii. Dominance is determined by type of molecular interaction and level of analysis 3. Interaction of Genes in Pathways a. General information i. Discovered by Beadle and Tatum in 1940s ii. Working with neurospora and arginine auxotrophs 1. Goes from precursorà(enzyme x)à ornithineà(enzyme y)àcitrullineà(enzyme z)àarginine a. Enzyme x from arg1 b. Enzyme y from arg2 c. Enzyme z from arg3 iii. Tyrosine metabolism and associated diseases [within synthetic pathway]: PKU, albinism, cretinism, tyrosinosis, alkaptonuria b. Complementation test: used to figure out if genes are interacting within a pathway (test for allelism) i. Complementation: the production of a wild type phenotype when two haploid genomes bearing different recessive mutations are united in the same cell ii. Result inferred from phenotype of double mutant 1. Mutants are crossed to bring two recessive mutations together in heterozygous form 2. If phenotype is mutant: mutations in same gene a. Fail to complement b. Both have loss of function 3. If phenotype is wild type: mutations are in different genes a. Each mutant contributes a normal gene at a different locus b. Complementation due to interaction of different proteins iii. Example: Harebell plant with mutant white and wild type blue iv. In haploid, complementation test cannot be used 1. Alternative: fusion resulting in heterokaryon: haploid nuclei from different strains occupy one cell (nuclei do not fuse) 2. Mimics a diploid state and allows complementation if; mutations in different genes conferring same mutant phenotype c. Interactions of Genes in Different Pathways i. Expected observable 9:3:3:1 ratio (dihybrid cross) ii. Two genes assort independently and do not interact 1. Gene products are synthesized independently in two distinct biochemical pathways iii. Example: skin pigment in corn snakes 1. O+: allele producing orange pigment. B+: allele producing black pigment 2. o+/o;b+/b x o+/o;b+/b à F2= a. 9 o+/-; b+/- : orange on black camoflauge b. 3 o+/-; b/b : orange c. 3 o/o; b+/- : black d. 1 o/o; b/b : albino 3. Pigment genes act independently at cellular level a. Precursoràb+à black pigment b. Precursorào+à orange pigment c. If presence of one mutant makes one pathway fail, other pathway is still active d. Interactions of Genes in Same Pathway i. Expected observable 9:7 ratio ii. Occurs when double mutant has same phenotype as single mutants [ one failure blocks entire pathway ] 1. Mutation observable if homozygous for recessive mutant allele of either or both genes 2. No matter which is absent, same pathway fails producing same phenotype 3. 3 genotypic classis produce same phenotype e. Interacting Genes in Same Pathway with Recessive Epistasis i. Expected observable 9:3:4 ratio 1. Double mutant expresses only one of the two single mutant phenotypes ii. Epistasis: allele of one gene eliminates expression of alleles of another gene, substituting its own phenotype 1. Indicative of biochemical or developmental sequential gene pathway 2. Overridden gene: hypostatic 3. Epistatic gene: overriding mutation that is upstream iii. Example: Blue Eyed Mary flower: upstream block 1. W+/w;m+/m x w+/w;m+/m a. 9 w+/-; m+/- Blue b. 3 w+/-; m/m Magenta c. 3 w/w; m+/m still white d. 1 w/w; m/m White 2. Colorlessà[w+]àmagentaà[m+]àblue a. Pink phenotype recessive to upstream white epistatic gene iv. Example: Labrador coats: developmentally downstream 1. Alleles B and b stand for black and brown coats (melonin) a. Allele e is epistatic to these alleles giving yellow coat b. B/-; e/e and b/b; e/e à yellow phenotype c. B/-;E/- and b/b;E/- à black and brown 2. This epistasis is not caused by upstream block a. Yellow labs still make black or brown pigment, seen in noses and lips b. Allele e prevents deposition of pigment in hairs c. Allele e, the epistatic gene, is developmentally downstream; f. Interacting Genes in Same Pathway with Dominant Epistasis i. Example: Fox Gloves 1. Two genes interact to determine petal coloration, unlinked a. One gene affects intensity of red color i. d: light red color of natural populations ii. D: mutant allele, darker red b. One gene affects in which cells the pigment is synthesized i. w: synthesis of pigment throughout petals in wild type ii. W: mutant allele, confines pigment to small throat spots 2. F2 of dihybrid cross a. 9 D/-; W/- à 9 white with spots i. not dark red because dominant W confines coloration to spots b. 3 d/d; W/- à 3 white with spots c. 3 D/-; w/w à 3 dark red d. 1 d/d;w/w à 1 light red 3. Dominant allele W is epistatic g. Suppression: interaction between two different biochemical or developmental pathways in which the product of one allele of a gene interferes with the product of another gene on a different pathway i. Suppressor: mutant allele of a gene that reverses the effect of a mutation on another gene, resulting in [near]wild type phenotype 1. Second mutation suppresses first mutation 2. Suppressor mutation alone; inactive ii. Numerous causes 1. Nonsense suppressors: mutation in tRNA results in anticodon that will bind to premature stop codon a. Suppressor allows translation to proceed past former block and make complete protein iii. Revertant: an allele with wild-type function arising from mutation of a mutant allele, caused by a complete reversal of the original event or by a compensatory second site mutation 1. Pseudorevertants: revertants that are double mutations in which one of the mutations is a suppressor [and can then be wild type] a. Can result in 13:3 ratio when suppressor has no phenotype b. Can result in 10:6 ratio when suppressor is like mutant h. Modifiers: modifier mutation at second locus changes degree of expression of a mutated gene at the first locus i. Synthetic lethals: two viable single mutants are intercrossed resulting in a double mutant that is lethal, manifested in 9:3:3 ratio 4. Penetrance and Expressivity a. Penetrance: the percentage of individuals within a given allele who exhibit the phenotype associated with that allele i. 100% penetrance= 100% of people with mutant genotype express mutant phenotype ii. Why would organism have genotype and not express corresponding phenotype? 1. Environment (cannot distinguish between wild type and mutant sometimes) 2. Interacting genes 3. Subtlety of mutant phenotype b. Expressivity: the degree to which a given allele is expressed at a phenotypic level; measures the intensity of the phenotype i. Affected by genetic background [interaction with rest of genome] and environment ii. Example: piebald spotting in beagles c. These phenomena make pedigree analysis and genetic counseling more difficult Chapter 7: DNA Structure and Replication - Establishing DNA as Genetic Material - Frederick Griffith: i. 1928 ii. Studying bacterium pneumococcus 1. Cause of pneumonia in humans, normally lethal in mice; however some have evolved to be less virulent 2. Used two strains a. S; smooth appearance; virulent b. R; rough appearance; mutant non virulent 3. Killed virulent cells by boiling, injected into mice, mice survived: cell carcasses do not cause death a. BUT mice injected with mixture of heat-killed virulent cells and living nonvirulent cells died i. Live cells could be recovered from these dead mice and were virulent ii. Cell debris of boiled virulent cells make live nonvirulent cells virulent (transformation) iii. But what caused the transformation? - Oswald Avery i. Approach: chemically destroy all major cellular components one by one and then find out if extract lost ability to transform 1. Polysaccharides, proteins, fats, DNA, RNA 2. No live S strains recovered when DNA was destroyed ii. When DNase from dog intestine was heat denatured, experiment failed - Hershey Chase: bacteriophage injecting DNA into E. coli - DNA Structure as Double Helix - 1953; James Watson and Francis Crick worked out 3d structure i. X-ray crystallography ii. Diameter of helix suggested that it was made of 2 polynucleotide strands; double helix 1. What kind of helix? a. Watson and Crick tried putting sugar and phosphate groups inside the molecule i. Didn’t work, nitrogenous bases on inside hydrogen bond - Chargaff’s Rule of base composition i. Before him DNA was thought to contain equal amounts of dinucleoutide ii. Studied base composition of different organisms 1. Total amount of pyrimidine nucleotides (T and C) equals amount of purine nucleotides (A and G) a. Amount of T= amount of A b. Amount of G= amount of C c. Amount of A and T does not equal amount of G and C - 4 Bases of DNA: i. Pyrimidines: Thymine and Cytosine 1. Single carbon rings ii. Purines: Adenine and Guanine 1. Double carbon rings iii. Structure of nucleotides: phosphate attached to deoxyribose sugar attached to nitrogenous base iv. Each base attached to 1’ carbon - The Double Helix i. Nucleotide strands held together by hydrogen bonding between nitrogenous bases 1. AT- 2 hydrogen bonds 2. CG- 3 hydrogen bonds 3. NH g2oups slightly positive (N pulls e- in) 4. O 2roups slightly negative (O pulls e- out) 5. Globally very strong, locally very weak ii. Sugar phosphate backbone 1. Phosphodiester linkage between 5’ carbon atom of one deoxyribose to 3’ carbon atom of another iii. Antiparallel strands: each strand oriented in opposite direction 1. 5’ end: phosphate group off of 5’ carbon 2. 3’ end: hydroxyl group off of 3’ carbon iv. Major grooves: more protein association v. Helical shape from pairing and stacking of bases 1. Right handed helix - Structure reflects function i. Complementary bases: self-replicating ii. One strand is template for synthesis of other strand - DNA Replication - Overview i. Prerequisite to cell division 1. Uses each strand as template for synthesis of complement a. Semi conservative: replicants made up of one old strand and one new strand 2. Leading strand synthesized continuously a. Lagging strand discontinuously 3. Nucleotides added at 3’ end of growing strand a. Template strand read 3’à5’ synthesizing new strand from 5’à3’ 4. Mitosis: replicated chromosomes partitioned into diploid nuclei of daughter cells 5. Meiosis: replicated chromosomes partitioned to haploid daughter cells (2 nuclear and 2 cell divisions) ii. Occurs in S Phase of cell cycle iii. Synthesis is bidirectional 1. Replicons have a. One origin of replication b. 2 diverging replication forms 2. At each fork, opposite strand is leading/lagging a. Same strand simultaneously leading/lagging strand - Enzymes Involved i. DNA polymerase: catalyzes DNA synthesis, requiring primers ii. Primase: synthesizes RNA primer iii. Gyrase: relaxes DNA double helix as it is unwound ahead of the replication fork (a tropoisomerase) 1. Removes extra twists [supercoils] 2. By breaking either single or double DNA strands, allowing DNA to rotate into relaxed molecule, then rejoins newly relaxed molecule iv. Helicase: unwinds DNA ahead of replication fork (breaks H-bonds) v. Ligase: joins lagging strand of DNA fragments together vi. Telomerase: elongates telomeric sequences on the ends of chromosomes (contains built in primer) - DNA Replication Fork i. Leading strand: nucleotides added towards replication fork ii. Lagging strand: nucleotides added from 5’ to 3’ away form replication fork 1. Synthesize short segments at a time; okazaki fragments (1000-2000 nucleotides) starting at newly available 5’ ends of template 2. Joined together by ligase - Repliosome: large nucleo-protein complex that coordinates activities at replication fork i. Require RNA primer: short chain of nucleotides 1. Synthesized by primase enzyme and set of proteins called primosome a. Complementary 2. Binds with template strand a. DNA polymerase can extend a chain but not start a chain, needs primer to bind to and begin replication 3. Leading strand only needs one primer a. Lagging strand needs one primer for each Okazaki fragment 4. DNA polymerase I replaces RNA primer with DNA ii. Remarkable speed: 2000 nucleotides/second in E.coli 1. 1000 nucleotides in each direction iii. DNA Pol III elongates DNA from primer 1. 2 DNA polymerases are single complex coupled together a. Dimers b. Physical connection coordinates synthesis of leading/lagging c. All together pol III holoenzyme: 2 catalytic cores and accessory proteins that connect iv. B-clamp: keeps DNA pol III attached to DNA strand v. Prokaryotic v Eukaryotic Repliosomes 1. Prokaryotic a. 13 components b. No nucleus c. No chromatin 2. Eukaryotic a. 27 components b. Need to disassemble nucleosomes and repackage c. Nucleus and chromatin - Prokaryotic Initiation of replication i. Begins from fixed origin OriC (1 per DNA molecule) 1. Continues until two forks merge 2. Origin is composed of DnaA boxes 3. Can be greater than or equal to one origin ii. Repliosome assembly 1. DnaA protein binds to specific 13 base pair sequence called DnaA box that is repeated 5x in OriC 2. This causes origin to unwind at cluster of A and T nucleotides (only 2 H-bonds in this pairing) 3. Additional DnaA proteins bind to newly unwound single stranded regions (coating origin) 4. Then two helicases; DnaB protein; bind and unwind helix 5. DNA pol III holoenzyme and primase recruited iii. DnaA – brings repliosome to correct place but is not part of replication machinery iv. Rolling circle replication 1. Plasmids a. One strand remains circular and is template for continuous replication b. Tail is template for discontinuous synthesis of okazaki fragments 2. Often continues beyond one chromosomal unit - Eukaryotic initiation of replication i. Problem of DNA packaging 1. Chromosomes exist as chromatin, basic unit is nucleosome; DNA wrapped around histone proteins 2. Need to be disassembled then reassembled in daughters a. Done by: i. Randomly distributing old histones to daughter molecules and ii. Delivering new histones in association with CAF-1; chromatin assembly factor 1 b. CAF-1 binds to histones and targets them to replication fork where they are assembled with new DNA i. CAF-1 and histones bind to PCNA; proliferating cell nuclear antigen that is eukaryotic B-clamp ii. Origin of Replication 1. Yeast a. 100-200 bp origins of replication i. Conserved DNA sequence with AT region that melts with initiatior proteins ii. About 400 ORIs iii. Double helices produced at each origin of replication elongate and join one another b. Replication occurs in S phase; link repliosome assembly to cell cycle i. 3 proteins required for assembly 1. ORC: origin recognition complex: binds to origin 2. Binding serves to recruit Cdc6 and Cdc1; cyclins ii. ORC with Cdc6 and Cdc1 recruit helicase making MCM complex 1. And recruit other components of repliosome iii. Cdc6 and Cdc1 synthesized during late mitosis and degraded after synthesis 1. Repliosome only assembled before S phase 2. Higher Eukaryotes a. Longer origins of replication (can be hundreds of thousands of nucleotides) i. Limited sequence similarity ii. ORC’s probably do not recognize DNA sequences 1. Harder to isolate origins of replication b. ORCs indirectly interact with origins by association with other protein complexes bound to chromosomes i. Euchromatin: less densely packed [gene rich], replicated earlier in S phase ii. Heterochromatin; tightly packed [less gene rich] replicated later in S phase iii. ORCs may have affinity for origins in open chromatin - Eukaryotic replication termination i. Continuous replication of leading strand can continue to very end of template ii. Lagging strands require primers ahead of process 1. When last primer is removed sequence is missing end of that strand 2. Internal gaps are filled 3. Terminal gap not filled following primer degradation; results in 3’ overhang on lagging strand 4. Solution: repetitive telomeric capping iii. Telomerase: enzyme that adds multiple copies of simple noncoding sequence (telomeres) to ends of chromosomes 1. Human chromosomes end in tandem repeats of 5’- TTAGGG-3’ [on 3’ end]! a. Extension of 3’ end/overhang gives DNA polymerase and primase template to fill in end of other DNA strand 2. Ciliate end in TTGGGG 3. Telomerase contains internal RNA template for repetitive sequence a. In humans, RNA sequence is 3’-AAUCCC-5’ i. Telomerase anneals to 3’ overhang b. Contains reverse transcriptase i. RNA as template for DNA 4. Telomerase activity high in germline gonads: make sure progeny receive long chromosomes a. Not as high in somatic cells b. As we get older, telomeres shorten (part of aging process) and telomerase efficiency declines 5. iv. Telomeres also preserve chromosomes by associating with proteins to form protective caps 1. Sequester 3’ single stranded overhang 2. Without, end would be seen as double stranded break by cell (and destroyed) 3. Tangle ends to make sure it does not fuse with other chromosomes a. WRN, TRF1, TRF2 v. Cancerous cells have telomerase activity Chapter 8: Transcription and Processing 1. About RNA a. Composed of phosphate, ribose sugar, and functional group i. One extra oxygen on the sugar when compared to DNA ii. Uracil functional group is pyrimidine replacing thymine 1. U instead of T b. Types of RNA i. Informational RNA (translated/messenger) 1. Messenger RNA; protein coding a. Translated into amino acid sequence 2. Primary transcripts in prokaryotes 3. Processed transcripts in eukaryotes a. Contain introns ii. Functional RNA 1. tRNA: transfer RNA: transport correct amino acid to mRNA in ribosome in translation 2. rRNA: ribosomal RNA: structural and catalytic component of ribosomes a. Enzymes are not the only catalysts 3. snRNA: small nuclear RNA: part of system that further processes RNA transcripts in eukaryotes a. Can unite with protein subunits to form ribonucleoprotein processing complex (the spliceosome) that removes introns b. Structural and catalytic component of spliceosome snRNPs 4. miRNAs: microRNAs: regulate amount of protein produced by eukaryotic genes a. Affect longevity of protein in cell 5. siRNA: small interfering RNA: inhibit production of viruses and prevent the spread of Tes*** a. Immune system: protect invaders of genome b. siRNA and piwi-interacting RNAs prevent spread of transposable elements to other chromosomal loci i. siRNA in plants ii. piRNA in animals 2. Transcription a. Overview i. DNA is template for transcription 1. Both strands can encode for genes a. Any one gene is encoded on only one strand b. Only one strand is template, but which strand varies with each gene 2. Transcription reads 3’à5’ along template, and complementary RNA strand is synthesized 5’à3’ a. Genes transcribed on opposite strands are transcribed in opposite directions ii. RNA polymerase: unwinds DNA and adds free nucleotides to growing strand 1. Transcript identical to nontemplate strand except T’s replaced with U’s 2. Prokaryotes have 1 RNA polymerase 3. Eukaryotes have 3 RNA polymerases a. RNA pol I: transcribes RNA genes b. RNA pol II: transcribes protein encoded genes (mRNA) c. RNA pol III: transcribes tRNA and 5S rRNA iii. Transcription and translation occur simultaneously in prokaryotes iv. WARNING: When DNA sequences are cited in scientific literature, the sequence of the nontemplate strand of DNA is presented 1. Because this is the same sequence as found in the RNA b. Steps i. Initiation 1. RNA synthesized 5’à3’ a. DNA read 3’à5’ 2. RNA polymerase binds to promoter: 5’ regulatory region a. Actually on 3’ end of template, but 5’ end of RNA b. Upstream of initiation site i. First DNA base transcribed; +1, and everything upstream is negative c. Promoter regions of e.coli i. Same RNA polymerase to bind, similar promoter sequences ii. Two regions of great similarity: -35 and -10 regions; consensus sequence: in agreement with most sequences [not necessarily identical] 1. TATA box 3. 5’ untranslated region: between promoter and initiation site 4. RNA polymerase holoenzyme scans DNA for promoter region a. 5 subunits of core enzyme b. One subunit: sigma factor i. Binds to -35 -10 region; correctly positions ii. Melts DNA around region for good binding iii. After bound, transcription begins and sigma factor disassociates iv. Different sigma factors recognize different promoter sequences 1. Same core enzymes can recognize and transcribe different sequences by association with different sigma factors ii. Elongation 1. Basic info a. Adds nucleotides to 3’ end of growing RNA b. Requires Mg+2 c. Energy from NTP substrates i. 2. RNA polymerase unwinds and rewinds DNA a. Transcription bubble: single stranded region iii. Termination 1. Occurs at 3’ end of gene beyond coding region, creates 3’ untranslated region 2. Stops at nucleotide signal for chain termination 3. E.coli method of direct termination a. Intrinsic b. Terminator region contains sequence with many GCs, hydrogen bond with each other resulting in hairpin loop i. GC bonding more stable c. Signals release of polymerase i. Unstable U’s following loop cause backtrack c. Prokaryotic v Eukaryotic Prokaryotic Eukaryotic • no nucleus - transcription and • transcription and processing in nucleus translation simultaneous - transported out for translation - - general transcription factors need to assemble at promoter before RNA pol transcribes - chromatin i. Transcription initiation in Eukaryotes 1. Require GTFs instead of sigma factor a. Required before core enzyme can bind b. Attract RNA pol II 2. Pre initiation complex: 6 GTFs and RNA pol II core [many protein subunits] 3. TATA box: promotor region 30 bp upstream from starting site a. TBP: TATA binding protein [part of a TF]: binds to TATA boxà attracts other GTFs and RNA pol II core forming PIC 4. Once transcription begins, RNA pol II disassociates from most GTFs 5. CTD: Carboxyl Tail Domain: protein tail on beta subunit of RNA pol II a. Located near where nascent RNA emerges b. Elongation phase begins with CTD has been phosphorylated by a GTF i. Potentially weakens connection of RNA pol II to proteins of PIC d. Cotranscriptional Processing of RNA i. Cotranscriptional: nascent RNA undergoes processing as it emerges from RNA pol II ii. FIRST: capped 1. Upon emergence, nascent mRNA has cap added to 5’ end 2. Caused by proteins interacting with carboxyl tail domain 3. Is a methyl guanosine residue linked by phosphate groups 4. Protects RNA from degradation and is required site for translation iii. THEN: cleaved at polyadenylation signal area iv. THEN: polyadenylation 1. Polyadenylation signal: AAUAAA sequence of mRNA from protein encoding regions; RNA cut 20 basepairs down a. Add 150-200 As to cut end; poly A tail v. LAST: spliced 1. Splicing: removal of introns and joining of exons 2. All introns have 5’GU and 3’AG recognition sequences 3. Alternative splicing: one gene can code for different proteins when spliced differently 4. Spliceosome: composed of 5 snRNPs [small nuclear ribonucleoproteins) à catalyze intron removal 3. Small Functional RNAs a. NOT snRNA or tRNA that are constitutively synthesized i. Instead, are miRNAs and siRNAs that are synthesized in response to cell’s changing environment b. miRNAs: micro RNAs: regulators of gene expression i. Basic ingo 1. Found in plant and animal genomes a. About 1000 miRNAs in genome 2. Repress gene expression 3. Constrained in size: 21-25 bp long a. Produced as longer strand that assumes stem-loop structure with mismatched base in stem b. Processed to smaller form in nucleus c. Moves to cytoplasm d. Dicer recognizes double stranded RNA dsRNA and cleaves to ~22 bp e. RISC/RNA induced silending complex splits dsRNA f. miRNA can now repress expression i. Repress translation ii. Promote degradation 4. 2002 Breakthrough of the Year c. siRNAs: small interfering RNAs: silence the gene that produces it i. Does not regulate other genes 1. Instead shuts off undesirable genetic elements to ensure genome stability (viral genes, transposable elements) ii. Gene silencing: selective shut off of gene by introducing dsRNA with sense (coding) region and antisense complementary RNA strand 1. Injecting dsRNA copies of a gene silences it [synthesized in lab] iii. Transgene: transformed gene, has been introduced into chromosomes of an organism in laboratory 1. Endogenous gene: normal gene at usual locus 2. In experiment, put DNA copy of gene normally found in genome into genome a. Result: cosupression: silences both expression of transgene and endogenous gene b. Two ways to generate dsRNA from transgene i. Antisense of transgene with mRNA of transgene ii. Antisense with mRNA of endogenous gene iv. How it works 1. SiRNA can arise from an antisense copy of any source of mRNA in the genome: endogenous genes, transgenes, viruses a. Most likely source is foreign DNA 2. Complimentarity between sense and antisense RNA produces dsRNA a. Recognized and cleaved by dicer b. Short sequences bound by RISC i. Unwinds into single strands ii. Targets RISC to complementary strand to degradeàPERFECT COMPLEMENTS d. Similar mechanisms generate siRNA and miRNA i. miRNA 1. Directs RISC to repress of translation ii. siRNA 1. Directs RISC to degrade mRNA a. Removes poly A tail 2. Perfect complementation Chapter 9: Proteins and Their Synthesis 1. Introductory Info About Proteins a. Basics i. Most abundant component of human body ii. Main determinants of biological form and function iii. Contain: C, H, O, N iv. Polymer (polypeptide chain) made of monomers known as amino acdis b. 7 Major Classes i. Structural: hair, cytoskeleton ii. Contractile: muscle, motile cells; produce movement iii. Storage: sources of amino acids; ie egg whites iv. Defense: antibodies, membrane proteins, complement proteins v. Transport: hemoglobin, membrane proteins vi. Signaling: hormones, membrane proteins vii. Enymatic: increase rate of biological reaction c. Constitution: Amino Acids i. All proteins made of 20 amino acids ii. Each amino acid is central carbon with 4 groups 1. Hydrogen atom 2. Amino group: NH2 3. Carboxyl group: COOH 4. Variable group R- distinguishing factor iii. Amino acids have different properties 1. Polar; hydrophilic 2. Nonpolar; hydrophobic 3. Electrically charges; acidic/basic iv. Be comfortable with 3 letter code of amino acids 1. Difficult a. Asn: asparagine b. Asp: aspartic acid c. Glu: glutamic acid d. Gln: glutamine 2. Protein Structure a. 4 Levels of Protein Structure i. Primary: linear sequence of amino acids 1. Carboxyl terminus and amino terminus 2. Amino acids linked by dehydration synthesis forming peptide bond 3. Alternating location of R groups ii. Secondary: 1. Caused by hydrogen bonding between residues a. And electrostatic and van der wall forces 2. Beta sheets and alpha helices iii. Tertiary: overall 3d shape 1. Caused by folding of secondary structures iv. Quaternary: relationship among multiple polypeptides of a protein b. Shape i. Globular: compact ii. Fibrous: linear iii. Important to function; active sites: where substrate binds on enzyme 3. Protein Synthesis: Translation a. Introductory Info i. DNAàmRNAàProtein ii. Players 1. mRNA 2. tRNA 3. Ribosome a. 95% of RNA you extract from cell is ribosomal b. Hundreds of copies of ribosomal subunits b. Genetic code i. Codon: section of DNA 3 base pairs long that encodes a single amino acid 1. With 4 nucleotides, a 2 letter code only gives 16 possible combinations (4 ) 3 2. A 3 letter code has 64 possible combinations (4 ) a. Some sequences code for the same amino acids 3. Alternate codon usage seen in yeast ii. Non overlapping iii. Degenerate: each triplet has a meaning 1. Frame shift mutations still produce amino acid iv. Stop codons: UAG, UGA, UAA c. tRNA i. Triplet anti codon that binds to complementary codon ii. Amino acid binds to 3’ attachment site on tRNA 1. AminoacyltRNA synthetase attaches the amino acids a. One synthetase for each different amino acid-20 i. Can be more than one tRNA for each amino acid ii. Makes tRNA charged (charged prior to entrance in A site of ribosomal RNA) b. Binding site for amino acid, binding site for tRNA, attaches to tRNA making it charged c. L shaped folded clover leaf; for all iii. Explanation of degeneracy 1. Some amino acids can be brought to ribosome by alternative tRNAs 2. Wobble: loose base pairing at 3’ end of mRNA and 5’ end of anti-codon a. 3 position of anti codon imperfectly matched, but there are Wobble RUles b. Same 2 letter code with different 3 letter for same amino acid c. Specific tRNA can bring amino acid to different codons d. I: Iosine: rare base found in tRNA often in anticodon position e. Serine: 6 different codons recognized by 3 different RNAs 3. Genomes can be more compact d. The ribosome i. Separated based on RNA density 1. Prokaryotic: 70s ribosome a. 50s large subunit b. 30s small subunit c. 3 rRNA molecules 2. Eukaryotic: 80s ribosome a. 60s large subunit b. 40s small subunit c. 4 rRNA molecules ii. Features 1. Small subunit: where mRNA actually binds a. A site: binds incoming charged aminoacyl-tRNA i. Decoding center b. P site: where polypeptide chain grows c. E site: deacylated tRNA released 2. Peptidyltransferase center in large subunit, where peptide bond formation is catalyzed e. The process i. Builds poly peptide chain starting with amine group 1. Reads mRNA template from 5’à3’ a. 5’ end of mRNA is amine end of polypeptide 2. Amino acids added to c-terminus of growing chain ii. Initiation 1. First amino acid in polypeptide is methionine, coded by AUG a. Not inserted by tRNA met [found in middle of sequence], but by tRNAmeti; initiator (special tRNA) b. 5’ UTR region adjacent to AUG 2. Prokaryotic; shine dalgarno a. Initiation codons preceded by Shine-Dalgarno Sequence; pairs with rRNA in 30s ribosomal subunit i. Can only pair when small subunit is disassociated form rest of ribosome b. 3 Initiation factors required for correct placement (IF1 and 2; associate with SMALL SUBUNIT, brings metTRNA releasing IF3) c. Initiation complex: 30s subunit, mRNA, and initiator tRNA i. Large subunit then associates causing release of initiation factors IF 1 and 2 3. Eukaryotic- 5’ cap a. Eukaryotic initiation factors remove secondary structure of mRNA, associate with 5’ cap, and with 40s subunit, and initiator tRNA to form initiation complex b. Once AUG codon recognized, 60s subunit joins and initiation factors dissociate iii. Elongation 1. Elongation factor Tu associates with amino-acyl tRNA making ternary complex; EF Tu released once codon/anticodon match has been made a. Facilitates bringing aminoacyl tRNA to A site 2. EF G appears to fit in A-site, pushes tRNAs in A and P to P and E and then leaves, opening up A site 3. Requires GTP iv. Termination 1. Stop codons do not get translated into protein a. No amino acids/ tRNA molecules recognize stop codon b. 3 stop codons: UAG, UGA, UAA 2. Release factors bind to the A site and promote the cleavage of the tRNA in the P site a. Stop codons recognized by tripeptides in RF proteins not an anticodon b. Water molecule gets into peptidyltransferase center c. Bacteria i. RF 1 and 2 recognize UAA and UAG ii. RF 2 also recognizes UGA iii. Both assisted by RF3 3. Ribosomal subunits separate 4. The Proteome: complete set of proteins in an organism, organ, tissue, or cell a. Alternative splicing produces related but distinct forms b. Interactome: all the protein interactions in an organism c. Ubiquitin: targets proteins for degradation d. Nuclear localization sequences: transport proteins to nucleus if necessary Chapter 11: Regulation of Gene Expression in Bacteria 1. Prokaryotic transcriptional regulation a. Promoter: determines where transcription begins i. RNA polymerase binds to promoter ii. Without promoter, gene cannot be transcribed b. DNA segments near promoter determine whether promoter driven transcription can take place i. Activator binding site 1. Positive regulation: activator on this site, starts transcription a. No activator, no transcription ii. Operator: repressor binding site 1. Negative regulation: repressor on this site stops transcription a. No repressor, transcription c. Repressor and activator require proteins to function: effectors i. Activator: effector needs to bind to allosteric site of activator for it to bind to DNA and start transcription ii. Repressor: effector binds to allosteric site to remove repressor and allow transcription 2. Lac Operon a. Basics i. Noble prize for Jacob and Monod ii. Model of negative and positive regulation in e.coli iii. Regulates utilization of lactose sugar b. i. Metabolism of lactose 1. Permease: brings lactose into cell 2. B-galactosidase: cleaves lactose into glucose and galactose ii. Lac structural genes (encode proteins) 1. LacZ: B-galactosidase 2. LacY: permease 3. LacA: transacetylase [not required for lactose metabolism] iii. Lac regulatory component 1. I: encodes for lac repressor 2. P: promoter site for RNA polymerase binding 3. O: operator site for repressor binding iv. Lac repressor 1. Tetramer of identical subunits 2. Single DNA binding site to dock with O 3. Four allosteric sites that bind with lactose and analogs a. Binding of lactose reduces repressor’s affinity for O, it is released, and transcription occurs b. In absence of lactose, transcription of Z, Y, A repressed 4. Example of induction: relief of repression by action of an inducer [lactose and analogs] a. After induction, genes are transcribed b. Induced: repressor removed from gene c. Regulatory mutations i. Z+ and Y+ wildtype alleles are dominant over Z- and Y- mutations 1. Produced when induced, even if heterozygous ii. P ; promoter mutation results in no transcription iii. Constitutive mutations: cause gene expression regardless of C - presence of inducer: O and I 1. O : make operator incapable of binding to repressor a. Cis-acting: affects only DNA on same molecule [same chromosome] b. As a heterozygote, repression still occurs in other chromosome 2. I-: cannot make a repressor a. I+ gene is Trans-acting: gene product regulates all structural lac operon genes regardless of location on same DNA molecule or different ones i. Having a functional I makes repressors for chromosome with dysfunctional I b. I- is constitutive c. I+ is dominant to I- d. I+ is inducible 3. I : super repressor mutations: causes repression even in the presence of an inducer s + - a. I is dominant over I (unlike I ) i. I proteins will bind to both operators in the cell regardless of inducer and I protein presence b. Alters allosteric site so it cannot bind to an inducer 3. Catabolite Repression: repression of transcription of lactose metabolizing genes in presence of glucose a. Glucose is preferred carbon source, induction only occurs when glucose is not available i. Low glucose causes production of cAMP(cyclic) ii. CAP: catabolite activator protein: cAMP binds to CAP and together, bind to promoter 1. Bends the DNA sequence and makes it easier for RNA polymerase to bind to promotor; bind next to each other 4. DNA binding sites a. Very different necessary for specificity b. DNA binding sites share rotational two fold symmetry i. Corresponds to symmetry within binding proteins composed of 2 or 4 identical subunits à allows tetramer binding, dimers, heterodimers 5. Metabolic pathways a. In prokaryotes, genes that encode enzymes in metabolic pathways are generally organized into operons b. Strong congruence between order of genes in operons and sequence in metabolic pathways i. Allows predictions 6. Arabinose Operon: dual positive and negative control a. Single DNA-binding protein is either repressor or activator : AraC Protein b. Structural genes: AraB, AraA, AraD: encode enzymes that break down arabinose sugar c. Transcription activated at AraI: initiator region with binding site for activator protein i. AraC encodes activator protein ii. When bound to arabinose, AraC protein binds to AraI 1. CAP-cAMP complex also needs to bind for transcription iii. In the absence of arabinose and CAP-cAMP, AraC protein binds AraI region to AraO region, repressing the operon 7. Tryptophan Operon: global control and fine tuned control; attenuation a. Genes in order needed in metabolic pathway (evolutionary pressure because used simultaneously) b. Global control i. When tryptophan is absent, trp gene expression high ii. When levels are high, trp gene expression low iii. Trp repressor (from gene trpR) binds to operator repressing expression 1. Only binds to operator in presence of high tryptophan amino acid 2. Prevents excess production of trp 3. Attenuation: level of transcription of an operon is reduced when the end product of the pathway is plentiful a. Regulated step is after initiation of transcription 8. Alternate sigma factors regulate large sets of genes a. Sigma factors bind ahead of promotor b. Coordinate expression of genes that need to be coordinated: reproduction or stress event c. Not physically connected in same operon d. Analys


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