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by: Steve Lubowitz


Steve Lubowitz
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This 31 page Class Notes was uploaded by Steve Lubowitz on Monday October 12, 2015. The Class Notes belongs to BIOL 358 at Idaho State University taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/222180/biol-358-idaho-state-university in Biological Sciences at Idaho State University.


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Date Created: 10/12/15
CELL STRUCTURE CELL CYCLE AND MITOSIS BIOL 358 Lecture 2 12 January 2006 CELL STRUCTURE Cell is basic unit of life Each cell contains genetic material to make new organism Two types of cells Prokaryotes Bacteria and Archaea Eukaryotes PROKARYOTIC CELL Bacteria Escherichia coll Unicellular Cell wall peptidoglycan Plasma membrane Single circular chromosome nucleoid No membranebound nucleus No membranebound organelles EUKARYOTIC CELL Membrane bound nucleus 2 or more chromosomes Cytoplasmic organelles NUCLEUS Double nuclear membrane with pores Chromatin Nucleolus dark structure in nucleus Chromosomes colored bodies EUKARYOTIC CHROMOSOMES Structure of Chromosomes 1 Centromere 2 Kinetochore 3 Arms a Slser mmmzlms CELL CYCLE AND MITDSIS dlvlslon m 1quot or2N cells Imerphase 1 2 Cell 1th and tun lnn ncmr have 3 Renllczllnn at DNA at mch mmrmsnme NTERPHASE The Cell Cycle 1G1Gap period that precedes DNA synthesis intense metabolic activity in cell preparing for DNA synthesis rnost organelles are duplicated can last from 5 to 9 hours depending upon species and cell type u The Cell Cycle 2 S phase synthesis DNA replication takes place can last from 7 to 8 hours chromosomes form sister chromatids G2 Gap preparation for cell division rnicrotubules for spindle formed centrioes duplicate can last from 2 to 3 hours CELL DIVISION Two stage process 1 Equal division of genetic material 2 Division of cytoplasm Animals cell membrane furrows Plants cell plate wall forms KARYOKINESIS 1 commuous process of cell division mas m m lesslhzn nne hnur m must Dells 2 can be amused mm discrete stages PMAT EAR LY P ROPHAS E m4gtUIgtwm gtZgtUIwm a nu CONSEQUENCES OF MITOSIS End products of each mitotic division 1 2 IMPORTANCE OF MITOSIS Prokaryotes Eukaryotes 1 2 LINKAGE CROSSING OVER AND CHROMOSOMAL MAPPING part III Genetics BIOL 358 Lectures 14 27 February 2006 CHIASMATA FREQUENCY Direct cytological evidence for crossing over Chromosome Length and Chiasmata Frequency Larger chromosomes with more chiasmata Average chiasmata frequency a measure of chromosome length map distance 0 fly A 7 r Meiau39 Numbaal Numbaorreur mam hmmasama miormotulA obsmedlal leal gtoooo0lt s s 2s W 4 s an m 2 3 w W I 25 25 Tmal ma 2T5 ACCURACY OF MAPPING Delectable recombination frequency may not accurately re ect relative physical distances between oci because39 Moltlole crossovers rnavnot oe detemahie detected recombination underestlmates crossover enlasrnatalrre oene Crossover at one slte reooees orooaollltv or anotner nearov event onvsleal lntenerenee reooees ranoorn reeornolnatlon rreooenev n E 3 lenotn or e rornosorne crossoverrreooenev ooe n t rnao olremlv oronlrorrnlv to onvsleal olstanee ertxand Double Crnssnverr Underesumates Deteets 02 erossovers 3rstxand Double Crnssnverr Underesb ates Deteets V2 erossovers 4rstxand Double Crnssnverr Fully estlma es Detects 22 erossovers E m m so we Maulrslmvcnlcw Genes lt 25 cM apart map Naccurately Ion At greater distances recombinatr underestimates map distance Coincidence and Interference Percentage of expected double crossovers Dcoem 1 Product of the recombination frequency etwe n eac pair 0 genes 22 4 7 hm v DCOW 02227 X 04337 0097 or 97 2 Generally there are fewer observed double crossovers nconhs than expected based upon the distance between the genes Coefficient of Coincidence 1 Expresses the reduction of double ers 2 Formula c Dcowgncow In the corn example nconhs 44 42 861109 0073 ncom 0091 C 00780097 0804 Only 80 as many double crossovers as that expected based on the map distances between the genes Interference 1 Caused by the decreased likelihood of a second crossover event occurring nearby a first may be due to mechanical stress imposed on chromatids due to presence of the fi chiasma may be due to the presence of the enzymes that participate in breaking and rejo strands during crossing overwhich Inh such events occurring nearby 2 Formula 1C For our example 1 0804 0196 or 196 interference IF I 10 or 100 Complete interference and a crossover at one place completely prevents adjacen crossovers No double recombinant phenotypes occur 10 gt gt 0 Positive interference fewer DCOobs than Most frequent type of interference 0 Negative interference more DCOobs than DCO exp Theoretically possible but rarely if ever observed Unequal Crossover Probability Different regions ofthe chromosome experience different recombination frequencies relative to physical distance Centromeres and Telomeres ends of chromosomes with fewer crossovers herefore recombination map does not represent re ative physI 39cal distance between genes only the relative ORDER of genes Thuqunmwannabmcmmnmnnvm Lane minimummm rm Wimmmm MECHANISM OF CROSSING OVER Does a physical exchange of genes between omolo ous chromosomes actually occur during crossing over 1Genetic re ombi 39 2 Presence fch 3 Demonstrated Curt 39 gests so smata suggests it does taneously b a simul sophil Harriet Creighton and Barbara McClintock in maize Creighton and McClintock 1931 e 39 39 Used mutant chromosome Mutant chrorn heter chr trans 9 in maize osome kno of o matin atone end anda location at the other end WK romosome its carried two mutant genes Kernel color c colored c colorless Kernel texture WX sta chywx waxy tcross with the heterozygote F2 progeny phenotypes Coloredstarchy Recombinant I Colorless Performed ates cwx w oloredwaxy Parental Colorlessstarchy Parental c waxy Reconl 39 w E l Evidence of a physical exchange pr kno th re essive kernels ould indicate that exchange took place 2 Cytologicall esence o 9 that h b but lacked thetranslocation e r fchromos ad the or had f t t anslocation but lacked the knob W The cytological evidence unequivocally demonstrated that a physical exchange m cur between the homologues PROKARYOTIC GENE REGULATION Genetics BIOL 358 Lecture 24 18 April 2006 REGULATION OF GENE EXPRESSION Why have gene control 1 Many more genes exist than products are needed at any one tim gene control functions to turn on and off genes as their products are needed saves energytranscription and translation are energetically expensive Response to the Environment 2 Prokaryotes need to adjust the enzymatic machinery of their cell to the immediate physical and nutritional environment bacterial cells garner all of their nutrition from 39ronment in i hey live are directly exposed to environmental changes need to turn specific genes on an off in response to changes in their environment Most frequent type of control 1 G Differential Transcription nes are only transcribed into mRNA when ir products ar needed are only produced when their 39 nvironment building blocks ofcertain products like amino acids are my 39 39 low within the cell kvelsitw agenmpussiun39 Ngulaledln Pmkaryutes lam s lltlvamulp mRNA new un lion mime EYEbull Types of Gene Expression 1 Constitutive expression transcribed continuously of housekeeping genes whose 39 nd in the cell gene is characteristic prod c e In constantdenia 2 Inducible expression or induction transcription is turned ON in the presence of an inducer in the environment such as presence ora substrate which will induce transcription ofthe gene that codes for the enxym e that uses that substrate Types of Gene Expression 3 Repressible expression or repression transcription is turned OFF by the presence of the productthatthe gene or genes produce called the corepressor such as when there is plenty ora particular amino acid its presencew l epres r S the genes whose enzymes are responsible for its production memmmmm lndunionnfenzymesyn lzs WW Wm M quot min WT POSI ive andor Negative Control 1 Regulator genes of other these gene producks can exen either positive control mechanisms or negative control mechanisms 2 Positive control mechanismsActivators transcription does not begin unless a regulator molecule directly skim ulates the initiation of ranscription 3 Negative control mechanismsRepressors gene expression continues until it is snm off by some type of regulator molecule WV e i i i Henressible System quots mu i mewlmnlvI i i 239 mirrmmumi Operons Coordinated Multigenic egulation operun 39 39 E w A e premeter lnr regulator gene Regulamr gene Promoter in E Structurai genes 3 o I Transtription Translation I Effector molemle R Repressnr O linuuur or co revressor mommumuien m mmmm n quot rmerquot quotnew mimlimlnpnmi rug 4239 19 Ill WWWW omi oquoto 8quot wquot m m 33 W MMHWUW INDUCTION OF LACTOSE METABOLISM IN E coli The Lac Operon Jacob and Monod 1961 1 Contains genes that code for enzymes in the 2 These genes are only transcribed in the presence of the inducer lactose 3 The induction of this operon can be under either negative or positive cont negative control through repressor Iacl protein positive control through CAPcAMP binding Organization of the lac operon 3 structural genes 1 IacZ codes for Bgalactosidase enzyme that breaks lactose into glucose and galactose 2 IacYcodes for Bgalactoside permease enzyme that facilitates diffusion of lactose into cell 3 IacA codes for Bgalactoside transacetylase enzyme whose function is unknown Regulatory Regions 1 Promoter 35 and 10 promoter sequences for initiation of transcription of lac operon genes CAP binding site binds CAP protein cyclicAMP complex also required for initiation of transcription 2 Operator Adjacent DNA sequence to promoter region Contains a protein binding domain where repressor l protein can bind 3 Repressor gene Iacl Iaclprotein produced by this gene is a regulator molecule that exerts negative control turns transcription off by binding to the operator region has its own promoter so can be transcribed independently of rest of operon E rcii hmmumme pro purf argD ntguiaior gene m39 c39nclno genez genev panel c uN39IMMW Wmmu39 I um 35m m m uu 3035 Iatveptesser Bgnlzdnsldixe B gallcwside gllinoside permease transaoeiyiase Negative control of lac operon 1 In absence of lactose iaci repressor binds to operator interferes with binding of RNA polymerase to lt15 promoter transcription is turned OFF negative control because transcription is turned off by presence of repressor Iacl Equot n presence of lactose iactose binds to receptor site on repressor Iacl protein cnanges binding site on repressor causing it to be released from operator now RNA polymerase can bind to lt15 promoter se uence transcription of operon is turned 0N iactose is the inducer because itrunctions to override negative control and induces transcription when iactose is all broken down notning bindsto ressor I so itwiii again turn OFF transcription Pos ve control of lac operon 1 Lactose alone is not enough to completely tu on transcriptio 2 Cataholite Activator Protein CAP exerts posl ve control lac promoter contains an additional binding site the CAP site CAP plus a small effector molecule cyclic AMP CAMP must be bound at this site binding of CAPcAMP is necessary to facilitate tight binding ofRNA polymeraseto as pr oter CAP site must be bound in orderrortranscription to beinduced If Adenylcyclase inncu a T 0 H k 07 on on 0H P uvluwv WWWquot ownquot lgquot uni r mm m Catabolite Repression gucose is preferred carbon source forthe cell intracellular concentration of cAMP is sensitive to lu se concentrations high glucose concentration prevents activation of adenylcyclase the enzyme that catalyzes conversion of ATP to cAMP high glucose concentrations result in low concentration of cAMP in cell no cAMP to bind to CAP protein CAP binding site is empty in promoter region transcription is turned OFF allows cell to utilize glucose first as a carbon source before lactose TRYPTOPHAN OPERON IN E coli Tryptophan synthesis in E coli Yanofsky 1 Controlled by operon with 5 structural genes encode enzymes that convert chorismic acid to tryptophan 2 Expression of trp operon is regulated at two levels repression which controls i ation of transcription attenuation which exerts ne a ve control causing premature termination of transciption Organization of the trp operon 1 5 structural genes trpE trpD trpC trpB trpA all code for enzymes in pathway of synthesis of tryptophan Repressor trpR regulatory gene expression is coded by separate promoter gene codes for repressor trpR protein repressor trpR and corepressor tryptophan bind to operator sequence and greatly decrease rate of transcription of operon N Organization of the trp operon 3 Promoter for operon contains binding sequenceror RNA polymerase doesn39t have a CAP site Operat r P nce activation of attenuator com pieteiy naits transcription n9 mmquot m n mm Repression of the trp operon In presence of some tryptophan tryptophan is a corepressor binds to repressor trpR protein tryptophanltrpR complex bind to operator sequence interrerewitn tight binding of RNA polym eraseto promoter greatiy reduce the rate ortranscription but do not nait it com pietei 2 In absence of tryptophan transcription of operon proceeds Attenuation Negative control of trp ape 1 Occurs by control of the terminatio t sc 39 near the end of the trpL leader e c rich halts transcription omeNA priorto rst structural ene orly activated in presence ofhigh levels of tryptophan in the cell Regulatory componenis of the mat region Lend Ptang an MD rt 539 pwwwltmnummmwcucnucwccuumm van1 may V mmw ml ludu 5 is2 39 Im in an no Raglan wcxwwmuumcmucwhum harm mm r ea t v rm Activation of the Attenuator script of attenuator sequence has alternate second ry structures in hairpin loop and terminate transcription by blocking RNA polymerase the other structure does not form hairpin loop tryptophan concentrations determine which structure is expressed absence of tryptophan no hairpin loop stru rewill form and transcription will proceed 2 in presence of tryptophan the hairpin loop structure will be form ed halting transcription Alluvial uzondaryslmnuveslarmed hmpr xransnipl ugsmma a pmd rm Ilplmnrlevmmulun u n h inmn mmquot Substrate 335 Immedim End mam ugtgtgtgtamp Enzyme End producl binding site subsum is nut Enzyme after an neriunn o s on End gram m mnfomln on bnund in e eunr molecule binding me GENE INTERACTION AND QUANTITATIVE INHERITANCE Genetics BIOL 358 Lecture 8 2 February 2006 Multiple Phenotypic Effects Pleiotropy Single gene affects the phenotype of more than one trait Example singed bristle mutation in Drosophila sn lsn Normal bristles fertile eggs sn lsn Normal bristles fertile eggs snlsn Short bristles infertile eggs GENOTYPE X ENVIRONMENT INTERACTIONS Phenotype is determined by interactions among alleles internal environment at single loci Dominance relations at different loci Epistasis Interactions of genes with external physical environment also influence trait expresssion Penetrance Not all individuals with a given genotype express its phenotype Degree of penetrance percentage of individuals with a given genotype who exhibit the trait lt100 incomplete penetrance Incomplete Penetrance Polydactyly A Dominant Trait with Incomplete A Penetrance Expressivity Not all individuals with a given press the phenotype in exactly the same w The extent to which the phenotype is expressed 39n an ind39 i ual of a given genotype varies with genetic and environmental background Variable Expressivity in Lobe Eye Mutation of Drosophila 7amp2 bx Eyeless Wild Type INHERITANCE OF QUANTITATIVE TRAITS Continuous variation of a trait Range of phenotypes in the population is continuously or normally distributed phenotypes do not occur in distinct classes examples include height weight skin color In umans crop yield e g and milk production diseasedrought tolerance etc in agriculture Quantitative Traits we call such traits quantitative because they have measurable values cm 9 kg etc 2m Height ofmale humans is continuously distributed Fewest individuals at extremes tails Frequency 8 Most individuals at the average mean VaIiance is the spread l Sigma 39 OfValues around the an Height in centimeters mmnnmmmm 5 W Polygenic Inheritance Expression of a quantitative trait is controlled by polygenic inheritance contribution of alleles at dozens to hundreds of loci controls the expression of the trait Alleles at different loci are inherited in an additive fashion each contributes the same small proportion to the phenotype phenotypic value is the sum of the effects of all alleles at all loci Example from snail shell diameter NA1 NA2 A2A2 B131 6 5 4 B132 5 4 3 3232 4 3 2 With the addition of every 1 allele 1 mm is added to the shell diameter Additive inheritance pattern produces considerable phenotypic variation in a trait each allele itself has a small effect but when all alleles are added over many loci a continuous range of variation is produced Quantitative genetic variation is also influenced by the environment diet soil nutrients maternal effects dominance temperature epistasis lfw dist Detection of quantitative inherita F2 No shortestt Not a 121 as would be expected from single ce gene codomina nce Long x ho All intermediate nguish from pattern of domIn e cross the F1s however now we can inguish nnal distribution of heights from o tallest Tobacco Flower Length Variation due to Environment r 39 4 Variation dueto i7 quot Environment V 39 1 Variation due to Genotype and Environment Variation due to 4 Environment inly Ma PHENOTYPIC VARIANCE Total Variance I 15 i in Population i E VT VP 5 I E l Both Genetic and a l Environmental c E m ts to Phenotype l l l l l mm i l V V V 7mm 1 T G E HERITABILITY u tIon due to genetic differences among individuals The proportion ofthe total phenotypic variance in the pop la39 Broadsense Heritability H2 VONT ldentical environments all differences are due to Genetics ldentical genotypes all differences are due to the Environment Narrowsense Heritability h2 The degree of resemblance between relatives due to shared genes Due only to additive effects of genes VA Genetic variance due to additive dominance and epistatic effects of genes VG vA vD v 2 h VANT Regression of offspring phenotypes on parental p enotypes Allows predictions of selected changes in traits Response to Selection R hZS SSH S selection differential h2 narrowsense Heritability mm R response to selection MW tun cinplgs Egg prrxluumn in pnulrn 39lEul length in 1mm Bul Dmrarlir39ln Mun l xi n mumwmmmOunnumnu Jrllrmnlml n H Lu nnn Hmliin Mean pupa weight mkrngrams m 3a 44 so an M an so umnanm on Dish Generations of selection


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