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by: Elna Nader
Elna Nader
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This 26 page Class Notes was uploaded by Elna Nader on Saturday September 12, 2015. The Class Notes belongs to GENE 3200 at University of Georgia taught by Staff in Fall. Since its upload, it has received 77 views. For similar materials see /class/202264/gene-3200-university-of-georgia in Genetics (Graduate Group) at University of Georgia.

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Date Created: 09/12/15
Ch 16 PROKARYOTIC GENE REGULATION GENE REGULATION IN GENERAL Some genes encode proteins always needed constitutively expressed Some products only needed at certain times or in certain environments regulation needed Can regulate transcription initiation elongation or termination RNA stability translation IET Protein stability or protein activity Allosteric feedback CTP UTP feedback on aspartic transcarbamylase Transcription regulation often involves control of initiation sigma factor expression but too few SF s so need more Lac Z Bgalactosidase Lac Y permease Lac A transacetylase all usually OFF Glucose repressor Lactose inducer Lac I DOMINANT to Lac 1 unable to bind DNA called Lac I REPRESSOR acts in trans When Lac I Lac I partial diploids made constitutive mutants isolated Oc Acts in cis Lac Is superrepressor unable to bind inducer DOMINANT to 1 and 1 KO promoter no expression of structural genes ZYA Lac I tetramer makes loop ofDNA when bound Lac 0 operator 23 bp binding site inverted repeat cAMP binds to CAP catabolite activator protein TERMS Inducible usually OFF turned ON in some circumstances Repressible usually ON turned OFF in some circumstances Positive regulator binding increases DNA expression Negative regulator binding decreases DNA expression Footprinting digest with DNase I gel electrophoresis autoradiography and look for gaps to find protein binding site Arabinose operon Induction and repression with same regulator araC is repressor in absence of arabinose araC binds to arall and araO forms uturn If coinducer present arabinose araC is inducer and forms dimer araC binds to araIl and ara12 Trp Operon Repressible under negative control trpR encodes repressor Corepressor is tryptophan causes allosteric change in TrpR and allows it to bind Early transcription termination trpR mutants still show some regulation by trp operon can give long and short transcripts low and hi trp respectively REQUIRES translation and transcription to cooccur and speed of translation to affect early transcription termination Uses leader sequence Contains four regions of complementary sequence two trp codons upstream of region 1 Regions 3 amp 4 can pair to form termination stem loop 1 amp 2 pair as well Formation of 23 stem loop prevents 34 loop transcription continues into structural genes If trp present charged tRNAtrp readily available allows ribosome to move quickly and 34 loop formed If trp absent tRNAtrp not available ribosome stalls and no 34 loop forms 23 loop instead ATTENUATION CANNOT OCCUR IN EUKARYOTES CH 17 EUKARYOTIC GENE REGULATION Control of RNA stability and processing more important in eukaryotes regulatory state can be passed to daughter cells Regulatory protein highly specific chromatin structure has role differential splicing polyadenylation and RNA transport occur in eukaryotes Operon coordinate control rare in eukaryotes miRNA s in both now Prokaryotes have 1 RNA pol euk have 3 RNA pol I 285 185 585 rRNA Pol II mRNA miRNA Pol III tRNA 5s rRNA snRNA Proximal promoter TATA box and CCAAT box bind basal transcription factors Distal enhancers bind activator and repressor TF s TATA B OX Binds TBP binding protein causes bend in DNA Has TAF s TBP associated factors TBP and TAF binding facilitates RNA pol II initiation and low levels of expression TATA box fround 35 bp upstream Consensus sequence TATAATAAT Common element aka 10 20 transcribed genes CAAT B OX 80 bp upstream GC box GGGCGG Octamer ATTTGCAT ENHANCER ELEMENTS Can bind multiple activators or repressors to form enhanceosome Can be distant up stream or downstream too Activators can interact with basal TF s to increase transcription 2 domains transcription activator domain to TF s and DNA binding domain Repressors Competition competition for enhancer between repressor and activator Quenching in type I repressor binds to DNA binding region ofactivator In type II repressor binds to activation domain ofactivator COMMON STRUCTURAL MOTIFS Homeodomain highly conserved domain 0f60 AA s found in TF s similar to HTH domains in bacterial TF s important in development TF s DNA binding POU domain DNA binding names for 3 TF s hybrid domain homeo and POU specific domain important in development Zinc finger domain DNA binding spacings around cysteine and histiding that allow zinc binding coordinates sequences into fingers which interdigitate into major groove of DNA helix Steroid TF s RNA pol III TF TFIIIA Steroid receptor TF cannot bind enhancer without SH steroid hormone 7 Winged helix domain DNA binding a helix and B sheets around helices forkhead TF s Leucine Zipper domain protein binding protein dimerization region leu spaced 7 AA s apart within ahelix regions DNA binding bottom protein binding top present in many TF s HelixLoopHelix protein binding involved in dimerization basic AA s on N terminus side of HLH domain where interaction with DNA sequence occurs HLH TF s that lack BASIC AA s are REPRESSORS that bind to the activator HLH proteins yo Cut that business off MYC MAX SYSTEM Myc is protooncogene expressed in cancers has DNA binding HLH and leucine zipper domains activates transcription Max has DNA binding HLH and leucine zippers no trans activation Max can form homodimers and bind DNA Myc cannot form homodimers Myc and Max can form heterodimer GAL REGULATION EXAMPLE yeast Several genes needed to metabolize galactose GAL1 7 10 Galactose induces expression of GAL structural genes regulated by ONE enhancer 3 Regulatory proteins GAL 4 is transcriptional activator GAL 80 is a corepressor GAL 3 and GAL 1 regulate GAL 80 When galactose absent GAL 80 is a quencher of GAL 4 blocks activation domain GAL1 and GAL3 bind galactose and bind to GAL 80 freeing activation domain leading to high levels of gene expression Once gal depleted GAL1 and GAL3 no longer bind 80 so it blocks GAL 4 activation INSULATORS Prevent regulation of nontarget genes Protein CTCF CCCTC binding factor binds to DNA sequences termed insulators and prevents adjacent gene whatever CTCF has 11 zinc fingers 15000 CTCF binding sites in human genome CHROMATIN EFFECTS Chromatin reduces binding to basal factor proteins and RNA pol II to very low levels Nterminal tails of H3 and H4 can be modified by methylation acetylation phosphorylation and ubiquitination Acetylation oflysine residue 14 on H3 is a marker of active transcription DNA methylation at C5 of cytosine in a CpG dinucleotide silences gene Heterochromatin is hyp ermethylated silences transcription Centromeres telomeres inactive X chr also silenced examples Chromatin remodeling nuclosomes repositionedremoved by remodeling complexes DNA at promoters and enhancers more accessible to TF s Example SWISNF remodeling complex uses ATP hydrolysis to alter nucleosome positioning INHERITANCE OF REGULATION Epigenetic changes changes in chromatin histone modifications or methylation state that are inherited across cell generations DNA SEQUENCE NOT ALTERED Genomic imprinting expression of a gene depends on maternal or paternal inheritance a few mammalian insect and plant genes Paternal imprinting maternal gene expressed paternal silent Maternal imprinting paternal gene expressed maternal silent Imprinting involves methylation an insulator that binds CTCF involved in reciprocal imprinting of Igf2 and H19 genes Methylating insulator means other gene gets expressed usually insulated Epigenetic imprints remain throughout life but in germ cells epigenetic imprints reset at each generation Meiosis imprints erased and new ones are set DISEASES OF IMPRINTING Small deletions in chromosome 15 PraiderWilli syndrome occurs when deletion inherited from father Angelman syndrome when inherited from mother Paternal imprinting receive genedeletion from mother and vice versa RNA STABILITY INTERFERENCE RNAi involves miRNA and siRNA act through same pathway Short around 2025 bases miRNA transcribed by RNA pol II capped and polyA d most contain no ORF Processing of miRNA Drosha and Argonaut recognize premiRNA and crops out stem loop miRNA then exported from nucleus picked up by Dicer dicer cuts premiRNA into miRNAmiRNA duplex and releases it RISC degrades one strand of duplex miRNA Modes ofinterference mRNA cleavage perfect match to miRNA miRISC cleaves RNases eat cleavage products translational repression miRNA not perfect match represses ribosome movement when quotboundquot CH 20 DEVELOPMENTAL GENETICS All ab out cell fate and cell differentiation Model organisms wellcharacterized easy to manipulate lots of mutants in stock genomes sequenced Yeast eukaryotic microbe enormous genetic tools development limited Arabidopsis thaliana short generation time small plant small genome Round worm limited number of cells transparent body Fruit y short generation time excellent genetic tools Mouse vertebrate short generation time small size Gain of function mutations usually dominant to WT Dominantnegative mutations usually dominant to WT deehhrrr Conditional mutations dominant or recessive eg temperature sensitive alleles Forward genetics phenotype to gene find 3 headed frog and find gene Reverse genetics gene to phenotype find random gene and make 3 headed frog Body plan cell fate established by toolkit genes Many are TF s and signal molecules Defined spatial and temporal expression patterns in embryos Highly conserved only small fraction of total genes Body plan established during one day of embryonic development by regulation of expression of small number of toolkit genes in drosophilia Arranged along AP axis and DV axis simple plan 14 segments total 3 head 3 thorax 8 abdominal Segmentation genes subdivide body into an array ofidentical body segments Homeotic genes assign a unique identity to each body segment Most mRNAs in the early embryos were transcribed in the oocyte before fertilization maternal genes Transcription ofmost zygotic genes does not begin until synctial blastoderm stage Multinucleate syncytium syncytial blastoderm cellular blastoderm SUMMARY LEADING TO SEGMENTATION IN DROSOPHILIA Morphogenic gradients along AP axis maternal effect genes Gradients make gap genes expressed only in certain broad regions of embryo Gap genes activate pairrule genes in series ofseven stripes gap genes dehrr Pair rule gene products restrict the expression of segment polarity genes to a series of 14 stripes one per segment pairrule genes Embryo is divided into 14 segmentsized units segment polarity genes Early bicoid gradient maternally expressed bicoid mRNA localizes at anterior pole of early embryo after fertilization gradient of bicoid protein forms along AP axis Bicoid hunchback caudal and nanos transcribed in egg but not translated until after fertilization see graph Bcd mRNA forms anterior to posterior gradient Nos mRNA forms a posterior to anterior gradient Proteins the same throughout embryo can have translation repression by other non commonthroughoutembryo proteins GAP GENES Nine gap genes total hunchback kruppel knirps giant are examples First ZYGOTIC segmentation genes to be expressed Transcription determined by concentration gradients of maternal effect proteins promoters of gap gens have binding sites with different affinities for maternal transcription factors eg some gap genes have low affinity binding for Bcd so are activated only in anterior region where Bcd levels are highest Gap genes encode transcription factors that regulate expression of other gap genes and pairrule genes PAIR RULE GENES Primary pairrule genes transcription depends on products of maternal genes and zygotic gap genes Example is Eve evenskipped Secondary pair rule genes Transcription depends on products of other pairrule genes such as Eve Example is fushitarazu th Each pairrule gene expressed in stripes in preblastoderm and blastoderm stages Two segment periodicity one stripe of expression for every two segments Eve has a regulatory region for each stripe Within 2 stripe regulatory region multiple binding sites for four transcriptional regulators Bcd and Hb activate Eve expression Kr and Gt repress Eve transcription Eve shows up where all four regulators in optimal combination SEGMENT IDENTITY GENES Pair rule genes activate them occur after cellularization of embryo after syncytial stages Simple diffusion ofTF s play no further role in development Wingless and engrailed maintain each other s expression Engrailed producing cells produce frizzled and hedgehog Wingless producing cells produce patch Hedgehog activates wingless production in cells expressing patch Wingless activates engrailed production in cells expressing frizzled Mutations in segment polarity genes cause deletion ofpart of each segment and its replacement by mirror image of different parts ofnext segment Hedgehog and vvingless are examples ofparacrine signaling ligand is secreted looks cyclic to me 9 gap genes 8 pair rule genes 17 segment polarity genes CELL DIFFERENTIATION Once cell fate determined particular cells differentiate into appropriate types Many of the genes involved are homeotichomeobox genes Homeotic genes Expression controlled by segment identity genes gap pr seg polarity GENE book notes CHAPTER 1 genes DNA regions that encode proteins Chromosomes DNA molecules carrying the genes Genome entire collection of chromosomes in each cell of an organism CHAPTER 2 Genes basic units of biological information 0 Region of DNA that encodes a specific protein or a particular type of RNA Heredity way genes transmit biochemical anatomical and behavioral traits from parents to offspring Genetics science ofheredity Themes from Mendel 0 Variation alternative forms ofa trait is widespread in nature 0 Observable variation is needed for following genes 0 Variation is not by chance but by genetic laws 0 Heredity laws apply equally to all sexually reproducing organisms To study genetics you need groups of organisms good records development of theoretical idea of origin Artificial selection purposeful control over mating by choice ofparents for the next generation Blended inheritance idea that parental traits become mixed and forever changed in the offspring Selffertilizationselfing both egg and pollen come from the same plant Crossfertilize Reasons pea plants were good examples 0 Organisms had both male and female organs in the same ower 0 Each generation produced large numbers ofindividuals within a short growing season Things Mendel did differently than people before him 0 He picked the pea plant He examined the inheritance of clearcut alternative forms of particular traits no intermediate forms I Discrete traits o Opposite of continuous traits Collected and perpetuated lines ofpeas that bred true Crossbred pairs to make hybrids Carefully controlled his matings Worked with large numbers of plants counted all offspring used numerical analysis and compared results with predictions from his models quantitative approach Purebreeding lines matings produce offspring carrying specific parental traits that remain constant from generation to generation Hybrids offspring of genetically different parents O 0000 Reciprocal crosses reversing traits of the male and female parents to control whether a particular trait was transmitted via the egg cell within the ovule or via a sperm cell within the pollen 0 Similar progeny in these tests disproves idea that one parent contributes more to the next generation but they contribute equally to inheritance Parental generation P First filial F1 progeny of the P generation Monohybrid crosses using hybrids for a single trait o Allowing F1 to selffertilize Second filial F2 progeny of the F1 generation 0 Creates a ratio of 31 Reappearance of the recessive trait disproves blending o If blending occurred one of the traits would be absent from both the F1 and F2 generations Genes units of inheritance from parents Alleles alternative forms of a single gene 0 Ex the gene for pea color has yellow and green alleles Monohybrids individuals having two different alleles for a single trait 0 Ex Mendel s hybrids of the F1 generation ofpea plants had a dominant allele and a recessive allele Recap of Mendel s experimentsreasoning Mendel started with two plants that were true breeding for the same trait but had different phenotypes For example yellow peas and green peas This means that one was homozygous dominant and one was homozygous recessive When the parents were bred the F1 all had the same phenotype of the dominant phenotype but they were heterozygous dominant Mendel proved this with the F2 generation because the recessive trait that was absent in the F1 generation reappeared in the F2 generation in a 31 ratio Some of the plants were again homozygous recessive and showed the trait that had been missing The other plants were either homozygous dominant or heterozygous and showed the dominant trait The law of segregation explains how genes are transmitted Gametes specialized cells that carry genes between generations 0 Eggs and sperm Zygote fertilized egg Law of segregation The two alleles for each trait separate segregate during gamete formation and then unite at random one from each parent at fertilization Segregation equal segregation in which one allele and only one allele of each gene goes to each gamete Product rule the probability of two or more independent events occurring together is the product of the probabilities that each event will occur by itself Sum rule the probability of either of two such mutually exclusive events occurring is the sum of their individual probabilities The segregation of dominant and recessive alleles during gamete formation and their random union at fertilization could explain 31 ratios from allowing hybrids to self fertilize Testcrosses establish genotype Phenotype observable characteristic Genotype actual pair of alleles present in an individual Homozygous two copies of the gene that determine a particular trait are the same 0 Ex YY or yy Heterozygous two different alleles for a trait or hybrid for a trait o Heterozygote phenotype determines what allele is dominant Testcross mating in which an individual showing the dominant phenotype is crossed with an individual expressing the recessive phenotype o If all offspring show the dominant phenotype then the unknown parent was a homozygous dominant o If some of the progeny shows the recessive trait and some show the dominant trait then the parent was heterozygous Dihybrid heterozygous for two genes at the same time Dihybrid cross produces parental types and recombinant types 0 Expected phenotypic ratio is 9331 in an F2 generation Recombinant types new phenotypic combinations that do not look like the parental types Independent assortment Law ofindependent assortment During gamete formation different pairs of alleles segregate independently of each other Multihybrid crosses matings between the F1 progeny of truebreeding parents that differed in three or more traits Page 27 good examples ofusing probability rules The larger the number of trials the more likely ratios will turn out as expected Two basic principles of gene transmission 0 Segregation of each gene s alleles during gamete formation followed by a random union at fertilization 0 Independent assortment of the alleles for two or more different genes Pedigree family history showing a family s relevant genetic features 0 Page 30 pedigree symbols Consanguineous mating mating between relatives Vertical pattern of inheritance tracing back through relatives of an affected individual there will be at least one affected person in each generation continuous line of family members with the disease 0 Evidence ofa dominant allele causing the trait 0 Ex Huntington s disease Most singlegene traits in humans are recessive Horizontal pattern of inheritance a trait is not seen in parents grandparents etc but several brothers or sisters in a generation may be affected 0 Indicates that the trait is recessive 0 Ex cystic fibrosis Carriers someone who has a dominant normal allele that masks the effects of the recessive abnormal one Dominant traits in a pedigree o Affected children have at least one affected parent 0 Show vertical pattern of inheritance trait shows up in every generation o 2 affected parents can produce unaffected children ifboth parents are heterozygotes Recessive traits in a pedigree o Affected individuals can be children of 2 unaffected carriers especially in consanguineous marriages o All children of the two affected parents should be affected 0 Rare recessive traits show a horizontal pattern ofinheritance the trait appears among members of one generation and is not seen in earlier generations 0 Recessive traits can show vertical pattern ofinheritance if the trait is common in the population CHAPTER 3 example of variation lentils Mendel s laws show a trait determined by a single gene with alternative alleles o Clearcut dominance and recessive Multifactorial traits determined by 2 or more factors 0 Multiple genes interacting with each other 0 One or more genes interacting with the environment 0 EX seed coat oflentils Singlegene inheritance o Allele pairs show deviations from complete dominance and recessiveness 0 Different gene forms are not limited to 2 alleles 0 One gene can determine more than one traite Multifactorial inheritance o Phenotype comes from I The interaction ofmore than 1 gene I The interaction of genes with the environment Complete dominance if F1 offspring show only one phenotype ofa parent then that allele is dominant 0 Like Mendel s experiments Incomplete dominance F1 hybrid does not resemble either purebleeding parent 0 EX late blooming and earlyblooming pea plants 0 Hybrids often have a phenotype that is intermediate between the purebreeding parents Neither parental allele is dominant or recessive because they both contribute to F1 phenotype 0 Ex ower color snapdragons If hybrid F1 s selfpollinate the F2 progeny shows a genotypic AND phenotypic ratio of 121 I Because heterozygotes don t look like either parent 0 One allele does not have enough function on its own to produce the full effect of the parent Codominant both traits of the parents show up equally in F1 hybrids 0 Each genotype has its own phenotype so ratio of F2 is 121 I 1 parent 2 F1 1 parent 0 O 0 ex distinguishing molecules of red blood cells I IA and IB that can be homozygous for one or heterozygous and have both but determines what surface sugars a red blood cell with show 0 Molecular level phenotypes are often determined by codominant alleles Summary of dominance stuff 0 Complete dominance F1 progeny look like one of the truebreeding parents 0 Incomplete dominance hybrids resemble neither parent and do not show either true breeding trait o Codominance phenotypes ofboth purebreeding lines are shown in the F1 hybrid o Dominance is determined by the phenotype of the F1 generation 0 F2 ratios I 31 complete dominance I 121 incomplete dominance or codominance The phenotypic dominance results have nothing to do with the allele segregation There can be more than two alleles for a gene 0 EX human blood types I 3 alleles IA IB andi I Type A IAIA IAi I Type B IBIB IBi I Type AB IAIB I Type O ii I Each person only gets two alleles so there are 6 possible genotypes Alleles are not inherently recessive or dominant but can only be classified when compared to other alleles New alleles come from mutations Mutations chance alterations of the genetic material that arise spontaneously in nature 0 Make it possible to follow gene transmission Alleles that give a survivalreproductive advantage to individuals so they can make more offspring are likely to increase in the population Allele frequency percentage of the total number of each allele out of the gene copies 0 Number ofindividuals X 2 number of copies ofa gene in a given population I Because each organism carries two copies of every gene Wildtype allele allele whose frequency is greater than 1 Mutant allele allele whose frequency is less than 1 0 Ex mouse fur has 14 mutant alleles that can change color Monomorphic a gene with only one wildtype allele 0 EX A in mice is the only wildtype allele for the agouti fur coat gene and has a frequency greater than 99 Polymorphic genes with more than one wildtype allele 0 EX ABO blood system IA IB and i are all wildtype alleles Pleiotropy phenomenon ofa single gene determining a number of distinct and seemingly unrelated characteristics Lethal alleles 0 Ex mouse light yellow coat color Ay I Yellow mice are heterozygous for the agouti gene and Ay mutation I Yellow has to be dominant to agouti even though agouti was dominant to at and a mutations for black color I Yellow yellow 2 yellow1 agouti 0 There is never a purebreeding yellow because two Ay genes is lethal Recessive lethal allele an allele that negatively affects the survival of a homozygote 0 Ex Ay in mouse fur color proves lethal if offspring receives two copies 0 Some may allow offspring to live past birth and die later because of genetic consequences I EX TaySachs disease in humans 0 These alleles are usually not passed on because the affected person dies before having children I Stay in population because of heterozygotes o Lateonset diseases mean the gene can be passed on by either homozygotes or heterozygotes I EX Huntington disease Sickle cell as an example 0 Multiple alleles 400 mutant alleles o Pleiotropy sickled cells have a lot of problems 0 Recessive lethality many homozygotes of the recessive mutant allele die very early in life 0 Different dominance relations Two genes can interact to determine one trait 0 Ex lentils with seed coat color I Both dominant alleles brown I One dominant allele tan or gray depending on which gene I No dominant alleles green I 9 brown 3 gray 3 tan 1 green in F2 from pure breeding tan and gray I F1 is brown I Each genotypic class determines a particular phenotype Complementary gene action 0 EX white owers give a purple F1 I F2 selfing gives ratio of9 purple 7 white I Two genes work to create the purple ower so there has to be a dominant allele ofboth A B Epistasis gene interaction in which the effects of an allele at one gene hide the effects of alleles at another gene 0 Epistatic gene allele doing the masking o Hypostatic gene gene being masked 0 Ex coat color in Labrador retrievers I A dominant B becomes black I A recessive bb homozygote is brown I Dominant E has no effect I Recessive ee homozygote is gold and hides any brown or black I 9 black 3 brown 4 gold Recessive epistasis the allele causing the epistasis is recessive 0 Can be identified by F2 with ratio 0f934 0 Ex the requirement of an ee homozygote to hide the effects of another gene Dominant epistasis the dominant allele of one gene hides the effects of another gene 0 Has phenotypic ratios of1231 or 133 0 Ex summer squash I B is epistatic to any Aa genotype and produces white squase I bb is recessive to any Aa genotype 0 A with bb gives yellow squash 0 aa with bb gives green squash I 12 white 3 yellow 1 green 0 ex feather color of leghorn chickens I 13 white birds 3 colored birds I For color there has to be at least one A and no B I Bis epistatic to A Table 32 Gene interactions Heterogeneous trait mutation at any one ofa number of genes that gives rise to the same phenotype 0 Ex human deafness 50 different genes can be have mutant alleles to cause deafness I Complementation offspring receiving two mutations express the wild type phenotype o The original mutations affected two different genes and for both genes the normal allele from one parent can provide what the mutant allele of the other parent cannot 0 Wild type allele ifparents are deaf but children can hear I Noncomplementation the two mutations independently alter the same gene 0 Offspring receiving two recessive mutant alleles express the mutant phenotype o Homozygous Mutant alleles ifparents are deaf and children are deaf Complementation cannot be used if either mutation is dominant to the wild type Genetic variations on the theme of multifactorial traits 0 Genes can interact to generate novel phenotypes 0 Dominant alleles of two interacting genes can both be necessary for the production of a particular phenotype 0 One gene s alleles can mask effects of another s Mutant alleles at one of two or more different genes can result in the same phenotype With incomplete dominance gene interactions can produce a gradient of shades or phenotypes 0 Ex owers from white to purple Factors that may alter the phenotypic expressing of genotype modifier genes environment chance 0 Penetrance how many members ofa population with a particular genotype show the expected phenotype occurrence 0 Complete 100 or incomplete Expressivity degree or intensity with which a particular genotype is expressed in a phenotype seriousness 0 Variable or unvarying Modifier genes alter the phenotypes produced by the alleles of other genes and have a more subtle secondary effect 0 In uence length ofa mouse tail Temperature can have an effect on phenotype 0 Ex color pattern of the Siamese cat I Enzyme for melanin is only functional at cooler areas of the cat s body feet face tail Temperature can affect survivability 0 Ex fruit y becomes paralyzed at too high temperature and then dies Conditional lethal lethal only under certain conditions 0 Permissive range 0 Restrictive Phenocopy change in phenotype arising from environmental gents that mimics the effects ofa mutation in a gene Discontinuous trait Continuous trait Polygenic controlled by multiple genes 0 Always multifactorial CHAPTER 4 Chromosomes transmit genetic information Chromosome theory of inheritance idea that chromosomes are carriers of genes Development depends of the type of genetic material and how much is there Meiosis accounts for segregation and independent assortment of genes Gametes sperm and egg 0 Each gamete contributes half the genetic material in a baby Mitosis nuclear division followed by cell division that results in two daughter cells that contain the same number and type of chromosomes as the original parent cell Meiosis nuclear division that generates egg or sperm cells containing half the number of chromosomes found in other cells in the organism Haploid cells that carry only a single set of chromosomes 0 EX gametes o n number of chromosomes in a haploid cell Diploid cells carrying two matching sets of chromosomes 0 EX zygote 0 2n number of chromosomes in a diploid cell Drosophila 0 I1 4 0 2n 8 Humans o n 23 0 2n 46 chromosomes are most visible right before the nucleus divides metaphase o Chromosomes have duplicated and condensed from thin threads into compact rodlike structures Sister chromatids identical halves ofa chromosome Centromere connects sister chromatids Metacentric chromosomes centromere is more or less in the middle Acrocentric chromosomes centromere is very close to one end Homologous chromosomeshomologs chromosomes that match in size shape and banding 0 Carry the same set of genes but possibly different alleles Nonhomologous chromosomes carry completely unrelated sets of genetic information Karyotype arrangement of stained chromosomes in pairs of decreasing size Autosomes chromosomes in matching pairs that are not gametes Each species carries a distinctive diploid number of chromosomes 0 Number of chromosomes does not always correlate with complexity or size of organism SeX chromosomes provide basis of seX determination 0 One seX has matching pair and other is unmatched 0 Human females XX 0 Human males XY 0 Near 11 ratio of males and females XXX 0 In humans almost normal female because they can handle it better than ies o Drosophila lethal OY o Humans and ies lethal o X chromosome is needed because it has essential genes that are not found on other chromosomes XXY o Humans klinefelter male sterile tall thin I Shows that two X chromosomes in the presence ofa Y are insufficient to produce a female 0 Flies normal female XX 0 Humans and ies normal female XY 0 Humans and ies normal male XO 0 Flies sterile male 0 Humans turner female sterile webbed neck 0 Flies normal male o Humans normal or almost normal male Drosophila seX is mainly determined by the ratio ofX chromosomes to autosomes o The presence of the Y chromosome is not a huge factor 0 Female ratio is 11 0 Males ratio is 12 0 The Y chromosome is needed for male fertility Differences in human and drosophila seX determination 0 Genes they carry on their seX chromosomes are not identical 0 Strategies used to deal with presence of additional seX chromosomes are not the same Chromatin DNA and protein 0 Surrounded by nuclear envelope 0 Condenses into threads to become the chromosome Nucleoli chromatin involved in manufacturing ribosome s Chromatid each rod making up a chromosome Mitosis results in two daughter cells that are genetically identical to the parent cell Cell cycle pattern of cell growth followed by division Interphase period between division in the cell cycle G0 resting part of interphase when cells will not usually divide again 0 EX mature human brain cells Centrosome organizing center near the nuclear envelope where microtubules radiate out into the cytoplasm Centrioles core of each centrosome Prophase chromosomes gradually begin to differentiate to begin mitosis o Nucleoli begin to break down and disappear 0 Cell metabolism shuts down so the cell s energy can be in chromosome movements and cellular division 0 Centrosomes move apart and become clearly distinguishable as separate entities I Go to opposite ends of the nucleus by microtubules Prophase in meiosis 0 Critical events I Condensation of chromatin I Pairing of homologous chromosomes I Reciprocal exchange of genetic information o Page 98 very in depth description 0 Synaptonemal compleX protein structure that aligns the homologs so corresponding genetic regions of the chromosome pair are lined up 0 Bivalent synapsed chromosome pair with two chromosomes 0 Recombination nodules structures along the synaptonemal compleX that allow for exchange ofpartes between nonsisterchromatids o Chiasmata sites where crossingover occurred Checkpoints moments at which the cell evaluates the results of previous steps and allow for sequential coordination of cellcycle events Meiosis 1 round of chromosome replication and 2 rounds of nuclear division 0 Division I parent nucleus divides to form two daughter nuclei I Centromeres that join chromatids stay intact during the whole division I Reductional division the number of chromosomes is reduced to onehalf the normal diploid number 0 Division II each of the two daughter nuclei divides 4 nuclei I Equational division each daughter cell has the same number of chromosomes as the parental cell present at the beginning of this division 0 End result 4 daughter cells that each have 11 number of chromosomes Figure 413 page 96 meiosis Crossing over results in recombination 0 Each chromosome pair may no longer be completely maternal or paternal Nondisjunction homologs ofa chromosome pair do not segregate during meiosis I Variation caused by independent assortment increases with the number of chromosomes in the genome Aspects of meiosis that contribute to genetic diversity in a population 0 Matter of chance towards which pole paternal and maternal homologs of each bivalent migrate during meiosis I o Reshuf ing of genetic information through crossingover ensures a greater amount of genetic diversity in gametes Table 43 page 102 compares meiosis and mitosis Gametogenesis gamete formation Oogenesis egg formation in humans 0 Meiosis I secondary oocyte and polar body 0 Only one of the products ofa single meiosis serves as a female gamete 0 Primary oocytes stay stuck in meiosis and a woman is born with all the oocytes she will ever produce 0 Ovulation released oocyte continues through metaphase of meiosis II o Ifoocyte is fertilized completes meiosis II Spermatogenesis production of sperm 0 Spermatogonia germ cells in the male testes where spermatogenesis occurs 0 Meiosis produces 4 equivalent haploid spermatids that will become sperm Evidence to support the chromosome theory ofinheritance o Phenotype of sexual identity is associated with the inheritance ofparticular chromosomes 0 Events of mitosis meiosis and gametogenesis ensure a constant number of chromosomes in the somatic cells ofa species 0 Inheritance of genes corresponds with inheritance of chromosomes in every detail 0 Transmission ofparticular chromosomes coincides with transmission of specific traits other than seX determination Crisscross inheritance males get trait from mom and daughters get trait from dad Hemizygous diploid cells have half the number of alleles carried by the female on her two X chromosomes 0 EX male drosophila for eye color gene Table 45 Pedigree patterns study this Sexlimited traits affect a structure or process that is found in only one seX Sexinlfuenced traits show up in both sexes but expression of traits may differ between the two sexes because of hormonal differences 0 Ex pattern baldness CHAPTER 5 Genetic linkage genes that travel together Ways recombinant progeny can result o Recombination of genes on the same chromosome during gamete formation 0 Independent assortment of genes on nonhomologous chromosomes Farther apart genes are greater probability of separation through recombination Syntenic located on the same gene Dihybrid crosses with ratios different from 1111 of F1 gametes indicate that two genes are on the same chromosome 0 Drosophila example with mutant eye color and body color 0 Males looklike mother because phenotype re ects the genotype of the one X chromosome they got from her 0 If the Drosophila genes assort independently then the dihybrid F1 females should make 4 kinds of gametes with four combinations of genes on X I Ratio should be 1111 0 Half parental types and half recombinant types 0 Counting male progeny tells if the ration is right because they get X from mom Parental types allele combinations seen in the original mother or father of the P generation Recombinant types reshuf ing of allele combinations so the combination is not seen in the P generation parents of the F1 females Parental classes combinations originally present in the P generation Numbers of parental and recombinant F2 progeny are equal in independent assortment o Doubly heterozygous F1 will make equal numbers of all 4 gamete types Linked genes when the number of F2 progeny with parental genotypes exceeds the number ofF2 progeny with recombinant genotypes Recombinants will not be as likely with alleles that are very tightly coupled Linkage does not have to be really tight To test linkage in autosomal traits there has to be a testcross between a doubly heterozygous parent and a double homozygous recessive parent 0 Offspring phenotypes then indicate which genes came from which parents 0 prarental classes outnumber recombinant classes autosomal genes are linked Linkage means 0 Genes do not assort independently 0 Can be sex chromosomes or autosomes 0 Genes are transmitted together more than 50 of the time X linked genes means that the genes simply reside on the same chromosome Linkage is never 100 may just have to observe more individuals to find recombinants 0 Linkage percentage indicates how often parental combinations travel together Independently assorting genes half of F2 progeny are parental classes and other half of F2 are recombinant classes Linked genes parental classes outnumber recombinant classes in F2 Chisquare test probability test that measures the goodness of hit between observed and predicted results 0 How often does an experimentally observed deviation from predictions ofa particular hypothesis occur solely by chance Accounts for sample size of an experimental population Must be calculated with actual data and not percentages Hypothesis must be precise p value numerical probability that observed experimental results represent a chance deviation from hypothesis predicted values I high probability likely that hypothesis explains data deviation is considered insignificant I Low probability observed deviation from what is expected is significant hypothesis is rejected just comparing percentages or ratios does not determine if observed values are that different from predicted values Absolute numbers re ect size of the experiment 0 If hypothesis is correct observed percentages should match predicted values Null hypothesis a hypothesis being tested that will either be accepted or rejected 0 Rejected if chisquare test shows that observed data differ significantly from what is expected Using the chi square test page 128129 0 What is the total number of offspring events analyzed How many different classes of offspring events are there In each class what is the number of offspring events observed Percentage of offspring expected for a class from the null hypothesis X total number offspring analyzed in the experiment 0 Use equation for each class of offspring and sum together individual results I Xquot2 2number observedNumber expectedquot2number expected 0 Degrees of freedom measure of the number of independently varying parameters in the experiment I ofdof one less than the number of classes I df N1 o chi square and df are used to determine p value I p value probability that a deviation from the predicted numbers at least as large as what is observed in the experiment will occur by chance 0 can be determined from a table of critical xquot2 values 0 evaluate significance of p value I p value probability that the null hypothesis is true and the alternative hypothesis is wrong 0000 000 I pgt005 in more than 5 of experiments of the same size even if the null hypothesis were true the observed data could show a deviation from the predictions of the null hypothesis 0 data is not significant for rejecting the null hypothesis I plt005 consider data showing deviation significant and reject null hypothesis null hypothesis ofindependent assortment predicts that parentalsrecombinants o If rejected it is likely that genes are linked Effects on p value with more data 0 Normally rises if the null hypothesis of no linkages is correct 0 Tends to fall if there is linkage Having a larger sample size reduces the risk ofpercentage deviations just happening by chance Chiasmata regions in which nonsiterchromatids of homologous chromosomes cross over each other Physical markers cytologically visible abnormalities that make it possible to keep track of specific chromosome parts from one generation to the neXt Genetic markers genes that could serve as points of reference in determining whether particular progeny were the result of recombination Genetic combination is associated with actual reciprocal exchange of segments between homologous chromosomes during meiosis Terminalization shifting of the chiasmata from the original position to a chromosome end or telomere During recombination the further genes are from each other the more likely a crossover will occur between them Recombination frequencies near 50 suggest that two genes are on different chromosomes or they lie far apart on the same chromosome Locus chromosomal position assigned to a gene Mapping process of determining a locus Linked genes 0 Parentalsgtrecombinants RFlt50 o Linked genes must by syntenic and sufficiently close together on the same chromosome so they do not assort independently Unlinked genes 0 Parentals recombinants RF50 o Occurs either when genes are on different chromosomes or when they are sufficiently far apart on the same chromosome Recombination frequency RF measure of the distance separating two genes along a chromosome Two point crosses crosses tracing two genes at a time Problems with two point crosses o Difficult to determine gene order if some gene pairs lie close together 0 Actual distances in the map do not always add up 0 Ignores double crossovers that may occur in the large interval between the y and r genes Three point crosses are more accurate than two point crosses 0 Look at pairs of genes and determine parental combinations and recombinant combinations 0 Determine distance apart by taking adding up amount ofprogeny for whichever traits are different than parentals and being comparedtotal progeny X 100 I Use the ones that have the gene you are looking at switched o The genes with the longest distance will be on the outside and the other gene must be in the middle Discrepancy in distances comes from rare double crossovers Recombination takes place during prophase I Have to establish the order of genes to determine where crossover events happen Double crossovers generate gametes where the gene in the middle has recombined relative to the parental combinations for the genes at the end Majority of offspring will be parental combinations 2 groups in the middle of middle sizes are from single crossover 2 smallest group double crossover events 0 have to adjust recombination frequency by adding double crossovers twice to make distance more accurate A number for a smaller distance is more accurate because there is less likely to be a double crossover than if it was a larger distance Chromosomal interference fact that the number of observed double crossovers is less than the number expected if the two exchanges are independent events the occurrence of one crossover reduces the likelihood that another crossover will occur in an adjacent part of the chromosome 0 Crossovers no occurring independently 0 May make sure that each homologous pair has a crossover event because homologous chromosome pairs without crossovers tend to segregate wrong 0 Lowers number of crossovers on large chromosomes 0 Raises the probability of crossovers occurring on small chromosomes 0 Interference 1 coefficient of coincidence Coefficient of coincidence ratio between the actual frequency of double crossovers observed in an experiment and the number of double crossovers expected on the basis of independent probabilities 0 Frequency observedfrequency expected 0 Provides a quantitative measure of the amount ofinterference in different chromosomal intervals Interference 0 0 Frequency of observed double crossovers equals expectations 0 Crossovers in adjacent regions ofa chromosome happen independently of each other Interference 1 complete 0 Coefficient of coincidence 0 o No double crossovers occur in the experimental progeny because one exchange prevents another In gametes formed by double crossovers gene whose alleles have recombined relative to the parental configurations of the other two genes must be the one in the middle 0 Determined from a three point cross Order of genes from mapping correlates with the order of those genes on the DNA molecule ofa chromosome Actual physical distancesamount of DNA separating genes does not show a correspondence to genetic map distances 0 Too much possibility of crossovers Chromosomes will never recombine more than 50 of the time no matter how far apart they are Recombination is not uniform over the length ofa single chromosome 0 Hotspots favored sites of recombination o Recombination deserts areas often near centromeres where few crossovers ever happen Recombination rates vary between species 0 Can also vary between males and females of a species I EX no recombination in male Drosophila Linkage group genes chained together by linkage relationships 0 Genes are syntenic if they can be linked together Total genetic distance sum of many short distances between genes Pairwise crosses between genes located at the two ends will not make more than 50 recombinant progeny DNA marker piece of DNA of known size representing a specific locus and comes in identifiable variations Fungi are good for genetic analysis because they hold the 4 haploid products of each meiosis in an ascus sac o Ascospores haploid cells that can germinate and survive as viable haploid individuals that grow and perpetuate themselves with mitosis I Phenotype directly shows genotype 0 Haploid cells come in two mating types so sexual reproduction is possible I Cells of opposite mating types can fuse diploid zygote I Stress can make diploid cells go into meiosis O Tetrad assemblage of 4 ascospores in a single ascus after meiosis I Unordered 4 meiotic products spores are arranged at random within the ascus I Ordered 4 pairs or 8 haplospores arranged in a line I If conditions promote meiosis the two unlinked genes will assort independently to produce equal frequencies of two different kinds of tetrads 0 Parental ditype PD tetrad that contains 4 parental class haploid cells 0 Nonparentalditype NPD tetrad that forms when the parental classes recombine to form two reciprocal nonparental combinations of alleles contains 4 recombinant spores Tetratype T tetrad that appears when cells undergo meiosis and carries four kinds ofhaploid cells two parental class and two recombinant class 0 Result of crossover between one of the two genes and the centromere of the chromosome it is located on 0 Classification is only based on number of parental and recombinant spores in the ascus PD NPD means that two genes are unlinked 0 Genes are either on different chromosomes or really far apart on the same chromosome PD gtgtNPD means that two genes are linked Linkage emergence ofmore parental types than recombinants among the progeny of a doubly heterozygous parent Double crossover involving two chromatids both crossovers affect the same two chromatids o Produces parentaltype progeny PD tetrad 0 Double crossover can happen with 24 strands No crossing over parental ditype Single crossover tetratype 2strand Double crossover parental ditype 3strand double crossover tetratype 4strand double crossover nonparentalditype Recombination frequency RF NPD 12T Total tetrads X 100 o Accurate for genes separated by small distances 0 Less reliable for more distant genes doesn t account for all types of double crossovers Low number of NPD shows that recombination happens after the chromosomes have replicated when there are four chromatids for each pair of homologs o Ifbefore then more tetrads would have NPD instead of T Meiotic recombination is almost always reciprocal 2 homologous chromosomes that are inverted images of each other 0 Unusual ratios can eXist but are quite rare 0 Recombination does not create new alleles Ordered tetrads allow for mapping of the centromere ofa chromosome relative to other genetic markers Octad eight haploid ascospores from the 4 haploid meiotic products dividing once by mitosis o Phenotype of each of the haploid cells can be determined 0 Can be analyzed as a tetrad because the products are lined up together Firstdivision segregation pattern each allele appears in spores located on only one side of an imaginary line through the middle of the ascus Seconddivision segregation pattern both alleles appear on the same side of the middle line but in different spores 0 Product ofa crossover between a gene and its centromere


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