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Biol 484 - Final Exam Notes (Full Course).pdf

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BIOL 484
Bruce Reed

BIOL484AdvancedEukaryoticGenetics MIDTERMREVIEW/STUDYQUESTIONS&KEY CONCEPTS BACKGROUND/INTRO (important concepts you should know): Mendel’s 5 rules (two ofwhich are referred to as “laws”). Which “law” is routinely broken and why?  1 law: law of segregation: during sex repro in hybrids with opposing traits, half of the gametes receive one factor (A) and half receive the other (a)  2nd law: law of independent assortment: factors associated with different traits segregate independently  During asexual reproduction, factors do not segregate  Inheritance of factors is bi-parental  Fertilization is random with respect to which factor is carried by a gamete The most frequently encountered exception, violating the second law of independent assortment is linkage because linkage causes some factors to segregate more with each other than others. Trisomy and things violate the first law of segregation Meiosis – stages of first meiotic prophase / timing and primary function of meiotic recombination. Stages of First Meiotic Prophase: 1. Leptotene – thin threads 2. Zygotene – homologue pairing first evident 3. Pachytene – meiotic recombination occurs & synaptonemal complex may be present 4. Diplotene – double thread – chiasmata visible 5. Diakinesis – continued chromatin condensation Bridges’ proof that genes are on chromosomes (chromosomal theoryof inheritance).  There was already an indication that genes paralleled chromosomes in some way.  To prove this, had to show that if there was a change in how chromosomes behaved, there is a corresponding change in inheritance patterns  Bridges found that when chromosome behavior was changed through XX + Y or X + XY pairing, inheritance patterns paralleled in X-lined mutants. The three point test cross – linked genes (a b c/ +++ x a b c/a b c) . Know how to correctly calculate map distances and quickly establish gene order.  (Single recombinants + 2x double recombinants)/ total The Chi square ( test and the limitations of the Chi square, ie. when/how can we use it?). Limitations:  Must be used on raw numerical data, not percentages  Cannot be used for experiments where the expected frequency of any phenotypic class is < 5  X^2 assumes that all progeny are equally viable Calculate:  Deviance is due to chance (n2ll hypothesis)  SUM((observed – expected) /expected) Relationship between physical genetic maps and recombination genetic maps – what does this tell us about recombination rates along the length of a chromosome?  Physical maps come from polytene chromosomes (arrive through a modified cell cycle) that form light and dark bands. These bands may reflect genes  very low recombination at the centromere and telomere locations (suppressed), highly repetitive heterochromatin What are the general characteristics of heterochromatin compared to euchromatin?  Constitutive Heterochromatin: highly repetitive, inactive, recombination reduced, gene density is low  Facultative Heterochromatin: only in certain situations and perhaps places in the body does it become inactive  Euchromatin: active DNA BACKGROUND FOR GENETICS IN MODEL GENETIC ORGANISMS: What general attributes (attribute = quality, property, or characteristic) make an organism highly suitable for genetic analysis in the lab?  Easy & inexpensive to keep in the lab (small)  Short generation time  Large number of progeny  Easy to control mating  Diploid – rather than haploid  Small number of chromosomes (linkage groups) Balancer chromosomes – for what are these used and what characteristics make a chromosome a “balancer chromosome”?  Balanced stock: genotype of the progeny must be the same as that of the parental  If you have a lethal mutation that survives in hetero but lethal in homo, to create a balanced stock, you need to get rid of the homo wild-type condition to leave all hetero mutations.  Balancer Chromosomes: useful for maintaining any mutation which is lethal or sterile when homozygous in the heterozygous condition o Carry multiple inversions to prevent crossovers o Carry a visible dominant marker Mutation / classes of mutation and genetic tests that help determine class (null, hypomorph, hypermorph, antimorph, neomorph)  Mutation: a mutation is a stable and heritable change in DNA sequence  Recessive mutation: m/+ appears wildtype and m/m displays phenotype  Dominant mutation: M/+ shows phenotype, mutant is dominant Classes:  Nullomorphic: a mutant with no remaining gene function o Phenotype: null = deletion o Test: protein nulls fail to produce protein product from antibody recognition o Classify: Df/m nul= m nu/m null  Hypomorphic: lower or residual activity compared to wildtype o Test: Df/m hypoor m /ml hypo> m hyp/m hypo  Hypermorphic: hyperactive protein or too much protein produced o Test: m hyp/m hype> m/+ > m hyp/Df  Antimorphic: produces protein that acts as a poison to the wildtype protein anti anti anti o Test: M / M >= M /Df o Should be able to reduce the effect of an antimorph by increasing wildtype dosage  Neomorphic: causes a gene to inappropriatelyactivate either in the incorrect place or at the incorrect time o Test: M ne/+ = M neo/+/+ etc… o You cannot reduce the effect of a neomorph by increasing the wildtype dosage In what different ways can mutations be associated with dominant phenotypes?  Haplo-insufficiency: single copy of wild-type gene is insufficient  Product of mutant gene is poisonous  Mutation causes gene to be misexpressed (at times or places that are weird)  Overexpression or over-activity In what different ways can mutations be induced or recovered in model genetic organisms in the laboratory setting?  Spontaneous  Irradiation: x-rays or gamma rays o Free radicals, DNA strand breaks o Best for producing gross chromosomal rearrangements  Chemical: alkylating agents and crosslinking agents cause transitions and transversions  Transposable genetic elements: preferred method. Molecular tag, known sequences. o can disrupt coding sequence, regulatory sequence, enhancer/promoter etc.  Targeted gene disruption (homologous recombination) MGO: DROSOPHILA: How has the Drosophila P element been engineered so as to be a useful method for random mutagenesis?  1) Control transposition (jumping) via genetic crosses: o P element mediated germline transformation: Using P elements that have inverted terminal repeats and gene of interest and marker (transformation plasmid) alongside P elements that have transposase functionality (delta-2-3 helper plasmid)  Defective elements can only be mobilized when crosses to a strain that produces transposase (delta-2-3) What is “plasmid rescue”? What is imprecise excision?  Plasmid rescue: method of inserting a transposon o 1) P element engineered to carry ORI and selectable marker o 2) engineer so the transposon is cut once within and one out o 3) re-ligate into plasmid and then replicate in E.coli  Imprecise Excision: o The P element may transpose imprecisely and remove some flanking DNA in the process creating a small Df Details of Drosophila as a model genetic organism (genome size, chromosome number, number ofgenes, generation time, care, setting up crosses, keeping balanced stocks, sex determination, polytene chromosomes…)  Crosses: requires collection of virgin females  Generation time: 9 days (@25 degrees)  Lifespan: 45 – 60 days  No. progeny: over 100 per single female  Genome Size: 120 Mb euchromatic, 60 Mb heterochromatic  Genes: 13,601 genes  Chromosome number: 2n = 8 (X;2;3;4 + Y)  Sex Determination: X chromosome to autosome ratio ( 1 = female, 0.5 = male)  Polytene Chromosomes: easy mapping of rearrangements, transposon movement detection  *There is no recombination during meiosis in male drosophila P element based techniques and methods (enhancer traps, GAL4/UAS, mosaics using FLP/FRT, PTT exon traps, GAL4 + GAL80 for marking mosaic clonal patches)  Enhancer Trapping: p element transposon can be used with delta-2-3 to insert in between an enhancer and coding region so that the enhancer expresses your gene of interest. You can look for tissue specificity  GAL4/UAS: by crossing GAL4 with UAS-goi, theGAL4 protein produced induces UAS to express your goi. EP element is a variation of this  Somatic mosaics: useful in bypassing lethality in early development and examine patches of mutant tissue later. Can also give you progenitor counts (can tell you how many cells were destined to form the tissue of interest at the time of somatic recombination)  FLP/FRT: FLP recombinase mediates site specific recombination between FRT sequences. You can incorporate these FRT sequences using P elements.  GAL80 blocks GAL4 protein production (creates mosaics since some are blocked)  You can insert GFP by using Transposon-PTT splicing sequences-GFP and inserting into genome in between introns randomly. The PTT and transposon and normal introns are spliced out leaving a chimeric protein see also the tutorial as it represents an application of much of the above MGO: C. ELEGANS Details of C. elegans as a model genetic organism (genome size, chromosome number, number ofgenes, generation time, food source, ,types of markers, setting up crosses, keeping balanced stocks, sex determination…)  Crosses: population is 99% “selfing” hermaphrodites  Generation time: 3 days  Lifespan: 2-3 weeks  No. progeny: 250 – 1000 per hermaphrodite  Balanced Stocks: balancer translocations  Genome Size: 97Mb  Genes: 19,099 genes  **Genome is smaller than drosophila but moregenes  Chromosome number: 2n =12 (1,2,3,4,5 + X)  Sex Determination: X chromosome to autosome ratio ( XX = hermaphrodite, XO = males)  **A third of everything drosophila but genes 20,000 and chrome num = 12 C elegans has holocentric chromosomes - what are holocentric chromosomes? What problem do holocentric chromosomes present during meiosis? How is this solved in C. elegans and what consequence does this have on the C. elegans recombination maps?  Holocentric chromosomes: differ in that their shape when being pulled to the centrioles. Microtubules can attach to anywhere on holocentric chromosomes. Appear straight and parallel to metaphase plate. No defined centromere*.  During meiosis, holocentric chromosomes are disastrous because the centrioles can tear the chromosomes. In C. elegans, they create localized microtubule attachment sites during meiosis through ONE crossover. This crossover allows the chromosomes to twist and align in a certain way to allow pulling at the tips.  Part of the solution is that there is only one crossover per chromosome towards one end! This explains the clustering of genes in the central region of C. elegans recomb maps In general, what strategy is used to clone a gene starting from a mutant and mutant phenotype in C. elegans?  Cosmid injection into synsitial area is the common method of gene cloning then rescue How is mosaic analysis achieved in C. elegans?  Using extrachromosomal arrays (cloned DNA with wild-type and marker) in the worm injected in synsitial area of worm  Homologous chromosomes have mutated gene  Extrachromosomal array anneals itself to the mutated gene to some of them and not others?  creates a mosaic  If you know the lineage, and see where GFP is functioning from the marker, you can further refine the pattern of expression  You know that wherever green shows, the gene is required for whatever phenotype In C. elegans screening for recessive mutations can be done in the F2 generation. Why is this not possible in Drosophila?  Works in worms because of hermaphroditism.  Herm can be exposed to mutagen, progeny can herm themselves  Don’t need to worry about original males contribution  In drosophila, you need a third generation to mate brothers and sisters from F2. What strategy was developed to screen for maternal effect lethal mutations in C. elegans.  You can screen for maternal effect lethal mutations using the bagofworms technique.  The rare plates will have 1/4 dead progeny that will not break out of mother. The others will break out MGO3 : YEAST Details of yeast as a model genetic organism (chromosome number, size of genome, number of genes, types of markers, mating system, recombination frequency…)  Generation Time: 90 min  Lifespan: mother produces 20 – 30 buds  No. progeny: unlimited  Crosses: haploid/diploid life cycle. Haploids can be crossed o Diploid yeast will sporulate only when deprived of nitrogen  Chromosome Number: 16 linear chromosomes  Number of Genes: 6607 (only3.8% have introns)  Size of Genome: 12 Mb Laboratory strains of yeast must be heterothallic – what does this mean?  Diploid yeast are always heterozygous at the mating-type locus (MATalpha and MATa) – like male and female (MAT a produces a pheromone and receptors for alpha pheromone and vice versa)  Heterothallic strains of yeast are stable and do not switch their MAT (do not randomly mate) Heterothallic strains do not have the HO gene.  Homothallic strains can switch mating types (activate their silent mating cassettes initiated by the HO (homothallism gene)) Homothallic strains carry at least one copy of this HO gene and grow as diploids instead of haploids and will mate What are the different type of shuttle vectors that are commonlyused in yeast genetics?  Ylp (integrative plasmids): o cannot replicate autonomously o One copy per cell o *needs bacterial origin of rep  YEp (episomal plasmids): o Replicate autonomously o Many copies per cell o *has own origin of rep  YCp (centromeric plasmids) o Replicate autonomously o One copy per cell o segregates in meiosis like a chromosome o *has yeast origin of rep o *has yeast chromosomal centromere!  All derived from pBR322 and can grow in E. coli. What is “cloning by complementation” in yeast?  Using some selectable marker on a plate (like the requirement of uracil)  Plate random YCp vectors with yfg and ura  Select colonies that can grow under restrictive conditions  The plasmid selected will have yfg because the only way the selectable marker is maintained is if the plasmid is repaired with yfg mutant that was removed Outline the strategy used to replace a yeast wild-type gene with a mutant gene (the two step gene replacement).  Integrate a Ylp plasmid with URA3 with mutant with the wildtype by one excision  The plasmid will be lost because no ORI and undergo recombination where some will lose the URA3.  Single copy will be remaining, either wild type or mutant  FOA medium negative select for cells expressing URA3 The yeast two hybrid assay: what is this used for and how does it work?  If your gene of interest is working together in some complex and you want to know what theother players are in that complex  Based on the functional reconstitution of an intact transcription factor that activates reporter gene expression  The “prey” bait protein has an activation domain and when fused with the DNA binding protein, will cause expression of reporter gene.  The only way the prey can fused with the dna binding protein is if there is a relationship and physical interaction between the gene of interest with its other players (one player is attached to the activation domain, another is attached to the dna binding domain)  This allows you to build protein-protein interaction maps What is the purpose of the sectoring yeast lethality assay? How does this work? What type of shuttle vector (YIp, YEp, YCp) does this assay require?  Requires a YCp shuttle vector to multiply onlyonce  YCp plasmid carries the gene mutant of interest and ADE2+  The areas that lose this plasmid turn red sectors, those that don’t turn white  Ade2- are red, ade3- are white, double mutant is white ADE3+ restores red  The plasmid carries ADE3+ and YFG2+ to rescue the mutant yfg2-  Another plasmid is transformed carring a second form of YFG2* without ADE3+  If the new mutant can rescue the original function, there will be sectors because the original YCp plasmid can be lost and still function. If it can’t it will be all red because the new one cannot rescue What is a YAC?  Yeast Artificial Chromosomes: allow the stable maintenance of extremely large linear DNA fragments  They are circular in form and have restriction cut sites  Have many selectable markers  Needs homologous recombination of insert and YAC arms to maintain  ARS1, CEN4 and TEL allow long-term propogation Tutorials: 1) You should be able to apply the binomial distribution (I will give you the equation) to predict the likelihood of the outcomes of simple genetics crosses. Something like: given 8 progeny from the cross Aa x Aa, what is the liklehood that exactly 2 progeny will be the recessive aa? P(r successful outcomes in N trials) or P(r number of a given genotype among N progeny) = N! . p . (1-p)-r r!(N-r)! 2) From the metastasis paper you should understand the design of the genetic screen and the main conclusion regarding how tumours associated with Ras activation become metastatic.  Lack of cell polarity and Ras causes metastasis  Function of Ras has no link to cell proliferation,growth and survival functions  eyFLP/FRT: to drive recombinase in the eye imaginal disc for proliferation  GAL4/UAS and GAL80: to promote Ras and GFP o Using UAS-Ras85D to cause hyper-proliferation o UAS-GFP for viewing expression FINALREVIEW/STUDYQUESTIONS&KEY CONCEPTS TETRAD ANALYSIS What, in general, is tetrad analysis? What is the prerequisite for tetrad analysis? In what organisms is it possible to perform tetrad analysis?  Tetrad analysis: the ability to isolate and test the four products of a single meiosis for genetic mapping. In yeast for example, you can isolate the 4 haploid spores to determine their genetics. From this, you can analyze the ratio between the different segregation types and determine linkage  A prerequisite for tetrad analysis is that the organism produces 4 distinct, separable haploid components  It is possible to perform tetrad analysis in yeast, neurospora crassa and Know how to identify the different tetrad types (parental ditype (PD), non-parental ditype (NPD), tetratype (TT)).  From Dihybrid unlinked unordered:  PD = AB AB ab ab  NPD = Ab Ab aB aB  Tetratype = AB aB Ab ab (from crossover at 1 meiotic division) In tetrad analysis how does one determine if two genes are linked/unlinked?  You will know if two genes are linked if the number of PD is approximately equal to the number of NPD because 50-50 chance of arrangement in either cis or trans. They are unlinked if there are many more NPD because these only come from recombination If two genes are not linked, but both are tightlylinked to their respective centromeres, what is the overall expected ratio of PD:NPD:TT?  In this case, crossovers need to happen between one of the genes and the centromere which is unlikely since they are tightly linked to their centromere. This is how you can determine separation between the genes  Expected ratio is 1:1 for PD:NPD Consider tetrad analysis of diploids resulting from a yeast mutant strain “a” crossed muta
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