LECTURE 6: TETRAD ANALYSISReading: Ch. 5, p. 132-140Problems: Ch. 5, solved problem III, 5-20, 5-24 – 5-27, 5-30 (mislabeled as the 2nd 5-29), 5-31---------First we went over “interference” (see notes from last lecture)--------TETRAD ANALYSIS IN FUNGIIn the diploid organisms that we’ve considered so far, each individual represents only one of fourpotential gametes that are produced from each parent in a single meiotic event. We don’t knowwhich of the other progeny in the cross are created by gametes produced by the same meioticevent. So we have to analyze large numbers of progeny and use statistics to establish linkage (orlack thereof) and to do mapping. By contrast, some yeast species house all four products of asingle meiosis in a sac called an ascus. The haploid cells are called ascospores (or haplospores)and can germinate and live in the haploid state, growing and dividing by mitosis. Thus,phenotype equals genotype in these haploid cells! We will review tetrad analysis in Baker’syeast, Saccharomyces cerevisiae; you should review ordered tetrad analysis in Neurosporacrassa on your own.Life cycle of Saccharomyces cerevisiae. The collection of four products of meiosis found ineach ascus is called a tetrad.Nomenclature conventions of Saccharomyces cerevisiae: Capital letters indicate dominant alleles (e.g. HIS4, or sometimes just +) Lower case letters indicate recessive alleles (e.g. his4) Usually, the wild-type allele is the dominant one (e.g. HIS4+)We characterize tetrads based upon the number of parental and recombinant spores theycontain. Consider a cross between an a-mating type haploid yeast strain of genotype HIS4 trp1with an α-mating type haploid yeast strain of the opposite genotype his4 TRP1. These two genesare unlinked. What are the possible kinds of tetrads that result when the resulting diploid strainundergoes meiosis?Parental ditype (PD): A tetrad containing 4 haploid cells of the parental class.Nonparental ditype (NPD): A tetrad containing 4 recombinant haploid cells (the two parentalclasses have recombined to form the reciprocal nonparental combination of alleles).Tetratype (T): A tetrad containing four kinds of haploid cells, two different parental class sporesand two different recombinant class spores. In crosses involving 2 unlinked genes, tetratypesarise when a crossover occurs between one of the two genes and its centromere.If PD = NPD, then the two genes are unlinked (either they are located on differentchromosomes, or they are far apart on the same chromosome). Think independent assortmenthere (50% recombination frequency). Since all Ts (no matter how many) are 50% parental and50% recombinant, then the only way that 50% of the total number of haploid progeny spores canbe 50% recombinant is if the number of PDs (100% parental class) equals the number of NPDs(100% recombinant).If PD >> NPD, then the two genes are linked. When PD >> NPD, the haploid spores of theparental class significantly outnumber the haploids of the recombinant class, a sign of linkage.Consider and example of two linked genes, ARG3 and URA2.P: arg3 ura2 x ARG3 URA2Diploid: arg3 ura2 / ARG3 URA2Meiotic products:PD (arg3 ura2; arg3 ura2; ARG3 URA2; ARG3 URA2) = 127NPD (arg3 URA2; arg3 URA2; ARG3 ura2; ARG3 ura2) = 3T: (arg3 ura2; arg3 URA2; ARG3 ura2; ARG3 URA2) = 70Recombination frequency, RF = [(NPD + 1/2T) / total number of tetrads] x 100. In our example,this is ([3 + 1/2(70)]/200) x 100 = 19 map units. We will modify the equation to obtain a betterestimate of map distance. If you draw out the possible crossover events between two linkedgenes, you can see the different tetrads that result; see Fig. 5.19.No crossovers --------> PDSingle crossover --------> TDouble crossover (2-strand) --------> PDDouble crossover (3-strand) --------> TDouble crossover (3-strand) --------> TDouble crossover (4-strand) --------> NPDYou can see how we can modify the equation to make it more accurate. Remember that half(2/4) the strands recombine if there is a single crossover event and that 4 strands recombine ifthere is a double crossover event (even if all of the strands don’t participate, some participatemore than once.)Map distance = (total rec. events / total tetrads) x 100 = [(1/2[SCO] + DCO) / total tetrads] x 100Map distance = ([1/2 (T – 2 NPD) + 4 NPD]/ total tetrads) x 100Map distance = (1/2 T + 3 NPD) / total tetrads x 100For our example above, map distance = ([1/2 (70) + 3 (3)] / 200) x 100 = 22 map unitsThis modified equation makes 2 assumptions: (1) there are no more than two crossovers in theinterval and (2) there is no chromosomal interference (all types of DCOs occur with equalfrequency. -----------We will cover the material below nest time----------ORDERED TETRADS AND GENE-CENTROMERE DISTANCEIn Neurospora crassa, meiosis occurs within the tight confines of a narrow ascus, resulting in theformation of ordered tetrads. Because of the precise positioning of each meiotic product withinthe ascus, one can infer the arrangement (and segregation) of each chromatid of homologouschromosomes during Meiosis I and II. This gives information about the distance between thegene and its centromere. (Meiosis II is followed by mitosis; each pair of genetically identicaldaugthers sits adjacent to one another. Each ascus is thus made of up 8 haploid ascospores.)Consider a gene required for ascospore color (ws+ gives black spores and ws gives whitespores):P: ws+ x wsDiploid: ws+ / ws (immediately undergoes meiosis)If no recombination between ws gene and the centromere occurs, then the resulting ascosporesare arranged in a neat array with black and white spores clearly segregated from one another,each type cleanly segregated to either side of the imaginary line separating the 4th and 5thascospores. This is called a first division segregation pattern. Since the daughters of themitotic division lie right next to one another, we can simplify the two possible configurations to:(ws+ ws+ ws ws)(ws ws ws+ ws+)If recombination occurs between ws and the centromere, then a second division segregationpattern is observed. Now, both types of spores are found on either side of the imaginary linebetween the 4th and 5th ascospores. Now there are four possible configurations:(ws+ ws ws+ ws)(ws ws+ ws+ ws)(ws+ ws ws ws+)(ws ws+ ws ws+)When an ascus shows a second division segregation pattern, we know
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