1、Ascomycetes (子囊菌)Ascomycetes are fungi that form their meiotic progeny in a asci (ascus,子囊).,This allows the products of meiosis (called a tetrad，四分子) to be recovered together.Some ascomycetes have order asci.,Unordered Tetrad,Ordered Tetrad,粗糙链孢霉,酿酒酵母,Our old friend, Neurospora,Neurospora is a hapl
2、oid organism that makes asci with 8 spores.,Neurospora Life cycle,粗糙链孢霉,同时的,教材p70,一种能产生分生孢子的特殊真菌菌丝,植物生长的,Tetrad Analysis（四分子分析）,Analysis of all four products of a single meiosis.,N. crassa spores are formed by a single mitotic division after the completion of meiosis.The meiotic progeny of N. crassa
3、 form in a asci (ascus), but they are present in a linear order that reflects meiotic division.Ordered tetrads in Neurospora allow mapping relative to the centromere.,Producing an Ordered Tetrad,性细胞,Neurospora MeiosisThis shows the process of meiosis in N. crassa.,Remember that the duplicated chromo
4、somes that are held together by the centromere segregate during the first meiotic division.Thus, a locus that is close to the centromere would often segregate during the first meiotic division.,MI Segregation Pattern,No crossover between gene and centromere,MII Segregation Pattern,Crossover between
5、gene and centromere,A second-division segregation Pattern, M2,Recombination between the gene of interest and the centromere would produce second division segregation tetrads.,Producing MII Segregation Patterns,教材p180 的4种交换型,Tetrads produced by second division segregation are related to the distance
6、from the centromere by this equation:,R = (1/2 second-divison / total ) x 100% = cM,Ordered Tetrads in Neurospora Allow Mapping Relative to the Centromere, MII/total x 100% = cM,Or,Genetic Analyses with Tetrads,Cross two haploid cells,Induce diploid to undergo meiosis,Genetic Analyses with Tetrads,P
7、roducing MII Segregation Patterns,The distance in cM between a and b can be estimated.,Parental Ditype (PD),Tetratype (TT),Nonparental Ditype (NPD),这没有完全概括教材的P185，请补充完整,Calculating Genetic Distances with Tetrad Analysis,Page 184 (教材上册),Page 184 (教材上册),Page 184 (教材上册),Distance from a to centromere = MII = (1+8+1) = 0.05 = 5 map units Total 100,Distance from b to centromere = MII = (18+8+1) = .135 = 13.5 map units Total 100,Distance from a to b = T + NPD = (18+1+1) +1 = .11 = 11 map unitsTotal 100,Page 184 (教材上册),
