Biology 97 Lecture 12 Class Notes
Biology 97 Lecture 12 Class Notes 61860
Irvine Valley College
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This 3 page Class Notes was uploaded by Idda Colcol on Sunday October 2, 2016. The Class Notes belongs to 61860 at Irvine Valley College taught by Amy McWhorter in Fall 2016. Since its upload, it has received 4 views. For similar materials see Genetics and Evolutionary Biology in Biology at Irvine Valley College.
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Date Created: 10/02/16
Bio 97 (#61860): Lecture 12 Class Notes 9/28/2016 (Powerpoint slides “Tetrad Analysis in Fungi” and “Bacterial Genetics I” I. Tetrad Analysis in Fungi A. spores are gametes in these organisms B. live as haploid individuals 1. haploid spores are called ascospores C. unique life cycle D. formed in groups of 4 E. model system F. tetrads refers to chromatids G. don’t need sexual reproduction to grow H. phenotypes are directly related to genotypes because they are haploid I. have opposite mating types II. Two Common Fungi A. saccharomyces cerevisiae 1. budding yeast fungus 2. simple organism to grow 3. survive as unicellular individuals, haploid, can grow as diploid 4. yeast are eukaryotic cells - linear chromosomes 5. vegetative life cycle - can survive as haploid 6. haploid daughters remain stuck together in casing called ascosproes B. neurospora crassa 1. common bread mold 2. has haploid part of its life cycle 3. haploid cells can germinate and form long chain-like structures 4. binucleate cell a) 2 haploid nuclei cell 5. meiosis II is not the end for these ascospores a) they undergo ﬁnal round of mitosis 6. ordered tetrad a) after mitosis, has octad III.Genetic Analysis in Fungi A. histidine and tryptophan are necessary amino acids for growth B. dominance and recessiveness can only be seen when they are diploid C. these cells can grow in medium source D. HIS4 and TRP1 are dominant IV. Tetrad Analysis in Fungi: Unlinked genes A. Parental ditype 1. refers to ascus 2. ascus all end up with parental combinations 3. this is because of how chromosomes aligned on metaphase plate in meiosis I, independent assortment B. Nonparental ditype 1. all 4 ascospores are non parental combinations 2. homologous pairs separate in meiosis I, sister chromatids separate in meiosis II C. out of all asci, we expect to see half parental and other half non-parental since they are unlinked D. tetratype 1. could have recombination event 2. reﬂection of recombination occurring along one of those chromosomes V. Tetrad Analysis in Fungi: Linked genes A. No crossing-over 1. all ascospores are parental ditties B. single crossover 1. all 4 ascospores have unique combinations that are different from each other C. double crossover 1. involves two strands 2. genes are so far apart, double crossover can occur 3. results in parental ditype D. double crossover in three strands 1. end up with tetratype arrangement of ascospores a) we arrived here through three strands of recombinations E. double crossover in four strands 1. only way to get a non parental ditype F. this shows that there will always be more PDs than NPDs, we don’t look at tetra types when considering linkage G. we can use recombination frequencies to determine distance 1. RF = NPD + 1/2 T 2. total tetrads x 100 VI. Ordered Tetrad Analysis A. neurospora produces B. they way they divide is linear C. have linear relationship between products of meiosis D. mitotic spindle lines perpendicular to ascospore E. enables us to see how meiosis occurs F. octad 1. every pair of cells in resulting octad representing one ascospore VII.Ordered Tetrad Analysis: M1 Segregation A. wildtype: produces pigment B. mutant: doesn’t produce C. M1 segregation pattern 1. pattern of ascospores arranged in ﬁnal ascus 2. how these alleles are distributed in ﬁnal 8 ascospores 3. if one allele is found on top half of one octad and second allele is found in bottom four then that reﬂects the ﬁrst meiosis segregation pattern a) meiosis 1 separated two alleles VIII.rdered Tetrad Analysis A. meiosis II ﬁnally separates wild types from mutant alleles from one another B. MII shows both alleles are present along dotted line C. frequency can be determined by mapping distance between locus of gene and its centromere, not another gene D. are these two genes linked? 1. look at PDs and NPDs IX. Bacterial Genetics I A. bacteria are great organisms to study B. basic processes C. prokaryotic cells that live as haploid individuals D. circular chromsome - plasmid E. helps us understand and study diseases X. Advantages of using Bacteria for genetic studies A. short doubling time B. every time bacteria divides, two cells are produced C. 1000 progeny cells in less than 4 hours D. 20 generations in 400 minutes E. keep exponentially increasing XI. Bacterial Clones A. streak plating B. dab of cells onto plate C. E. coli like body temperature D. incubate them at this temperature, they will duplicate and grow E. end up with colonies visible on plate
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