Biology 97 Lecture 10 Class Notes
Biology 97 Lecture 10 Class Notes 61860
Irvine Valley College
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This 3 page Class Notes was uploaded by Idda Colcol on Wednesday September 21, 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: 09/21/16
Bio 97 (#61860): Lecture 10 Class Notes 9/21/2016 (Powerpoint slides 1-28) I. Mendel’s Principle of Independent Assortment A. 4 different gametes produced B. 50% alignment pattern 1. equal proportions II. Gene Linkage A. gene linkage 1. genes close together on same chromosome tend to be inherited together and are passed as a unit 2. they do not sort independently III.Demonstration of Linkage and Recombination A. mutations can be inherited dominantly or recessively 1. curly allele is dominant to wild type 2. Cy = curly wings 3. Cy+ = straight wings 4. capital C tells me curly mutation is dominant to wild type 5. purple allele is recessive to wild type IV. Morgan’s Test Cross A. F1:dihybrid ﬂy 1. heterozygous a) phenotype: wildtype, wildtype B. Test cross 1. gives us output of what gametes are produced in F1 2. we get recessive from test parent 3. we would expect 1:1:1:1 gametes 4. but that’s not what we get a) we have disproportionately high number of individuals who inherited purple and vestigial together and wild type and wild type together (1) call these parental phenotypes (a) combination of alleles in same way inherited from parents b) tester strain always gives us recessive allele 5. What can the numbers of each phenotypic class tell us? a) it is a direct read out of gametes produced by F1 individual V. Morgan’s test cross progeny phenotype ratio reﬂects the F1 gamete ratio A. visualize gametes produced B. parental inheritance pattern is much more signiﬁcant than recombinant inheritance pattern C. recombinant gamete 1. passed from parents to offspring in unique way VI. Demonstration of Linkage and Recombination A. How do we explain the appearance of the minority class of non parental combinations? 1. Because of the recombination that can take place VII.Crossing Over A. regions of homologous chromosomes break and rejoin together B. tetrad synapses occur in prophase I C. occurs in three dimensional space D. any chromatid can break and rejoin, but when recombination occurs between non-sister chromatids that is when recombination takes place E. what makes them homologous chromosomes 1. genes found on same loci VIII.olecular Description of Crossing Over A. happens only during prophase I of meiosis I B. we end up making 4 unique gametes IX. Recombination Frequency A. helps us determine distance B. usually expressed as percentage C. recombination frequency 1. frequency with which crossing over between two genes is affected by distance between those two genes X. Recombination and Genetic Distance A. the closer the genes, the less likely for recombination to occur (low frequency) B. example 1. blindly have scissors and cut a chromosome 2. less likely to cut between A and B and more likely to cut between B and C XI. Recombination and Genetic Distance A. maps units are not consistently measurable from organism to organism 1. one map unit is distance between two genes XII.Linkage Mapping A. enables us to place genes in order B. what makes them homologous is that genes are found on same loci XIII.wo Point Cross: Our Previous Example A. recombination frequency for ﬂies in morgan’s test cross back a few slides 1. 305/2839 = .107 x 100% = 10.7% XIV.Using Map Distances A. What percent of progeny from test cross will have Ab phenotype? 1. 28 mu between A and B 2. working backwards 3. determine gametes a) gamete tester parent will produce ab b) gametes from heterozygote (1) AB (2) ab (3) aB (4) Ab 4. parental is AB and ab 5. recombinants are Ab and aB a) we know that the map unit between A and B is 28 mu, so the percentage of all recombinants is 28% and percentage of all parental is 100-28 = 72% (1) since half are Ab and half are aB, half of 28% is 14%. (a)14% is Ab (b)14% is aB (2) half of parental are AB and half are ab (a)36% is AB (b)36% is ab XV. Recombinants A. look at what is inherited and what gametes are produced XVI.Linkage Map A. 4 chromosomes in drosophila B. roughoid eyes in chromosome 3 is 70 mu from ebony body 1. the distance between two genes that is greater than 50 assort virtually independently or appear to assort independently 2. these two genes are so far apart that it is guaranteed there will be a cross over event between two genes XVII.Two approaches to determine gene order using three markers A. DCO = double cross over event 1. cross over in region 1 between A and B and second cross over in region 2 from B to C 2. double cross over event is less likely than single cross over event 3. chance of DCO is product of single cross over in region 1 times single cross over in region 2 B. this helps us with ordering genes XVIII.Eight possible gamete types of three-point cross A. offspring are all recombinations XIX.Determining Gene Order: Method I A. have to calculate all 3