Midterm#1 comments#2 So far, ~ 50 % of exams graded, wide range of results: 5 perfect scores (200 pts) Lowest score so far, 25 pts Partial credit is given if you get part of the answer right Tests will be returned this Thursday Discussion section this week with go over genetic fundamentals to help you solve problems- you must understand these fundamentals to pass the course. Overview- chapter 6 In meiosis, recombinant products with new combinations of parental alleles are generated by: independent assortment (segregation) of alleles on nonhomologous chromosomes. crossing-over in premeiotic S between nonsister homologs. In dihybrid meiosis, 50% recombinants indicates either that genes are on different chromosomes or that they are far apart on the same chromosome. Recombination frequencies can be used to map gene loci to relative positions; such maps are linear. Crossing-over involves formation of DNA heteroduplex. Crossing-over No loss of genetic material, just formation of new chromatids Parental chromatids are noncrossover products Recombinant chromatids are always products of crossing-over
Consequences of crossing-over Frequency of recombinant gametes is 0-50%, depending on frequency of meiocytes with crossing-over Results in deviation from 1:1:1:1 in testcrosses parental combination is most frequent recombinant combination is rarest Allows drawing of linkage maps based on recombination frequencies (RF) Linkage maps Two genes (loci) on the same chromosome can exhibit linkage = co-inheritance The closer two genes are to each other, the more likely they are to show linkage why? Because crossing over occurs by chance and the more DNA that separates two genes, the greater the likelihood of a cross over (=recombination event) Groups of genes can be mapped using cross overs to show their relative order on a chromosome or part of a chromosome ( = linkage group) Distance between genes is determined by map units = recombination frequencies, 1 mu= 1% recombination between 2 genes Is recombination perfect for maps? Some cross overs will restore the parental phase of genes used as markers, as the last diagram showed In an experiment where you follow 3 markers, a double cross over between the 2 outside markers will produce double recombinants that have the parental allelic combinations - So, recombination underestimates the physical distance (length of DNA) between the markers, and small intervals are used to sum big distances.
Recombination frequency (RF) Experimentally determined from frequency of recombinant phenotypes in testcrosses Roughly proportional to physical length of DNA between loci Greater physical distance between two loci, greater chance of recombination by crossing-over 1% recombinants = 1 map unit (m.u.) 1 m.u. = 1 centimorgan (cm) Linkage maps # observed 140 50 60 150 RF is (60+50)/400=25%, clearly less than 50% Map is given by: A B 25 m.u. Mapping RF analysis determines relative gene order RF between same two loci may be different in different strains or sexes RF values are roughly additive up to 50% multiple crossovers essentially uncouple loci, mimicking independent assortment Maps based on RF can be combined with molecular and cytological analyses to provide more precise locations of genes
What do we need to do trihybrid mapping? This is a linkage analysis or testcross for assigning relative gene distance (in recombination frequency) and gene order using 3 markers (loci, genes) at once, here vestigial, purple, and black One parent will be heterozygous for 3 different genes (construct this genotype by breeding) The other parent will be homozygous recessive for those same genes (find or construct this one too) There are 2x2x2 =8 gametic genotypes that are possible, and we have to be able to infer the genotype from the phenotype ( that is why we use the triple recessive in the testcross) Trihybrid testcross Sometimes called three-point testcross Determines gene order as well as relative gene distances 8 categories of offspring for linked genes, significant departure from 1:1:1:1:1:1:1:1 Works best with large numbers of offspring, as in fungi, Drosophila Analysis of trihybrid testcross data Identify pairs of parental and recombinant offspring parental (noncrossover); most abundant double crossovers; least abundant single crossovers; intermediate abundance identify on the basis of reciprocal combinations of alleles Determine gene order by inspection (the parental gene order yields double crossovers by switching middle genes) Calculate RF for single crossovers, adding double crossovers each time Draw map
Genetic maps Useful in understanding and experimenting with the genome of organisms Available for many organisms in the literature and at Web sites Maps based on RF are supplemented with maps based on molecular markers, segments of chromosomes with different nucleotide sequences Tomato genotypes and the CEO Your sister s high end restaurant serves expensive beaked, yellow tomatoes, and you want to corner the market in Hawaii, make lots of money, and retire at 30! From Bio 375 you ve learned that those two markers should be on different chromosomes beaked on 1, yellow on 2 You ask if these are really unlinked genes, so you plant a garden from seeds you gathered in a greenhouse experiment of crossing, and now have parentals and recombinants, but the numbers aren t exact and you don t want to waste your money. What to do? Chi-square test Statistical analysis of goodness of fit between observed data and expected outcome (null hypothesis) Calculates the probability of chance deviations from expectation if hypothesis is true 5% cutoff for rejecting hypothesis may therefore reject true hypothesis statistical tests never provide certainty, merely probability Example of a test? Start with a cross between A/A.B/B and a/a.b/b, in our case this could be non-beaked, red x beaked, yellow plants. Obtain a dihybrid, A/a.B/b (non-beaked, and red), testcross to a beaked, yellow plant (a/a.b/b) Get 500 progeny, classified as 142 A.B (non-beaked, red) 133 a.b (beaked, yellow) 113 A.b (non-beaked, yellow) 112 a.b (beaked, red) Add the classes; parentals = 275 ( 142 +133) recombinants = 225 ( 113 + 112), so RF= 225/500 =45%, and is <50%, so are these genes linked?