Discussion 3/2/071234++aa1234a+a+Crossover in G2+aa+1423++aa1324Mitotic RecombinationQ. What are the probabilities of these two events?Q. What happens if there is a gene conversion event in G2, so the a+ onchromatid 1 is converted to the a allele?mitosisanaphase1234++aa1234a+a+Crossover in G2+aa+1423++aa1324Mitotic RecombinationQ. What are the probabilities of these two events? A. 0.5 eachQ. What happens if there is a gene conversion event in G2, so the a+ onchromatid 1 is converted to the a allele? A. 1 of daughter cells ishomozygous and 1 heterozygous, regardless of how centromeres segregate.Pedigree with Huntington’s Disease (Dominant H)and Molecular Marker (RFLPFill in genotypes for both loci.RFLP = restriction fragment length polymorphism: some individuals in a populationhave a particular restriction site; others lack it.Can be due to difference in single base pair (single nucleotide polymorphism = SNP), or toinsertions or deletions.Now to see whether individual is AA, Aa, or as, isolate a small sample of DNA, amplify this region,restrict the amplified DNA with HaeIII, run on gel to separate fragments and estimate their sizes.genotype fragmentsR R 0.9Rr 0.9, 0.5, 0.4rr 0.5, 0.4.Mutations that are due to a transposable element can be detected by PCR of the region andchecking its size on a gel.Hh12hh11hh12hh11Hh12hh11hh12I didn’t fill in all the genotypes in generation 3; except shown, all affected are Hh 12 and allnormal are hh 11.Any pattern?Is the H locus linked to the RFLP locus?Hh11Hh11Hh11hh11hh11hh11Hh12Hh12Hh12Most Hh are 1/2Most hh are 11Suggests what?Hh12hh11hh12hh11Hh12hh11hh12Hh11Hh11Hh11hh11hh11hh11Hh12Hh12Hh12Most Hh are 1/2Most hh are 11Suggests H is linked to 2 and h is linked to 1What is recombination frequency between H locus and the molecular marker locus?Hh12hh11hh12hh11Hh12hh11hh12Hh11Hh11Hh11hh11hh11hh11Hh12Hh12Hh12Most Hh are 1/2Most hh are 11Suggests H is linked to 2 and h is linked to 125 people are offspring of h1/h1 × H2/h1. If a crossover occurred between the two genes in theirH2/h1 parents, it would produce gametes H1 and h2.Of those 25 people, 5 had a recombinant chromosome. Recombination frequency = 5/25 = 0.2.Genes are 20 map units apart.Hh12hh11hh12hh11Hh12hh11hh12Hh11Hh11Hh11hh11hh11hh11Hh12Hh12Hh12Chi-square ProblemO E O - E (O-E)2 (O-E)2/E81819!Sum of O-E must be 0!What is the model or hypothesis?What is the expected ratio?Chi-square = 3.00 Degrees of freedom = 2 - 1 = 1 P ≈ 0.08 > 0.05Probability of this big a deviation from expected just by chance is reasonably high.Hypothesis not rejected.•Find the parentals and double crossover types.•Find the parentals and double crossover types.•Find which gene is in the middle.PDDP•Find the parentals and double crossover types.•Find which gene is in the middle. Now we know arrangement of genes inheterozygous parent.•What was genotype of other parent?PDDPLz Su Gllz su gl•Find the parentals and double crossover types.•Find which gene is in the middle. Now we know arrangement of genes inheterozygous parent. What was genotype of other parent?PDDPLz Su Gllz su gllz su gllz su gl•Find the parentals and doublecrossover types.•Find which gene is in the middle.Now we know arrangement ofgenes in heterozygous parent.•Calculate pairwise recombinationfrequencies.PDDPLz Su Gllz su gllz-glRecombinants areLz gl and lz GlLz Su Gllz su glLz Su Gllz su gl•Find the parentals and doublecrossover types.•Find which gene is in the middle.Now we know arrangement ofgenes in heterozygous parent.•Calculate pairwise recombinationfrequencies.PDDPLz Su Gllz su gllz-su33 4 24079176/740 = 0.238 0.107 0.147lz-gl33594440176gl-su59 4 244109•Find the parentals and double crossovertypes.•Find which gene is in the middle. Nowwe know arrangement of genes inheterozygous parent.•Calculate pairwise recombinationfrequencies. NOTE THE PATTERNS!•Make mapPDDPLz Su Gllz su gllz-su33 4 24079lz-gl33594440176gl-su59 4 244109lz su gl 10.7 14.7lz su gl0 10.7 ?Slide 150.238 0.107 0.147•Find the parentals and double crossovertypes.•Find which gene is in the middle. Nowwe know arrangement of genes inheterozygous parent.•Calculate pairwise recombinationfrequencies. NOTE THE PATTERNS!•Make mapPDDPLz Su Gllz su gllz-su33 4 240790.238 0.107 0.147lz-gl33594440176gl-su59 4 244109lz su gl 10.7 14.7lz su gl0 10.7 24.7SSSFSFSFSS SFS