BIO 305 1st Edition Lecture 5 Outline of Last Lecture I. Intro to Sex ChromosomesII. Morgan’s Experiment – Different Phenotypic Ratios Among SexIII. Bridges’ Experiment – Non-disjunction As Proof Of Chromosomal TheoryIV. Three Cases of Sex-Linked DiseasesOutline of Current Lecture I. Chromosome Mapping in EukaryotesII. Three-Point Mapping: A Few Rules and One Big ExampleIII. InterferenceIV. Map DistanceV. Sample QuestionsCurrent LectureI. Chromosome Mapping in EukaryotesCrossover – recall that crossover occurs in the Diplonema stage of Prophase IChiasmata – cytologically visible X-shaped regions where crossing over has occurred between a synapsed pair of homologous chromosomesWhen there are two genes on two different homologous pairs of chromosomes, and they undergo independent assortment, there are four possibilities of gametes produced:AB, Ab, aB, abWhat if all genes were on the same chromosome?They would be inherited together, and this is called “complete linkage”No exchange occurs: AB AB ab abWhat if there was linkage between a single pair of homologs / between two non-sister chromatids?Frequencies would no longer be 1:1:1:1, you get…2 Parental (non-crossover) gametes: AB, ab2 Recombinant gametes: Ab, aBRecombinant rate - the proportion of recombinant gametes generatedmaximum possible rate is 50%, when genes are furthest apart as possibleWhen two genes are close together, the recombination rate approaches 0  means that more likely that they are on the same chromosome aka the same “linkage group”Let’s look at Complete linkage of two traits:Parental generation = (bw hv+) x (bw+ hv)F1 generation = a double heterozygous gene:bw hv+bw+ hva) If you self F1 x F1: You get a 1:2:1 ratio of phenotypesb) If you test-cross F1 x Parent: You get a 1:1 ratio of phenotypesNow let’s look at an incomplete linkage resulting when genes are close to each other:F1 generation = (yw) x (y+w+) (gender doesn’t matter)F2 generation = the frequency of the parental types (yw or y+w+) are 98.7%the frequency of the recombinant types (y+w or yw+) are 1.3%What about an incomplete linkage, when the genes are far apart from each other:F1 generation = (yw) x (y+w+)F2 generation = the frequency of the parental types (yw or y+w+) are 62.8%the frequency of the recombinant types (y+w or yw+) are 37.2%The higher frequency of recombinance = distance between genes is larger = incomplete linkageT. H. Morgan proposed crossing over and recombinant gametes, his student A. H. Sturtevant was the first to discover gene mapping on a linear chromosome:Extensive maps were made for the X and autosomal chromosomes of drosophila in Morgan’s lab (despite the norm, no crossing over occurs in drosophila males)Why does distance apart affect the rate of recombination?The farther apart two genes are on segments of two sister chromatids, the probability that the genotype is changed after the chromosome crosses over in one or more locations increasesThe percentage of tetrads involved in an exchange between two genes = 2*recombination rateMultiple Exchanges: Notice that for double-crossover gametes, the middle gene is the one that is different from the genes on either end: this is important if asked what the gene order is because what really matters is “which is the middle gene”?In general, double crossover rate is lower than single crossover rate:What is the probability a double crossover between genes A, B, and C if you know the probabilities of their individual crossovers (10%, 20%)?Single crossover between A & B: 10%, Single crossover between B & C: 20%If two events independent, double-crosser = 10% * 20% = 2% chanceII. Three Point Mapping – A Few Rules and an Example:1) Organism producing the crossover gametes must be heterozygous at all loci underconsideration2) Cross must be constructed so that genotype can be determined by phenotype3) A large sample is required to obtain all crossover classesEXAMPLE: A recessive mutant-type female is crossed over with a wild type male. Be able to determine what the F1 generation results look like. A large sample size should be given to you for the F2 generation.Because you have visualized what the F1 cross looks like, you must consider all the possible crossovers and non-crossovers. Non-crossovers should look like their parents and should also bethe majority of the observed number. Indeed, 94.4% are either y w ec / y+ w+ ec+Now consider single crossovers combinations. This can occur between (y and w) or (w and ec). Look for pairs of observed numbers (in decreasing order) that look similar to each other. You willsee that the matching observed pairs correspond with two possible combinations of a crossoverbetween y/w or with the two possible combinations in a crossover of w/ecFinally, a double crossover is possible in a 3 point map and as expected, the two combinations of crossovers has the lowest chance of occurring as it has the smallest observed numberHow do we determine gene order?1) There are three possible orders2) Determine the arrangement of alleles along each homolog of the heterozygous parent giving rise to non-crossover and crossover gametes3) Using double crossover to determine the gene order (see example below)4) Confirm with the single crossoverEx) If you cross a wild-type and a mutant that you know have gene ABC and you want to know what order it is, you look at the F1 generation and see two types of phenotypes as a resultof a double crossover: AC or B. You know that B is in the second gene (whether ABC or CBA) because the two possibilities for F1 genotypes due to a double crossover are: A+ B C+ = AC phenotypeor A B+ C = just B phenotype  the gene in the middleHow do we determine distance between genes?The “map distance” between two genes = % of any crossovers involving the geneEx: The map distance between y and w = SCObetweenYW + DCO involving YW= 1.5% + .06% = 1.56%III. InterferenceInterference is when the crossover in one region reduces a crossover in nearby regionsHow to calculate interference when you know % of DCOs between genes ABC, and knowing each map distance between them?1) Find expected DCO rate, assuming events are independent2) Find Coefficient of Coincidence (C) = observed value /expected value3) Interference (I) = 1 - CExample: Map distance between A and B is 22.3 cM, between B and C is 43.4 cM, saw that 7.8% genotypes were double crossovers. ANS: Expected double crossover rate (independent events) =