Gene Mapping PDF

Genetic and Cytological Mapping of Chromosomes A linkage/genetic/crossover map provides a diagrammatic representation of relative distances between linked genes of a chromosome American geneticists Alfred Sturtevant (1891-1970) and Thomas Hunt Morgan (1866-1945) are considered to be the pioneers of...

Genetic and Cytological Mapping of Chromosomes A linkage/genetic/crossover map provides a diagrammatic representation of relative distances between linked genes of a chromosome American geneticists Alfred Sturtevant (1891-1970) and Thomas Hunt Morgan (1866-1945) are considered to be the pioneers of genetic mapping Fig: Alfred Sturtevant Fig: Thomas Hunt Morgan Genetic and Cytological Mapping of Chromosomes Its Significance By helping pinpoint the location of genes on a chromosome, genetic mapping assists: Study of genetic diseases, their causes and preventive measures Selection of desirable traits in breeding programs Study of phylogeny between 2 or more genetically close species Fig: Scientist studying genetic Biotechnological practices such as gene cloning, genetic map of mice modification, etc. Fig: Logo of the Human Genome Project The Human Genome Project (1990-2006) initiated by the US government was the largest collaborative biological project, involved in the genome mapping of humans, which has helped us understand phylogenetic evolution of humans far better Construction of a linkage map It was first found in Drosophila that the rate of crossing-over of genes and the distances between them were related i.e., shorter the distance between the genes, greater the chance of crossing-over between them How is Map Distance Determined? Genes are always present in a linear form in chromosomes, i.e. they are present in straight lines Customary units of light microscopy cannot be used, as the distance between genes are much smaller An arbitrary unit, called the Map Unit, is preferred by most geneticists The Morgan Unit is also used, in honor of TH Morgan, although Map Unit is more preferred The Map Unit (Centimorgan) If, on average, 1% of sample chromosomes combine/crossover, then the distance between them is 1 Map Unit. If 100% recombination frequency is seen, distance is 1 Morgan Unit. 1 Map Unit = 100 Morgan Unit =1 centimorgan (cM) If an F1 hybrid having genotype Ab/aB If the map distance between 2 gene loci produces 8% of crossover gametes AB B and C is 12 cM, then 12% gametes of and ab, then distance between A and B genotype BC/bc should be crossover is 16 cM. types. Each chiasma produces 50% crossover products, so the distance between each chiasma is 50cM. Thus, Total length of map = mean number of chiasmata × 50 cM Two-Point/Dihybrid Test Cross It is used to calculate the percentage of crossing over in an F1 dihybrid by crossing them with a double recessive parent Let, an F1 dihybrid Ac/ac is to be tested, it is crossed with a double recessive parent ac/ac Results of F2 generation: 37% dominant genes at both gene loci (AC/ac) 37% recessive genes at both gene loci (ac/ac) 13% dominant at 1st and recessive at 2nd gene loci (Ac/ac) 13% recessive at 1st and dominant at 2nd gene loci (aC/ac) Here, the groups Ac/ac and ac/ac are formed by crossing over, which means Percentage of crossing over = 13% + 13% = 26% Distance betn gene loci A and C = 26 cM Three-Point/Trihybrid Test Cross Because, double crossover usually do not occur between genes less than 5 cM apart, three- point test crosses are used. It utilizes the presence of a marker gene between two genes to ascertain the linear distance between them. Let, distance between genes A and C is tested using marker B, i.e. ABC/abc and abc/abc are tested, results of F2 generation: Parental Type Region I Region II Double Crossover 36% ABC/abc 9% Abc/abc 4% Abc/abc 1% AbC/abc 36% abc/abc 9% aBC/abc 4% abC/abc 1% aBc/abc Total=72% Total=18% Total=8% Total=2% NOTE: Double Crossovers may not always be detected in a Trihybrid Cross. Three-Point/Trihybrid Test Cross To ascertain the distance between A and C, distance between A and B, then B and C is required. For A to B distance Single crossovers of region I and double crossovers are counted Total crossovers=18%+2%=20% For B to C distance Single crossovers of region II and double crossovers are counted Total crossovers=8%+2%=10% Parental Type Region I Region II Double Crossover 36% ABC/abc 9% Abc/abc 4% Abc/abc 1% AbC/abc For A to C distance 36% abc/abc 9% aBC/abc 4% abC/abc 1% aBc/abc Total crossovers=20%+10%=30% Total=72% Total=18% Total=8% Total=2% Total distance=30 Map Units/cM Determination of Gene Order If genes A, B, C are tested, then, they can be in any one of the following orders: A-B-C B-A-C C-A-B Gene order is pretty simple to ascertain. Let us understand the following examples: Let, A-B distance=12 cM, B-C distance=7 cM, A-C distance=5 cM Case I. Assume C is in the middle (order A-C-B) A 5cM C C 7cM B Here 5+7=12 So, C must be in the middle A 12cM B Case II. Assume A is in the middle (order B-A-C) B 12cM A A 5cM C B 7cM C Here 12+5≠7 So, A cannot be in the middle