Specialized Mapping Finding Chromosomal Locations Using Tetrad Analysis To Study Genetic Distances (see Tetrad Analysis Web Module for Chapter 7 on Text Web Site http://whfreeman.com/pierce/)
Physical Chromosome Mapping Somatic-Cell Hybridization Using human-rodent somatic cell hybrids to study the location of genes on chromosomes
Human-mouse hybrid cells with different numbers of human chromosomes (blue).
Physical Chromosome Mapping How can we determine which chromosome carries a specific gene? In human-mouse hybrid cells, a 1:1 correspondence exists between the presence of the enzymatic activity for the gene and the presence of the chromosome carrying the gene.
Problem 4, Page 2-2 Hybrid cells containing human and mouse chromosomes were analyzed. The grid on the left shows the presence or absence of each of four human chromosomes in hybrid cell lines A through D. The grid on the right shows the presence or absence of human enzyme activity in each of the cell lines. Assign the gene for each enzyme to the chromosome that carries the gene. Human Enzyme Human Chromosome ADH PEP HexA GAPDH A + - B C D 5 7 11 18 A - + B C D Hybrid Cell Line Hybrid Cell Line
Problem 4, Page 2-2 A + - B C D 5 7 11 18 A - + B C D Human Chromosome Human Enzyme ADH PEP HexA GAPDH A + - B C D 5 7 11 18 A - + B C D Hybrid Cell Line Hybrid Cell Line
Problem 4, Page 2-2 A + - B C D 5 7 11 18 A - + B C D Human Chromosome Human Enzyme ADH PEP HexA GAPDH A + - B C D 5 7 11 18 A - + B C D Hybrid Cell Line Hybrid Cell Line
Problem 4, Page 2-2 A + - B C D 5 7 11 18 A - + B C D Human Chromosome Human Enzyme ADH PEP HexA GAPDH A + - B C D 5 7 11 18 A - + B C D Hybrid Cell Line Hybrid Cell Line
Problem 4, Page 2-2 A + - B C D 5 7 11 18 A - + B C D Human Chromosome Human Enzyme ADH PEP HexA GAPDH A + - B C D 5 7 11 18 A - + B C D Hybrid Cell Line Hybrid Cell Line
Physical Chromosome Mapping How can we determine which portion of a chromosome carries a specific gene? If the enzymatic activity is present in a cell line with an intact chromosome but missing from a line with a deletion in that chromosome, the gene for the enzyme is in the deleted region.
Enzyme Activity Present with intact Chromosome 4 Gene is located on the short arm of Chromosome 4, in the region missing from Cell line 3 Enzyme Activity Present with intact Chromosome 4 Absent without Chromosome 4 Absent when short arm of Chromosome 4 is deleted
Analysis of all four products of a single meiosis Two Types of Tetrads Tetrad Analysis Analysis of all four products of a single meiosis Two Types of Tetrads Ordered Tetrad Unordered Tetrad
Producing an Ordered Tetrad
Genetic Analyses with Tetrads Cross two haploid cells a b a+ b+ X a b a+ b+ Induce diploid to undergo meiosis
Genetic Analyses with Tetrads a b a+ b+ a b a+ b+ Parentals a b a+ b+ X Recombinants a b+ a+ b
MI Segregation Pattern A first-division segregation pattern, MI No crossover between gene and centromere
MII Segregation Pattern Crossover between gene and centromere
Types of Tetrads a b a+ b+ a b+ a+ b a+ b a+ b a a+ a+ a a a+ a+ a+ Parental Ditype (PD) Non-parental Ditype (NPD) Tetratype (T) MI pattern (both genes adjacent) MII Pattern (at least two alleles separated) a b a+ b+ a b+ a+ b a+ b a+ b a a+ a+ a a a+ a+ a+ a a
Producing MII Segregation Patterns
Producing MII Segregation Patterns
Calculating Genetic Distances with Tetrad Analysis Unordered Ordered (Linear) Example Yeast Neurospora Gene-Gene Distance RF= 1/2T + NPD total Gene-Centromere Cannot be determined ½ MII/total
Problem 2, Page 2-1 In a Neurospora cross of ab x a+b+, the following classes and numbers of tetrads were produced. Neurospora produces ordered tetrads that undergo a single mitosis after formation. Pairs of spores are listed below for simplicity.
Problem 2, Page 2-1 ab a+b+ ab+ a+b 71 1 18 8 Type For a For b
Problem 2, Page 2-1 ab a+b+ ab+ a+b 71 1 18 8 Type For a For b PD MI Type For a For b PD MI NPD MI T MI MII T MII MI PD MII NPD MII T MII
Problem 2, Page 2-1 Distance from a centromere = ½ MII = ½ (1+8+1) = 0.05 = 5 map units Total 100 Distance from b centromere = ½ MII = ½ (18+8+1) = .135 = 13.5 map units Total 100 Distance from ab = ½ T + NPD = ½ (18+1+1) +1 = .11 = 11 map units Total 100
Problem 2, Page 2-1 b 13.5 mu 5 mu a a b 11 mu a b 5 mu 8.5 mu Best solution