1. Specialized Mapping Finding Chromosomal Locations
Using Tetrad Analysis To Study Genetic Distances (see Tetrad Analysis Web Module for Chapter 7 on Text Web Site

2. Physical Chromosome Mapping
Somatic-Cell Hybridization
Using human-rodent somatic cell hybrids to study the location of genes on chromosomes

3. Human-mouse hybrid cells with different
numbers of human chromosomes (blue).

4. 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.

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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.

11. 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

12. 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

13. Producing an Ordered Tetrad

14. Genetic Analyses with Tetrads
Cross two haploid cells
a b
a+ b+ X
a b
a+ b+
Induce diploid to undergo meiosis

15. Genetic Analyses with Tetrads
a b
a+ b+
a b
a+ b+
Parentals
a b
a+ b+ X
Recombinants
a b+
a+ b

16. MI Segregation Pattern
A first-division
segregation
pattern, MI
No crossover between gene and centromere

17. MII Segregation Pattern
Crossover between gene and centromere

18. 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

19. Producing MII Segregation Patterns

20. Producing MII Segregation Patterns

21. 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

22. 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.

23. Problem 2, Page 2-1 ab
a+b+
ab+
a+b 71 1 18 8
Type
For a
For b

24. 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

25. Problem 2, Page 2-1 Distance from a centromere =
½ MII = ½ (1+8+1) = 0.05 = 5 map units
Total
Distance from b centromere =
½ MII = ½ (18+8+1) = .135 = 13.5 map units
Total
Distance from ab =
½ T + NPD = ½ (18+1+1) +1 = .11 = 11 map units
Total

26. 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