1. Eukaryotic Chromosome Mapping
Using Genetic Recombination to Estimate Distances Between Genes

2. Linked Genes Mendel’s experiments Linked Genes Gene location
Genes on separate chromosomes
Genes on the same chromosome
Gamete types
Equal numbers of all possible allele combinations
More parental combinations than recombinant combinations

3. Independent Assortment vs. Gene Linkage
Example from Drosophila
Red eyes, x Pink eyes
Beige body Ebony body
RRBB rrbb
F1: Red eyes, Beige body
RrBb

4. Independent Assortment vs. Gene Linkage
Testcross: cross to individual of known genotype
F1:Red eyes X Pink eyes
Beige body Ebony body
RrBb rrbb

5. Independent Assortment vs. Gene Linkage
F2 phenotype
Number of Offspring
Expected for Unlinked Genes
Red eyes
Beige body
398
250
Pink eyes
Ebony body
382
108
112

6. Independent Assortment vs. Gene Linkage
F1:Red eyes X Pink eyes
Beige body Ebony body
RrBb rrbb RB Rb rB rb
RrBb
Red
Beige
Rrbb
Ebony
rrBb
Pink
rrbb rb

7. Independent Assortment vs. Gene Linkage
If genes are linked:
Red eyes, x Pink eyes Beige body Ebony body
F1: Red eyes, Beige body
R B
r b
R B
r b
Coupling
or Cis
Configuration

8. Independent Assortment vs. Gene Linkage
F1: Red eyes, Beige body
R B
r b X
R b
Four types of gametes are produced Parental Recombinant
R B
r b
r B

9. Independent Assortment vs. Gene Linkage
F1:Red eyes X Pink eyes
Beige body Ebony body
R B r b
r b r b
R B
r b
R b
r B
R B
r b
R b
r B
r b
r b
r b
r b
r b

10. Independent Assortment vs. Gene Linkage
F2 phenotype
Number of Offspring
Chromosome arrangement
Red eyes
Beige body
398
RB//rb
Parental
Pink eyes
Ebony body
382
rb//rb
108
Rb//rb
Recombinant
112
rB//rb

11. Genetic Map Units 1% recombination = 1 map unit = 1 centimorgan
x 100 = 22%
1000
These genes are located 22 map units apart on the same chromosome.

12. Limits of Genetic Mapping
Genes that are 50 map units apart will appear to assort independently.
The calculated distance between any TWO genes on the same chromosome should be less than 50 map units.

13. Predicting Gamete Frequencies for Linked Genes
Red eyes, x Pink eyes Ebony body Beige body
F1: Red eyes, Beige body
R b
r B
R b
r B
Repulsion
or Trans
Configuration

14. Predicting Gamete Frequencies for Linked Genes
F1: Red eyes Beige body
R b
r B
The genes are 22 map units apart, therefore we expect 22% recombinant gametes and 78% parental gametes.
R B
R b
0.11
0.39
r b
0.11
r B
0.39
0.22 recombinants
0.78 parentals

15. Using a Three-point Testcross to Determine Genetic Distance
A cross between two parental strains is used to produce a tri-hybrid (heterozygous for three genes).
The tri-hybrid is crossed to an organism that is homozygous recessive for all three genes.
Eight classes of offspring are analyzed to determine recombination frequencies.

16. Problem 1, Page 2-1
In corn, a strain homozygous for the recessive alleles a (green), d (dwarf) and rg (normal leaves) was crossed to a strain homozygous for the dominant alleles of each of these genes, namely A (red), D (tall) and Rg (ragged leaves). Offspring of this cross were then crossed to plants that were green, dwarf and had normal leaves. The following phenotypic classes were observed.

17. Problem 1, Page 2-1 Offspring Resulting from Three-Point Testcross 265
red, tall, ragged
green, dwarf, normal
red, tall, normal
green, dwarf, ragged
red, dwarf, normal
green, tall, ragged
red, dwarf, ragged
green, tall, normal
265
275 24 16 90 70
120
140

18. Problem 1, Page 2-1 A Red a Green D Tall d Dwarf Rg Ragged leaves rg
Normal leaves

19. Problem 1, Page 2-1 With Arbitrary Gene Order
A D Rg
a d rg X
a d rg
A D Rg F1
a d rg
A D Rg
Testcross X F2

20. Problem 1, Page 2-1 With Arbitrary Gene Order
a d rg
A D Rg X F2
Parentals: A D Rg
a d rg
Recombinants: A d rg
a D Rg
A D rg
a d Rg
A d Rg
a D rg
a d rg
Used as a genetic
background to see
the contribution from
the tri-hybrid.

21. Problem 1, Page 2-1 Determine which classes are parentals
The two parental classes will represent the largest number of offspring in the F2 generation.
Information on the parents may be given in the problem description itself.
Parentals: red, tall, ragged
green, dwarf, normal

22. Problem 1, Page 2-1 Determine which classes are double recombinants
Double recombinants have two crossovers: one between the first and middle gene and one between the middle and third gene
These will be the two smallest classes.
Double Recombinants: red, tall, normal
green, dwarf, ragged

23. Problem 1, Page 2-1 A rg D A Rg D a rg d a Rg d X
Determine the gene order
The middle gene is the one that changes places in the double recombinants when compared to the parental combinations.
A rg D
a Rg d
Red, tall, normal
Green, dwarf, ragged
A Rg D
a rg d X

24. Problem 1, Page 2-1 This shows why other gene orders are incorrect.
A d Rg
A D Rg
Red, dwarf, ragged X X
a d rg
a D rg
Green, tall, normal
D a Rg
D A Rg
Green, tall, ragged X X
d a rg
d A rg
Red, dwarf, normal

25. Contribution of F1 parent
Problem 1, Page 2-1
Assign genotypes to all classes
Use correct gene order
Contribution of F1 parent
red, tall, ragged
green, dwarf, normal
red, tall, normal
green, dwarf, ragged
red, dwarf, normal
green, tall, ragged
red, dwarf, ragged
green, tall, normal P
A Rg D
265
275 24 16 90 70
120
140 P
a rg d DC
A rg D DC
a Rg d
A Rg
A rg d
A Rg
a Rg D
Rg D
A Rg d
Rg D
a rg D

26. A rg d A Rg D a rg d a Rg D A rg D A Rg D a rg d a Rg d
Problem 1, Page 2-1
Recombination between A and Rg
Single Crossovers
A rg d
a Rg D
Red, dwarf, normal
Green, tall, ragged
A Rg D
a rg d X 90 70
Double Crossovers
A rg D
a Rg d
Red, tall, normal
Green, dwarf, ragged
A Rg D
a rg d X 24 16
Total = 200
Recombination = (200/1000) x 100 = 20 %

27. A Rg d A Rg D a rg d a rg D A rg D A Rg D a rg d a Rg d
Problem 1, Page 2-1
Recombination between Rg and D
Single Crossovers
A Rg d
a rg D
Red, dwarf, ragged
Green, tall, normal
A Rg D
a rg d X
120
140
Double Crossovers
A rg D
a Rg d
Red, tall, normal
Green, dwarf, ragged
A Rg D
a rg d X 24 16
Total = 300
Recombination = (300/1000) x 100 = 30 %

28. Problem 1, Page 2-1 Two maps are possible: A Rg D
20 map units map units or
D Rg A
30 map units map units

29. Interference
Interference: crossover in one region inhibits crossover in an adjacent region
Interference = 1 – (coefficient of coincidence)
Coefficient of coincidence =
Observed double crossovers
Expected double crossovers

30. Calculating Interference
Coefficient of coincidence =
Observed double crossovers =
Expected double crossovers
= 40 = 0.667
0.2 x 0.3 x
Interference =
1–(coefficient of coincidence) =
= 0.333