1. Chapter 5 Genetic Linkage and Chromosome Mapping
Jones and Bartlett Publishers © 2005

2. Terms to learn:
linkage linkage group
recombination
synteny
cis (coupling) slash (/)
trans (repulsion) +/ wild type
map unit
centiMorgan

3. Two genes on nonhomologous chromosomes (unlinked genes) yield 4 kinds of gametes in equal proportions

4. Recombinant gametes are created by recombination (crossing over) between homologous chromosomes

5. Recombination frequency
How often recombination occurs depends on how far apart the genes are:
If genes are close together, there will be very low recombination rates.

6. Another way of writing this is with a slash (“/”):
Two genes on the same chromosome (linked genes) can exist in two possible configurations
Another way of writing this is with a slash (“/”):
w m+ / w+ m and w m / w+ m+
= w + /+ m and w m / + +

7. Recombination frequency
How often recombination occurs depends on how far apart the genes are:
If genes are close together, there will be very low recombination rates.
If genes are farther apart, there will be higher rates of recombination.

8. Recombination frequency
R = proportion of recombinants = proportion of recombinant gametes produced by a heterozygote
= # recombinant gametes/ # total gametes x 100%
= # recombinant progeny / # total progeny x 100% (in a test cross).

9. Recombination frequency
Recombination can be expressed in four equivalent ways:
Frequency of recombination ( )
Percent recombination (0-50%)
Map distance in map units
A map distance in centiMorgans (cM) where 1 map unit = 1 cM.

10. Crossing over must occur between 2 genes to produce recombinant gametes
Here the crossing over did not occur between the 2 genes. As a result, all four gametes are nonrecombinant (parental combinations)

11. The genetic distance between 2 genes is expressed in map units (% recombination)

12. Drosophila example Normal (p+) vs. purple eye (p)
Normal (v+) vs. variegated wings (v)
Testcross p+v+/pv x pv/pv
Without recombination, all progeny will be p+v+/pv or pv/pv
If unlinked, all four progeny classes should be equal:
p+v+/pv, p+v/pv, pv+/pv, pv/pv
( = + +/p v, +v/pv, p+/pv, pv/pv)

13. Drosophila example But this is what was obtained: p+v+/pv 495 p+v/pv 7
R = recombinants/total = (7+8)/1000 = 1.5
So, these two genes are 1.5 cM apart.

14. Chi-square analyses for linkage
Statistical tests can be made to determine if observed data could be sampling variation, or are indeed evidence of linkage.
Independent segregation of each allele can be tested as well.
Degrees of freedom and X2 values are additive.

15. (Page 181)

16. Recombination frequency
Recombination frequency is usually greater than the actual proportion of recombinants, unless the two genes are very close together.
This is because double crossovers make genes appear to be closer than they are.

17. Linkage groups
If the recombination rates are measured for many pairs of genes that are linked, a linkage map can be made.
This linkage group map is a chromosome map, but underestimates actual distance due to double crossovers.

18. Two recombinations between a pair of genes result in 4 nonrecombinant gametes

19. Additivity of map distances can be used for creating genetic maps
For the 3 genes rb, y and cv with the map distances between rb-y and rb-cv shown in (A) above, there are 2 possible genetic maps. The distance y-cv of 1.3 map units yields map (B) and y-cv distance of 13.7 map units yields map (C).

20. A genetic map of chromosome 10 of corn

21. Use of cytologically marked chromosomes shows that crossing over involves breakage and reunion of chromosomes