Chapter 5 Genetic Linkage and Chromosome Mapping Jones
- Slides: 21
Chapter 5 Genetic Linkage and Chromosome Mapping Jones and Bartlett Publishers © 2005
Terms to learn: linkage recombination synteny cis (coupling) trans (repulsion) map unit centi. Morgan linkage group slash (/) +/ wild type
Two genes on nonhomologous chromosomes (unlinked genes) yield 4 kinds of gametes in equal proportions
Recombinant gametes are created by recombination (crossing over) between homologous chromosomes
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.
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 = w + /+ m and w m / w + m+ and wm/++
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.
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).
Recombination frequency • Recombination can be expressed in four equivalent ways: • • Frequency of recombination (0. 00 -0. 50) Percent recombination (0 -50%) Map distance in map units A map distance in centi. Morgans (c. M) where 1 map unit = 1 c. M.
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)
The genetic distance between 2 genes is expressed in map units (% recombination)
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)
Drosophila example • • But this is what was obtained: p+v+/pv 495 p+v/pv 7 pv+/pv 8 pv/pv 490 R = recombinants/total = (7+8)/1000 = 1. 5 So, these two genes are 1. 5 c. M apart.
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 X 2 values are additive.
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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.
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.
Two recombinations between a pair of genes result in 4 nonrecombinant gametes
Additivity of map distances can be used for creating genetic maps For the 3 genes rb, y and cv with the map distances between rby 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 ycv distance of 13. 7 map units yields map (C).
A genetic map of chromosome 10 of corn
Use of cytologically marked chromosomes shows that crossing over involves breakage and reunion of chromosomes
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