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Linkage & Recombination. Tutorial #2 by Ma’ayan Fishelson. Crossing Over. Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over . New combinations are obtained, called the crossover products. Recombination During Meiosis. Recombinant gametes. - PowerPoint PPT Presentation
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Tutorial #2by Ma’ayan Fishelson
Crossing Over
• Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over.
• New combinations are obtained, called the crossover products.
Recombination During Meiosis
Recombinant gametes
Linkage• 2 genes on separate chromosomes assort
independently at meiosis.
• 2 genes far apart on the same chromosome can also assort independently at meiosis.
• 2 genes close together on the same chromosome pair do not assort independently at meiosis.
• A recombination frequency << 50% between 2 genes shows that they are linked.
Two Loci Inheritance
Recombinant
2 1A AB B
a ab b
A aB b 3 4
a ab b
A ab b 5 6
A aB b
Linkage Maps• Let U and V be 2 genes on the same chromosome.
• In every meiosis, chromatids cross over at random along the chromosome.
• If the chromatids cross over between U & V, then a recombinant is produced.
The farther apart U & V are the greater the chance that a crossing over would occur between them the greater the chance of recombination between them.
Recombination Fraction
Linkage) No(5.0)ionRecombinat(0)Linkage( P
• The recombination fraction between two loci is the percentage of times a recombination occurs between the two loci.
• is a monotone, nonlinear function of the physical distance separating between the loci on the chromosome.
Centimorgan (cM)
• 1 cM (or 1 genetic map unit, m.u.) is the distance between genes for which the recombination frequency is 1%.
Interference
• Crossovers in adjacent chromosome regions are usually not independent. This interaction is called interference.
• A crossover in one region usually decreases the probability of a crossover in an adjacent region.
tsrecombinan double of # expected
tsrecombinan double of # observerd1ce(I)Interferen
Building Genetic Maps• At first: only genes with variant alleles
producing detectably different phenotypes were used as markers for mapping.
• Problem: the chromosomal intervals between the genes were too large the resolution of the maps wasn’t high enough.
• Solution: use of molecular markers (a site of heterozygosity for some type of silent DNA variation not associated with any measurable phenotypic DNA variation).
Linkage Mapping by Recombination in
Humans.• Problems:
– It’s impossible to make controlled crosses in humans.
– Human progenies are rather small.
– The human genome is immense. The distances between genes are large on average.
Lod Score for Linkage Testing by Pedigrees
The results of many identical matings are combined to get
a more reliable estimate of the recombination fraction.
1. Calculate the probabilities of obtaining a set of results in a family on the basis of (a) independent assortment and (b) a specific degree of linkage.
2. Calculate the Lod score = log(b/a).
A Lod score of 3 is considered convincing support for a specific recombination fraction.A Lod score of 3 is considered convincing support for a specific recombination fraction.
Question #1• In rabbits, black (B) is dominant to brown (b), while
full color (C) is dominant to chinchilla (c). The genes controlling these traits are linked.
• The following cross was made: rabbits heterozygous for both traits that express black, full color, with rabbits that are brown, chinchilla. The following results were obtained:
o31 brown, chinchillao35 black, fullo16 brown, fullo19 black, full
• Determine the genotype of the heterozygous parents, and the map distance between the 2 genes.
Question #2
On chromosome 3 in Drosophila, there are the following mutations:• Lyra (Ly) and Stubble (Sb) which are
dominant mutations. • A recessive mutation with bright red eyes
(br).
A female heterozygous for the 3 mutations was mated to a male homozygous for the bright redmutation.
Question #2 (cont.)
The following data was obtained:
Draw a chromosomemap for the three genes.
PhenotypeNumber
Ly Sb br404- - -422Ly - -18 - Sb br16Ly - br75
-Sb- 59Ly Sb- 4
- -br2
Question #3 - Detecting & Measuring Linkage in
Humans
1 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5