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Tutorial #2 by Ma’ayan Fishelson. Crossing Over Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over. New combinations.

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Presentation on theme: "Tutorial #2 by Ma’ayan Fishelson. Crossing Over Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over. New combinations."— Presentation transcript:

1 Tutorial #2 by Ma’ayan Fishelson

2 Crossing Over Sometimes in meiosis, homologous chromosomes exchange parts in a process called crossing-over. New combinations are obtained, called the crossover products.

3 Recombination During Meiosis Recombinant gametes

4 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.

5 Two Loci Inheritance Recombinant 2 1 A B a b A a B b 34 a b A a b 56 A a B b

6 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.

7 Recombination Fraction 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.

8 Centimorgan (cM) 1 cM (or 1 genetic map unit, m.u.) is the distance between genes for which the recombination frequency is 1%.

9 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.

10 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).

11 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.

12 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.

13 Question #1 In rabbits, black (B) is dominant to brown (b), while full color (F) is dominant to chinchilla (f). 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, chinchilla o35 black, full o16 brown, full o19 black, chinchilla Determine the genotype of the heterozygous parents, and the map distance between the 2 genes.

14 Question #1-Solution The information given about the phenotypes of the parents and the dominance relationships amongst these alleles indicates that this cross was a testcross: BbFf x bbff We’re given the numbers for each of the expected phenotypic classes amongst the offspring: 31 bbff 34 BbFf 16 bbFf 19 Bbff The latter two classes, which have considerably fewer members, must be the non- parentals. Hence we can calculate the recombination frequency as ( ) / ( ) or 35%. An examination of the genotypes of the parental classes indicate that the heterozygous parents must have had one chromosome with the B & F allele on it and the other chromosome with the b & f allele. We can represent this as BF/bf or BF bf.

15 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 red mutation.

16 Question #2 (cont.) The following data was obtained : Total: 1000  Draw a chromosome map for the three genes. NumberPhenotype 404Ly Sb br Ly Sb br 75Ly - br 59- Sb - 4Ly Sb br

17 Question #2 – (solution highlights) NumberPhenotype 404Ly Sb br Ly Sb br 75Ly - br 59- Sb - 4Ly Sb br Number of recombinations Genes = 168 Ly/Sb = 40 Ly/br = 140 Sb/br

18 Question #3 - Detecting & Measuring Linkage in Humans

19 Question #3 - (solution highlights) TotalDiseasedPhenotype 81OO 00BB 119BO 30AO


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