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 Recombinantgametes
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.
A A B B a a b b A a B b a a b b A a b b A a B b 1 2 3 4 5 6 Two Loci Inheritance Recombinant
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. 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.
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.
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.
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 arecombined to get a more reliable estimate of the recombination fraction. • 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. • Calculate the Lod score = log(b/a). 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 (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: • 31 brown, chinchilla • 35 black, full • 16 brown, full • 19 black, chinchilla • Determine the genotype of the heterozygous parents, and the map distance between the 2 genes.
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 (16 + 19) / (31 + 34 + 16 + 19) 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 BFbf.
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.
Question #2 (cont.) The following data was obtained: Total: 1000 • Draw a chromosome map for the three genes.
1 2 1 2 3 4 6 7 8 9 10 11 12 13 14 15 16 5 1 2 4 5 3 Question #3 - Detecting & Measuring Linkage in Humans