# Genetic Mapping - Genes located close together on the chromosome assort with one another more regularly (linkage) than genes farther apart, which would have a greater probability of breaking and rejoining (less linkage) - The rate of crossing over could be used as a tool for determining the distance between two genes or loci, and was used to *map the relative positions of genes* on chromosomes ??: - Linkage: 50% to 100% - When the distance between two genes exceeds 50 cM, the genes are considered to assort independently (“randomly”) - One map unit = 1 cM = 1% recombination/crossing over (can go over 100%) - = ~250k bp in Drosophila - = ~1 million bp in humans (variable) - The farther apart two genes are on a chromosome, the greater the chance of crossing over. At 50% (50 cM) genes assort independently and show 50% linkage (never less) - Even when two genes are at the far ends of a very long chromosome (which would provide more opportunities for crossing over), they sort together at least 50% of the time because multiple crossovers can occur (it is essentially random beyond the 50 cM range) - single or odd # crossovers result in crossover gametes - double or even # crossovers result in parental gametes - For mapping, traits must be within 50% (50 cM) in theory - <25% (25 cM) in practice (must be closer) due to double crossovers, and since offspring must be produced and counted **Using Markers to Map Genes** - If a trait is within 50cM of a known trait/marker, it can be mapped to EITHER side of that loci - For exact place, a second marker must be used, spaced ~50 cM apart - ![[Pasted image 20240201064357.png|300]] - Markers Needed = Round up (Mb/50) - 1 - 1 Mbp (megabase pair) = 1 million base pairs ## The Two-Factor Cross - One of the arms from each chromosome must cross over during meiosis, meaning that at least 1 cross over event will always occur **The Two-Factor Cross** - Three individual, two factor crosses can be performed (A vs C vs B → AC vs BC vs AB) to find the relative placement of each gene - Account for Double Crossovers - ![[Pasted image 20240201065726.png|300]] - $AB=AC+BC-2(AC)(BC)$ - 0.10 + 0.25 - 2(0.10)(0.25) = 0.30 (Single Crossover) - 0.05 (Double Crossover) ![[Pasted image 20240201065917.png|300]] - A three-factor cross can be performed instead, which only requires a single cross (rather than 3) - The least frequent pair must be the double crossover - ![[Pasted image 20240201070717.png|300]]