Two linked genes are separated by a distance such that exact

Two linked genes are separated by a distance such that exactly 4 percent of cells in meiotic prophase have one crossover between the genes, and 96 percent of the cells in meiotic prophase have no crossovers between the two genes. What is the map distance between these two genes? Since half of the chromosomes from a single exchange meiosis will be recombinants, and half will be non-recombinants, there will be 2% recombinant chromosomes, or 2 cM. A triply heterozygous parent was testcrossed to produce a map of three linked genes. The following results were obtained: FGH 4 FGh 41

Solution

2) Exactly 4% of the cells have crossover between the genes and 96% of the genes have no crossovers. That means , two type of offsprings or alleles are favoured at the expense of other two. It is assumed that this is a Di hybrid cross leading to 4 outcomes of 25%, each in case of crossover, but instead a percentage of 4-96 is obtained. This can be further divided such that 2 of the offsprings have 96% chance I.e. They are linked, providing a 48% favourability for the two linked genes each.

The other two outcomes , forming out of crossovers have a percentage of 4, I.e. The two recombinant outcomes will have 2% each, as a chance of crossover. Again, this implies for linkage between the other two genes.

As known 1 centiMorgan = 1% chance of recombination between the loci = ~1Million bps. Therefore, here, 4% implies of 4cM , is the distance between the alleles, that can be measured by the frequency of recombination/crossover between the alleles and 2cM distance between the chromosomes, if studied for each chromosome.