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By finding recombination frequencies for many gene pairs, we can make linkage maps that show the order and relative distances of the genes on the chromosome.
Step 2: Convert recombination frequency to map distance: Map Distance (cM) = RF × 100. Map Distance = 0.15 × 100 = 15 cM. Thus, the map distance between Gene A and Gene B is 15 centimorgans. Limitations and Challenges. While the process outlined above is relatively straightforward, it’s essential to be aware of some limitations and challenges:
A genetic map shows the map distance, in cM, that separates any two loci, and the position of these loci relative to all other mapped loci. The genetic map distance is roughly proportional to the physical distance, i.e., the amount of DNA between two loci.
Map distance = (total rec. events / total tetrads) x 100 = [(1/2[SCO] + DCO) / total tetrads] x 100 Map distance = ([1/2 (T – 2 NPD) + 4 NPD]/ total tetrads) x 100 Map distance = (1/2 T + 3 NPD) / total tetrads x 100 For our example above, map distance = ([1/2 (70) + 3 (3)] / 200) x 100 = 22 map units
Calculate the map distance between loci given the phenotypes of offspring or predict phenotypes of offspring given the recombination frequency between loci. Use the distance to construct genetic maps based on data from two-point or three-point testcrosses.
Analyzing the number of offspring of each phenotype will allow us to determine the order of the genes on the chromosome and the map distances between them. The first thing to do in solving a three-factor cross is to count the number of phenotypes seen in the offspring.
The recombination frequency is therefore a measure of the distance between two genes. If you work out the recombination frequencies for different pairs of genes, you can construct a map of their relative positions on the chromosome (Figure 5.17).
