4 In a threepoint linkage analysis the RF values for genes A

4. In a three-point linkage analysis, the RF values for genes A, B and C were determined as the following: A-B (RF=35%), B-C (RF=32%), A-C(RF=50%). Please explain if this analysis identifies the gene located in the middle and describes the chromosome distance between each pair of the three genes in cM.

5. Consider the following cross examining four unlinked genes. Identify what proportion of progeny will be phenotypically identical to parent 2? Parent 1: A/a; B/b; D/d; e/e ´ Parent 2: A/a; b/b; D/d; E/e

6. In the following cross, the genes are inherited independently except for A and B, which are tightly linked and show zero recombination: A/A . b/b; D/D; e/e; F/F X a/a . B/B; d/d; E/E; f/f. Following selfing of the F1 hybrid what proportion of the F2 individuals will be phenotypically like either parent?

Solution

Recombination frequency helps to findout the crossingover frquency and again distance between genes is directly proportional to RF. As 1% RF equals to 1 map unit or 1cM, A-B (RF=35%) , the distance between A and B is 35 m.u. or cM B-C (RF=32%) , the distance between B and C is 32 m.u. or cM A-C(RF=50%) , the distance between A and C is 50 m.u. or cM C..................................A.................B ..............50.............................35........ AB and AC are linked genes as they have RF below 50% and RF 50% of AC describes that genes are located on different chromosomes.