Question

genetics-please show all work

8. (9 pts) A series of two-point crosses were carried out among six loci (a, b, c, d, e and f), producing the following recombination frequencies. The six loci map to three different linkage groups. Map the six loci, showing their linkage groups (you can use a single line for each linkage group), the order of the loci in each linkage group, and the distances between the loci of each group. If there is only one locus in a linkage group, you do not need to determine order or distance. 50 50 50 Loci Percent Recombination a and b a and c a and d a and e 25 a andf b and cT band d 50 bande bandf cand d cand e candf d and e d and f e and 50 50

Answer 1

f b + +

Since a and e have a lesser percentage of recombination, these two are likely to be in the same linkage group, so is e and f. But since a and f have 50% recombination they do not exhibit any genetic linkage but in the same linkage group since both of them exhibit genetic linkage to e. Similarly, c and d are also of the same linkage group. But b, having exhibited no genetic linkage at all belong to one single linkage group.

Percentage recombination between two loci indicates their map distance. That is 1 % recombination = 1 m.u.

So, Distance between a & e = 25 m.u.

  Distance between a & f = 50 m.u.

  Distance between e & f = 35 m.u.

  Distance between c & d = 18 m.u.