Eight genes designated ad b c ca pr sc v and wg were mapped

Eight genes, designated ad, b, c, ca, pr, sc, v, and wg, were mapped by two point crosses. For each pair of genes examined, the total percentage of non parental phenotypes observed in a test cross of a F1 dihybrid female with double recessive male testers is given.

Cross Genes % of Non parental types

1 pr-b 12

2 pr-vg 24

3 pr-sc 50

4 pr-v 50

5 pr-ca 50

6 b-vg 12

7 b-c 22

8 vg-c 10

9 vg-ad 15

10 c-ad 5

11 c-v 50

12 sc-v 2

13 sc-ca 50

A. HoW MANY LINKAGE GROUPS WERE REVEALED?

B. Construct genetic map of each linkage group, showing gene and distances. It is just from a single experiment

Solution

Linkage is a table for a species or population that shows the position of its genes and markers relative to each other in terms of recombination manner, genetic map will based on the frequencies of recombination between markers at time of crossover. If greater the frequency of recombination between two genetic markers, the further apart they will be assumed and if lower the frequency of recombination between the markers, the smaller the physical distance between them. Unit used to determine genetic linkage is the centimorgan (cM) which is equivalent to percentage of cross.

2.   pr-vg     24 = 24%

3.   pr-sc     50 = 50%

4.   pr-v       50 = 50%

5.   pr-ca     50 = 50%

6.   b-vg      12 = 12%

7.   b-c        22 = 22%

8.   vg-c      10 = 10%

9.   vg-ad    15 = 15%

10.       c-ad   5 = 5%

11.       c-v     50 = 50%

12.       sc-v    2 = 2%

13.       sc-ca 50 = 50%

Linkage is present when fewer than 50%

A.Total number of linkage group present there are Five

B.1. At very small distances, crossover is very rare, and most gametes are parental.

2. As the distance between two genes increases, crossover frequency increases. More recombinant gametes, fewer parental gametes.