Consider the next testcross data using the genes E S and Z T

Consider the next test-cross data, using the genes E, S and Z: The distribution of offspring phenotypes from the cross is: Total number of offspring: Add up the number of offspring in each crossover class, as well as total offspring, before continuing. What is the order of these genes (E, S and Z)? What distances separate adjacent genes in this group? Draw a map of the linkage group ESZ- then fit this map into that of the other genes (Problems 1 & 2):

Solution

a. The order of these genes are E Z S. Because the frequency of double cross over will be less and in this case the number of ESz and esZ recombinants are less it is only 1 and 2 respectively. Hence the gene Z is in the middle.

b. Total number of progeny = 31+33+7+9+1+2+5+7 = 94

To find the distance between E and Z = (5+7+1+2)/94 x 100 = 15.9%

To find the distance between Z and S = (7+9+1+2)/94 x 100 = 20.2%

Hence the distance between the genes are E ---- 15.9---- Z -------20.2--- S