Question

In Drosophila, the dwarp mutation leads to flies with small bodies and warped wings, the rumpled mutation causes bristles to be deranged, the pallid mutation makes wings pale, while raven leads to dark eyes and bodies. Females heterozygous for all 4 genes are crossed to males from a true-breeding dwarp rumpled pallid raven stock. The 1000 progeny obtained were as follows:

            pallid                                       3

            pallid, raven                            428

            pallid, raven, rumpled             48

            pallid, rumpled                        23

            dwarp, raven                           22

            dwarp, raven, rumpled            2

            dwarp, rumpled                       427

            dwarp                                      47

Indicate the best map for these four genes and calculate interference values where appropriate.

Answer 1

Solution:

Suppose Drosophilla females of heterozygous genotype AaBbCc for all genes are crossed to males of genotype aabbcc from a true breeding led to 1000 progengy of the following types. Consider genotypes for gametes are pallid, raven (ABC), dwarp, rumpled (abc), pallid, raven, rumpled (aBC), dwarp (Abc), pallid, rumpled (ABc), dwarp, raven (abC), pallid (AbC) and dwarp, raven, rumpled (aBc)

pallid 3

            pallid, raven                            428

            pallid, raven, rumpled             48

            pallid, rumpled                        23

            dwarp, raven                           22

            dwarp, raven, rumpled            2

            dwarp, rumpled                       427

            dwarp                                      47

First rearrange the above given data as follow:

pallid, raven                            428 (Prental gametes)

dwarp, rumpled                       427 (Prental gametes)

pallid, raven, rumpled             48 (Single crossover between A and B)

dwarp                                      47 (Single crossover between A and B)

pallid, rumpled                        23 (Single crossover between B and C)

dwarp, raven                           22 (Single crossover between B and C)

pallid 3 (Double crossover)

dwarp, raven, rumpled            2   (Double crossover)

So we can deduce order of genes - ABC.

Single crossover frequency between A and B = (48 + 47+ 3+ 2) / 1000

= 100 /1000

= 0.1

Single crossover frequency between B and C = (23 + 22 + 3+ 2) / 1000

= 50/ 1000

= 0.05

Thus, expected double crossover frequency = 0.1 x 0.05 = 0.005

Number of expected double crossovers = 0.005 x 1000 = 50

And the actual double crossovers only 3+2 = 5 (observed double crossover

The coefficient of coincidence   = 5 / 50 = 0.1

Interference = 1 -  Coefficient of coincidence

= 1 - 0.1

Interference    = 0.9

  

A) Gene map genes - gene order is ABC = 15 cM

For A to B = (48 + 47+ 3+ 2 ) / 1000 X 100 = 10 cM

For B to C = (23 + 22 + 3+ 2) / 1000 X 100 = 5 cM

This map is A 10cM B 5 cM C

B) Interference = 1 -  Coefficient of coincidence

= 1 - 0.1

  Interference    = 0.9 ( value indicates partial interference)