Given data is: Phenotype of Testcross Genotype of Gametes Number of Progeny Wild

Question

Given data is:

Phenotype of Testcross Genotype of Gametes Number of Progeny
   Wild type                         Lz Gl Su                         286
   Sugary                              Lz Gl su                            4
   Glossy, sugary                   Lz gl su                            40
   Lazy, glossy, sugary            lz gl su                            272
   Lazy                                  lz Gl Su                            33
   Lazy, sugary                     lz Gl su                              44
   Glossy                              Lz gl Su                            59
   Lazy, glossy                        lz gl Su                            2

Total progeny = 740



Recombinant frequency between Lz and Gl = 40 + 33+ 44+ 59 = 176 / 740 = 0.24

Recombinant frequency between Gl and Su = 4 + 44+ 59+ 2 = 109 / 740 = 0.14

Recombinant frequency between Lz and Su = 4+ 40+ 33+ 2 = 79 / 740 = 0.10

From above frequencies the map distances are

Lz----------Su---------------Gl

ß--0.10--àß----0.14---à

ß--------0.24-------------à


Actual double recombinant frequency = 44 + 59 = 103

Expected double recombinant frequency = 176 / 740 * 109 / 740 = 0.035 = 25.9 per 740.


coefficient of coincidence = 103 / 25.9 = 3.97


Interference = 3.97 – 1 = 2.97

               :::::QUESTION::::: Using the data from question 1 and the following information, can you fit these 4 genes into chromosome 3 of these flowering plants you have been studying:

a) The chromosome is only 52 m.u. long,
b) Gene Fr is flanking at one end, Gene K is on the opposite and gene Lz is directly in the middle,
c) Gene K and Gl almost never have recombination events take place between them,
d) Gene MM is 20 m.u. away from Su. E) Gene Bs is 10 m.u. away from MM and not far from Fr.

Map the whole chromosome with all 7 genes. Given data is:

Phenotype of Testcross Genotype of Gametes Number of Progeny
   Wild type                         Lz Gl Su                         286
   Sugary                              Lz Gl su                            4
   Glossy, sugary                   Lz gl su                            40
   Lazy, glossy, sugary            lz gl su                            272
   Lazy                                  lz Gl Su                            33
   Lazy, sugary                     lz Gl su                              44
   Glossy                              Lz gl Su                            59
   Lazy, glossy                        lz gl Su                            2

Total progeny = 740



Recombinant frequency between Lz and Gl = 40 + 33+ 44+ 59 = 176 / 740 = 0.24

Recombinant frequency between Gl and Su = 4 + 44+ 59+ 2 = 109 / 740 = 0.14

Recombinant frequency between Lz and Su = 4+ 40+ 33+ 2 = 79 / 740 = 0.10

From above frequencies the map distances are

Lz----------Su---------------Gl

ß--0.10--àß----0.14---à

ß--------0.24-------------à


Actual double recombinant frequency = 44 + 59 = 103

Expected double recombinant frequency = 176 / 740 * 109 / 740 = 0.035 = 25.9 per 740.


coefficient of coincidence = 103 / 25.9 = 3.97


Interference = 3.97 – 1 = 2.97

               :::::QUESTION::::: Using the data from question 1 and the following information, can you fit these 4 genes into chromosome 3 of these flowering plants you have been studying:

a) The chromosome is only 52 m.u. long,
b) Gene Fr is flanking at one end, Gene K is on the opposite and gene Lz is directly in the middle,
c) Gene K and Gl almost never have recombination events take place between them,
d) Gene MM is 20 m.u. away from Su. E) Gene Bs is 10 m.u. away from MM and not far from Fr.

Map the whole chromosome with all 7 genes.

Explanation / Answer

According to given data the map of whole genome is Fr------Bs------MM------Lz------Su------Gl------K

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