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In maize, liguleless (l) is recessive to having ligules (L), and dwarf (d) is recessive to tall (D). These two genes are linked on chromosome 3. A cross is made and the F2 consisted of the following progeny:

F2 phenotype Number of progeny
Ligules, tall 6156
Liguleless, tall 1344
Ligules, dwarf 1344
Liguleless, dwarf 1156

a. In this cross, are these genes in coupling or repulsion? Why?
b. What were the phenotype and the genotype of the P and F1 generations?
c. Calculate the percent recombination between these two genes.

Answer :

Answer:

given,

LL║ll  and DD║dd,

after crossing over,    Ld║Ld  ×   lD║lD

LLdd × llDD

in F1 generation, they will give, LLll, DDdd, LLDD, lldd → LlDd

in eF2 generation, the offsprings will be, LLll  ×  DDdd

LLDD ,LLdd, LLDd, llDD, lldd, llDd, LlDD, Lldd, LlDd

1. yes, these genes are coupling.

2. → phenotypes of parental genes = ligules, liguleless, tall and dwarf

    genotypes of parental genes= LLdd, llDD i.e; heterozygous

 →   phenotypes of F1 generation = 25% Ligules, 25% tall, 25% ligules-tall, 25% liguleless-dwarf

      genotypes of F1 generation = heterozygous ; LlDd that is ligules-tall

3. recombination frequency= no of recombinant / total progeny × 100%

                                              = (1344+1344) / 6156+1344+1344+1156  × 100%

                                              = 2688/ 10000 × 100%

                                              = 26.88 %

that means linkage happened

and the map distance is 26.88 map unit.

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