Question

Find the distance in linkage map units of the recombinants (recombinants/total)x100: Mating: AaBb x aabb A...

Find the distance in linkage map units of the recombinants (recombinants/total)x100:

Mating: AaBb x aabb

A = albino, a = non-albino, B = Blue, b = white

Offspring:

(AaBb) Albino / Blue = 1120

(aaBb) Non-albino / blue = 100

(Aabb) Albino / white = 150

(aabb) Non-Albino / white= 1130

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Answer #1

Answer:

AaBb x aabb

  • the offsprings of the above cross are parental type( non recombinants) or non parental type ( recombinants).
  • The recombinants have the lowest frequency, therefore non hairy/ early and hairy/ late are recombinant progeny.
  • The distance between the hairness gene and flower fertility gene = (number of recombinant offsprings/ total offsprings) x 100
  • = ( 150 + 100 / 1120 + 150 + 100 + 1130) x 100
  • = 250 / 2500 x 100
  • = 10 map units
  • 11 &12) The distance in map units does not tell the actual physical distance in basepairs between the genes.
  • It also does not tell the base composition of the two genes.

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