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two genes are located on the same chromosome and are known to be 12 mu apart. An AABB individual was crossed to an aabb individual to produce AaBb offsoring. the AaBb offspring was then crossed with aabb individuals. if this cross produces 1000 offsprings, what are the predicted numbers of offsprings with each of the four genotypes: AaBb,Aabb,aaBb, and aabb?

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Answer:

440 AaBb, 60 Aabb, 60 aaBb, and 440 aabb

Explanation:

Genes A and B are 12 mu apart. 1 mu = 1% recombination frequency.

Initial cross:

AB/AB x ab/ab

      AB/ab

Offspring testcross:

AB/ab x ab/ab

The homozygous recessive individual can only produce ab gametes.

The F1 individual can produce 4 types of gametes:

  • AB, parental
  • ab, parental
  • Ab, recombinant
  • aB, recombinant

The genes are separated by 12 map units, so there is 12% recombination between them. There are 2 possible recombinant gametes, so each of them will appear in 6% of the progeny. The non-recombinant gametes (100% - 12% = 88%) are the parentals, and there are two of them so each will appear in 44% of the progeny.

Among 1000 offspring then, the expected numbers of each genotype will be:

  • AB/ab 0.44x1000= 440
  • ab/ab 0.44x1000= 440
  • Ab/ab 0.06x1000= 60
  • aB/ab 0.06x1000= 60

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