Question

Using the test cross data in the table below, answer the following questions. Be sure to answer all three parts! phenotype co
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Answer #1

Determine gene order:

  • PDB - 756
  • pDB - 59
  • PdB - 414
  • PDb - 21
  • pdB - 23
  • pDb - 417
  • Pdb - 60
  • pdb - 750

The offspring is larger for the gene PDB (756) and pdb (750). So, it is the parental phenotype because of the preservation of the parental phenotype.

The phenotypes which are the smallest are PDb (21) and pdB (23). They are the double reciprocal because it requires a cross between Pand D and D and B.

Compare the parental phenotype with the double reciprocal. In this case the double cross PDb is similar to the parent phenoype PDB and the double cross pdB is similar to parental pdb. It suggest that B allele can presen with both alleles P or p and can be also present wit D and d. Thus, B/b allele should be in the middle.

SO, the position should be P-B-D or p-b-d.

Gene map:

  • The parents can have PBD and pbd.
  • The single cross overs are pDB, PdB, pDb and Pdb.
  • It can be written as pBD, PBd, pbD and Pbd.
  • the reciprocals are in pair: pBD and Pbd , PBd and pbD.
  • Compare the cross over with the parent (PBD/abd).
  • Recombination betrween B and P: pBD and Pbd.
  • Recombination between B and D: PBd and pdD.
  • Recombination of B and P: pBD (59) + Pbd (60) + PDb (21) + pdB (23) = 163

The reombination is in 2500 total offspring = 163 / 2500 = 0.065cM

  • Recombination of B and P: PBd (414) + pbD (417) + PDb (21) + pdB (23) = 875

The reombination is in 2500 total offspring = 875 / 2500 = 0.35cM

1. The parental phenotype: PDB (756) and pdb (750)

2. The middle gene is B/b.

3. The distance between P/p and B/b is 0.065cM and B/b and D/d is 0.35cM.

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