Question

Question 33 10 pts Using the test cross data in the table below, answer the following questions. Be sure to answer all three parts! count phenotype PDB PDB PdB PDb pdB pDb Pdb pdb total 756 59 414 21 23 417 60 750 2500 1. Which genotypes are non-recombinants? (2 points) 2. Which gene is in the middle? (2 points) 3. Determine the genetic distances between each gene on the "outside” of the group of three to the gene in the “middle.” Show your work, and indicate your answer in centimorgans. (6 points)

Answer 1

According to the question, the phenotype and counts are

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

1) The non-recombinants: The phenotypes present in the large number must be the parent because the phenotype is preserved. So, the phenotype must be PDB and pdb.

2) Determination of the order:

• The parent phenotype: PDB and pdb
• The double cross phenotype: It should be less in number because the double-cross is rare and they are PDb and pdB.
• Here, when the parent phenotype and double-cross phenotype is compared then it can be observed that B/b is present with P/p and D/d.

Thus, B/b should be in the middle.

3) Genetic distance:

• The single cross over are pDB (59), PdB (414), PdB (417), and Pdb (60).
• Make the recombinant pairs are pDB and Pdb; PdB and pDb.
• So to calculate the distance, pDB, and Pdb are formed by the recombination between P/p and B/b.
• The double reciprocals are PDb and pdB.

The distance between P/p and B/b: pDB + Pdb + PDb + pdB = 59 + 60 + 21 + 23 = 163

The population is out of 2500 = 163/2500 = 0.065

The distance between P/p and B/b is 0.065 cM.

• So to calculate the distance, PdB and pDb are formed by the recombination between D/d and B/b.

The double reciprocals are PDb and pdB.

The distance between D/d and B/b: PdB + pDb + PDb + pdB = 414 + 417 + 21 + 23 = 875

The population is out of 2500 = 875/2500 = 0.35

The distance between P/p and D/d is 0.35 cM.

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