The answers are highlighted in red How do you solve this pro

The answers are highlighted in red. How do you solve this problem?

Mutants 1 and 2 are two point mutations of rIIA. Mutant 1 does not overlap with mutant 2. Mutants 3 and 4 are two deletion mutations of rIIB. Mutant 3 overlaps with mutant 4. Mutant 5 is a deletion that overlaps mutants 2 and 4, but does not overlap mutants 1 and 3. For each of the following crosses, you perform a high m.o.i infection of both mutants on E. coli strain B. Then you harvest the lysate and perform a high m.o.i infection on E. coli strain K. Finally, you dilute this second lysate and perform low m.o.i infections on separate plates with E. coli B and E. coli K. Predict the results by circling ONE answer per cross. Select \"Equal Number\" if the E. coli B and E. coli K plates will have some plaques in equal proportions; select \"No Plaques\" if both plates will show 0 plaques; select \"Can\'t tell\" if the information provided is insufficient to predict the result with certainty.

Solution

Principle of this project
1. Bacteriophage with rII strain infect E.Coli B and form large plaques but No growth in E.coli K strain.
2. Recombination occur between two point mutation strains.
3. Recombination occur between two deletion mutation strains if this mutation do not overlap.
4. Recombination does not take place if point mutation and deletion mutation overlaps.
5. Recombination does not occur between two deletion mutation strains if this mutations are overlap.

1. Bacteriophage of two strains - one contain Mutant 1 and other contains Mutant 2. If both strains infect on E.Coli B recombination happen between Mutant 1 and Mutant 2 of rIIA and reverse back its activity. So, it lysis the E.coli B strain. Then lysate of Strain B no effect on E.coli strain K based on principle 1. So, More plaques seen on Strain B.

2. 2 point mutation and 5 deletion and its overlap between 2. According to principle 4 no recombination happen and no activity of the bacteriophage and no plaque.

3. 3-Deletion and 4 deletion but overlap so no recombination happen between them. Based on principle 5. so no plaque.

4. 3-Deletion and 5-deletion but both are no overlap between them. So recombination takes place reverse its activity and infect the B strains. Then lysate of Strain B no effect on E.coli strain K. So, More plaques seen on Strain B.