Question

Can someone please walk me through how to do this?

Name = deletions 7this tes whether 10. Seven non-reverting mutations at the nii locus of T4 were used to co-infect E.coli strain B at high multiplicity of infection. The lysates from each co-infection were then in spotted onto a lawn of strain K to test whether they could infect and lyse strain K. The results of this test are shown in the following table, where "+" indicates lysis of strain K 1 2 3 4 5 recombin can be genera 2 - - + + - 6 a. Draw a map that indicates which part of the ril locus is missing in each mutation. (Guidelines are provided to aid accuracy of your drawing.) PITA TIL B b, d, e this is a ag b. Five pt. mutations (a-e) were tested for their ability to lyse strain K when co-infected contest with each of the seven non-reverting mutations. The results are shown below by componda hafi known that mutation a is located in A. Use this information to locate the boundary between A and B Show this on the map above, label the genes and place the point mutations into the correct genes.

Answer 1

Looks like the question is already solved. I assume that you want to know how it was solved. if I am right, please read below. Otherwise, get back in comments for additional queries.

1. The complementation tests are done to identify the location of genes on chromosomes, called locus. The Benzer experiments used several deletion mutants (non-revertants) and coinfection lysate was used to test whether the two deletion mutants can complement each other and lyse the host.
- Co-culture of deletion mutant 1 gave + ( lysis) with mutant 4 and 6. This means the mutation in 1 is not at the same place as 4 and 6.
- Mutant 2 gave + with 4,5,6,7 but not with 3. hence deletion 2 is present away from 4,5,6,7 and is in same place as deletion 3.
- Deletion 4 is negative with 5,6,7. So the mutation 4 lies within other deletion 5,6 and7.
- Deletion 7 is positive with 2 and 6 only. Hence, the mutation 7 shares location with 5,6,6,1 and 3.\
From the above logic, the given deletion map is drawn ( in the question image).

Now, about point mutations.
There are 5 point mutations, a to e. Use same logic as above. If a point mutation and deletion mutant are able to complement (+), then point mutation is not present in the deleted region. If they cannot complement (-) then the mutation is present in the same deleted segment.
Point mutations a and c are + with deletions 4 and 6. Hence they do not belong to those deleted region. This region is rIIA. Other point mutations b,d and e are found on deleted regions 4, 5 and 6, and this is rII B.