Seven deletion mutations (1 to 7 in the table below) are tested for their abilit
ID: 174422 • Letter: S
Question
Seven deletion mutations (1 to 7 in the table below) are tested for their ability to form wild-type recombinants with five point mutations (a to e). The symbol “+” indicates that wild type recombination occurs, and “-“ indicates that wild types are not formed. Use the data to construct a genetic map of the order of point mutations, and indicate the seqment deleted by each deletion mutation
DELETION
DELETION
DELETION
DELETION
DELETION
DELETION
DELETION
POINT MUTATION
1
2
3
4
5
6
7
a
-
+
-
-
+
+
-
b
+
+
+
-
+
-
-
c
+
+
+
+
-
-
-
d
-
+
+
-
+
-
-
e
+
-
-
-
+
+
-
DELETION
DELETION
DELETION
DELETION
DELETION
DELETION
DELETION
POINT MUTATION
1
2
3
4
5
6
7
a
-
+
-
-
+
+
-
b
+
+
+
-
+
-
-
c
+
+
+
+
-
-
-
d
-
+
+
-
+
-
-
e
+
-
-
-
+
+
-
Explanation / Answer
The key principle to solve such question is that point mutations can ONLY recombine with the deletions that do not extend past the mutations.
1. In our case, deletion 7 produces NO wildtype recombinants with any point mutation, hence it must be outside the mutation region.
2. Deletion 4, however only recobines with C, hence C must be right side of the deletion.
3. Deetion 6, can only combine with a and e, but not with c, so a/e re present beforec. but very close to it.
going by same logic, the order of point mutation: d, b, a/e/c
Segments deleted: deletion 2,5 delete region bere d.
deletion 1 deletes region d.
deletion 3,4,6 delete d and b .
deletion 7 deleted none of the above regions.
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