- The following scheme represents deletions (1-4) in the rll locus of phage T4 from a common reference point:
(The bars represent the extent of deletion in each case)
Four point mutations (a to d) are tested against four deletions for their ability (+) or inability to give wild type (rll+) recombinants.
The results are summarized below:
| a | b | c | d | |
| 1 | + | + | + | + |
| 2 | + | + | + | – |
| 3 | + | – | + | – |
| 4 | – | – | + | – |
Based on the above the predicted order of the point mutations is:
(1) b-d-a-c (2) d-b-a-c
(3) d-b-c-a (4) c-d-a-b
Introduction
Deletion mapping of the rII locus of phage T4 is a classical method used to map point mutations by testing their recombination with a series of overlapping deletions.
In this CSIR NET Life Sciences question, four deletions (1–4) of increasing length from a common left reference point are crossed with four point mutations (a–d), and the presence (+) or absence (−) of wild‑type recombinants is used to infer the linear order of these point mutations.
Step‑by‑step solution
1. Principle of deletion mapping
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A deletion removes a continuous stretch of DNA; any point mutation lying within that deleted stretch cannot recombine with it to give wild‑type because there is no corresponding sequence present. Hence, such a pair shows no wild‑type recombinants (−).
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If the point mutation lies outside the deleted region, recombination can restore a wild‑type sequence and wild‑type recombinants are obtained (+).
In the problem, deletions 1–4 extend progressively further to the right from the same left start point; thus, deletion 1 is the shortest and deletion 4 is the longest.
2. Writing the +/− table
From the question (rII locus of phage T4, four deletions vs four point mutations):
| Deletion | a | b | c | d |
|---|---|---|---|---|
| 1 | + | + | + | + |
| 2 | + | + | + | − |
| 3 | + | − | + | − |
| 4 | − | − | + | − |
Here “+” means point mutation is outside that deletion; “−” means it lies within that deletion’s span.
3. Deducing relative positions of a, b, c and d
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Mutation c
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Shows “+” with all four deletions (1–4).
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Therefore, c lies to the right of the rightmost end of deletion 4, i.e., beyond the longest deletion.
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Mutation d
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Shows “−” with deletions 2, 3 and 4, but “+” with deletion 1.
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Hence, d lies to the right of deletion 1’s end, but within the common region shared by deletions 2, 3 and 4.
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Mutation b
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Shows “−” with deletions 3 and 4, but “+” with deletions 1 and 2.
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Therefore, b lies beyond the end of deletion 2 but within the overlap of deletions 3 and 4.
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Mutation a
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Shows “−” only with deletion 4 and “+” with deletions 1, 2 and 3.
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Thus, a lies beyond the end of deletion 3, but within the extra part included only in deletion 4 (not in 3).
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Putting these constraints along a left‑to‑right line:
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Region covered by deletions 2, 3, 4 (overlap) contains d.
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Region covered by deletions 3 and 4 only (but not 2) contains b.
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Region covered only by deletion 4 beyond deletion 3 contains a.
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Region beyond deletion 4 contains c.
This gives the order (left → right):
d − b − a − c
So the required order is d‑b‑a‑c.
4. Checking each option
| Option | Order given | Correct or not? | Reason |
|---|---|---|---|
| (1) | b‑d‑a‑c | Incorrect | Places b to the left of d, but b must lie in a more rightward region (deletions 3 and 4 only) than d (deletions 2, 3, 4). |
| (2) | d‑b‑a‑c | Correct | Matches the deduced left‑to‑right positions: d in 2–4 overlap, b in 3–4 overlap, a only in 4, c beyond all deletions. |
| (3) | d‑b‑c‑a | Incorrect | Puts c between b and a, but c lies outside all deletions, hence must be rightmost, beyond a. |
| (4) | c‑d‑a‑b | Incorrect | Places c at the extreme left and inside deletions, contradicting its “+” result with all deletions, and misplaces a and b relative to their deletion overlaps. |
Therefore, Option (2) d‑b‑a‑c is the correct answer.
