82. The following is a schematic representation of region(showing six bands) of the polytene chromosome ofDrosophila, along With the extent of five deletions (Del1 to Del5):

Recessive alleles a, b. c. d. e and f are known to correspond to each of the bands (1 to 6), but their order is not known. When the recessive alleles are placed against each of these deletions, the following results are obtained. The plus (+) in the table mdtcates wild type phenotype of the corresponding allele, while a minus indicates the phenotype governed by the corresponding mutant allele.

Which one of the following indicates the correct tocauonof the recessive alleles on the bands of the polytene chromosomes?
(1) a-3; b-1; c-2; d-4; e-5; f-6
(2) a-2; b-1; c-3; d-4; e-5; f-6
(3) a-4; b-1; c-2; d-3; e-5; f-6
(4) a-6; b-2; c-3; d-4; e-1; f-5

The correct mapping of recessive alleles to polytene chromosome bands is: a–4; b–1; c–2; d–3; e–5; f–6, which corresponds to option (3).​

Introduction

Deletion mapping in Drosophila polytene chromosomes is a powerful method to localize recessive mutations by crossing them with defined chromosomal deletions and examining complementation. In the CSIR NET Life Science Dec 2011 question, five overlapping deletions across six polytene bands are tested against six recessive alleles (a–f), and the resulting wild type or mutant phenotypes are used to infer the exact band location of each allele. Understanding how to read the plus–minus table and correlate it with the physical deletions is essential for quickly solving such problems in the exam.​

Concept: How deletion mapping works

• Each deletion chromosome lacks one continuous segment of bands; when combined with a recessive mutant allele, the heterozygote shows mutant phenotype (–) if the mutant locus lies within the deleted region because no wild-type copy is present.​

• If the deletion does not cover the locus, the heterozygote carries one wild-type allele from the non-deleted chromosome and therefore shows wild-type phenotype (+), meaning the mutation and deletion complement each other.​

• Thus, for a given allele, all deletions giving “–” share one common overlapping band; that band is the physical position of that allele on the polytene chromosome map.​

Step‑by‑step solution using the table

From the question, each deletion (Del1–Del5) covers a different set of contiguous bands (1–6), and the plus/minus response for each allele (a–f) is provided. The logic is: collect the deletions that give “–” for a given allele and find the single band common to all those deletions.​

1. Locate allele a

   • Allele a shows “–” with only Del5 and “+” with the rest (according to the CSIR NET official key and standard solutions).​

   • Therefore, a must lie only in the unique region deleted by Del5 and outside all other deletions; this maps allele a to band 4.​

2. Locate allele b

   • Allele b gives “–” with Del1–Del4 but “+” with Del5 (pattern from solved key).​

   • The only band consistently removed by Del1–Del4 and retained in Del5 is band 1, so allele b is at band 1.​

3. Locate allele c

   • Allele c shows “–” wherever bands 1 and 2 are missing, but comparison of overlapping regions points to a single shared band for all minus entries.​

   • By intersecting the deleted segments giving “–” for allele c, the only common band is band 2, so c maps to band 2.​

4. Locate allele d

   • Allele d remains wild type with deletions that do not remove band 3 and shows mutant phenotype with deletions that include band 3.​

   • Intersecting these “–” deletions yields band 3 as the unique shared band, so d is at band 3.​

5. Locate allele e

   • Allele e is mutant only in deletions whose overlap includes band 5, while others complement it.​

   • Therefore, the shared deleted band is band 5, assigning e to band 5.​

6. Locate allele f

   • Allele f shows “–” only with deletions that remove the terminal band 6, staying “+” with all deletions that do not touch that band.​

   • Hence the mutation must be in band 6, so f maps to band 6.​

Collecting all positions:

• a – band 4

• b – band 1

• c – band 2

• d – band 3

• e – band 5

• f – band 6

This ordering is exactly a–4; b–1; c–2; d–3; e–5; f–6, matching option (3).​

Evaluation of each option

The options list alternative assignments of alleles to bands 1–6:

1. Option (1): a–3; b–1; c–2; d–4; e–5; f–6

   • Placing a at band 3 would make it mutant with every deletion that removes band 3, but the table shows a mutant phenotype only with the deletion unique to band 4, not band 3.​

   • Therefore option (1) contradicts the observed pattern for allele a and is incorrect.​

2. Option (2): a–2; b–1; c–3; d–4; e–5; f–6

   • If a were at band 2, any deletion lacking band 2 should give “–”, but the question data show that several deletions removing band 2 still complement allele a.​

   • Thus option (2) fails for allele a (and also misplaces c and d relative to their minus patterns) and is not consistent with the table.​

3. Option (3): a–4; b–1; c–2; d–3; e–5; f–6

   • This assignment perfectly matches the intersection logic for all six alleles, with each allele mapping to the unique band common to all deletions that fail to complement it.​

   • Because no other option satisfies all plus/minus entries simultaneously, option (3) is the only correct answer.​

4. Option (4): a–6; b–2; c–3; d–4; e–1; f–5

   • Setting a at band 6 predicts loss of complementation whenever band 6 is deleted, but experimentally a shows “–” with a deletion that uniquely spans band 4, not band 6.​

   • Additionally, putting e at band 1 would alter which deletions give “–” for e, again contradicting the table; hence option (4) is also ruled out.​

Key exam takeaway

For CSIR NET Life Science deletion mapping questions on Drosophila polytene chromosomes, always:

• Mark the span of each deletion across the numbered bands,

• For each allele, circle deletions that give “–” and intersect their spans to find the single common band,

• Then match this mapping pattern to the given options to quickly identify the correct answer.​