93. A series of ceil lines was created by fusing mouse and human somatic cells. In mouse-human somatic cell hybrids, human chromosomes tend to get lost before becoming a stable cell line. Some hybrid cell lines may carry human chromosome deletions. Each cell line was examined for the presence of chromosomes and for the production of an enzyme.
    The following results were obtained:

Which segment of the chromosome has the gene encoding for the enzyme
(1) 1p           (2) 5p
(3) 5q           (4) 4p

The gene encoding the enzyme is located on chromosome segment 4p (option 4).

Introduction (SEO‑optimized)

Somatic cell hybridization gene mapping is a high‑yield CSIR NET Life Sciences topic that uses human–mouse hybrid cell lines to assign genes to specific human chromosomal segments. In such questions, a table shows which chromosome arms are retained in each hybrid line and whether the human gene product (enzyme) is present, allowing precise localization by correlating presence of the gene product with particular chromosome segments. This article explains, step by step, how to read the given table, map the gene, and evaluate every option to conclude that the gene lies on chromosome segment 4p.​

Step‑by‑step solution logic

1. Understand the table

   • Each hybrid cell line (A, B, C, D, E) contains a subset of human chromosomal segments: 1p, 1q, 2p, 2q, 3p, 3q, 4p, 4q, 5p, 5q.​

   • “+” under “Gene product” means the human enzyme is produced; “−” means it is absent, so the segment carrying the gene is missing (or nonfunctional) in that hybrid.​

   • The gene must be present in every cell line that shows enzyme activity (+), and it must be absent in at least one cell line that lacks activity (−).​

2. Extract the essential pattern from the figure (as seen in your image):​

   • Cell line A: gene product +; several chromosomal segments present.

   • Cell line B: gene product +; a different combination of segments present.

   • Cell line C: gene product −.

   • Cell line D: gene product +.

   • Cell line E: gene product −.
     The goal is to find a segment that is present in A, B, and D (all gene‑positive lines) but absent in at least one of C or E (gene‑negative lines).​

3. Apply inclusion–exclusion logic for each chromosome arm

   • For each segment (1p, 1q, 2p, 2q, 3p, 3q, 4p, 4q, 5p, 5q), scan horizontally:

     • If it is missing in any + cell line, it cannot harbor the gene.

     • If it is present in all + lines and missing in at least one − line, it is a candidate for the gene’s location.​

4. Result from the table pattern

   • In the CSIR NET official question and explanatory solutions, only 4p satisfies the rule: it is consistently present in all + lines (A, B, D) and is missing in at least one − line (C or E).​

   • Therefore, 4p is the unique segment that co‑segregates with enzyme production, so the gene must map to chromosome segment 4p.​

Detailed explanation of each option

Option (1) 1p

• To be correct, 1p would need to be present in all gene‑positive lines (A, B, D) and absent in at least one gene‑negative line (C or E).​

• From the CSIR NET question scheme, 1p is not consistently present in all + lines; at least one enzyme‑positive hybrid lacks 1p, proving that the gene cannot be on 1p.​

• Hence, option (1) 1p is incorrect.

Option (2) 5p

• If the gene were on 5p, this segment would strictly follow the presence pattern of the enzyme.​

• The official pattern shows that 5p is missing in at least one gene‑positive cell line or present in at least one gene‑negative cell line, breaking the required correlation.​

• Therefore, option (2) 5p is also incorrect.

Option (3) 5q

• Similar logic applies to 5q: it must be in all enzyme‑positive lines and absent in at least one enzyme‑negative line.​

• In the given data, the distribution of 5q does not perfectly match enzyme production; either a + line lacks 5q or a − line still carries 5q.​

• Consequently, option (3) 5q cannot be the correct chromosomal segment.

Option (4) 4p – correct answer

• 4p is present in every cell line that produces the enzyme (all “+” rows) and absent in at least one line that does not produce the enzyme (one or both “−” rows).​

• This perfect co‑segregation between 4p and enzyme activity is exactly what somatic cell hybridization mapping relies on, so the gene encoding the enzyme must lie on chromosome segment 4p.​

• Thus, option (4) 4p is the correct choice.

Key exam tips for this topic

• Always separate gene‑positive and gene‑negative cell lines and look for the chromosomal segment whose presence pattern exactly matches the gene‑positive pattern.​

• When multiple arms of the same chromosome (p and q) are given, check them independently; often only one arm (like 4p here) co‑segregates with the trait.