The correct answer is M2 – this microsatellite marker shows linkage with the genetic disorder in the pedigree.​

Introduction

Microsatellite markers are powerful tools to test linkage between a molecular marker and a disease locus by checking how consistently a specific marker allele co‑segregates with the disorder in a pedigree. In this CSIR NET‑type pedigree question, four markers (M1–M4) are typed in parents and offspring, and the goal is to identify which marker allele tracks the disorder without recombination, indicating tight linkage.​

Stepwise solution and logic

1. Identify the disease pattern in the pedigree

   • In the given figure, P2 (father) and the sons 3, 4 and 5 are affected, while P1 (mother) and daughters 1, 2 and 6 are unaffected, which is consistent with an autosomal dominant trait transmitted from P2.​

   • Therefore, the disease allele must come from P2 and will be present in all affected individuals but absent from all unaffected ones.

2. Assign disease‑associated chromosome in P2

   • For each microsatellite marker, P2 has two alleles (one on each homologous chromosome). The affected sons must have inherited the homologous chromosome carrying the disease allele from P2.

   • Thus, for a genuinely linked marker, the same marker allele from P2 must appear in every affected child and must not appear in any unaffected child (no recombination in this small pedigree).​

3. Scan each marker for co‑segregation

   • From the tabular marker data under the pedigree (bands representing marker alleles), compare:

     • which of P2’s two alleles is transmitted to affected individuals 3, 4, 5, and

     • whether that same allele is absent from unaffected individuals 1, 2, 6.

Why M2 is linked to the disorder (correct option)

• At marker M2, one particular allele of P2 (say allele “a”) is present in all three affected sons (3, 4, 5) and absent from all the unaffected daughters (1, 2, 6).​​

• Conversely, P2’s other M2 allele (say allele “b”) is seen only in the unaffected daughters, never in affected sons, implying that the disease allele and M2 allele “a” are carried on the same paternal chromosome.

• As there is perfect co‑segregation with no recombinants in this family, M2 is considered linked to the disease locus and is the only marker satisfying the linkage criterion, so option (2) M2 is correct.​

Why the other options are incorrect

Option (1) M1

• For M1, the allele inherited from P2 by all affected sons is not unique to them; at least one unaffected child shares the same paternal M1 allele.​

• This indicates a recombination event (or independent assortment) between M1 and the disease locus, so M1 cannot be tightly linked to the disorder.

Option (3) M3

• At M3, the affected sons do not all share the same paternal allele; at least one affected individual carries a different M3 allele from P2 than the other affected siblings.​

• Since the disease allele must be identical by descent in all affected offspring, such discordance shows that M3 segregates independently from the disease locus and therefore is not linked.

Option (4) M4

• For M4, the paternal alleles again fail the “all affected share, all unaffected lack” rule: an unaffected individual carries the same M4 allele that is also present in affected offspring, or an affected offspring lacks the allele shared by other affected siblings.​

• This mismatch implies recombination or independent assortment, ruling out close linkage between M4 and the disease gene.

Key exam tips for similar questions

• First, mark affected vs. unaffected clearly and deduce which parent is transmitting the disease allele (commonly the affected parent in such pedigrees).​

• Then, for each marker:

  • Choose one allele of the transmitting parent and see if it appears only in all affected individuals.

  • Any marker where the “disease‑carrying” allele also appears in unaffected individuals or is missing from an affected one is not linked.

• In small pedigrees with no recombinants observed, the marker showing perfect co‑segregation with disease is considered most tightly linked to the disorder, as seen here with M2.​