Q.39 IgM is a pentamer. Upon visualisation of a pure IgM by electrophoresis
on a reducing polyacrylamide gel, the number of protein bands to be
visible on the gel would be:

1. 5
2. 10
3. 2
4. 4

   The correct answer is 2 protein bands. Pure IgM under reducing polyacrylamide gel electrophoresis (PAGE) dissociates into heavy (μ) chains (~70 kDa) and light chains (~25 kDa), producing exactly two distinct bands.

   IgM Structure Breakdown

   Native IgM is a pentamer composed of 5 monomer units + 1 J-chain. Each monomer has 2 identical heavy chains (μ-chains) and 2 identical light chains (κ or λ).

   Total components:

   • 5 monomers × 2 μ-chains = 10 heavy chains

   • 5 monomers × 2 light chains = 10 light chains

   • 1 J-chain (~15 kDa)

   Reducing PAGE effect: SDS + reducing agent (DTT/β-mercaptoethanol) breaks all disulfide bonds, denatures proteins, and separates by molecular weight only.

   Protein Bands Observed

   Band	Protein	Molecular Weight	Origin
   Band 1	μ heavy chain	~70 kDa	10 identical μ-chains from all monomers
   Band 2	Light chain	~25 kDa	10 identical light chains (κ or λ)

   J-chain (~15 kDa) typically migrates with dye front or not resolved distinctly due to small size and poor staining.

   Option Analysis

   • 5 bands: Wrong. Assumes 5 different proteins; ignores identical chains within monomers.

   • 10 bands: Wrong. Assumes all 20 chains (10H + 10L) separate individually; ignores identical molecular weights.

   • 2 bands: CORRECT. Only 2 distinct molecular weights: μ-chain (70 kDa) + light chain (25 kDa).

   • 4 bands: Wrong. Might assume J-chain + heavy + light + something else; J-chain not distinctly visible.

   Pure IgM pentamer electrophoresis on reducing polyacrylamide gel reveals exactly 2 protein bands, a classic immunology question testing antibody structure knowledge. This comprehensive guide explains why 2 is correct, analyzes all MCQ options, and provides exam-solving strategies.

   IgM Pentamer Architecture

   text

   Native IgM Pentamer = 5 × (μ₂L₂ monomers) + 1 J-chain
= 10 μ-chains + 10 light chains + 1 J-chain


   Molecular weights:

   • μ heavy chain: 70 kDa

   • Light chain (κ/λ): 25 kDa

   • J-chain: 15 kDa

   Reducing PAGE Mechanism

   1. SDS denatures proteins + imparts uniform negative charge

   2. Reducing agent (DTT/β-ME) cleaves all disulfide bonds

   3. Separation by size only – identical proteins = single band

   text

   Pentamer → 10 identical μ-chains (70 kDa) + 10 identical L-chains (25 kDa)
↓
2 distinct bands


   Why Exactly 2 Bands?

   text

   Band 1 (70 kDa): All 10 μ heavy chains migrate together
Band 2 (25 kDa): All 10 light chains migrate together
J-chain (15 kDa): Runs with dye front, not resolved


   Complete Option Elimination

   Option	Bands	Why Wrong
   A. 5	5	Thinks 5 monomers = 5 bands; ignores identical chains
   B. 10	10	Counts individual chains; forgets identical MW
   C. 2	2	✅ CORRECT: μ-chain + light chain only
   D. 4	4	Includes J-chain + assumes heterogeneity

   Exam Strategy

   Pattern recognition: “Reducing gel + antibody” = count distinct polypeptide chains by MW

   • IgG: 2 bands (γ + L)

   • IgM: 2 bands (μ + L)

   • IgA dimer: 2 bands (α + L)

   Key visualization:

   text

   Native IgM Pentamer ──(reducing SDS-PAGE)──> μ-chain band + Light chain band
(950 kDa) (70 kDa) (25 kDa)


   Memory trick: Reducing conditions = “2 bands rule” for all immunoglobulin heavy + light chain pairs.