In the Gram staining technique, the decolourisation step with acetone-alcohol is critical. Gram-negative bacteria lose the crystal violet-iodine complex and appear red (safranin-stained) because:

• A Gram-negative outer membrane is dissolved by acetone-alcohol, widening pores in the thin peptidoglycan layer through which the dye complex escapes
• B Gram-negative peptidoglycan contains D-glutamic acid instead of L-glutamic acid, reducing dye binding
• C Crystal violet binds more tightly to teichoic acid in Gram-positive cell walls, whereas Gram-negative LPS does not bind crystal violet
• D Lipid-rich outer membrane of Gram-negatives is extracted by acetone, creating large channels for dye loss; thick peptidoglycan in Gram-positives retains the complex after lipid extraction ✓

Explanation

The differential staining in Gram's technique is primarily explained by cell wall structure. Gram-negative bacteria have a lipopolysaccharide-phospholipid outer membrane surrounding a thin peptidoglycan layer. Acetone-alcohol (a lipid solvent) dissolves the outer membrane, creating large gaps through which the crystal violet-iodine (mordant-dye) complex readily diffuses out. Gram-positive bacteria have a very thick peptidoglycan layer (20–80 nm) without an outer membrane; upon decolourisation, the peptidoglycan dehydrates and the pores shrink, trapping the crystal violet-iodine complex inside. Teichoic acids contribute to the negative charge of Gram-positive cell walls but are NOT the primary retention mechanism for crystal violet. Safranin then counterstains decolourised Gram-negative cells red.

Reference: Ananthanarayan & Paniker's Textbook of Microbiology, 11th ed.

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Written and medically reviewed by the StethoPrep medical team.