Which organism is inhibited by bile, produces a black pigment, and ferments glucose, sucrose, and lactose?

Which organism is inhibited by bile, produces a black pigment, and ferments glucose, sucrose, and lactose?

• A. Bacteroides fragilis
• B. Campylobacter ureolyticus
• C. Porphyromonas gingivalis
• D. Prevotella melaninogenica

Answer: (D)

This item tests how pigment production, bile sensitivity, and carbohydrate fermentation help identify anaerobic oral bacteria. Prevotella melaninogenica is known for forming black pigmented colonies due to melanin production, and it is inhibited by bile, meaning it won’t grow in the presence of bile. It also ferments multiple sugars, including glucose, lactose, and sucrose, which is a distinctive metabolic pattern for this organism. Together, the combination of a black pigment, bile inhibition, and the ability to ferment glucose, lactose, and sucrose points to Prevotella melaninogenica. In contrast, the other organisms either do not produce the same pigmented colonies, are bile resistant, or do not ferment all three sugars in this pattern, so they do not fit this profile as well.

This item tests how pigment production, bile sensitivity, and carbohydrate fermentation help identify anaerobic oral bacteria. Prevotella melaninogenica is known for forming black pigmented colonies due to melanin production, and it is inhibited by bile, meaning it won’t grow in the presence of bile. It also ferments multiple sugars, including glucose, lactose, and sucrose, which is a distinctive metabolic pattern for this organism. Together, the combination of a black pigment, bile inhibition, and the ability to ferment glucose, lactose, and sucrose points to Prevotella melaninogenica. In contrast, the other organisms either do not produce the same pigmented colonies, are bile resistant, or do not ferment all three sugars in this pattern, so they do not fit this profile as well.