A catalase-negative gram-positive coccus is isolated from a urine sample of a hospitalized patient. The bacterium produced a black pigment on bile-esculin agar and formed acid from glucose in the presence of 6.5% NaCl. What is the most likely identification?

• A. Abiotrophia sp.
• B. Enterococcus faecalis
• C. Group B streptococci
• D. Group D streptococci

Answer: (B)

A key distinction here is that Enterococcus species combine bile-esculin positivity with the ability to grow in high salt (6.5% NaCl). A catalase-negative Gram-positive coccus from a hospitalized patient’s urine fits Enterococcus as a common nosocomial cause of UTIs. The black pigment on bile-esculin agar means the organism can hydrolyze esculin in the presence of bile, forming the dark complex. Among Gram-positive cocci, this esculin hydrolysis is shared by enterococci and some group D streptococci, but only enterococci tolerate and grow in 6.5% NaCl, which distinguishes Enterococcus from the non-enterococcal group D streptococci. The combination of these traits—catalase-negative, bile-esculin positive with blackening, and growth in 6.5% NaCl—points most strongly to Enterococcus faecalis as the responsible organism. Abiotrophia would require special media and does not show this bile-esculin/salt profile, and group B streptococci do not typically grow in 6.5% NaCl, so they’re less likely given this test pattern.