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Biochemistry · Translation: chloramphenicol and the 50S subunit
Sample itemA premature neonate treated with chloramphenicol for sepsis develops abdominal distension, vomiting, ashen-grey colour and circulatory collapse. At the ribosomal level, which step of bacterial protein synthesis does this drug block?
Original practice item, not a recalled exam question. Nothing here is recorded.
The answer and the reasoning
C. Peptide bond formation by peptidyl transferase on the 50S subunit
Chloramphenicol binds the 50S subunit and inhibits peptidyl transferase activity of the 23S rRNA, so peptide bonds are not formed. Neonates, especially premature ones, have immature glucuronidation and cannot clear the drug, which accumulates and causes grey baby syndrome. Mitochondrial ribosomes resemble bacterial ones, which contributes to bone marrow toxicity.
Option by option
A. Translocation of peptidyl-tRNA by elongation factor G
Macrolides block translocation on the 50S subunit; chloramphenicol acts on peptide bond formation itself.
B. Initiation complex formation with the 30S subunit and formylmethionyl-tRNA
Aminoglycosides and related agents interfere with initiation at the 30S subunit.
C. Peptide bond formation by peptidyl transferase on the 50S subunit (correct)
Inhibition of peptidyl transferase halts chain elongation.
D. Binding of aminoacyl-tRNA to the A site of the 30S subunit
Tetracyclines block aminoacyl-tRNA binding at the A site of the 30S subunit.
Source note: Original NextPrep sample item — pending expert review. This item is original practice material awaiting independent expert review, so confirm important facts in a standard textbook.
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