A new bactericidal antibiotic, Brooklamycin, has a Miminum Inhibitory Concentrat

Question

A new bactericidal antibiotic, Brooklamycin, has a Miminum Inhibitory Concentration of 1 nM (nanoMolar). If you use that concentration of Streptomycin to kill 10^9 bacterium in 1 ml, how many molecules of the antibiotic are necessary to kill each cell? You will make this assumption that each of the molecules used are "lost" in their deadly interaction with the bacterium. The molecular weight of Brooklamycin is the same as Streptomycin (also linked to Brooklyn College as you know) if you wanted to know. As usual be concise and show your rationale and your work.

Explanation / Answer

Given that 109 bacteria present in 1 mL of medium are killed by Brooklamycin at an MIC of 1 nM.

Number of molecules of Brooklamycin present in this volume of medium = 6.022×1023 × 10-9 × 10-3 = 6.022×1011

(6.022×1023 is Avogadro's number, 10-9 is the factor for nanomolar concentration and 10-3 is the factor for volume in mL)

Given, that the number of bacteria = 109

Therefore, number of molecules required to kill each bacterial cell = (6.022 × 1011)/109 = 6.022 × 102 ~ 602

Hence, 602 molecules of Brooklamycin are required per bacterial cell for killing it.

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