Question

Antibiotic-resistant bacteria have an enzyme, penicillinase, that catalyzes the decomposition of the antibiotic. The molecular mass of penicillinase is 31200 g/mol. The turnover number of the enzyme at 28 °C is 2.00 × 103 s–1. If 4.10 μg of penicillinase catalyzes the destruction of 2.83 mg of amoxicillin, an antibiotic with a molecular mass of 364 g/mol, in 29.6 seconds at 28 °C, how many active sites does the enzyme have? Assume that the enzyme is fully saturated under the conditions described above.

Answer 1

Answer:

The number of  active sites enzyme have is 1.

Explanation:

Mass of penicillinase = 4.10 μg =$4.10\times 10^{-6} g$

1 g = 1000000 μg

The turnover number of the enzyme at 28 °C = $2.00\times 10^3 s^{-1}$

Moles of penicillinase =

$\frac{4.10\times 10^{-6} g}{31200 g/mol}=1.31410\times 10^{-10} mol$

Mass of antiboitic-amoxicillin =2.83 mg =$2.83\times 10^{-3} g$

Moles of amoxicillin =

$\frac{2.83\times 10^{-3} g}{364 g/mol}=7.7747\times 10^{-6} mol$

Moles of reactant which are converted into product per second:

$1.31410\times 10^{-10} mol\times 2.00\times 10^3 s^{-1}$

=$2.6282\times 10^{-7} mol/s$

Moles of product converted in 29.6 seconds:

$2.6282\times 10^{-7} mol/s\times 29.6 s=7.779472\times 10^{-6} mol$

Number of sites:

$=\frac{\text{moles of product}}{\text{moles of reactant}}$

$=\frac{7.779472\times 10^{-6} mol}{7.7747\times 10^{-6} mol}=1.00$