Download 27-36

27-36
Antibiotic-resistant bacteria have an enzyme, penicillase, that catalyzes the composition
of the antibiotic. The molecular mass of penicillase is 30,000g/mol. The turnover
number of the enzyme at 28°C is 2,000 s-1. If 6.4μg of penicillase catalyzes the
destruction of 3.11mg of amoxicillin, an antibiotic with a molecular mass of 364 g/mol,
in 20s at 28°C, how many active sites does the enzyme have?
From the lecture notes on Monday, we know that the turnover or catalytic constant kcat is
equal to Vmax/[E]. From the data given above, we can determine the velocity at which
amoxicillin is being catalyzed. This is done by changing the amount of amoxicillin
catalyzed into moles and dividing by time(in this case it will be 20s).
Calculation:
3.11e  6 gamoxicillin 
6.4e  6 gpenicilla se 
v
1mol
 8.54e  6molamoxicillin
364 g
1mol
 2.13e  10molespenicillase
30,000 g
moles 8.54e  6moles amoxicilli n

 4.27e  7mol / s
time
20s
Since there is an excess amount of substrate we can say that the v obtained above is equal
to our Vmax. So the following calculation can be done.
Calculation:
Vmax  v
Vmax
[E]
V
4.27e  7 mol / s
[ E ]  max 
 2.135e  10mol penicillas e
k cat
2000s 1
k cat 
Since the description of the turnover number is maximum rate per mole of enzyme active
sites one can say that penicillase only has one active site, which is figured out by taking
the following ratio:
[E]
moles needed to catalyze certain amount of amoxicilli n
2.135e  10molpenicil lase

 1.00activesites
2.13e  10molpenicil lase
# active sites 