Question

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kd. The amount of penicillin hydrolyzed in 1 minute in a 10-ml solution containing 10^-9 g of purified penicillinase was measured as a function of the concentration of penicillin. A plot of 1/V_0 versus 1/[S] indicated the V_max=6.8 times^- 10 mol minute^-1. Assuming one active site per enzyme molecule, the turnover number for penicillinase is
(A) 337 s^-1
(B) 337 min^-1
(C) 600 hour^-1
(D) 33 hour^-1 600 s^-1

Answer 1

Answer:

The turnover number is the maximum substrate quantitiy converted to product per enzyme and per second. It can be calculate as follows:

$k_c = \frac{V_{max}}{[E]}$ with $[E]$ active enzyme concentration.

In this case we have Vmax an data to calculate $[E]$

$moles of E=\frac{10^{-9}g}{29600 g/mol}=3.38*10^{-14}$

$[E]=\frac{3.38*10^{-14}}{0.01 L}=3.38*10^{-12} M$

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-12}}=201.28 min^{-1}$

it is not like any option.

If we assume that $V_{max}$ have the non usual units of $mol L^{-1} min^{-1}$ and it is $6.8*10^-10 mol min^{-1}$

So we need divide by the moles of E (in place of [E])

Now $k_c=\frac{6.8*10^-10}{3.38*10^{-14}}=20128 min^{-1} = 335 s{-1}$

(pass from $min^{-1}$ to $s^{-1}$ dividing by 60)