Question

The turnover number is defined as the maximum number of substrate molecules that can be converted into product molecules per unit time by an enzyme molecule. The concentration of enzyme active sites is not necessarily equal to the concentration of enzyme molecules, because some enzyme molecules have more than one active site. If the enzyme molecule has one active site, the turnover number is given by turnover number = R max [ E ] t = k 2 ( R max is often written as V max ) If the enzyme molecule has more than one active site, then [ E ] t is multiplied by the number of active sites to determine its effective concentration. Determine the value of the turnover number of the enzyme carbonic anhydrase, given that R max for carbonic anhydrase equals 249 μmol ⋅ L − 1 ⋅ s − 1 and [ E ] t = 2.35 nmol ⋅ L − 1 . Carbonic anhydrase has a single active site.

Answer 1

Answer:

The value of turnover number is $1.06\times10^5 s^{-1}$

Explanation:

As the given data

$R_{max}=249 \mu\, mol/Ls=249 \times 10^{-6} mol/Ls$

$[E]_t=235 n \, mol\L=235 \times 10^{-9} mol/L$

As per the given formula of turnover number for a single active site enzyme is given as

$n_{turnover}=\frac{R_{max}}{[E]_t}$

By substituting values

$n_{turnover}=\frac{R_{max}}{[E]_t}\\n_{turnover}=\frac{249 \times 10^{-6} mol/Ls}{2.35 \times 10^{-9} mol/L}\\n_{turnover}=1.06\times10^5 s^{-1}$

So the value of turnover number is $1.06\times10^5 s^{-1}$