Question

You determine the molecular weight of an enzyme to be 75 kDa. By assaying 5 g of enzyme under saturating [S], you determine the Vmax to be 1.68 moles per second. Calculate the turnover number

Answer 1

Answer:

The turnover number is $k_t = 25200s^{-1}$

Explanation:

from the question we are told that

               The molecular weight(Molar mass) of an enzyme is $w_z = 75KDa = 75 *10^3 g/mol$

               The mass of the enzyme is $m_z = 5 \mu g$

               The maximum velocity is $V_{max} = 1.68 \mu moles/sec$

The objective of this solution is  to calculate the Turnover number

   Generally number of moles of a molecule is given as

                        $No \ of \ mole = \frac{mass}{molar \ mass}$

                                            $=\frac{5 *10^{-6}}{75*10^3} = 6.67*10^{-5} \mu mol$

The turnover number mathematically represented as

                          $k_t = \frac{max \ velocity }{number \ of \ moles}$

                              $=\frac{1.68*10^{-6}}{6.67*10^{-5}}$

                             $= 25200s^{-1}$