Question regarding enzyme kinetics

Question

Under conditions of saturating amounts of substrate the rate of formation of product is 50 nanomoles per ml of reaction volume every minute. [Hint: what do the words saturating amounts of substrate imply?] The reaction contains 10 nanomole enzyme per ml of reaction volume. The KM of the enzyme is 300 microM determine : a) the concentration of enzyme in microM? b) the Vmax of the reaction in units of microM/s c) The maximum number of molecules of product catalysed by each molecule of enzyme in units of s-1 d) The catalytic efficiency of the enzyme?

Answer 1

a) concentration of enzyme in uM:  10 nmole enzyme/ml = 10,000 nmole/liter = 10 uM

b) Vmax in uM/s:  50 nmoles/ml/min = 50 umoles/L/min = 50 umoles/L/60 sec = 0.833 umol/L/sec = 0.833 uM/s

c) max. number molecules = k_cat = V_max/[E_t] = 0.833 uM/s/10 uM = 0.0833 s^-1

d) catalytic efficiency = k_cat/K_m = 0.0833 s-1/300 uM = 0.00278 uM^-1s^-1 = 2.78x10^-3 uM^-1s^-1

 

NOTES:  The hint about saturated amounts indicates that the enzyme is operating at V_max under these conditions, so V_max can be taken as 50 nmoles/ml/min.  Also, (c) is essentially asking for enzyme turnover, or k_cat. Finally, catalytic efficiency is basically how well the enzyme functions at low [S] where according to Michaelis-Menton

v = V_max[S]/K_m + [S] and at low [S] this becomes v ~ V_max[S]/K_m