Question

1. Catalase (MW250,000Da) has one of the highest turnover numbers we know of.

It catalyzes the decomposition of hydrogen peroxide as shown in the reaction

                                    (2H₂O₂à2H₂O + O₂)

If 15.0 mg of pure catalasecatalyzes the decomposition of 0.40 g of H₂O₂in 1 min at 37 °C at its Vmax, what is the turnover number (kcat) of catalase(in units of sec^-1)?

2. Plot a graph of Abs vs. time at different concentrations of enzyme and replot a figure and find out the optimal enzyme concentration range that would be linear.

Time (sec)	Absorbance [Enzyme] = 0.01mg/ml]	Absorbance [Enzyme] = 0.015mg/ml]	Absorbance [Enzyme] = 0.02mg/ml]
0	0	0	0
15	0.096	0.2	0.3
30	0.21	0.38	0.56
45	0.39	0.57	0.8
60	0.55	0.79	1
120	1.1	1.5	1.8

3. Calculate the Km, Vmax and Kcat for an enzyme (final protein concentration in the cuvet = 10mg/ml) with a molecular weight = 60kDa from the following data points.

[S], mM	Vo = mmol/min
0	0
10	0.2
20	0.4
30	0.5
40	0.6
50	0.7

Answer 1

After using the curve fitting technique the linearity of the graph was determined by goodness of fit, the values of R² for the three concentrations are 0.9995, 0.9991, and 0.9999 respectively. If the value of R²=1 then the graph is absolutely linear. Therefore when the concentration of enzyme is 0.02mg/ml the graph would be linear.

From the LB plot, -1/Km is -0.1

Therefore, Km=1/0.1=10,

1/Vmax=0.5, therefore 1/vmax is 1/0.5=2mmolemin-1