Question

cat 2 14. You measure the kinetics of an enzyme E as a function of substrat using the double-reciprocal (Lineweaver-Burk) plot (Figure maintained constant at a level of 1 M (-106 M) e concentration and plot the data below). The enzyme concentration is S]. Vo. 2.9 4.4 5.4 5.8 6.2 6.4 0.15 00s 10 1/Is]. 10* M From these data, determine Vmax, KM, kcat and the turnover number for the enzyme. Determine these parameters by two different methods: (1) from the double-reciprocal plot and (2) from the table (or you can plot the hyperbolic curve). Compare the results. What is the reason for the difference betweern these two methods? Which is more accurate? (8) 2) v mox 6m 0.1T a5.88 Cat Tne double reciprocal plot orients dato into a much more ace drat e taSni﹄ . How So

Answer 1

Attached is the graph of the Double Reciprocal Plot required.

First we calculate the 1/[S] and the 1/Vo values and then plot them on the x and y axes respectively.

If your instructor insists on a graph paper, Pls plot the same on the sheet and note the x-axis and y-axis scales

Now in the graph the intercept made by the graph on the -ve x-axis = -1/k_m

Vo 1/[S 1/Vo 0.345 0.263 0.227 0.200 0.185 0.172 0.161 0.156 0.149 2.9 0.500 0.333 0.250 0.200 0.167 0.143 0.125 Double Reciprocal Plot 3.8 4 0.400 5.4 5.8 6.2 6.4 6.7 0.350 0.300 0.250 9 10 0.100 0.200 0.150 0.1 0.050 -0.400 0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 1/[S

From the graph -1/k_m = -0.21 (approx)

Therefore k_m = 1/0.21 = 4.76*10^-6M

1/V_max is the intercept made by the line on the y-axis = 0.1

=> V_max = 1/0.1 = 10*10^-6M/min