Question

and Vmax 2 points for interpreting the gapl Name (can be on graph paper or Excel. It is easier on Excel). This is worth 10 points, 2 points for determining the correct data, 2 points for the graph 4 points for them 1. Use this experimental data to plot a lineweaver burke graph 2. Use the graph to determine the Km and Vmax of the reactions 3. Be sure to keep the same number of gigs in your answer and state the units of the answer 4. Put the answer in the blue and pink boxes 5. Interpret the graph and results competitive, noncompetitive or uncompetitive inhibition V nmol/min V nmol/min with 0.5 Wubstrate UM no inibitor m inbaltor 0.10 066 0201 10 114 160 200 Km (no inhibitor) Vmas Vmax withibito

Answer 1

lineweaver-burk equation  is

$\small \frac{1}{V_{0}}$ = $\small \frac{Km}{V_{max}[S]}$ +  $\small \frac{1}{V_{max}}$

now lineweaver-burk plot at no inhibitor

В C D E F G Н К м 1 0.1 0.66 10 1.51 2 0.2 5 1 1,6 0.746 0.4 1.34 2.5 1.4 1.25 y 0.10x0.50 R2 1.00 0.8 1.6 0.625 0.555 5 2 1,8 0.5 1.2 1/Is] 1/v 6 [S] VO 1 7 1/V0.8 no inhibitor 8 9 0.6 10 0.4 11 0.2 12 0 13 0 2 4 6 10 12 14 15 1/IS] 16 17

bestline equation is

y = 0.10 x + 0.50

so, $\small \frac{1}{V_{max}}$ = 0.50

or, V_max = (1/.0.50) = 2.00 nmol/min.

$\small \frac{Km}{V_{max}}$ = 0.10

or, K_m = V_max * 0.10 = (2 *0.10) = 0.20 $\small \mu$M

now the plot in presence of inhibitor

М A C D E F G 1 0.1 0.4 10 2.5 2 0.2 0.66 5 1.515 3 0.4 1 2,5 1 4 0.8 1.34 1,25 0.746 y 0.20x0.50 R2 1.00 2.5 5 2 1,65 0.5 0.606 6 [S] 1/[s] 1/v 2 7 inhibitor 8 1.5 9 10 1 11 12 0.5 13 14 15 0 2 4 6 8 10 12 16 NO

best line equation is

y = 0.20 x + 0.5

or, V_max = (1/.0.50 ) = 2.00 nmol/min.

$\small \frac{Km}{V_{max}}$ = 0.20

or, K_m = V_max * 0.20 = (2 *0.20) = 0.40 $\small \mu$M

Therefore, in presence of inhibitor V_max is unchanged and K_m has increased, so type of inhibition is competitive .

Competitive inhibitor : The structure of inhibitor is similar to substrate so it competes with the substrate so such kind of inhibition is overcome by increasing the substrate concentration.