Question

using Excel.

61 MATHEMATICAL For the following aspartase reaction (see Ques- tion 28) in the presence of the inhibitor hydroxymethylaspar- tate, determine KM and whether the inhibition is competitive or noncompetitive. V, Inhibitor Present (same arbitrary units) [S] (molarity) V, No Inhibitor (arbitrary units) OIV0.010 e1x 10 5 X 104 10.026 0.040 0.086 0.120 0.092 0.136 0.150 2.5 X 10-3 5 x 10 0.142 0.165 d) that enzymes can be re-

Answer 1

lineweaver burk plot in absence of inhibitor

m 3 1 1/S 1/V 2 10000 38.46 2000 10.87 4 666.67 7.35 400 6.67 2006.06 7 11/5 1/1 + 5 y = 0.003x + 5.057 R2 = 0.998 6 NO inhibitor 0 2000 4000 8000 10000 12000 6000 1/[S]

lineweaver burk plot in presence of inhibitor

2 A 10000 2000 666.67 400 200 1/5 - 100 25 11.63 8.33 7.04 4 5 6 1/ y = 0.009x +5.213 R2 = 0.999 1N inhibitor 0 2000 4000 6000 8000 10000 12000 1/[5]

lineweaver burk equation is

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

hence intercept = $\small \frac{1}{V_{max}}$

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

from the above plots trendline equation in absence of inhibitor is

y = 0.003x + 5.057

then, $\small \frac{1}{V_{max}}$ = 5.057

or, Vmax= 0.1977 ( arbitary units)

now, $\small \frac{Km}{V_{max}}$ = 0.003

or, Km = 0.003 *V_max = 0.003 * 0.1977 = 5.9*10^-4 M

from the above plots trendline equation in presence of inhibitor is

y = 0.009 x + 5.213

then, $\small \frac{1}{V_{max}}$ = 5.213

or, Vmax= 0.1918 ( arbitary units)

now, $\small \frac{Km}{V_{max}}$ = 0.009

or, Km = 0.009 *V_max = 0.009 * 0.1918 = 1.72*10^-3 M

so, in presence of inhibitor V_max is almost unchanged but K_m has increased . Therefore it is competitive inhibitor.

competitive inhibitor : The structure is similar to substrate so they competes with the substrate this kind of  inhibition can be overcome by increasing substrate concentration.