Question

7) The activity of L-aspartate 4-carboxylase can be assayed by following the evolution of CO2 from L-aspartate: L-aspartate → L-alanine + CO2 The following results were obtained without an inhibitor and with an inhibitor: Initial velocity (nmol CO2 min) No Inhibitor With Inhibitor Concentration of L-aspartate (umol L 25 33.3 50 100 200 17.0 21.3 27.8 41.7 52.6 5.7 8.1 14.7 25.0 i) Calculate the Km and Vmax of the reaction; ii) What type of inhibition is this, justify your answer? iii) Calculate K, for the inhibitor if its concentration is 10 pimol L. (Show your work and show the Lineweaver-Burk plots)

Answer 1

y = 5.4545x + 0.0133 M/T y = 1.1437x +0.013-.. 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 1/[S] No inhibitor . With inhibitor .... Linear (No inhibitor) ........ Linear (With inhibitor) ..

i - According to the Lineweaver-Burk equation

1/Vo= (KM/V mar 1/S +1/V mar

Comparing this equation with Y=aX + c

where a is the slope and c is the intercept

1/V max = intercept

so V max = 1/intercept

For Uninhibited State.

y = 1.1437x + 0.013

1/Vmax = 0.013

Vmax = 1/0.013= 76.9 umol/min

Km/Vmax is equal to slope

Km/76.9 = 1.1437

Km = 76*1.1437

= 87.97 umolL^-1

For inhibited state

y = 5.4546x + 0.0133

1/Vmax = 0.0133

Vmax = 1/0.0133= 75.6 umol/min

Km/Vmax is equal to slope

Km/75.6 = 5.4546

Km = 75.6 * 5.4546

= 334.93 umolL^-1

ii - Type of inhibition is competitive as Vmax is almost same but Km is increased.

iii- Kmapp = Km (1 + [I]/Ki)

Kmapp = 334.93 umolL^-1

Km = 87.97 umolL^-1

I = 10 umolL^-1

334.93 = 87.97 (1+ 10/Ki)

334/87.97 = (Ki +10)/Ki

3.8 = (Ki +10)/Ki

3.8Ki = 10+Ki

10 = 3.8Ki -Ki

10 = 2.8Ki

Ki = 10/2.8 =

Ki = 3.571 umolL^-1

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