Question

14. (The three sentences minimum rule will not be applicable to the following items) Determine the values of KM and Vmax for the decarboxylation of a B-keto acid given the following data. Substrate Concentration (mol L') Velocity (mM min') 2.500 0.588 1.000 0.500 0.714 0.417 0.526 0.370 0.250 0.256

Answer 1

from Line weaver Burk eanation we get vo - Vmax [s]- Vmax ... where, Vo= Initial velocity of the reaction. Vmax= maximum velocity of the reaction. (S) = substrate concentration km² Michaelis constant. Now, if we plot 25 on a axis 4 / on y axis. we cuill get a linear canation. Now, compairing with general lineal equation. ya mrt – – –. lii) where, slope fm) = km , Intercept(b) = Image Now flotting values of y ou in axis & swooning axis we get following lanation. y = 061734 + 2.4 673 ... (ii) [Imolt=1m) Now compairing (i), (ii) f(ii) . - 1h 673 . Vrax = 14 678 = 0.682 mm km= 0.6173 xVmax Vmax "KM = 0.6173 Vmax =06173X0.682 = 0.421 CM

1/vo(min mM-1) 1.700680272 2.5 2 1S)(moll-1) vo(mMmin-1) 1/[S](Lmol-1) 0.588 0.4 0.5 1 0.714 0.417 1.400560224 0.526 0.37 1.901140684 0.25 0.256 2.398081535 2.702702703 3.90625 Line Weaver Burk plot 4.5 13 in coin 15 16 1/vo(min mM-1) y= 0.6173x + 1.4673 1.5 17 1 18 19 0.5 20 -2 -1 -0.5 O 3 21 1/[S](Lmol-1) 22 23 24 25